[ { "idx": 1, "question": "What are the parameters used to quantitatively describe crystal structures?", "answer": "Unit cell parameters." }, { "idx": 2, "question": "What are the parameters used to qualitatively describe crystal structures?", "answer": "Symmetry axes, symmetry center, crystal system, lattice." }, { "idx": 3, "question": "According to the nature of bonding forces, what are the types of bonding interactions in crystals?", "answer": "The bonding interactions in crystals can be classified into ionic bonds, covalent bonds, metallic bonds, van der Waals bonds, and hydrogen bonds." }, { "idx": 4, "question": "The intercepts of a crystal plane on the x, y, and z axes are a/3, b/2, and c, respectively. Determine the Miller indices of this crystal plane.", "answer": "h:k:l=3:2:1, the Miller indices of this crystal plane are (321)." }, { "idx": 5, "question": "What are the two types of voids in the closest packing of equal spheres?", "answer": "The closest packing of equal spheres includes hexagonal close packing and face-centered cubic close packing." }, { "idx": 6, "question": "A crystal plane has intercepts of 2a, 3b, and 6c on the X, y, and z axes respectively. Find the Miller indices of this crystal plane.", "answer": "h:k:1=2:3:6=3:2:1, the Miller indices of this crystal plane are (321)." }, { "idx": 7, "question": "What are the characteristics of covalent bonds?", "answer": "The characteristics of covalent bonds are directionality and saturation, and the bonding force is also very strong." }, { "idx": 8, "question": "What are the characteristics of metallic bonds?", "answer": "Metallic bonds are non-directional and non-saturated covalent bonds, where the binding force is the electrostatic Coulomb force between ions." }, { "idx": 9, "question": "Calculate the number of atoms in a face-centered cubic unit cell", "answer": "Number of atoms 4" }, { "idx": 10, "question": "What are the characteristics of hydrogen bonds?", "answer": "A hydrogen bond is a bond formed between two atoms with relatively high electronegativity, and it possesses saturation." }, { "idx": 11, "question": "What are the characteristics of van der Waals bonds?", "answer": "Van der Waals bonds are formed through molecular forces, and the molecular forces are very weak." }, { "idx": 12, "question": "How many tetrahedral voids and octahedral voids are there around a sphere?", "answer": "There are 8 tetrahedral voids and 6 octahedral voids around a sphere." }, { "idx": 13, "question": "When n equal-sized spheres are closely packed, how many tetrahedral voids and octahedral voids can be formed?", "answer": "When n equal-sized spheres are closely packed, 2n tetrahedral voids and n octahedral voids can be formed." }, { "idx": 14, "question": "What are the characteristics of ionic bonds?", "answer": "The characteristics of ionic bonds are the absence of directionality and saturation, with very strong bonding forces." }, { "idx": 15, "question": "Calculate the coordination number in a face-centered cubic unit cell", "answer": "Coordination number 6" }, { "idx": 16, "question": "How are unequal-sized spheres packed?", "answer": "In the close packing of unequal-sized spheres, it can be considered that the larger spheres are first closely packed as equal-sized spheres, and then the smaller spheres are filled into the gaps according to their sizes. Slightly larger small spheres fill the octahedral voids, while slightly smaller small spheres fill the tetrahedral voids, forming the close packing of unequal-sized spheres." }, { "idx": 17, "question": "Write the coordinates of all nodes on the unit parallelepiped of the face-centered cubic lattice.", "answer": "All nodes on the unit parallelepiped of the face-centered cubic lattice are: (000), (001), (100), (101), (110), (010), (011), (111), (1/2 0 1/2), (0 1/2 1/2), (1/2 1/2 0), (1 1/2 1/2), (1/2 1 1/2)." }, { "idx": 18, "question": "Calculate the packing coefficient in a face-centered cubic unit cell", "answer": "Atomic packing coefficient APC=(4×(4/3)πr^3)/(2√2r)^3=74.05%" }, { "idx": 19, "question": "Calculate the number of atoms in a hexagonal close-packed unit cell", "answer": "Number of atoms 6" }, { "idx": 20, "question": "Calculate the packing coefficient in a hexagonal close-packed unit cell", "answer": "Atomic packing coefficient APC=(6×(4/3)πa^3)/((3√3a^2/2)√8)=74.05%" }, { "idx": 21, "question": "Calculate the coordination number in a hexagonal close-packed unit cell", "answer": "Coordination number 6" }, { "idx": 22, "question": "According to the radius ratio relationship, what is the coordination number when Si4+ coordinates with O2-? Given rO2-=0.132nm, rSi4+=0.039nm.", "answer": "For Si4+, r+/r-=0.039/0.132=0.295; based on the relationship between cation coordination number and cation-anion radius ratio, the coordination number is 4." }, { "idx": 23, "question": "Based on the radius ratio relationship, determine the coordination number of K+ when coordinated with O2-. Given rO2-=0.132nm and rK+=0.131nm.", "answer": "For K+, r+/r-=0.131/0.132=0.99; according to the relationship between cation coordination number and the radius ratio of positive to negative ions, the coordination number is 8." }, { "idx": 24, "question": "The metal magnesium atoms form a hexagonal close packing, and its density is measured to be $1.74\\\\mathrm{g/cm}^{3}$. Find the volume of its unit cell.", "answer": "Let the volume of the unit cell be V, and the relative atomic mass be M. Then the unit cell volume $$V={\\\\frac{n M}{M_{0}\\\\rho}}={\\\\frac{6\\\\times24}{6.023\\\\times10^{23}\\\\times1.74}}=1.37\\\\times10^{-22}\\\\quad\\\\mathrm{\\\\quad\\\\quad}$$" }, { "idx": 25, "question": "In CaO, the unit cell parameter a=0.480 nm, and the anions and cations are in contact with each other. Find the radius of O2-. The radius of Ca2+ is known to be 0.107 nm.", "answer": "In CaO, a=0.480 nm, and the anions and cations are in contact with each other. a=2(r+ + r-), therefore rO2-=(0.480 nm / 2) - 0.107 nm = 0.097 nm." }, { "idx": 26, "question": "According to the closest packing principle, the higher the space utilization, the more stable the structure. The space utilization of the diamond structure is very low (only 34.01%), why is it also very stable?", "answer": "The closest packing principle is based on the spherical symmetry and non-directionality of the electron cloud distribution of particles, so it only applies to typical ionic crystals and metallic crystals, and cannot be used to measure the stability of atomic crystals. In addition, the number of single bonds in diamond is 4, meaning each atom is surrounded by 4 single bonds (or atoms), forming a three-dimensional structure through covalent bonding of tetrahedra sharing vertices. Therefore, although the space utilization of the diamond structure is very low (only 34.01%), it is still very stable." }, { "idx": 27, "question": "According to the radius ratio rule, what is the coordination number when Mg2+ coordinates with O2-? Given rO2-=0.132nm, rMg2+=0.078nm.", "answer": "For Mg2+, r+/r-=0.078/0.132=0.59; based on the relationship between cation coordination number and cation-anion radius ratio, the coordination number is 6." }, { "idx": 28, "question": "In MgS, the unit cell parameter a=0.5201 nm, and the anions are in contact with each other. Calculate the radius of S2-.", "answer": "In MgS, a=0.5201 nm, and the anions are in contact with each other, a=2√2 r-, thus rS2-=0.5201 nm / (2√2) = 0.177 nm." }, { "idx": 29, "question": "Calculate whether the valence of I- in CdI2 crystal is saturated?", "answer": "In CdI2 crystal, the coordination number (CN) of Cd2+ is 6. I is connected to three Cd2+ on the same side, and the coordination number (CN) of I is 3. Therefore, CN ∑i(z+/CN)=1=1z−1, meaning the valence of I− is saturated." }, { "idx": 30, "question": "Why is the transformation temperature between different series of quartz variants much higher than that between variants of the same series?", "answer": "The transformation within the same series of quartz is a displacive transformation, which does not involve the breaking and rebuilding of bonds in the crystal structure, but only the adjustment of bond lengths and angles, requiring lower energy and being rapidly reversible. In contrast, transformations between different series are reconstructive, involving the breaking of old bonds and the formation of new ones, thus requiring higher energy and proceeding at a slower rate. Therefore, the transformation temperature between different series of quartz is much higher than that between variants of the same series." }, { "idx": 31, "question": "In CaS, the unit cell parameter a=0.567 nm, and the anions and cations are in contact with each other. Calculate the radius of Ca2+. The radius of S2- is known to be 0.177 nm.", "answer": "In CaS, a=0.567 nm, and the anions and cations are in contact with each other, a=2(r+ + r-), therefore rCa2+=(0.567 nm / 2) - 0.177 nm = 0.107 nm." }, { "idx": 32, "question": "Compare the unit cell parameter values obtained from density calculations with those calculated from ionic radii.", "answer": "Calculated from ionic radii: a=2(r₊+r₋)=0.414 nm ∴ a₀ < a" }, { "idx": 33, "question": "In MgO, the unit cell parameter a=0.420 nm, and the anions and cations are in contact with each other. Calculate the radius of Mg2+. The radius of O2- is known to be 0.097 nm.", "answer": "In MgO, a=0.420 nm, and the anions and cations are in contact with each other. a=2(r+ + r-), therefore rMg2+=(0.420 nm / 2) - 0.097 nm = 0.113 nm." }, { "idx": 34, "question": "According to the rules of ionic compound formation, when all octahedral interstitial sites are filled, what valence cations should be inserted in the interstitial sites? Please provide examples.", "answer": "The ratio of cation to anion valences should be 1:1, such as NaCl." }, { "idx": 35, "question": "Lithium fluoride (LiF) has a NaCl-type structure, and its measured density is 2.6 g/cm³. Based on this data, calculate the unit cell parameter.", "answer": "Assuming the volume of the unit cell is V and the relative atomic mass is M, for the NaCl-type structure, n=4. Then the unit cell volume V = (n M)/(N0ρ) = (4×26)/(6.023×10²³×2.6) = 6.64×10⁻²³ cm³. Therefore, the unit cell parameter: a₀ = ³√V = 0.405 nm." }, { "idx": 36, "question": "Calculate the ratio of octahedral void number to O2- number", "answer": "The ratio of octahedral void number to O2- number is 1:1" }, { "idx": 37, "question": "Calculate the ratio of the number of tetrahedral voids to the number of O2- ions", "answer": "The ratio of the number of tetrahedral voids to the number of O2- ions is 2:1" }, { "idx": 38, "question": "MgO and CaO both belong to the NaCl-type structure, but when they react with water, CaO is more reactive than MgO. Please explain.", "answer": "Because ${r_{i\\\\parallel_{g}}}^{2+}$ and ${r_{C a}}^{21}$ are different, $r_{\\\\tt C a2+}>r_{\\\\tt B_{\\\\tt B}2+}$, making the structure of CaO looser than that of $\\\\mathrm{Mg0}$, allowing $\\\\mathrm{H}_{2}\\\\mathrm{0}$ to enter more easily, hence more reactive." }, { "idx": 39, "question": "Calculate whether the valence of O2- in CaTiO3 crystal is saturated?", "answer": "In CaTiO3 crystal, the coordination number of Ca2+ CN=12, the coordination number of Ti4+ CN=6, and the coordination number of O2- CN=6, so CN ∑i(z+/CN)=2=|z−|, that is, the valence of O2- is saturated." }, { "idx": 40, "question": "According to the rules of ionic valence, when all tetrahedral interstitial sites are filled, what valence cations should be filled in the interstitial sites, and provide an example.", "answer": "The ratio of cation to anion valences should be 1:2, such as CaF2." }, { "idx": 41, "question": "According to the电价规则, when half of the tetrahedral interstitial sites are filled, what valency cations should be inserted into the interstitial sites, and provide an example to illustrate.", "answer": "The ratio of cation to anion valency should be 1:1, such as ZnS." }, { "idx": 42, "question": "According to the电价规则, when half of the octahedral interstitial sites are filled, what valence cations should be inserted into the interstices, and provide examples to illustrate.", "answer": "The valence ratio of anions to cations should be 1:2, such as TiO2." }, { "idx": 43, "question": "According to the radius ratio relationship, what is the coordination number when Al3+ coordinates with O2-? Given rO2-=0.132nm, rAl3+=0.057nm.", "answer": "For Al3+, r+/r-=0.057/0.132=0.43; based on the relationship between cation coordination number and the radius ratio of positive to negative ions, the coordination number is 6." }, { "idx": 44, "question": "Barium titanate is an important ferroelectric ceramic, and its crystal type is perovskite structure. What lattice does it belong to?", "answer": "It belongs to the cubic crystal system" }, { "idx": 45, "question": "Barium titanate is an important ferroelectric ceramic with a perovskite crystal structure. What are the coordination numbers of the ions in this structure?", "answer": "The coordination numbers of Ba²⁺, Ti⁴⁺, and O²⁻ are 12, 6, and 6, respectively." }, { "idx": 46, "question": "What is the basis for the classification of silicate crystals?", "answer": "Silicate crystals are mainly classified based on the arrangement and bonding manner of [Si04] in the structure." }, { "idx": 47, "question": "Into which categories can silicate crystals be classified?", "answer": "Silicate crystals can be specifically classified into five categories: island, group, chain, layer, and framework." }, { "idx": 48, "question": "What are the structural characteristics of island silicates?", "answer": "The structural characteristics of island silicates: [Si04] shares 0 O2-, the shape is tetrahedral, the complex anion group is [Si04], and the Si:O ratio is 1:4." }, { "idx": 49, "question": "MnS has three polycrystalline forms, two with the NaCl-type structure and one with the cubic ZnS-type structure. When transforming from the cubic ZnS-type structure to the NaCl-type structure, what is the percentage change in volume? Given that for CN=6, ${r_{\\\\mathrm{{Mn}}}}^{2+}=0.08\\\\mathrm{{nm}}$, $\\\\ensuremath{r_{\\\\mathrm{s}}}^{2-}=0.184\\\\mathrm{nm}$; and for CN=4, ${r_{\\\\mathrm{Mn}}}^{2+}\\\\mathrm{=}0.073\\\\mathrm{nm}$, ${r_{\\\\mathrm{S}}}^{2-}{=}0.167\\\\mathrm{nm}$.", "answer": "For the cubic ZnS-type structure: $a1={\\\\frac{4}{\\\\sqrt{2}}}r_{S}^{2-}={\\\\frac{4}{\\\\sqrt{2}}}\\\\times0.167=0.472\\\\mathrm{nm}$. For the NaCl-type structure: $a_{2}=2(r_{\\\\mathrm{Mn}}^{2+}+r_{\\\\mathrm{S}}^{2-})=2(0.08+0.184)=0.384\\\\mathrm{nm}$. Thus, the volume change is: $V={\\\\frac{a1^{3}-a2^{3}}{a1^{3}}}={\\\\frac{0.472^{3}-0.384^{3}}{0.472^{3}}}=46.15\\\\%$." }, { "idx": 50, "question": "Barium titanate is an important ferroelectric ceramic with a perovskite crystal structure. Does this structure obey Pauling's rules? Please discuss in detail.", "answer": "This structure obeys Pauling's rules. Pauling's first rule—the coordination polyhedron rule: rO²⁻=0.132nm, rTi⁴⁺=0.064nm, rBa²⁺=0.143nm. For TiO₆, rTi⁴⁺/rO²⁻=0.064/0.132=0.485, coordination number is 6. For BaO₁₂, rBa²⁺/rO²⁻=0.143/0.132=1.083, coordination number is 12. This complies with Pauling's first rule. Pauling's second rule—the electrostatic valence rule: the anion charge Z=∑(zi⁺/CNi), thus the O²⁻ ion charge=(2/12)×4+(4/6)×2=2, which equals the O²⁻ ion charge, so it complies with Pauling's second rule. Furthermore, according to the perovskite-type structure, its coordination polyhedra do not share edges or faces, and the structural situation also complies with Pauling's fourth rule—the rule of different polyhedron connections and Pauling's fifth rule—the parsimony rule. Therefore, the perovskite structure obeys Pauling's rules." }, { "idx": 51, "question": "The chemical handbook lists the density of NH4Cl as 1.5 g/cm3. X-ray data indicate that NH4Cl has two crystal structures: one is the NaCl-type structure with a=0.726 nm, and the other is the CsCl structure with a=0.387 nm. To which crystal type does the above density value belong? (The NH4+ ion occupies the crystal lattice as a single unit).", "answer": "Solution: If NH4Cl has the NaCl structure, the density can be calculated using the formula: rho = n / (N_A * V) * M. Thus, the calculation yields: rho = 4 / (6.02 * 10^23) * 52.5 / (0.726 * 10^-7)^3 = 0.912 g/cm3. If NH4Cl has the CsCl structure, the density can be calculated using the formula: rho = n / (N_A * V) * M. Thus, the calculation yields: rho = 1 / (6.02 * 10^23) * 52.5 / (0.387 * 10^-7)^3 = 1.505 g/cm3. The calculation shows that the density of the NaCl-type NH4Cl is close to the value given in the chemical handbook, so the density corresponds to the NaCl crystal type." }, { "idx": 52, "question": "What are the structural characteristics of chain silicates?", "answer": "The structural characteristics of chain silicates: [Si04] shares 2 or 3 O2-, shapes include single chains and double chains, the complex anion groups are [Si206]4 and [Si40], with Si:O ratios of 1:3 and 4:11 respectively." }, { "idx": 53, "question": "What are the structural characteristics of island silicates?", "answer": "The structural characteristics of island silicates: [Si04] shares 1 or 2 O2-, with shapes including double tetrahedrons, three-membered rings, four-membered rings, and six-membered rings. The complex anion groups are [Si20], [Sis0]6, [Si4012]8, and [SisO18]12, with Si:O ratios of 2:7, 1:3, 1:3, and 1:3, respectively." }, { "idx": 54, "question": "What are the structural characteristics of layered silicates?", "answer": "The structural characteristics of layered silicates: [Si04] shares 3 O2-, the shape is a planar layer, the complex anion group is [Si4010], and the Si:O ratio is 4:10." }, { "idx": 55, "question": "What are the structural characteristics of framework silicates?", "answer": "The structural characteristics of framework silicates: [SiO4] shares 4 O2-, forming a skeleton [(AlxSi4)O], with the complex anion group being [SiO4]4, and the Si:O ratio is 1:4." }, { "idx": 56, "question": "What are the characteristics of silicate crystal structures? How are their chemical formulas characterized?", "answer": "Silicate crystal structures are very complex, but different structures share the following common characteristics: (1) The $\\\\mathrm{Si}^{4+}$ ions in the structure are located at the center of tetrahedra formed by $\\\\mathrm{O}^{2-}$ ions, constituting the basic structural unit of silicate crystals, the $[\\\\mathrm{SiO}_{4}]$ tetrahedron. The Si-O-Si bond forms a bent line with varying angles, generally around $145^{\\\\circ}$. (2) Each vertex of the $[\\\\mathrm{SiO}_{4}]$ tetrahedron, i.e., the $\\\\mathrm{O}^{2-}$ ion, can be shared by at most two $[\\\\mathrm{SiO}_{4}]$ tetrahedra. (3) Two adjacent $[\\\\mathrm{SiO}_{4}]$ tetrahedra can only share vertices and cannot share edges or faces. (4) The $\\\\mathrm{Si}^{4+}$ ion at the center of the $[\\\\mathrm{SiO}_{4}]$ tetrahedron can be partially replaced by $\\\\mathrm{Al}^{3+}$ ions. This substitution, known as isomorphous replacement, does not significantly alter the structure itself but greatly changes the properties of the crystal, offering possibilities for material modification. The chemical formulas of silicates are primarily characterized by the following two methods: (1) Oxide representation method All oxides constituting the silicate crystal are written out in a certain proportion and order, starting with monovalent alkali metal oxides, followed by divalent and trivalent metal oxides, and finally $\\\\mathrm{SiO}_{2}$. (2) Inorganic complex salt representation method All ions constituting the silicate crystal are written out in a certain proportion and order, with the relevant complex anions enclosed in brackets. The order starts with monovalent and divalent metal ions, followed by $\\\\mathrm{Al}^{3+}$ ions and $\\\\mathrm{Si}^{4+}$ ions, and finally $\\\\mathrm{O}^{2-}$ ions and $\\\\mathrm{OH}^{-}$ ions. The advantage of the oxide representation method is that it clearly reflects the chemical composition of the crystal, allowing for laboratory synthesis based on this formulation. The inorganic complex salt method, on the other hand, provides a more intuitive reflection of the structural type of the crystal, enabling predictions about its structure and properties. The two representation methods can be converted into each other." }, { "idx": 57, "question": "What type of silicate structure does Mg2[SiO4] belong to?", "answer": "Island" }, { "idx": 58, "question": "What type of silicate structure does K[AlSi3O8] belong to?", "answer": "Framework" }, { "idx": 59, "question": "What type of silicate structure does CaMg[Si2O6] belong to?", "answer": "Single chain" }, { "idx": 60, "question": "What type of silicate structure does Mg3[Si4O10](OH)2 belong to?", "answer": "Layered (double-chain)" }, { "idx": 61, "question": "What type of silicate structure does Ca2Al[AlSiO7] belong to?", "answer": "Group (double tetrahedron)" }, { "idx": 62, "question": "Based on the projection diagram of Mg2[SiO4] on the (100) plane, answer: Is the valence of O2- saturated?", "answer": "O2- is bonded to 3 [Mg] and 1 [SiO], N∑i(z+/CN)=2=1z-1, so O2- is saturated" }, { "idx": 63, "question": "Based on the projection of Mg2[SiO4] on the (100) plane, answer: what is the number of molecules in the unit cell?", "answer": "z=4" }, { "idx": 64, "question": "Graphite, talc, and kaolinite have layered structures. Explain the differences in the structure of graphite and the resulting differences in properties.", "answer": "Compared to talc and kaolinite, in graphite, the carbon atoms in the same layer undergo SP2 hybridization, forming large π bonds, and each layer has a hexagonal network structure. Due to the large interlayer gaps, electrons can move within the same layer, enabling electrical conductivity. The layers are held together by intermolecular forces, making graphite relatively soft." }, { "idx": 65, "question": "Based on the projection diagram of Mg2[SiO4] on the (100) plane, answer: What fraction of the tetrahedral and octahedral voids are occupied by Si4+ and Mg2+?", "answer": "Si4+ occupies tetrahedral voids = 1/8, Mg2+ occupies octahedral voids = 1/2" }, { "idx": 66, "question": "Describe the basic types of dislocations and their characteristics.", "answer": "There are two main types of dislocations: edge dislocation and screw dislocation. Characteristics of edge dislocation: the slip direction is perpendicular to the dislocation line, denoted by the symbol ⊥, with an extra half-plane of atoms. Characteristics of screw dislocation: the slip direction is parallel to the dislocation line, the plane perpendicular to the dislocation line is not flat, presenting a spiral shape, hence called screw dislocation." }, { "idx": 67, "question": "Graphite, talc, and kaolinite have layered structures. Explain the differences between the structures of talc and kaolinite and the resulting differences in properties.", "answer": "The main difference between talc and kaolinite is that talc is a 2:1 type structure with Mg replacing Al3+, where the octahedral layer is a trioctahedral structure, while kaolinite is a 1:1 type dioctahedral structure." }, { "idx": 68, "question": "In silicate crystals, why can Al3+ partially replace Si4+ in the silicon-oxygen framework?", "answer": "Al3+ can form [AlO4]5- with O2-. Al3+ and Si4+ are in the second period and have similar properties, making it easy for Al3+ to enter the silicate crystal structure and undergo isomorphous substitution with Si4+. Due to compliance with Pauling's rules, only partial substitution is possible." }, { "idx": 69, "question": "What is the effect of Al3+ replacing Si4+ on the composition of silicates?", "answer": "The replacement of Si4+ by Al3+ is a partial substitution. When Al3+ replaces Si4+, the structural units [AlSiO4][AlSiO5] lose electrical neutrality, resulting in excess negative charge. To maintain electrical neutrality, some larger cations with lower charges such as K+, Ca2+, and Ba2+ will enter the structure." }, { "idx": 70, "question": "Asbestos minerals such as tremolite $\\mathrm{Ca_{2}M g_{5}[S i_{4}O_{11}]}$ (OH)2 exhibit a fibrous crystal habit, while talc Mgs[Si4Oo](OH)2 displays a platy crystal habit. Please explain this phenomenon.", "answer": "Tremolite has a double-chain structure, where the Si-O bonds within the chains are much stronger than the Ca-O and Mg-O bonds between the chains. Therefore, it easily cleaves along the weaker interchain bonding sites, resulting in a fibrous habit. Talc has a layered structure composed of two [SiO4] layers with a brucite-like layer in between. The layers are held together by weak van der Waals forces, and due to the weak intermolecular forces, talc readily cleaves along these weak bonding planes to form platy crystals." }, { "idx": 71, "question": "What are the characteristics of non-stoichiometric compounds?", "answer": "Characteristics of non-stoichiometric compounds: The formation and defect concentration of non-stoichiometric compounds are related to the nature and pressure of the atmosphere; they can be regarded as solid solutions of higher-valent and lower-valent compounds; the defect concentration is related to temperature, which can be seen from the equilibrium constant; non-stoichiometric compounds are all semiconductors." }, { "idx": 72, "question": "Using the电价规则, explain that when Al3+ replaces Si4+ in the framework, it usually does not exceed half, otherwise the structure will become unstable.", "answer": "Assuming Al3+ replaces half of the Si4+, then O2- is connected to one Si4+ and one Al3+. The electrostatic bond strength of the cation = 3/4×1 + 4/4×1 = 7/4. The charge number of O2- is -2, and the difference between the two is 1/4. If the replacement exceeds half, the difference will inevitably be >1/4, causing structural instability." }, { "idx": 73, "question": "Why are non-stoichiometric compounds all n-type or p-type semiconductor materials?", "answer": "Due to anion vacancies and interstitial cations leading to an excess of metal ions, metal-excess (n-type) semiconductors are formed; cation vacancies and interstitial anions leading to an excess of anions form anion-excess (p-type) semiconductors." }, { "idx": 74, "question": "Explain the meaning of the symbol V_{Na}", "answer": "Sodium atom vacancy" }, { "idx": 75, "question": "What is the effect on crystal stability after forming a solid solution?", "answer": "Stabilizes the crystal lattice and prevents certain polymorphic transitions." }, { "idx": 76, "question": "What is the effect on lattice activation after forming a solid solution?", "answer": "Activating the lattice, after forming a solid solution, the lattice structure undergoes certain distortion and is in a high-energy activated state, which is conducive to chemical reactions." }, { "idx": 77, "question": "What are the factors affecting the formation of substitutional solid solutions?", "answer": "Factors influencing the formation of substitutional solid solutions include: (1) Ionic size: The 15% rule - 1. If (R1-R2)/R1 > 15%, discontinuous. 2. If ≤15%, continuous. 3. If >40%, solid solution cannot form. (2) Ionic valence: Same valence leads to continuous solid solution formation. (3) Crystal structure factors: Same structure of matrix and impurity results in continuous solid solution. (4) Field strength factor. (5) Electronegativity: Small difference favors solid solution formation, while large difference leads to compound formation." }, { "idx": 78, "question": "Based on the projection diagram of Mg2[SiO4] on the (100) plane, answer: How many types of coordination polyhedra are there in the structure, and what are the connection modes between the various coordination polyhedra?", "answer": "There are two types of coordination polyhedra, [SiO4] and [MgO6]. [MgO6] octahedra in the same layer share edges, such as 59[MgO] and 49[MgO6] sharing edges 7502- and 2702-. [MgO6] octahedra in different layers share vertices, such as 1[MgO] and 51[MgO6] sharing vertex 2202-. [MgO] and [SiO4] in the same layer share vertices, such as T[MgO] and 7[SiO4] sharing vertex 2202-. [MgO6] and [SiO4] in different layers share edges, such as T[MgO] and 43[SiO4] sharing 2802- and 2802-." }, { "idx": 79, "question": "What is the effect of forming a solid solution on the mechanical properties of materials?", "answer": "Solid solution strengthening; the dissolution of solute atoms increases the strength and hardness of the solid solution. The strength and hardness of the solid solution are often higher than those of the constituent elements, while the plasticity is lower." }, { "idx": 80, "question": "What is the effect on the physical properties of materials after forming a solid solution?", "answer": "The electrical, thermal, magnetic, and other physical properties of solid solutions also change continuously with composition, but generally not in a linear relationship." }, { "idx": 81, "question": "Explain the meaning of the symbol V_{Cl}^*", "answer": "Chlorine ion vacancy, with a single positive charge" }, { "idx": 82, "question": "Explain the meaning of the symbol (V_{Na}'V_{Cl}^*)", "answer": "An associated center formed by the nearest Na vacancy and Cl vacancy" }, { "idx": 83, "question": "Explain the meaning of the symbol Ca_{i}^{* *}", "answer": "Ca2+ is located at the interstitial site of the lattice" }, { "idx": 84, "question": "Explain the meaning of the symbol Ca_{K}^*", "answer": "Ca2+ occupies the K site, with a unit positive charge" }, { "idx": 85, "question": "Explain the meaning of the symbol Ca_{Ca}", "answer": "Ca atom located at the Ca atom site" }, { "idx": 86, "question": "Explain the meaning of the symbol V_{Na}'", "answer": "Sodium ion vacancy, with a single negative charge" }, { "idx": 87, "question": "Write the defect reaction equation for NaCl dissolving into CaCl2 to form a vacancy-type solid solution", "answer": "NaCl > Naca + Cla + Va* NaC" }, { "idx": 88, "question": "Write the defect reaction equation for the formation of Schottky defects in NaCl", "answer": "$Rsh$ $+V_{cl}$" }, { "idx": 89, "question": "Write the defect reaction equation for the formation of a vacancy-type solid solution when CaCl2 dissolves in NaCl", "answer": "CaCl2 > CaNa' + 2Clci + Va'" }, { "idx": 90, "question": "What are the factors affecting the formation of interstitial solid solutions?", "answer": "Factors influencing the formation of interstitial solid solutions include: (1) Size of impurity particles: The smaller the added atoms, the easier it is to form a solid solution, and vice versa. (2) Crystal (matrix) structure: The size of ions is closely related to the crystal structure, and to a certain extent, the size of the interstitial sites plays a decisive role. Generally, the larger the voids in the crystal, the looser the structure, and the easier it is to form a solid solution. (3) Valence factor: When foreign impurity atoms enter the interstitial sites, they inevitably cause an imbalance in the valence of the crystal structure. This can be compensated by generating vacancies, partial substitution, or changes in the valence state of ions to maintain valence balance." }, { "idx": 91, "question": "Write the defect reaction equation for the formation of Frenkel defects (Ag entering interstitial sites) in AgI", "answer": "AgAg $V_{Ag}$ $+Ag_{i}$" }, { "idx": 92, "question": "If the partial pressure of surrounding oxygen is increased, how will the density of the non-stoichiometric compound Fe1-xO change? Increase or decrease? Why?", "answer": "202(g) 0o+Vre’ +2h* k=[O][Vr’’][h']/PO21/2=4[00][Vr’′]3/PO1/2 [Vre′′]PO²1/6, PO[Vr]↓" }, { "idx": 93, "question": "Both MgO (NaCl-type structure) and Li2O (anti-fluorite-type structure) are based on the cubic close packing of oxygen, and the cations are in the gaps of this arrangement. Why are Frenkel-type point defects the main defects in Li2O?", "answer": "Li occupies octahedral voids. The octahedral voids are larger, making it easier for Li ions to move from their normal lattice sites to interstitial positions, forming Frenkel defects (simultaneous presence of cation vacancies and interstitial cations)." }, { "idx": 94, "question": "The density of MgO is $3.58\\\\mathrm{g/cm^{3}}$, and its lattice parameter is $0.42\\\\mathrm{nm}$. Calculate the number of Schottky defects per unit cell of MgO.", "answer": "Let the number of molecules per unit cell of defective $\\\\mathrm{MgO}$ be ${\\\\bf X}$, the unit cell volume $\\\\mathrm{V}{=}$ (4.20)°, $\\\\scriptstyle{\\\\mathbf{x}}={\\\\mathbf{p}}$ VNO/M=3.96. The number of Schottky defects per unit cell $:=4-\\\\tt X=0.04$." }, { "idx": 95, "question": "The Schottky defect formation energy of Mg0 crystal is 84 kJ/mol. Calculate the defect concentration of this crystal at 1000K and 1500K.", "answer": "$\\\\mathrm{n}/\\\\mathrm{N}{=}\\\\mathrm{exp}$ (-E/2RT), $\\\\mathrm{R}{=}8.314$, $\\\\mathrm{T}{=}1000\\\\mathrm{k}$ :n/N=6. $4\\\\times{10}^{-3}$ $\\\\mathrm{T}=1500\\\\mathrm{k}$ $\\\\mathrm{n}/\\\\mathrm{N}{=}3.5\\\\times{10}^{-2}$" }, { "idx": 96, "question": "In the non-stoichiometric compound Fe_xO, Fe^{3+}/Fe^{2+}=0.1. Find the vacancy concentration in Fe_xO.", "answer": "Fe2O3 →2FeFe +3O0+V_Fe. Let y be the concentration of Fe^{3+}, then 2y =0.1→2y=0.1-0.3y→y=0.1/2.3=0.0435. The vacancy concentration [V_Fe''] = y = 0.0435." }, { "idx": 97, "question": "In the non-stoichiometric compound Fe_xO, Fe^{3+}/Fe^{2+}=0.1, find the value of x in Fe_xO.", "answer": "Let y be the concentration of Fe^{3+}, y=0.0435. x=1-y=1-0.0435=0.9565, thus the chemical formula is Fe_0.9565O." }, { "idx": 98, "question": "For edge dislocations, what are the characteristics of the dislocation line direction, Burgers vector, and dislocation motion direction?", "answer": "Edge dislocation: the dislocation line is perpendicular to the Burgers vector, and the dislocation line is perpendicular to the direction of dislocation motion." }, { "idx": 99, "question": "When two edge dislocations with the same sign meet on the same slip plane, will they repel or attract each other?", "answer": "Repel, tensile stress overlaps, compressive stress overlaps." }, { "idx": 100, "question": "For screw dislocations, what are the characteristics of the dislocation line direction, Burgers vector, and dislocation motion direction?", "answer": "Screw dislocation: the dislocation line is parallel to the Burgers vector, and the dislocation line is parallel to the direction of dislocation motion." }, { "idx": 101, "question": "Both MgO (NaCl-type structure) and Li2O (anti-fluorite-type structure) are based on the cubic close packing of oxygen, and the cations are located in the interstitial sites of this arrangement. Why are Schottky defects the predominant point defects in MgO?", "answer": "Mg occupies the tetrahedral sites. The tetrahedral sites are relatively small, making it difficult for Mg ions to migrate within the lattice to form Frenkel defects. Therefore, they are more inclined to form Schottky defects (paired cation and anion vacancies)." }, { "idx": 102, "question": "If the partial pressure of surrounding oxygen is increased, how will the density of the non-stoichiometric compound Zn1+xO change? Increase or decrease? Why?", "answer": "Zn (g) →Zni· +e Zn (g) +1/2O2=ZnO Zn:+e’+1/2O→ZnO [ZnO]=[e'] PO2↑,[Zni]↓" }, { "idx": 103, "question": "What effect will grain boundaries have on the movement of dislocations? Can it be predicted?", "answer": "Grain boundaries hinder the movement of dislocations." }, { "idx": 104, "question": "Grain boundaries can be divided into small-angle grain boundaries and large-angle grain boundaries. Can large-angle grain boundaries be described by arrays of dislocations?", "answer": "No, in large-angle grain boundaries, the atomic arrangement is close to a disordered state, and the distance between dislocations may only be 1 or 2 atoms in size, which is not applicable to large-angle grain boundaries." }, { "idx": 105, "question": "Taking the dissolution of solute A0 in solvent B2O3 as an example, compare the chemical compositions of solid solutions, compounds, and mechanical mixtures.", "answer": "
Comparison itemSolid solutionCompoundMechanical mixture
Chemical compositionB2xAxO X （x=0~2）AB2O4AO+B2O3
" }, { "idx": 106, "question": "From the perspective of chemical composition and phase composition, compare the differences between solid solutions and mechanical mixtures.", "answer": "
Solid solutionMechanical mixture
Formation reasonFormed by atomic-scale 'dissolution'Powder mixing
Number of phasesUniform single phaseMultiphase
Chemical compositionUncertainAs many chemical compositions as there are mixtures
" }, { "idx": 107, "question": "Explain the similarities between solid solutions, lattice defects, and non-stoichiometric compounds", "answer": "Solid solutions, lattice defects, and non-stoichiometric compounds are all point defects, which are crystal structure defects, and they are single-phase homogeneous solids with structures identical to the host crystal phase." }, { "idx": 108, "question": "Explain the differences between solid solutions, lattice defects, and non-stoichiometric compounds", "answer": "Thermal defects - intrinsic defects; solid solutions - extrinsic defects; non-stoichiometric compounds - caused by changes in environmental atmosphere properties and pressure." }, { "idx": 109, "question": "Al2O3 forms a limited solid solution in MgO, with approximately 18wt% Al2O3 dissolved in MgO at the eutectic temperature of 1995℃, assuming the unit cell size change of MgO is negligible. Estimate the density change when Al3+ acts as a substitutional ion.", "answer": "The defect reaction is: Al2O3→MgO 2Al•Mg + 3O×o + V''Mg. Taking 100g of sample as the basis: mAl2O3 = 18/102 = 0.176 mol, mMgO = 82/40.3 = 2.035 mol. The molecular formula after solid solution is Mg2.035Al0.352O2.563. The density change is ρ/ρMgO = (0.176×102 + 2.035×40.3)/(2.563×40.3) = 0.968. The density after solid solution is less than that before solid solution." }, { "idx": 110, "question": "In MgO-Al2O3 and PbTiO3-PbZrO3, which pair forms a limited solid solution, and why?", "answer": "MgO-Al2O3 forms a limited solid solution because the ionic radii of Mg2+ and Al3+ differ significantly, and the crystal structure types of MgO (NaCl-type structure) and Al2O3 (corundum-type structure) are quite different." }, { "idx": 111, "question": "For MgO, Al2O3, and Cr2O3, the cation-anion radius ratios are 0.47, 0.36, and 0.40 respectively. Is the solid solubility in the MgO-Cr2O3 system expected to be limited or unlimited? Why?", "answer": "The solid solubility between MgO and Cr2O3 is limited. Reason: different structure types, MgO has a NaCl-type structure while Cr2O3 has a corundum structure. Although (0.47-0.40)/0.47=14.89%<15%, continuous solid solution still cannot be formed." }, { "idx": 112, "question": "Al2O3 forms a limited solid solution in MgO, with approximately 18wt% Al2O3 dissolved in MgO at the eutectic temperature of 1995℃, assuming the unit cell size change of MgO is negligible. Estimate the density change when Al3+ acts as an interstitial ion.", "answer": "The defect reaction is: Al2O3→MgO 2Al••i + 3O×o + 3V''Mg. Taking 100g of the sample as the basis: mAl2O3 = 18/102 = 0.176 mol, mMgO = 82/40.3 = 2.035 mol. The molecular formula after solid solution is Mg2.035Al0.352O2.563. The density change is ρ/ρMgO = (0.176×102 + 2.035×40.3)/(2.563×40.3) = 0.968. The density after solid solution is less than that before solid solution." }, { "idx": 113, "question": "A certain NiO is non-stoichiometric. If the ratio of $\\\\mathrm{Ni^{3+}/Ni^{2+}}$ in NiO is $10^{-4}$, how many charge carriers are there per $\\\\mathrm{m}^{3}$?", "answer": "Let the non-stoichiometric compound be $\\\\mathrm{Ni_{1}O}$. The reaction is: $Ni_{2}O_{3}\\\\xrightarrow{2\\\\sqrt{6}O}2^{1\\\\sqrt{1}i_{2}}Ni_{3}O_{0}+\\\\mathrm{V_{Ni}^{''}}$. The ratio $\\\\mathrm{Ni^{3+}/Ni^{2+}}=2y/(1-3y)=10^{-4}$. Solving gives $y=5\\\\times10^{-5}$, $x=1-y=0.99995$, leading to Ni$_{0.99995}$O. The number of charge carriers per $\\\\mathrm{m}^{3}$ is equal to the vacancy concentration: $[\\\\mathrm{V_{Ni}^{''}}]=y/\\\\Omega(1+x)=2.5\\\\times10^{-5}$." }, { "idx": 114, "question": "Explain why only substitutional solid solutions can achieve complete mutual solubility between the two components, while interstitial solid solutions cannot.", "answer": "(1) The interstitial sites in a crystal are limited, with an impurity accommodation capacity of $\\zeta10\\%$; (2) The formation of interstitial solid solutions generally increases the lattice constant, and when this increase reaches a certain extent, the lattice becomes unstable and dissociates; substitutional solid solutions are formed by the exchange of positions between ions of the same type, which does not affect the bonding, thus allowing the formation of continuous solid solutions." }, { "idx": 115, "question": "For MgO, Al2O3, and Cr2O3, the radius ratios of cations to anions are 0.47, 0.36, and 0.40, respectively. Is it possible for Al2O3 and Cr2O3 to form continuous solid solutions? Why?", "answer": "Al2O3 and Cr2O3 can form continuous solid solutions, because: 1) They have the same crystal structure type, both belonging to the corundum structure. 2) (0.40-0.36)/0.40=10%<15%" }, { "idx": 116, "question": "List a concise table comparing solid solutions, lattice defects, and non-stoichiometric compounds", "answer": "
ClassificationFormation reasonFormation conditionDefect reactionChemical formulaSolubility, defect concentration
Thermal defectSchottky FrenkelThermal fluctuationT>OkVM M+Vx 0MX MXOnly controlled by temperature
M>M+V
Solid solutionInfinite, finite, substitution, interstitialImpurity dissolutionSize, electronegativity, valence, structureNone: controlled by temperature With: impurity amount < solubility limit controlled by temperature impurity amount > solubility limit controlled by solubility limit
Non-stoichiometric compoundCation vacancy anion interstitial cation interstitial anion vacancyChanges in environmental atmosphere nature and pressureFe1x0 UO2x Zn1x0 TiO2-x[h'] pl16 8 20 [Zni]oPo P-1/6
" }, { "idx": 117, "question": "Taking the dissolution of solute A0 in solvent B2O3 as an example, compare the phase composition of solid solutions, compounds, and mechanical mixtures.", "answer": "
Comparison itemSolid solutionCompoundMechanical mixture
Phase compositionHomogeneous single phaseSingle phaseTwo phases with interface
" }, { "idx": 118, "question": "Chemical analysis of pyrite yields two possible compositions based on the Fe/S ratio from the analytical data: Fe1-xS and FeS1-x. The former implies a defect structure with Fe vacancies, while the latter indicates Fe substitution. How can experimental methods determine whether the mineral belongs to Fe1-xS (Fe vacancy defect structure)?", "answer": "Fe1-xS contains Fe vacancies and is a non-stoichiometric compound, exhibiting h'P-type semiconductor properties. By measuring its semiconductor characteristics, if the results show h'P-type semiconductor behavior, it can be confirmed that the mineral composition is Fe1-xS." }, { "idx": 119, "question": "In MgO-Al2O3 and PbTiO3-PbZrO3, which pair forms an infinite solid solution, and why?", "answer": "PbTiO3-PbZrO3 forms an infinite solid solution. Although the ionic radii of Ti4+ and Zr4+ differ significantly (approximately 15.28%), they both possess the ABO3 perovskite-type structure, and both Ti4+ and Zr4+ occupy the octahedral voids. These voids are relatively large, allowing the radii of the inserted cations to vary within a certain range without causing structural changes." }, { "idx": 120, "question": "Chemical analysis of pyrite, based on the Fe/S ratio from the analytical data, yields two possible compositions: Fe1-xS and FeS1-x. The former implies a defect structure with Fe vacancies, while the latter indicates substitution of Fe. How can experimental methods determine that the mineral belongs to FeS1-x (a defect structure with S vacancies)?", "answer": "In FeS1-x, there is an excess of metal ions and the presence of S2- vacancies, which manifests as an N-type semiconductor. By measuring its semiconductor properties, if the results show it is an N-type semiconductor, then the mineral composition can be confirmed as FeS1-x." }, { "idx": 121, "question": "Use experimental methods to identify SiO2 glass", "answer": "Use X-ray detection. SiO2 glass - isotropic." }, { "idx": 122, "question": "Use experimental methods to identify crystalline SiO2", "answer": "Use X-ray detection. Crystalline SiO2—particles are arranged regularly in three-dimensional space, exhibiting anisotropy." }, { "idx": 123, "question": "What are the characteristics of the polymer structure of silicate melts?", "answer": "The characteristics of the polymer structure of silicate melts include: being composed of silicon-oxygen tetrahedra as the basic units forming polymers of varying sizes; undergoing three stages during formation: differentiation, polycondensation, and equilibrium; accompanying deformation during polycondensation, where chain polymers undergo rotation and bending, layered polymers experience wrinkling and warping, and framework polymers exhibit increased thermal defects with changes in the Si-O-Si bond angle; ultimately reaching an equilibrium state of polymerization and depolymerization under the influence of time and temperature." }, { "idx": 124, "question": "Use experimental methods to identify SiO2 melt", "answer": "Use X-ray detection. SiO2 melt—internal structure is framework-like, short-range ordered, long-range disordered." }, { "idx": 125, "question": "Use experimental methods to identify silica gel", "answer": "Use X-ray detection. Silica gel—loose and porous." }, { "idx": 126, "question": "Describe the structural and property characteristics of quartz melt", "answer": "
StructureFramework structure, long-range disorder
PropertiesHigh viscosity, large surface tension
" }, { "idx": 127, "question": "What are the factors affecting the viscosity of the melt?", "answer": "The main factors affecting the viscosity of the melt: temperature and the composition of the melt. The increase in the content of alkaline oxides drastically reduces the viscosity. As the temperature decreases, the viscosity of the melt increases exponentially." }, { "idx": 128, "question": "What is the process of polymer structure formation in silicate melts?", "answer": "The formation of polymers is based on the silicon-oxygen tetrahedron as the fundamental unit, forming aggregates of varying sizes. It can be divided into three stages: Initial stage: The breakdown of quartz, where the framework [Si0_{4}] fractures, forming polymers of varying degrees of polymerization in the melt. Middle stage: Polycondensation accompanied by deformation—linear polymers tend to rotate around the Si-O axis while bending, layered polymers cause the layers themselves to wrinkle and warp, framework polymers exhibit increased thermal defects, and the Si-O-Si bond angles change. [Si0_{4}]Na_{4} + [Si_{2}0_{7}]Na_{6} → [Si_{3}0_{10}]Na_{8} + Na_{2}O (short bond) 3[Si_{3}0_{10}]Na_{8} → [Si_{6}0_{18}]Na_{12} + 2Na_{2}O (six-membered ring). Final stage: Within a certain time and temperature range, polymerization and depolymerization reach equilibrium. The Na_{2}O released by polycondensation can further erode the quartz framework, breaking it down into oligomers, and this cycle continues until the system reaches a breakdown-polycondensation equilibrium." }, { "idx": 129, "question": "Analyze the reason why monovalent alkali metal oxides reduce the viscosity of silicate melts.", "answer": "Generally, alkali metal oxides (Li2O, Na2O, K2O, Rb2O, Cs2O) can reduce melt viscosity. These cations, due to their small charge, large radius, and weak interaction with O2-, provide 'free oxygen' in the system, increasing the O/Si ratio. This causes the original silicon-oxygen anion groups to depolymerize into simpler structural units, thereby reducing the activation energy and decreasing viscosity." }, { "idx": 130, "question": "Describe the structural and property characteristics of quartz crystals", "answer": "
StructureRegular ordered arrangement, long-range order
Properties
" }, { "idx": 131, "question": "The above data were obtained under constant pressure. If obtained under constant volume, do you think the activation energy would change? Why?", "answer": "If obtained under constant volume, the activation energy would not change. Because activation energy is the energy required for liquid particles to undergo linear motion. It is related to the melt composition and the degree of [Si04] polymerization in the melt." }, { "idx": 132, "question": "What are the structural differences between crystalline SiO2, SiO2 glass, silica gel, and SiO2 melt?", "answer": "Crystalline SiO2—particles are arranged regularly in three-dimensional space, exhibiting anisotropy; SiO2 glass—isotropic; silica gel—loose and porous; SiO2 melt—internal structure is framework-like, with short-range order and long-range disorder." }, { "idx": 133, "question": "Describe the structural and property characteristics of Na2O·2SiO2 melt", "answer": "
StructureSix-membered or eight-membered rings, long-range disorder
PropertiesHigh electrical conductivity, high surface tension
" }, { "idx": 134, "question": "Can glass be formed by rapid quenching at 1050°C?", "answer": "At this temperature, rapid quenching can form glass." }, { "idx": 135, "question": "CeO2 has a fluorite structure. When 15 mol% CaO is added to form a solid solution, the measured density of the solid solution is d = 7.01 g/cm³, and the unit cell parameter is a = 0.5417 nm. Determine through calculation which type of solid solution is formed. The atomic weights are Ce 140.12, Ca 40.08, O 16.00.", "answer": "For the CaO-CeO2 solid solution, from the perspective of maintaining electrical neutrality, it can form either an oxygen vacancy solid solution or a solid solution with Ca²⁺ embedded in the anion interstitial sites. The solid solution equations are as follows: For substitutional solid solution, x = 0.15, 1 - x = 0.85, 2 - x = 1.85, so the chemical formula of the substitutional solid solution is Ca0.15Ce0.85O1.85. Since CeO2 has a fluorite structure, the number of unit cell molecules Z = 4, and the unit cell contains three types of ions: Ca²⁺, Ce⁴⁺, and O²⁻. The mass of the unit cell is: W = ΣWi = [4 × (0.15/1) × MCa²⁺ + 4 × (0.85/1) × MCe⁴⁺ + 8 × (1.85/2) × MO²⁻] / 6.022 × 10²² = 102.766 × 10⁻²² g. The calculated density dR = W/V = 0.6465 g/cm³. For interstitial solid solution, the chemical formula is Ca2Ce1-yO2. Compared with the given composition Ca0.15Ce0.85O1.85, the O²⁻ content differs: Ca0.15Ce0.85O1.85 → Ca0.15×2/1.85Ce0.85×2/1.85O2. y = 0.15 × 2 / 1.85, 1 - y = 0.85 × 2 / 1.85 ⇒ y = 0.15 / 1.85. The chemical formula of the interstitial solid solution is Ca0.15×2/1.85Ce1.7/1.85O2. Similarly, the calculated density dg = W/V = 7.033 g/cm³. The measured density is d = 7.01 g/cm³, which is close to d2. Therefore, an interstitial solid solution is formed, with interstitial Ca²⁺ ions present." }, { "idx": 136, "question": "The viscosity of SiO2 melt is 10^14 Pa·s at 1000°C and 10^7 Pa·s at 1400°C. What is the activation energy for viscous flow of SiO2 glass?", "answer": "According to the formula: η=η0exp(ΔE/RT)\\nAt 1000°C, η=10^14 Pa·s, T=1000+273=1273 K\\nAt 1400°C, η=10^7 Pa·s, T=1400+273=1673 K\\nSolving the two equations simultaneously yields: η0=5.27×10^-16 Pa·s\\nΔE=713.5 kJ/mol" }, { "idx": 137, "question": "The melt viscosity is 10^7 Pa·s at 727°C and 10^3 Pa·s at 1156°C. At what temperature will it be 10^6 Pa·s?", "answer": "According to logη=A+B/T, at 727°C, η=10^7 Pa·s, the equation gives: log10^7=A+B/(727+273) (1). At 1156°C, η=10^3 Pa·s, the equation gives: log10^3=A+B/(1156+273) (2). Solving equations (1) and (2) simultaneously yields A=-6.32 and B=13324. When η=10^6 Pa·s, log10^6=-6.32+13324/(t+273), solving gives t=808.5°C." }, { "idx": 138, "question": "A melt has a viscosity of 310Pa·s at 1300℃ and 10^7Pa·s at 800℃. What is its viscosity at 1050℃?", "answer": "According to logη=A+B/(T+273), at 1300℃, η=310Pa·s, the formula gives: log310=A+B/(1300+273)①; at 800℃, η=10^7Pa·s, the formula gives: log10^7=A+B/(800+273)②. Solving equations ① and ② simultaneously yields A=−7.2, B=15219.6. When t=1050℃, logη=−7.2+15219.6/(1050+273), solving gives η=20130.5Pa·s." }, { "idx": 139, "question": "The viscosity of Pyrex glass is 10^9 Pa·s at 1400°C and 10^13 Pa·s at 840°C. To facilitate forming, approximately what temperature is required for the glass to reach a viscosity of 10^5 Pa·s?", "answer": "According to the formula: η = η0 exp(ΔE / (R T)). Given η0 = 11.22 Pa·s and ΔE = 254.62 kJ/mol. When η = 10^5 Pa·s, 10^5 = 11.22 exp(254.62 × 1000 / ((273 + t) × 8.314)). Solving gives t = 3094.2°C." }, { "idx": 140, "question": "The viscosity of Pyrex glass is 10^9 Pa·s at 1400℃ and 10^13 Pa·s at 840℃. What is the activation energy for viscous flow?", "answer": "According to the formula: η = η0 exp(ΔE / (R T)). At 1400℃, η = 10^9 Pa·s, T = 1400 + 273 = 1673 K, 10^9 = η0 exp(ΔE / (1673 × 8.314)). At 840℃, η = 10^13 Pa·s, T = 840 + 273 = 1113 K, 10^13 = η0 exp(ΔE / (1113 × 8.314)). Solving the two equations simultaneously yields: η0 = 11.22 Pa·s, ΔE = 254.62 kJ/mol." }, { "idx": 141, "question": "From the following two glaze formulas, determine the difference in their melting temperatures? Explain the reason. Glaze formula 1: 0.2K2O 0.2Na2O 0.4CaO 0.2PbO 0.3Al2O3 2.1SiO2; Glaze formula 2: 0.2K2O 0.2MgO 0.6CaO 1.1Al2O3 10.0SiO2", "answer": "(2) The melting temperature of glaze formula 1 > the melting temperature of glaze formula 2" }, { "idx": 142, "question": "A borosilicate glass used for sealing lamps has an annealing point of 544°C and a softening point of 780°C. Calculate the activation energy for viscous flow of this glass.", "answer": "According to the formula: η=η0exp(ΔE/RT). At the annealing point of 544°C, η=1.0×10^12 Pa·s, T=544+273=817K; at the softening point of 780°C, η=4.5×10^6 Pa·s, T=780+273=1053K. Solving the simultaneous equations gives: η0=1.39×10^-12 Pa·s, ΔE=373.13 kJ/mol." }, { "idx": 143, "question": "From the following two glaze formulas, determine the difference in surface tension between them? Explain the reason. Glaze formula 1: 0.2K2O 0.2Na2O 0.4CaO 0.2PbO 0.3Al2O3 2.1SiO2; Glaze formula 2: 0.2K2O 0.2MgO 0.6CaO 1.1Al2O3 10.0SiO2", "answer": "(3) Difference in surface tension: The surface tension of glaze formula 1 < the surface tension of glaze formula 2, because the O/Si ratio of glaze formula 1 is less than that of glaze formula 2, and glaze formula 1 contains PbO and B2O3, which can reduce surface tension." }, { "idx": 144, "question": "A borosilicate glass used for sealing lighting lamps has an annealing point of 544°C, a softening point of 780°C, and a viscous flow activation energy of 373.13 kJ/mol. Determine its working range.", "answer": "The working temperature range viscosity is generally 10^3~10^7 Pa·s. According to the formula T=ΔE/(R·ln(η/η0)): when η=10^3 Pa·s, T=1033.6K=760.6°C; when η=10^7 Pa·s, T=1038.9°C. Therefore, the working temperature range is 760.6°C~1038.9°C." }, { "idx": 145, "question": "The working range of a certain type of glass is from $870^{\\\\circ}\\\\mathrm{C}$ ($\\\\eta=10^{6}\\\\mathrm{{Pa}\\\\cdot{}}$) to $1300^{\\\\circ}\\\\mathrm{C}$ ($\\\\eta=10^{2.5}\\\\mathrm{{Pa}\\\\cdot\\\\mathrm{{s}}}$). Estimate its annealing point ($\\\\eta=10^{12}\\\\mathrm{{Pa}\\\\cdot\\\\Delta s}$)?", "answer": "According to the formula: $\\\\mathfrak{H}=\\\\mathfrak{N}_{0}\\\\exp(\\\\frac{\\\\Delta\\\\mathcal{E}}{R T})$ \\n\\nAt $870^{\\\\circ}\\\\mathrm{C}$, $\\\\eta=10^{6}\\\\mathrm{{Pa}\\\\cdot\\\\mathrm{{s}}}$, T=870+273=1143K, $10^{6}=1.57\\\\times10^{-7}\\\\exp[\\\\frac{280.16\\\\times1000}{1143\\\\times8.314}]$ (1) \\nAt $1300^{\\\\circ}\\\\mathrm{C}$, $\\\\eta=10^{2.5}\\\\mathrm{{Pa}\\\\cdot\\\\mathrm{{s}}}$ (2) \\nBy solving equations (1) and (2) simultaneously, we get: $\\\\mathfrak{V}_{0}=1.57\\\\times\\\\ensuremath{10^{-7}}\\\\mathrm{Pa}\\\\cdot\\\\mathrm{s}$, $\\\\Delta E=280.16\\\\mathrm{kJ/mol}$ \\nWhen $\\\\eta=10^{12}\\\\mathrm{{Pa}\\\\cdot\\\\mathrm{{s}}}$, $10^{12}=1.57\\\\times10^{-7}\\\\exp[\\\\frac{280.16\\\\times1000}{(273+t)\\\\times8.314}]$ \\nSolving gives $t=505.15^{\\\\circ}\\\\mathrm{C}$" }, { "idx": 146, "question": "What are the kinetic factors affecting the glass formation process?", "answer": "The key factor affecting glass formation is the cooling rate of the melt. Whether the melt crystallizes or forms glass is related to the degree of supercooling, viscosity, nucleation rate, and crystal growth rate." }, { "idx": 147, "question": "What are the crystallochemical factors affecting the glass formation process?", "answer": "The crystallochemical factors affecting glass formation include: the size and arrangement of complex anion groups, bond strength, and bond type." }, { "idx": 148, "question": "From the following two glaze formulas, determine the difference in viscosity between the two? Explain the reason. Glaze formula 1: 0.2K2O 0.2Na2O 0.4CaO 0.2PbO 0.3Al2O3 2.1SiO2; Glaze formula 2: 0.2K2O 0.2MgO 0.6CaO 1.1Al2O3 10.0SiO2", "answer": "(1) Difference in viscosity\\nFor glaze formula 1:\\n∵(K2O+Na2O+CaO+PbO)/Al2O3=(0.2+0.2+0.4+0.2)/0.3=3.33>1, Al3+ acts as a network-forming ion,\\nR1=(0.2+0.2+0.4+0.2+0.3×3+2.1×2+0.5×3)/(0.3×2+2.1+0.5×2)=2.05\\nX1=2×2.05-4=0.1\\nY1=4-0.1=3.9\\nFor glaze formula 2:\\n∴(K2O+MgO+CaO)/Al2O3=(0.2+0.2+0.6)/1.1=0.910<1, Al3+ is considered a network-modifying ion\\nR2=(0.2+0.2+0.6+1.1×3+10×2)/10.0=2.43\\nX2=2×2.43-4=0.86\\nY2=4-0.86=3.14\\nThat is: Y of glaze formula 1 > Y of glaze formula 2, so at high temperatures, the viscosity of glaze formula 1 > the viscosity of glaze formula 2." }, { "idx": 149, "question": "A borosilicate glass used for sealing lighting lamps has an annealing point of 544°C, a softening point of 780°C, and a viscous flow activation energy of 373.13 kJ/mol. Determine its melting range.", "answer": "The viscosity range for melting is generally 10~100 Pa·s. According to the formula T=ΔE/(R·ln(η/η0)): when η=10 Pa·s, T=1516.0K=1243.0°C; when η=100 Pa·s, T=1406.6K=1133.6°C. Therefore, the melting temperature range is 1133.6°C~1243.0°C." }, { "idx": 150, "question": "Compare which of the two types of glass has higher viscosity at high temperature?", "answer": "V1>V2, Glass No.1 has higher viscosity at high temperature." }, { "idx": 151, "question": "Calculate the structural parameters of Glass No. 2, with the composition of Na2O 10 mol%, CaO 0 mol%, Al2O3 20 mol%, SiO2 60 mol%, B2O3 10 mol%", "answer": "No. 2: Z=4, (Na2O+CaO)/Al2O3=(10+0)/20=0.5<1, Al3+ is considered as a network modifier ion. R1=(10+60+120+30)/(60+20)=2.75. X2=2R-Z=1.5. Y2=4-1.5=2.5." }, { "idx": 152, "question": "Calculate the structural parameters and non-bridging oxygen fraction of the glass $\\mathrm{_{6}(1)N a_{2}O\\bullet S i0_{2}}$", "answer": "Z=4, R=3/1=3, X=2R-Z=6-4=2, Y=8-2R=8-6=2. Non-bridging oxygen %=2/(1+2)=66.7%" }, { "idx": 153, "question": "Which substances can form non-crystalline solids (NCS)?", "answer": "Melts and glassy solids can form non-crystalline solids." }, { "idx": 154, "question": "Calculate the structural parameters and non-bridging oxygen fraction of the glass $\\mathrm{(3)Na_{2}0\\bullet_{}1/3A l_{2}O_{3}\\bullet S i O_{2}}$", "answer": "Al2O3>1, Al3+ is considered as a network-forming ion. Z=4, R=(1+1+2)/(2*(2/3)+1)=2.4, X=2R-Z=4.8-4=0.8, Y=4-0.8=3.2. Non-bridging oxygen %=0.8/(1.6+0.8)=33.3%" }, { "idx": 155, "question": "Calculate the structural parameters of Glass No. 1, with the composition of Na2O 20 mol%, CaO 10 mol%, Al2O3 10 mol%, SiO2 60 mol%, B2O3 0 mol%", "answer": "No. 1: Z=4, (Na2O+CaO)/Al2O3=(20+10)/10=3>1, Al3+ is considered as a network-forming ion. R1=(20+10+30+120)/(20+60)=2.25. X1=2R-Z=0.5. Y1=4-0.5=3.5." }, { "idx": 156, "question": "Calculate the structural parameters and non-bridging oxygen fraction of the glass with the composition $\\mathrm{(4)18Na_{2}0\\cdot10C a0\\cdot72S i0_{2}}$ (wt%)", "answer": "Na2O: 18%, 0.290 mol, 17.4 mol%; CaO: 10%, 0.179 mol, 10.7 mol%; SiO2: 72%, 1.200 mol, 71.9 mol%. Z=4, R=(17.4+10.7+71.9*2)/71.9=2.39, X=2R-Z=0.78, Y=4-0.78=3.22. Non-bridging oxygen %=0.78/(1.61+0.78)=32.6%" }, { "idx": 157, "question": "For glass No. 1 with the composition of Na2O 8 wt%, Al2O3 12 wt%, and SiO2 80 wt%, use the glass structural parameters to explain the magnitude of its viscosity at high temperatures?", "answer": "For No. 1: Z=4 Na2O/Al2O3=8.16/7.47>1 Al2O3 is considered as a network-forming ion R=(8.16+7.47×3+84.37×2)/(7.47×2+84.37)=2.007 X1=2R−Z=0.014 Y1=4−X=3.986 Y1=3.986, therefore, the viscosity of No. 1 at high temperatures is high." }, { "idx": 158, "question": "A glass composition (wt%) is Na2O 14%, CaO 13%, SiO2 73%, with a density of 2.5 g/cm3. If soda ash, limestone, and quartz sand are used as raw materials to melt this glass with 1000 kg of quartz sand, how much of the other two raw materials are needed?", "answer": "With 1000 kg of quartz sand, the required soda ash (Na2CO3) is: (14/73)×1000×(106/62)=327.88 kg. The required limestone (CaCO3) is: (13/73)×1000×(100/56)=318.00 kg." }, { "idx": 159, "question": "For glass No. 2 with the composition of Na2O 12 wt%, Al2O3 8 wt%, and SiO2 80 wt%, use the glass structural parameters to explain the magnitude of its viscosity at high temperatures?", "answer": "For No. 2: Z=4; Na2O/Al2O3=12.09/4.86>1; Al2O3 is considered as a network-forming ion; R=(12.09+4.86×3+83.05×2)/(4.86×2+83.05)=2.08; X2=2R−Z=0.16; Y2=4−X=3.84; Y2=3.84, therefore No. 2 has a relatively low viscosity at high temperatures." }, { "idx": 160, "question": "Calculate the structural parameters and non-bridging oxygen fraction of the glass (2) Na2O•CaO•Al2O3•SiO2", "answer": "Na2O+CaO/(2*Al2O3)=2>1, Al is considered as a network-forming ion. Z=4, R=(1+1+3+2)/(2+1)=2.33, X=2R-Z=4.66-4=0.66, Y=4-0.66=3.34. Non-bridging oxygen %=0.66/(1.67+0.66)=28.3%" }, { "idx": 161, "question": "Does this composition have a tendency to form glass? Why?", "answer": "This composition has a tendency to form glass because the structure maintains a three-dimensional framework at this point, and the viscosity of the glass is still relatively high, making it easy to form glass." }, { "idx": 162, "question": "Among the following three substances, which one is the least likely to form glass, and why? (1) Na2O·2SiO2; (2) Na2O·SiO2; (3) NaCl", "answer": "(3) NaCl is the least likely to form glass. NaCl lacks a network structure and is a typical ionic crystal, making it difficult to form glass." }, { "idx": 163, "question": "Among the following three substances, which one is most likely to form glass, and why? (1) Na2O·2SiO2; (2) Na2O·SiO2; (3) NaCl", "answer": "(1) Na2O·2SiO2 is most likely to form glass. Calculations show that R1=2.5, Y1=3. At high temperatures, (1) has high viscosity and is prone to form glass." }, { "idx": 164, "question": "Briefly describe the differences in structure and properties between tempered glass and annealed glass.", "answer": "Glass that eliminates and balances internal stresses caused by temperature gradients is called annealed glass, which is less prone to breakage and easier to cut. Tempering involves heating the product to near its softening temperature to fully anneal the glass, followed by rapid cooling (quenching). This creates uniform internal stresses, resulting in pre-compressive stresses on the glass surface, thereby increasing resistance to bending, impact, and torsional deformation." }, { "idx": 165, "question": "What are the methods for forming non-crystalline solids (NCS)?", "answer": "Non-crystalline solids can be obtained by supercooling melts and glasses." }, { "idx": 166, "question": "A glass composition (wt%) is Na2O 14%, CaO 13%, SiO2 73%, with a density of 2.5 g/cm3. Calculate the atomic packing factor (AFP) and structural parameter values of this glass?", "answer": "The average molecular weight of the glass GW=0.14×62+0.13×56+0.73×60.02=59.77. The number of atoms in 1Å3 is n=2.5×10-24×6.02×1023/59.77=0.252 atoms/Å3. The volume occupied by atoms in 1Å3 V=0.0252×4/3π[0.14×2×0.983+0.13×1.063+0.73×0.393+(0.14+0.13+0.73+0.13+0.13)]×1.323=0.4685. AFP=0.46. Structural parameters: Na2O wt% 14, CaO wt% 13, SiO2 wt% 73; mol Na2O 0.23, CaO 0.23, SiO2 1.22; mol% Na2O 13.7, CaO 13.7, SiO2 72.6. R=(13.7+13.7+72.6×2)/72.6=2.38. Z=4. X=2R-Z=2.38×2-4=0.76. Y=Z-X=4-0.76=3.24." }, { "idx": 167, "question": "How much CaO needs to be added to 100g of SiO2 to achieve an O:Si ratio of 2.5?", "answer": "46.67" }, { "idx": 168, "question": "By adding $20\\%\\\\mathrm{B}_{2}0_{3}$ to $\\\\mathrm{Si0_{2}}$, calculate the $0:\\\\mathrm{Si}$ ratio of the melt.", "answer": "S $\\\\frac{{\\\\cal O}}{\\\\bar{\\\\cal S}\\\\bar{\\\\imath}}=\\\\frac{1\\\\times2+0.2\\\\times3}{1+0.2\\\\times2}=1.86$" }, { "idx": 169, "question": "If 50 mol% Na2O is added to SiO2, what is the O:Si ratio?", "answer": "O/Si = (0.5 + 1 × 2)/1 = 2.5" }, { "idx": 170, "question": "If 10mol% Na2O is added to SiO2, what is the calculated O:Si ratio?", "answer": "O:Si = (0.1 + 1 × 2)/1 = 2.1" }, { "idx": 171, "question": "When Na2O is added to SiO2, making O/Si=2.5, is the crystallization ability enhanced or weakened?", "answer": "Because O/Si increases, the viscosity decreases, and the crystallization ability is enhanced." }, { "idx": 172, "question": "Adding 50mol% Na2O to SiO2, can this ratio form a glass? Why?", "answer": "Yes, it can form a glass. When 50mol% Na2O is added, although the continuous SiO network skeleton becomes relaxed, it still maintains a three-dimensional network structure, allowing the formation of a glass." }, { "idx": 173, "question": "How much Na2O should be added to SiO2 to make the O/Si ratio of the glass equal to 2.5? Assume the amount of SiO2 is ymol.", "answer": "Let xmol of Na2O be added, then O/Si=(x+2y)/y=2.5. Solving gives x=y/2, meaning when the molar ratio of the two is 1:2, O/Si=2.5." }, { "idx": 174, "question": "What is surface tension?", "answer": "Surface tension: the surface constriction force acting perpendicularly on a unit length line segment or the work required to increase an object's surface area by one unit; σ= force/total length (N/m)" }, { "idx": 175, "question": "What is surface energy?", "answer": "Surface energy: Under constant temperature, pressure, and composition, the non-volume work required to reversibly increase the surface area of a substance is called surface energy; J/m2=N/m" }, { "idx": 176, "question": "What is the difference between surface tension and surface energy in the liquid state?", "answer": "Liquid: cannot withstand shear stress, the work done by external forces manifests as an expansion of surface area, because surface tension and surface energy are numerically the same" }, { "idx": 177, "question": "Explain based on the structure why the mixture of alkali and wollastonite with 0/Si=3 crystallizes instead of forming glass?", "answer": "When 0/Si=3, Y=2. For silicate glasses, a three-dimensional network cannot be formed when Y=2 because the number of bridging oxygens shared between tetrahedra is less than 2, and the structure mostly consists of tetrahedral chains of varying lengths. Therefore, the mixture of alkali and wollastonite with 0/Si=3 crystallizes instead of forming glass." }, { "idx": 178, "question": "Based on the structure, explain why a mixture of alkali and silica with 2<0/Si<2.5 can form glass?", "answer": "When 2<0/Si<2.5, 390^{\\\\circ},B T\\\\cup\\\\gamma_{\\\\mathrm{SRE}}\\\\times1875^{\\\\circ}$$ So it cannot wet." }, { "idx": 198, "question": "From an atomic scale perspective, explain the differences in the bonding effects of welding", "answer": "Welding: A process where two or more materials (of the same or different types) are permanently joined by heating, applying pressure, or both, to achieve atomic bonding. The conditions during welding include: base material type, plate thickness, groove shape, joint form, restraint state, ambient temperature and humidity, cleanliness, as well as the type and diameter of the welding wire (or electrode), welding current, voltage, welding speed, welding sequence, deposition method, and gun (or electrode) manipulation method determined based on the above factors. If the weld groove or surface has impurities such as oil (paint), water, or rust, they can cause defects like pores, inclusions, slag, or cracks in the weld, posing hazards and potential risks to the welded joint." }, { "idx": 199, "question": "Heat treat Si3N4 at the eutectic temperature and measure its thermal etching groove angle as 60°, calculate the grain boundary energy of Si3N4.", "answer": "γss=2γsvcos(ψ/2)=2×900cos30°=1558.8mN/m" }, { "idx": 200, "question": "The eutectic of the MgO-Al2O3-SiO2 system is placed on a Si3N4 ceramic plate. At the eutectic temperature, the surface tension of the liquid phase is 900 mN/m, the interfacial energy between the liquid and solid is 600 mN/m, and the measured contact angle is 70.52°. Find the surface tension of Si3N4.", "answer": "Yv = Ycosθ + Y = 900cos70.52° + 600 = 900 mN/m" }, { "idx": 201, "question": "From an atomic scale perspective, explain the differences in the bonding effects of sintering", "answer": "Sintering: It is a high-temperature treatment process that imparts material properties. The diffusion of atoms to contact points causes bonding between particles, and further diffusion ultimately fills the remaining pores and increases the density of the material. It involves heating powder or powder compacts to a temperature below the melting point of their basic components, followed by cooling to room temperature using specific methods and rates. The result of sintering is the bonding between powder particles, increasing the strength of the sintered body, transforming the aggregate of powder particles into a coalescence of grains, thereby obtaining products or materials with desired physical and mechanical properties. Low-temperature pre-sintering stage: During this stage, metal recovery, volatilization of adsorbed gases and moisture, decomposition, and removal of forming agents in the compact mainly occur. Medium-temperature sintering stage: In this stage, recrystallization begins to appear. Within the particles, deformed grains are restored and reorganized into new grains, while surface oxides are reduced, and sintering necks form at particle interfaces. High-temperature sintering completion stage: In this stage, diffusion and flow proceed sufficiently and near completion, forming a large number of closed pores that continue to shrink, reducing pore size and total porosity, significantly increasing the density of the sintered body." }, { "idx": 202, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the trend in potential changes of clay after adsorbing the following different cations (use arrows to show: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Al3+ < Ba2+ < Sr2+ < Ca2+ < Mg2+ < NH4+ < K+ < Na+ < Li+" }, { "idx": 203, "question": "Silver plating is required on alumina ceramic parts. It is known that at 1000°C, γ_{Al2O3(s)} = 1.00 mN/m, γ_{Ag(L)} = 0.92 mN/m, and γ_{Ag(L)-Al2O3(s)} = 1.77 mN/m. Can liquid silver wet the surface of alumina ceramic parts? What methods can be used to improve the wettability between them?", "answer": "cosθ = (γ_{sv} - γ_{sl}) / γ_{lv} = -0.837, θ = 146.8° > 90°, cannot wet. To silver plate ceramic components, the surface of the ceramic must first be ground and polished to improve the wettability between the ceramic and the silver layer." }, { "idx": 204, "question": "Analyze the scope of action of firmly bound water and loosely bound water in micelles", "answer": "Firmly bound water surrounds the clay particles, forming an integral whole with them and moving together in the medium, with a thickness of approximately 3 to 10 water molecules. Loosely bound water is located on the periphery of firmly bound water, with poorer orientation. As they are farther from the clay particle surface, the binding force between them is smaller." }, { "idx": 205, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation pattern of the ion exchange capacity of clay after adsorbing the following different cations (use arrows to represent: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "Li+ < Na+ < K+ < NH4+ < Mg2+ < Ca2+ < Sr2+ < Ba2+ < Al3+ < H+" }, { "idx": 206, "question": "Explain the differences between structural water, bound water (tightly bound water, loosely bound water), and free water in clay.", "answer": "Structural water in clay refers to the water within the clay structure. Since clay particles are generally negatively charged and water is a polar molecule, when clay particles are dispersed in water, under the influence of the negative electric field on the clay surface, water molecules align in a specific orientation around the clay particles, forming hydrogen bonds with the oxygen and hydroxyl groups on the surface, with their negative ends facing outward. A negatively charged surface forms around the first layer of water molecules, attracting a second layer of water molecules. The attractive force of the negative electric field on water molecules weakens with increasing distance from the clay surface, causing the arrangement of water molecules to transition gradually from ordered to disordered. The inner layer of water molecules arranged in an ordered manner is called tightly bound water, which surrounds the clay particles and moves together with them as a whole in the medium, with a thickness of approximately 3 to 10 water molecules. The outer layer of water molecules, which is less ordered, is called loosely bound water. Due to their greater distance from the clay particle surface, the binding force between them is weaker. The water beyond the loosely bound water is called free water." }, { "idx": 207, "question": "Analyze the influence of firmly bound water and loosely bound water on process performance", "answer": "The density of bound water is high, the heat capacity is small, the dielectric constant is small, and the freezing point is low, which makes it different from free water in physical properties. The amount of clay-water combination can be judged by measuring the heat of wetting. The state and quantity of clay combined with these three types of water will affect the process performance of the clay-water system. When the water content of the clay is constant, if the bound water decreases, the free water increases, resulting in a reduced volume of clay particles that are easier to move, thus lowering the slurry viscosity and improving fluidity. When the amount of bound water is high, the water film is thick, facilitating the sliding between clay particles, which enhances plasticity." }, { "idx": 208, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation pattern of combined water content in clay after adsorbing the following different cations (use arrows to represent: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Al3+ < Ba2+ < Sr2+ < Ca2+ < Mg2+ < NH4+ < K+ < Na+ < Li+" }, { "idx": 209, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation trend in slurry stability when clay adsorbs the following different cations (use arrows to show: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Al3+ < Ba2+ < Sr2+ < Ca2+ < Mg2+ < NH4+ < K+ < Na+ < Li+" }, { "idx": 210, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation pattern of slurry fluidity for clay adsorbed with the following different cations (use arrows to denote: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Al3+ < Ba2+ < Sr2+ < Ca2+ < Mg2+ < NH4+ < K+ < Na+ < Li+" }, { "idx": 211, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation trend of the thixotropy of clay slurry when adsorbing the following different cations (use arrows to represent: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Li+ < Na+ < K+ < NH4+ < Mg2+ < Ca2+ < Sr2+ < Ba2+ < Al3+" }, { "idx": 212, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation pattern of plasticity in clay lumps after adsorbing the following different cations (use arrows to denote: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "Li+ < Na+ < K+ < NH4+ < Mg2+ < Ca2+ < Sr2+ < Ba2+ < Al3+ < H+" }, { "idx": 213, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation pattern of the filtrate loss of clay slurry adsorbed with the following different cations (use arrows to represent: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Al3+ < Ba2+ < Sr2+ < Ca2+ < Mg2+ < NH4+ < K+ < Na+ < Li+" }, { "idx": 214, "question": "According to Figure 5-13 and Table 5-2, the atomic densities on different crystallographic planes (110), (100), and (111) of a face-centered cubic lattice are different. Please answer, on which plane will the solid-gas surface energy be the lowest? Why?", "answer": "According to the formula for surface energy $\\\\frac{\\\\underline{{L}}^{s}\\\\mathcal{E}_{\\\\underline{{\\\\bar{L}}}}}{N_{0}}[1-\\\\frac{n_{i}^{s}}{n_{i}^{\\\\delta}}]=(\\\\Delta\\\\zeta7)_{s,v}\\\\bullet\\\\underline{{\\\\zeta}}^{s}=\\\\gamma_{s o}^{0}$, where $\\\\gamma_{50}^{0}$ is the solid surface energy, $E_{L}$ is the lattice energy, $N_{0}$ is Avogadro's number, and $Z$ is the number of atoms per 1m² surface. $n_{i}^{\\\\delta}$ and $n_{i}^{s}$ represent the number of nearest neighbor atoms for the i-th atom in the bulk and on the surface, respectively. In a face-centered cubic crystal, $n_{i}^{\\\\delta}=12$; $n_{i}^{s}$ is 6 on the (111) plane, 4 on the (100) plane, and 2 on the (110) plane. Substituting these values into the formula gives $\\\\gamma_{s o(110)}^{0}=0.550\\\\frac{E_{z}}{N_{0}}(1-\\\\frac{2}{12})=0.46\\\\frac{E_{z}}{N_{0}}$,\\n$\\\\gamma_{s o(100)}^{0}=0.785\\\\frac{E_{L}}{N_{0}}(1-\\\\frac{4}{12})=0.53\\\\frac{E_{L}}{N_{0}}$,\\n$\\\\gamma_{s o(111)}^{0}=0.907\\\\frac{E_{z}}{N_{0}}(1-\\\\frac{6}{12})=0.45\\\\frac{E_{z}}{N_{0}}$.\\nThus: $\\\\gamma_{_{50(100)}}^{0}>\\\\gamma_{_{50(110)}}^{0}>\\\\gamma_{_{50(111)}}^{0}$.\\n\\nTherefore, the (111) close-packed plane has the lowest surface energy." }, { "idx": 215, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation pattern of the casting time of clay slurry when adsorbing the following different cations (use arrows to represent: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Al3+ < Ba2+ < Sr2+ < Ca2+ < Mg2+ < NH4+ < K+ < Na+ < Li+" }, { "idx": 216, "question": "Many properties of clay are related to the types of adsorbed cations. Indicate the variation pattern of the green body formation rate when the clay adsorbs the following different cations (use arrows to represent: small—large) H+ Al3+ Ba2+ Sr2+ Ca2+ Mg2+ NH4+ K+ Na+ Li+", "answer": "H+ < Al3+ < Ba2+ < Sr2+ < Ca2+ < Mg2+ < NH4+ < K+ < Na+ < Li+" }, { "idx": 217, "question": "What measures can be taken in production to increase or decrease the plasticity of clay to meet the needs of forming processes?", "answer": "In production, the specific surface area of mineral components can be increased to enhance the capillary force, thereby increasing plasticity; the valence of adsorbed cations can also be increased or decreased to alter the plasticity of clay; or the particle size of minerals can be reduced to increase contact points, thereby enhancing the plasticity of clay;" }, { "idx": 218, "question": "Using Na2CO3 and Na2SiO3 to dilute the same type of clay slurry (mainly composed of kaolinite mineral), compare the differences in the casting rate of the two slurries when the same amount of electrolyte is added.", "answer": "The addition of Na2CO3 basically has no effect on the casting rate of the clay, while the addition of Na2SiO3 increases the casting rate of the slurry." }, { "idx": 219, "question": "Using Na2CO3 and Na2SiO3 to dilute the same type of clay slurry (mainly composed of kaolinite mineral), compare the differences in thixotropy of the two slurries when the same amount of electrolyte is added.", "answer": "Adding Na2CO3 basically has no effect on the thixotropy of the clay, while adding Na2SiO3 reduces the thixotropy of the slurry." }, { "idx": 220, "question": "What are the factors affecting the plasticity of clay?", "answer": "The influencing factors include: 1. Mineral composition, different mineral compositions result in varying interparticle forces. 2. Types of adsorbed cations, higher valence cations improve plasticity. 3. Particle size and shape, finer particles with larger specific surface area and more contact points between particles increase plasticity. 4. Water content, etc." }, { "idx": 221, "question": "Using Na2CO3 and Na2SiO3 to dilute the same type of clay slurry (mainly composed of kaolinite minerals), compare the differences in the body density of the two slurries when the same amount of electrolyte is added.", "answer": "Adding Na2CO3 basically has no effect on the body density of the clay, while adding Na2SiO3 increases the body density of the slurry." }, { "idx": 222, "question": "What is the practical significance of Gibbs phase rule?", "answer": "Applying the phase rule can conveniently determine the degrees of freedom of an equilibrium system." }, { "idx": 223, "question": "In the phase diagram of the SiO2 system, find two examples of reversible polymorphic transformations", "answer": "Reversible polymorphic transformations: β-quartz ↔ α-quartz, α-quartz ↔ tridymite" }, { "idx": 224, "question": "Using Na2CO3 and Na2SiO3 to dilute the same type of clay (mainly composed of kaolinite mineral) slurry, respectively, compare the differences in fluidity of the two slurries when the same amount of electrolyte is added.", "answer": "Adding Na2CO3 basically has no effect on the fluidity of the clay, while adding Na2SiO3 increases the fluidity of the slurry." }, { "idx": 225, "question": "What is Gibbs phase rule?", "answer": "The phase rule is a fundamental law of phase equilibrium derived by Gibbs based on thermodynamic principles, also known as Gibbs phase rule, used to describe the relationship between the number of degrees of freedom, the number of components, and the number of phases in a system at phase equilibrium. The general mathematical expression is F=C-P+2, where F is the number of degrees of freedom, C is the number of components, P is the number of phases, and 2 represents the two variables of temperature and pressure." }, { "idx": 226, "question": "In the SiO2 phase diagram, identify two examples of irreversible polymorphic transformations", "answer": "Irreversible polymorphic transformations: β-cristobalite←→β-quartz, V-tridymite←→β-quartz" }, { "idx": 227, "question": "In the Al2O3-SiO2 system, the liquidus line of SiO is very steep, so why is it necessary to strictly prevent the mixing of Al2O3 in silica bricks, otherwise the refractoriness of silica bricks will be greatly reduced?", "answer": "Al2O3 in the SiO2 system can significantly increase the amount of liquid phase, leading to a substantial decrease in the eutectic point temperature, thereby greatly reducing the refractoriness of silica bricks. Therefore, it is essential to strictly prevent the mixing of Al2O3." }, { "idx": 228, "question": "At this temperature, after prolonged holding to reach equilibrium, what is the phase composition of the system?", "answer": "After prolonged holding at 1595°C, the system consists of liquid phase and A3S2, with L% = 21.8%." }, { "idx": 229, "question": "In the CaO-SiO2 system, the liquidus line of SiO2 is very steep. Why can a small amount of CaO be added as a mineralizer in the production of silica bricks without reducing their refractoriness?", "answer": "When a small amount of CaO is added to SiO2, at the eutectic point of 1436°C, the liquid phase content is 2/37=5.4%. The increase in liquid phase content is not significant and does not reduce the refractoriness of silica bricks, so a small amount of CaO can be added as a mineralizer." }, { "idx": 230, "question": "When the clay mineral kaolinite (Al2O3•2SiO2•2H2O) is heated to 600°C, it decomposes into water vapor and Al2O3•2SiO2. What happens when it is further heated to 1595°C?", "answer": "When heated to 1595°C, A3S2 is formed." }, { "idx": 231, "question": "At what temperature does the clay completely melt?", "answer": "Complete melting means the solid phase completely disappears, which should be at the temperature where the 33% line intersects with the liquidus line." }, { "idx": 232, "question": "When the system generates 40% liquid phase, what temperature should be reached?", "answer": "The temperature at which 40% liquid phase is generated needs to be determined by referring to the phase diagram, and the specific temperature is not provided in the answer." }, { "idx": 233, "question": "Compare the characteristics of the eutectic point in ternary invariant points and write its equilibrium relationship", "answer": "The eutectic point is a type of invariant point where several crystalline phases simultaneously precipitate from the melt upon cooling or melt together upon heating. The phase equilibrium relationship is L(E)⇄A+B+C" }, { "idx": 234, "question": "Cool a melt containing MgO and $\\\\mathrm{{Al}_{2}\\\\mathrm{{O}_{3}}}$ to a certain temperature, then filter out the precipitated crystals and analyze the remaining liquid phase. It is found that the liquid phase contains 65% MgO, and the amount of liquid phase is 70% of the total system. Determine the composition of the original melt.", "answer": "$\\\\mathrm{MgO\\\\ 45.5\\\\%}$; $\\\\mathrm{Al_{2}O_{3}\\\\ 54.5\\\\%}$" }, { "idx": 235, "question": "Compare the characteristics of the singular peritectic point in ternary invariant points, and write its equilibrium relationship", "answer": "If the invariant point is at the intersection position, it is a singular eutectic point; if it is at the conjugate position, it is a double peritectic point. The phase transformation relationship is L_{(g)}+A⇌D+C, L_{(g)}+A+B⇌S" }, { "idx": 236, "question": "Compare the characteristics of transition points in ternary invariant points and write their equilibrium relationships", "answer": "The transition point is where the properties change from peritectic to eutectic, with no corresponding triangle, and the compositions of the three crystalline phases in equilibrium lie on a straight line" }, { "idx": 237, "question": "Compare the characteristics of the double transition point in the ternary invariant point and write its equilibrium relationship", "answer": "If the invariant point is at the intersection position, it is a single eutectic point; if it is at the conjugate position, it is a double transition point. The phase transformation relationship is L_{(g)}+A⇌D+C, L_{(g)}+A+B⇌S" }, { "idx": 238, "question": "From an atomic scale perspective, explain the differences in the bonding mechanism of glass-to-metal sealing.", "answer": "The sealing of glass and metal is governed by many factors. The most important factors affecting the seal are the following four aspects: (1) Oxidation of the metal: Before the hermetic sealing of glass and metal, the metal is usually moderately oxidized, placing the metal oxide between the glass and the metal. This step is essential for sealing and is also a microscopic control method for glass sealing. The oxidation treatment of the metal is the most critical factor affecting the sealing of glass and metal, and the oxidation mechanism is a key issue in sealing technology. (2) Thermal expansion coefficients of glass and metal: The consistency of the thermal expansion coefficients of glass and metal is a macroscopic control method for achieving good sealing. When the glass melt is sealed with the metal, the glass at high temperature has sufficient viscous fluidity. It deforms while simultaneously contracting along the thermal contraction curve of the metal. However, as the temperature decreases, the glass gradually loses its fluidity, causing it to separate from the metal's thermal contraction curve. This change is continuous and depends on the cooling rate, making it impossible to determine a specific temperature at which the glass separates from the metal's contraction curve. For convenience, a specific temperature TM is often used to represent the state when the glass is fixed to the metal. This means that when T > Tμ, the glass has complete fluidity and does not generate stress. When T < Ti1, the sealing glass contracts along its inherent thermal contraction curve. This specific temperature TM is called the solidification temperature, which is very close to the strain temperature of the glass. At any temperature T, a contraction difference ∠d is generated between the glass and metal, producing stress proportional to △d in the seal. When the stress exceeds the strength limit of the glass, the glass is damaged, affecting the hermeticity of the seal. Below the solidification temperature Ty, the relative relationship of the thermal contraction curves essentially reflects the matching degree of the expansion coefficients, i.e., the contraction difference between the glass and metal starting from Ty should be: △d = (ag - am)(T - T) (Equation 1-1). In Equation 1-1, ag and am represent the expansion coefficients of the glass and metal, respectively, from TM to the matching temperature T. To eliminate permanent stress in the glass, the seal must be annealed. Proper annealing is crucial for sealing quality. After annealing, the seal should not be cooled rapidly because the metal has better thermal conductivity than the glass, causing the metal to cool faster. When the expansion coefficients of the metal and glass are the same, this difference in cooling rates causes the metal to contract more than the glass. Once the glass loses fluidity, the metal must cool within a narrow range, and the length changes affect the adhesion between the glass and metal. If the rapid cooling starts at a temperature above the lower limit of the glass's annealing temperature, the glass will be in a stretched state. To prevent this tensile stress and to ensure the glass is slightly pressurized, the metal part is often heated separately with a gas flame after sealing. (3) Strength of the glass and interfacial diffusion: On the basis of matching the thermal expansion coefficients of the glass and metal, improving the mechanical strength of the glass, especially its tensile strength, is beneficial when the seal is subjected to thermal shock, thermal stress due to temperature gradients, or external forces during use. Generally, the compressive strength of glass can be very high, reaching 600~1500 MPa, while the tensile strength is extremely low, only about 10% of the compressive strength. In practice, only tensile strength poses problems. If possible, using crystallized glass for sealing is an effective way to improve the tensile strength of the glass, typically achieving 3~5 times or even more than 5 times the original tensile strength. It must be noted that the presence of a large number of bubbles (especially clustered bubbles) at the sealing interface is highly detrimental, as it can reduce mechanical strength and cause chronic leaks. Gas dissolved in the metal being released during sealing heating is one cause of bubbles, which is rare in metals like tungsten, molybdenum, and platinum but more common in nickel, iron, and their alloys. To eliminate this factor, metals produced by vacuum smelting are preferred, or the metal can be pre-degassed by heating in a vacuum or hydrogen atmosphere. Another cause of bubbles is carbon, especially in the surface layer of the metal, which oxidizes into carbon dioxide gas during sealing, forming bubbles. This is more severe in nickel, iron, and their alloys than in other metals. The solution is to anneal the metal in wet hydrogen or a vacuum to remove gas and carbon, thereby preventing bubble formation. The annealing temperature is generally 900~1100°C, and the duration depends on the thickness of the metal. Near the sealing interface, there are two scenarios: one involves direct mutual diffusion at the interface between two different materials, such as in glass-to-glass optical grinding seals or metal-to-metal welding. The other involves placing a different material at the interface to facilitate sealing, such as in glass-to-metal or ceramic-to-metal seals where direct mutual diffusion is difficult. In this case, melting diffusion occurs. However, whether mutual diffusion or melting diffusion occurs, the composition and properties near the interface differ from those in the bulk material, and this variation significantly affects the stability of the seal. (4) Shape, size, and surface roughness of the seal: The magnitude and distribution of stress within the seal are influenced by its shape and size. When the stress exceeds the strength limit of the sealing material, the seal is inevitably damaged. Generally, after sealing, if the seal is only a component of an electron tube or vacuum device and requires further processing or sealing with another similar device, it must undergo additional heat treatment or mechanical force. The temporary stress caused by external forces will combine with the thermal contraction stress between the materials. The combined effect of the permanent stress from thermal contraction differences and temporary stress from various causes is considerable. If the influence of shape and size on stress is overlooked, the stability of the seal may be compromised. For example, in production, the plasticity of thin metal edges can be utilized to reduce stress, or even the elasticity of the metal can be used for sealing. Additionally, adhesion tests have shown that if the substrate has uniform凹凸 (凹凸 refers to surface roughness or unevenness), the adhesion is good, which positively affects the sealing." }, { "idx": 239, "question": "Concentration gradient causes diffusion, does diffusion always proceed from high concentration to low concentration? Why?", "answer": "Diffusion is caused by gradient differences, and concentration gradient is only one type of gradient difference. When another gradient difference, such as stress gradient, has a greater influence than the concentration gradient, diffusion can proceed from low concentration to high concentration." }, { "idx": 240, "question": "A petrographic analysis of a certain Portland cement revealed that the flux mineral CA precipitated as a crystalline phase before CAF. Is this formulation a high-alumina formulation (P>1.38) or a high-iron formulation (P<1.38)?", "answer": "It is a high-alumina formulation (P>1.38)." }, { "idx": 241, "question": "Compare the characteristics of polymorphic transition points in ternary invariant points and write their equilibrium relationships", "answer": "Polymorphic transition points are the temperature points at which two or three crystal forms undergo crystal transformation" }, { "idx": 242, "question": "How should the cooling rate in the clinkering zone be controlled for high-alumina formulations (P>1.38)?", "answer": "For formulations with an alumina modulus P>1.38, when the melt cools past the KT2 boundary, the liquid phase will reabsorb C3S and precipitate C2S and CsA. Therefore, if the clinker is cooled slowly in the clinkering zone, bringing the cooling process close to equilibrium, some C3S will be reabsorbed, which is detrimental to cement quality. Hence, for formulations with P>1.38, rapid cooling should be employed during the cooling process in the clinkering zone." }, { "idx": 243, "question": "The batching of high-alumina cement is usually selected within the range of the CA phase region, and it is often produced by firing to melting and then cooling. The main mineral of high-alumina cement is CA, and C2AS has no hydraulic properties, so it is desirable that the cement does not contain C2AS. Therefore, what range of batching should be chosen within the CA phase region, and why (note that complete equilibrium cannot be achieved during production, and independent crystallization processes may occur)?", "answer": "High-alumina cement has the characteristic of rapid hardening and is widely used in the defense industry. The batching range is Al2O3 35%~55%, CaO 35%~45%, SiO2 5%~10%, Fe2O3 0~15%. This can prevent the formation of C2AS." }, { "idx": 244, "question": "How should the cooling rate of high iron composition (P<1.38) in the burning zone be controlled?", "answer": "For compositions with an alumina modulus P<1.38, the melt cooling and crystallization process passes through the WT2 boundary, where the liquid phase absorbs C2S back and precipitates CsS and C4AF. Therefore, the slower the clinker cooling rate, the closer it approaches equilibrium, and the more fully C2S is absorbed back. The content of CS in the clinker will increase, which is beneficial for cement quality. Thus, for clinker with P<1.38, the cooling rate in the burning zone should be appropriately slowed down." }, { "idx": 245, "question": "The diffusion coefficient values of CaO at 1145°C and 1650°C", "answer": "As shown in the figure, the diffusion coefficient values of CaO at 1145°C and 1650°C are D1 and D2, respectively (specific values need to be read from the figure)." }, { "idx": 246, "question": "The diffusion coefficient values of Al2O3 at 1393°C and 1716°C", "answer": "As shown in the figure, the diffusion coefficient values of Al2O3 at 1393°C and 1716°C are D3 and D4, respectively (specific values need to be read from the figure)." }, { "idx": 247, "question": "Provide a full explanation of the various property value estimates you made in the calculation. The Schottky defect formation energy of CaO is known to be 6eV.", "answer": "The property values used in the calculation include: Schottky defect formation energy ΔHf = 6 eV, converted to joules as 6 × 1.6 × 10^-19 J; Avogadro's constant NA = 6.23 × 10^23 mol^-1; gas constant R = 8.314 J/(mol·K); melting point temperature T = 2600℃ = 2873 K. These values are used to calculate the Schottky defect concentration [V′′Ca] = exp(-ΔHf / 2RT)." }, { "idx": 248, "question": "Given the diffusion data of hydrogen in face-centered cubic iron as D_Ni = 0.0063 exp(-10300 × 4.18 / RT) cm²/s, calculate the diffusion coefficient at 1000°C.", "answer": "Substituting T=1000°C into the equation yields D_H=3.6×10^-5 cm²/s." }, { "idx": 249, "question": "What concentration of trivalent ions is required to ensure that the diffusion of Ca2+ in CaO remains non-intrinsic up to the melting point of CaO (2600°C)? The Schottky defect formation energy of CaO is known to be 6eV.", "answer": "The defect reaction for doping M3+ is as follows: M2O3 → 2M′′Ca + V′′Ca + 3OO. When CaO is at its melting point, the concentration of Schottky defects is: [V′′Ca] = exp(-ΔHf / 2RT) = exp(-6 × 1.6 × 10^-19 × 6.23 × 10^23 / (2 × 8.314 × 2873)) = 3.6 × 10^-6. Therefore, to ensure that the diffusion of Ca2+ in CaO remains non-intrinsic up to the melting point of CaO (2600°C), the concentration of M3+ must be [M3+] = [M′′Ca] = 2[V′′Ca] > [V′′Ca], i.e., [M3+] > 2 × 3.6 × 10^-6 = 7.2 × 10^-6." }, { "idx": 250, "question": "Calculate the diffusion activation energy Q and D0 value of Ca2+ in CaO", "answer": "According to D=D0exp(-Q/RT), the ratio of diffusion coefficients of CaO at 1145°C and 1650°C can be used to calculate Q and D0. The specific calculation process is: D1/D2 = exp[-Q/R(1/T1 - 1/T2)], where T1=1418K(1145°C), T2=1923K(1650°C). Solving this equation yields the Q value, and substituting the diffusion coefficient at either temperature gives D0." }, { "idx": 251, "question": "Given the diffusion data of nickel in face-centered cubic iron as D_Ni = 4.1 exp(-64000 × 4.18 / RT) cm²/s, calculate the diffusion coefficient at 1000°C.", "answer": "Substituting T=1000°C into the equation yields D_Ni=4.35×10^-11 cm²/s." }, { "idx": 252, "question": "Calculate the diffusion activation energy Q and D0 value of Al3+ in Al2O3", "answer": "According to D=D0exp(-Q/RT), the ratio of diffusion coefficients of Al2O3 at 1393℃ and 1716℃ can be used to calculate Q and D0. The specific calculation process is: D3/D4 = exp[-Q/R(1/T3 - 1/T4)], where T3=1666K(1393℃), T4=1989K(1716℃). Solving this equation yields the Q value, and substituting it into the diffusion coefficient at any temperature gives D0." }, { "idx": 253, "question": "Explain the difference in diffusion coefficients between hydrogen and nickel in face-centered cubic iron.", "answer": "Compared to nickel atoms, hydrogen atoms are much smaller and can diffuse more easily through the gaps in face-centered cubic iron." }, { "idx": 254, "question": "When Zn2+ diffuses in ZnS, the diffusion coefficient at 563°C is 3×10−4cm2/s, and at 450°C it is 1.0×10−4cm2/s. Calculate the activation energy for diffusion and D0.", "answer": "Referring to 7-4, Q=48856J/mol, D0=3×10−15cm2/s." }, { "idx": 255, "question": "Given that the activation energy Q for Zn2+ diffusion in ZnS is 48856 J/mol, and D0=3×10−15 cm2/s, calculate the diffusion coefficient at 750°C.", "answer": "Substituting T=1023K into D=D0exp(−Q/RT) yields D1023=9.6×10−4 cm2/s." }, { "idx": 256, "question": "The diffusion coefficients of carbon in titanium measured at different temperatures are 2×10−9cm2/s (736℃), 5×10−9cm2/s (782℃), and 1.3×10−8cm2/s (838℃). Please calculate the activation energy for diffusion.", "answer": "From the previous step, it is known that Q=2342787J/mol." }, { "idx": 257, "question": "The diffusion coefficients of carbon in titanium at different temperatures were experimentally measured as 2×10−9cm2/s (736℃), 5×10−9cm2/s (782℃), and 1.3×10−8cm2/s (838℃). Given the activation energy for diffusion Q=2342787J/mol, determine the diffusion coefficient of carbon at 500℃.", "answer": "Substituting T=773K into D=D0exp(−Q/RT), with Q=2342787J/mol known, D0 is derived from D̅1=2×10−9cm2/s and T1=1009K as D0=2×10−9exp(2342787/(8.31×1009)). The calculated D500℃=1.87×10−10cm2/s." }, { "idx": 258, "question": "The diffusion coefficients of carbon in titanium were experimentally measured at different temperatures as 2×10−9cm2/s (736℃), 5×10−9cm2/s (782℃), and 1.3×10−8cm2/s (838℃). Please determine whether the experimental results conform to D̅=D̅0exp(−ΔG/RT).", "answer": "Substituting D̅1=2×10−9cm2/s, D̅2=5×10−9cm2/s, D̅3=1.3×10−8cm2/s, T1=1009K, T2=1055K, T3=1111K into D=D0exp(−ΔG/RT) and following the method used in 7-4 yields Q1=2342787J/mol. Similarly, substituting D̅2, D̅3, T2, T3 gives Q2=2342132J/mol. It can be concluded that the experiment conforms to D=D0exp(−ΔG/RT)." }, { "idx": 259, "question": "Calculate the diffusion coefficient of carbon in α-Fe (body-centered cubic), given the diffusion coefficient formula as D=0.0079exp[-83600/RT] cm²/s, temperature T=800°C (1073K), and gas constant R=8.314 J/(mol·K).", "answer": "Substitute T=1073K into the formula D=0.0079exp[-83600/(8.314×1073)] cm²/s, and calculate to obtain D_α=6.7×10^-7 cm²/s." }, { "idx": 260, "question": "In a certain material, the grain boundary diffusion coefficient and volume diffusion coefficient of a certain particle are $\\mathrm{D}_{\\mathrm{sb}}{=}2.00\\times{10}^{-10}\\mathrm{exp(-}$ 19100/RT) $\\mathrm{cm}^{2}/\\mathrm{s}$ and $\\mathrm{D}_{\\mathrm{v}}{=}1.00\\times\\mathrm{10}^{-4}\\mathrm{exp}$ (－38200/RT) cm²/s, respectively. Determine the temperature ranges in which the grain boundary diffusion coefficient and volume diffusion coefficient dominate.", "answer": "When the grain boundary diffusion coefficient dominates, $\\mathrm{D}_{\\mathrm{gb}}>\\mathrm{D}_{\\mathrm{v}}$, i.e., $2.00\\times10^{-10}\\exp(-\\frac{19100}{R T})$ $1.00\\times10^{-4}\\exp(-\\frac{38200}{R T})$. Thus, T<1455.6K; when T>1455.6K, the volume diffusion coefficient dominates." }, { "idx": 261, "question": "In the fabrication of silicon semiconductor devices, boron is often diffused into silicon single crystals. If at a temperature of 1600K, the boron concentration on the surface of the silicon single crystal is kept constant (constant source semi-infinite diffusion), and the boron concentration at a depth of $10^{-3}\\\\mathrm{cm}$ from the surface is required to be half of the surface concentration (erfc =0.5 $\\\\frac{x}{2\\\\sqrt{D t}}$ 0.5), how long is needed (given $\\\\boldsymbol{D_{1600^{*}C}}=8\\\\times10^{-12}\\\\mathrm{{cm}^{2}/{s}}$ when 2√Dt, 2√D# )?", "answer": "This model can be regarded as a one-dimensional diffusion problem of a semi-infinite rod, which can be solved using the Gaussian error function. $\\\\frac{C_{0}-C}{C_{0}-C_{1}}=\\\\mathrm{erf}(\\\\frac{x}{2\\\\sqrt{D t}})$ where $\\\\mathcal{C}_{1\\\\mathrm{~=~}0}$, $C=0.5C_{0}$, so there is $\\\\cot(\\\\frac{x}{2\\\\sqrt{D t}})$, 2=0.5. Substituting $x=10^{-3}\\\\mathrm{cm}$ and $D=8\\\\times10^{-12}\\\\mathrm{{cm}^{2}/\\\\mathrm{{s}}}$ yields $t=1.25\\\\times10^{5}\\\\mathrm{~s~}$." }, { "idx": 262, "question": "Explain the difference in diffusion coefficients of carbon in α-Fe and γ-Fe.", "answer": "The structure of the diffusion medium has a significant impact on diffusion. α-Fe has a body-centered cubic structure, while γ-Fe has a face-centered cubic structure. The body-centered cubic structure is more open than the face-centered cubic structure. A more open structure results in lower diffusion resistance and a higher diffusion coefficient." }, { "idx": 263, "question": "Calculate the diffusion coefficient of carbon in γ-Fe (face-centered cubic), given the diffusion coefficient formula as D=0.21exp[-141284/RT] cm²/s, temperature T=800°C (1073K), and gas constant R=8.314 J/(mol·K).", "answer": "Substitute T=1073K into the formula D=0.21exp[-141284/(8.314×1073)] cm²/s, and calculate to obtain D_γ=2.1×10^-8 cm²/s." }, { "idx": 264, "question": "The diffusion activation energies of carbon, nitrogen, and hydrogen in body-centered cubic iron are 84 kJ/mol, 75 kJ/mol, and 13 kJ/mol, respectively. Analyze and explain this difference.", "answer": "The atomic radii of carbon, nitrogen, and hydrogen decrease sequentially. The smaller the atomic radius, the easier it is for the atom to diffuse through the gaps in body-centered cubic iron, and the lower the corresponding diffusion activation energy." }, { "idx": 265, "question": "Analyze the reasons why the diffusion coefficient of anions is generally smaller than that of cations in ionic crystals.", "answer": "In ionic crystals, anions generally form close packing, while cations fill the tetrahedral or octahedral voids. Therefore, cations diffuse more easily. If anions diffuse, the crystal packing arrangement must be altered, disrupting the structural framework of the ionic crystal, which results in greater resistance. Hence, in ionic crystals, the diffusion coefficient of anions is generally smaller than that of cations." }, { "idx": 266, "question": "What are the advantages, disadvantages, and applicable conditions of the Jander equation?", "answer": "The Jander equation has good adaptability in the initial stage of the reaction, but the Jander model assumes that the cross-section of spherical particles remains unchanged. Therefore, it is only applicable to cases with low conversion rates in the initial stage of the reaction. Both equations are only applicable to stable diffusion conditions." }, { "idx": 267, "question": "What are the advantages, disadvantages, and applicable conditions of the Ginstling equation?", "answer": "The Ginstling equation takes into account the fact that the reaction cross-sectional area changes during the reaction process, thus the Ginstling equation has a wider range of application and can be suitable for the initial and middle stages of the reaction. Both equations are only applicable to steady-state diffusion conditions." }, { "idx": 268, "question": "Explain from the perspectives of structure and energy why $\\mathrm{D}_{\\rightleftarrows\\mathrm{m}}>\\mathrm{D}_{\\rightleftarrows\\mathrm{m}}>\\mathrm{D}$ within the crystal.", "answer": "Under the influence of surface forces, the particles on the solid surface undergo polarization, deformation, rearrangement, and cause lattice distortion, making the surface structure different from the interior and placing the surface in a higher energy state. The internal particles of the crystal are arranged periodically, with each particle's force field being symmetrical. The activation energy required for particle migration on the surface is smaller than that within the crystal, resulting in a larger diffusion coefficient. Similarly, the arrangement of particles at grain boundaries differs from the interior, being disordered and containing defects such as vacancies and dislocations, placing them in a state of stress distortion with higher energy. The activation energy required for particle migration at grain boundaries is smaller than that within the crystal, leading to a larger diffusion coefficient. However, compared to the interior of the crystal, particles at grain boundaries are influenced by two grains to reach an equilibrium state, adopting a transitional arrangement with lower energy than surface particles. Thus, the resistance to particle migration is greater, resulting in $\\mathrm{D}_{\\mathrm{~\\rightmoon~}}\\mathrm{\\approx}{}$ the surface." }, { "idx": 269, "question": "For spherical Al2O3 particles with a diameter of 1μm surrounded by excess MgO particles, the formation of spinel is observed. At a constant temperature, 20% of the Al2O3 reacts in the first hour. Calculate the time required for complete reaction using the Jander equation.", "answer": "Calculation using the Jander equation: [1-(1-G)^(1/3)]^2 = kt ⇒ k = [1-(1-G)^(1/3)]^2 / t Substituting the reaction time of 1h and reaction progress of 20% given in the problem, we obtain k = [1-(1-0.2)^(1/3)]^2 / 1 = 5.138×10^-3 h^-1 Therefore, the time required for complete reaction (G=1) is t = 1/k = 1/(5.138×10^-3) = 194.62 h" }, { "idx": 270, "question": "When measuring the decomposition rate of alumina monohydrate, it was found that during the isothermal reaction, the mass loss increased linearly with time up to about 50%, and the rate of mass loss was less than the linear law when exceeding 50%. The rate increases exponentially with temperature. Is this a diffusion-controlled reaction or an interface first-order reaction-controlled reaction?", "answer": "According to the analysis of the kinetic equations for some important solid-phase reactions in Table 8-2 and the G-t/t0.5 curves for various types of reactions in Figure 8-22, it is known from the problem statement that when G ≤ 50%, G-t shows a linear relationship, and when G > 50%, G-t is less than the linear law. This is a diffusion-controlled reaction, and G² = kt." }, { "idx": 271, "question": "When the temperature increases from 451°C to 493°C, the rate increases by a factor of 10. Calculate the activation energy of this process. (Analyze using Table 8-2 and Figure 8-22)", "answer": "k=c exp(-Q/RT), and since G²=kt, substituting T₁=451°C=724K, T₂=493°C=766K, G₁=G, G₂=10G, we get k₂/k₁=G₂²/G₁²=exp(-Q/RT₂)/exp(-Q/RT₁)=100. Solving for Q gives Q=RT₁T₂ln(k₂/k₁)/(T₂-T₁)=8.314×724×766×ln(100)/(766-724)=505.561×10³ J/mol." }, { "idx": 272, "question": "What is phase transition?", "answer": "Phase transition is the mutual transformation between different phases of a material system." }, { "idx": 273, "question": "The formation reaction of mullite from Al2O3 and SiO powders is diffusion-controlled and conforms to the Jander equation. What effective measures should be taken to accelerate the formation of mullite?", "answer": "All factors favorable for diffusion can be employed to accelerate the formation of mullite: reducing particle size, using reactive reactants (such as Al2O3·3H2O), applying appropriate pressure, etc." }, { "idx": 274, "question": "Spherical Al2O3 particles with a diameter of 1μm are surrounded by excess MgO particles, and the formation of spinel is observed. At a constant temperature, 20% of the Al2O3 reacts in the first hour. Use the Ginstling equation to calculate the time required for complete reaction.", "answer": "Calculation using the Ginstling equation:\\n\\n1-(2/3)G-(1-G)^(2/3) = kt ⇒ k = [1-(2/3)G-(1-G)^(2/3)] / t\\n\\nSimilarly, substituting the reaction time of 1h and reaction progress of 20% from the problem, we get\\n\\nk = [1-(2/3)×0.2-(1-0.2)^(2/3)] / 1 = 4.893×10^-3 h^-1\\n\\nTherefore, for complete reaction (G=1),\\n\\n1-(2/3)G-(1-G)^2 = kt ⇒ kt = 1/3\\n\\nThus, the time required for complete reaction is t = 1/(3k) = 1/(3×4.893×10^-3) = 68.12 h" }, { "idx": 275, "question": "Analyze the influence of the chemical composition and structure of reactants on solid-phase reactions", "answer": "The greater the interaction force between particles in the reactants, the lower the reaction capability; in the same reaction system, the solid-phase reaction rate is related to the proportion between the reactants; the special role of mineralizers." }, { "idx": 276, "question": "The reaction of forming mullite from Al2O3 and SiO powders is controlled by diffusion and conforms to the Jander equation. The experiment was conducted under constant temperature conditions. When the reaction proceeded for 1 hour, it was measured that 15% of the reactants had reacted. In how much time will all the reactants be converted into products?", "answer": "According to the Jander equation, [1-(1-0.15)^(1/3)]^2=k×1 ⇒ k=0.00278. The time required for the reaction to complete (G=1) is t=1/k=1/(2.78×10^(-3))=359.63h." }, { "idx": 277, "question": "Analyze the effect of reaction temperature on solid-phase reactions", "answer": "The higher the temperature, the stronger the thermal motion of particles, and the enhanced reaction and diffusion capabilities." }, { "idx": 278, "question": "Analyze the effect of mineralizers on solid-state reactions", "answer": "The larger the lattice energy, the more complete and stable the structure, and the lower the reaction activity. Adding mineralizers can enhance solid-state reactions." }, { "idx": 279, "question": "Analyze the influence of pressure and atmosphere on solid-phase reactions", "answer": "For reactions between two solid phases, increasing pressure helps enlarge the contact area between particles, accelerates the mass transfer process, and increases the reaction rate; for solid-phase reactions involving liquid or gas phases, raising pressure does not show a positive effect and may even be counterproductive." }, { "idx": 280, "question": "Analyze the influence of particle size and distribution on solid-phase reactions", "answer": "The smaller the particle size, the faster the reaction rate; in the same reaction system, due to differences in material size, the reaction rate will be governed by different kinetic regimes; the presence of a small amount of larger-sized particles can significantly delay the completion of the reaction process." }, { "idx": 281, "question": "Analyze the effect of strain energy on the kinetics of solid-state phase transformations", "answer": "Strain energy can influence the kinetics of phase transformation processes." }, { "idx": 282, "question": "According to the phase transition mechanism, what types can it be divided into?", "answer": "According to the phase transition mechanism, it can be divided into diffusion-type phase transition, non-diffusion-type phase transition, and semi-diffusion-type phase transition. Phase transitions that rely on long-distance diffusion of atoms or ions are called diffusion-type phase transitions. Non-diffusion-type phase transitions refer to the movement of atoms or ions, but the relative displacement does not exceed the atomic spacing." }, { "idx": 283, "question": "If magnesium aluminate spinel is to be synthesized, the available raw materials are MgCO, Mg(OH)2, MgO, Al2O3·3H2O, γ-Al2O3, α-Al2O3. From the perspective of increasing the reaction rate, which raw materials should be selected? Please explain the reason.", "answer": "It is better to use MgCO3, Mg(OH)2, and Al2O3·3H2O as raw materials. This is because MgCO3 and Mg(OH)2 can undergo thermal decomposition during the reaction, and Al2O3·3H2O undergoes dehydration and crystal transformation, resulting in nascent or amorphous substances with larger specific surface areas and lattice defects, thereby enhancing the reaction activity and accelerating the solid-state reaction." }, { "idx": 284, "question": "Analyze the effect of component changes on the driving force of solid-state phase transformation", "answer": "The driving force for phase transformation is the difference in volume free energy between the new and old phases at the phase transition temperature (ΔG̃τ), and ΔG<0 is a necessary condition for the formation of the new phase. When two components mix to form a solid solution, the free energy of the mixed system changes. The magnitude of the driving force for phase transformation can be determined through the free energy-composition curve." }, { "idx": 285, "question": "Analyze the effect of strain energy on the thermodynamics of solid-state phase transformations", "answer": "Strain energy can influence the magnitude of the phase transformation driving force." }, { "idx": 286, "question": "What is the difference between martensitic transformation and nucleation-growth transformation?", "answer": "In the nucleation-growth process, there is a diffusion transformation, the composition of the parent phase and the crystalline phase can be the same or different, the transformation speed is relatively slow, and there is no obvious start and end temperature." }, { "idx": 287, "question": "What are the characteristics of martensitic transformation?", "answer": "Martensitic transformation is a first-order nucleation and growth phase transformation in which substitutional atoms undergo diffusionless shear displacement (uniform or non-uniform), resulting in shape change and surface relief, and possesses the characteristics of invariant plane strain. Features: it has shear uniformity and regularity, no atomic diffusion occurs, the transformation speed is fast, the transformation occurs within a certain range, and there is a large shear-type elastic strain energy." }, { "idx": 288, "question": "Analyze the effect of surface energy on the kinetics of solid-state phase transformations", "answer": "Surface energy can influence the kinetics of phase transformation processes." }, { "idx": 289, "question": "Analyze the influence of surface energy on the shape of new phases", "answer": "Surface energy can influence the shape of new phases." }, { "idx": 290, "question": "Analyze the effect of strain energy on the shape of the new phase", "answer": "Strain energy can influence the shape of the new phase." }, { "idx": 291, "question": "Analyze the influence of surface energy on the thermodynamics of solid-state phase transformations", "answer": "The surface of a substance has surface tension α. To reversibly increase the surface area dA under constant temperature and pressure, work αdA is required. Since the work required equals the increase in the system's free energy, and this increase is due to the enlargement of the system's surface area, it is termed surface free energy or surface energy. Surface energy can influence the magnitude of the phase transformation driving force." }, { "idx": 292, "question": "Analyze the effect of undercooling variation on the driving force during solid-state phase transformation", "answer": "Undercooling is the difference between the critical phase transformation temperature and the actual transformation temperature. The thermodynamic condition for phase transformation nucleation requires undercooling. The relationship between the driving force ΔG and undercooling ΔT is: ΔGg=-Lν(ΔT/T0)T, which further illustrates the thermodynamic condition for nucleation." }, { "idx": 293, "question": "What are the characteristics of the microstructure and properties obtained by spinodal decomposition?", "answer": "The microstructure obtained by spinodal decomposition typically exhibits a quasi-periodic and interconnected composition modulation structure or a sponge-like organization, which is uniformly fine and can only be resolved under an electron microscope." }, { "idx": 294, "question": "What is homogeneous nucleation?", "answer": "Homogeneous nucleation occurs in a uniform medium, where the probability of nucleation is the same throughout the entire medium, independent of interfaces or defects." }, { "idx": 295, "question": "Please analyze the influence of temperature on the thermodynamics and kinetics of phase transitions.", "answer": "When the temperature decreases, the degree of undercooling increases, the nucleation barrier decreases, and the nucleation rate increases until reaching the maximum value; when the temperature continues to decrease, the liquid phase viscosity increases, and the diffusion rate of atoms or molecules decreases. Both excessively high and low temperatures are unfavorable for nucleation and growth rates, and only at a certain temperature can the maximum nucleation and growth rates be achieved." }, { "idx": 296, "question": "What effect does a nucleating agent have on the critical nucleus radius r* during melt crystallization?", "answer": "Using a nucleating agent can reduce ?s, thus r* decreases." }, { "idx": 297, "question": "Given that iron has a body-centered cubic lattice with a lattice constant a=0.305 nm, determine how many unit cells the critical nucleus consists of at an undercooling of 100°C.", "answer": "Nucleus volume = (4/3)×3.14×(1.775×10⁻⁹)³ = 2.34×10⁻²⁶ m³\\nUnit cell volume = (0.305×10⁻⁹)³ = 2.83×10⁻²⁹ m³\\nNumber of unit cells = 2.34×10⁻²⁶/2.83×10⁻²⁹ = 8.25×10⁸" }, { "idx": 298, "question": "The atomic weight of iron is 55.84, its density is 7.3g/cm³, its melting point is 1593°C, its heat of fusion is 11495J/mol, and the solid-liquid interface energy is 2.04×10⁻⁵J/cm². Calculate the critical nucleus size at an undercooling of 10°C.", "answer": "ΔGᵥ = - (7.3×10³)/(55.84×10⁻³) × (11495×283)/1876 = -2.267×10⁸ J/m³\\nr* = - (2×2.04×10⁻⁵)/(-2.267×10⁸) = 1.8×10⁻⁷ m" }, { "idx": 299, "question": "Why is a certain degree of undercooling or overheating required for phase transformation in the nucleation-growth mechanism? Under what conditions is undercooling needed, and under what conditions is overheating needed?", "answer": "From the thermodynamic formula ΔG=ΔH-TΔS, at equilibrium, ΔG_V=ΔH-T_mΔS=0, ΔS=ΔH/T_m. T: equilibrium phase transition temperature; ΔH: heat of phase transition. At temperature T, the system is in a non-equilibrium state, then ΔG=ΔH-TΔS≠0. ΔG=ΔH(T_m-T)/T_m=ΔHΔT/T_m. For the phase transition to occur spontaneously, ΔG<0 must be satisfied, i.e., ΔTΔH<0. Therefore, ΔT≠0 must be achieved for the phase transition to occur. For exothermic processes such as crystallization and condensation, ΔH<0, then ΔT>0, T>0, undercooling is required. For endothermic processes such as evaporation and melting, ΔH>0, then ΔT<0, T>0, overheating is required." }, { "idx": 300, "question": "What are the differences and similarities between spinodal decomposition and precipitation decomposition?", "answer": "Spinodal decomposition involves the segregation through diffusion, decomposing a single solid solution into two solid solutions with the same structure as the parent phase but different compositions. Precipitation decomposition is the process of precipitating a second phase from a supersaturated solid solution. Their main differences are as follows: (1) Spinodal decomposition belongs to continuous phase transformation. It is a solid-state phase transformation without a thermodynamic energy barrier or nucleation. Precipitation decomposition is a nucleation-growth type phase transformation, with a thermodynamic energy barrier and a nucleation process. (2) In the early stages of spinodal decomposition, compositional fluctuations within the parent phase are gradually established, and the compositions of the two phases change continuously over time following a sinusoidal distribution pattern, eventually reaching the equilibrium phase composition. Once the nucleus of precipitation decomposition forms in the parent phase, its composition is that of the equilibrium phase, with little subsequent change. (3) Spinodal decomposition occurs uniformly in the parent phase; precipitation nuclei generally form at crystal defects. (4) The amplification process in spinodal decomposition occurs through uphill diffusion. The formation of precipitate nuclei occurs through downhill diffusion. (5) The two segregated regions in spinodal decomposition have an indistinct coherent interface. The precipitate phase and the parent phase have a distinct interface. (6) The microstructure of spinodal decomposition is regular, while that of precipitation decomposition is less uniform. Similarity: Both proceed through solute diffusion." }, { "idx": 301, "question": "What is heterogeneous nucleation?", "answer": "Heterogeneous nucleation occurs at heterogeneous interfaces, such as container walls, bubble interfaces, or on foreign substances (impurities or nucleating agents)." }, { "idx": 302, "question": "The atomic weight of iron is 55.84, its density is 7.3g/cm³, its melting point is 1593°C, its heat of fusion is 11495J/mol, and the solid-liquid interface energy is 2.04×10⁻⁵J/cm². Calculate the critical nucleus size at an undercooling of 100°C.", "answer": "ΔGᵥ = - (7.3×10³)/(55.84×10⁻³) × (11495×373)/1876 = -2.98×10⁸ J/m³\\nr* = - (2×2.04×10⁻⁵)/(-2.98×10⁸) = 1.775×10⁻⁷ m" }, { "idx": 303, "question": "During the melt cooling and crystallization process, given the solid-liquid interface energy γ_sl=5×10^-6 J/cm^2 and the unit volume free energy change △Gv=2090 J/cm^3 at 900°C, calculate the critical nucleus radius.", "answer": "The critical nucleus radius r* = -2γ_sl/△Gv = -2×5×10^-6/2090 = 4.78×10^-10 m = 0.478 nm" }, { "idx": 304, "question": "During the melt cooling and crystallization process, given the solid-liquid interface energy γ_sl=5×10^-6 J/cm^2 and the unit volume free energy change △Gv=418 J/cm^3 at 1000°C, calculate the critical nucleus radius.", "answer": "The critical nucleus radius r* = -2γ_sl/△Gv = -2×5×10^-6/418 = 2.39×10^-9 m = 2.39 nm" }, { "idx": 305, "question": "Given that iron has a body-centered cubic lattice with a lattice constant a=0.305nm, calculate the number of unit cells that make up the critical nucleus at an undercooling of 10°C.", "answer": "Nucleus volume = (4/3)×3.14×(1.8×10⁻⁹)³ = 2.44×10⁻²⁶ m³\\nUnit cell volume = (0.305×10⁻⁹)³ = 2.83×10⁻²⁹ m³\\nNumber of unit cells = 2.44×10⁻²⁶/2.83×10⁻²⁹ = 8.61×10⁸" }, { "idx": 306, "question": "During the melt cooling and crystallization process, given the solid-liquid interfacial energy γ_sl=5×10^-6 J/cm^2 and the unit volume free energy change △Gv=2090 J/cm^3 at 900°C, calculate the energy required for the phase transition.", "answer": "The energy required for the phase transition △G* = 16πγ_sl^3/3(△Gv)^2 = 16×3.14×(5×10^-6)^3/3×(2090)^2 = 3.24×10^-19 J" }, { "idx": 307, "question": "During the melt cooling and crystallization process, given the solid-liquid interface energy γ_sl=5×10^-6 J/cm^2 and the unit volume free energy change △Gv=418 J/cm^3 at 1000°C, calculate the energy required for the phase transition.", "answer": "The energy required for the phase transition △G* = 16πγ_sl^3/3(△Gv)^2 = 16×3.14×(5×10^-6)^3/3×(418)^2 = 1.19×10^-17 J" }, { "idx": 308, "question": "When forming a cubic nucleus with edge length a in the liquid phase, determine the critical free energy change ΔG*", "answer": "ΔGa* = -64γzs³/ΔGv² · ΔGv + 6 × 16γzs²/ΔGv² = 32γzs³/ΔGv²" }, { "idx": 309, "question": "When forming a cubic nucleus with edge length a in the liquid phase, determine the critical nucleus edge length a*", "answer": "From ΔGa = ΔGv + ΔGs = a³ΔGv + 6a²γzs, setting dΔGa/da = 0, we obtain 3a²ΔGv + 12aγzs = 0. Therefore, a* = -4γzs/ΔGv" }, { "idx": 310, "question": "Why is the ΔG* of a cube greater than that of a sphere?", "answer": "When forming nuclei of the same volume, a³ = (4/3)πr³ ⇒ a > r. The surface area of a cube 6a² > the surface area of a sphere 4πr², therefore ΔG*cube > ΔG*sphere" }, { "idx": 311, "question": "Are there differences in the atomic stacking modes and packing densities between face-centered cubic and hexagonal close-packed metals? Please explain.", "answer": "FCC stacks in the ABCABC sequence, while HCP stacks in the ABABAB sequence; there is no difference in packing density, both are 0.74." }, { "idx": 312, "question": "What is a crystal face family?", "answer": "A crystal face family refers to the combination of crystal faces with the same arrangement of atoms or molecules in a crystal. Due to symmetry relationships, there are often more than one type of such faces." }, { "idx": 313, "question": "Which crystal planes are included in the cubic {111} plane family?", "answer": "The cubic {111} plane family includes four planes: (111), (111), (111), (111)." }, { "idx": 314, "question": "What is the (100) interplanar spacing of a face-centered cubic metal? (a is the lattice constant)", "answer": "The interplanar spacing d=a/2" }, { "idx": 315, "question": "Compare interstitial solid solution and interstitial phase", "answer": "Commonality: In both, the alloying elements are in interstitial positions and are themselves very small in size. Differences: Interstitial solid solution is a solid solution that retains the crystal structure of the solvent and has very low solubility, exhibiting tough and good plastic properties; interstitial phase is an intermediate phase (size-factor compound), with A and B atoms in proportional quantities, exhibiting hard properties and poor plasticity." }, { "idx": 316, "question": "The melting point of copper T_m=1385K, at an undercooling of △T=0.2T_m, crystalline copper is obtained through homogeneous nucleation. Calculate the critical nucleus radius at this temperature. (ΔH=1628J/cm³, γ=1.77×10⁻⁵J/cm²)", "answer": "From ΔG_v = - (ΔH × ΔT) / T_m = - (1628 × 0.2 × 1385) / 1385 = -325.6 J/cm³. The critical nucleus radius γ* = - (2 × γ) / ΔG_v = - (2 × 1.77 × 10⁻⁵) / (-325.6) = 1.087 × 10⁻⁷ cm ≈ 1.087 nm." }, { "idx": 317, "question": "How do dislocations enter crystals?", "answer": "The increase in dislocations mainly relies on deformation, where dislocations continuously generate within grains through nucleation and multiplication." }, { "idx": 318, "question": "Explain the meaning of interstitial solid solution", "answer": "An interstitial solid solution is a type of solid solution that retains the crystal structure of the solvent and has very low solubility. The alloying elements are located in the interstitial positions and are themselves very small in size. In terms of properties, it exhibits good toughness and ductility." }, { "idx": 319, "question": "Why is the strength of solid solutions often higher than that of pure metals?", "answer": "Because the different sizes of the two types of atoms in the alloy cause lattice distortion, which hinders dislocation movement and results in solid solution strengthening." }, { "idx": 320, "question": "What is the angle between the [100] and [111] crystallographic directions in face-centered cubic metals?", "answer": "The angle between crystallographic planes cosφ=1/√3; φ=54.7°" }, { "idx": 321, "question": "What is the reason for the increase in strength caused by the proliferation of line defects and planar defects in crystals?", "answer": "The reason is that the increase in both types of defects significantly hinders the movement of dislocations, thereby enhancing strength." }, { "idx": 322, "question": "If a crystal has a high density of line defects (dislocations) or planar defects (grain boundaries, twin boundaries, etc.), its strength will significantly increase. What are these phenomena called?", "answer": "Called strain hardening and grain boundary strengthening (or fine-grain strengthening)." }, { "idx": 323, "question": "How to increase the number of dislocations?", "answer": "The number of dislocations can be increased by large deformation methods." }, { "idx": 324, "question": "Explain the meaning of interstitial phase", "answer": "Interstitial phase is an intermediate phase (size-factor compound), and the number of A and B atoms is proportional. The alloy components are all in interstitial positions, with very small sizes themselves. In terms of properties, it exhibits high hardness and poor plasticity." }, { "idx": 325, "question": "Calculate the number of atoms in the critical nucleus for copper with a face-centered cubic crystal structure. (a=0.3615nm)", "answer": "The volume of the critical nucleus V = (4/3) × π × (γ*)³ = (4/3) × 3.14 × (1.087 × 10⁻⁷)³ = 5.39 × 10⁻²¹ cm³. The unit cell volume V_cell = a³ = (0.3615 × 10⁻⁷)³ = 4.72 × 10⁻²³ cm³. The number of unit cells N = V / V_cell = 5.39 × 10⁻²¹ / 4.72 × 10⁻²³ ≈ 114. The number of atoms in a face-centered cubic unit cell is 4, so the number of atoms in the critical nucleus = 114 × 4 = 456." }, { "idx": 326, "question": "How is grain refinement achieved?", "answer": "Grain refinement can be achieved by adding heterogeneous nucleating agents or using high cooling rates during solidification, or through large deformation, recrystallization, or phase transformation methods." }, { "idx": 327, "question": "How to improve the degree of grain refinement?", "answer": "The degree of grain refinement can be improved by adding heterogeneous nucleation agents or using high cooling rates during solidification, or through large deformation, recrystallization, or phase transformation methods." }, { "idx": 328, "question": "What are the main mechanisms of metal plastic deformation at room temperature?", "answer": "The main deformation mechanisms are slip and twinning." }, { "idx": 329, "question": "Under the same degree of supercooling, compare the critical radius, critical nucleation work, and critical nucleus volume between homogeneous nucleation and heterogeneous nucleation. Which is larger?", "answer": "The critical radius is the same; the critical nucleation work is higher for homogeneous nucleation; the critical nucleus volume is also larger for homogeneous nucleation." }, { "idx": 330, "question": "Why is the liquid/solid interface front of an alloy more prone to undercooling during solidification compared to that of a pure metal?", "answer": "The interface front of an alloy exhibits constitutional undercooling, where solute enrichment at the front raises the local melting point, making undercooling more likely to occur." }, { "idx": 331, "question": "What is the main difference between slip and twinning?", "answer": "The shear displacement produced by slip is an integer multiple of the atomic spacing, while that produced by twinning is a fraction of the atomic spacing; this leads to a series of other differences." }, { "idx": 332, "question": "What are the morphological differences between typical metals (such as iron) and typical non-metals (such as silicon, graphite) when grown individually in the liquid phase?", "answer": "Because they are rough interface (iron) and smooth interface (silicon, etc.) respectively, the former forms uniform equiaxed crystals or dendrites, while the latter forms regular polygons with angular shapes." }, { "idx": 333, "question": "Give the basic conditions (driving force) for metal recrystallization.", "answer": "There must be a certain amount of deformation stored energy and a certain temperature." }, { "idx": 334, "question": "What is the main difference between recrystallization and crystallization?", "answer": "Recrystallization is only a microstructural change without structural transformation, driven by deformation stored energy; crystallization is the process of forming crystals from amorphous liquid, gas, or solid states." }, { "idx": 335, "question": "What are the upper and lower yield point effects (in pure iron or low-carbon steel)? What are the reasons?", "answer": "Significant work hardening occurs during deformation at low temperatures (or high strain rates); a balance between hardening and softening (dynamic recovery) appears during deformation at medium temperatures (or medium strain rates); a distinct softening stage (dynamic recrystallization) occurs during deformation at high temperatures (or low strain rates)." }, { "idx": 336, "question": "What is the main difference between recrystallization and solid-state phase transformation?", "answer": "Recrystallization is only a microstructural change without structural change, and the driving force is deformation stored energy; solid-state phase transformation is a structural change between solid/solid phases." }, { "idx": 337, "question": "What are the main characteristics of allotriomorphic transformation?", "answer": "Allotriomorphic transformation is primarily a phase change that occurs in pure components in the solid state, with no compositional changes, controlled by short-range diffusion processes." }, { "idx": 338, "question": "What is the main difference between crystallization and solid-state phase transformation?", "answer": "Crystallization is the process of forming crystals from amorphous liquid, gas, or solid non-crystalline states; solid-state phase transformation is the structural change between solid/solid phases. The driving forces for both processes are the chemical free energy difference." }, { "idx": 339, "question": "What are the main characteristics of martensitic transformation?", "answer": "Martensitic transformation is a diffusionless, shear-type phase transformation that occurs in both pure metals and alloys, and is controlled by interface processes." }, { "idx": 340, "question": "What are the main characteristics of precipitation transformation?", "answer": "Precipitation occurs in alloys, involves compositional changes, and is primarily controlled by long-range diffusion." }, { "idx": 341, "question": "Briefly describe the effect of deformation amount on metal properties under uniaxial compression?", "answer": "As the deformation amount increases, strength and hardness improve, while plasticity decreases." }, { "idx": 342, "question": "Briefly describe the effect of deformation amount on the microstructure of metals under uniaxial compression (including changes in grain shape and dislocation substructure)?", "answer": "From a lateral observation, as the deformation amount increases, the grains change from equiaxed to elongated, and the dislocations within the grains increase, forming dislocation tangles, subgrain boundaries, or new high-angle grain boundaries." }, { "idx": 343, "question": "Briefly describe the effect of deformation temperature on metal properties under uniaxial compression.", "answer": "As the deformation temperature increases, the rates of both strength increase and plasticity decrease slow down." }, { "idx": 344, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? How many atoms are there in an FCC unit cell?", "answer": "The number of atoms in an FCC unit cell is 4." }, { "idx": 345, "question": "Briefly describe the effect of deformation temperature on metal microstructure (including changes in grain shape and dislocation substructure) under uniaxial compression.", "answer": "As the deformation temperature increases, the rate at which grains elongate slows down due to enhanced thermal activation, subgrain boundaries form more rapidly, subgrain size tends to stabilize, and even dynamic recrystallization structures may appear." }, { "idx": 346, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? What is the close-packed plane of an FCC unit cell?", "answer": "Close-packed plane {111}." }, { "idx": 347, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? What is the close-packed direction of an FCC unit cell?", "answer": "Close-packed direction <110>." }, { "idx": 348, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? What is the coordination number of an FCC unit cell?", "answer": "Coordination number 12." }, { "idx": 349, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? What are the interstitial positions and their number in an FCC unit cell?", "answer": "Interstitial positions (octahedral interstitial at the body center and equivalent positions) and number 4." }, { "idx": 350, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? What is the stacking sequence of an FCC unit cell?", "answer": "The stacking sequence is ABCABC." }, { "idx": 351, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? What is the packing density of an FCC unit cell?", "answer": "Packing density 0.74." }, { "idx": 352, "question": "(Taking the face-centered cubic unit cell as an example) What are the common parameters used to describe the characteristics of a crystal structure (unit cell)? What is the atomic radius of an FCC unit cell?", "answer": "The atomic radius is √2a/4." }, { "idx": 353, "question": "What is the main structural difference between (metal-based) solid solutions and intermediate phases?", "answer": "Solid solutions retain the crystal structure of the pure metal, while the structure of intermediate phases is generally different from that of both constituent elements." }, { "idx": 354, "question": "What are the main differences in properties between (metal-based) solid solutions and intermediate phases?", "answer": "Solid solutions have good plasticity and toughness, while intermediate phases exhibit high strength but poorer toughness." }, { "idx": 355, "question": "What is the main difference in bonding nature between (metal-based) solid solutions and intermediate phases?", "answer": "The atoms in solid solutions are primarily bonded by metallic bonds, while intermediate phases are mainly bonded by covalent and ionic bonds." }, { "idx": 356, "question": "How does the dissolution of another element in a pure metal (assuming no new phase is formed) cause changes in properties due to microstructural changes?", "answer": "The strength increases due to solid solution strengthening, while the plasticity decreases; the electrical resistance generally increases." }, { "idx": 357, "question": "After dissolving another element into a pure metal (assuming no new phase is formed), what microstructural changes will occur?", "answer": "It causes lattice distortion, and the lattice constant will change; local segregation or ordering may occur, and even a superlattice can form." }, { "idx": 358, "question": "What are the microscopic mechanisms of diffusion?", "answer": "The main microscopic mechanisms of diffusion are the interstitial mechanism and the substitutional mechanism." }, { "idx": 359, "question": "How does the interaction between point defects and dislocations affect mechanical properties?", "answer": "At this point, the dislocations are pinned and difficult to move, increasing strength and producing upper and lower yield point effects." }, { "idx": 360, "question": "Under normal circumstances, which mechanism diffuses faster?", "answer": "Diffusion via the interstitial mechanism is faster, as interstitial atoms are smaller in size and do not require the presence of vacancies." }, { "idx": 361, "question": "Write a specific slip system for a face-centered cubic metal", "answer": "Such as (111)[110]" }, { "idx": 362, "question": "For an alloy that has solidified with microscopic non-equilibrium segregation, what measures can be taken to accelerate diffusion and homogenize the alloy?", "answer": "Heating and annealing, deformation followed by annealing, or increasing vacancy concentration through high-energy particle irradiation to enhance diffusion (though this is uneconomical and impractical)." }, { "idx": 363, "question": "Describe which solution of the second law of diffusion is applied in this process?", "answer": "The sinusoidal solution can describe the concentration distribution during diffusion." }, { "idx": 364, "question": "Why do point defects (such as interstitial atoms or substitutional atoms) and line defects (such as dislocations) interact?", "answer": "Point defects cause distortion, increasing local energy and creating an elastic strain field nearby; dislocations also exhibit this behavior, but the stress field state around dislocations varies at different positions, with some being compressive stress and others tensile stress. Point defects will aggregate onto dislocations to reduce strain energy, thereby lowering the system's energy. Dislocations that adsorb solute atoms represent a stable configuration." }, { "idx": 365, "question": "What are the types of deformation textures?", "answer": "Textures are divided into sheet textures and fiber textures." }, { "idx": 366, "question": "During the deformation of polycrystals, under a certain amount of deformation, why do some grains exhibit single slip while others exhibit multiple slips?", "answer": "Hard-oriented stress axes induce multiple slips, such as <111>, <100>, <110> directions, while soft-oriented stress axes correspond to single slip, such as the <123> direction." }, { "idx": 367, "question": "What is deformation texture?", "answer": "The deformation process causes the grains to rotate, and finally certain crystallographic directions within each grain tend to become parallel. This preferred orientation due to deformation is called deformation texture." }, { "idx": 368, "question": "During the deformation of polycrystals, under a certain amount of deformation, why do some grains exhibit large slip amounts while others show small slip amounts?", "answer": "When the orientation factor of a slip system in a grain relative to the force axis is large, slip initiates first; when the orientation factor is small, slip initiates later, resulting in different deformation amounts among grains." }, { "idx": 369, "question": "How to represent the fiber texture?", "answer": "The fiber texture is represented as ." }, { "idx": 370, "question": "Why is recrystallization annealing often required in actual production?", "answer": "Continuous plastic deformation of the material causes severe work hardening, making further processing difficult; at the same time, the deformed structure is unstable, with poor toughness and plasticity, resulting in unsatisfactory service performance; moreover, the formation of strong deformation texture is also undesirable." }, { "idx": 371, "question": "Which locations may be preferred nucleation sites?", "answer": "Preferred nucleation sites include: original grain boundaries, newly formed high-angle grain boundaries during deformation or those gradually formed through subgrain growth, and the vicinity of second-phase particles." }, { "idx": 372, "question": "Schottky defect", "answer": "A vacancy defect formed by displaced atoms migrating to the outer surface under thermal equilibrium conditions" }, { "idx": 373, "question": "What are the characteristics or features of recrystallization nucleation sites?", "answer": "Locally high dislocation density/deformation stored energy, or significant differences in dislocation density; near high-mobility high-angle grain boundaries." }, { "idx": 374, "question": "Briefly describe the laws of changes in material microstructure and properties during recovery and recrystallization annealing", "answer": "As the annealing temperature increases or the annealing time prolongs, dislocation tangles in the deformed structure evolve into subgrains, which merge and grow; recrystallization nucleation and growth occur in areas of uneven deformation, with equiaxed grains replacing elongated deformed grains; followed by normal grain growth; in terms of properties, strength and hardness decrease, electrical resistance decreases; plasticity and toughness improve, and density increases. These processes are more pronounced during the recrystallization stage than during the recovery stage." }, { "idx": 375, "question": "Up-hill diffusion", "answer": "Driven by the chemical potential gradient, solute diffuses from areas of low concentration to areas of high concentration." }, { "idx": 376, "question": "Center of gravity rule", "answer": "For an alloy in three-phase equilibrium, its composition point must lie at the centroid position of the conjugate triangle." }, { "idx": 377, "question": "同质异构体", "answer": "Chemical compositions are the same but form different crystal structures due to different thermodynamic conditions." }, { "idx": 378, "question": "Habit plane", "answer": "During solid-state phase transformation, the new phase often begins to form on certain crystallographic planes of the parent phase, which are called habit planes." }, { "idx": 379, "question": "In the diamond structure, carbon is connected by (5) bonds, and the coordination number is (6).", "answer": "(5) covalent; (6) 4" }, { "idx": 380, "question": "There are two types of the most closely packed crystal structures: one is (1), with (2) atoms in each unit cell; the other is (3), with (4) atoms in each unit cell.", "answer": "(1) ABC; (2) 4; (3) AB; (4) 2(or 6)" }, { "idx": 381, "question": "The driving force for solid-state phase transformation is (10), while the resistances are (11) and (12).", "answer": "(10) free energy difference between new and old phases; (11) interface energy; (12) strain energy" }, { "idx": 382, "question": "Coincidence site lattice", "answer": "Consider two identical and coinciding lattices $L_{1}$ and $L_{2}$. After rotating or translating $L_{2}$ relative to $L_{1}$, the lattices formed by the coinciding positions of the two lattices constitute a new periodic superlattice." }, { "idx": 383, "question": "The positional relationship between the screw dislocation line and the Burgers vector is (7), the positional relationship between the edge dislocation line and the Burgers vector is (8), and the dislocation whose dislocation line intersects obliquely with the Burgers vector is (9).", "answer": "(7) parallel; (8) perpendicular; (9) mixed dislocation" }, { "idx": 384, "question": "The higher the diffusion temperature, the more conducive it is to diffusion.", "answer": "√" }, { "idx": 385, "question": "In the composition triangle of the A-B-C ternary system, for all alloys whose composition points lie on a line parallel to the AB side, the content of the (13) component is a fixed value.", "answer": "(13) C" }, { "idx": 386, "question": "From a diffusion perspective, the mobility of low-angle grain boundaries is lower compared to that of high-angle grain boundaries.", "answer": "√" }, { "idx": 387, "question": "The principle of zone refining is based on (16).", "answer": "(16) Redistribution of solute during directional solidification of solid solution (fractional crystallization)" }, { "idx": 388, "question": "When stretching a single crystal, the slip plane is most prone to slip when it turns to an angle of (14) with the external force axis.", "answer": "(14) 45°" }, { "idx": 389, "question": "The decrease in volume free energy during the formation of a critical nucleus can only compensate for 1/3 of the newly added surface energy.", "answer": "×" }, { "idx": 390, "question": "Both edge dislocations and screw dislocations have climb and glide motions.", "answer": "×" }, { "idx": 391, "question": "Both thermoplastic and thermosetting plastics can be reused.", "answer": "×" }, { "idx": 392, "question": "In a binary system, the transformation that occurs at a certain temperature, L1 = L2 + α, is called (15) transformation.", "answer": "(15) monotectic transformation" }, { "idx": 393, "question": "Common strengthening methods for metallic materials include (17), (18), (19), and (20).", "answer": "(17) Solid solution strengthening; (18) Grain refinement strengthening; (19) Second phase strengthening; (20) Work hardening" }, { "idx": 394, "question": "The brittle phase is dispersed in granular form in the matrix of another phase, which is a microstructure state that is more beneficial to the strength and toughness of the material.", "answer": "√" }, { "idx": 395, "question": "Grain boundaries with a misorientation angle less than $2^{\\circ}$ between adjacent grains are called high-angle grain boundaries.", "answer": "×" }, { "idx": 396, "question": "If the arrangement of atoms in the parent phase of an alloy is ordered, then after martensitic transformation, the arrangement of atoms in the martensite becomes disordered.", "answer": "×" }, { "idx": 397, "question": "What is the driving force for grain boundary migration?", "answer": "The driving forces for grain boundary migration are: the stored energy of deformation and the chemical potential difference across the grain boundary caused by grain boundary curvature." }, { "idx": 398, "question": "What is spinodal decomposition?", "answer": "Spinodal decomposition is a special form of solid solution, where a single solid solution decomposes into two solid solutions with the same structure as the parent phase but different compositions through a diffusion clustering mechanism." }, { "idx": 399, "question": "Are the geometric conditions for the following dislocation reaction satisfied? Among them: b1=a/2[110], b2=a/6[12¯1], b3=a/6[211]", "answer": "Geometric conditions: after the reaction b2+b3=a/6[12¯1]+a/6[211]=a/6[330]=a/2[110] before the reaction b1=a/2[110] the geometric conditions are satisfied" }, { "idx": 400, "question": "What are the characteristics of the modulated structure?", "answer": "The modulated structure exhibits a periodic pattern, with high dispersion, uniform distribution, and high connectivity." }, { "idx": 401, "question": "Explain the conditions for spinodal decomposition.", "answer": "Conditions for spinodal decomposition: In a binary alloy phase diagram with a miscibility gap, the composition free energy curve has a range where ∂²G/∂x²<0, the temperature is sufficiently high for solute atoms to diffuse. (The decrease in free energy must be sufficient to overcome gradient energy and strain energy. This point is not required to be answered and will not be penalized.)" }, { "idx": 402, "question": "What are the main factors affecting grain boundary migration?", "answer": "The main factors affecting grain boundary migration rate: 1 solute atoms; 2 second-phase particles; 3 temperature; 4 orientation of grains on both sides of the grain boundary." }, { "idx": 403, "question": "van der Waals bond", "answer": "van der Waals bond: A physical bond formed by intermolecular attractive forces generated by instantaneous dipole moments and induced dipole moments." }, { "idx": 404, "question": "Grain boundary", "answer": "Grain boundary: The region of atomic misalignment at the interface where two grains meet." }, { "idx": 405, "question": "Kirkendall effect", "answer": "Kirkendall effect: A phenomenon in substitutional solid solutions where the relative diffusion of atoms of two components at different rates causes the migration of marker planes." }, { "idx": 406, "question": "Tie line", "answer": "Tie line: the line connecting the composition points of two equilibrium phases." }, { "idx": 407, "question": "Solid solution", "answer": "Solid solution: When foreign components enter the crystal structure, occupying part of the host crystal phase's lattice sites or interstitial positions, while still maintaining a single crystal phase, such a crystal is called a solid solution." }, { "idx": 408, "question": "Is the energy condition for the following dislocation reaction satisfied? Where: b1=a/2[110], b2=a/6[12¯1], b3=a/6[211]", "answer": "Energy condition: after the reaction |b2|^2+|b3|^2=(a^2/6^2)[1^2+2^2+(-1)^2]+(a^2/6^2)[2^2+1^2+1^2]=a^2/6+a^2/6=a^2/3 before the reaction |b1|^2=(a^2/2^2)(1^2+1^2)=a^2/2 the energy condition is satisfied and the dislocation reaction can proceed." }, { "idx": 409, "question": "Dislocation climb", "answer": "Dislocation climb: The movement of an edge dislocation perpendicular to the slip plane." }, { "idx": 410, "question": "Eutectoid transformation", "answer": "Eutectoid transformation: The process in which a single solid phase simultaneously precipitates two new solid phases with different compositions and crystal structures is called eutectoid transformation." }, { "idx": 411, "question": "Constitutional supercooling", "answer": "Constitutional supercooling: During crystallization, the redistribution of solid and liquid phase compositions leads to the formation of a supercooled region in the liquid phase near the solid-liquid interface. This phenomenon is called constitutional supercooling." }, { "idx": 412, "question": "Habit plane", "answer": "Habit plane: During solid-state phase transformation, the new phase often begins to form on certain crystallographic planes of the parent phase, and these planes are called habit planes." }, { "idx": 413, "question": "1. In the ionic crystal structure, the positive and negative ions form (1)", "answer": "coordination polyhedron" }, { "idx": 414, "question": "For a carbon steel containing 0.1% carbon, carburized at 930‰, the carburized layer thickness is 0.4mm after 3 hours. Someone wants to obtain a 0.8mm carburized layer and plans to use 6 hours. Is this person's plan correct? Why?", "answer": "$$ c{\\left(\\begin{array}{l l}{x,t}\\end{array}\\right)}=c_{1}+{\\left(\\begin{array}{l}{c_{8}-c_{1}}\\end{array}\\right)}{\\left(1-\\operatorname{erf}{\\frac{x}{2{\\sqrt{D t}}}}\\right)} $$ According to the problem, $c(\\textit{x}_{1},\\textit{t}_{1})=c(\\textit{x}_{2},\\textit{t}_{2})c_{s}$, and $c_{1}$ is a constant. Therefore, $$ \\operatorname{erf}{\\frac{x_{1}}{2{\\sqrt{D t_{1}}}}}=\\operatorname{erf}{\\frac{x_{2}}{2{\\sqrt{D t_{2}}}}} $$ Hence, $$ {\\frac{x_{1}}{2{\\sqrt{D t_{1}}}}}={\\frac{x_{2}}{2{\\sqrt{D t_{2}}}}}\\quad{\\frac{x_{1}}{\\sqrt{t_{1}}}}={\\frac{x_{2}}{\\sqrt{t_{2}}}},\\quad{\\frac{0.4}{\\sqrt{3}}}={\\frac{0.8}{\\sqrt{t_{2}}}} $$ $$ t_{2}={\\left({\\frac{0.8}{0.4}}\\times{\\sqrt{3}}\\right)}^{2}\\mathbf{h}=12\\mathbf{h} $$ Thus, the person's plan is incorrect." }, { "idx": 415, "question": "1. In the ionic crystal structure, the distance between positive and negative ions depends on (2)", "answer": "The sum of positive and negative ion radii" }, { "idx": 416, "question": "Multiple slip", "answer": "Multiple slip: When the resolved shear stress on several slip systems is equal and simultaneously reaches the critical resolved shear stress, the phenomenon of simultaneous slip occurs." }, { "idx": 417, "question": "3. The factors influencing the formation of substitutional solid solutions are _ (8)", "answer": "Electronegativity" }, { "idx": 418, "question": "2. In polymer chains, the different spatial forms of molecules due to (4) are called conformations", "answer": "Internal rotation of single bonds" }, { "idx": 419, "question": "2.The property of polymers that allows them to change conformation is called (5)", "answer": "Flexibility" }, { "idx": 420, "question": "3. The factors influencing the formation of substitutional solid solutions include _ (6)", "answer": "Ionic size" }, { "idx": 421, "question": "1.In the ionic crystal structure, the coordination number depends on the (3) of the positive and negative ions", "answer": "radius ratio" }, { "idx": 422, "question": "3. The factors influencing the formation of substitutional solid solutions include _ (7)", "answer": "Crystal structure type" }, { "idx": 423, "question": "3. The factors influencing the formation of substitutional solid solutions include _ (9)", "answer": "Electron concentration factor" }, { "idx": 424, "question": "5. The two basic characteristics of martensitic transformation are (12)", "answer": "Coherent shear" }, { "idx": 425, "question": "6. Common methods for strengthening metal materials include: (15)", "answer": "Dispersion strengthening" }, { "idx": 426, "question": "5. The two basic characteristics of martensitic transformation are (11)", "answer": "Diffusionless transformation" }, { "idx": 427, "question": "6. Common methods for strengthening metal materials include: (13)", "answer": "Grain refinement strengthening" }, { "idx": 428, "question": "6.Common methods for strengthening metal materials include: (14)", "answer": "Solid solution strengthening" }, { "idx": 429, "question": "4.The expression for the relationship between the diffusion coefficient, diffusion activation energy, and diffusion temperature is (10)", "answer": "D=D0exp(-Q/RT)" }, { "idx": 430, "question": "7. The typical ingot structure usually consists of (17)", "answer": "Surface fine grain zone" }, { "idx": 431, "question": "7.The typical ingot structure usually consists of (18)", "answer": "columnar crystal zone" }, { "idx": 432, "question": "Point defects are thermodynamically stable defects, and a certain number of equilibrium defects exist in crystals at a given temperature, also known as intrinsic defects.", "answer": "(√)" }, { "idx": 433, "question": "6.Common methods for strengthening metal materials include: (16)", "answer": "Deformation" }, { "idx": 434, "question": "7.Typical ingot structure usually has (19)", "answer": "Central equiaxed crystal zone" }, { "idx": 435, "question": "Non-spontaneous nucleation still requires structural fluctuations, compositional fluctuations, and energy fluctuations.", "answer": "(√)" }, { "idx": 436, "question": "The eutectic transformation occurs in systems where the liquid phase is completely miscible and the solid phase is completely immiscible.", "answer": "(×)" }, { "idx": 437, "question": "Polymers with simple structure, high regularity, and good symmetry are not prone to crystallization.", "answer": "(×)" }, { "idx": 438, "question": "The reason why the carbon solubility of austenite is higher than that of ferrite is because the crystal interstices of austenite are larger.", "answer": "(√)" }, { "idx": 439, "question": "8.The driving force for the coarsening of precipitates is (20)", "answer": "The Gibbs free energy difference between different particles" }, { "idx": 440, "question": "In a ternary phase diagram, the degree of freedom at the eutectic temperature point is 0. At this time, it is a three-phase equilibrium.", "answer": "(×)" }, { "idx": 441, "question": "During diffusion, solute atoms always migrate from high concentration to low concentration.", "answer": "(×)" }, { "idx": 442, "question": "In the crystal structure of iodides, iodine occupies the corners of the cube and the body-centered position, so its structure type is body-centered lattice.", "answer": "(×)" }, { "idx": 443, "question": "The yield strength σs of crystalline materials changes with the variation of the tensile axis relative to the crystal orientation.", "answer": "(√)" }, { "idx": 444, "question": "Generally speaking, during solid-state phase transformations, crystal defects in the parent phase can promote the formation of the new phase.", "answer": "(√)" }, { "idx": 445, "question": "What is the total cost for carburizing 500 gears at 900°C (1173K) for 10 hours, given the cost is 1000 yuan per hour?", "answer": "Total cost = 1000 yuan/h * 10 h = 10000 yuan" }, { "idx": 446, "question": "What is the total cost for carburizing 500 gears at 1000°C (1273K) for 3.299 hours, given the cost is 1500 yuan per hour?", "answer": "Total cost = 1500 yuan/h * 3.299 h = 4948.5 yuan" }, { "idx": 447, "question": "What is the equivalent time required to achieve the same carburization depth at 1000°C (1273K) as 10 hours at 900°C (1173K), given Q=32900 cal/mol and R=1.987 cal?", "answer": "t_1273 = D_1173 * t_1173 / D_1273 = 10 * exp[-32900/(1.987*1173)] / exp[-32900/(1.987*1273)] = 10 * exp(-14.1156) / exp(-13.0068) h = 10 * exp(-1.5089) h = 10 * 0.3299 h = 3.299 h" }, { "idx": 448, "question": "Explain the reason for cold deformation strengthening of pure metals using dislocation theory", "answer": "Strengthening reason: Dislocation intersection produces kinks and jogs, dislocation reactions produce immobile dislocations, and dislocation multiplication increases dislocation density." }, { "idx": 449, "question": "At which temperature is the carburization cost lower for processing 500 gears to achieve the same depth?", "answer": "The cost at 1000°C (4948.5 yuan) is lower than at 900°C (10000 yuan), so 1000°C is more cost-effective." }, { "idx": 450, "question": "What effect does a rough interface have on the crystal growth mode?", "answer": "For a rough interface, continuous growth occurs." }, { "idx": 451, "question": "Changes occurring during recrystallization annealing", "answer": "Changes occurring: During annealing, the dislocation density decreases, deformed grains transform into equiaxed grains, residual stresses are eliminated, and strength and hardness are reduced." }, { "idx": 452, "question": "Approximate process parameters for recrystallization annealing", "answer": "Process parameters: Heating temperature T=0.4Tm plus 100~200Ω, heating time depends on the specific furnace load and workpiece size." }, { "idx": 453, "question": "What is a rough interface?", "answer": "Rough interface: It is a solid-liquid interface where the atomic arrangement on the solid-phase surface is uneven and rough, displaying no crystallographic plane characteristics." }, { "idx": 454, "question": "What is a smooth interface?", "answer": "Smooth interface: It is another type of solid-liquid interface where the atoms on the solid-phase interface are arranged into a flat atomic plane, specifically a certain crystallographic plane." }, { "idx": 455, "question": "How to completely eliminate work hardening? Explain the heat treatment method used", "answer": "Method to eliminate work hardening: recrystallization annealing." }, { "idx": 456, "question": "Can the perfect dislocation $\\pmb{b}_{1}$ in a face-centered cubic crystal decompose into partial dislocations $b_{2}, b_{3}?$ Given ${\\pmb b}_{1}=\\frac{a}{2}[\\stackrel{-}{1}10], {\\pmb b}_{2}=\\frac{a}{6}[\\stackrel{-}{1}2\\stackrel{-}{1}], {\\pmb b}_{3}=\\frac{a}{6}[\\stackrel{-}{2}11],$ state the reasons.", "answer": "Geometric condition: $\\begin{array}{l}{{\\vec{b}_{1}=a/2\\cdot\\left[\\stackrel{\\rightharpoonup}{1}10\\right]}}\\ {{}}\\ {{\\vec{b}_{2}+\\stackrel{\\rightharpoonup}{b}_{3}=a/6\\cdot\\left[\\stackrel{\\rightharpoonup}{3}30\\right]=a/2\\cdot\\left[\\stackrel{\\rightharpoonup}{1}10\\right]}}\\end{array}$ Energy condition: $\\mid\\vec{b}_{1}\\mid^{2}=(a/2\\sqrt{1+1+0})^{2}=a^{2}/2$ $|\\vec{b}_{2}|^{2}+|\\vec{b}_{3}|^{2}=(a/6\\sqrt{1+4+1})^{2}+\\bigl(a/6\\sqrt{4+1+1}^{2}=a^{2}/3
Plate martensiteLenticular shape
Lath martensiteLong columnar shape with elliptical cross-section
" }, { "idx": 462, "question": "Screw dislocation", "answer": "A dislocation where the dislocation line is parallel to the Burgers vector." }, { "idx": 463, "question": "Explain the differences in metallographic structure between plate martensite and lath martensite", "answer": "
Plate martensiteThe initially formed martensite needles traverse the austenite grains, while subsequently formed ones gradually become smaller. The plates are not parallel to each other. Those with C%>1.4% have a midrib
Lath martensiteThe original austenite grains are divided into several blocks, each consisting of several lath bundles, and each lath bundle is composed of many parallel lath martensites
" }, { "idx": 464, "question": "Explain the difference in substructure between plate martensite and lath martensite", "answer": "
Plate martensiteThere are a large number of twins in the middle of the martensite, and dislocations at the edges. The midrib is fine twins
Lath martensiteThere are high-density dislocations inside the martensite, forming dislocation cells
" }, { "idx": 465, "question": "Habit plane", "answer": "During solid-state phase transformation, the new phase often forms along specific atomic planes of the parent phase. The crystallographic plane of the parent phase that is parallel to the main plane of the new phase is called the habit plane." }, { "idx": 466, "question": "Sorbite", "answer": "A product of medium-temperature pearlite transformation, composed of lamellar ferrite and cementite, with smaller interlamellar spacing and thinner layers." }, { "idx": 467, "question": "Cross-slip", "answer": "The slip process transitions from one slip plane to another along the same slip direction." }, { "idx": 468, "question": "Frenkel defect", "answer": "The displacement of an atom to an interstitial site in the crystal lattice is called a Frenkel defect." }, { "idx": 469, "question": "Pseudoeutectoid transformation", "answer": "Pseudoeutectoid transformation: During non-equilibrium transformation, hypoeutectoid or hypereutectoid alloys near the eutectoid composition point exhibit a fully eutectoid microstructure upon completion of the transformation." }, { "idx": 470, "question": "Perfect dislocation", "answer": "A dislocation whose Burgers vector equals a lattice vector is called a perfect dislocation." }, { "idx": 471, "question": "Coordination polyhedron", "answer": "The polyhedron formed by connecting the centers of atoms or ions directly bonded to a central atom or ion is called the coordination polyhedron of that atom or ion." }, { "idx": 472, "question": "4. According to the different atomic arrangement structures at the interface, what three types of interfaces can phase interfaces in solids be divided into?", "answer": "Coherent, semi-coherent, incoherent" }, { "idx": 473, "question": "1. How many independent slip systems must be activated for plastic deformation in polycrystalline materials?", "answer": "5" }, { "idx": 474, "question": "The number-average relative molecular mass of polymers ( $\\overline{{{\\cal M}_{n}}}$ )", "answer": "The relative molecular mass of polymers weighted by number, $\\widetilde{M_{n}}=\\sum_{i}{N_{i}M_{i}}/{\\sum{N_{i}}}$, where $N_{i}$ is the molar fraction of molecules with molecular weight $M_{i}$." }, { "idx": 475, "question": "2. During solid-state phase transformation nucleation, when the nucleus morphology is disk-shaped, it has the minimum what energy and the maximum what?", "answer": "strain energy, interface" }, { "idx": 476, "question": "Explain the differences in mechanical properties between plate martensite and lath martensite", "answer": "
Plate martensiteHard and brittle
Lath martensiteStrong and tough
" }, { "idx": 477, "question": "3. In solid-state crystals, diffusion can be classified into which two types based on whether new phase structures are formed during atomic diffusion?", "answer": "Atomic diffusion and reactive diffusion" }, { "idx": 478, "question": "Eutectoid transformation", "answer": "The process in which a single solid phase simultaneously precipitates two new solid phases with different compositions and crystal structures is called eutectoid transformation." }, { "idx": 479, "question": "7. What are the main structural types (subtypes) of silicate crystals?", "answer": "Island, group (or ring), chain, layer, framework" }, { "idx": 480, "question": "6. In polymers, the different spatial forms of molecules caused by what are called conformations, and the property of polymers to change conformations is called what?", "answer": "Single bond internal rotation, flexibility" }, { "idx": 481, "question": "What is the ground-state electron configuration of Cr (atomic number 24)?", "answer": "1s22s22p63s23p63d54s1" }, { "idx": 482, "question": "9.Linear polymers can be reused and are also called what kind of plastic; cross-linked polymers cannot be reused and are called what kind of plastic?", "answer": "Thermoplastic, thermosetting" }, { "idx": 483, "question": "8. Intermetallic compounds typically include what controlled by electronegativity, close-packed phases primarily governed by atomic size, and what mainly controlled by electron concentration?", "answer": "Normal valence compounds, topologically close-packed phases, electron compounds" }, { "idx": 484, "question": "In the bainitic transformation, neither Fe nor C atoms undergo diffusion.", "answer": "√" }, { "idx": 485, "question": "In the solid-state crystal diffusion process, the direction of diffusing atom migration is always from high concentration to low concentration.", "answer": "√" }, { "idx": 486, "question": "The finer the grain size, the higher the strength and hardness of the crystal, and the better the plasticity and toughness.", "answer": "√" }, { "idx": 487, "question": "During solid-state phase transformation, crystal defects in the parent phase hinder the formation of nuclei of the new phase.", "answer": "√" }, { "idx": 488, "question": "In a ternary phase diagram, the degree of freedom at the eutectic temperature point is 0. At this time, it is a three-phase equilibrium.", "answer": "×" }, { "idx": 489, "question": "Component crystal types differ, but under specific conditions, they can also form infinitely miscible solid solutions.", "answer": "×" }, { "idx": 490, "question": "Polymers with simple structures, high regularity, and good symmetry are prone to crystallization.", "answer": "√" }, { "idx": 491, "question": "Two edge dislocations with opposite signs and parallel dislocation lines on the same slip plane will move away from each other due to their interaction.", "answer": "×" }, { "idx": 492, "question": "What is aging?", "answer": "Aging: The precipitation process of a supersaturated solid solution." }, { "idx": 493, "question": "In the cesium chloride structure, chlorine occupies the corner positions and the body center position of the cube, so its structure type is body-centered lattice.", "answer": "×" }, { "idx": 494, "question": "What are the main morphological types of martensitic transformation products in steel?", "answer": "(2) Morphology of transformation products: lath martensite (1 point); acicular martensite. (1 point)" }, { "idx": 495, "question": "The reason why the carbon solubility of austenite is higher than that of ferrite is that the interstitial spaces in the austenite crystal structure are larger.", "answer": "√" }, { "idx": 496, "question": "What processes does aging typically undergo?", "answer": "Aging process: 1) Formation of GP zones. 2) Formation of a series of metastable intermediate phases that maintain coherency or semi-coherency with the parent phase. 3) Formation of stable precipitates. 4) Coarsening and growth of stable precipitates." }, { "idx": 497, "question": "What are the intracrystalline substructures of martensitic transformation products in steel?", "answer": "(3) Intracrystalline substructure: Lath martensite mainly consists of high-density dislocations (1 point); acicular martensite has high-density twins with some dislocations at the edges. (1 point)" }, { "idx": 498, "question": "What are the main characteristics of martensitic transformation in steel?", "answer": "(1) Transformation characteristics: $\\textcircled{1}$ Diffusionless, neither Fe nor C atoms diffuse. (2 points) $\\textcircled{2}$ Coherent shear, with habit plane and orientation relationship between the new phase and parent phase. (2 points) $\\textcircled{3}$ Non-isothermal nature. (1 point) $\\textcircled{4}$ Incompleteness of martensitic transformation. (1 point)" }, { "idx": 499, "question": "What are the Burgers vector and dislocation properties of the generated dislocation?", "answer": "The Burgers vector of the generated dislocation $\\widehat{b}_{3}$ is $a/2~[110]$, which is a pure edge dislocation. The slip plane (001) is not the slip plane of the face-centered cubic structure, so it is a sessile dislocation." }, { "idx": 500, "question": "What effect will this dislocation reaction have on the plastic deformation of the metal?", "answer": "The dislocation reaction generates a sessile dislocation $\\pmb{a}/2$ [110], which will hinder the movement of dislocations on the (111) and (111) planes, resulting in significant strain hardening macroscopically." }, { "idx": 501, "question": "What is the reason for age hardening?", "answer": "Reason for age hardening: Dispersion precipitation of metastable phases, maintaining coherent or semi-coherent interfaces with the matrix phase, hindering dislocation movement in the matrix phase. 1) For deformable second-phase particles, dislocations cut through the second-phase particles, destroying the coherent or semi-coherent relationship between the second-phase particles and increasing the phase boundary area, while the second-phase particles themselves possess high strength. 2) For non-deformable second-phase particles, dislocations bypass the second-phase particles, leaving dislocation loops around them, increasing the dislocation line length and enhancing the resistance of the second-phase particles to subsequent dislocation movement." }, { "idx": 502, "question": "Configuration of polymer chains", "answer": "Configuration of polymer chains: The spatial geometric arrangement of atoms in polymers fixed by chemical bonds." }, { "idx": 503, "question": "How to overcome constitutional supercooling in single crystal growth?", "answer": "To overcome constitutional supercooling, increase the temperature gradient and reduce the growth rate R." }, { "idx": 504, "question": "What is constitutional supercooling? Explain its cause.", "answer": "Due to the addition of a second component, the solidification point of the melt decreases at the solid-liquid interface, while the melt away from the interface maintains a higher melting point. Due to the presence of impurities, there is a distribution coefficient ko= Cs/CL at the solid-liquid interface, where the impurity concentration is higher than the average concentration C0. The concentration changes with distance as: CL=C0(1+(1-k0)/k0 e^(-R X/D)), where D is the diffusion coefficient, resulting in TL=Tm-mLC0(1+(1-k0)/k0 e^(-R X/D)). When the cooling rate of the melt is high, the liquid temperature falls below the solidification temperature, causing droplets to be surrounded by crystals and forming scattered particles. This phenomenon is called constitutional supercooling." }, { "idx": 505, "question": "Explain the reason why the dislocation $b_{1}=a/2\\\\big[10\\\\Bar{1}\\\\big]$ in the (111) plane of a face-centered cubic lattice metal and the dislocation $b_{2}=a/2[011]$ in the (111) plane can undergo a dislocation reaction.", "answer": "The dislocation reaction $a/2{\\\\left[10{\\\\stackrel{\\\\leftarrow}{1}}\\\\right]}+a/2{\\\\left[011\\\\right]}\\\\rightarrow a/2{\\\\left[110\\\\right]}$ can occur for the following reasons: $\\\\textcircled{1}$ It satisfies the structural condition $\\\\vec{b}_{1}+\\\\vec{b}_{2}=\\\\vec{b}_{3}$. $\\\\textcircled{2}$ It satisfies the energy condition: $a^{2}/2+a^{2}/2>a^{2}/2_{\\\\circ}$." }, { "idx": 506, "question": "Magnesium oxide and aluminum oxide (solute) form a substitutional solid solution, write the defect equation.", "answer": "Defect equation: Al2O3→MgO 2AlMg· + VMg'' + 3OOc" }, { "idx": 507, "question": "Dislocation slip", "answer": "Dislocation slip: Under a certain stress, the movement of dislocations where the dislocation line moves along the slip plane." }, { "idx": 508, "question": "Can magnesium oxide and aluminum oxide (solute) form a continuous solid solution? Explain the reason.", "answer": "They cannot form a continuous solid solution due to different crystal structures." }, { "idx": 509, "question": "Active oxygen", "answer": "Active oxygen: oxygen in silicate structures with partially unsaturated valence." }, { "idx": 510, "question": "Homogeneous nucleation", "answer": "Homogeneous nucleation: The formation of nuclei in a homogeneous phase relying on conditions such as its own structural fluctuations and energy fluctuations." }, { "idx": 511, "question": "Equilibrium distribution coefficient", "answer": "Equilibrium distribution coefficient: The ratio of the solute concentration in the solid phase to that in the liquid phase during equilibrium solidification." }, { "idx": 512, "question": "What are the main influencing factors of constitutional supercooling?", "answer": "The main influencing factors of constitutional supercooling include: melt cooling rate, impurity concentration C0, partition coefficient k0, diffusion coefficient D, growth rate R, and temperature gradient." }, { "idx": 513, "question": "Uphill diffusion", "answer": "Uphill diffusion: Under the condition of chemical potential difference as the driving force, atoms diffuse from low concentration positions to high concentration positions." }, { "idx": 514, "question": "Stored energy", "answer": "Stored energy: A small portion of the energy consumed during cold deformation exists within the deformed crystal in the form of elastic strain energy and structural defect energy, which is called stored energy." }, { "idx": 515, "question": "In face-centered cubic metals, can the dislocation b3 generated by the dislocation reaction b1 + b2 = b3 move on the slip plane?", "answer": "The slip plane of face-centered cubic metals is {111}, and b3 is perpendicular to this crystal plane, so it cannot slip." }, { "idx": 516, "question": "Pseudo-eutectoid", "answer": "Pseudo-eutectoid: During non-equilibrium transformation, hypoeutectoid or hypereutectoid alloys near the eutectoid composition point form a completely eutectoid microstructure by the end of the transformation." }, { "idx": 517, "question": "How do crystal structure and solid solution type affect atomic diffusion in crystalline solids?", "answer": "In crystal structures with smaller packing density, the activation energy for diffusion is lower, making diffusion easier to occur; in crystal structures with lower symmetry, the anisotropy of the diffusion coefficient is more pronounced; the activation energy for diffusion in interstitial solid solutions is much smaller than that in substitutional solid solutions, facilitating easier diffusion." }, { "idx": 518, "question": "Recrystallization", "answer": "Recrystallization: The process in which deformed materials, upon heating, undergo the replacement of deformed grains by strain-free grains." }, { "idx": 519, "question": "How does temperature affect atomic diffusion in crystalline solids?", "answer": "The higher the temperature, the greater the diffusion coefficient and the faster the diffusion rate." }, { "idx": 520, "question": "In face-centered cubic metals, what types of dislocations are b1 = [101], b2 = (a/6)[1̄2̄1], b3 = (a/3)[1̄1̄1̄] respectively?", "answer": "b1: unit dislocation; b2: Shockley partial dislocation; b3: Frank partial dislocation." }, { "idx": 521, "question": "Habit plane", "answer": "Habit plane: During solid-state phase transformations, the new phase often forms along specific crystallographic planes of the parent phase, which are called habit planes." }, { "idx": 522, "question": "How do crystal defects affect atomic diffusion in crystalline solids?", "answer": "The diffusion coefficient along grain boundaries is much greater than that of bulk diffusion; diffusion activation energy is smaller when occurring along dislocation pipes, thus dislocations accelerate diffusion." }, { "idx": 523, "question": "What is aging?", "answer": "Aging refers to the process in which a supersaturated solid solution is retained at room temperature or higher temperature for a period of time, resulting in the precipitation of a second phase from the matrix." }, { "idx": 524, "question": "What changes occur in the microstructure of metals after plastic deformation?", "answer": "After plastic deformation of metals, in terms of microstructure morphology, the originally equiaxed grains are elongated along the deformation direction. Under large deformation, grain boundaries may even appear fibrous. If hard and brittle second-phase particles or inclusions are present, they often distribute in a banded pattern along the deformation direction." }, { "idx": 525, "question": "What changes occur in the properties of metal after plastic deformation?", "answer": "In terms of properties, cold-deformed metal will undergo work hardening, manifested as a significant increase in strength and a noticeable decrease in plasticity." }, { "idx": 526, "question": "In face-centered cubic metals, can the dislocation reaction b1 + b2 = b3 proceed, where b1 = [101], b2 = (a/6)[1̄2̄1], b3 = (a/3)[1̄1̄1̄]?", "answer": "Geometric condition: b1 + b2 = (a/2)[101̄] + (a/6)[1̄2̄1] = (a/3)[1̄1̄1̄], b3 = (a/3)[1̄1̄1̄], satisfying the geometric condition; Energy condition: after the reaction |b3|² = (a²/9) × (√(1+1+1))² = a²/3, before the reaction Σ|bi|² = (a²/4) × (√(1+0+1))² + (a²/36) × (√(1+4+1))² = 2a²/3 > after the reaction |b3|²; satisfying the energy condition, the reaction b1 + b2 → b3 can proceed." }, { "idx": 527, "question": "2. The symmetry elements of the internal structure of a crystal are (3), (4), and (5) in addition to c, Pλ, Ln, and Lin.", "answer": "(3) Translation axis; (4) Glide plane; (5) Screw axis" }, { "idx": 528, "question": "After plastic deformation of metal, what changes occur in its structure?", "answer": "Microstructurally, the density of defects (vacancies and dislocations) significantly increases. Due to the multiplication of dislocations during deformation and the intersection and interaction of dislocations during movement, dislocation tangles form, and the annihilation of dislocations with opposite signs results in a cellular structure. As the deformation increases, the number of dislocation cells increases, their size decreases, and the stored energy within the crystal rises." }, { "idx": 529, "question": "1. Kaolinite belongs to the layered silicate structure, with one layer being (1) and another layer being (2).", "answer": "(1) silicon-oxygen tetrahedral layer; (2) aluminum-oxygen octahedral layer" }, { "idx": 530, "question": "How does the third component affect atomic diffusion in crystalline solids?", "answer": "Depending on the nature of the added third component, some promote diffusion while others hinder it." }, { "idx": 531, "question": "5. When Ca0 is doped into ZrO2, where Ca2+ ions replace Zr4+ ions, due to the requirement of electrical neutrality, this substitution simultaneously creates a (9) defect, which can be represented by the (10) defect reaction equation.", "answer": "(9) vacancy; (10) CaO→ZrO2CaZr′′+V0..+ O0" }, { "idx": 532, "question": "Explain why the strength of polycrystalline materials increases with decreasing grain size.", "answer": "During plastic deformation of polycrystalline materials, coarse grains have a higher number of dislocations piled up at grain boundaries, creating a larger stress field that can activate dislocation sources in adjacent grains, allowing deformation to continue. In contrast, fine grains have fewer dislocations piled up at grain boundaries, requiring a greater external force to activate dislocation sources in adjacent grains for deformation to proceed. Therefore, finer-grained materials need a larger external force to undergo plastic deformation, meaning the strength of the crystal increases with decreasing grain size. (5 points)" }, { "idx": 533, "question": "4. The term 'segment' is used to describe the (8) property of polymers.", "answer": "(8) flexibility" }, { "idx": 534, "question": "8. Methods to strengthen metal materials include (16) strengthening, (17) strengthening, (18) strengthening, (19) strengthening.", "answer": "(16) solid solution; (17) dislocation; (18) fine grain; (19) dispersion (or precipitate particles)" }, { "idx": 535, "question": "Syndiotactic", "answer": "Syndiotactic: A polymer in which the substituents are alternately attached to opposite sides of the main chain plane." }, { "idx": 536, "question": "What is the reason for age strengthening?", "answer": "The reasons for age strengthening are as follows: First, when disc-shaped metastable phases precipitate and have a certain orientation relationship with the parent phase, large elastic strains are generated within the matrix, which can significantly strengthen the alloy. Second, when the alloy undergoes deformation, the interaction between dispersed particles and dislocations strengthens the alloy. If the precipitate particles are deformable, as dislocations cut through them, work is required to increase the surface area of the particles, thereby increasing the resistance to dislocation motion and strengthening the alloy. If the precipitate particles are strong and incoherent with the matrix, dislocation lines find it difficult to cut through the particles and will bypass them under applied stress, leaving dislocation loops. For the alloy to continue deforming, the stress exerted by the particles as dislocation lines bypass them must be overcome, requiring further increase in the applied stress, thus resulting in age strengthening of the alloy." }, { "idx": 537, "question": "Non-uniform nucleation", "answer": "Non-uniform nucleation: The nucleation process during the cooling of a molten liquid that occurs on a certain interface within the parent phase." }, { "idx": 538, "question": "7. Small-angle grain boundaries are composed of dislocations, among which symmetric tilt boundaries are composed of (14) dislocations, and twist boundaries are composed of (15) dislocations.", "answer": "(14) edge; (15) screw" }, { "idx": 539, "question": "Polymorph", "answer": "Polymorph: Crystals of the same chemical composition that form different structures under varying thermodynamic conditions." }, { "idx": 540, "question": "9.Linear polymers can be reused, also known as (20); cross-linked polymers cannot be reused and are called thermosetting plastics.", "answer": "(20) Thermoplastic" }, { "idx": 541, "question": "Deformation texture", "answer": "Deformation texture: As the amount of plastic deformation increases, the phenomenon where a certain crystallographic orientation of the deformed polycrystal tends to become consistent." }, { "idx": 542, "question": "Peritectic transformation", "answer": "Peritectic transformation: At a certain temperature, a liquid phase with a specific composition interacts with a solid phase of a specific composition to form a new solid phase with another specific composition." }, { "idx": 543, "question": "Pearlite", "answer": "Pearlite: The product of eutectoid transformation in iron-carbon alloys, it is a lamellar mixture of eutectoid ferrite and eutectoid cementite." }, { "idx": 544, "question": "3. In a ternary phase diagram, the eutectic point is (6) phase equilibrium, and the degree of freedom is (7).", "answer": "(6) four; (7) 0" }, { "idx": 545, "question": "Li Sheng", "answer": "Li Sheng: The uniform shear process in which one part of a crystal moves relative to another along specific crystallographic planes and directions under the action of shear stress." }, { "idx": 546, "question": "2. In face-centered cubic packing, the number of atoms in the unit cell is (5), it has (6) _ tetrahedral voids, (7) _ octahedral voids.", "answer": "(5) 4; (6) 8; (7) 4" }, { "idx": 547, "question": "Spinodal decomposition", "answer": "Spinodal decomposition: A transformation in which a solid solution decomposes into two solid solutions with the same structure as the parent phase but different compositions through uphill diffusion." }, { "idx": 548, "question": "Recrystallization", "answer": "Recrystallization: The process in which plastically deformed metal is subsequently heated, and nucleation and growth of strain-free equiaxed grains gradually replace the deformed grains." }, { "idx": 549, "question": "Aging", "answer": "Aging: The subsequent process of solute atom precipitation from a supersaturated solid solution at room temperature or above." }, { "idx": 550, "question": "1. Crystals can be classified into (1) crystal families, (2) crystal systems according to symmetry, with a total of (3) point groups and (4) space groups.", "answer": "(1) 3; (2) 7; (3) 32; (4) 230" }, { "idx": 551, "question": "6. The driving force for nucleation in solid-state phase transformation is (11), and the main resistances are (12) and (13).", "answer": "(11) The difference in free energy between the new phase and the parent phase; (12) Interface energy; (13) Strain energy" }, { "idx": 552, "question": "4. The driving force for nucleation in solid-state phase transformation is (11), and the main resistances are (12) and (13).", "answer": "(11) The difference in free energy between the new phase and the parent phase; (12) interface energy; (13) strain energy" }, { "idx": 553, "question": "3. The three crystal zones of an ingot refer to (8), (9), and (10).", "answer": "(8) surface fine grain zone; (9) columnar crystal zone; (10) central equiaxed coarse grain zone" }, { "idx": 554, "question": "5.After recrystallization is completed, grain growth can be divided into (14) grain growth and (15) grain growth.", "answer": "(14) normal; (15) abnormal" }, { "idx": 555, "question": "7.During martensitic transformation, the new phase often begins to form on a specific crystallographic plane of the parent phase, which is called (20).", "answer": "(20) habit plane" }, { "idx": 556, "question": "Given: When the grain size of annealed pure iron is 16 per $\\mathbf{\\chi}_{\\mathbf{m}\\mathbf{m}^{2}}$, the yield strength $\\sigma_{\\mathsf{s}}=100\\mathrm{N/\\Omega}$ $\\mathbf{m}\\mathbf{m}^{2}$; when the grain size is 4096 per $\\prime_{\\mathbf{m}\\mathbf{m}^{2}}$, $\\sigma_{\\mathrm{s}}=250\\mathrm{N}/\\mathrm{mm}^{2}$. Find the value of yield strength $\\pmb{\\sigma_{\\S}}$ when the grain size is 256 per $\\mathbf{m}\\mathbf{m}^{2}$.", "answer": "According to the Hall-Petch formula: $\\sigma_{\\mathrm{s}}=\\sigma_{0}+K_{\\mathrm{y}}d^{-1/2}$, the yield strength $\\sigma_{\\mathbf{s}}$ of the material is calculated from the average grain size $^d$. Expressing the grain size in terms of the radius of an equal-area circle: $$d_{1}=(4A_{1}/\\pi)^{1/2}d_{2}=(4A_{2}/\\pi)^{1/2}$$$A_{1}$ and $A_{2}$ are the grain areas. Thus, $$\\sigma_{\\mathrm{s1}}=\\sigma_{0}+K_{\\mathrm{y}}d_{1}^{-1/2}\\sigma_{\\mathrm{s2}}=\\sigma_{0}+K_{\\mathrm{y}}d_{2}^{-1/2}$$Given that $A_{1}=1/16\\mathrm{mm}^{2}$, $A_{2}=1/4096\\mathrm{mm}^{2}$, $\\sigma_{\\mathrm{s}1}=100\\mathrm{N/mm}^{2}$, $\\sigma_{\\mathrm{s}2}=250\\mathrm{N}/\\mathrm{mm}^{2}$, substituting into the above equations yields $K_{\\mathrm{y}}=25{\\sqrt{2}}\\pi^{-1/4}\\mathrm{N}/\\mathrm{mm}^{3/2}$, $\\pmb{\\sigma_{0}}=50\\mathbf{N}/\\mathbf{m}\\mathbf{m}^{2}$. Then, with $A_{3}=1/256\\mathrm{mm}^{2}$, we find $\\sigma_{\\mathbf{s}3}=150\\mathbf{N}/\\mathbf{m}\\mathbf{m}^{2}$. (10 points)" }, { "idx": 557, "question": "6. Methods for strengthening metal materials include (16) strengthening, (17) strengthening, (18) strengthening, and (19) strengthening.", "answer": "(16) solid solution; (17) dislocation; (18) fine grain; (19) dispersion (or precipitate particles)" }, { "idx": 558, "question": "What are the factors affecting atomic diffusion in crystalline solids? Briefly explain the influence of temperature.", "answer": "Temperature. The higher the temperature, the greater the diffusion coefficient and the faster the diffusion rate." }, { "idx": 559, "question": "What are the factors affecting atomic diffusion in crystalline solids? Briefly explain the influence of the third component.", "answer": "The third component. Depending on the nature of the added third component, some promote diffusion while others hinder it." }, { "idx": 560, "question": "What are the factors affecting atomic diffusion in crystalline solids? Briefly explain the influence of crystal defects.", "answer": "Crystal defects. The diffusion coefficient along grain boundaries is much larger than that of bulk diffusion; the activation energy for diffusion along dislocation pipes is smaller, thus dislocations accelerate diffusion." }, { "idx": 561, "question": "What are the factors affecting atomic diffusion in crystalline solids? Briefly explain the influence of crystal structure and solid solution type.", "answer": "Crystal structure and solid solution type. In crystal structures with smaller packing density, the diffusion activation energy is lower, making diffusion easier to occur; in crystal structures with lower symmetry, the anisotropy of the diffusion coefficient is significant; the diffusion activation energy in interstitial solid solutions is much smaller than that in substitutional solid solutions, facilitating easier diffusion." }, { "idx": 562, "question": "Analyze the changes in vacancies and dislocations in deformed metals during the recrystallization stage and their effects on properties.", "answer": "Recrystallization stage: With the rearrangement of atoms during recrystallization, the dislocation density significantly decreases, the complex dislocation interactions generated during deformation disappear, the work hardening phenomenon vanishes, and the various properties of the deformed metal return to their pre-deformation state." }, { "idx": 563, "question": "Briefly explain the disadvantages brought by the work hardening phenomenon", "answer": "Disadvantages brought by the work hardening phenomenon: causes difficulty in plastic deformation. (3 points)" }, { "idx": 564, "question": "Analyze the changes of vacancies and dislocations in deformed metals during the recovery stage and their effects on properties.", "answer": "Recovery stage: The vacancy concentration decreases to approach the equilibrium concentration, leading to a reduction in resistivity and an increase in crystal density. The mutual annihilation of dislocations with opposite signs on the same slip plane causes a slight decrease in dislocation density. During high-temperature recovery, edge dislocations undergo polygonization through slip and climb movements, resulting in the retention of work hardening. The strength and hardness slightly decrease, while residual stresses are essentially eliminated." }, { "idx": 565, "question": "How to eliminate work hardening phenomenon", "answer": "Subsequent recrystallization treatment is adopted to eliminate work hardening phenomenon. (3 points)" }, { "idx": 566, "question": "Briefly explain the benefits brought by the work hardening phenomenon", "answer": "The benefits brought by the work hardening phenomenon: During the deformation process, the dislocation density increases, and interactions between dislocations occur (such as dislocation intersection, dislocation entanglement, etc.), leading to dislocation pile-up, thereby strengthening the metal material. It also makes the deformation of the metal material more uniform. (4 points)" }, { "idx": 567, "question": "Given the diffusion constant of carbon in γ-Fe D0=2.0×10^-5m^2/s and the activation energy for diffusion Q=140×10^3J/mol, calculate the diffusion coefficient of carbon in γ-Fe at 927℃.", "answer": "The diffusion coefficient of carbon in γ-Fe at 927℃: 15.99×10^-12m^2/s" }, { "idx": 568, "question": "Given the diffusion constant of carbon in γ-Fe D0=2.0×10^-5m^2/s and the diffusion activation energy Q=140×10^3J/mol, calculate the diffusion coefficient of carbon in γ-Fe at 870℃.", "answer": "The diffusion coefficient of carbon in γ-Fe at 870℃: 7.94×10^-12m^2/s" }, { "idx": 569, "question": "In a face-centered cubic crystal, a dislocation with Burgers vector b1= [101] meets a dislocation with b2= [121]. Can a dislocation reaction occur to form b3? Write the equation.", "answer": "Geometric condition: b3=b1+b2= a/2[101]+a/6[121]=a/6[222]=a/3[111]" }, { "idx": 570, "question": "Given the diffusion constant of carbon in γ-Fe D0=2.0×10^-5m^2/s and the activation energy for diffusion Q=140×10^3J/mol, how long is required to achieve the same carburizing thickness at 870°C as that obtained by carburizing for 10 hours at 927°C? (Ignore the difference in solubility of carbon in γ-Fe at different temperatures)", "answer": "20.12 hours are required." }, { "idx": 571, "question": "When $\\\\mathrm{Yb}\\\\mathrm{F}_{3}$ and $\\\\mathbf{NaF}$ are doped into $\\\\mathbf{CaF}_{2}$ crystals, if both ytterbium ions and sodium ions replace the cation positions, please write the defect reaction equation.", "answer": "$\\\\mathrm{YbF}_{3}+\\\\mathrm{NaF}\\\\xrightarrow{\\\\mathrm{CaF}_{2}}\\\\mathrm{Yb}_{\\\\mathrm{Ca}}^{*}+\\\\mathrm{Na}_{\\\\mathrm{Ca}}^{\\\\prime}+4\\\\mathrm{F}_{\\\\mathrm{Fc}}$" }, { "idx": 572, "question": "In a face-centered cubic crystal, indicate which types of dislocations b1, b2, and b3 belong to respectively", "answer": "b1 is a perfect dislocation, b2 is a Shockley partial dislocation, and b3 is a Frank partial dislocation." }, { "idx": 573, "question": "How do dislocations strengthen metallic materials?", "answer": "Dislocations can strengthen metallic materials." }, { "idx": 574, "question": "Give two applications of solidification theory in the crystallization of metallic materials, and briefly explain them", "answer": "Single crystal preparation. Using directional solidification technology to control the heat flow direction, enabling the metal to grow along a single crystal orientation, resulting in single crystals without grain boundaries, which are used in critical components such as high-temperature alloy blades." }, { "idx": 575, "question": "Give one application of solidification theory in the crystallization of metallic materials, and briefly explain it", "answer": "Ingot structure control. By controlling the cooling rate and temperature gradient during solidification, the grain size and distribution of the ingot can be adjusted, thereby improving the mechanical properties of the metallic material." }, { "idx": 576, "question": "Give three applications of solidification theory in the crystallization of metallic materials, and provide a brief explanation for each", "answer": "Amorphous material preparation. By extremely rapid cooling (on the order of 10^6 K/s), the metal melt skips the crystallization stage and directly forms an amorphous structure, obtaining high-strength, corrosion-resistant amorphous alloys." }, { "idx": 577, "question": "In a face-centered cubic crystal, a dislocation with Burgers vector b1= [101] meets a dislocation with b2= [121]. Explain the reason whether the dislocation reaction can occur.", "answer": "Energy condition: b1^2+b2^2= (a^2/4)(1+0+1)+(a^2/36)(1+4+1)=a^2/2+a^2/6=2a^2/3, b3^2= (a^2/9)(1+1+1)=a^2/3, b3^2" }, { "idx": 592, "question": "Explain the term: grain boundary segregation", "answer": "Grain boundary segregation: The enrichment phenomenon of solute atoms or impurity atoms at grain boundaries due to the difference in distortion energy between the grain interior and the grain boundaries or the presence of vacancies." }, { "idx": 593, "question": "The packing density of FCC structure is", "answer": "0.74" }, { "idx": 594, "question": "The stacking sequence of the close-packed planes in an FCC structure is", "answer": "ABCABC…" }, { "idx": 595, "question": "In the FCC structure, when atoms are regarded as rigid spheres with radius r, the relationship between the atomic radius and the lattice constant a is", "answer": "r=a" }, { "idx": 596, "question": "The number of atoms in the unit cell of an FCC structure is", "answer": "4" }, { "idx": 597, "question": "The necessary condition for forming an ordered solid solution is", "answer": "Slow cooling rate" }, { "idx": 598, "question": "The close-packed plane of FCC structure is", "answer": "{111}" }, { "idx": 599, "question": "The necessary condition for forming an ordered solid solution is", "answer": "a certain chemical composition" }, { "idx": 600, "question": "The necessary condition for forming an ordered solid solution is", "answer": "The mutual attraction between dissimilar atoms is greater than that between similar atoms" }, { "idx": 601, "question": "The coordination number of FCC structure is", "answer": "12" }, { "idx": 602, "question": "When a disordered solid solution transforms into an ordered solid solution, the general trend in the changes of alloy properties is that strength and hardness", "answer": "increase" }, { "idx": 603, "question": "In the Fe-Fe3C phase diagram, what is the carbon content below which the material is considered steel?", "answer": "2.11%C" }, { "idx": 604, "question": "When a disordered solid solution transforms into an ordered solid solution, the general trend in the change of alloy properties is: electrical conductivity", "answer": "decreases" }, { "idx": 605, "question": "When a disordered solid solution transforms into an ordered solid solution, the general trend in the change of alloy properties is: plasticity", "answer": "decreases" }, { "idx": 606, "question": "In the Fe-Fe3C phase diagram, what carbon content is greater than that of cast iron?", "answer": "2.11%C" }, { "idx": 607, "question": "The carbon content of ledeburite is", "answer": "4.3%" }, { "idx": 608, "question": "The crystal structure of austenite is", "answer": "FCC" }, { "idx": 609, "question": "At room temperature, the equilibrium microstructure of hypoeutectoid steel is", "answer": "ferrite and pearlite" }, { "idx": 610, "question": "During the equilibrium solidification of alloys, the maximum carbon content in austenite is", "answer": "2.11%" }, { "idx": 611, "question": "What is the carbon content of pearlite", "answer": "0.77%" }, { "idx": 612, "question": "What are the two basic phases that make up the room-temperature equilibrium structure of iron-carbon alloys", "answer": "Ferrite (α)" }, { "idx": 613, "question": "From what is Fe3CI precipitated", "answer": "Liquid phase" }, { "idx": 614, "question": "At room temperature, the equilibrium microstructure of hypereutectoid steel is", "answer": "pearlite and Fe3CII" }, { "idx": 615, "question": "What are the two basic phases that make up the room-temperature equilibrium microstructure of iron-carbon alloys?", "answer": "Cementite (Fe3C)" }, { "idx": 616, "question": "From what is Fe3CII precipitated", "answer": "Austenite" }, { "idx": 617, "question": "According to the effect of cooling rate on metal microstructure, how should the solidification be controlled to obtain an amorphous structure?", "answer": "To obtain an amorphous structure, the molten metal should be cooled at an extremely rapid rate." }, { "idx": 618, "question": "Cold working and hot working", "answer": "Cold working and hot working: Generally distinguished based on the recrystallization temperature of metal materials. Processing above the recrystallization temperature is called hot working, while processing below the recrystallization temperature and at room temperature is called cold working." }, { "idx": 619, "question": "Crystal zone and zone axis", "answer": "Crystal zone and zone axis: The collective term for various sets of crystal planes parallel to the same crystal direction is called a crystal zone, and the crystal direction parallel to these sets of crystal planes is called the zone axis." }, { "idx": 620, "question": "Jog and kink", "answer": "Jog and kink: During the movement of dislocations, certain dislocation intersection lines are formed after intersecting with other dislocations. If the dislocation line after intersection lies on the original slip plane of the dislocation, this dislocation line is called a kink. If the dislocation line after intersection is perpendicular to the original slip plane of the dislocation, this dislocation line is called a jog." }, { "idx": 621, "question": "Peritectoid reaction and eutectoid reaction", "answer": "Peritectoid reaction and eutectoid reaction: The process in which two solid phases react to form one solid phase is called a peritectoid reaction, while the reaction in which one solid phase decomposes into two other solid phases is called a eutectoid reaction." }, { "idx": 622, "question": "From what is Fe3CIII precipitated", "answer": "Ferrite" }, { "idx": 623, "question": "The carbon content of Fe3CI, Fe3CII, Fe3CIII is", "answer": "6.69%" }, { "idx": 624, "question": "According to the influence of cooling rate on metal microstructure, how should the solidification be controlled to obtain metastable phases?", "answer": "To obtain metastable phases, the cooling rate must far exceed the equilibrium cooling rate." }, { "idx": 625, "question": "Columnar crystals and equiaxed crystals", "answer": "Columnar crystals and equiaxed crystals: During the crystallization process of metal crystals, the elongated grains that preferentially grow along the heat dissipation direction are called columnar crystals, while if the grains grow without a preferred direction and the growth rates in all directions are roughly equal, the resulting grains are called equiaxed crystals." }, { "idx": 626, "question": "Explain the effect of grain boundaries on material deformation", "answer": "Grain boundaries strengthen metal materials at room temperature but weaken them at high temperatures." }, { "idx": 627, "question": "According to the solidification theory, analyze the characteristics of typical ingot structure", "answer": "The typical ingot structure features fine equiaxed crystals in the outer layer, columnar crystals extending inward from the surface, and coarse equiaxed crystals in the center." }, { "idx": 628, "question": "What final heat treatment process can be used for overhead copper conductors (requiring certain strength) after processing for outdoor use? Why?", "answer": "Overhead copper conductors for outdoor use require high strength, and recovery annealing is generally used to eliminate stress while retaining certain strength." }, { "idx": 629, "question": "How to improve the strength of solid solution alloys?", "answer": "Solid solution strengthening, work hardening, grain refinement strengthening, precipitation strengthening." }, { "idx": 630, "question": "What final heat treatment process can be used for decorative wires for indoor electric lights after processing? Why?", "answer": "Decorative wires for indoor electric lights need to be easily deformable and have high electrical conductivity. Recrystallization annealing can be used to soften them and achieve high electrical conductivity." }, { "idx": 631, "question": "Give two examples to illustrate the phenomenon of uphill diffusion in metals", "answer": "Examples of uphill diffusion in metals include the formation of Cottrell atmospheres and the spinodal decomposition process." }, { "idx": 632, "question": "Briefly explain the effect of atomic diffusion on materials", "answer": "The effects of atomic diffusion on materials include processes such as grain nucleation and growth in casting, grain boundary movement, oxidation, etc., all of which involve atomic diffusion." }, { "idx": 633, "question": "Explain why solid solution alloys can grow in a dendritic manner under a positive temperature gradient during solidification, while pure metals cannot form dendritic crystals.", "answer": "Due to solute redistribution causing constitutional supercooling, solid solution alloys can grow dendritically even under a positive temperature gradient during solidification; whereas pure metals require a negative temperature gradient to form dendritic crystals." }, { "idx": 634, "question": "Explain the influence of grain boundaries on material properties", "answer": "Grain boundaries affect various aspects of materials, possessing grain boundary energy, influencing the shape of second phases in polycrystalline materials. Grain boundaries can move, exhibit grain boundary segregation, undergo changes in grain boundary charge, bear load transfer effects, have low melting points and are prone to overheating. Grain boundaries serve as easy diffusion paths, facilitate nucleation, and are susceptible to corrosion." }, { "idx": 635, "question": "Metal glass", "answer": "Metal glass: refers to a solid formed when metal solidifies from a liquid state, retaining the same structure as the liquid metal" }, { "idx": 636, "question": "Grain boundary segregation", "answer": "Grain boundary segregation: The enrichment phenomenon of solute atoms or impurity atoms at grain boundaries due to the difference in distortion energy between the grain interior and grain boundaries or the presence of vacancies." }, { "idx": 637, "question": "Intermetallic compounds", "answer": "Intermetallic compounds: Compounds formed between metals and metals, or between metals and certain non-metals, whose structure and composition differ from those of the pure metals forming the intermetallic compounds. They generally exhibit characteristics such as high melting points, high hardness, and significant brittleness." }, { "idx": 638, "question": "Cottrell atmosphere", "answer": "Cottrell atmosphere: the phenomenon of solute atoms aggregating around an edge dislocation. This atmosphere can impede dislocation motion, resulting in effects such as solid solution strengthening." }, { "idx": 639, "question": "Divorced eutectic", "answer": "Divorced eutectic: In alloys with eutectic reactions, if the composition is far from the eutectic point, due to the large amount of primary crystals and the small amount of eutectic, the same phase in the eutectic as the primary crystals attaches to the primary crystals and grows, while the other phase in the eutectic appears separately distributed, causing the eutectic structure to lose its characteristic organizational features." }, { "idx": 640, "question": "Twinning", "answer": "Twinning: An important mode of plastic deformation in crystals. Under shear stress, one part of the crystal undergoes uniform shear relative to another part along specific crystallographic planes and directions, resulting in different orientations of adjacent crystal regions that are mirror-symmetrical across the twinning plane (twin boundary)." }, { "idx": 641, "question": "Explain the main factors affecting the diffusion coefficient in the equation", "answer": "The main factors affecting the diffusion coefficient in the equation include temperature, crystal structure, crystal defects, type of solid solution, properties of the diffusing element, and concentration of the diffusing component." }, { "idx": 642, "question": "Deformation texture", "answer": "Deformation texture: After plastic deformation, the originally differently oriented grains in the polycrystal become essentially aligned, forming a preferred orientation of the grains. The resulting crystal structure with preferred orientation is called texture. If the texture is generated during plastic deformation, it is referred to as deformation texture." }, { "idx": 643, "question": "Determine whether the dislocation reaction a/6[2\\overline{1}\\overline{1}] + a/6[\\overline{1}21] → a/6[110] can proceed in a face-centered cubic crystal?", "answer": "With reference to the geometric and energy conditions, the dislocation reaction can proceed." }, { "idx": 644, "question": "If the leading dislocations of two extended dislocations undergo the reaction a/6[2\\overline{1}\\overline{1}] + a/6[\\overline{1}21] → a/6[110], what effect will it have on the properties of face-centered cubic metals?", "answer": "The dislocations become immobile after the reaction, affecting the work hardening mechanism and fracture properties of the crystal." }, { "idx": 645, "question": "Reaction diffusion", "answer": "Reaction diffusion: The diffusion accompanied by chemical reactions that leads to the formation of new phases is called reaction diffusion. For example, when infiltrating elements diffuse from the metal surface into the interior and their concentration exceeds the solubility limit, new phases appear." }, { "idx": 646, "question": "Write the expression for the unsteady-state diffusion equation", "answer": "The expression for the unsteady-state diffusion equation is: ∂C/∂t = ∂(D(∂c/∂x))/∂x. If the diffusion coefficient D does not vary with concentration, it can be simplified to ∂C/∂t = D(∂²c/∂x²)." }, { "idx": 647, "question": "Point out the main factors affecting the recrystallization temperature of metals after cold deformation", "answer": "The main factors affecting the recrystallization temperature of metals after cold deformation are: degree of deformation, trace impurities and alloying elements, metal grain size, heating time, heating rate" }, { "idx": 648, "question": "What are the main measures to obtain fine recrystallized grains, and why?", "answer": "The main measures to obtain fine recrystallized grains are: increasing the degree of cold deformation, adding trace alloying elements, increasing the heating rate, and using fine-grained metals" }, { "idx": 649, "question": "Describe the main strengthening mechanisms that can be employed to strengthen Al-5%Cu alloy", "answer": "The strengthening mechanisms for Al-5%Cu alloy are solid solution strengthening, precipitation strengthening, work hardening, and grain refinement strengthening" }, { "idx": 650, "question": "What is the relationship between the corrosion of engineering materials and bonding?", "answer": "The essence of corrosion in engineering materials is the formation and destruction of bonds." }, { "idx": 651, "question": "Describe the main strengthening mechanisms that may be employed to strengthen the Al-5%Al2O3 composite material", "answer": "The strengthening mechanisms for the Al-5%Al2O3 composite material are work hardening, fine grain strengthening, and dispersion strengthening" }, { "idx": 652, "question": "For industrial pure copper ingots with low stacking fault energy, after hot breakdown rolling at $T=0.5T_{\\\\xi\\\\sharp,\\\\xi}$ temperature, what methods can be adopted in the final process to obtain semi-hard products, and why?", "answer": "To obtain semi-hard products, cold working followed by recovery annealing, or appropriate cold deformation after complete recrystallization annealing can be adopted." }, { "idx": 653, "question": "Describe the main strengthening mechanisms that can be employed for industrial pure aluminum", "answer": "The main strengthening mechanisms for industrial pure aluminum are work hardening and grain refinement strengthening" }, { "idx": 654, "question": "For industrial pure copper ingots with low stacking fault energy, after hot breakdown rolling at $T=0.5T_{\\\\xi\\\\sharp,\\\\xi}$ temperature, what methods can be used in the final process to obtain soft-state products, and why?", "answer": "To obtain soft-state products, cold working followed by recrystallization annealing can be used." }, { "idx": 655, "question": "What is work hardening?", "answer": "The phenomenon where the flow stress of a metal increases with increasing strain during cold working is called work hardening." }, { "idx": 656, "question": "For industrial pure copper ingots with low stacking fault energy, after hot breakdown rolling at $T=0.5T_{\\\\xi\\\\sharp,\\\\xi}$ temperature, what methods can be adopted in the final process to obtain hard-state products, and why?", "answer": "To obtain hard-state metal after breakdown, cold working can be performed, with the mechanism being work hardening." }, { "idx": 657, "question": "How do the types of material bonding bonds affect the physical properties of materials?", "answer": "The stronger the bonding bond, the higher the melting point, the smaller the thermal expansion coefficient, and the greater the density. Metals have luster, high electrical and thermal conductivity, good mechanical strength and plasticity, and a positive temperature coefficient of resistance, which are related to the metallic bonds of metals. Ceramics and polymers are generally non-conductive in the solid state, which is related to their non-metallic bonding." }, { "idx": 658, "question": "What are the adverse effects of texture on metal materials?", "answer": "Texture induces anisotropy in metals, and anisotropy often causes difficulties in the processing and use of metals. Deep-drawn metal cups may develop ears. Polycrystalline metal rods with texture can significantly elongate after repeated heating and cooling cycles." }, { "idx": 659, "question": "How do the types of material bonding affect the mechanical properties of materials?", "answer": "The hardness of crystalline materials is related to the bonding of the crystals. Generally, crystals bonded by covalent bonds, ionic bonds, and metallic bonds have higher hardness than those bonded by molecular bonds. The greater the bonding energy between the bonds, the higher the elastic modulus. The strength of engineering materials is also somewhat related to the bonding energy. Generally, higher bonding energy results in higher strength. The plasticity of materials is also related to the type of bonding. Materials bonded by metallic bonds exhibit good plasticity, while those bonded by ionic or covalent bonds have difficulty undergoing plastic deformation, hence ceramic materials exhibit very poor plasticity." }, { "idx": 660, "question": "List the various possible mechanisms of work hardening", "answer": "The mechanisms of work hardening are mainly the interactions between dislocations, such as the formation of jogs, dislocation intersection and entanglement, the interaction between moving dislocations and forest dislocations, the formation of Lomer-Cottrell locks, and the formation of solute atom atmospheres due to the interaction between solute atoms and dislocations." }, { "idx": 661, "question": "Choose any material and describe its possible uses and processing methods.", "answer": "Aluminum alloy is used to manufacture packaging containers such as beverage cans and food cans. For example, beverage companies like Coca-Cola and Pepsi extensively use aluminum alloy cans, which offer excellent sealing, corrosion resistance, and recyclability, ensuring beverage quality protection and convenience for consumers." }, { "idx": 662, "question": "How does temperature affect the diffusion coefficient?", "answer": "As temperature increases, the diffusion coefficient increases sharply." }, { "idx": 663, "question": "What are the beneficial applications of texture in metallic materials?", "answer": "In some cases, it is necessary to deliberately obtain certain textures to utilize their anisotropy, such as silicon steel sheets for transformers, high-voltage electronic aluminum foils for precision electronic capacitors, and interstitial-free steels used in automotive and mechanical sectors that require superior deep-drawing performance. For heat-treatable non-strengthening aluminum alloys, work hardening through cold deformation can be employed to strengthen the alloy." }, { "idx": 664, "question": "Explain the industrial applications of work hardening", "answer": "Industrial applications of work hardening: During processing, the resistance of metal to plastic deformation continuously increases, making the metal brittle and necessitating multiple intermediate annealing processes, which requires more power consumption for cold working of the metal; work hardening provides the metal matrix with a certain ability to resist accidental overload; appropriate combination of work hardening and plastic deformation enables uniform plastic deformation of the metal, and some processing methods require the metal to have a certain degree of work hardening; work hardening is also one of the important means to strengthen metals, and for some pure metals, work hardening is a key method to improve strength; some components continuously harden on the surface under working conditions, meeting requirements for impact and wear resistance on the surface; after work hardening, the plasticity of materials decreases, improving machinability of materials such as low-carbon steel; the final properties of products can be controlled through cold working." }, { "idx": 665, "question": "What are the methods for controlling the texture of metal materials?", "answer": "One method to control texture is by regulating the processing and heat treatment regimes to obtain a processed structure with only slight texture and a fine-grained recrystallized structure. Additionally, altering the production process of rolled plates, combining intermediate recrystallization annealing during cold working with final recovery annealing can control texture. The cross-rolling method can also be employed. Controlling the hot rolling process of aluminum alloy sheets, multi-directional cross cold rolling, and corresponding annealing can reduce the formation of ears." }, { "idx": 666, "question": "How do crystal defects affect the diffusion coefficient?", "answer": "The diffusion rate of atoms along line defects (dislocations) and planar defects (grain boundaries and free surfaces, etc.) is much greater than that of bulk diffusion within the crystal." }, { "idx": 667, "question": "Briefly explain the role of the second phase during cold plastic deformation", "answer": "The primary deformation modes of duplex alloys remain slip and twinning. Due to differences in the type, quantity, size, shape, distribution characteristics of the second phase in the alloy, as well as its interfacial bonding with the matrix, the influence of the second phase on plastic deformation is complex. If both phases in the duplex alloy are plastic, and the size and deformability of the second phase are similar to those of the matrix, the deformation of the alloy depends on the volume fraction of the two phases. If one phase is plastic and the other is hard and brittle, the plastic deformation of the alloy mainly depends on the presence of the hard and brittle phase. If the second phase is coarse or large-needle-shaped, deformation occurs only in the matrix, and the alloy's plasticity will not be high. If the second phase is continuously distributed along the grain boundaries of the solid solution, the alloy becomes very brittle and almost incapable of plastic deformation." }, { "idx": 668, "question": "How does the anisotropy of crystals affect the diffusion coefficient?", "answer": "The anisotropy of crystals also affects the diffusion coefficient, but this anisotropy gradually decreases with increasing temperature." }, { "idx": 669, "question": "How does allotropic transformation affect the diffusion coefficient?", "answer": "In metals with allotropic transformation, the diffusion coefficient changes significantly with the alteration of crystal structure, and the diffusion coefficient of solute atoms in BCC crystal structure is larger than that in FCC." }, { "idx": 670, "question": "What is the difference in diffusion activation energy between interstitial atoms and substitutional atoms?", "answer": "The diffusion activation energy of interstitial atoms is much smaller than that of substitutional atoms." }, { "idx": 671, "question": "How does solute concentration affect the diffusion coefficient?", "answer": "In many solid solution alloys, the diffusion coefficient of the solute increases with increasing concentration." }, { "idx": 672, "question": "Explain the basic concept of dislocations", "answer": "Dislocations are a type of arrangement defect in crystal structures, which can be divided into edge dislocations, screw dislocations, and mixed dislocations." }, { "idx": 673, "question": "How does the difference in properties between the diffusing element and the solvent metal affect the diffusion coefficient?", "answer": "The greater the difference in properties between the diffusing element and the solvent metal, the larger the diffusion coefficient." }, { "idx": 674, "question": "What is the effect of a third element on the diffusion of components in a binary alloy?", "answer": "The third element that forms compounds with the components in the alloy hinders the diffusion of solute atoms, thereby reducing the diffusion coefficient." }, { "idx": 675, "question": "Explain the effect of grain size on the mechanical properties of metal materials at room temperature", "answer": "Room temperature strengthening. Fine grain strengthening can improve the strength and hardness of metal materials. The smaller the grain size, the more grain boundaries there are, and the greater the resistance to dislocation movement." }, { "idx": 676, "question": "Explain the diffusion mechanism", "answer": "The diffusion mechanisms mainly include: interstitial diffusion where solute atoms diffuse through interstices in interstitial solid solutions, and the vacancy mechanism where atoms exchange with vacancies to achieve diffusion in substitutional solid solutions." }, { "idx": 677, "question": "Explain uphill diffusion", "answer": "Uphill diffusion is a diffusion process where the diffusing element moves from a region of low concentration to a region of high concentration; examples include the formation of various solute atom atmospheres and the formation of eutectoid reaction products, both of which are uphill diffusion." }, { "idx": 678, "question": "How does the effect of solute elements on the melting point of alloys influence the diffusion coefficient?", "answer": "Solute elements that can lower the melting point of alloys (or cause a decrease in the liquidus line) will increase the diffusion coefficient; conversely, they will decrease the diffusion coefficient." }, { "idx": 679, "question": "How to control the microstructure and properties of materials through deformation and heating?", "answer": "Deformation can increase dislocation density, thereby enhancing material strength; heating can promote dislocation rearrangement and recrystallization, thereby refining grains and improving material properties." }, { "idx": 680, "question": "How to control the microstructure and properties of materials through deformation and heating?", "answer": "Deformation and heating can serve as effective means to control the microstructure and properties of materials. For example, cold deformation can lead to work hardening, fibrous microstructure, internal stress, texture, and other changes in physical and chemical properties. Heating cold-deformed metals can achieve the desired microstructure and properties through the changes in microstructure and properties caused by recovery and recrystallization." }, { "idx": 681, "question": "Explain the effect of grain size on the high-temperature mechanical properties of metallic materials", "answer": "High-temperature weakening. Grain boundaries become weak points at high temperatures, excessively fine grains reduce high-temperature creep resistance, and appropriately coarsening grains is beneficial for high-temperature performance." }, { "idx": 682, "question": "How to control the grain size of materials in production", "answer": "1. Use inoculation treatment or rapid cooling to refine grains during casting; 2. Control rolling process parameters to refine the microstructure during hot working; 3. Utilize alloy elements to pin grain boundaries and inhibit grain growth during recrystallization annealing; 4. Coordinate the relationship between cold deformation amount and heating temperature according to the recrystallization diagram." }, { "idx": 683, "question": "All symmetry elements present in macroscopic crystals must pass through the center of the crystal", "answer": "√" }, { "idx": 684, "question": "The symmetry characteristic of the cubic crystal system is having four 3-fold axes", "answer": "√" }, { "idx": 685, "question": "The core of microscopic symmetry elements is the translation axis", "answer": "√" }, { "idx": 686, "question": "Amorphous silicon prepared by CVD method also possesses the four general characteristics of glass", "answer": "×" }, { "idx": 687, "question": "Summarize the applications of diffusion in materials science", "answer": "The roles of atomic diffusion in materials include: nucleation and growth during crystal solidification; component undercooling in alloys; composition homogenization, retention of high-temperature microstructure characteristics during peritectic reaction non-equilibrium solidification, nucleation during solid-state phase transformation, grain boundary nucleation, grain boundary movement, grain boundary segregation, high-temperature creep, oxidation, welding, chemical heat treatment (such as carburization, nitriding, etc.), powder metallurgy, coating, and various other aspects." }, { "idx": 688, "question": "Order-disorder transformation refers to the transformation between crystals and non-crystals", "answer": "×" }, { "idx": 689, "question": "A solid solution is an amorphous solid that has dissolved impurity components", "answer": "×" }, { "idx": 690, "question": "Most solid-phase reactions are controlled by diffusion rates", "answer": "×" }, { "idx": 691, "question": "At low temperatures, the diffusion that generally occurs in crystals is intrinsic diffusion", "answer": "√" }, { "idx": 692, "question": "In the wetting of solid-liquid interfaces, increasing the roughness of the solid surface necessarily improves wetting.", "answer": "×" }, { "idx": 693, "question": "Summarize the role of dislocations in materials", "answer": "Dislocations can greatly influence the performance of materials, and their roles in materials are diverse: (1) The plastic deformation of metallic materials is accomplished through dislocation motion. (2) Dislocations have a distorted stress field around them, which can strengthen materials through mechanisms such as cutting or bypassing second-phase particles. Increasing dislocation density during cold working can also strengthen materials, as can the formation of Cottrell atmospheres. Additionally, mutual intersections during dislocation motion or the formation of jogs and Lomer-Cottrell locks can strengthen materials, thereby affecting their strength. (3) Dislocations influence the precipitation of second phases and have an impact on solid-state phase transformations such as nucleation mechanisms during recrystallization. (4) The areas around dislocations serve as preferential diffusion pathways." }, { "idx": 694, "question": "Sintering is a process that involves various physical and chemical changes", "answer": "×" }, { "idx": 695, "question": "The collection of all symmetry elements in a crystal structure is called", "answer": "The crystal structure with space group $\\bf{F m}3m$ belongs to the (2) crystal family and (3) crystal system" }, { "idx": 696, "question": "In the cubic crystal system, the possible types of space lattices are (a) P, I, C (b) P, I, F (c) P, C, F (d) F, I, C", "answer": "The unit cell is (1)" }, { "idx": 697, "question": "Ionic crystals typically reduce their surface energy through the polarization deformation and rearrangement of surface ions. Among the following ionic crystals, the one with the smallest surface energy is (a) CaF2 (b) PbF2 (c) PbI2 (d) BaSO4 (e) SrSO4", "answer": "The main reasons for clay charging are: (11) _, (12) _, and (13) _" }, { "idx": 698, "question": "(a) Symmetry type (b) Point group (c) Collection of microscopic symmetry elements (d) Space group", "answer": "When a small amount of $\\\\mathbf{CaO}$ is added to the $\\\\mathrm{Th}\\\\mathbf{O}_{2}$ lattice to form a solid solution, write the possible defect reaction equations and solid solution formulas: defect reaction equation (7) and corresponding solid solution formula (8). Defect reaction equation (9) and corresponding solid solution formula (10)" }, { "idx": 699, "question": "In the face-centered cubic crystal structure, the close-packed plane is (a) {001} plane (b) {011} plane (c) {111} plane", "answer": "The main reasons for clay charging are: (11) (12) and (13)" }, { "idx": 700, "question": "The four structural parameters $z$, $R$, $X$, and $Y$ of the $\\mathrm{Na}{2}\\mathrm{{\\bfO}}\\cdot\\mathrm{Ca}\\mathrm{{\\bfO}}\\cdot\\mathrm{Al}{2}\\mathrm{{\\bfO}}{3}\\cdot2\\mathrm{SiO}{2}$ glass", "answer": "$,1.{\\surd};2.{\\surd};3.{\\searrow};4.\\times;5.\\times;6.\\times;7.{\\surd};8.\\times;9.\\times;10.\\times_{\\mathrm{~o~}}$ II. 1. b; 2. d; 3. a; 4. a; 5. c; 6. c; 7. (A) c、(B) a; 8. a、c; 9." }, { "idx": 701, "question": "In the thermodynamic relation of diffusion coefficient, $\\\\left(1+\\\\frac{\\\\partial{\\\\ln{\\\\gamma_{i}}}}{\\\\partial{\\\\ln{N_{i}}}}\\\\right)$ is called the thermodynamic factor of diffusion coefficient. In non-ideal mixing systems: when the thermodynamic factor of diffusion coefficient $>0$, the diffusion result causes the solute to (A); when the thermodynamic factor of diffusion coefficient $<0$, the diffusion result causes the solute to (B). (a) segregation occurs (b) concentration remains unchanged (c) concentration tends to be uniform", "answer": "On the curve of glass properties changing with temperature, there are two characteristic temperatures (18) and (19), and the viscosities corresponding to these two characteristic temperatures are (20) and (21) respectively." }, { "idx": 702, "question": "Network former", "answer": "A substance with single bond energy $\\geq335\\mathrm{kJ/mol}$ that can form glass alone." }, { "idx": 703, "question": "In the phase transformation of quartz, the one that belongs to reconstructive transformation is (a) $\\\\alpha\\\\cdot$ quartz $=====\\\\alpha\\\\cdot$ tridymite (b) $\\\\alpha\\\\cdot$ quartz $\\\\mathbf{\\\\mu}=====\\\\beta\\\\mathbf{\\\\sigma}$ quartz (c) $\\\\alpha\\\\cdot$ tridymite $=====\\\\alpha\\\\cdot$ quartz (d) $\\\\alpha\\\\cdot$-cristobalite $\\\\begin{array}{r}{======\\\\beta.}\\\\end{array}$ cristobalite", "answer": "Intrinsic diffusion is the migration of particles caused by (22), and the activation energy of intrinsic diffusion consists of two parts: (23) and (24). The relationship between the diffusion coefficient and temperature is given by: (25)." }, { "idx": 704, "question": "First-order phase transition", "answer": "During the phase transition, the chemical potentials of the two phases are equal, but the first-order partial derivatives of the chemical potential are not equal. A first-order phase transition involves latent heat and changes in volume." }, { "idx": 705, "question": "In the sintering process, the mass transfer method that only changes the pore shape without causing shrinkage of the green body is (a) surface diffusion (b) flow mass transfer (c) evaporation-condensation (d) grain boundary diffusion", "answer": "Martensitic transformation has the following characteristics: (26), (27), (28), and (29)_, etc. IV. Term Explanation (15 points) (Choose five questions; if all are answered, only the first five will be graded) 1. Network former 2. First-order phase transition 3. Schottky defect 4. Polymorphism 5. Wetting 6. Homogeneous nucleation 7. Non-stoichiometric structural defects 8. Grain growth" }, { "idx": 706, "question": "Polymorphism", "answer": "Substances with the same chemical composition can form crystals with different structures under different thermodynamic conditions" }, { "idx": 707, "question": "What are the characteristics of pore changes during secondary recrystallization", "answer": "During secondary recrystallization, pores are encapsulated within the grains" }, { "idx": 708, "question": "Characteristics of grain growth", "answer": "Grain growth is the process in which the average grain size of a strain-free material continuously increases during heat treatment without altering its distribution. Within the green body, grain sizes grow uniformly; during grain growth, pores remain at grain boundaries or grain boundary junctions." }, { "idx": 709, "question": "Grain growth", "answer": "is the process in which the average grain size of a strain-free material continuously increases during heat treatment without altering its distribution" }, { "idx": 710, "question": "Schottky defect", "answer": "When the lattice undergoes thermal vibration, some atoms with sufficient energy leave their equilibrium positions, jump to the surface of the crystal, and leave vacancies at the original normal lattice points." }, { "idx": 711, "question": "Wetting", "answer": "When the Gibbs free energy of the system decreases after a solid comes into contact with a liquid, it is called wetting." }, { "idx": 712, "question": "What are the process characteristics of secondary recrystallization", "answer": "It is an abnormal growth process where a few large grains grow at the expense of fine grains. It is the abnormal growth of individual grains." }, { "idx": 713, "question": "Non-stoichiometric structural defects", "answer": "The composition of certain compounds deviates from their stoichiometric ratio depending on the nature of the surrounding atmosphere and the magnitude of pressure. Such compounds are called non-stoichiometric compounds. The defects arising from the deviation in composition from the stoichiometric ratio are called non-stoichiometric structural defects." }, { "idx": 714, "question": "There are countless translational axes in the crystal structure", "answer": "√" }, { "idx": 715, "question": "Pauling's rules apply to all crystal structures", "answer": "×" }, { "idx": 716, "question": "The ratio of the intercepts of a crystal plane on the three coordinate axes must be a simple integer ratio", "answer": "√" }, { "idx": 717, "question": "The two most closely packed arrangements of equal-sized spheres", "answer": "There are mainly cubic closest packing and hexagonal closest packing; the cubic closest packing arrangement has the close-packed planes arranged in the sequence ABCABC..., while the hexagonal closest packing arrangement has the close-packed planes arranged in the sequence ABAB..." }, { "idx": 718, "question": "Types and quantitative relationship of voids", "answer": "When equal-sized spheres are closely packed, the main types of voids are octahedral voids and tetrahedral voids. When $\\pmb{n}$ spheres are closely packed, there will be $\\textbf{\\em n}$ octahedral voids and $2n$ tetrahedral voids" }, { "idx": 719, "question": "The dislocation line of an edge dislocation is parallel to the slip direction.", "answer": "×" }, { "idx": 720, "question": "Generally speaking, grain boundaries are the main diffusion channels for pores to reach the exterior of the sintered body.", "answer": "√" }, { "idx": 721, "question": "What is the relationship between secondary recrystallization and the particle size of the raw material", "answer": "Secondary recrystallization is also related to the particle size of the raw material" }, { "idx": 722, "question": "Feldspar is a silicate crystal with a layered structure", "answer": "×" }, { "idx": 723, "question": "For some materials, no matter how fast the cooling rate is, it is impossible to form a glass", "answer": "×" }, { "idx": 724, "question": "The flow of clay slurry belongs to plastic flow", "answer": "√" }, { "idx": 725, "question": "The Jander equation has a wider applicable range than the Ginstling equation", "answer": "×" }, { "idx": 726, "question": "Non-stoichiometric structural defects are a type of impurity defect caused by doping", "answer": "×" }, { "idx": 727, "question": "At low temperatures, the diffusion that generally occurs in crystals is extrinsic diffusion", "answer": "√" }, { "idx": 728, "question": "Most solid-phase reactions are controlled by the diffusion rate.", "answer": "×" }, { "idx": 729, "question": "During the cooling process of the melt, the greater the degree of undercooling, the greater the overall crystallization rate", "answer": "√" }, { "idx": 730, "question": "A space lattice is formed by arranging in space with a regular repetition. (a) atoms (b) ions (c) geometric points (d) molecules", "answer": "c" }, { "idx": 731, "question": "In a narrow sense, the phase transition process is a physical change process", "answer": "√" }, { "idx": 732, "question": "In the cubic crystal system, the possible types of space lattices are (a) P, I, C (b) P, I, F (c) P, C, F (d) F, I, C", "answer": "b" }, { "idx": 733, "question": "The crystal structure with point group $\\\\bf F m3m$ belongs to the crystal system. (a) cubic (b) hexagonal (c) tetragonal (d) orthorhombic", "answer": "a" }, { "idx": 734, "question": "In the $\\\\mathbf{ABO}_{3}$ (perovskite) type structure, B ions occupy (a) tetrahedral voids (b) octahedral voids (c) cubic voids (d) trigonal prismatic voids", "answer": "b" }, { "idx": 735, "question": "Montmorillonite belongs to which type of silicate structure. (a) Island silicate structure (b) Layered silicate structure (c) Chain silicate structure (d) Framework silicate structure", "answer": "b" }, { "idx": 736, "question": "What property must the medium exhibit for slurry peptization? (a) Acidic (b) Alkaline (c) Neutral", "answer": "b" }, { "idx": 737, "question": "In the non-stoichiometric compound $\\\\operatorname{Cd}_{1+x}0$, the lattice defect present is (a) anion vacancy (b) cation vacancy (c) anion interstitial (d) cation interstitial", "answer": "d" }, { "idx": 738, "question": "Which valence state of cations must be present in the exchange of originally adsorbed cations in clay for mud peptization? (d) Monovalent (e) Divalent (f) Trivalent", "answer": "d" }, { "idx": 739, "question": "The essence of crystals", "answer": "is the periodic repetition of particles in three-dimensional space" }, { "idx": 740, "question": "What are the two types of arrangements in the closest packing of equal-sized spheres?", "answer": "(2) Cubic close packing; (3) Hexagonal close packing" }, { "idx": 741, "question": "In the closest packing of equal-sized spheres, what is the packing arrangement of cubic close packing?", "answer": "(4) ABCABC……" }, { "idx": 742, "question": "Ionic crystals usually reduce their surface energy through the polarization deformation and rearrangement of surface ions. Among the following ionic crystals, the one with the smallest surface energy is (a) $\\\\mathrm{CaF}{2}$ (b) $\\\\mathrm{Pb}\\\\mathrm{F}{2}$ c) $\\\\mathrm{Pb}\\\\mathbf{I}{2}$ (d) $\\\\mathbf{BaSO{4}}$ (e) SrsO4", "answer": "c" }, { "idx": 743, "question": "In the closest packing of equal large spheres, what is the packing arrangement of hexagonal close packing?", "answer": "(5) ABAB……" }, { "idx": 744, "question": "The basis for classifying silicate crystals is (6)", "answer": "[ $\\mathrm{Si}0{4}$ connection method" }, { "idx": 745, "question": "In the face-centered cubic crystal structure, the close-packed plane is (7)", "answer": "{111} plane" }, { "idx": 746, "question": "After the interaction between water and clay, what can be formed around the clay colloidal particles as the distance increases: (8) what?", "answer": "Firmly bound water" }, { "idx": 747, "question": "What is one of the main factors that determines the concentration of component defects?", "answer": "Doping amount" }, { "idx": 748, "question": "What are the two main factors that determine the concentration of component defects?", "answer": "Solid solubility" }, { "idx": 749, "question": "On the curve of glass properties changing with temperature, what is the first characteristic temperature?", "answer": "Tg" }, { "idx": 750, "question": "In the ${\\\\bf R O}\\\\mathrm{-SiO}{2}$ system, the order of the size of the immiscibility regions for: (1) $\\\\mathbf{Mg0-SiO}{2}$; (2) $\\\\mathbf{CaO}\\\\mathbf{-SiO}{2}$; (3) $\\\\mathbf{SrO}\\\\mathbf{-SiO}{2}$; (4) BaO$\\\\mathrm{SiO}_{2}$ is (a)$\\\\left(1\\\\right)>\\\\left(2\\\\right)>\\\\left(3\\\\right)>\\\\left(4\\\\right)$(b)(4)>(3)>(2)>(1)(c)$\\\\left(2\\\\right)>\\\\left(1\\\\right)>\\\\left(3\\\\right)>\\\\left(4\\\\right)$ (d $)(3)>(4)>(2)>(1)$", "answer": "a" }, { "idx": 751, "question": "On the curve of glass properties changing with temperature, what is the second characteristic temperature?", "answer": "Tr" }, { "idx": 752, "question": "As a result of what reason, 'compositional defects' are inevitably produced in the crystal structure?", "answer": "Doping of non-equivalent ions" }, { "idx": 753, "question": "Methods to prevent secondary recrystallization", "answer": "Methods to prevent secondary recrystallization: control sintering temperature, sintering time, control the uniformity of raw material particle size, and introduce sintering additives." }, { "idx": 754, "question": "After the interaction between water and clay, what can be found around the clay colloidal particles as the distance increases: (10) what?", "answer": "Free water" }, { "idx": 755, "question": "After water interacts with clay, what can be found around the clay colloidal particles as the distance increases: (9) what?", "answer": "Loosely bound water" }, { "idx": 756, "question": "Briefly describe the characteristics of grain growth", "answer": "Characteristics of grain growth: Grain growth is a process during heat treatment of a strain-free material where the average grain size continuously increases without altering its distribution. Within the bulk, grain sizes grow uniformly; during grain growth, pores remain at grain boundaries or their junctions." }, { "idx": 757, "question": "Causes of secondary recrystallization", "answer": "Causes of secondary recrystallization: uneven particle size of raw materials, excessively high sintering temperature, and too fast sintering rate." }, { "idx": 758, "question": "Briefly describe the characteristics of secondary recrystallization", "answer": "Characteristics of secondary recrystallization: it is an abnormal growth process where a few large grains grow at the expense of fine grains. It involves the abnormal growth of individual grains; during secondary recrystallization, pores are enclosed within the grains; secondary recrystallization is also related to the particle size of the raw material." }, { "idx": 759, "question": "What is intrinsic diffusion?", "answer": "Intrinsic diffusion: refers to the migration phenomenon caused by vacancies originating from the intrinsic thermal defects of the crystal. The activation energy of intrinsic diffusion consists of two parts: the vacancy formation energy and the particle migration energy. At high temperatures, intrinsic diffusion predominates." }, { "idx": 760, "question": "In the orthorhombic crystal system, the (001) plane must be perpendicular to the (110) plane", "answer": "√" }, { "idx": 761, "question": "What is extrinsic diffusion?", "answer": "Extrinsic diffusion: It is a migration phenomenon caused by vacancies generated from the doping of inequivalent impurity ions. The activation energy of extrinsic diffusion only includes the migration energy of mass points, and extrinsic diffusion dominates at low temperatures." }, { "idx": 762, "question": "What are the characteristics of extrinsic diffusion?", "answer": "The activation energy of extrinsic diffusion only includes the energy of particle migration, and extrinsic diffusion dominates at low temperatures." }, { "idx": 763, "question": "In macroscopic crystals, the symmetry elements present must all pass through the center of the crystal", "answer": "√" }, { "idx": 764, "question": "What are the similarities between solid-phase sintering and liquid-phase sintering?", "answer": "The driving force for sintering is surface energy in both cases, and the sintering process consists of stages such as particle rearrangement, pore filling, and grain growth." }, { "idx": 765, "question": "What are the characteristics of intrinsic diffusion?", "answer": "The activation energy of intrinsic diffusion consists of two parts: vacancy formation energy and particle migration energy, and intrinsic diffusion dominates at high temperatures." }, { "idx": 766, "question": "Diamond, graphite, and carbon nanotubes are allotropes of carbon", "answer": "√" }, { "idx": 767, "question": "What are the differences between solid-phase sintering and liquid-phase sintering?", "answer": "Due to the faster mass transfer rate by flow compared to diffusion, liquid-phase sintering has a higher densification rate and requires a lower sintering temperature. Additionally, the rate of the liquid-phase sintering process is also related to factors such as the amount of liquid phase, its properties (viscosity, surface tension, etc.), the wetting condition between the liquid and solid phases, and the solubility of the solid phase in the liquid phase. The factors influencing liquid-phase sintering are more complex than those in solid-phase sintering." }, { "idx": 768, "question": "The symmetry characteristic of the cubic crystal system is having four 3-fold axes", "answer": "√" }, { "idx": 769, "question": "Amorphous silicon prepared by CVD method also has the four general characteristics of glass", "answer": "×" }, { "idx": 770, "question": "A solid solution is an amorphous solid that has dissolved impurity components", "answer": "×" }, { "idx": 771, "question": "As long as the temperature is above 0K, point defects always exist in crystals", "answer": "√" }, { "idx": 772, "question": "Fick's first law is applicable to solving unstable diffusion problems", "answer": "×" }, { "idx": 773, "question": "In the cubic crystal system, the possible types of space lattices are F, I, and C", "answer": "×" }, { "idx": 774, "question": "Order-disorder transformation refers to the transformation between crystals and non-crystals", "answer": "×" }, { "idx": 775, "question": "The critical radius of the nucleus $r_{k}$ decreases with increasing $\\Delta T$, making the phase transition easier to proceed.", "answer": "√" }, { "idx": 776, "question": "In the wetting of solid-liquid interfaces, increasing the roughness of the solid surface necessarily improves wetting.", "answer": "×" }, { "idx": 777, "question": "Briefly describe the conditions for forming continuous substitutional solid solutions", "answer": "The conditions for forming continuous substitutional solid solutions are: ionic size factor, the radius difference between the two substituting ions (r1-r2)/r1<15%; the two components must have exactly the same crystal structure type; continuous substitutional solid solutions can only form when the ionic valences are the same or when the total valence of composite substituting ions is equal; similar electronegativity and polarization properties" }, { "idx": 778, "question": "During the growth of crystals, there must be the presence of undercooling", "answer": "×" }, { "idx": 779, "question": "For the same system, the nucleation barrier for heterogeneous nucleation ≤ the nucleation barrier for homogeneous nucleation.", "answer": "√" }, { "idx": 780, "question": "Liquid phase sintering refers to a sintering process in which a liquid phase is present.", "answer": "×" }, { "idx": 781, "question": "1. Network former", "answer": "Network former: substances with single bond energy ≥335kJ/mol that can form glass independently." }, { "idx": 782, "question": "2. First-order phase transition", "answer": "First-order phase transition: During the phase transition, the chemical potentials of the two phases are equal, but the first-order partial derivatives of the chemical potential are not equal. A first-order phase transition involves latent heat and volume changes." }, { "idx": 783, "question": "3. Schottky defect", "answer": "Schottky defect: When the lattice undergoes thermal vibration, some atoms with sufficient energy leave their equilibrium positions, migrate to the surface of the crystal, and leave vacancies at the original normal lattice sites." }, { "idx": 784, "question": "6.Polymorphism", "answer": "Polymorphism: Substances with the same chemical composition can form crystals with different structures under different thermodynamic conditions." }, { "idx": 785, "question": "4.Secondary recrystallization", "answer": "Secondary recrystallization: is an abnormal growth process where a few large grains grow at the expense of fine grains." }, { "idx": 786, "question": "5. Inverse spinel structure", "answer": "Inverse spinel structure: It belongs to the cubic crystal system, where oxygen ions can be considered as arranged in a cubic close packing. Divalent cation A fills the octahedral voids, while trivalent cation B fills half of the octahedral voids and half of the tetrahedral voids." }, { "idx": 787, "question": "7. Lattice points", "answer": "Space lattice: A series of equivalent points identified in the crystal structure must be arranged in a periodic repetition in three-dimensional space. A series of geometric points arranged in a periodic repetition in three-dimensional space is called a space lattice." }, { "idx": 788, "question": "8.Sintering", "answer": "Sintering: A process in which one or more solid powders are formed and then heated to a certain temperature, causing them to shrink and become a dense, hard sintered body below the melting point temperature. Alternatively: Due to the mutual attraction of molecules (or atoms) in solids, heating causes the powder particles to bond, and through material migration and diffusion, the powder gains strength, leading to densification and recrystallization—this process is called sintering." }, { "idx": 789, "question": "4. Briefly describe the characteristics of diffusionless phase transformation.", "answer": "Characteristics of diffusionless phase transformation: shape change caused by uniform shear (change in crystal morphology); the new phase has the same chemical composition as the parent phase; it can occur at low temperatures with a fast transformation rate; there is a certain orientation relationship between the new phase and the parent phase." }, { "idx": 790, "question": "In a simple cubic crystal, if the direction of the dislocation line is [001] and $b=a$ [110], determine what type of dislocation this belongs to.", "answer": "nan" }, { "idx": 791, "question": "2. The concentration of structural defects in non-stoichiometric compounds is related to the nature and pressure of the surrounding atmosphere. Analyze how the density of the non-stoichiometric compounds Fe1-xO and Zn1+xO will change if the partial pressure of the surrounding oxygen atmosphere is increased.", "answer": "If the partial pressure of the surrounding oxygen atmosphere is increased, the value of x in the non-stoichiometric compound Fe1-xO increases, leading to an increase in the concentration of cation vacancies, which results in a decrease in the density of Fe1-xO. Similarly, if the partial pressure of the surrounding oxygen atmosphere is increased, the value of x in the non-stoichiometric compound Zn1+xO decreases, leading to a reduction in the concentration of interstitial cations, which also results in a decrease in the density of Zn1+xO." }, { "idx": 792, "question": "Explain what type of solid solution compound austenite in carbon steel belongs to", "answer": "Austenite is an interstitial solid solution of carbon in γ-Fe" }, { "idx": 793, "question": "Explain what type of solid solution compound Fe3C belongs to", "answer": "Fe3C is an interstitial compound, belonging to the interstitial phase with a complex structure" }, { "idx": 794, "question": "1. Briefly describe the characteristics of grain growth and secondary recrystallization, as well as the causes of secondary recrystallization and methods to prevent it.", "answer": "Characteristics of grain growth: Grain growth is a process in which the average grain size of a strain-free material continuously increases during heat treatment without changing its distribution. The grain size grows uniformly within the body; during grain growth, pores remain at grain boundaries or grain boundary intersections. Characteristics of secondary recrystallization: It is an abnormal growth process where a few large grains grow at the expense of fine grains. It involves the abnormal growth of individual grains; during secondary recrystallization, pores are trapped inside the grains; secondary recrystallization is also related to the particle size of the raw material. Causes of secondary recrystallization: Non-uniform particle size of the raw material, excessively high sintering temperature, and too fast sintering rate. Methods to prevent secondary recrystallization: Control sintering temperature and time, ensure uniformity of raw material particle size, and introduce sintering additives." }, { "idx": 795, "question": "Explain what type of solid solution compound ME2Si belongs to", "answer": "ME2Si is an intermetallic compound" }, { "idx": 796, "question": "Explain what type of solid solution compound Cu3Sn belongs to", "answer": "Cu3Sn is an electron compound, an intermetallic compound with a specific electron concentration" }, { "idx": 797, "question": "What factors influence the formation of amorphous metals? Why", "answer": "Viscosity of liquid metal: the higher the viscosity, the more difficult the atomic diffusion, making it easier to retain the liquid metal structure." }, { "idx": 798, "question": "In a simple cubic crystal, if the direction of the dislocation line is [112] and the Burgers vector is $b=a$ [110], determine the type of this dislocation.", "answer": "Since the direction of the dislocation line is perpendicular to the direction of the Burgers vector, this dislocation is an edge dislocation." }, { "idx": 799, "question": "3. Briefly describe the main types and characteristics of solid-phase sintering and liquid-phase sintering, as well as the similarities and differences between solid-phase sintering and liquid-phase sintering.", "answer": "Main types and characteristics of solid-phase sintering: Evaporation-condensation mass transfer, caused by the vapor pressure difference in different parts, does not result in shrinkage of the green body during the sintering process. Diffusion mass transfer, caused by the vacancy concentration difference in different parts of the particles, is the main method of mass transfer for most solid materials during sintering. Main types and characteristics of liquid-phase sintering: Flow mass transfer, which is the main method of mass transfer for most silicate materials during sintering, can be divided into viscous flow and plastic flow. Dissolution-precipitation mass transfer, where both solid and liquid phases exist, and the solid phase is soluble in the liquid phase. Similarities between solid-phase sintering and liquid-phase sintering: The driving force for sintering is surface energy, and the sintering process consists of stages such as particle rearrangement, pore filling, and grain growth. Differences: Due to the faster rate of flow mass transfer compared to diffusion, liquid-phase sintering has a higher densification rate and lower sintering temperature. Additionally, the rate of the liquid-phase sintering process is also related to factors such as the amount and properties of the liquid phase (viscosity, surface tension, etc.), the wetting condition between the liquid and solid phases, and the solubility of the solid phase in the liquid phase. The factors influencing liquid-phase sintering are more complex than those for solid-phase sintering." }, { "idx": 800, "question": "Please briefly describe the similarities and differences between interstitial solid solutions, interstitial phases, and interstitial compounds?", "answer": "Similarity: small atoms dissolve. Differences: interstitial solid solutions retain the solvent (large atoms) lattice; interstitial phases and interstitial compounds alter the large atoms lattice, forming a new lattice. Interstitial phases have simple structures; interstitial compounds have complex structures." }, { "idx": 801, "question": "What are the characteristics of various alloy strengthening methods?", "answer": "Grain refinement strengthening, solid solution strengthening, multiphase strengthening, dispersion strengthening (age hardening), and work hardening." }, { "idx": 802, "question": "What is the physical significance of the critical nucleus?", "answer": "The physical significance of the critical nucleus: the smallest embryo that can grow spontaneously (or, a nucleus with a radius equal to rk)." }, { "idx": 803, "question": "What are the methods of alloy strengthening?", "answer": "Grain refinement strengthening, solid solution strengthening, multiphase strengthening, dispersion strengthening (age hardening), and work hardening." }, { "idx": 804, "question": "If it is an edge dislocation, determine the crystallographic plane indices of the half-atom plane and the crystallographic direction indices of the insertion direction.", "answer": "The crystallographic plane indices of the half-atom plane are (110)." }, { "idx": 805, "question": "What are the sufficient conditions for the formation of a critical nucleus?", "answer": "The sufficient conditions for the formation of a critical nucleus: (1) Formation of an embryo with r≥rk; (2) Acquisition of nucleation work with A≥A* (critical nucleation work)." }, { "idx": 806, "question": "Please briefly describe the main factors affecting diffusion", "answer": "The main factors affecting diffusion: (1) temperature; (2) crystal structure and type; (3) crystal defects; (4) chemical composition." }, { "idx": 807, "question": "Briefly describe the growth mechanism of pure metal crystals", "answer": "The crystal growth mechanism refers to the microscopic growth mode of crystals, which is related to the liquid-solid interface structure. For substances with rough interfaces, since about $50\\\\%$ of the atomic positions on the interface are vacant, these vacancies can accept atoms. Therefore, liquid atoms can individually enter the vacancies and connect with the crystal, causing the interface to advance perpendicularly along its normal direction, resulting in continuous growth. For crystals with smooth interfaces, growth does not occur through the attachment of individual atoms but rather through homogeneous nucleation, forming a two-dimensional nucleus one atomic layer thick on the crystallographic facet interface, creating a step between the new nucleus and the original interface. Individual atoms can then fill in the step, allowing the two-dimensional nucleus to grow laterally. Once the layer is filled, a new two-dimensional nucleus forms on the new interface, and the process repeats. If the smooth interface of the crystal has an exposed screw dislocation, the interface becomes a spiral surface, forming a step that never disappears. Atoms attach to the step, enabling crystal growth." }, { "idx": 808, "question": "What are the limitations of strain hardening as a strengthening method?", "answer": "Strain hardening does not fundamentally change the properties of the alloy. Under certain conditions, such as temperature increase, strain hardening may be lost due to the occurrence of recrystallization." }, { "idx": 809, "question": "According to the position of solute atoms in the lattice, into how many categories can solid solution phases be divided?", "answer": "Based on the position of solute atoms in the lattice, solid solutions can be divided into substitutional solid solutions, such as the α phase in Al-Cu alloys; interstitial solid solutions, such as the α phase in Fe-C alloys." }, { "idx": 810, "question": "What is the significance of strain hardening in production?", "answer": "Strain hardening, also known as work hardening, can improve the strength and hardness of alloys. For example, cold-drawn steel wires are strengthened by utilizing the strain hardening effect. Especially for some alloys that cannot be strengthened by heat treatment, strain hardening is a very important strengthening method." }, { "idx": 811, "question": "During non-equilibrium solidification of solid solution alloys, sometimes microsegregation forms, and sometimes macrosegregation forms. What is the reason for this?", "answer": "Microsegregation occurs within the range of a single grain and is caused by the slow diffusion rate during crystallization. Macrosegregation occurs throughout the entire part or sample and is caused by the redistribution of solute atoms during crystallization." }, { "idx": 812, "question": "What are the necessary conditions for an alloy to undergo precipitation hardening?", "answer": "The necessary condition for precipitation hardening to occur is the availability of a supersaturated solid solution to facilitate the precipitation of a second phase." }, { "idx": 813, "question": "For a face-centered cubic crystal with a movable slip system of (111)[110], if the slip is caused by an edge dislocation, indicate the direction of the dislocation line.", "answer": "Direction of the dislocation line: [112]" }, { "idx": 814, "question": "For a face-centered cubic crystal with a movable slip system of (111)[110], please indicate the Burgers vector of the unit dislocation causing the slip.", "answer": "Burgers vector; b=a/2 [110]" }, { "idx": 815, "question": "For a face-centered cubic crystal with a movable slip system of (111)[110], if the slip is caused by an edge dislocation, indicate the direction of motion of the dislocation line.", "answer": "The direction of motion of the dislocation line is parallel to the Burgers vector. F=τb=9.899×10^-11MN/m" }, { "idx": 816, "question": "Explain the meaning of the rolling texture {110}<112> in face-centered cubic alloy α-brass.", "answer": "It is a sheet texture. The {110}<112> texture indicates that the {110} plane is parallel to the rolling plane, and the <112> direction is parallel to the rolling direction." }, { "idx": 817, "question": "How many equivalent slip systems can be simultaneously activated when a face-centered cubic metal single crystal is stretched along [111]? And specifically write the indices of each slip system.", "answer": "When an FCC-structured crystal is stretched along the [111] direction, there are 6 equivalent slip systems, which are: (111)[011], (111)[110], (111)[011], (111)[101], (111)[101], (111)[110]." }, { "idx": 818, "question": "How many equivalent slip systems can be simultaneously activated when a face-centered cubic metal single crystal is stretched along the [001] direction? Also, specifically write the indices of each slip system.", "answer": "According to the 'mirror rule,' using the standard projection of the cubic crystal, it can be determined that when an FCC-structured crystal is stretched along the [001] axis, there are a total of 8 equivalent slip systems. These are: (111)[011], (111)[101], (111)[011], (111)[101], (111)[101], [111][011], (111)[101], (111)[011]." }, { "idx": 819, "question": "What is the Kirkendall effect in the process of alloy diffusion?", "answer": "Kirkendall effect: As shown in Figure 1-4, a very thin molybdenum wire is applied as a marker on an FCC-structured α-brass (Cu+30%Zn) rod, and then copper is plated on the brass, enclosing the molybdenum wire between the brass and copper. Diffusion is carried out at a certain temperature. The molybdenum wire serves only as a marker and does not participate in the diffusion throughout the experiment. The diffusing components are copper and zinc, which form a substitutional solid solution. In the diffusion process of the substitutional solid solution, the markers placed at the original interface move towards the direction of the low-melting-point element, and the displacement distance exhibits a parabolic relationship with time. The reason for this phenomenon is that the low-melting-point component diffuses faster, while the high-melting-point component diffuses slower. This unequal atomic exchange results in the Kirkendall effect." }, { "idx": 820, "question": "How many types of two-dimensional lattices are there? Indicate their types and illustrate with diagrams.", "answer": "There are 5 types of two-dimensional lattices, namely oblique, hexagonal, rectangular, centered rectangular, and square, as shown in Table 1-1." }, { "idx": 821, "question": "What are the main characteristics of Shockley partial dislocations in FCC crystals? Including the dislocation name, Burgers vector, dislocation nature, formation method, and motion state.", "answer": "Shockley partial dislocation; Burgers vector is 1 <112> 6; dislocation nature is screw-type, edge-type, mixed-type; formation method is can only be formed through local slip of the crystal; motion state is even edge-type partial dislocations can only slip and cannot climb. Even screw-type partial dislocations cannot cross-slip." }, { "idx": 822, "question": "What are the main characteristics of Frank partial dislocations in FCC crystals? Including the dislocation name, Burgers vector, dislocation nature, formation method, and motion state.", "answer": "Frank partial dislocation; Burgers vector is 1/3 <111>; the nature of the dislocation is edge-type; the formation method involves inserting or removing a layer of {111} close-packed plane in a local region of the crystal; the motion state is that it can only climb, not glide." }, { "idx": 823, "question": "What are the main characteristics of the Shockley partial dislocation in the extended dislocation of an FCC crystal? Include the dislocation name, Burgers vector, dislocation nature, formation method, and motion state.", "answer": "Shockley partial dislocation in the extended dislocation; Burgers vector is 1 [112] 6 [121] 6; dislocation nature is mixed-type; formation method is the decomposition of a perfect dislocation into two parallel Shockley partial dislocations of screw-type, edge-type, and mixed-type, with a stacking fault region in between; motion state is glide-only. Climb or cross-slip can only occur after constriction." }, { "idx": 824, "question": "What are the main characteristics of perfect dislocations in FCC crystals? Including dislocation name, Burgers vector, dislocation nature, dislocation formation mode, and motion state.", "answer": "Perfect dislocations (relatively rare in actual crystals); Burgers vector is 1 <110> 2; dislocation nature is screw-type, edge-type, mixed-type; dislocation formation mode is local slip or local displacement; motion state is can slip or climb." }, { "idx": 825, "question": "What are the main characteristics of the L-C sessile dislocation in FCC crystals? Including the dislocation name, Burgers vector, dislocation nature, formation method, and motion state.", "answer": "L-C sessile dislocation (with multiple configurations); Burgers vector is for example—[110] 6; dislocation nature is screw-type, edge-type, mixed-type; formation method is obtained through Shockley partial dislocation synthesis (or dislocation reaction); motion state is immobile." }, { "idx": 826, "question": "Using the standard projection diagram of a cubic crystal, explain the formation reason of the rolling texture {110}<112> in face-centered cubic alloy α-brass.", "answer": "α-brass has an FCC structure, and its slip system is {111}<10\\overline{1}>. Under the action of tensile force along the rolling direction, the crystal slips and rotates. In the crystallographic coordinate system, if the tensile axis T1 is located in the 001-101-111 orientation triangle, the initial slip system is (111)[011], and the tensile axis turns toward the [011] direction, reducing the angle λ between the tensile axis and the slip direction. When the force axis reaches the common edge of the two orientation triangles, i.e., T2, double slip begins, and the slip system (111)[101] is also activated. The tensile axis turns both toward the [011] direction and the [101] direction, resulting in rotation along the common edge. When it reaches the [112] direction, since [101], [112], and [011] lie on the same great circle, the two λ angles simultaneously decrease to the minimum value, making [112] the final stable position. Thus, the <112> direction tends to align with the rolling direction. On the rolling plane, under the action of compressive force, if the compressive axis P1 is located in the 001-011-111 orientation triangle, the initial slip system is (111)[101], and the compressive axis turns toward the (111) plane, reducing the angle φ between the compressive axis and the slip plane. When the force axis reaches the common edge of the two orientation triangles, i.e., P2, double slip begins, and the slip system (111)[101] is also activated. The compressive axis turns both toward the (111) plane and the (111) plane, resulting in rotation along the common edge. When it reaches the (011) plane, since (111), (011), and (111) lie on the same great circle, the two φ angles simultaneously decrease to the minimum value, making (011) the final stable position. Thus, the (011) plane tends to become parallel to the rolling plane. As a result, the {110} plane becomes parallel to the rolling plane, and the <112> direction becomes parallel to the rolling direction." }, { "idx": 827, "question": "What insights do the Kirkendall experiment results provide?", "answer": "The practical significance of the Kirkendall effect includes: ① Revealing the intrinsic connection between macroscopic diffusion laws and microscopic diffusion mechanisms, which is universal; ② Directly refuting the exchange mechanism of substitutional solid solution diffusion and supporting the vacancy mechanism; ③ Each component in the diffusion system has its own diffusion coefficient; ④ The Kirkendall phenomenon often produces side effects, such as incomplete shrinkage leading to Kirkendall pores, etc. These side effects often have adverse impacts in practice and should therefore be controlled." }, { "idx": 828, "question": "Indicate the second main mechanism of alloy strengthening and explain its strengthening reason", "answer": "Precipitation strengthening and dispersion strengthening. The compounds of alloying elements and matrix elements obtained through phase transformation processes in alloys, as well as hard particles mechanically mixed into the matrix material, both cause alloy strengthening, referred to as precipitation strengthening and dispersion strengthening, respectively. The effects of precipitation strengthening and dispersion strengthening are much greater than solid solution strengthening. When dislocations encounter second phases during movement, they need to cut through (small-sized particles in precipitation strengthening and particles in dispersion strengthening) or bypass (large-sized particles in precipitation strengthening) the second phases. Therefore, the second phases (precipitates and dispersoids) hinder dislocation motion." }, { "idx": 829, "question": "Given that the recrystallization activation energy of a Cu-30%Zn alloy is 250 kJ/mol, and it takes 1 hour for this alloy to complete recrystallization at a constant temperature of 400°C, calculate how many hours it will take for this alloy to complete recrystallization at a constant temperature of 390°C.", "answer": "According to the formula, t2/t1 = exp[-Q/R(1/T1 - 1/T2)] = exp[-250×10^3/8.314×(1/(400+273) - 1/(390+273))] = 1.962. Therefore, t2 = t1×1.962 = 1.962 h." }, { "idx": 830, "question": "Indicate the first main mechanism of alloy strengthening and explain its strengthening reason", "answer": "Solid solution strengthening. The alloy element atoms dissolved in the lattice interstices or nodes, due to their size differing from the matrix atoms, generate a certain stress field that hinders dislocation movement; Cottrell atmospheres and Suzuki atmospheres, the former being interstitial atoms preferentially distributed in the tensile stress region of edge dislocations in BCC metals, producing a pinning effect on dislocations, the latter being alloy elements preferentially distributed in the stacking fault region of extended dislocations in FCC metals, reducing stacking fault energy, expanding the stacking fault region, making the slip of extended dislocations more difficult." }, { "idx": 831, "question": "What is the practical significance of the Kirkendall effect?", "answer": "The practical significance of the Kirkendall effect includes: ① It reveals the intrinsic connection between macroscopic diffusion laws and microscopic diffusion mechanisms, demonstrating universality; ② It directly refutes the exchange mechanism of substitutional solid solution diffusion and supports the vacancy mechanism; ③ Each component in a diffusion system has its own diffusion coefficient; ④ The Kirkendall phenomenon often produces side effects, such as incomplete shrinkage leading to Kirkendall voids, which can have adverse impacts in practice and should therefore be controlled." }, { "idx": 832, "question": "Indicate the third main mechanism of alloy strengthening and explain the reason for its strengthening", "answer": "Grain boundary strengthening. According to the Hall-Petch formula, the relationship between the yield point σs and the grain diameter d is σs=σ0+K d^(-1/2). The essence is that additional stress is required for dislocations to cross grain boundaries. Therefore, low-temperature steels often adopt a fine-grained structure." }, { "idx": 833, "question": "Explain the processes and characteristics of recovery, recrystallization, and grain growth when cold-deformed metal is heated.", "answer": "When cold-deformed metal is heated, three processes occur sequentially: recovery, recrystallization, and grain growth. Their respective characteristics are as follows:\\n\\n(1) Characteristics of the recovery process\\n\\n① The microstructure does not change during recovery, and the elongated grains of the deformed state are still retained.\\n\\n② The recovery process completely eliminates the macroscopic (Type I) stresses caused by deformation and largely eliminates the microscopic (Type II) stresses.\\n\\n③ During recovery, the mechanical properties generally change little, with hardness and strength only slightly decreasing, plasticity slightly increasing, and certain physical properties changing significantly, such as resistivity decreasing markedly and density increasing.\\n\\n④ The stored energy from deformation is partially released during the recovery stage.\\n\\n(2) Characteristics of the recrystallization process\\n\\n① The microstructure changes, transforming from the elongated grains of cold deformation into new equiaxed grains.\\n\\n② The mechanical properties change drastically, with strength and hardness sharply decreasing and plasticity increasing, returning to the state before deformation.\\n\\n③ The stored energy from deformation is fully released during recrystallization, lattice distortions (Type III stresses) are eliminated, and dislocation density decreases.\\n\\n(3) Characteristics of the grain growth process\\n\\n① The grains grow larger.\\n\\n② It causes changes in some properties, such as decreases in strength, plasticity, and toughness.\\n\\n③ Along with grain growth, other structural changes occur, such as recrystallization texture." }, { "idx": 834, "question": "Indicate the fourth main mechanism of alloy strengthening and explain the reason for its strengthening", "answer": "Order strengthening. The dislocations in ordered alloys are superdislocations. To cause plastic deformation in the metal, both partial dislocations of the superdislocation need to move simultaneously, thus requiring greater external stress. The bonding force between dissimilar element atoms is stronger than that between similar element atoms, so the ordered arrangement of dissimilar atoms gives ordered alloys higher strength." }, { "idx": 835, "question": "Explain the formation reasons of the surface fine grain zone in typical ingot structure", "answer": "The surface fine grain zone has numerous nucleation sites, fast cooling rate, and large undercooling, growing dendritically in various directions, thus forming fine, equiaxed crystals. Due to the rapid crystallization of the fine grain zone, the released latent heat of crystallization cannot dissipate in time, causing the temperature at the liquid-solid interface to rise sharply. This quickly halts the development of the fine grain zone, resulting in a very thin shell of fine grain zone." }, { "idx": 836, "question": "Explain the formation reason of the central equiaxed crystal zone in typical ingot structure", "answer": "The central equiaxed crystal zone forms due to the non-directional growth of crystal nuclei in the uniformly cooled central liquid. The crystal nuclei originate from two pathways: one is exogenous nuclei, including detached surface grains and broken dendrites; the other is endogenous nuclei, including homogeneous and heterogeneous nucleation in the undercooled liquid of the central region." }, { "idx": 837, "question": "Write the close-packed plane, interplanar spacing of close-packed planes, close-packed direction, and minimum unit length of the close-packed direction for an FCC crystal.", "answer": "
FCC111}√3 a 3<110>√2 2 a
" }, { "idx": 838, "question": "Write the close-packed plane, interplanar spacing of close-packed planes, close-packed direction, and minimum unit length of the close-packed direction for BCC crystals.", "answer": "
BCC110}√2 D 2<111>√3 2 a
" }, { "idx": 839, "question": "Write the indices of the crystal plane family with an interplanar spacing of $0.1246\\\\mathrm{nm}$ in nickel (Ni) crystal. The lattice constant of nickel is $0.3524\\\\mathrm{nm}$.", "answer": "The crystal structure and lattice of nickel (Ni) are face-centered cubic (FCC). The formula for the interplanar spacing in a cubic crystal is: $$d={\\\\frac{a}{\\\\sqrt{h^{2}+k^{2}+l^{2}}}}$$Thus, $$h^{2}+k^{2}+l^{2}=\\\\frac{a^{2}}{d^{2}}=\\\\frac{0.3524^{2}}{0.1246^{2}}\\\\approx8$$Since $\\\\smash{\\\\boldsymbol{h}_{3},\\\\boldsymbol{k}_{3},\\\\boldsymbol{l}}$ are all integers, the possible values for $\\\\smash{\\\\boldsymbol{h}_{3},\\\\boldsymbol{k}_{3},\\\\boldsymbol{l}}$ are: 0, 2, 2. Therefore, The indices of the crystal plane family that meet the requirement are {220}." }, { "idx": 840, "question": "Write the close-packed plane, interplanar spacing of close-packed planes, close-packed direction, and the minimum unit length of the close-packed direction for HCP (M(a)>√8/3) crystals.", "answer": "
HCP{0001}1 C 2<1120>D
" }, { "idx": 841, "question": "Indicate and explain the reverse transformation phenomenon in the characteristics of martensitic transformation", "answer": "When martensite is heated at a sufficiently rapid rate, it can directly transform back to the high-temperature phase without decomposition." }, { "idx": 842, "question": "Explain the formation reason of columnar crystal zone in typical ingot structure", "answer": "After the formation of the fine-grained zone, the mold wall temperature increases, heat dissipation slows down, the cooling rate of the liquid decreases, the undercooling reduces, and nucleation no longer occurs. The crystals with fast growth rates in the fine-grained zone can develop along the direction opposite to heat dissipation, which is perpendicular to the mold wall. Their lateral growth is hindered due to mutual interference, thus forming columnar crystals with well-developed primary axes." }, { "idx": 843, "question": "Indicate and explain the surface relief and shear coherency in the characteristics of martensitic transformation", "answer": "Martensitic transformation produces uniform shear or lattice shear, causing structural changes, resulting in surface relief phenomena on the specimen. The interface between martensite and the parent phase is a coherent interface." }, { "idx": 844, "question": "What are the dislocation name, Burgers vector, dislocation nature, formation method, and motion state of perfect dislocations in FCC crystals?", "answer": "Perfect dislocation (relatively rare in actual crystals); 1 <110> 2; screw type, edge type, mixed type; local slip or local displacement; can slip or climb" }, { "idx": 845, "question": "Point out and explain the strict orientation relationship between the new phase and the parent phase in the characteristics of martensitic transformation", "answer": "The main orientation relationships between martensite and the parent phase are: 1 K·S relationship; 2 G-T relationship; 3 Nishiyama relationship." }, { "idx": 846, "question": "Indicate and explain the internal substructure in the characteristics of martensitic transformation", "answer": "In addition to lattice shear, martensitic transformation also involves lattice-invariant shear, accomplished through slip or twinning, resulting in the formation of dislocation or twin substructures within the martensite." }, { "idx": 847, "question": "Write the reaction equation for the decomposition of a perfect dislocation into an extended dislocation in a face-centered cubic (FCC) crystal, and analyze the possibility of the reaction.", "answer": "Taking the (111) plane as an example, there are three reaction equations for the decomposition of a perfect dislocation into an extended dislocation on this plane: \\n(1) ${\\\\frac{1}{2}}[110]{-}{\\\\frac{1}{6}}[121]+{\\\\frac{1}{6}}[21\\\\overline{{1}}]$ Geometric condition: ${\\\\frac{1}{2}}[110]={\\\\frac{1}{6}}[121]+{\\\\frac{1}{2}}[21\\\\bar{1}]$ Energy condition: $\\\\left({\\\\frac{\\\\sqrt{2}}{2}}\\\\right)^{2}={\\\\frac{1}{2}}>\\\\biggl[\\\\left({\\\\frac{\\\\sqrt{6}}{6}}\\\\right)^{2}+\\\\left({\\\\frac{\\\\sqrt{6}}{6}}\\\\right)^{2}\\\\biggr]={\\\\frac{1}{3}}$ Therefore, it meets the geometric and energy conditions for dislocation reactions, and this dislocation reaction can proceed. \\n(2) $\\\\frac{1}{2}[011]{\\\\rightarrow}\\\\frac{1}{6}[121]+\\\\frac{1}{6}[\\\\overline{{{1}}}12]$ Geometric condition: ${\\\\frac{1}{2}}[011]={\\\\frac{1}{6}}[121]+{\\\\frac{1}{2}}[\\\\overline{{{1}}}12]$ Energy condition: $\\\\left({\\\\frac{\\\\sqrt{2}}{2}}\\\\right)^{2}={\\\\frac{1}{2}}>\\\\Big[\\\\left({\\\\frac{\\\\sqrt{6}}{6}}\\\\right)^{2}+\\\\left({\\\\frac{\\\\sqrt{6}}{6}}\\\\right)^{2}\\\\Big]={\\\\frac{1}{3}}$ Therefore, it meets the geometric and energy conditions for dislocation reactions, and this dislocation reaction can proceed. \\n(3) $\\\\frac{1}{2}[\\\\overline{{1}}01]{\\\\rightarrow}\\\\frac{1}{6}[\\\\overline{{1}}12]+\\\\frac{1}{6}[\\\\overline{{2}}\\\\overline{{1}}1]$ Geometric condition: $\\\\frac{1}{2}[\\\\overline{{1}}01]=\\\\frac{1}{6}[\\\\overline{{1}}12]+\\\\frac{1}{6}[\\\\overline{{2}}\\\\overline{{1}}1]$ Energy condition: $\\\\left({\\\\frac{\\\\sqrt{2}}{2}}\\\\right)^{2}={\\\\frac{1}{2}}>\\\\Big[\\\\left({\\\\frac{\\\\sqrt{6}}{6}}\\\\right)^{2}+\\\\left({\\\\frac{\\\\sqrt{6}}{6}}\\\\right)^{2}\\\\Big]={\\\\frac{1}{3}}$ \\nTherefore, it meets the geometric and energy conditions for dislocation reactions, and this dislocation reaction can proceed." }, { "idx": 848, "question": "Point out and explain the diffusionless characteristic of martensitic transformation", "answer": "The transformation process involves no compositional change, with all participating atoms moving in a coordinated manner, maintaining unchanged relative positions between neighboring atoms, and the relative displacement being less than one atomic spacing." }, { "idx": 849, "question": "What are the dislocation name, Burgers vector, dislocation nature, formation method, and motion state of the L-C sessile dislocation in FCC crystals?", "answer": "L-C sessile dislocation (with multiple configurations); for example—[110] 6; screw-type, edge-type, mixed-type; formed by the synthesis of Shockley partial dislocations (or dislocation reactions); cannot move" }, { "idx": 850, "question": "Indicate and explain the habit plane and its invariance in the characteristics of martensitic transformation", "answer": "Martensite forms on certain crystallographic planes of the parent phase, and these planes are called habit planes. The habit plane is an undistorted and non-rotating plane." }, { "idx": 851, "question": "What are the dislocation name, Burgers vector, dislocation nature, formation method, and motion state of the Shockley partial dislocation in FCC crystals?", "answer": "Shockley partial dislocation; 1 <112> 6; screw-type, edge-type, mixed-type; can only be formed through local slip of the crystal; even edge-type partial dislocations can only slip, not climb. Even screw-type partial dislocations cannot cross-slip." }, { "idx": 852, "question": "What are the dislocation name, Burgers vector, dislocation nature, formation method, and motion state of the Frank partial dislocation in FCC crystals?", "answer": "Frank partial dislocation; 1 <111> 3; edge type; formed by inserting or removing a layer of {111} close-packed plane in a local region of the crystal; can only climb, cannot glide" }, { "idx": 853, "question": "What are the dislocation name, Burgers vector, dislocation nature, formation method, and motion state of the Shockley partial dislocation in the extended dislocation of an FCC crystal?", "answer": "Shockley partial dislocation in the extended dislocation; 1 [112] 6 [121] 6; glissile; formed by the decomposition of a perfect dislocation into two parallel Shockley partial dislocations of screw, edge, or mixed type, with a stacking fault region in between; can only glide. Climb or cross-slip is possible only after constriction." }, { "idx": 854, "question": "When the vacancy equilibrium concentration in germanium crystals decreases by six orders of magnitude from $600\\\\%$ to $300\\\\mathrm{\\\\textperthousand}$, calculate the vacancy formation energy in germanium crystals (Boltzmann constant $k=8.617\\\\times10^{-5}\\\\mathrm{eV}/\\\\mathrm{K})$.", "answer": "The equilibrium concentration formula for Schottky defects is: $$\\\\begin{array}{r l}&{\\\\mathrm{atsing~sefiell}\\\\eta+\\\\mathrm{fight}\\\\mathcal{R}\\\\Sigma\\\\Delta t:}\\\\ &{\\\\qquad\\\\overline{{C}}_{\\\\mathrm{s}}\\\\leq e\\\\Bigg\\\\{-\\\\frac{\\\\Delta C_{s}}{H\\\\Gamma}\\\\Bigg\\\\}=\\\\mathrm{evel}\\\\Bigg(-\\\\frac{\\\\Delta C_{s}}{H\\\\Gamma}\\\\Bigg)=\\\\mathrm{evel}\\\\Bigg(-\\\\frac{\\\\Delta C_{s}}{H\\\\Gamma}\\\\Bigg)}\\\\ &{\\\\mathrm{atsin}}\\\\ &{\\\\frac{\\\\Delta C_{s}}{H\\\\mathcal{N}_{x}\\\\Gamma}=-\\\\mathrm{in}\\\\overline{{C}}_{s},}\\\\ &{\\\\frac{\\\\Delta C_{s}}{\\\\Delta W_{x}}\\\\Bigg(\\\\frac{1}{T_{2}}-\\\\frac{1}{T_{1}}\\\\Bigg)=\\\\mathrm{in}\\\\overline{{C}}_{\\\\mathrm{s}}-\\\\mathrm{in}\\\\overline{{C}}_{\\\\mathrm{s}}}\\\\ &{\\\\Delta G_{s}=\\\\frac{\\\\mathrm{in}\\\\overline{{C}}_{\\\\mathrm{s}}-\\\\mathrm{in}\\\\overline{{C}}_{\\\\mathrm{s}}}{\\\\frac{1}{T_{2}}-\\\\frac{1}{T_{1}}}(k\\\\chi_{x})}\\\\ &{\\\\qquad-\\\\frac{1}{\\\\frac{\\\\mathrm{fim}}{2}-\\\\frac{1}{T_{2}}}\\\\frac{\\\\kappa_{\\\\mathrm{B}}\\\\lambda_{1}\\\\Gamma\\\\kappa_{\\\\mathrm{m}}t^{-1}}{-\\\\frac{1}{\\\\omega_{0}+2/3}}}\\\\ &{\\\\qquad\\\\frac{1}{3(\\\\omega+2/3)}-\\\\frac{1}{6(\\\\omega+2/3)}}\\\\ &{\\\\qquad\\\\nu\\\\neq\\\\ell\\\\geq\\\\frac{1}{\\\\mathrm{in}\\\\mathcal{L}_{x}\\\\Bigg\\\\{1,2,3\\\\}\\\\Gamma\\\\kappa_{\\\\mathrm{m}}\\\\mathcal{R}_{x}\\\\Bigg\\\\}.}\\\\end{array}$$ That is, $10^{5}\\\\mathrm{J}^{\\\\cdot}\\\\mathrm{mol}^{-1}$ in germanium crystals." }, { "idx": 855, "question": "Explain one of the main mechanisms of alloy strengthening from a microstructural perspective: solid solution strengthening, and provide examples", "answer": "Solid solution strengthening occurs when alloying atoms dissolved in the lattice interstices or lattice points, due to their different sizes compared to the matrix atoms, create stress fields that hinder dislocation movement. Examples include Cottrell atmospheres and Suzuki atmospheres. The former refers to interstitial atoms preferentially distributing in the tensile stress region of edge dislocations in BCC metals, pinning the dislocations. The latter involves alloying elements preferentially distributing in the stacking fault regions of extended dislocations in FCC metals, reducing stacking fault energy, expanding the stacking fault region, and making the slip of extended dislocations more difficult." }, { "idx": 856, "question": "Explain the second main mechanism of alloy strengthening from the microstructure perspective: precipitation strengthening and dispersion strengthening, and provide examples", "answer": "Alloy strengthening is caused by compounds of alloying elements and matrix elements obtained through phase transformation processes and hard particles mechanically mixed into the matrix material, which are referred to as precipitation strengthening and dispersion strengthening, respectively. The effects of precipitation strengthening and dispersion strengthening are far greater than those of solid solution strengthening. When dislocations encounter second phases during movement, they need to cut through (small-sized particles in precipitation strengthening and particles in dispersion strengthening) or bypass (large-sized particles in precipitation strengthening) the second phases. Therefore, the second phases (precipitates and dispersoids) hinder dislocation movement." }, { "idx": 857, "question": "When an aluminum (Al) single crystal is stretched along the [123] direction to induce plastic deformation, determine the rotation law and rotation axis of the crystal during double slip.", "answer": "During double slip, the specimen axis rotates toward [101], with the rotation axis being n1=[11¯2]×[10¯1]=[¯1¯1¯1], and also rotates toward [011], with the rotation axis being n2=[11¯2]×[01¯1]=[111]. Therefore, the resultant rotation axis is n=n1+ n2=[000], meaning the crystal orientation no longer changes and is only elongated." }, { "idx": 858, "question": "Tensile deformation is applied along the [123] direction of an aluminum (Al) single crystal to induce plastic deformation. Please determine the double slip systems.", "answer": "When point F1 moves to point F2, double slip occurs. The double slip systems are:" }, { "idx": 859, "question": "Tensile deformation is applied along the [123] direction of an aluminum (Al) single crystal to induce plastic deformation. Determine the rotation rule and rotation axis.", "answer": "During single slip, the specimen axis should rotate towards [101], and the rotation axis is [12¯3]×[10¯1]=[¯1¯1¯1]." }, { "idx": 860, "question": "Explain one of the main mechanisms of alloy strengthening from the microstructure perspective: order strengthening, and provide an example", "answer": "In ordered alloys, dislocations are superdislocations. To induce plastic deformation in the metal, both partial dislocations of the superdislocation must move simultaneously, requiring greater external stress. The bonding force between atoms of different elements is stronger than that between atoms of the same element, so the ordered arrangement of dissimilar atoms imparts higher strength to ordered alloys. For example, high-strength nickel-based superalloys contain ordered precipitate phases Ni(Ti, Al), whose strengthening mechanisms include general precipitation strengthening and order strengthening." }, { "idx": 861, "question": "Stretch the aluminum (Al) single crystal along the [123] direction to induce plastic deformation, determine the final orientation of the crystal.", "answer": "The final orientation of the crystal is F2, namely [112]." }, { "idx": 862, "question": "Tensile deformation is applied along the [123] direction of an aluminum (Al) single crystal to induce plastic deformation. Determine the crystal orientation and shear strain at the onset of double slip.", "answer": "It can be calculated as follows: L=l+γ(l⋅n)b Substituting into the above equation: [m] =[123] + [[123][][10] √3 1√2 Thus, v=1+γ4/√6=2, W=−3−γ4/√6 Solving gives v=2, w=−4, γ=√6/4 That is, the orientation at the onset of double slip is [112], and the shear strain is √6/4." }, { "idx": 863, "question": "Explain one of the three main mechanisms of alloy strengthening from a microstructural perspective: grain boundary strengthening, and provide an example", "answer": "According to the Hall-Petch formula, the relationship between the yield point σs and the grain diameter d is σs=σ0+k d^(-1/2). The essence is that additional stress is required for dislocations to cross grain boundaries. Therefore, steels for low-temperature applications often adopt a fine-grained structure." }, { "idx": 864, "question": "There are two edge dislocations 1 and 2 separated by a distance d, with Burgers vectors b₁ and b₂ respectively. Given the stress field of the edge dislocation as σ_x, σ_y, σ_z, τ_xy, determine the climb force on dislocation 2 due to dislocation 1.", "answer": "The climb force f_{y,12} = σ_x b₂ = - (G b₁ b₂) / (2π(1-ν)) * (d(3x² + d²)) / (x² + d²)²" }, { "idx": 865, "question": "There are two edge dislocations 1 and 2 separated by a distance d, with Burgers vectors b₁ and b₂, respectively. Given the stress field of an edge dislocation as σ_x, σ_y, σ_z, τ_xy, determine the glide force on dislocation 2 caused by dislocation 1.", "answer": "The glide force f_{x,12} = -τ_{xy}b₂ = - (G b₁ b₂) / (2π(1-ν)) * (x(x² - d²)) / (x² + d²)²" }, { "idx": 866, "question": "Write the expression for the equilibrium concentration of point defects and indicate the physical meaning of each parameter.", "answer": "The Schottky defect concentration Cs = exp(-ΔGs/RT), where Cs is the equilibrium concentration of Schottky point defects; ΔGs is the molar Gibbs free energy of formation for Schottky point defects; T is the thermodynamic temperature (K) of the system; R is the molar gas constant. The Frenkel defect concentration CF = exp(-ΔGF/RT), where CF is the equilibrium concentration of Frenkel point defects; ΔGF is the molar Gibbs free energy of formation for Frenkel point defects; T is the thermodynamic temperature (K) of the system; R is the molar gas constant." }, { "idx": 867, "question": "There are two single-crystal Zn rods (Zn has an HCP structure with c/a=1.86), with the axial direction [0001]. Now, under room temperature conditions, they are stretched along the axial direction. Analyze the possible deformation modes and write the corresponding slip systems or twinning systems (if activated).", "answer": "When the Zn rod with the axial direction [0001] is stretched along the axial direction, the axial elongation occurs, which is opposite to the length change trend during twinning, so twinning is not possible. During stretching, the external force is along the axial direction [0001], perpendicular to the slip plane (0001), so the Schmid factor is 0, and slip cannot occur. Therefore, when the Zn rod with the axial direction [0001] is stretched along the axial direction, the Zn rod will fracture in a brittle manner." }, { "idx": 868, "question": "Write the expression for the relationship between the diffusion coefficient and temperature, and indicate the physical meaning of each parameter.", "answer": "D=D0exp(-Q/RT), where D0 is the diffusion constant or frequency factor; Q is the diffusion activation energy (J/mol); T is the thermodynamic temperature of the system (K); R is the molar gas constant." }, { "idx": 869, "question": "There are two single-crystal Zn rods (Zn has an HCP structure with c/a=1.86), with the axial direction [1010]. Now, under room temperature conditions, they are stretched along the axial direction. Analyze the possible deformation modes and write the corresponding slip systems or twin systems (if activated).", "answer": "When the Zn rod with the axial direction [1010] is stretched along the axial direction, on one hand, the [0001] crystal direction will shorten, which is consistent with the length change during twinning, so twinning can occur. The twin systems are (1012)[1011] and (1012)[1011]. On the other hand, according to Schmid's law, it can be concluded that the Zn rod does not undergo slip. However, the result of twinning can change the orientation, thereby enabling slip, so the Zn rod will exhibit certain plasticity." }, { "idx": 870, "question": "What are the three typical zones of a metal ingot structure?", "answer": "Chill zone, columnar crystal zone, and equiaxed crystal zone." }, { "idx": 871, "question": "What insights do the results of the Kirkendall experiment provide?", "answer": "1. It reveals the intrinsic connection between macroscopic diffusion laws and microscopic diffusion mechanisms, which is universal; 2. It directly refutes the exchange mechanism of substitutional solid solution diffusion and supports the vacancy mechanism; 3. Each component in a diffusion system has its own diffusion coefficient; 4. The Kirkendall phenomenon often produces side effects, such as incomplete shrinkage leading to Kirkendall voids, which generally have adverse impacts in practice and thus should be controlled." }, { "idx": 872, "question": "Many precipitate phases have specific orientation relationships with the parent matrix. Analyze the reasons for the formation of such orientation relationships.", "answer": "To achieve low interfacial energy, the atoms on the primary interface between the parent phase and the new phase must have a good matching relationship, which can only be realized under certain orientation conditions. Therefore, specific orientation relationships exist between the precipitate phase and the parent phase." }, { "idx": 873, "question": "What is the Kirkendall effect", "answer": "On an FCC-structured α-brass (Cu+30%Zn) rod, very fine molybdenum wires are applied as markers, and then copper is plated on the brass, enclosing the molybdenum wires between the brass and copper. Diffusion is carried out at a certain temperature. The molybdenum wires serve only as markers and do not participate in the diffusion throughout the experiment. The diffusing components are copper and zinc, which form a substitutional solid solution. During the diffusion process in a substitutional solid solution, the markers placed at the original interface move toward the direction of the lower melting point element, with the displacement distance following a parabolic relationship with time. The reason for this phenomenon is that the lower melting point component diffuses faster, while the higher melting point component diffuses slower. This unequal atomic exchange results in the Kirkendall effect." }, { "idx": 874, "question": "Propose three technical approaches to achieve the aforementioned zone enlargement, and briefly explain the theoretical basis.", "answer": "1. Add refiners to promote heterogeneous nucleation; 2. Apply mechanical vibration to reduce dendrites and provide new nuclei; 3. Perform ultrasonic treatment or electromagnetic field treatment." }, { "idx": 875, "question": "What is the Kirkendall effect? Please explain it using diffusion theory.", "answer": "Kirkendall effect: In the diffusion process of substitutional solid solutions, markers placed at the original interface move towards the direction of the low-melting-point element, with the movement rate following a parabolic relationship with time. The Kirkendall effect negates the exchange mechanism of diffusion in substitutional solid solutions and confirms the vacancy mechanism; different components in the system have different partial diffusion coefficients. Relatively speaking, the low-melting-point component diffuses faster, while the high-melting-point component diffuses slower. This unequal atomic exchange results in the Kirkendall effect." }, { "idx": 876, "question": "To obtain a casting with uniform composition and performance, which zone should be expanded?", "answer": "To obtain a casting with uniform composition, the equiaxed crystal zone should be expanded." }, { "idx": 877, "question": "Determine whether the following dislocation reaction can occur in FCC and confirm the reaction direction in the absence of external force: 1/6[112]+1/6[110]⇔1/3[111]", "answer": "It can occur, to the left." }, { "idx": 878, "question": "During the aging treatment of alloy A, metastable phases often precipitate first. Discuss the reasons why the equilibrium phase does not precipitate based on thermodynamic and kinetic theories.", "answer": "The initial precipitation of metastable phases occurs because the interfacial energy between the equilibrium phase and the parent phase is high, whereas that between the metastable phase and the parent phase is low. Interfacial energy is the primary energy barrier for nucleation during solid-state phase transformation. Overcoming this nucleation barrier for the equilibrium phase often requires significant undercooling before noticeable nucleation can occur in the system. From a kinetic perspective, at a certain degree of undercooling, if the metastable phase forms much faster than the equilibrium phase, which forms much more slowly, the alloy will preferentially form metastable phases during aging treatment." }, { "idx": 879, "question": "Two batches of industrial pure aluminum samples were rolled with the same deformation amount on June 5 and June 9, respectively, and then subjected to recrystallization annealing simultaneously on the 9th. The experiment found that for the same annealing time, the temperatures required to complete recrystallization were different for the two batches of samples. Which batch do you think has a higher recrystallization temperature, and why?", "answer": "The sample deformed on June 5 may have undergone partial recovery during storage, as the stored deformation energy decreases. To initiate recrystallization, it requires higher activation energy compared to the sample that was not stored. Therefore, the sample from June 5 has a higher recrystallization temperature." }, { "idx": 880, "question": "Briefly describe Pauling's rules", "answer": "Pauling's rules: $\\textcircled{1}$ The coordination polyhedron rule: A coordination polyhedron of anions is formed around a cation. The distance between the cation and anion depends on the sum of their ionic radii, while the coordination number depends on the ratio of their ionic radii. $\\textcircled{2}$ The electrostatic valence rule: In forming an ionic bond, the number of valence electrons given by the cation equals the number received by the anion. $\\textcircled{3}$ The rule of polyhedron linkage: Coordination polyhedra tend to share vertices, followed by edges and faces. The higher the cation's valence and the lower its coordination number, the more pronounced this effect. $\\textcircled{4}$ In crystals containing more than one type of cation, the coordination polyhedra of anions around cations with higher valence and smaller coordination numbers tend to share vertices. $\\textcircled{5}$ The types of coordination polyhedra in a crystal tend to be minimal." }, { "idx": 881, "question": "Given that the nearest neighbor atomic spacing in a diamond unit cell is 0.1544nm, determine the lattice constant a of diamond.", "answer": "The nearest neighbor atomic spacing is 0.1544nm, i.e., the bond length d = 0.1544nm. In the diamond structure, the relationship between the bond length and the lattice constant is d = (a√3)/4, thus a = (4d)/√3 = (4 × 0.1544nm)/1.732 ≈ 0.3566nm." }, { "idx": 882, "question": "Given that the nearest neighbor atomic spacing in a diamond unit cell is 0.1544nm, determine the coordination number C.N. of diamond.", "answer": "In the diamond structure, each carbon atom forms covalent bonds with 4 nearest neighbor carbon atoms, so the coordination number C.N. = 4." }, { "idx": 883, "question": "A compression test is conducted along the axial direction of an FCC metal single crystal specimen with a cross-sectional area of 10cm². The critical resolved shear stress is known to be 0.1kgf/mm², and the initial orientation of the rod axis is [215]. Please determine the double slip systems.", "answer": "Double slip systems: (111)[011] + (111)[011]." }, { "idx": 884, "question": "Determine whether the following dislocation reaction can occur in FCC and confirm the reaction direction in the absence of external forces: 1/2[10−1]⇔1/6[2−1−1]+1/6[11−1]", "answer": "It can occur, to the right." }, { "idx": 885, "question": "A single crystal specimen of an FCC metal with a cross-sectional area of 10cm² is subjected to a compression test along its axial direction. Given that the critical resolved shear stress is 0.1kgf/mm² and the initial orientation of the rod axis is [215], determine the crystal orientation (i.e., the direction of the rod axis) when double slip begins.", "answer": "The crystal orientation when double slip begins is: [102]." }, { "idx": 886, "question": "A single crystal test bar of an FCC metal with a cross-sectional area of 10 cm² is subjected to a compression test along the axial direction. The critical resolved shear stress is known to be 0.1 kgf/mm², and the initial orientation of the bar axis is [215]. Please determine the initial slip system (specific indices are required).", "answer": "Initial slip system: (111)[011]." }, { "idx": 887, "question": "Term explanation: Lattice distortion", "answer": "In local regions, atoms deviate from their normal lattice equilibrium positions, causing lattice distortion." }, { "idx": 888, "question": "Given that the nearest neighbor atomic spacing in a diamond unit cell is 0.1544nm, calculate the packing density ξ of diamond.", "answer": "The diamond unit cell contains 8 atoms (each of the 8 corner atoms contributes 1/8, each of the 6 face-centered atoms contributes 1/2, and each of the 4 internal atoms contributes 1). The atomic radius r = d/2 = 0.0772nm. The packing density ξ = (total volume of atoms)/(volume of unit cell) = [(8 × (4/3)πr³)]/a³ = [8 × (4/3) × 3.1416 × (0.0772nm)³]/(0.3566nm)³ ≈ 0.34." }, { "idx": 889, "question": "Determine whether the following dislocation reaction can proceed in FCC and confirm the reaction direction in the absence of external forces: 1/3[112]+1/2[111]⇔1/6[11−1]", "answer": "The reaction cannot proceed." }, { "idx": 890, "question": "If diffusion occurs in a Cu-Al diffusion couple, in which direction will the interface marker move?", "answer": "When diffusion occurs in a Cu-Al diffusion couple, the interface marker will move toward the Al side." }, { "idx": 891, "question": "A single crystal test bar of FCC metal with a cross-sectional area of 10cm² is subjected to a compression test along the axial direction. The critical resolved shear stress is known to be 0.1kgf/mm², and the initial orientation of the bar axis is [215]. Please determine the final stable orientation of the crystal (assuming the test bar does not fracture before reaching the stable orientation).", "answer": "The final stable orientation of the crystal: [101]." }, { "idx": 892, "question": "Term explanation: Burgers vector", "answer": "An important vector describing the characteristics of a dislocation, which comprehensively reflects the magnitude and direction of the total distortion within the dislocation region; it also represents the amount of relative sliding of the crystal after the dislocation has swept through." }, { "idx": 893, "question": "A single crystal test bar of an FCC metal with a cross-sectional area of 10cm² is subjected to a compression test along the axial direction. Given that the critical resolved shear stress is 0.1kgf/mm² and the initial orientation of the bar axis is [215], determine the axial pressure P at the onset of double slip (without considering physical hardening).", "answer": "P = (0.1×1000)/((√10/5)×(√10/5))×9.8N = 2450N." }, { "idx": 894, "question": "Term explanation: phase diagram", "answer": "A graphical representation describing the conditions for phase equilibrium or the coexistence relationships of phases; it can also be referred to as the geometric trajectory of thermodynamic parameters at equilibrium." }, { "idx": 895, "question": "A single crystal test bar of FCC metal with a cross-sectional area of 10 cm² is subjected to a compression test along the axial direction. The critical resolved shear stress is known to be 0.1 kgf/mm², and the initial orientation of the bar axis is [215]. Please answer the rotation law and rotation axis when slip begins.", "answer": "Since it is compression, the rotation law is: F→[111], and the rotation axis is [215]×[111]=[633]=[211]." }, { "idx": 896, "question": "Term explanation: deformation texture", "answer": "The phenomenon of crystallographic orientation preference occurring during polycrystalline deformation is called deformation texture." }, { "idx": 897, "question": "Term explanation: Degree of supercooling", "answer": "In the phase transformation process, after cooling to a certain temperature below the phase transition point, the transformation occurs. The difference between the equilibrium phase transition temperature and this actual transformation temperature is called the degree of supercooling." }, { "idx": 898, "question": "Term explanation: Slip system", "answer": "In a crystal, the combination of a slip plane and a slip direction on that plane is called a slip system." }, { "idx": 899, "question": "Write the mathematical expression of Fick's first law and explain its significance.", "answer": "In one dimension, J=-D(dc/dx). J: diffusion flux, the amount of substance diffusing per unit time through a unit area, g/s·m²; D: diffusion coefficient, m²/s; dc/dx: concentration gradient, g/m³. Its significance is that the amount of substance diffused is proportional to the concentration gradient of the substance and in the opposite direction." }, { "idx": 900, "question": "Term explanation: Twinning", "answer": "Twinning refers to the shear process in which a crystal undergoes deformation by forming twins under stress." }, { "idx": 901, "question": "What are the general structural zones of an ingot? List their names.", "answer": "It is divided into three crystal zones: chill zone, columnar crystal zone, and central equiaxed crystal zone." }, { "idx": 902, "question": "Briefly describe the changes in mechanical properties of deformed metals during annealing", "answer": "Hardness and strength decrease, elongation increases." }, { "idx": 903, "question": "Term explanation: secondary recrystallization", "answer": "The phenomenon of abnormal growth of a few grains occurring when normal growth is inhibited after the completion of recrystallization." }, { "idx": 904, "question": "Briefly describe the changes in stored energy during the annealing process of deformed metals", "answer": "The stored energy is gradually released." }, { "idx": 905, "question": "Briefly describe the changes in physical properties of deformed metal during annealing", "answer": "Electrical resistance decreases, density increases. The changes in various properties during recrystallization are much more pronounced than during recovery." }, { "idx": 906, "question": "Briefly describe the factors affecting the crystalline structure of ingots.", "answer": "Factors affecting the crystalline structure of ingots: 1 Liquid superheat, the smaller the better; 2 Solidification temperature range, the larger the better, which is beneficial for dendrite fragmentation; 3 Temperature gradient, the smaller the more conducive to equiaxed crystals; 4 Alloy with low melting point, small temperature gradient; 5 Stirring or adding inoculants." }, { "idx": 907, "question": "Briefly describe the changes in the microstructure of deformed metal during the annealing process", "answer": "As the annealing temperature increases or the time prolongs, subgrain coalescence and growth occur, followed by recrystallization nucleation and growth. The deformed grains with high dislocation density and elongated shape are replaced by equiaxed recrystallized grains free of dislocations, and then normal grain growth takes place." }, { "idx": 908, "question": "Briefly describe the similarities between solid-state phase transformation and liquid-state phase transformation", "answer": "Both are phase transformations, consisting of nucleation and growth. The critical radius and critical nucleation work have the same forms. The transformation kinetics are also the same." }, { "idx": 909, "question": "Briefly describe the differences between solid-state phase transformation and liquid-state phase transformation", "answer": "The nucleation resistance includes an additional strain energy term, leading to an increase in the critical radius and nucleation work for solid-state phase transformation; the new phase can appear in a metastable manner, with coherent, semi-coherent interfaces, specific orientation relationships, and heterogeneous nucleation." }, { "idx": 910, "question": "Briefly describe the factors affecting diffusion.", "answer": "1. Temperature. It follows the relationship D=D0e^(-Q/RT), where an increase in T leads to an increase in D; 2. Interfaces, surfaces, and dislocations. These serve as fast diffusion paths; 3. Third elements. They can have varying effects on binary diffusion, such as Mo and W reducing the diffusion coefficient of C in γ-Fe, Co and Si accelerating the diffusion of C, and Mn having little effect; 4. Crystal structure. In low-symmetry crystal structures, diffusion anisotropy exists, such as in hexagonal crystals where the diffusion coefficients parallel and perpendicular to the basal plane (0001) differ; 5. Melting point. Within the same alloy system, diffusion is slower in alloys with higher melting points and faster in those with lower melting points at the same temperature." }, { "idx": 911, "question": "What is the difference in the driving force for nucleation during recrystallization and solidification?", "answer": "The driving force for nucleation during solidification is the chemical potential difference between the new and old phases, while the driving force for recrystallization is only the stored energy of deformation." }, { "idx": 912, "question": "What is the difference in orientation relationship between growth during recrystallization and growth during solidification?", "answer": "During solidification growth, there is no orientation relationship with the parent phase, whereas during recrystallization growth, a specific orientation relationship may exist." }, { "idx": 913, "question": "What is the difference in nucleation sites between recrystallization and solidification processes?", "answer": "Solidification is often homogeneous nucleation; recrystallization nucleation occurs in existing deformation inhomogeneous regions, such as near grain boundaries, shear bands, deformation bands, and around second-phase particles." }, { "idx": 914, "question": "5. Secondary recrystallization", "answer": "A phenomenon where the normal growth process is inhibited after recrystallization, leading to abnormal growth of a few grains." }, { "idx": 915, "question": "Briefly describe the inverse segregation in the macrosegregation of ingots", "answer": "Inverse segregation: It still follows the distribution coefficient relationship, but after the formation of a large number of dendrites, the solute-enriched liquid phase will flow backward along the interdendritic channels to the vicinity of the initially solidified ingot surface, resulting in an abnormal composition distribution from the surface to the center." }, { "idx": 916, "question": "Briefly describe the gravity segregation in the macrosegregation of ingots", "answer": "Gravity segregation: During solidification, the solid phase and the liquid phase have different densities, causing deposition or floating, which results in uneven composition between the lower and upper ends of the ingot, also on a macroscopic scale." }, { "idx": 917, "question": "Briefly describe normal segregation in the macro-segregation of ingots", "answer": "Normal segregation: refers to the phenomenon where the initially solidified portion has lower solute content and the later solidified portion has higher solute content, according to the alloy's distribution coefficient (assuming k0<1). Due to the large size of the ingot, the composition varies from the surface to the center, resulting in non-uniformity, and the segregation occurs on a macroscopic scale, known as macro-segregation." }, { "idx": 918, "question": "4.Deformation texture", "answer": "The phenomenon of crystallographic orientation preference occurring during polycrystalline deformation." }, { "idx": 919, "question": "3. Congruent solidification and incongruent solidification", "answer": "Solidification without compositional change is called congruent solidification; conversely, solidification accompanied by compositional change is called incongruent solidification." }, { "idx": 920, "question": "2.Grain boundaries and interfacial energy", "answer": "A grain boundary is the interface between grains of the same composition and structure. Atoms at the interface are in a broken-bond state and possess excess energy. The average excess energy per unit area of the interface is called interfacial energy." }, { "idx": 921, "question": "Analyze the crystallization process of an iron-carbon alloy with a carbon content of 0.12%", "answer": "For steel with 0.12% carbon, when cooled from the liquid phase, δ-ferrite forms first. The compositions of the solid and liquid phases change according to the solidus and liquidus lines on the phase diagram. At the peritectic temperature of 1495°C, partial peritectic reaction occurs; the new phase γ-austenite nucleates on the pre-existing δ-ferrite and grows into both the δ-ferrite and liquid phase. After the reaction, the microstructure consists of two phases: δ-ferrite + γ-austenite. The relative amount of δ-ferrite is: (0.17-0.12)/(0.17-0.09)=62.5%. Upon further cooling, single-phase austenite is obtained." }, { "idx": 922, "question": "What are the characteristics of the space lattice in the tetragonal crystal system?", "answer": "The lattice parameters of the tetragonal crystal system are a≠b≠c, α=β=γ=90°." }, { "idx": 923, "question": "Briefly describe the types of defects in metal crystals", "answer": "According to size, they can be divided into: point defects, such as solute, impurity atoms, vacancies; line defects, such as dislocations; planar defects, such as various grain boundaries, phase boundaries, surfaces, etc.; volume defects, such as pores, bubbles, etc. Volume defects are absolutely harmful to material properties." }, { "idx": 924, "question": "1. Edge dislocation and screw dislocation models", "answer": "Cut the upper half of the crystal, insert half a crystal plane, and then bond it together; in this way, within a certain range near the centerline corresponding to the edge end, the atoms undergo regular misalignment. Its characteristic is that the upper half is under compression, and the lower half is under tension. This is the same as the scenario caused by edge dislocations in real crystals, known as the edge dislocation model. Similarly, cut the front half of the crystal, use the edge end as the boundary to make the left and right parts undergo a relative shear of one atomic spacing up and down, and then bond them together. At this time, near the boundary line between the sheared and unsheared regions, the atomic misalignment is similar to that of a real screw dislocation, known as the screw dislocation model." }, { "idx": 925, "question": "Briefly describe the characteristics of the eutectic reaction in a binary system and calculate the degrees of freedom when its phases are in equilibrium.", "answer": "The eutectic reaction is: the liquid phase simultaneously solidifies into two solid phases of different compositions, which grow cooperatively and generally form a lamellar structure. When the eutectic reaction occurs, the degrees of freedom are 0, meaning the compositions of the three phases are fixed, and the temperature is also fixed." }, { "idx": 926, "question": "Briefly describe the characteristics of the peritectic reaction in a binary system and calculate the degrees of freedom at equilibrium for each phase.", "answer": "The peritectic reaction is: a liquid phase reacts with a solid phase to form another solid phase, where the newly formed solid phase envelops the original solid phase. The reaction requires diffusion within the solid phase and proceeds relatively slowly. When a peritectic reaction occurs, the degrees of freedom are 0, meaning the compositions of the three phases are fixed, and the temperature is also fixed." }, { "idx": 927, "question": "Briefly describe the characteristics of the eutectoid reaction in a binary system and calculate the degrees of freedom when its phases are in equilibrium.", "answer": "The eutectoid is similar to the eutectic, except that the parent phase is a solid phase, i.e., one solid phase simultaneously transforms into two other solid phases of different compositions. When the eutectoid reaction occurs, the degrees of freedom are 0, meaning the compositions of the three phases are fixed, and the temperature is also fixed." }, { "idx": 928, "question": "What are the characteristics of the space lattice in the cubic crystal system?", "answer": "The lattice parameters of the cubic crystal system are a=b=c, α=β=γ=90°." }, { "idx": 929, "question": "Briefly describe the meaning of Fick's first law, write its expression, and indicate the physical meanings of the letters.", "answer": "Fick's first law J=-D(dc/dx) J: diffusion flux, which is the amount of substance passing through a unit area per unit time, g/m²·s; D: diffusion coefficient, m²/s; dc/dx: concentration gradient, g/m⁴." }, { "idx": 930, "question": "Briefly explain the meaning of Fick's second law, write its expression, and indicate the physical meanings of the letters.", "answer": "Fick's second law ∂c/∂t=∂/∂x(D∂c/∂x) ∂c/∂t: rate of concentration change over time, g/m³·s; ∂J/∂x: gradient of flux, g/m³·s." }, { "idx": 931, "question": "What is the method of nucleation during the recrystallization process?", "answer": "Recrystallization nucleation often occurs in existing regions of deformation inhomogeneity, such as near grain boundaries, shear bands, deformation bands, and around second-phase particles." }, { "idx": 932, "question": "What is a lattice parameter?", "answer": "Lattice parameters are the fundamental parameters that describe the geometric shape of a unit cell, consisting of six parameters: the three edge lengths a, b, c and the three angles α, β, γ between them." }, { "idx": 933, "question": "6.Hardenability and hardening capacity", "answer": "Hardenability refers to the ability of an alloy to be quenched into martensite, mainly related to the critical cooling rate, with its magnitude represented by the depth of the hardened layer. Hardening capacity refers to the maximum hardness that can be achieved after quenching, primarily related to the carbon content of the steel." }, { "idx": 934, "question": "Briefly describe the role and influence of grain boundaries on the plastic deformation of polycrystals", "answer": "The slip planes on both sides of the grain boundary are not parallel. On one hand, grain boundaries act as obstacles to dislocation motion, causing pile-ups and strengthening; meanwhile, they necessitate the occurrence of multiple slip systems near the grain boundaries to accommodate the deformation on both sides." }, { "idx": 935, "question": "Briefly describe the characteristics of phase transformation resistance in solid-state phase transformations", "answer": "The phase transformation resistance includes an additional term for strain energy." }, { "idx": 936, "question": "What is the main difference between nucleation and growth during solidification and nucleation and growth during recrystallization?", "answer": "The driving force for nucleation and growth during solidification is the chemical potential difference between the new and old phases, while the driving force for nucleation and growth during recrystallization is solely the stored deformation energy. Nucleation during solidification is often homogeneous; nucleation during recrystallization typically occurs in existing regions of deformation inhomogeneity, such as near grain boundaries, shear bands, deformation bands, or around second-phase particles. During solidification growth, there is no orientation relationship with the parent phase, whereas during recrystallization growth, there may be a certain orientation relationship." }, { "idx": 937, "question": "What is primary banded structure?", "answer": "Primary banded structure is formed by dendrites during solidification and becomes banded after hot rolling; under CuCl2 etching, the dark bands are dendrite trunks (with less solute), while the light bands are often interdendritic regions enriched with impurities, also known as original bands." }, { "idx": 938, "question": "What is secondary banded structure?", "answer": "Secondary banding is a product of solid-state transformation. Under nitric acid alcohol etching, the white bands are proeutectoid ferrite, and the black bands are pearlite, also known as micro banded structure. Secondary banding only occurs on the basis of primary banding." }, { "idx": 939, "question": "Analyze the causes of primary banded structure formation.", "answer": "The formation of primary banded structure is related to dendrite formation during solidification and the hot rolling process, where the dark bands are dendrite trunks (with fewer solutes) and the white bands are interdendritic regions enriched with impurities." }, { "idx": 940, "question": "Analyze the causes of secondary banded structure formation.", "answer": "The formation of secondary banded structure is the result of solid-state transformation. The white bands are proeutectoid ferrite, and the black bands are pearlite, which must form on the basis of primary bands. These bands are all related to element segregation and the special distribution of inclusions." }, { "idx": 941, "question": "Briefly describe the characteristics of nucleation in solid-state phase transformations", "answer": "Non-uniform nucleation, with specific orientation relationships, often forming coherent or semi-coherent interfaces." }, { "idx": 942, "question": "Briefly describe the characteristics of growth in solid-state phase transformations", "answer": "The occurrence of habit phenomena, i.e., there is a precipitation sequence; special/regular microstructural morphologies, such as lamellar, acicular." }, { "idx": 943, "question": "Briefly describe the characteristics of metastable phases in solid-state phase transformations", "answer": "The appearance of metastable phases serves to reduce the resistance to phase transformation." }, { "idx": 944, "question": "Describe the mechanism of crystal growth during solidification", "answer": "There are three growth mechanisms: continuous growth, two-dimensional nucleation and lateral growth via steps, and growth via screw dislocations." }, { "idx": 945, "question": "Briefly describe the macroscopic characteristics of the solidification process", "answer": "The macroscopic characteristics during solidification are: a certain degree of undercooling is required, and significant latent heat of crystallization is released." }, { "idx": 946, "question": "Briefly describe the role and influence of grain size on the plastic deformation of polycrystals", "answer": "When grains are fine, the entire grain can deform more uniformly with fewer combinations of slip systems; when grains are coarse, different regions within the grain cannot coordinate with each other during deformation, requiring the activation of different combinations of slip systems, often leading to grain 'fragmentation,' where a large grain is 'fragmented' into several parts with different slip systems activated in each part. The effect on performance follows the Hall-Petch relationship σs=σ0+Kd−1/2, meaning the finer the grains, the more grain boundaries there are, and the more significant the hindrance of grain boundaries to moving dislocations, resulting in a greater increase in strength." }, { "idx": 947, "question": "What is a solid solution?", "answer": "A single homogeneous solid formed by solute atoms dissolving into the solvent lattice in atomic form; the lattice type of the solvent is retained." }, { "idx": 948, "question": "List one method to increase the number of nuclei during the solidification process, and briefly analyze the reason for the increase in the number of nuclei.", "answer": "Increase the cooling rate to enhance the degree of undercooling. The phase transformation driving force is increased, thereby improving the nucleation rate." }, { "idx": 949, "question": "What changes occur in the structure, mechanical properties, and physical properties of a solid solution compared to its pure solvent component?", "answer": "Structural changes in solid solutions: lattice distortion, changes in lattice constants, segregation and short-range ordering, and even the formation of ordered solid solutions. Changes in mechanical properties: increased hardness and strength, decreased plasticity. Changes in physical properties: increased electrical resistance, decreased conductivity." }, { "idx": 950, "question": "List another method to increase the number of nuclei during solidification, and briefly analyze the reason for the increase in the number of nuclei.", "answer": "Add inoculants. They provide a large number of heterogeneous nucleation sites, increase the nucleation rate, and reduce the nucleation barrier." }, { "idx": 951, "question": "List the third method to increase the number of nuclei during solidification, and briefly analyze the reason for the increase in the number of nuclei.", "answer": "Mechanical or electromagnetic stirring. Breaking dendrites to increase the number of nuclei or enhancing thermal activation energy drop, which improves the nucleation rate." }, { "idx": 952, "question": "Briefly describe the crystallization process of steel with a carbon content of 0.25%", "answer": "The liquid phase first solidifies into ferrite, and a peritectic reaction occurs at 1495°C to form γ-austenite; upon further cooling, the remaining liquid phase precipitates γ-austenite again, which then transforms into single-phase austenite." }, { "idx": 953, "question": "Briefly describe the meaning of continuous precipitation", "answer": "Continuous precipitation: as the new phase forms, the composition of the parent phase continuously changes from a supersaturated state to a saturated state." }, { "idx": 954, "question": "What are the factors affecting solid solubility?", "answer": "The factors affecting solid solubility include: 1. Atomic size factor. When the relative difference in atomic diameters between the solvent and solute is less than ±15%, there is a large substitutional solubility. 2. Electronegativity factor. The smaller the electronegativity difference between the solvent and solute, the greater the solubility; generally, a difference less than 0.4∼0.5 results in higher solubility. 3. Electron concentration factor. This has two aspects: one is the atomic valence effect, meaning for the same solvent metal, the higher the valence of the solute, the lower the solubility; the other is the relative valence effect, meaning the solubility of a high-valence solute in a low-valence solvent is higher than the opposite scenario." }, { "idx": 955, "question": "Briefly describe the meaning of discontinuous precipitation", "answer": "Discontinuous precipitation: also known as cellular precipitation, where an interface exists between the parent phase and the newly formed α phase. Across this interface, the parent phase discontinuously changes from supersaturated to saturated state, and the lattice parameter also changes discontinuously." }, { "idx": 956, "question": "Briefly describe the solid-state phase transformation process of steel with a carbon content of 0.25%", "answer": "When cooled to about 800°C, proeutectoid α-ferrite precipitates from γ-austenite; at 727°C, the eutectoid reaction occurs, forming pearlite, and finally a ferrite plus pearlite structure is obtained." }, { "idx": 957, "question": "Briefly describe metallic bond", "answer": "The bonding force generated by the electrostatic interaction between free electrons and atomic nuclei." }, { "idx": 958, "question": "Briefly describe the critical resolved shear stress for slip", "answer": "The minimum resolved shear stress required to initiate slip system movement; it is a constant value, related to the intrinsic properties of the material and independent of the orientation of external forces." }, { "idx": 959, "question": "Briefly describe intermediate phases", "answer": "Phases formed between components in an alloy, which have structures different from those of the pure components. Located in the intermediate region of the phase diagram." }, { "idx": 960, "question": "Briefly describe martensitic transformation", "answer": "The process follows a diffusionless, shear-type phase transformation." }, { "idx": 961, "question": "Briefly describe the Bravais lattice", "answer": "A lattice that considers not only the shape of the unit cell but also the positions of the lattice points (or: considers not only rotational symmetry but also translational symmetry, forming all lattice points after centering)." }, { "idx": 962, "question": "Briefly describe the slip system", "answer": "A slip system in a crystal refers to a combination of a slip plane and a slip direction on that plane." }, { "idx": 963, "question": "According to the size of defects relative to the crystal dimensions and their range of influence, into which categories can defects be classified? Briefly describe the characteristics of these categories.", "answer": "There are four categories: Point defects: Very small in all three dimensions, such as solute atoms, interstitial atoms, and vacancies. Line defects: Very small in two dimensions, but large in the third dimension, potentially extending through the entire crystal, referring to dislocations. Planar defects: Very small in one dimension but large in the other two dimensions, such as grain boundaries and phase boundaries. Volume defects: Relatively large in all three dimensions but not excessively so, such as second-phase particles and microscopic voids." }, { "idx": 964, "question": "Briefly describe segregation", "answer": "The non-uniformity of chemical composition in alloys." }, { "idx": 965, "question": "Briefly describe dislocations", "answer": "It is a type of linear defect in crystals, characterized by a regular misalignment of atoms along a line; this defect is described by both a line direction and a Burgers vector." }, { "idx": 966, "question": "Briefly describe secondary recrystallization", "answer": "The phenomenon of abnormal growth of a few grains that occurs when the normal growth process is inhibited after recrystallization is completed." }, { "idx": 967, "question": "Briefly describe the recrystallization temperature", "answer": "The minimum temperature at which a deformed metal just completes recrystallization within a certain time (generally 1h)." }, { "idx": 968, "question": "What is the structural model that constitutes a small-angle grain boundary?", "answer": "The structural model of a small-angle grain boundary is the dislocation model, for example, a symmetric tilt grain boundary is described by a set of parallel edge dislocations." }, { "idx": 969, "question": "What is a solid solution?", "answer": "A single homogeneous solid formed by solute atoms dissolving into the solvent lattice in an atomic state; the lattice type of the solvent is retained." }, { "idx": 970, "question": "Discuss the relationship between crystal structure and space lattice.", "answer": "The relationship between the two can be described as 'space lattice $^+$ basis $\\asymp$ crystal structure.' There are only 14 types of space lattices, while the basis can be infinitely varied, thus resulting in an infinite variety of specific crystal structures." }, { "idx": 971, "question": "The interface between grains in single-phase metals or alloys is generally called grain boundaries, which are usually divided into two categories: low-angle grain boundaries and high-angle grain boundaries. What is the basis for this classification?", "answer": "The classification is based on the misorientation between the grains on either side of the boundary. Boundaries with a misorientation <15° are called low-angle grain boundaries, while those >15° are called high-angle grain boundaries." }, { "idx": 972, "question": "How does temperature affect atomic diffusion in metals or alloys?", "answer": "Temperature follows the relationship D=Doe-Q/RT, where an increase in temperature accelerates diffusion." }, { "idx": 973, "question": "What effects do interfaces, surfaces, and dislocations have on atomic diffusion in metals or alloys?", "answer": "Interfaces, surfaces, and dislocations act as fast diffusion paths." }, { "idx": 974, "question": "How does the third component affect binary diffusion?", "answer": "The third component can have different effects on binary diffusion, such as Mo and W reducing the diffusion coefficient of C in γ-Fe; Co and Si accelerating the diffusion of C; Mn and Ni having little effect." }, { "idx": 975, "question": "What are the characteristics of nucleation in the solid-state phase transformation of metals?", "answer": "Predominantly heterogeneous nucleation; exhibits specific orientation relationships; phase interfaces are often coherent or semi-coherent." }, { "idx": 976, "question": "How does melting point affect atomic diffusion within the same alloy system?", "answer": "Within the same alloy system, at the same temperature, diffusion is slower in alloys with higher melting points and faster in those with lower melting points." }, { "idx": 977, "question": "Analyze and discuss the characteristics of microstructural and property changes during the recovery and recrystallization process of cold-worked metals or alloys after plastic deformation.", "answer": "As the annealing temperature increases or time prolongs, subgrains merge and grow, recrystallization nucleation and growth occur, and equiaxed recrystallized grains with no (or low-density) dislocations replace elongated deformed grains with high dislocation density, followed by normal grain growth. The stored energy is gradually released, most notably during the recrystallization stage; hardness and strength decrease, while elongation increases; electrical resistance decreases and density improves. The changes in various properties during recrystallization are much more pronounced than during recovery." }, { "idx": 978, "question": "What are the main factors affecting the solubility of solid solutions?", "answer": "The factors affecting the solubility of solid solutions include:\\n1. Atomic size factor. When the relative difference in atomic diameters between the solvent and solute is less than ±15%, it favors higher solubility in substitutional solid solutions; when the relative difference in diameters between the two elements exceeds 41%, it favors higher solubility in interstitial solid solutions.\\n2. Electronegativity factor. The smaller the electronegativity difference between the solvent and solute, the greater the solubility, generally less than 0.4–0.5 for higher solubility.\\n3. Electron concentration factor. This has two aspects: one is the atomic valence effect, where in the same solvent metal, the higher the valence of the solute, the lower the solubility; the other is the relative valence effect, where the solubility is higher when a high-valence solute dissolves into a low-valence solvent compared to the opposite scenario." }, { "idx": 979, "question": "What is the difference in phase transformation resistance between the solid-state phase transformation of metals and the crystallization process of metals?", "answer": "There is an additional term of strain energy in the phase transformation resistance." }, { "idx": 980, "question": "Discuss the effect of aging temperature on the precipitation sequence of Al 4.5%Cu alloy", "answer": "Increasing the aging temperature accelerates precipitation but reduces supersaturation and the driving force for phase transformation, potentially leading to direct precipitation of the equilibrium θ phase and weakening age-hardening capability; conversely, too low aging temperature prolongs the time required to achieve optimal performance." }, { "idx": 981, "question": "How does crystal structure affect atomic diffusion in metals or alloys?", "answer": "In crystal structures with low symmetry, there is anisotropy in diffusion coefficients, such as in hexagonal crystals where the diffusion coefficients parallel and perpendicular to the basal plane (0001) differ." }, { "idx": 982, "question": "What are the characteristics of the phase transformation products in the solid-state phase transformation of metals?", "answer": "There are metastable phases." }, { "idx": 983, "question": "Taking the Al-4.5%Cu alloy as an example, analyze the decomposition process of supersaturated solid solution (precipitation sequence).", "answer": "After solution treatment of the Al-4.5%Cu alloy, aging at the optimal temperature of ~150°C will result in the precipitation sequence: GP zones are regions enriched with copper atoms; θ' is a metastable phase with a tetragonal structure, disk-shaped, precipitating along the {100} planes of the matrix, possessing coherent/semi-coherent interfaces with specific orientation relationships to the matrix; θ is a stable phase with a tetragonal structure, irregular in shape." }, { "idx": 984, "question": "What is the driving force for solid-state phase transformation?", "answer": "The driving force for solid-state phase transformation is the free energy difference between the new and old phases." }, { "idx": 985, "question": "Both allotropic transformation and recrystallization transformation occur through nucleation and growth mechanisms. What are the differences between them?", "answer": "Allotropic transformation is a phase transition process, where the derivative of a certain thermodynamic quantity becomes discontinuous; recrystallization transformation is merely the reformation of grains and not a phase transition process." }, { "idx": 986, "question": "What are the characteristics of solid-state phase transformation in metals in terms of growth?", "answer": "Exhibits habit plane phenomena, with specific microstructural morphologies such as plate-like or needle-like." }, { "idx": 987, "question": "How to eliminate or improve macrostructure defects in steel ingots?", "answer": "Macro defects (chemical inhomogeneity, physical inhomogeneity, and structural inhomogeneity) are often interrelated. Generally, it is desirable to have as many and fine central equiaxed crystals as possible. Methods such as adding inoculants, increasing cooling rate, and enhancing liquid movement (e.g., electromagnetic stirring, mechanical stirring) can be used to refine grains and eliminate columnar crystals. In this way, macro segregation, shrinkage cavities, and gas bubbles associated with columnar/dendritic crystal zones are significantly improved." }, { "idx": 988, "question": "What are the common macrostructural defects in steel ingots?", "answer": "Macro defects include: macrosegregation (such as normal segregation, inverse segregation, gravity segregation), banded structure, as well as shrinkage cavities, porosity, and blowholes. Strictly speaking, it also includes the structural inhomogeneity of the three-crystal zone." }, { "idx": 989, "question": "What is the driving force of recovery recrystallization?", "answer": "The driving force of recovery recrystallization is deformation stored energy." }, { "idx": 990, "question": "Please briefly describe the basic conditions for the crystallization of binary alloys.", "answer": "Thermodynamic condition: $\\\\Delta G<0$; structural condition: $r>r^{*}$; energy condition: $A>\\\\Delta G_{\\\\mathrm{max}}$; composition condition." }, { "idx": 991, "question": "Describe the common internal and external interfaces in metal crystals.", "answer": "They include grain boundaries, phase boundaries, surfaces, twin boundaries, and stacking faults. Grain boundaries are the interfaces between grains of the same type; phase boundaries are the interfaces between phases with different structures and compositions; surfaces are the interfaces between crystals and the atmosphere or external environment; twin boundaries are new interfaces formed after twinning, which are special high-angle grain boundaries and can be coherent or semi-coherent; low-energy stacking faults are new interfaces formed within a single-phase crystal due to anomalous changes in stacking sequence, which are also low-energy interfaces with energy levels similar to twin boundaries." }, { "idx": 992, "question": "Please briefly describe the microscopic mechanisms of diffusion.", "answer": "Substitutional mechanism: including vacancy mechanism and direct exchange or ring exchange mechanisms, among which the vacancy mechanism is the primary one, while direct exchange and ring exchange mechanisms require high activation energy and generally only occur at high temperatures. Interstitial mechanism: including interstitial mechanism and interstitialcy mechanism, among which the interstitial mechanism is the primary one." }, { "idx": 993, "question": "What are the factors that affect diffusion?", "answer": "The main factors affecting diffusion include: temperature (the higher the temperature, the faster the diffusion rate); crystal structure and type (including packing density, solid solubility, anisotropy, etc.); crystal defects; chemical composition (including concentration, third element, etc.)." }, { "idx": 994, "question": "Please briefly describe the mechanism of low-temperature recovery and its driving force", "answer": "Low-temperature mechanism: corresponds to the disappearance of vacancies. Driving force: release of stored energy (mainly lattice distortion energy) generated during cold deformation." }, { "idx": 995, "question": "What is the difference between kinks and jogs produced when two dislocations intersect?", "answer": "The intersection of dislocations belongs to the interaction between dislocations, resulting in a bend on each other's dislocation line with a magnitude and direction equal to its Burgers vector. This bend is called a kink or a jog. A kink is the bend produced after intersection that lies on the original slip plane and does not affect the movement of the dislocation, making it easy to disappear. A jog is the bend that does not lie on the original slip plane and affects the slip of the dislocation." }, { "idx": 996, "question": "Please briefly describe the mechanism of medium-temperature recovery and its driving force", "answer": "Medium-temperature mechanism: corresponds to the slip (rearrangement, annihilation) of dislocations. Driving force: the release of stored energy (mainly lattice distortion energy) generated during cold deformation." }, { "idx": 997, "question": "Please briefly describe the mechanism of high-temperature recovery and its driving force", "answer": "High-temperature mechanism: corresponds to polygonization (dislocation slip + climb). Driving force: release of stored energy (mainly lattice distortion energy) generated during cold deformation." }, { "idx": 998, "question": "Please analyze the characteristics and mechanisms of work hardening.", "answer": "Work hardening: is the strengthening caused by the multiplication of dislocations due to plastic deformation." }, { "idx": 999, "question": "Please analyze the characteristics and mechanisms of grain refinement strengthening.", "answer": "Grain refinement strengthening: It is caused by the increase in the number of grains and the decrease in their size, which increases the resistance to continuous dislocation slip, leading to strengthening. At the same time, the dispersion of slip also enhances plasticity. This strengthening mechanism is the only one that can simultaneously increase both strength and plasticity." }, { "idx": 1000, "question": "Assuming that the slip system that can be activated in a face-centered cubic crystal is (111)[011], if the slip dislocation is a pure edge dislocation, please indicate the direction of its dislocation line.", "answer": "The direction of the dislocation line for a pure edge dislocation is perpendicular to $\\vec{b}$ and lies on the slip plane, which is [211]." }, { "idx": 1001, "question": "Assuming that the active slip system in a face-centered cubic crystal is (111)[011], if the slip dislocation is a pure screw dislocation, what is the direction of the dislocation line?", "answer": "The dislocation line of a pure screw dislocation is parallel to $\\\\vec{b}$, which is [011]" }, { "idx": 1002, "question": "Please analyze the characteristics and mechanisms of solid solution strengthening.", "answer": "Solid solution strengthening: Due to solute atoms impeding dislocation motion. Includes elastic interaction (Cottrell atmosphere), electrical interaction (Suzuki atmosphere), and chemical interaction." }, { "idx": 1003, "question": "Assuming that the active slip system in a face-centered cubic crystal is (111)[011], please provide the Burgers vector of the unit dislocation for the slip.", "answer": "The Burgers vector of the unit dislocation is ${\\vec{b}}={\\frac{a}{2}}$ [011]" }, { "idx": 1004, "question": "Please analyze the characteristics and mechanisms of dispersion strengthening (age strengthening).", "answer": "Dispersion strengthening (age strengthening): It is the strengthening caused by fine dispersed second phases hindering dislocation motion. This includes the shearing mechanism and the bypassing mechanism." }, { "idx": 1005, "question": "Please analyze the characteristics and mechanisms of multiphase strengthening.", "answer": "Multiphase strengthening: When the relative content of the second phase is on the same order of magnitude as the matrix, strengthening occurs. The degree of strengthening depends on the quantity, size, distribution, and morphology of the second phase, and if the strength of the second phase is lower than that of the matrix, it may not contribute to strengthening." }, { "idx": 1006, "question": "If a piece of iron is heated from room temperature 20°C to 850°C, then cooled very quickly to 20°C, calculate the change in the number of vacancies before and after the treatment (assuming the energy required to form 1 mole of vacancies in iron is 104600J).", "answer": "$$ \\\\cdot\\\\frac{c_{850\\\\mathrm{PC}}}{c_{20\\\\mathrm{PC}}}=\\\\frac{A\\\\mathrm{e}^{-\\\\frac{\\\\Delta E}{k\\\\cdot(850+273)}}}{A\\\\mathrm{e}^{-\\\\frac{\\\\Delta E}{k\\\\cdot(20+273)}}}=\\\\mathrm{e}^{-\\\\frac{104675}{1123\\\\times8.31}\\\\frac{104675}{293\\\\times8.31}}=6.3\\\\times10^{13}$$" }, { "idx": 1007, "question": "From the perspective of bonding, analyze the characteristics of metallic materials", "answer": "Metallic materials: Primarily bonded by metallic bonds, most metals exhibit high strength and hardness, along with good plasticity." }, { "idx": 1008, "question": "From the perspective of bonding, analyze the characteristics of polymer materials", "answer": "Polymer materials: primarily covalent bonds within molecules, with intermolecular forces mainly consisting of molecular bonds and hydrogen bonds" }, { "idx": 1009, "question": "What are the common characteristics of the two major types of ceramic crystalline phases?", "answer": "(1) The bonding is primarily ionic, with a certain proportion of covalent bonding; (2) They have a definite composition that can be expressed by an exact chemical formula; (3) They exhibit typical non-metallic properties." }, { "idx": 1010, "question": "From the perspective of bonding types, analyze the characteristics of composite materials", "answer": "Composite materials: artificial combinations of the above three basic materials, with a wide variety of bonding types and significant performance differences." }, { "idx": 1011, "question": "What are the two major categories of ceramic crystal phases?", "answer": "Oxide ceramics and silicate ceramics." }, { "idx": 1012, "question": "What are the characterization methods for dislocation density?", "answer": "There are two methods: volume density, which is the length of dislocation lines per unit volume; and surface density, which is the number of dislocation lines perpendicularly passing through a unit area." }, { "idx": 1013, "question": "From the perspective of bonding, analyze the characteristics of ceramic materials", "answer": "Ceramic materials: primarily covalent and ionic bonds, hard, brittle, difficult to deform, high melting point." }, { "idx": 1014, "question": "Compare the activation energies of substitutional diffusion and interstitial diffusion.", "answer": "In comparison, the activation energy of interstitial diffusion is smaller." }, { "idx": 1015, "question": "For cold-rolled pure copper sheets, if high strength is required, what heat treatment should be performed?", "answer": "To maintain high strength, low-temperature annealing should be performed to allow only recovery and eliminate residual stresses." }, { "idx": 1016, "question": "For cold-rolled pure copper sheets, what heat treatment should be performed if further cold rolling to reduce thickness is required?", "answer": "To continue cold deformation, high-temperature annealing should be conducted to induce recrystallization and soften the microstructure." }, { "idx": 1017, "question": "What is the physical meaning of diffusion activation energy?", "answer": "The physical meaning of diffusion activation energy is the energy barrier that must be overcome during atomic jumps, which is the obstruction posed by surrounding atoms." }, { "idx": 1018, "question": "Analyze the effect of cold plastic deformation on the microstructure of alloys", "answer": "1) Microstructure: 1 Formation of fibrous structure, grains are elongated along the deformation direction; 2 Formation of dislocation cells; 3 Grain rotation forms deformation texture." }, { "idx": 1019, "question": "Given a certain low-carbon steel with $\\sigma_{0}=64\\mathrm{kPa}$, $K=393.7\\mathbf{k}\\mathbf{Pa}\\cdot\\mu\\mathbf{m}^{\\frac{1}{2}}$, and a grain diameter of $50\\mu\\mathrm{m}$, what is the yield strength of this low-carbon steel?", "answer": "According to the Hall-Petch formula: $$\\sigma_{\\mathrm{s}}=\\sigma_{0}+K d^{-{\\frac{1}{2}}}=(64+393.7\\times50^{-{\\frac{1}{2}}})\\mathbf{kPa}=(64+55.68)\\mathbf{kPa}=119.7\\mathbf{kPa}$$" }, { "idx": 1020, "question": "Given that for carbon diffusion in γ-Fe, D0=2.0×10^(-5)m²/s, Q=1.4×10^5J/mol, gas constant R=8.314J/(mol·K), calculate the diffusion coefficient D927°C at 927°C.", "answer": "D927°C = D0 * exp(-Q / (R * T)) = 2.0×10^(-5) * exp(-1.4×10^5 / (8.314 * (927 + 273))) = 2.0×10^(-5) * exp(-1.4×10^5 / (8.314 * 1200)) = 2.0×10^(-5) * exp(-14.033) ≈ 2.0×10^(-5) * 8.315×10^(-7) ≈ 1.663×10^(-11) m²/s" }, { "idx": 1021, "question": "Calculate the packing density of the most densely packed plane in a BCC crystal.", "answer": "The most densely packed plane in BCC is the {110} plane, and its area is: $A=a\\\\times{\\\\sqrt{2}}a={\\\\sqrt{2}}a^{2}$ \\n\\nThe area occupied by atoms on the {110} plane (two atoms) is: $$A^{\\\\prime}=2\\\\times\\\\pi R^{2}=2\\\\pi{\\\\left(\\\\frac{\\\\sqrt{3}}{4}a\\\\right)}^{2}=\\\\frac{3}{8}\\\\pi a^{2}$$ The packing density: $d=\\\\frac{A^{\\\\prime}}{A}=\\\\frac{3}{8\\\\sqrt{2}}\\\\pi=0.8332$" }, { "idx": 1022, "question": "Given that for carbon diffusion in γ-Fe, D0=2.0×10^(-5)m²/s, Q=1.4×10^5J/mol, gas constant R=8.314J/(mol·K), calculate the diffusion coefficient D1027°C at 1027°C.", "answer": "D1027°C = D0 * exp(-Q / (R * T)) = 2.0×10^(-5) * exp(-1.4×10^5 / (8.314 * (1027 + 273))) = 2.0×10^(-5) * exp(-1.4×10^5 / (8.314 * 1300)) = 2.0×10^(-5) * exp(-12.953) ≈ 2.0×10^(-5) * 2.394×10^(-6) ≈ 4.788×10^(-11) m²/s" }, { "idx": 1023, "question": "Analyze the effect of cold plastic deformation on the energy of the alloy system", "answer": "(4) The system energy includes two parts: 1 The lattice distortion caused by a large number of defects generated during cold deformation increases the distortion energy; 2 The microscopic and macroscopic internal stresses caused by uneven deformation between grains and different parts of the workpiece. These two parts are collectively referred to as stored energy, with the former being the primary. The changes in microstructure and properties caused by cold deformation prepare the alloy for subsequent recovery and recrystallization in terms of both structure and energy." }, { "idx": 1024, "question": "Analyze the effect of cold plastic deformation on the mechanical properties of alloys", "answer": "(2) Mechanical properties: The dislocation density increases, dislocations become entangled with each other, and the resistance to movement increases, resulting in work hardening." }, { "idx": 1025, "question": "Calculate the multiple of change in the diffusion coefficient when the temperature increases from 927°C to 1027°C.", "answer": "Multiple of change = D1027°C / D927°C = (4.788×10^(-11)) / (1.663×10^(-11)) ≈ 2.88 times" }, { "idx": 1026, "question": "Analyze the effect of cold plastic deformation on the physical and chemical properties of alloys", "answer": "(3) Physical and chemical properties: The changes are complex, mainly affecting electrical conductivity, thermal conductivity, chemical activity, chemical potential, etc." }, { "idx": 1027, "question": "Analyze the characteristics of grain boundaries from the structural features of grain boundaries", "answer": "Structural features of grain boundaries: The atomic arrangement is relatively disordered and contains a large number of defects. Characteristics of grain boundaries: (3) Numerous defects such as dislocations and vacancies—high grain boundary diffusion rate. (6) Enrichment of trace elements and impurities." }, { "idx": 1028, "question": "Analyze the characteristics of grain boundaries from the perspective of their energy features", "answer": "Energy features of grain boundaries: Atoms have higher energy and stronger activity compared to the grain interior. Characteristics of grain boundaries: (1) Grain boundary—distortion—grain boundary energy—transformation to a lower energy state—grain growth, grain boundary straightening—reduction in grain boundary area. (2) Hindering dislocation movement—increase in σb—grain refinement strengthening. (4) High grain boundary energy and complex structure—easily meeting the conditions for solid-state phase transformation—primary site for solid-state phase transformation. (5) Poor chemical stability—grain boundaries are prone to corrosion." }, { "idx": 1029, "question": "Dislocation", "answer": "Dislocation: A one-dimensional or linear defect in a crystal, known as a dislocation." }, { "idx": 1030, "question": "Coordination number", "answer": "Coordination number: In a crystal structure, the number of atoms that are adjacent to and equidistant from any given atom." }, { "idx": 1031, "question": "Crystal face family", "answer": "Crystal face family: Groups of equivalent crystal faces with the same symmetrical relationship (atomic arrangement and distribution, interplanar spacing) are called a crystal face family, denoted by hll." }, { "idx": 1032, "question": "Reaction diffusion", "answer": "Reaction diffusion: The phenomenon of forming a new phase through diffusion." }, { "idx": 1033, "question": "Heterogeneous nucleation", "answer": "Heterogeneous nucleation: The formation of crystal nuclei in liquid metal preferentially occurs on the surfaces of foreign substances (mold walls or impurities) or at locations with temperature inhomogeneity." }, { "idx": 1034, "question": "Compositional supercooling", "answer": "Compositional supercooling: During the solidification of an alloy solution, the theoretical solidification temperature remains unchanged, and the degree of supercooling is entirely determined by the distribution of solute components. This type of supercooling is called compositional supercooling." }, { "idx": 1035, "question": "Pseudoeutectic", "answer": "Pseudoeutectic: The eutectic structure obtained from an alloy of non-eutectic composition is called pseudoeutectic." }, { "idx": 1036, "question": "Recrystallization", "answer": "Recrystallization: The process by which deformed metal, under certain heating conditions, forms a new strain-free grain structure through the formation and subsequent movement of new mobile high-angle grain boundaries." }, { "idx": 1037, "question": "Please indicate the method and principle of vibration refinement for grain refinement in metallic materials.", "answer": "Vibration refinement. By solidifying the molten metal under vibration, the nucleation rate of the liquid phase is increased on one hand, and the growing crystals are fragmented on the other hand, thereby providing more crystallization nuclei, thus achieving the purpose of grain refinement." }, { "idx": 1038, "question": "Law of the center of gravity", "answer": "Law of the center of gravity: In a ternary alloy phase diagram, if $R$ decomposes into three phases $\\alpha,~\\beta,$ $\\gamma$, then the concentration point of $R$ phase must be located at the center of gravity of $\\Delta\\alpha\\beta\\gamma$ (which is the weight center of gravity of the three phases, not the geometric center of the triangle). Moreover, the weight of $R$ phase has the following relationship with the weights of $\\alpha$, $\\beta$, and $\\gamma$ phases: $$ \\begin{array}{r l}&{w_{R}\\times R d=w_{\\alpha}\\times\\alpha d}\\ &{w_{R}\\times R e=w_{\\beta}\\times\\beta e}\\ &{w_{R}\\times R f=w_{\\gamma}\\times\\gamma f}\\end{array}$$" }, { "idx": 1039, "question": "Cross-slip", "answer": "Cross-slip: The process of moving from one slip plane to another while the Burgers vector of the dislocation line remains unchanged." }, { "idx": 1040, "question": "Please indicate the method of adding nucleating agents to refine the grain size of metal materials and its principle.", "answer": "Adding nucleating agents. Before the melt solidifies, some fine nucleating agents are added and dispersed in the melt to serve as ready-made substrates required for heterogeneous nucleation, thereby significantly increasing the number of nuclei and markedly refining the grain size." }, { "idx": 1041, "question": "Analyze the reason why the solubility of carbon in austenite is greater than that in ferrite.", "answer": "Austenite is a face-centered cubic crystal, while ferrite is a body-centered cubic crystal. In both face-centered cubic and body-centered cubic structures, carbon atoms are located in their octahedral interstitial sites. The size of the octahedral interstitial site in face-centered cubic is: $0.535\\\\mathring{\\\\mathbf{A}}^{\\\\odot}$, while in body-centered cubic, it is: 0.129A. Therefore, it can be seen that the solubility of carbon in austenite is much greater than that in ferrite." }, { "idx": 1042, "question": "Please list more than three methods to improve the strength of metal materials and explain their principles.", "answer": "(1) Grain refinement strengthening. The principle is: increasing the number of grains to enhance the hindering effect of grain boundaries on moving dislocations, thereby achieving strengthening. (2) Solid solution strengthening. This involves dissolving solute atoms into the base metal, causing lattice distortion in the base metal, which inhibits the activity of dislocation sources and improves the strength of the base metal. (3) Work hardening. When a crystal undergoes deformation, dislocations inside the crystal pile up or become entangled, making them difficult to move, thereby strengthening the crystal." }, { "idx": 1043, "question": "Write all the crystallographic directions included in the <110> direction family in the cubic crystal system.", "answer": "[110], [101], [011], [110], [101], [011], totaling 6 crystallographic directions." }, { "idx": 1044, "question": "Write all the crystallographic directions included in the <123> direction family in the cubic crystal system.", "answer": "[123], [132], [231], [213], [321], [312], [123], [132], [231], [213], [321], [312], [123], [132], [231], [213], [321], [312], [123], [132], [231], [213], [321], [312], totaling 24 crystallographic directions." }, { "idx": 1045, "question": "Analyze the reasons for the formation of the iron-carbon dual phase diagram from a thermodynamic perspective", "answer": "Thermodynamic analysis reveals that the total free energy of graphite is lower than that of Fe3C. Therefore, graphite is the more stable phase, while Fe3C is a metastable phase. Hence, the Fe-Fe3C system is commonly referred to as the metastable system, and the Fe-C system is called the stable system." }, { "idx": 1046, "question": "Analyze the equilibrium crystallization process of the Fe-0.4%C alloy and indicate the phase composition at room temperature", "answer": "When C%=0.4%, the Fe-C alloy cools to room temperature, and the phase constituents are α+Fe3C" }, { "idx": 1047, "question": "Please indicate the methods and principles of refining metal grain size by increasing undercooling.", "answer": "Increase undercooling. Since the number of grains is directly proportional to the nucleation rate N and inversely proportional to the grain growth rate Vg, increasing N/Vg can refine the grains. Moreover, when undercooling is increased, although both N and Vg increase, the rate of increase in N is faster than that in Vg. Therefore, N/Vg increases with the increase in undercooling, meaning the grains become finer." }, { "idx": 1048, "question": "Analyze the cause of the iron-carbon dual phase diagram from a kinetic perspective", "answer": "From a kinetic analysis, since Fe3C contains 6.69% carbon, while graphite contains 100% carbon, and the carbon content of commonly used steel materials is less than 5%. Thus, the compositional fluctuation required to form the graphite phase is much larger than that for Fe3C, meaning that forming graphite nuclei is much more difficult than forming Fe3C nuclei." }, { "idx": 1049, "question": "Analyze the equilibrium crystallization process of the Fe-0.4%C alloy and indicate the microstructure constituents at room temperature", "answer": "When C%=0.4%, the microstructure constituents of the Fe-C alloy upon cooling to room temperature are P+α" }, { "idx": 1050, "question": "Calculate the content of pearlite in the microstructure of an Fe-0.4%C alloy", "answer": "The content of pearlite in the microstructure: P%=[(0.4-0.0218)/(0.77-0.0218)]×100%=50.55%" }, { "idx": 1051, "question": "Calculate the graphite content in Fe-3.6%C alloy", "answer": "The graphite content is: wG = (3.6 - 0.68) / (100 - 0.68) × 100% ≈ 2.94%." }, { "idx": 1052, "question": "Analyze the crystallization process of Fe-3.6%C alloy to obtain a pearlite matrix gray iron structure", "answer": "For the Fe-3.6%C alloy, to obtain a pearlite matrix gray cast iron structure, according to the Fe-graphite phase diagram, the crystallization process is as follows: L↔γA, L↔(γ+G)3, γ↔G, γ↔(α+Fe3C)JE." }, { "idx": 1053, "question": "If the aforementioned alloy undergoes directional solidification in a 100cm long horizontal round mold, assuming no diffusion in the solid phase during solidification and complete mixing of the liquid composition, and the lines in the phase diagram can be simplified as straight lines, calculate the length of the ledeburite structure at the end of solidification. It is known that the length of the δ phase is 28.75cm, the length of the γ phase is 70.08cm, and the total length is 100cm.", "answer": "The length of the ledeburite structure is: (100-28.75-70.08)cm=1.17cm." }, { "idx": 1054, "question": "Coordination number", "answer": "Coordination number: The number of nearest neighbor and equidistant atoms surrounding each atom in a crystal lattice." }, { "idx": 1055, "question": "If the above alloy undergoes directional solidification in a horizontal round mold with a length of 100 cm, assuming no diffusion in the solid phase during solidification and complete mixing of the liquid composition, and the lines in the phase diagram can be simplified as straight lines, calculate the length of the δ phase at the end of solidification. Given C0=0.4, K0=0.09/0.53=0.1698.", "answer": "Using the formula: Cs=K0C0(1-Z/L)^(K0-1). Substituting the known conditions into the formula gives: 0.09=0.1698×0.4×(1-Z1/100)^(0.1698-1). Calculation yields: 0.09/0.06792=(1-Z1/100)^(-0.8302). Further calculation: 1.325=(1-Z1/100)^(-0.8302). (1-Z1/100)^0.8302=1/1.325=0.7547. 1-Z1/100=0.7125. Z1=28.75 cm." }, { "idx": 1056, "question": "If the aforementioned alloy undergoes directional solidification in a horizontal cylindrical mold with a length of 100 cm, assuming no diffusion in the solid phase during solidification and complete mixing of the liquid composition, and the lines in the phase diagram can be simplified as straight lines, calculate the length of the γ phase at the end of solidification. Given L2=(100-28.75)cm=71.25cm, C0'=0.53, K0'=2.11/4.3=0.4907.", "answer": "Using the formula: Cs=K0C0(1-Z/L)^(K0-1). Substituting the known conditions into the formula gives: 2.11=0.4907×0.53×(1-Z2/71.25)^(0.4907-1). Calculation yields: 2.11/0.26007=(1-Z2/71.25)^(-0.5093)=8.1132. (1-Z2/71.25)^0.5093=1/8.1132=0.12326. 1-Z2/71.25=0.0164. Z2=70.08 cm." }, { "idx": 1057, "question": "Uphill diffusion", "answer": "Uphill diffusion: The migration of solute atoms in the opposite direction of the concentration gradient, that is, from a low-concentration region to a high-concentration region, is called uphill diffusion." }, { "idx": 1058, "question": "Interstitial phase", "answer": "Interstitial phase: A compound formed between non-metal atoms (X) and transition metal atoms, when the ratio of atomic radii of the metal element (M) to the non-metal element (X) $R_{\\\\mathrm{X}}/R_{\\\\mathrm{M}}<0.59$, the non-metal atoms are located in the interstices of the metal atoms and exhibit a relatively simple crystal structure, it is called an interstitial phase." }, { "idx": 1059, "question": "Critical nucleus", "answer": "Critical nucleus: In a metal liquid, an embryo with a radius larger than the critical radius rk can stably grow and is called a critical nucleus." }, { "idx": 1060, "question": "Phase boundary contact rule", "answer": "Phase boundary contact rule: In a phase diagram, the difference in the number of phases between adjacent phase regions is 1." }, { "idx": 1061, "question": "Write all the crystal planes included in the {110} crystal plane family in the cubic crystal system", "answer": "(110)(101)(011)(110)(101)(011), totaling 6." }, { "idx": 1062, "question": "Write all the crystal planes included in the {111} crystal plane family in the cubic crystal system", "answer": "(111)(111)(1 11)(111), totaling 4." }, { "idx": 1063, "question": "Write all the crystal planes included in the {112} crystal plane family of the cubic crystal system", "answer": "(112)(112)(1 12)(11 2) + (121)(121)(1 21)(12 1) + (211)(211)(2 11)(21 1), totaling 12." }, { "idx": 1064, "question": "If the slip plane of a body-centered cubic crystal is {110} and the slip direction is [111], write out the specific slip systems.", "answer": "(110)[1 11], (10 1)[1 11], (011)[1 11]" }, { "idx": 1065, "question": "Write all the crystal planes included in the {123} crystal plane family in the cubic crystal system", "answer": "(123)(123)(123)(123) + (132)(132)(132)(132) + (213)(213)(213)(213) + (231)(231)(231)(231) + (312)(312)(312)(312) + (321)(321)(321)(321), totaling 24." }, { "idx": 1066, "question": "If the slip plane of a body-centered cubic crystal is {112} and the slip direction is [111], write out the specific slip systems.", "answer": "(112)[1 11], (121)[1 11], (21 1)[1 11]" }, { "idx": 1067, "question": "Given that copper has an FCC structure with an atomic radius of $0.1278\\\\mathrm{nm}$, calculate its density. (The atomic weight of copper is 63.5, and the Avogadro constant is $0.602\\\\times10^{24}$)", "answer": "Given that the Cu lattice has an FCC structure, meaning the unit cell contains 4 atoms, to find its density ρ, we only need to calculate the ratio of the mass of the atoms in the unit cell to the volume of the unit cell: $$\\\\rho={\\\\frac{n M}{N_{A}(2{\\\\sqrt{2}}r)^{3}}}={\\\\frac{4\\\\times63.5\\\\times10^{-3}}{0.602\\\\times10^{24}\\\\times(2{\\\\sqrt{2}}\\\\times0.1278\\\\times10^{-9})^{3}}}\\\\mathrm{kg/m^{3}}\\\\approx8933\\\\mathrm{kg/m^{3}}$$" }, { "idx": 1068, "question": "Describe the phase transformation process, products, and reaction conditions along the HJB horizontal line in the phase diagram", "answer": "The HJB horizontal line undergoes a peritectic reaction: L_B + δ_H ⇌ γ_J; product: austenite (A)" }, { "idx": 1069, "question": "If the slip plane of a body-centered cubic crystal is {123} and the slip direction is [111], write out the specific slip systems.", "answer": "(123)[1 11], (213)[1 11], (231)[1 11], (31 2)[1 11], (132)[1 11], (32 1)[1 11]" }, { "idx": 1070, "question": "Assuming carbon atoms are all located in the octahedral interstices of γ-Fe, calculate the maximum proportion of octahedral interstices occupied by carbon atoms in γ-Fe. (The atomic weight of iron is 55.85, and the Avogadro constant is 0.602×10^24)", "answer": "In the γ-Fe phase, the maximum carbon concentration C_max = 2.11%. Given that γ-Fe has an FCC structure with n = 4, there are 400 Fe atoms per 100 unit cells. The mass fraction of Fe is: w_Fe = 1 - 2.11% = 97.89%. The total mass is: M = (400 × 55.85) / 0.9789 = 22821.5. The number of carbon atoms is: n_c = (M × C_max) / M_c = (22821.5 × 0.0211) / 12 = 40. Thus, there are 40 carbon atoms per 100 unit cells." }, { "idx": 1071, "question": "Describe the phase transformation process, products, and reaction conditions along the ECF line in the phase diagram", "answer": "The ECF line undergoes a eutectic reaction: L_C ⇌ γ_E + Fe3C; the product is: ledeburite (Ld)" }, { "idx": 1072, "question": "Describe the phase transformation process, products, and reaction conditions along the PSK line in the phase diagram", "answer": "The PSK line undergoes a eutectoid reaction: γ_S ⇌ α_P + Fe3C; product: pearlite (P)" }, { "idx": 1073, "question": "Density", "answer": "Density: The ratio of the volume occupied by atoms in a unit cell to the volume of the unit cell." }, { "idx": 1074, "question": "Non-steady state diffusion", "answer": "Non-steady state diffusion: diffusion where both the concentration gradient and the diffusion flux vary with time and distance." }, { "idx": 1075, "question": "Equilibrium distribution coefficient", "answer": "Equilibrium distribution coefficient: At a certain temperature, the ratio of the compositions of the solid and liquid phases when the two phases are in equilibrium, i.e., $k_{0}=C_{\\\\mathrm{S}}/$ $C_{\\\\mathrm{{L}}}$" }, { "idx": 1076, "question": "Critical nucleation work", "answer": "Critical nucleation work: the increment of free energy when forming a critical nucleus." }, { "idx": 1077, "question": "Calculate the relative content of each phase when an alloy containing 0.40% C is cooled to room temperature under equilibrium conditions.", "answer": "Relative content of each phase: w_α=(6.69-0.40)/(6.69-0.0008)×100%=94%; w_Fe3C=1-w_α=1-94%=6%" }, { "idx": 1078, "question": "Calculate the relative content of each microstructural constituent in an alloy containing 0.40% C when it is cooled to room temperature under equilibrium conditions.", "answer": "The relative content of each microstructural constituent: w_P=(0.40-0.0218)/(0.77-0.0218)×100%=50.5%; w_a=1-w_P=1-50.5%=49.5%" }, { "idx": 1079, "question": "Extended dislocation", "answer": "Extended dislocation: In FCC crystals, a perfect dislocation is difficult to exist due to its high energy and will automatically decompose into two partial dislocations and a stacking fault in between, forming the so-called extended dislocation configuration." }, { "idx": 1080, "question": "Write all the crystal planes included in the {110} plane family of the cubic crystal system.", "answer": "(110), (101), (011), (110), (101), (011)." }, { "idx": 1081, "question": "Write all the crystal planes included in the {111} crystal plane family in the cubic crystal system.", "answer": "(111), (1̄11), (11̄1), (111̄)." }, { "idx": 1082, "question": "Write all the crystal planes included in the {112} plane family of the cubic crystal system.", "answer": "(112), (121), (211), (112), (121), (211), (112), (121), (211), (112), (121), (211)." }, { "idx": 1083, "question": "Write all the crystal planes included in the {123} crystal plane family of the cubic crystal system.", "answer": "(123), (123), (123), (123), (132), (132), (132), (132), (213), (213), (213), (213), (231), (231), (231), (231), (312), (312), (312), (312), (321), (321), (321), (321)." }, { "idx": 1084, "question": "Are the shapes of the room-temperature tensile curves of low-carbon steel and aluminum alloy the same?", "answer": "The shapes of the room-temperature tensile curves of low-carbon steel and aluminum alloy are not the same." }, { "idx": 1085, "question": "Calculate the relative content of each phase when an alloy containing 0.40% C is cooled to room temperature under equilibrium conditions.", "answer": "The room temperature microstructure of an alloy containing 0.4% C consists of ferrite and cementite. According to the lever rule, the relative content of ferrite = (6.69-0.40)/6.69=94%, and the relative content of cementite =1-94%=6%." }, { "idx": 1086, "question": "For annealed low-carbon steel with a grain size of NA=16 grains/mm², its yield strength σs=100MPa; when NA=4096 grains/mm², σs=250MPa. Calculate the yield strength when NA=250 grains/mm².", "answer": "According to the Hall-Petch formula: $\\sigma_{\\mathrm{s}}=\\sigma_{0}+K d^{-1/2}$. Since $N_{A}$ is inversely proportional to the square of the grain diameter $\\pmb{d}$, we have: Furthermore, it can be solved that $$\\begin{array}{c}{{\\displaystyle{\\sigma_{\\mathrm{s}}=100}{\\mathrm{MPa}}=\\sigma_{0}+K{d_{1}}^{-1/2}}}\\ {{\\displaystyle{\\sigma_{\\mathrm{s}}=250}{\\mathrm{MPa}}=\\sigma_{0}+K{d_{2}}^{-1/2}}}\\ {{\\displaystyle{\\sigma_{\\mathrm{s}}=\\sigma_{0}+K{d_{3}}^{-1/2}}}}\\ {{\\displaystyle{d_{2}}/d_{1}=1/16\\qquadd_{3}/d_{1}=4/5~\\sqrt{10}}}\\ {{\\displaystyle{\\sigma_{\\mathrm{s}}=149}{\\mathrm{MPa}}}}\\end{array}$$" }, { "idx": 1087, "question": "What are the characteristics of the room temperature tensile curve of low-carbon steel?", "answer": "Low-carbon steel exhibits distinct upper and lower yield points due to the presence of Cottrell atmospheres." }, { "idx": 1088, "question": "Assuming carbon atoms are all located in the octahedral interstices of γ-Fe, at 1000°C, 1.7 weight percent of carbon atoms are dissolved in γ-Fe. Calculate how many carbon atoms are present in 100 unit cells and determine the proportion of octahedral interstices occupied by carbon atoms (the atomic weight of iron is 55.85, and the atomic weight of carbon is 12.01).", "answer": "γ-Fe has a face-centered cubic structure, with 4 Fe atoms and 4 octahedral interstices per unit cell. Let the number of carbon atoms in one unit cell be $A$, then: Solving gives $A=0.32$ $$\\frac{100A\\times12.01}{100\\times4\\times55.85+100A\\times12.01}=1.7\\%$$ Thus, the number of carbon atoms in 100 unit cells is: $0.32\\times100=32$ The proportion of octahedral interstices occupied by C atoms: $$A/4=0.32/4=0.08=8\\%$$" }, { "idx": 1089, "question": "Non-steady state diffusion", "answer": "Non-steady state diffusion: diffusion where both the concentration gradient and diffusion flux vary with time and distance" }, { "idx": 1090, "question": "Critical resolved shear stress", "answer": "Critical resolved shear stress: When $\\sigma_{0}=\\sigma_{\\mathfrak{s}}$, the crystal begins to slip, and the resolved shear stress in the slip direction at this point is called the critical resolved shear stress." }, { "idx": 1091, "question": "What are the characteristics of the room temperature tensile curve of aluminum alloy?", "answer": "Aluminum alloy does not exhibit a distinct yield point phenomenon." }, { "idx": 1092, "question": "Coordination number", "answer": "Coordination number: In a crystal structure, the number of neighboring atoms equidistant from any given atom." }, { "idx": 1093, "question": "Cottrell atmosphere", "answer": "Cottrell atmosphere: The yield point observed in certain plastic materials is related to the presence of trace interstitial elements in the metal. Under suitable temperature and time conditions, these interstitial atoms aggregate on the expanded side of the dislocation line. As a result, a cluster of interstitial atoms forms near the dislocation line, creating what is known as the Cottrell atmosphere." }, { "idx": 1094, "question": "Divorced eutectic", "answer": "Divorced eutectic: For some hypoeutectic or hypereutectic alloys with compositions far from the eutectic point, the amount of primary crystals is large while the eutectic amount is small. During eutectic transformation, if the phase adjacent to the primary crystals in the crystal grows attached to the primary crystals, while the remaining phase exists separately at the grain boundaries of the primary crystal grains, causing the eutectic structure characteristics to disappear. This type of separated two-phase eutectic is called divorced eutectic." }, { "idx": 1095, "question": "Cross-slip can enable multiple slip systems in crystals", "answer": "Cross-slip is the process where a dislocation line transfers from one slip plane to another." }, { "idx": 1096, "question": "Interstitial solid solutions and interstitial compounds both belong to interstitial phases.", "answer": "The Burgers vectors of edge dislocations and screw dislocations are perpendicular and parallel to the direction vector of the dislocation line, respectively. For a curved dislocation line, its Burgers vector remains constant, but the properties vary along different sections depending on the geometric relationship between the direction vector of the dislocation line and the Burgers vector." }, { "idx": 1097, "question": "For metals or alloys without phase transformation in the solid state, their grain size cannot be changed unless remelted.", "answer": "For metals or alloys without phase transformation in the solid state, their grain size can also be changed through plastic deformation and recrystallization treatment without remelting." }, { "idx": 1098, "question": "The Burgers vectors of edge dislocations and screw dislocations are perpendicular and parallel to the direction vectors of the dislocation lines, respectively, so a curved dislocation line has different properties.", "answer": "The Burgers vectors of edge dislocations and screw dislocations are perpendicular and parallel to the direction vectors of the dislocation lines, respectively. For a curved dislocation line, its Burgers vector remains constant, but the properties vary along the line as the geometric relationship between the direction vector of the dislocation line and the Burgers vector changes." }, { "idx": 1099, "question": "Calculate the planar density of the (100) plane in a face-centered cubic crystal, given the atomic radius r=(√2/4)a", "answer": "For the (100) plane of a face-centered cubic crystal, there are 2 atoms. The planar density is calculated as: 2×π(√2/4 a)^2 / a^2 = 2π×1/8 = π/4 = 0.785" }, { "idx": 1100, "question": "Calculate the planar density of the (110) plane in a face-centered cubic crystal, given the atomic radius r=(√2/4)a", "answer": "For the (110) plane of a face-centered cubic crystal, there are 2 atoms. The planar density is calculated as: 2×π(√2/4 a)^2 / (√2 a×a) = 2π×1/8 / √2 = √2/8 π = 0.56" }, { "idx": 1101, "question": "A metal test bar with a length of 20cm and a cross-sectional area of 4cm² is fixed at the upper end and subjected to a tensile force of 980N at the lower end. Find the angle θ at which the shear stress is maximum.", "answer": "The angle θ at which the shear stress is maximum is 45°." }, { "idx": 1102, "question": "Calculate the planar density of the (111) plane in a face-centered cubic crystal, given the atomic radius r=(√2/4)a", "answer": "For the (111) plane of a face-centered cubic crystal, there are 2 atoms. The planar density is calculated as: 2π(√2/4 a)^2 / (1/2 √2 a×√2 a×sin60°) = 2π×1/8 / (√3/2) = π/(2√3) = 0.90" }, { "idx": 1103, "question": "A metal test bar with a length of 20cm and a cross-sectional area of 4cm² is fixed at the upper end, and a tensile force of 980N is applied at the lower end. Calculate the normal stress and shear stress on a plane that forms a 0° angle with the central axis of the metal test bar.", "answer": "When θ=0°: normal stress σθ=0; shear stress τθ=0." }, { "idx": 1104, "question": "A metal test bar with a length of 20cm and a cross-sectional area of 4cm² is fixed at the upper end and subjected to a tensile force of 980N at the lower end. Calculate the normal stress and shear stress on a plane that forms a 30° angle with the central axis of the metal test bar.", "answer": "When θ=30°: normal stress σθ=612500Pa; shear stress τθ=1060881Pa." }, { "idx": 1105, "question": "A metal test bar with a length of 20cm and a cross-sectional area of 4cm² is fixed at the upper end and subjected to a tensile force of 980N at the lower end. Calculate the normal stress and shear stress on a plane that forms a 45° angle with the central axis of the metal test bar.", "answer": "When θ=45°: normal stress σθ=1225000Pa; shear stress τθ=1225000Pa." }, { "idx": 1106, "question": "A metal test bar with a length of 20cm and a cross-sectional area of 4cm² is fixed at the upper end and subjected to a tensile force of 980N at the lower end. Determine the general expressions for the normal stress and shear stress on a plane that forms an angle θ° with the central axis of the metal test bar.", "answer": "Normal stress σθ=σsin²θ; shear stress τθ=(σ/2)sin2θ, where σ=P/A=980N/4cm²=2450000Pa." }, { "idx": 1107, "question": "Indicate which plane among (100), (110), and (111) in a face-centered cubic crystal is the close-packed plane", "answer": "The (111) plane has the highest packing density (0.90), so the close-packed plane in a face-centered cubic crystal is the (111) plane" }, { "idx": 1108, "question": "Interstitial compound", "answer": "Interstitial compound: When small solute atoms such as C, H, O, N, etc. are distributed in the interstices of the solvent atomic structure, altering the crystal structure of the solvent atoms, and the composition can generally be represented by a chemical formula, such a phase is called an interstitial compound." }, { "idx": 1109, "question": "Number of atoms per unit cell", "answer": "Number of atoms per unit cell: In the spatial structure of a crystal, a unit cell is adjacent to 26 other unit cells. Therefore, the atoms at the 8 corners, 12 edges, and 6 faces of the unit cell are shared with other unit cells, and only the atoms inside the unit cell entirely belong to this unit cell. The sum of the fractional contributions of the shared atoms and the actual number of atoms belonging to this unit cell is called the number of atoms per unit cell." }, { "idx": 1110, "question": "Degree of supercooling", "answer": "Degree of supercooling: Under rapid cooling conditions, when a metallic material cools to its melting point temperature, it does not solidify immediately but begins to solidify at a lower temperature. The difference between the actual solidification temperature and the melting point temperature is referred to as the degree of supercooling." }, { "idx": 1111, "question": "Kirkendall effect", "answer": "Kirkendall effect: Several turns of thin molybdenum wire are wound around a brass bulk material as marker lines, then a thick layer of pure copper is plated on it, and it is placed in a high-temperature annealing furnace for long-term annealing. After several weeks, it is found that the marker lines have moved a certain distance towards the brass side. This experimental result is called the Kirkendall effect." }, { "idx": 1112, "question": "The Burgers vector of edge dislocations and screw dislocations changes with the direction vector of the dislocation line.", "answer": "Wrong! The Burgers vector of an edge dislocation is perpendicular to the direction vector of the dislocation line; the Burgers vector of a screw dislocation is parallel to the direction vector of the dislocation line." }, { "idx": 1113, "question": "Both austenite and pearlite are solid solutions of carbon in $\\alpha-\\mathrm{\\bfFe}$ with a body-centered cubic structure.", "answer": "Wrong! Austenite is a solid solution of carbon in $\\gamma.$ -Fe with a face-centered cubic structure, while pearlite is a two-phase structure composed of ferrite and cementite." }, { "idx": 1114, "question": "Strain aging", "answer": "Strain aging: After the first stretching, if a second stretching is performed immediately, the yield stage does not appear on the stress-strain curve. However, if the low-carbon steel specimen after the first stretching is left at room temperature for a period of time before the second stretching, the yield stage reappears on the stress-strain curve. Nevertheless, the yield strength upon re-yielding is higher than that during the initial yielding. This experimental phenomenon is called strain aging." }, { "idx": 1115, "question": "The diffusion of solute atoms always proceeds from regions of high concentration to regions of low concentration.", "answer": "Wrong! The accurate statement is that the diffusion of solute atoms always proceeds from regions of high chemical potential to regions of low chemical potential. Only in this way can the driving force of reactive diffusion be satisfactorily explained." }, { "idx": 1116, "question": "Both martensitic transformation and bainitic transformation are diffusionless solid-state phase transformations.", "answer": "Wrong! Martensitic transformation is a diffusionless solid-state phase transformation, while bainitic transformation involves both a diffusionless shear process and a diffusive atomic transport process, they are not the same." }, { "idx": 1117, "question": "In the iron-carbon phase diagram, the horizontal line where the eutectoid reaction occurs has two-phase regions both above and below it, which does not conform to the phase region contact rule.", "answer": "Wrong! The horizontal line where the eutectoid reaction occurs in the iron-carbon phase diagram represents a three-phase equilibrium reaction, which is actually a three-phase region. Thus, the difference in the number of phases between the two-phase regions above and below the horizontal line is still 1, still conforming to the phase region contact rule." }, { "idx": 1118, "question": "In terms of the lattice constant, what is the atomic diameter in an FCC crystal structure?", "answer": "The close-packed direction in an FCC structure is [110]. In the unit cell, there are two atoms along the [110] direction, so the atomic diameter is (√2/2)a." }, { "idx": 1119, "question": "Why cannot the lever rule be used to analyze the relative amounts of phases in a vertical section of a ternary equilibrium phase diagram?", "answer": "The vertical section of a ternary equilibrium phase diagram is a pseudo-binary phase diagram, and the composition points of the equilibrium phases cannot be determined on this vertical section. Therefore, the lever rule cannot be used to calculate the relative amounts of phases on the vertical section of a ternary equilibrium phase diagram." }, { "idx": 1120, "question": "Describe the characteristics of grain growth at temperatures of 760‰ and 870‰ and their engineering applications.", "answer": "At 760‰ held for 1 hour, the grain size only grows to 0.0516mm. Comparing 0.0516mm with 0.05mm indicates that there is little change in this case, and the grains have basically not grown. However, at 870‰ held for 1 hour, the grain size grows to 0.069mm. Comparing 0.069mm with 0.05mm shows a 37% increase in grain size, indicating significant growth in this case. Therefore, when heating at relatively high temperatures, special attention must be paid to grain growth." }, { "idx": 1121, "question": "Phase", "answer": "Phase: In alloys, a homogeneous component with the same state of aggregation, the same crystal structure, essentially the same composition, and clearly defined interfaces separating it from other parts." }, { "idx": 1122, "question": "If the carbon content is 0.8% in high-carbon steel with an initial grain size of 0.05mm, calculate the grain size and its growth value after holding at 870‰ for 1 hour. The formula D^(1/n) - D0^(1/n) = c t can be used for calculation, where at 870‰, the constant c is 2×10^(-8), n is 0.2, D and D0 are grain sizes in mm, and t is time in min.", "answer": "At 870‰ for 1h: D^(1/0.2) - D0^(1/0.2) = 2×10^(-8)×60; D^5 - D0^5 = 1.2×10^(-6); D^5 = 0.05^5 + 1.2×10^(-6) = 3.125×10^(-7) + 12×10^(-7) = 15.13×10^(-7); D = (15.13×10^(-7))^(1/5) mm = 0.069mm." }, { "idx": 1123, "question": "If the carbon content is 0.8% in high-carbon steel with an original grain size of 0.05mm, calculate the grain size and its growth value after holding at 760‰ for 1 hour. The formula D^(1/n) - D0^(1/n) = c t can be used for calculation, where at 760‰, the constant c is 6×10^(-16), n is 0.1, D and D0 are grain sizes in mm, and t is time in minutes.", "answer": "Holding at 760‰ for 1h: D^(1/0.1) - D0^(1/0.1) = 6×10^(-16)×60; D^10 - D0^10 = 3.6×10^(-14); D^10 = 3.6×10^(-14) + D0^10 = 0.05^10 + 3.6×10^(-14) = 9.77×10^(-14) + 3.6×10^(-14) = 13.4×10^(-14); D = (13.4×10^(-14))^(1/10) = 5.16×10^(-2) mm = 0.0516mm." }, { "idx": 1124, "question": "Use the phase rule to explain why a ternary alloy can exhibit four-phase equilibrium", "answer": "The phase rule formula is f=c-p+1, where f, c, p represent degrees of freedom, number of components, and number of phases, respectively. For a ternary alloy, when f reaches its minimum value of 0, the number of phases p is 4, indicating that four-phase equilibrium can occur. However, at this point, the degrees of freedom f is 0, meaning the compositions of all phases and the temperature are fixed." }, { "idx": 1125, "question": "Why can the lever rule be used to analyze the relative content of each phase in the horizontal section of a ternary equilibrium phase diagram", "answer": "On the horizontal section (i.e., the four-phase equilibrium plane) of a ternary equilibrium phase diagram, the composition of each phase and the temperature are all determined, so the lever rule can be used to analyze the relative content of each phase." }, { "idx": 1126, "question": "Equilibrium distribution coefficient", "answer": "Equilibrium distribution coefficient: Solid solution alloys exhibit selective crystallization during the solidification process. Therefore, at a certain temperature under equilibrium conditions, the ratio of the solid phase composition to the liquid phase composition is called the equilibrium distribution coefficient. This parameter reflects the distribution coefficient of solute between the solid and liquid phases and the extent to which the solute affects the melting point of the alloy." }, { "idx": 1127, "question": "Packing density", "answer": "Packing density: It represents the ratio of the volume occupied by atoms in a unit cell to the volume of the unit cell, and is a parameter to measure the tightness of atomic arrangement. The higher the packing density, the tighter the atomic arrangement in the crystal, and the denser the crystal structure." }, { "idx": 1128, "question": "In terms of the lattice constant, what is the atomic diameter in the HCP crystal structure?", "answer": "The close-packed direction in the HCP structure is [11 20], along which one atom is arranged, so the atomic diameter is a." }, { "idx": 1129, "question": "Solid solution", "answer": "Solid solution: refers to an alloy phase formed by dissolving atoms of other components (solute) into the crystal lattice of one component (solvent) of the alloy, characterized by retaining the solvent's lattice type with other component atoms located at lattice points or interstitial sites." }, { "idx": 1130, "question": "In terms of lattice constant, what is the atomic diameter in a BCC crystal structure?", "answer": "The close-packed direction in BCC structure is [111]. In the unit cell, there are two atoms along the [111] direction, so the atomic diameter is (√3/2)a." }, { "idx": 1131, "question": "Divorced eutectic", "answer": "Divorced eutectic: For hypoeutectic and hypereutectic alloys with composition points near the two ends of the eutectic transformation line, the primary crystals are abundant and the eutectic structure is scarce after crystallization. Moreover, the phase in the eutectic that is the same as the primary crystal combines with the primary crystal, pushing the other phase in the eutectic to the grain boundary, resulting in a non-equilibrium structure where the two phases of the eutectic are separated." }, { "idx": 1132, "question": "The typical crystal structure types of metals are (1), (2), and (3). What are their names?", "answer": "(1) fcc; (2) bcc; (3) hcp;" }, { "idx": 1133, "question": "Reaction diffusion", "answer": "Reaction diffusion: During the process of solid-state diffusion, if the solubility of the diffusing element in the metal is limited, as the number of diffusing atoms increases and the concentration of the diffusing atoms exceeds the saturation solubility, a solid solution or intermediate phase different from the original phase is formed. This causes the metal surface layer to divide into two regions: one where a new phase appears and another where it does not. The process of forming a new phase through diffusion is called reaction diffusion." }, { "idx": 1134, "question": "What is the movement of dislocations perpendicular to the slip plane called?", "answer": "Climb" }, { "idx": 1135, "question": "According to the phase rule, what is the maximum number of equilibrium phases in a ternary system?", "answer": "4" }, { "idx": 1136, "question": "Solid solution strengthening", "answer": "Solid solution strengthening: When a solid solution is formed, the solvent lattice becomes distorted due to the dissolution of solute atoms. The stress field of the solute atoms interacts with dislocations, hindering their movement and increasing the resistance to dislocation motion. This results in a critical resolved shear stress much higher than that of pure metals, making the activation of slip systems more difficult compared to pure metals. Consequently, the material's resistance to plastic deformation increases, leading to higher hardness and strength, while plasticity and toughness decrease. This phenomenon is known as solid solution strengthening." }, { "idx": 1137, "question": "According to the phase rule, what is the degree of freedom when the ternary system has the maximum number of equilibrium phases?", "answer": "Zero" }, { "idx": 1138, "question": "What is the movement of dislocations on the slip plane called?", "answer": "Slip" }, { "idx": 1139, "question": "According to the phase rule, what does the maximum number of equilibrium phases in a ternary system appear as on the phase diagram?", "answer": "Plane" }, { "idx": 1140, "question": "What kind of motion cannot be performed by a screw dislocation?", "answer": "Climb" }, { "idx": 1141, "question": "The typical crystal structure types of metals are (1), (2), and (3), and what are their corresponding coordination numbers?", "answer": "(4) 12; (5) 8; (6) 12;" }, { "idx": 1142, "question": "What is the slip plane of face-centered cubic metals?", "answer": "{111}" }, { "idx": 1143, "question": "What is the slip direction of face-centered cubic metals?", "answer": "<110>" }, { "idx": 1144, "question": "How many slip systems can be formed in face-centered cubic metals?", "answer": "12" }, { "idx": 1145, "question": "The unit of diffusion flux is (24)", "answer": "(24)1/cm2·s" }, { "idx": 1146, "question": "The symbol for diffusion flux is (25), indicating that diffusion proceeds from high concentration to low concentration.", "answer": "(25) negative sign" }, { "idx": 1147, "question": "The first law of diffusion applies to steady-state diffusion, and its mathematical expression can be written as (23).", "answer": "(23) J=-Ddc/dx" }, { "idx": 1148, "question": "What is the conceptual difference in recrystallization?", "answer": "Recrystallization refers to the transformation of a substance from one structure to another in the solid state, which is a solid-state phase transition process." }, { "idx": 1149, "question": "What is the conceptual difference in crystallization?", "answer": "Crystallization—The process by which a metal transitions from a liquid to a solid state is called solidification, and since the solid metal is a crystal, solidification is also referred to as crystallization." }, { "idx": 1150, "question": "What is constitutional supercooling?", "answer": "During the solidification of a solid solution alloy, the distribution of solute in the liquid phase changes, which alters the alloy's melting point. Even if the actual temperature distribution remains unchanged, the degree of supercooling at the solid-liquid interface front will vary. Therefore, the supercooling of a solid solution alloy is determined by both the changing alloy melting point and the actual temperature distribution. This type of supercooling caused by changes in liquid phase composition is called constitutional supercooling." }, { "idx": 1151, "question": "What effect does constitutional supercooling have on crystal growth morphology?", "answer": "During the crystallization of solid solutions, constitutional supercooling significantly influences the morphology of crystal growth, leading to the formation of cellular structures or even dendritic crystals even under a positive temperature gradient. Specifically, without constitutional supercooling, the interface advances in a planar manner; with slight constitutional supercooling, the interface becomes cellular; and with substantial constitutional supercooling, the interface turns dendritic." }, { "idx": 1152, "question": "What are the differences and connections among crystallization, recrystallization, and grain growth?", "answer": "The differences and connections among the three: Crystallization and recrystallization involve phase transformation processes, while grain growth does not; crystallization, recrystallization, and grain growth are all processes of nucleation and growth. The driving force for crystallization and recrystallization is the free energy difference between the reactant phase and the product phase, whereas for grain growth, it is the stored energy. After grain growth, strength and hardness decrease while ductility and toughness improve, whereas recrystallization involves allotropic transformation." }, { "idx": 1153, "question": "Briefly describe the differences between solid solution alloys and pure metals during the crystallization process.", "answer": "During crystallization, the interface of pure metals is rough and grows under a positive temperature gradient. Since heat is dissipated through the solid mold wall as the crystal grows, the solid-liquid interface is isothermal. If dynamic undercooling is achieved, the interface moves forward. If a small protrusion locally extends into the overheated liquid phase, it will be melted, and the interface remains straight, resulting in planar crystal growth. During solid solution crystallization, compositional undercooling occurs, creating a compositionally undercooled zone ahead of the solid-liquid interface. If any small protrusion extends into this undercooled zone, it will continue to grow due to the undercooling. The interface cannot remain straight and stable, leading to the formation of dendritic crystals." }, { "idx": 1154, "question": "What is the conceptual difference in recrystallization?", "answer": "Recrystallization—when a metal that has undergone cold working is heated to a certain temperature, new distortion-free equiaxed grains are regenerated within the deformed structure, and the properties return to the softened state before cold working. This process still belongs to the solid-state phase." }, { "idx": 1155, "question": "Analyze the changes of vacancies and dislocations during the low-temperature recovery stage and their effects on properties", "answer": "During the low-temperature recovery stage, the primary phenomenon is the disappearance of vacancies. The large number of vacancies generated after cold deformation are drastically reduced through various mechanisms, including vacancy migration to surfaces or grain boundaries, recombination of vacancies with interstitial atoms, interaction between vacancies and dislocations, and aggregation of vacancies into vacancy clusters." }, { "idx": 1156, "question": "Explain the reason why the strength is higher and the plasticity is better when the grain size is smaller during plastic deformation of polycrystalline metals.", "answer": "The reason why the strength is higher and the plasticity is better when the grain size is smaller during plastic deformation of polycrystalline metals is as follows: Due to the fine grains, the slip planes available for dislocation pile-up in each grain are shorter, and the number of piled-up dislocations $_{n}$ is also smaller. The stress concentration caused by dislocation pile-up is small and dispersed, making it more difficult to activate dislocation sources in adjacent grains, thus resulting in higher yield strength. With fine and numerous grains, under the same external force, the number of grains in favorable slip orientations increases, allowing more grains to participate in slip. The slip amount is dispersed among various grains, and the stress concentration is small. This reduces the likelihood of cracking during metal deformation, enabling a larger plastic deformation before fracture." }, { "idx": 1157, "question": "Analyze the changes of vacancies and dislocations during the medium-temperature recovery stage and their effects on properties", "answer": "During the medium-temperature recovery stage, the increase in temperature facilitates the glide of dislocations. Dislocations with opposite signs on the same slip plane attract and annihilate each other, not only reducing the number of dislocations within subgrains but also significantly decreasing the tangled dislocations in cell walls. The rearrangement becomes more ordered, and the cell walls become clearer, forming recovered subgrains. This stage is primarily characterized by the glide of dislocations, leading to dislocation recombination, the annihilation of dislocations with opposite signs, and the growth of subgrains." }, { "idx": 1158, "question": "Analyze the changes of vacancies and dislocations during the high-temperature recovery stage and their effects on properties", "answer": "During the high-temperature recovery stage, the kinetic conditions for dislocation motion are more sufficient, and slip is accompanied by climb, causing the dislocation densities on multiple slip planes to tend to equalize. The interactions between dislocations lead to a uniform distribution of dislocations on the same slip plane with roughly equal spacing, forming regularly arranged dislocation walls perpendicular to the slip plane, a process known as polygonization. The dislocation walls formed by polygonization constitute small-angle grain boundaries, which divide the original grains into several subgrains." }, { "idx": 1159, "question": "Annealing", "answer": "Annealing: A heat treatment process in which a metal or its alloy is heated above the phase transition temperature, held for a period of time, and then cooled at a relatively slow rate to obtain a nearly equilibrium microstructure is called annealing." }, { "idx": 1160, "question": "Density", "answer": "Density: Represents the ratio of the volume occupied by atoms in a unit cell to the volume of the unit cell. It is a parameter that measures the tightness of atomic arrangement. The higher the density, the tighter the atomic arrangement in the crystal, and the more compact the crystal structure." }, { "idx": 1161, "question": "A cylindrical face-centered cubic single crystal with a diameter of 1mm has its tensile axis aligned with the [001] direction of the crystal. Determine its equivalent slip systems.", "answer": "When the crystal is FCC and the force axis is in the [001] direction, among the 12 slip systems, the [110] crystal direction is perpendicular to the force axis, while the others have equal angles with the force axis, and 4 slip systems are also equivalent. Therefore, the equivalent slip systems are 8 in total. Specifically, they are (111)[101], (111)[011], (111)[101], (111)[011], (111)[10 1], (111)[011], (11 1)[011], and (11 1)[101]." }, { "idx": 1162, "question": "Equilibrium distribution coefficient", "answer": "Equilibrium distribution coefficient: During the crystallization of solid solution alloys, selective crystallization occurs, resulting in a certain ratio between the solid phase composition and the liquid phase composition. When equilibrium is reached at a certain temperature, the ratio of the solid phase composition to the liquid phase composition is called the equilibrium distribution coefficient. It reflects the distribution coefficient of solute between the solid and liquid phases and the extent to which the solute affects the melting point of the alloy." }, { "idx": 1163, "question": "Solid solution", "answer": "Solid solution: refers to an alloy phase formed by dissolving atoms of other components (solute) into the crystal lattice of one component (solvent) of the alloy, characterized by retaining the solvent's lattice type with solute atoms occupying lattice sites or interstitial positions." }, { "idx": 1164, "question": "Divorced eutectic", "answer": "Divorced eutectic: For hypoeutectic and hypereutectic alloys with composition points near the two ends of the eutectic transformation line, the primary crystals are abundant and the eutectic structure is scarce after crystallization. Moreover, the same phase in the eutectic as the primary crystal combines with the primary crystal, pushing the other phase in the eutectic to the grain boundary, resulting in a non-equilibrium structure where the two phases of the eutectic are separated." }, { "idx": 1165, "question": "Hardenability", "answer": "Hardenability: refers to the tendency of steel to obtain martensite structure during quenching (i.e., the ability of steel to be hardened)." }, { "idx": 1166, "question": "Phase", "answer": "Phase: In an alloy, a homogeneous component that has the same state of aggregation, the same crystal structure, essentially the same composition, and is distinctly separated from other parts by clear boundaries." }, { "idx": 1167, "question": "The typical crystal structure types of metals are (1), (2), and (3)", "answer": "(1) FCC; (2) BCC; (3) HCP" }, { "idx": 1168, "question": "A cylindrical face-centered cubic single crystal with a diameter of 1mm has its tensile axis aligned with the [001] direction of the crystal. If the tensile force at yield is 0.5N, what is the resolved shear stress on the slip system?", "answer": "Ω=cosλcosφ=√2/2×1/√3=√6/6 τ=P/AΩ=0.5/π(1/2)^2×√6/6MPa=0.26MPa" }, { "idx": 1169, "question": "What type of solid solution is ferrite?", "answer": "Interstitial" }, { "idx": 1170, "question": "The typical coordination numbers for metal crystal structures should be (4), (5), and (6)", "answer": "(4) 12; (5) 8; (6) 12" }, { "idx": 1171, "question": "Martensite is a supersaturated solid solution of carbon in what?", "answer": "α-Fe" }, { "idx": 1172, "question": "What type of solid solution is formed when the radius of solute atoms differs significantly from that of solvent atoms?", "answer": "Interstitial" }, { "idx": 1173, "question": "What type of solid solution can be formed when the atomic radius of the solute is similar to that of the solvent?", "answer": "Substitutional" }, { "idx": 1174, "question": "What is the slip plane of face-centered cubic metals?", "answer": "{111}" }, { "idx": 1175, "question": "What is the slip direction of face-centered cubic metals?", "answer": "<110>" }, { "idx": 1176, "question": "How many slip systems can be formed in face-centered cubic metals?", "answer": "12" }, { "idx": 1177, "question": "What are the two metallographic morphologies of martensite in quenched steel?", "answer": "Lath martensite and acicular martensite" }, { "idx": 1178, "question": "The solid-liquid interface of crystals is divided into smooth interface and rough interface. According to the growth speed from slow to fast, the growth methods are (13), (14), and (15) in order.", "answer": "(13) Two-dimensional nucleation growth; (14) Crystal defect growth; (15) Vertical growth" }, { "idx": 1179, "question": "The two basic types of dislocations are (22) and (23)", "answer": "(22) edge dislocation; (23) screw dislocation" }, { "idx": 1180, "question": "During austenite nucleation, not only (10) fluctuations and (11) fluctuations are required, but also (12) fluctuations are needed", "answer": "(10) structure; (11) energy; (12) composition" }, { "idx": 1181, "question": "The relationship between the direction of the dislocation line of an edge dislocation and the Burgers vector is (24)", "answer": "(24) perpendicular" }, { "idx": 1182, "question": "Ferritic stainless steel is (27)", "answer": "(27) 1Cr17" }, { "idx": 1183, "question": "The typical steel grade of martensitic stainless steel is (25）", "answer": "（25）1Cr13" }, { "idx": 1184, "question": "The typical steel grade of austenitic stainless steel is (26)", "answer": "(26) 1Cr18Ni9Ti" }, { "idx": 1185, "question": "Uphill diffusion refers to the diffusion of atoms from (28) to (29). The reason for uphill diffusion is the existence of (30) in the alloy system.", "answer": "(28) low concentration; (29) high concentration; (30) chemical potential gradient." }, { "idx": 1186, "question": "What is work hardening?", "answer": "The phenomenon where metallic materials exhibit continuously increasing strength and hardness while plasticity and toughness decrease with the amount of deformation during plastic deformation is called 'work hardening'. The reason for this phenomenon is that the increase in external force causes continuous dislocation multiplication, and the mutual intersection and reaction of dislocations make the movement of dislocations more difficult." }, { "idx": 1187, "question": "Discuss the characteristics exhibited by pure metals during the solidification process and explain the reasons?", "answer": "During crystallization, the interface of a pure metal is rough and grows under a positive temperature gradient. Since heat is dissipated through the solid mold wall as the crystal grows, the solid-liquid interface remains isothermal. If a dynamic undercooling is achieved, the interface advances forward. If a small protrusion locally extends into the overheated liquid phase, it will be melted, keeping the interface straight, and the crystal grows in a planar manner." }, { "idx": 1188, "question": "Discuss the characteristics exhibited by solid solution alloys during the solidification process and explain the reasons?", "answer": "During the crystallization of a solid solution, constitutional supercooling occurs, leading to the formation of a constitutionally supercooled zone ahead of the solid-liquid interface. If any small protrusion appears on the interface, it will extend into the constitutionally supercooled zone and continue to grow by obtaining supercooling. As a result, the interface cannot remain flat and stable, leading to the formation of dendritic crystals." }, { "idx": 1189, "question": "What is the main difference between the first type and the second type of temper brittleness?", "answer": "The first type of temper brittleness is irreversible, while the second type is reversible. The first type of temper brittleness is caused by the precipitation of discontinuous thin-shell carbides along the interfaces of martensite laths or plates during martensite decomposition. The second type of temper brittleness is caused by the segregation of impurity elements such as Sb, Sn, P, and As to the original austenite grain boundaries during tempering." }, { "idx": 1190, "question": "What is the first type of temper brittleness?", "answer": "The brittleness that occurs during tempering between 250~400°C is called low-temperature temper brittleness, also known as the first type of temper brittleness. The first type of temper brittleness is irreversible temper brittleness, which is caused by the discontinuous thin-shell-like carbides precipitated along the interface of martensite laths or plates during martensite decomposition, reducing the fracture strength of grain boundaries and making them the path for crack propagation, thus leading to brittle fracture." }, { "idx": 1191, "question": "How to solve the difficulties caused by work hardening in subsequent processing?", "answer": "Recrystallization annealing can be used to eliminate the difficulties caused by work hardening in subsequent processing." }, { "idx": 1192, "question": "What is the second type of temper brittleness?", "answer": "The brittleness that occurs during tempering between 450~650℃ is called high-temperature temper brittleness, also known as the second type of temper brittleness. The second type of temper brittleness is reversible. The main reason for its occurrence is that impurity elements such as Sb, Sn, P, and As segregate to the original austenite grain boundaries during tempering, weakening the atomic bonding force at the austenite grain boundaries and reducing the grain boundary fracture strength." }, { "idx": 1193, "question": "Analyze the changes of vacancies during the low-temperature recovery stage and their impact on properties", "answer": "During the low-temperature recovery stage, the main manifestation is the disappearance of vacancies. The large number of vacancies generated after cold deformation are drastically reduced through mechanisms such as vacancies migrating to surfaces or grain boundaries, recombination of vacancies with interstitial atoms, interaction of vacancies with dislocations, and aggregation of vacancies into vacancy clusters." }, { "idx": 1194, "question": "The typical crystal structures of metals are (1), (2), and (3)", "answer": "(1) body-centered cubic; (2) face-centered cubic; (3) close-packed hexagonal" }, { "idx": 1195, "question": "Analyze the changes in dislocations during the high-temperature recovery stage and their impact on properties", "answer": "During the high-temperature recovery stage, the kinetic conditions for dislocation motion are more sufficient, involving both slip and climb, which leads to a tendency for the dislocation densities on multiple slip planes to become equal. The interactions between dislocations cause the dislocations on the same slip plane to distribute uniformly with roughly equal spacing, forming regularly arranged dislocation walls perpendicular to the slip plane, a process known as polygonization. The dislocation walls formed by polygonization are small-angle grain boundaries, which divide the original grain into several subgrains." }, { "idx": 1196, "question": "In homogeneous nucleation, assuming the nucleus shape is a cube with side length a, given σ and ΔGv, find the critical nucleus side length", "answer": "The critical nucleus side length is (-4σ)/ΔGv" }, { "idx": 1197, "question": "Analyze the changes in dislocations during the medium-temperature recovery stage and their impact on properties", "answer": "During the medium-temperature recovery stage, the increase in temperature facilitates the glide of dislocations. Dislocations of opposite signs on the same slip plane attract and annihilate each other, not only reducing the number of dislocations within the subgrains but also significantly decreasing the tangled dislocations in the cell walls. This leads to a rearrangement into a more regular configuration, resulting in clearer cell walls and the formation of recovery subgrains. In this stage, the primary phenomena are the glide of dislocations, leading to their recombination, the annihilation of dislocations of opposite signs, and the growth of subgrains." }, { "idx": 1198, "question": "What kind of motion can an edge dislocation undergo?", "answer": "Slip" }, { "idx": 1199, "question": "What other motion can an edge dislocation undergo?", "answer": "Climb" }, { "idx": 1200, "question": "When a copper single crystal is stretched, if the force axis is in the [001] direction and the critical resolved shear stress is 0.64 MPa, what tensile stress is required to initiate slip in the crystal?", "answer": "Since copper is face-centered cubic, the slip plane is {111} and the slip direction is <110>. Given that the force axis is [001], the four slip directions [110] and [1-10] are excluded, leaving eight equivalent slip systems. γ = 45°. cosλ = cos45° = √2/2. cosφ = |1×0 + 1×0 + 1×0| / (√(1² + 1² + 1²) × √(0² + 0² + 1²)) = √3/3. The formula is σ = τ / Ω = 0.64 / (cosλ cosφ) = 1.57 MPa." }, { "idx": 1201, "question": "What kind of motion can a screw dislocation undergo?", "answer": "Slip" }, { "idx": 1202, "question": "Why can screw dislocations only undergo glide motion?", "answer": "Because it does not have a fixed half-atom plane" }, { "idx": 1203, "question": "The corresponding number of atoms per unit cell is (4), (5), and (6)", "answer": "(4) 2; (5) 4; (6) 6" }, { "idx": 1204, "question": "During the crystallization of a solid solution, nucleation requires both (13) fluctuations and (14) fluctuations, as well as (15) fluctuations.", "answer": "(13) structure; (14) energy; (15) composition" }, { "idx": 1205, "question": "In homogeneous nucleation, assuming the nucleus shape is a cube with side length a, given σ and ΔGv, find the critical nucleation work", "answer": "The critical nucleation work is 32σ³/ΔGv²" }, { "idx": 1206, "question": "The characteristics of polycrystalline plastic deformation are (18)ity, (19)ity, and (20)ity.", "answer": "(18) non-simultaneous; (19) coordinated; (20) non-uniform" }, { "idx": 1207, "question": "When a crystal undergoes tensile slip, the slip plane is driven to rotate, attempting to align parallel to the (16), while the slip direction also rotates to coincide with the (17) direction.", "answer": "(16) direction of the force axis; (17) maximum resolved shear stress" }, { "idx": 1208, "question": "Burgers vector", "answer": "Burgers vector: A physical quantity used to describe the lattice distortion caused by dislocations. The magnitude of this vector represents the strength of the dislocation, indicating the total extent of lattice distortion, while its direction represents the direction of lattice point distortion. Generally, the larger this vector is, the greater the degree of crystal distortion." }, { "idx": 1209, "question": "Phase", "answer": "Phase: In alloys, a homogeneous component with the same state of aggregation, the same crystal structure, essentially the same composition, and clearly defined boundaries separating it from other parts." }, { "idx": 1210, "question": "What is the equilibrium distribution coefficient?", "answer": "Equilibrium distribution coefficient: During the crystallization of solid solution alloys, selective crystallization occurs. Therefore, under equilibrium conditions at a certain temperature, the ratio of the solid phase composition to the liquid phase composition is called the equilibrium distribution coefficient." }, { "idx": 1211, "question": "What is the physical meaning of the equilibrium partition coefficient?", "answer": "This parameter reflects the partition coefficient of solute between solid and liquid phases and the degree of influence of solute on the alloy's melting point." }, { "idx": 1212, "question": "Divorced eutectic", "answer": "Divorced eutectic: In hypoeutectic and hypereutectic alloys with composition points near the two ends of the eutectic transformation line, the primary crystals are abundant and the eutectic structure is scarce after crystallization. Moreover, the same phase in the eutectic as the primary crystal combines with the primary crystal, pushing the other phase in the eutectic to the grain boundary, resulting in a non-equilibrium structure where the two phases of the eutectic are separated." }, { "idx": 1213, "question": "Constitutional supercooling", "answer": "Constitutional supercooling: During the solidification of a solid solution alloy, the distribution of solute in the liquid phase changes, which in turn alters the melting point of the alloy. Even if the actual temperature distribution remains unchanged, the degree of supercooling at the solid-liquid interface front will vary. Therefore, the supercooling of a solid solution alloy is determined by both the changing alloy melting point and the actual temperature distribution. This type of supercooling caused by changes in liquid phase composition is referred to as constitutional supercooling." }, { "idx": 1214, "question": "Solid solution strengthening", "answer": "Solid solution strengthening: When a solid solution is formed, the solvent lattice is distorted due to the dissolution of solute atoms, and the stress field of the solute atoms interacts with dislocations, hindering their movement. This increases the resistance to dislocation motion, making the critical resolved shear stress much higher than that of pure metals. The activation of slip systems becomes more difficult compared to pure metals, resulting in increased resistance to plastic deformation, higher hardness and strength, and reduced plasticity and toughness. This phenomenon is known as solid solution strengthening." }, { "idx": 1215, "question": "Why is carburization chosen in γ-Fe around 930‰, where the diffusion coefficient increases with carbon content?", "answer": "In γ-Fe, the diffusion coefficient increases with carbon content" }, { "idx": 1216, "question": "Why can a larger concentration gradient be formed in γ-Fe near 930‰ during carburization?", "answer": "A larger concentration gradient can be formed" }, { "idx": 1217, "question": "Why does metal slip occur on the closest-packed planes and directions?", "answer": "The slip in metal crystals occurs on the closest-packed planes and directions under external force because the atomic spacing is the smallest and the bonding force is the strongest on these planes, while the distance between two adjacent closest-packed planes is the largest and the bonding force is the weakest. It can be inferred that the resistance to slip is the smallest between the closest-packed atomic planes, requiring the least external force. Therefore, the closest-packed atomic planes and directions become the slip planes and slip directions for crystal slip." }, { "idx": 1218, "question": "Why is carburization performed in γ-Fe around 930‰, where the diffusion coefficient increases with temperature at high temperatures?", "answer": "At high temperatures, the diffusion coefficient increases with temperature, so high temperatures are chosen. However, excessively high temperatures can cause significant grain growth in austenite." }, { "idx": 1219, "question": "Dislocations belong to crystal defects and also to line defects.", "answer": "~\\\\surd~" }, { "idx": 1220, "question": "Peritectic transformation refers to the transformation in which a liquid phase interacts with a solid phase, enveloping the original solid phase to form another new solid phase.", "answer": "~\\surd~" }, { "idx": 1221, "question": "Materials with different crystal lattices have different slip systems, thus their tendency to deform also varies.", "answer": "~\\surd~" }, { "idx": 1222, "question": "The higher the dislocation density, the larger the Burgers vector of the corresponding dislocation.", "answer": "~\\\\times~" }, { "idx": 1223, "question": "Generally, the higher the activation energy for atomic diffusion in a crystal, the larger the diffusion coefficient and the faster the diffusion rate.", "answer": "~\\\\times~" }, { "idx": 1224, "question": "Stacking faults result from the local presence of an extra half atomic plane in the crystal lattice.", "answer": "~\\\\times~" }, { "idx": 1225, "question": "Metal solidification occurs through two processes: nucleation of the solid phase and grain growth. During the grain growth process, the liquid-solid interface of pure metals (such as Fe, Ni, Cu, Au, etc.) generally remains smooth.", "answer": "~\\\\times~" }, { "idx": 1226, "question": "Non-equilibrium crystallization refers to the crystallization of an alloy under actual cooling conditions where the cooling rate is relatively high, deviating from equilibrium conditions. Under non-equilibrium conditions, the composition of the newly formed solid solution after a homogeneous transformation is inhomogeneous.", "answer": "~\\\\surd~" }, { "idx": 1227, "question": "What is the essential difference between recrystallization and secondary recrystallization?", "answer": "Recrystallization refers to the process in which a cold-worked metal is heated to a certain temperature, and new distortion-free equiaxed grains are regenerated within the deformed structure, restoring the properties to the softened state before cold working. Secondary recrystallization refers to the phenomenon where, after recrystallization annealing, the metal is held at a higher temperature or for a longer time, causing a very few grains to rapidly consume other grains and grow, resulting in the entire metal being composed of a few exceptionally large grains that are tens to hundreds of times larger than those after recrystallization. The essential difference between the two: Recrystallization is a process of nucleation and growth, while secondary recrystallization is only a growth process. The driving force for recrystallization is stored energy, whereas for secondary recrystallization, it is interfacial energy. After recrystallization, strength and hardness decrease while plasticity and toughness increase, whereas after secondary recrystallization, the material's strength, plasticity, and toughness all decrease." }, { "idx": 1228, "question": "Due to the varying degrees of tightness in the regular arrangement of atoms in different directions, the spacing between atoms and the bonding forces between them differ, resulting in different properties in different directions. Therefore, most metals and alloys also exhibit anisotropic characteristics.", "answer": "~\\\\times~" }, { "idx": 1229, "question": "Iron-carbon alloy equilibrium phase diagram: (1) Contains peritectic transformation. (2) Contains eutectic transformation. (3) Contains congruent transformation. (4) Contains monotectic transformation. (5) Contains precipitation transformation of solid solution.", "answer": "(1)(2)(3)(5)" }, { "idx": 1230, "question": "Solid-state diffusion: (1) The diffusion of iron atoms in steel belongs to self-diffusion and is unrelated to the concentration gradient. (2) The diffusion of alloying elements in steel belongs to hetero-diffusion; in the absence of a chemical gradient, it is related to the concentration gradient. (3) The first law of diffusion is only applicable to steady-state diffusion conditions, hence it cannot be used to accurately determine the diffusion coefficient of carbon during the carburization process of steel. (4) Up-hill diffusion is solely related to the chemical gradient of the diffusing element and is unrelated to the concentration gradient. (5) The multiplication of dislocations is achieved through the diffusion movement of atoms.", "answer": "(1)(4)(5)" }, { "idx": 1231, "question": "Steel with a carbon content below the composition of point $S$ in the Fe-C phase diagram cannot obtain a completely pearlitic structure under any composition or condition.", "answer": "~\\\\times~" }, { "idx": 1232, "question": "Austenite in steel: \\n\\n(1) Can be represented as a $\\upgamma$-solid solution with dissolved carbon atoms. \\n\\n(2) Can be represented as an $\\alpha$-solid solution with dissolved carbon atoms. \\n(3) Exhibits good plasticity both at room temperature and high temperatures. \\n(4) Has ferromagnetism. \\n(5) Has paramagnetism.", "answer": "(1) (3)(5)" }, { "idx": 1233, "question": "The typical crystal structure types of metals are (1), (2), and (3)", "answer": "(1) FCC; (2) BCC; (3) HCP" }, { "idx": 1234, "question": "Fracture characteristics of metals: (1) One form of brittle fracture is transgranular cleavage. (2) Cleavage fractures mainly exhibit river patterns. (3) The propagation direction of micro-cracks can be determined by the flow direction of the rivers, i.e., the convergence direction of the rivers indicates the crack propagation direction. (4) Some materials exhibit tongue patterns in their ductile fracture surfaces. (5) Ductile fractures mainly manifest as intergranular dimple fracture forms.", "answer": "(2) (3) (5)" }, { "idx": 1235, "question": "The typical coordination numbers for metal crystal structure types are (4), (5), and (6)", "answer": "(4) 12; (5) 8; (6) 12" }, { "idx": 1236, "question": "In polycrystals, grain boundaries are classified into large-angle and small-angle grain boundaries. What is the meaning of the angle?", "answer": "The misorientation between adjacent grains" }, { "idx": 1237, "question": "What are the types of small-angle grain boundaries classified according to their characteristics?", "answer": "Twist, tilt, and coincidence" }, { "idx": 1238, "question": "In polycrystals, grain boundaries are divided into large-angle and small-angle grain boundaries. What is the typical identification angle for distinguishing between large-angle and small-angle grain boundaries?", "answer": "10°" }, { "idx": 1239, "question": "When an embryo appears in an undercooled liquid, what can the total free energy change ΔG be written as?", "answer": "ΔG=VΔGV+σS" }, { "idx": 1240, "question": "What does the size of the critical nucleus radius depend on?", "answer": "ΔGV and σ" }, { "idx": 1241, "question": "When dΔG/dr=0, what is the obtained 'r' value called?", "answer": "Critical nucleus radius" }, { "idx": 1242, "question": "What does a decrease in 'r' imply about the nucleation rate?", "answer": "Increase" }, { "idx": 1243, "question": "According to the phase rule, what is the maximum number of equilibrium phases in a binary system?", "answer": "17" }, { "idx": 1244, "question": "According to the phase rule, what is the degree of freedom when the number of equilibrium phases in a binary system is at its maximum?", "answer": "18" }, { "idx": 1245, "question": "What is the movement of an edge dislocation on the slip plane called?", "answer": "Slip" }, { "idx": 1246, "question": "According to the phase rule, what is the maximum number of equilibrium phases in a binary system represented on the phase diagram?", "answer": "19" }, { "idx": 1247, "question": "What is the movement of an edge dislocation perpendicular to the slip plane called?", "answer": "Climb" }, { "idx": 1248, "question": "What kind of motion cannot be performed by a screw dislocation?", "answer": "Climb" }, { "idx": 1249, "question": "What are the usual slip planes for face-centered cubic metals?", "answer": "{111}" }, { "idx": 1250, "question": "What is the slip direction of face-centered cubic metals?", "answer": "<110>" }, { "idx": 1251, "question": "How many slip systems can be formed in face-centered cubic metals?", "answer": "12" }, { "idx": 1252, "question": "Under what conditions is the first law of diffusion only applicable?", "answer": "Steady state" }, { "idx": 1253, "question": "What is the relationship between diffusion flux and concentration gradient?", "answer": "Proportional" }, { "idx": 1254, "question": "What is the basic condition expressed by the first law of diffusion?", "answer": "dc/dt=0" }, { "idx": 1255, "question": "What is the direction of diffusion flow from what concentration to what concentration?", "answer": "From high concentration to low concentration" }, { "idx": 1256, "question": "What effect does constitutional supercooling have on crystal growth morphology?", "answer": "After constitutional supercooling occurs, even under a positive temperature gradient, crystals will grow in a cellular manner during solidification. When the supercooling degree further increases, dendritic crystal growth will occur." }, { "idx": 1257, "question": "What is annealing twin? Explain its formation mechanism.", "answer": "Annealing twin is a type of twin, which refers to the formation of lamellar twins with straight interfaces inside grains after cold deformation and recrystallization annealing in certain face-centered cubic metals, such as Cu, Ni, nickel alloys, and austenitic steels. These twin interfaces are on the {111} planes. Since they appear only after annealing, they are called annealing twins." }, { "idx": 1258, "question": "What is the kinking mechanism in the analysis of the mechanisms and manifestations of plastic deformation in materials?", "answer": "When a crystal can neither slip nor undergo twinning, it can undergo plastic deformation through kinking. Its characteristic is that the orientation of the kinked crystal undergoes asymmetric changes. The dislocation mechanism refers to the process where, during plastic deformation, edge dislocations of the same sign accumulate in other regions due to dislocation motion. The convergence of dislocations generates bending stress, causing the crystal lattice to bend and kink, thereby forming a kinked region." }, { "idx": 1259, "question": "What is constitutional supercooling?", "answer": "Constitutional supercooling refers to the phenomenon in solid solution alloys during the cooling process, where the distribution of solute in the liquid phase changes, causing the alloy's melting point to vary accordingly. Even if the actual temperature distribution remains unchanged, the degree of supercooling at the solid-liquid interface front will change. The supercooling of solid solution alloys is determined by both the varying alloy melting point and the actual temperature distribution. This type of supercooling, formed due to changes in liquid phase composition, is called constitutional supercooling. Its characteristic is that the alloy's melting point is lowest at the interface, with the smallest degree of supercooling. As the distance from the S/L interface increases, the degree of supercooling increases instead, until the boundary layer of solute accumulation disappears at a certain distance, and the supercooled region subsequently vanishes." }, { "idx": 1260, "question": "What is the slip mechanism in the analysis of the mechanisms and manifestations of plastic deformation in materials?", "answer": "Slip is the relative movement of a crystal along the slip direction on the slip plane under the action of shear stress. Its dislocation mechanism is as follows: Due to the periodicity of the crystal lattice structure, when a dislocation moves along the slip plane, the energy at the dislocation center also undergoes periodic changes, causing the dislocation movement to encounter lattice resistance. However, in actual crystals, at a certain temperature, when the dislocation line moves from an energy valley position to an adjacent energy valley position, it does not simultaneously overcome the energy peak along its entire length. With the assistance of thermal activation energy, part of it can overcome the peak first, while the dislocation line forms a dislocation kink. Subsequently, the movement of the dislocation, aided by the kinked dislocation line, can easily shift sideways, resulting in a significant reduction in the stress required for the entire dislocation line to slip. During the slip process, in addition to lattice resistance, the resistance generated by the interaction between dislocations, the kinks and jogs formed after dislocation intersections, and the resistance caused by interactions between dislocations and other crystal defects all contribute to resistance, leading to crystal strengthening during slip." }, { "idx": 1261, "question": "What is the twinning mechanism in the analysis of the mechanisms and manifestations of plastic deformation in materials?", "answer": "Twinning is a mode of plastic deformation that occurs when a crystal cannot undergo slip. It also involves the relative motion of the crystal along the twinning direction and twinning plane under shear stress. The difference from slip is that twinning is a uniform shear deformation, and the two parts of the twin crystal form a mirror-symmetric relationship. The dislocation mechanism lies in the fact that after the sliding of a partial dislocation in the crystal, parallel and adjacent crystal planes undergo stacking faults, resulting in the formation of a twin." }, { "idx": 1262, "question": "Coordination number", "answer": "Coordination number: It is one of the physical quantities reflecting the tightness of atomic arrangement, referring to the number of nearest and equidistant atoms around any atom in the lattice. Generally, the larger the coordination number, the denser the crystal arrangement structure." }, { "idx": 1263, "question": "Burgers vector", "answer": "Burgers vector: A physical quantity used to describe the lattice distortion caused by dislocations. The magnitude of this vector represents the strength of the dislocation, indicating the extent of total lattice distortion, while its direction represents the orientation of the total lattice distortion. Generally, the larger this vector, the greater the degree of crystal distortion." }, { "idx": 1264, "question": "What is a twin crystal? Explain its formation mechanism.", "answer": "A twin crystal refers to two crystals (or two parts of one crystal) forming a mirror-symmetric orientation relationship along a common crystal plane, and these two parts are called twin crystals. The formation of twin crystals is due to uniform shear occurring throughout the twin region during twinning deformation, where the relative displacement of each crystal layer is caused by the movement of a partial dislocation. Taking face-centered cubic crystals as an example, the formation of twin crystals requires the generation of stacking faults. For instance, face-centered cubic crystals are stacked in the order of ABCABC··· on the 111 plane, which can be represented as △△△△△⋯. If the stacking sequence is reversed starting from a certain layer, becoming ABCACBACBA·.……., i.e., △△△∇∇∇∇⋯, the upper and lower parts of the crystal form a mirror-symmetric twin relationship. It can be observed that the ···CAC part corresponds to a stacking fault, followed by stacking in the reversed order, which still belongs to the normal FCC stacking sequence but is exactly opposite to the sequence of the crystal part before the fault, thus forming a symmetric relationship. This process is caused by the movement of a partial dislocation (Shockley partial dislocation)." }, { "idx": 1265, "question": "From the perspective of bonding, discuss why metals generally exhibit higher plasticity or ductility compared to inorganic non-metals.", "answer": "The bonding in metallic materials is primarily metallic bonding, while transition metals exhibit a mix of metallic and covalent bonding. Inorganic non-metallic materials are mainly bonded through ionic and covalent bonds. Metallic bonding is characterized by electron delocalization, where free electrons can move throughout the crystal, giving the bond no saturation or directionality. Due to the absence of saturation and directionality, each atom can potentially bond with more atoms and tends to form low-energy close-packed structures. When metals undergo deformation under force, the relative positions of atoms change without breaking the metallic bonds, which grants metals excellent ductility. Ionic bonding involves ions as the bonding units, where the bond is maintained by electrostatic attraction between positive and negative ions. This bond is strong, making it difficult for free electrons to move. Covalent bonds are formed between two or more atoms with similar electronegativity through shared electron pairs. Each bond has a fixed orientation, coordination numbers are relatively small, and the bonds are strong. Shared electron pairs between adjacent atoms cannot move freely, giving covalent bonds directionality and saturation, which generally results in poor plasticity and toughness." }, { "idx": 1266, "question": "Reaction diffusion", "answer": "Reaction diffusion: During the process of solid-state diffusion, if the solubility of the infiltrating element in the metal is limited, as the number of diffusing atoms increases and the concentration of the infiltrating atoms exceeds the saturation solubility, a solid solution or intermediate phase different from the original phase is formed. This causes the metal surface layer to divide into two layers—one where a new phase appears and another where it does not. The process of forming a new phase through diffusion is called reaction diffusion." }, { "idx": 1267, "question": "Divorced eutectic", "answer": "Divorced eutectic: For hypoeutectic and hypereutectic alloys with composition points near the two ends of the eutectic transformation line, after crystallization, the primary phase is abundant in the microstructure while the eutectic is scarce. Moreover, the same phase in the eutectic as the primary phase combines with the primary phase, pushing the other phase in the eutectic to the grain boundary, resulting in a non-equilibrium microstructure where the two phases of the eutectic are separated." }, { "idx": 1268, "question": "Critical resolved shear stress", "answer": "Critical resolved shear stress: Whether slip occurs in a specific slip system of a crystal depends on the magnitude of the resolved shear stress in the slip direction within the slip plane. When the resolved shear stress reaches a certain critical value, slip can begin. This stress is known as the critical resolved shear stress, which is the minimum resolved shear stress required to activate a slip system. The critical resolved shear stress of a material depends on its intrinsic properties but is also related to factors such as temperature and material purity." }, { "idx": 1269, "question": "When a spherical embryo with radius r appears in an undercooled liquid, what is the critical nucleus radius?", "answer": "-2σ/ΔGv" }, { "idx": 1270, "question": "When a spherical crystal embryo with radius r appears in an undercooled liquid, what happens to the number of nucleation events as r* decreases?", "answer": "Increases" }, { "idx": 1271, "question": "The solubility of substitutional solid solutions is related to atomic size factor, (4), electron concentration factor, and (5).", "answer": "4) electronegativity factor; (5) crystal structure factor" }, { "idx": 1272, "question": "Equilibrium distribution coefficient", "answer": "Equilibrium distribution coefficient: Solid solution alloys exhibit selective crystallization during the solidification process. Therefore, at a certain temperature under equilibrium conditions, the ratio of the solid phase composition to the liquid phase composition is called the equilibrium distribution coefficient. This parameter reflects the distribution coefficient of solute between the solid and liquid phases and the extent to which the solute affects the melting point of the alloy." }, { "idx": 1273, "question": "The typical crystal structure types of metals are fcc, bcc, and hcp, with coordination numbers of (1), (2), and (3) respectively.", "answer": "(1) 12; (2) 8; (3) 12" }, { "idx": 1274, "question": "When a spherical embryo with radius r appears in an undercooled liquid, what is the critical nucleation work?", "answer": "-1/3σS = 16πσ³/3ΔGv²" }, { "idx": 1275, "question": "Intermediate phase", "answer": "Intermediate phase: Refers to a new phase formed by the interaction between alloy components when the solid solubility limit of the solid solution is exceeded. This phase has a lattice structure and properties completely different from any of the components and exhibits metallic characteristics. Since it is often located in the middle of the phase diagram, it is also called an intermediate phase." }, { "idx": 1276, "question": "During the crystallization process of solid solution alloys, the laws of nucleation and grain growth are followed, but unlike pure metals, what additional fluctuation is required during nucleation?", "answer": "composition" }, { "idx": 1277, "question": "During the crystallization process of solid solution alloys, in what process does the composition fluctuation during nucleation occur?", "answer": "Variable temperature" }, { "idx": 1278, "question": "What is the growth mode of crystals related to?", "answer": "Interface structure" }, { "idx": 1279, "question": "During the crystallization process of a solid solution alloy, what kind of diffusion is always accompanied?", "answer": "Heterogeneous atoms (solute atoms)" }, { "idx": 1280, "question": "What is the crystal growth morphology related to?", "answer": "Interface structure" }, { "idx": 1281, "question": "The growth morphology of crystals is related to what distribution at the interface front?", "answer": "Temperature gradient" }, { "idx": 1282, "question": "Solid solution strengthening", "answer": "Solid solution strengthening: When a solid solution is formed, the solvent lattice becomes distorted due to the dissolution of solute atoms. The stress field of the solute atoms interacts with dislocations, hindering their movement and increasing the resistance to dislocation motion. This results in a critical resolved shear stress much higher than that of pure metals, making the activation of slip systems more difficult compared to pure metals. Consequently, the material's resistance to plastic deformation increases, leading to higher hardness and strength, while plasticity and toughness decrease. This phenomenon is known as solid solution strengthening." }, { "idx": 1283, "question": "According to the phase rule, the maximum number of equilibrium phases when the degrees of freedom in a ternary system are zero is", "answer": "4" }, { "idx": 1284, "question": "According to the phase rule, when the degrees of freedom in a ternary system are zero, it is represented on the phase diagram as", "answer": "plane" }, { "idx": 1285, "question": "When a single crystal undergoes plastic deformation, the common methods are (15), (16), and (17).", "answer": "(15) Slip; (16) Twinning; (17) Kinking" }, { "idx": 1286, "question": "The negative sign of the diffusion flux indicates that diffusion proceeds from high concentration to low concentration, and the driving force for diffusion in this process is (20)", "answer": "Chemical potential gradient (chemical force)" }, { "idx": 1287, "question": "After cold deformation, the control of grain size after recrystallization is related to the degree of cold deformation, original grain size, recrystallization temperature, and impurities, among others. Among these, what is (21)?", "answer": "Degree of cold deformation" }, { "idx": 1288, "question": "The first law of diffusion applies to steady-state diffusion, and its mathematical expression can be written as (19)", "answer": "J=-Ddc/dx" }, { "idx": 1289, "question": "Dynamic recovery and dynamic recrystallization refer to deformation at high temperatures, where (18) occurs simultaneously during the deformation process.", "answer": "(18) work hardening and softening" }, { "idx": 1290, "question": "After cold deformation, the control of grain size after recrystallization is related to the degree of cold deformation, original grain size, (22), and impurities, etc. What is (22)?", "answer": "Recrystallization temperature" }, { "idx": 1291, "question": "When a crystal is subjected to external forces, the moving dislocations inside it will intersect, resulting in the formation of jogs and kinks. What is the length of these jogs and kinks the same as that of the intersecting dislocations?", "answer": "The magnitude of the Burgers vector" }, { "idx": 1292, "question": "If the slip of the jog is inconsistent with the slip of the main dislocation line, the main dislocation line will drag the jog to produce climb motion, resulting in what phenomenon?", "answer": "Jog hardening" }, { "idx": 1293, "question": "What does secondary recrystallization conceptually refer to?", "answer": "Secondary recrystallization refers to the phenomenon where, after recrystallization annealing, a metal subjected to higher temperatures or prolonged holding times experiences a few grains rapidly consuming other grains and growing, resulting in the entire metal being composed of a small number of exceptionally large grains that are tens to hundreds of times larger than the post-recrystallization grains." }, { "idx": 1294, "question": "What does recrystallization conceptually refer to as a process?", "answer": "Recrystallization refers to the process in which a substance transforms from one structure to another in the solid state, i.e., an allotropic transformation reaction." }, { "idx": 1295, "question": "Why does pure metal grow in a planar manner during solidification under a positive temperature gradient?", "answer": "During crystallization, the interface of pure metal is rough and grows under a positive temperature gradient. As the crystal grows, heat is dissipated through the solid mold wall, making the solid-liquid interface isothermal. If dynamic undercooling is achieved, the interface moves forward. If there is a small protrusion locally on the interface extending into the superheated liquid phase, the protrusion will be melted, so the interface remains straight; the crystal grows in a planar manner." }, { "idx": 1296, "question": "What does recrystallization conceptually refer to as a process?", "answer": "Recrystallization—the process in which a metal that has undergone cold working is heated to a certain temperature, leading to the formation of new undistorted equiaxed grains within the deformed structure, restoring its properties to the softened state prior to cold working." }, { "idx": 1297, "question": "What are the differences in the nucleation and growth processes among recrystallization, primary recrystallization, and secondary recrystallization?", "answer": "Recrystallization and primary recrystallization involve nucleation and growth processes; secondary recrystallization only involves a growth process." }, { "idx": 1298, "question": "Why do solid solution alloys usually grow in a dendritic manner during solidification under a positive temperature gradient?", "answer": "During the crystallization of solid solutions, constitutional supercooling occurs, leading to a constitutionally supercooled zone ahead of the solid-liquid interface. If any small protrusion exists on the interface, it will extend into the constitutionally supercooled zone and continue to grow as it gains supercooling. Therefore, the interface cannot remain flat and stable, resulting in dendritic growth." }, { "idx": 1299, "question": "Discuss the theory of grain refinement strengthening from the mechanism of plastic deformation in materials.", "answer": "In the strengthening theories of materials, grain refinement strengthening is the only method that can simultaneously improve both the strength and toughness of materials. The specific explanation is as follows: Due to the fine grains, the slip planes available for dislocation pile-up are shorter, and the number of piled-up dislocations is smaller. The stress concentration caused by dislocation pile-up is dispersed among various grains, thereby increasing the yield strength. On the other hand, because the grains are fine, under the same external force, there are more grains in favorable slip directions, and the stress is distributed among the grains. Even under large plastic deformation, the material maintains its good performance without cracking, thus improving the toughness. In summary, grain refinement strengthening can enhance the comprehensive mechanical properties of materials." }, { "idx": 1300, "question": "What are the differences in the driving forces among recrystallization, primary recrystallization, and secondary recrystallization?", "answer": "The driving force for recrystallization is the free energy difference between the new and old phases, for primary recrystallization it is the stored energy, and for secondary recrystallization it is the interfacial energy." }, { "idx": 1301, "question": "Why do metal crystals always slide along the slip plane and slip direction during the slip process?", "answer": "The slip of metal crystals is the result of dislocation movement, and the dislocation movement is mainly subjected to the resistance of the crystal lattice. In the crystal, the greater the interplanar spacing between the densest atomic planes, and the smaller the atomic spacing in the most closely packed direction, the smaller the sliding resistance when dislocations slide along the slip plane and slip direction. Therefore, metal crystals always slide along the slip plane and slip direction during the slip process." }, { "idx": 1302, "question": "What are the differences in the effects of recrystallization and secondary recrystallization on material properties?", "answer": "After recrystallization, strength and hardness decrease while plasticity and toughness improve; after secondary recrystallization, the material's strength, plasticity, and toughness all decrease." }, { "idx": 1303, "question": "What is the structure of austenite?", "answer": "Austenite: An interstitial solid solution of carbon in γ-Fe, with a face-centered cubic structure." }, { "idx": 1304, "question": "What are the differences in the phase transformation processes among recrystallization, primary recrystallization, and secondary recrystallization?", "answer": "Recrystallization involves a phase transformation process, while primary recrystallization and secondary recrystallization do not involve a phase transformation process." }, { "idx": 1305, "question": "What type of structure is cementite?", "answer": "Cementite: an interstitial compound formed by iron and carbon, belonging to the orthorhombic system." }, { "idx": 1306, "question": "At room temperature, bending a lead plate makes it increasingly harder, but after some time, bending it again makes the lead plate as soft as it was initially. Why? Please explain this phenomenon.", "answer": "At room temperature, bending a lead plate makes it increasingly harder due to work hardening, which increases the strength and hardness of the lead plate. After some time, because the melting point of lead is relatively low, recrystallization softening occurs at room temperature, causing its hardness to decrease again." }, { "idx": 1307, "question": "What kind of structure is ferrite?", "answer": "Ferrite: an interstitial solid solution of carbon in α-Fe, with a body-centered cubic structure." }, { "idx": 1308, "question": "The solubility of substitutional solid solutions is related to atomic size factor, (1), electron concentration factor, and (2).", "answer": "(1) electronegativity factor; (2) crystal structure factor" }, { "idx": 1309, "question": "A decrease in $r^{}$ means the number of nucleation sites (4).", "answer": "increases" }, { "idx": 1310, "question": "When a spherical embryo with radius $r$ appears in an undercooled liquid, the resulting critical nucleus radius is (3).", "answer": "$\\\\frac{-2\\\\sigma}{\\\\Delta G_{V}}$" }, { "idx": 1311, "question": "Homogeneous nucleation of crystal nuclei relies on (5) providing atomic clusters with $r>r^{*}$ to act as nuclei.", "answer": "Structural fluctuations" }, { "idx": 1312, "question": "Homogeneous nucleation of crystal nuclei relies on (6) to provide nucleation work equivalent to the interfacial energy (7).", "answer": "energy fluctuation; 1/3" }, { "idx": 1313, "question": "Dynamic recovery and dynamic recrystallization refer to the simultaneous occurrence of (15) during deformation.", "answer": "(15) softening and strain hardening." }, { "idx": 1314, "question": "What kind of diffusion always accompanies the crystallization process of solid solution alloys?", "answer": "Dissimilar atoms" }, { "idx": 1315, "question": "What mainly occurs during the high-temperature recovery of cold-worked metals?", "answer": "Polygonization" }, { "idx": 1316, "question": "What is the main phenomenon during the low-temperature recovery of cold-worked metals?", "answer": "The disappearance of point defects" }, { "idx": 1317, "question": "Solid solution", "answer": "Solid solution: refers to an alloy phase formed by dissolving atoms of other components (solute) into the crystal lattice of one component (solvent) of the alloy, characterized by retaining the solvent's lattice type, with other component atoms located at lattice points or interstitial sites." }, { "idx": 1318, "question": "In the crystallization process of solid solution alloys, during which process does the composition fluctuation occur during nucleation?", "answer": "Variable temperature" }, { "idx": 1319, "question": "During the crystallization process of solid solution alloys, they follow the laws of nucleation and grain growth, but unlike pure metals, what additional fluctuation is required during nucleation?", "answer": "Composition" }, { "idx": 1320, "question": "During the plastic deformation of metals, after twinning occurs, the crystal orientations on both sides of the twin plane exhibit (11), and the crystal undergoes (12) shear.", "answer": "(11) symmetric relationship; (12) uniform" }, { "idx": 1321, "question": "Briefly describe the growth mechanism of pure metal crystals and its relationship with the microstructure of the solid/liquid interface.", "answer": "The solid/liquid interface of pure metal crystals is generally a rough interface. Therefore, for pure metal crystals, during the growth process, they usually grow according to the vertical growth mode of the rough interface. Since nearly $50\\\\%$ of the positions on the interface are vacant, the addition of liquid-phase atoms is not restricted by position, which facilitates continuous filling of atoms, thereby enabling the solid/liquid interface to grow rapidly along the normal direction." }, { "idx": 1322, "question": "Reaction diffusion", "answer": "Reaction diffusion: During the process of solid-state diffusion, if the solubility of the infiltrating element in the metal is limited, as the number of diffusing atoms increases and the concentration of the infiltrating atoms exceeds the saturation solubility, a solid solution or intermediate phase different from the original phase is formed. This causes the metal surface layer to divide into two layers—one where new phases appear and another where they do not. The process of forming new phases through diffusion is called reaction diffusion." }, { "idx": 1323, "question": "Phase", "answer": "Phase: In alloys, a homogeneous component with the same state of aggregation, the same crystal structure, essentially the same composition, and clearly defined interfaces separating it from other parts." }, { "idx": 1324, "question": "Dendritic segregation", "answer": "Dendritic segregation: It is a type of microsegregation in materials, where under non-equilibrium cooling conditions, the newly formed solid solution grains after a homogeneous transformation exhibit non-uniform composition within. The initially crystallized core contains more high-melting-point component atoms, while the later crystallized outer regions contain more low-melting-point component atoms. Typically, solid solution crystals grow in a dendritic manner, resulting in the dendrite arms containing more high-melting-point components and the interdendritic regions containing more low-melting-point component atoms. This leads to compositional inhomogeneity within the same grain." }, { "idx": 1325, "question": "Constitutional supercooling", "answer": "Constitutional supercooling: During the solidification of a solid solution alloy, the distribution of solute in the liquid phase changes, which alters the melting point of the alloy. Even if the actual temperature distribution remains unchanged, the degree of supercooling at the solid-liquid interface front will vary. Therefore, the supercooling of a solid solution alloy is determined by both the changing melting point of the alloy and the actual temperature distribution. This type of supercooling caused by changes in liquid phase composition is referred to as constitutional supercooling." }, { "idx": 1326, "question": "Explain the source and morphological characteristics of Fe3CI", "answer": "Fe3CI: originates from the liquid phase, appearing as white elongated strips." }, { "idx": 1327, "question": "Explain the origin and morphological characteristics of Fe3CII", "answer": "Fe3CII: Originates from the austenite precipitation reaction, generally distributed in a network-like pattern along grain boundaries." }, { "idx": 1328, "question": "According to the reaction temperature from high to low, sequentially write the reaction formulas of the three-phase equilibrium reactions in the Fe-Fe3C alloy system", "answer": "L0.53 + δ0.09 → γ0.17 (1495°C)\\nL4.3 → γ2.11 + Fe3C (1148°C)\\nγ0.77 → α0.0218 + Fe3C (727°C)" }, { "idx": 1329, "question": "Explain the origin and morphological characteristics of Fe3CIII", "answer": "Fe3CIII: Originates from the precipitation reaction of ferrite, generally distributed in a network pattern along grain boundaries." }, { "idx": 1330, "question": "Explain the origin and morphological characteristics of Fe3C eutectoid", "answer": "Fe3C eutectoid: The cementite in the eutectoid product pearlite, generally distributed in a lamellar form." }, { "idx": 1331, "question": "Explain the origin and morphological characteristics of Fe3C eutectic", "answer": "Fe3C eutectic: The cementite in the eutectic product ledeburite, which generally exists as the matrix." }, { "idx": 1332, "question": "Explain the temperature range and alias of the first type of temper brittleness", "answer": "The brittleness that occurs when tempering between 250~400Ω is called low-temperature temper brittleness, also known as the first type of temper brittleness." }, { "idx": 1333, "question": "Explain the reversibility of the second type of temper embrittlement", "answer": "The second type of temper embrittlement is reversible." }, { "idx": 1334, "question": "Explain the reversibility of the first type of temper embrittlement", "answer": "The first type of temper embrittlement is irreversible temper embrittlement." }, { "idx": 1335, "question": "Briefly describe one of the essential reasons for the high hardness of martensite: the characteristics of its crystal structure", "answer": "The crystal structure of martensite is body-centered tetragonal, lacking closely packed slip planes that facilitate dislocation movement, thus making dislocation slip difficult." }, { "idx": 1336, "question": "Explain the cause of the first type of temper brittleness", "answer": "The first type of temper brittleness occurs due to the precipitation of discontinuous thin-shell-like carbides along the interfaces of martensite laths or plates during martensite decomposition, which reduces the fracture strength of grain boundaries, making them the path for crack propagation and thus leading to brittle fracture." }, { "idx": 1337, "question": "Explain the temperature range and alias of the second type of temper brittleness", "answer": "The brittleness that occurs when tempering between 450~650°C is called high-temperature temper brittleness, also known as the second type of temper brittleness." }, { "idx": 1338, "question": "Explain the cause of the second type of temper brittleness", "answer": "The cause of the second type of temper brittleness is the segregation of impurity elements such as Sb, Sn, P, and As to the original austenite grain boundaries during tempering, which weakens the atomic bonding force at the austenite grain boundaries and reduces the grain boundary fracture strength. This is the main reason for the occurrence of the second type of temper brittleness." }, { "idx": 1339, "question": "Briefly describe the third essential reason for the high hardness of martensite: grain boundary strengthening mechanism", "answer": "Whether it is the laths of lath martensite or the plates of plate martensite, they are all very fine, and the grain boundary strengthening mechanism also plays a significant role." }, { "idx": 1340, "question": "Briefly describe the second essential reason for the high hardness of martensite: the solid solution strengthening mechanism", "answer": "Ferrite typically contains only 0.03% carbon atoms, while the carbon content in martensite is the same as that of the material itself. Therefore, martensite contains a large number of supersaturated carbon atoms, making solid solution strengthening one of the primary mechanisms for the high hardness of martensite." }, { "idx": 1341, "question": "How does temperature affect the diffusion coefficient?", "answer": "The higher the temperature, the faster the diffusion." }, { "idx": 1342, "question": "How do crystal defects affect the diffusion coefficient?", "answer": "Grain boundaries, dislocations, and vacancies all have an impact on diffusion." }, { "idx": 1343, "question": "How does the concentration of a solid solution affect the diffusion coefficient?", "answer": "The higher the concentration, the easier the diffusion." }, { "idx": 1344, "question": "How does the type of solid solution affect the diffusion coefficient?", "answer": "Different solid solutions have different atomic diffusion and mechanisms." }, { "idx": 1345, "question": "How does crystal structure affect the diffusion coefficient?", "answer": "Different structures result in different diffusion coefficients." }, { "idx": 1346, "question": "How does chemical composition affect the diffusion coefficient?", "answer": "The addition of chemical elements hinders diffusion." }, { "idx": 1347, "question": "Briefly describe the fourth essential reason for the high hardness of martensite: the transformation strengthening mechanism", "answer": "During martensitic transformation, substructures with high lattice defect density are created within the crystal. The high-density dislocation networks in lath martensite and the fine twins in plate martensite both hinder dislocation motion, thereby causing strengthening through transformation." }, { "idx": 1348, "question": "Briefly describe the fifth essential reason for the high hardness of martensite: the aging strengthening mechanism", "answer": "After the formation of martensite, carbon and alloy element atoms diffuse and segregate or precipitate to dislocations or other crystal defects, pinning the dislocations and making them difficult to move, thereby increasing the hardness and strength of martensite through aging strengthening." }, { "idx": 1349, "question": "A stress of 70MPa is applied in the [001] direction of an FCC crystal. Which slip system in the crystal will activate first?", "answer": "The resolved shear stress on the (111)[10-1] slip system is 28.58MPa, while the (1-11)[110] slip system does not slip. Therefore, the (111)[10-1] slip system activates first." }, { "idx": 1350, "question": "Analyze the essential similarities between work hardening, fine grain strengthening, solid solution strengthening, and second-phase strengthening", "answer": "Similarities: All involve hindering dislocation movement, increasing the resistance to dislocation glide, thereby strengthening the material." }, { "idx": 1351, "question": "Analyze the fundamental differences of grain refinement strengthening", "answer": "Grain refinement strengthening: increases grain boundaries, enlarges the range of dislocation pile-up." }, { "idx": 1352, "question": "A stress of 70 MPa is applied in the [001] direction of an FCC crystal. Determine the resolved shear stress on the (1-11)[110] slip system.", "answer": "Substitute into the formula: τ=σcosφcosλ, where cosλ=|a u+b v+c w|/(√(a²+b²+c²)√(u²+v²+w²)), cosφ=|a h+b k+c l|/(√(a²+b²+c²)√(h²+k²+l²)). For the (1-11)[110] slip system, since cosλ=0, this slip system does not slip under the applied stress." }, { "idx": 1353, "question": "According to solidification theory, what is the fundamental principle of grain refinement by vibration and stirring?", "answer": "Vibration and stirring. Vibration and stirring can input additional energy into the liquid to provide nucleation work, promoting nucleation. On the other hand, they can cause crystallized crystals to fragment under the impact of liquid flow, increasing the number of nuclei." }, { "idx": 1354, "question": "According to solidification theory, what is the fundamental principle of refining grain size by increasing undercooling?", "answer": "Increasing undercooling. Grain size depends on the relative relationship between nucleation rate and grain growth velocity. When undercooling is significantly high, the increase rate of nucleation surpasses that of grain growth, thus enhancing undercooling ensures N > G and widens the gap between them, leading to grain refinement." }, { "idx": 1355, "question": "According to solidification theory, what is the fundamental principle of grain refinement through modification treatment?", "answer": "Modification treatment. That is, adding a modifier to the molten metal before pouring to promote heterogeneous nucleation and increase the number of nuclei, thereby refining the grains." }, { "idx": 1356, "question": "Describe the behavior and manifestations of crystal defects in metals during the recovery stage and the corresponding material properties, and explain the driving force that promotes the movement of these crystal defects in this stage.", "answer": "Recovery: Line defects remain largely unchanged, point defects significantly decrease, mechanical properties remain unchanged, while corresponding physicochemical properties alter; the driving force is stored energy." }, { "idx": 1357, "question": "Analyze the fundamental differences in solid solution strengthening", "answer": "Solid solution strengthening: solute atoms gather along dislocations and pin them." }, { "idx": 1358, "question": "Analyze the essential differences in second-phase strengthening", "answer": "Second-phase strengthening: The dispersed strengthening phase particles force dislocations to cut through or bypass the strengthening phase particles, and the additional work required constitutes the dislocation mechanism of dispersed phase strengthening." }, { "idx": 1359, "question": "Analyze the effect of cold deformation degree on recrystallized grain size.", "answer": "When the cold deformation degree is less than the critical deformation degree, the grains after recrystallization basically remain in the state before cold deformation. Due to the minimal stored energy, recrystallization does not actually occur, so the cold deformation degree has no relation to the size of recrystallized grains. When the cold deformation degree increases to the critical deformation degree, the stored energy from cold deformation is sufficient to drive recrystallization. However, because the overall deformation degree is small and uneven, only a few areas with higher deformation degrees can form nuclei and grow. At this point, since $G>>N$, only a few nuclei form and rapidly grow, resulting in the largest grain size after recrystallization. Beyond the critical deformation degree, the cold deformation degree is inversely proportional to the recrystallized grain size. When the deformation degree reaches a certain level, the grain size remains essentially unchanged." }, { "idx": 1360, "question": "A stress of 70MPa is applied in the [001] direction of an FCC crystal. Calculate the resolved shear stress on the (111)[10-1] slip system.", "answer": "Substitute into the formula: τ=σcosφcosλ, where cosλ=|a u+b v+c w|/(√(a²+b²+c²)√(u²+v²+w²)), cosφ=|a h+b k+c l|/(√(a²+b²+c²)√(h²+k²+l²)). For the (111)[10-1] slip system, substituting the values gives τ=70×(1/√3)×(1/√2)=28.58MPa." }, { "idx": 1361, "question": "At what temperature or temperature range is eutectic cementite formed through what reaction? And write the reaction equation.", "answer": "Eutectic cementite: 1148°C, eutectic reaction, L→(γ+Fe3C) eutectic." }, { "idx": 1362, "question": "Describe the behavior and manifestations of crystal defects in metals during the cold deformation stage and the corresponding material properties, and explain the driving force that promotes the movement of these crystal defects at this stage.", "answer": "Cold deformation: Crystal defects including point defects, line defects, and planar defects all increase significantly, leading to work hardening and corresponding changes in physical and chemical properties; the driving force is the externally applied resolved shear stress." }, { "idx": 1363, "question": "Describe the behavior and manifestations of crystal defects in metals during the recrystallization stage and the corresponding material properties, and explain the driving force that promotes the movement of these crystal defects at this stage.", "answer": "Recrystallization: Line defects significantly decrease, and the material exhibits softening; the driving force is stored energy." }, { "idx": 1364, "question": "Describe the behavior and manifestations of crystal defects in metals during the grain growth stage and the corresponding material properties, and explain the driving force that promotes the movement of these crystal defects at this stage.", "answer": "Grain growth: planar defects significantly decrease; the driving force is the total interfacial free energy." }, { "idx": 1365, "question": "At what temperature or temperature range is tertiary cementite formed, and through what reaction? Also, write the reaction equation.", "answer": "Tertiary cementite: <727°C, secondary precipitation reaction, α→Fe3CIII." }, { "idx": 1366, "question": "At what temperature or temperature range is primary cementite formed, and through what reaction? Also, write the reaction equation.", "answer": "Primary cementite: >1148°C, eutectic reaction, L→Fe3C1." }, { "idx": 1367, "question": "Analyze the fundamental differences in work hardening", "answer": "Work hardening: Possible mechanisms include dislocation pile-up, forest dislocation resistance, and the consumption of external force work through the formation of jogs." }, { "idx": 1368, "question": "What types of cementite may exist in iron-carbon alloys?", "answer": "Primary cementite, eutectic cementite, secondary cementite, eutectoid cementite, tertiary cementite." }, { "idx": 1369, "question": "What are the differences in morphology, size, and distribution of primary cementite?", "answer": "Regular strip shape, coarse, distributed on the ledeburite matrix." }, { "idx": 1370, "question": "At what temperature or temperature range is eutectoid cementite formed through what reaction? And write the reaction equation.", "answer": "Eutectoid cementite: 727°C, eutectoid reaction, γ→(α+Fe3C) eutectoid." }, { "idx": 1371, "question": "At what temperature or temperature range is secondary cementite formed, and through what reaction? Also, write the reaction equation.", "answer": "Secondary cementite: 1148~727°C, secondary precipitation reaction, γ→Fe3CII." }, { "idx": 1372, "question": "What are the differences in morphology, size, and distribution of eutectic cementite?", "answer": "Acts as the continuous matrix of ledeburite." }, { "idx": 1373, "question": "What are the differences in morphology, size, and distribution of eutectoid cementite?", "answer": "Regular lamellar, smaller, alternately distributed with lamellar ferrite to form pearlite." }, { "idx": 1374, "question": "What are the differences in morphology, size, and distribution of tertiary cementite?", "answer": "Discontinuous granular, small amount, distributed at ferrite grain boundaries." }, { "idx": 1375, "question": "What effect does eutectic cementite have on the mechanical properties of iron-carbon alloys?", "answer": "Increases hardness but reduces toughness." }, { "idx": 1376, "question": "What are the differences in morphology, size, and distribution of secondary cementite?", "answer": "It is distributed at the original austenite grain boundaries that have transformed into pearlite. When the amount is small, it appears as discontinuous granular; when the amount is large, it forms a continuous network." }, { "idx": 1377, "question": "What effect will primary cementite have on the mechanical properties of iron-carbon alloys?", "answer": "Increases hardness but reduces toughness." }, { "idx": 1378, "question": "What effect does eutectoid cementite have on the mechanical properties of iron-carbon alloys?", "answer": "Increases hardness and strength but reduces toughness." }, { "idx": 1379, "question": "What effect will tertiary cementite have on the mechanical properties of iron-carbon alloys?", "answer": "Increases hardness and strength but reduces toughness." }, { "idx": 1380, "question": "What effect does secondary cementite have on the mechanical properties of iron-carbon alloys?", "answer": "Increases hardness but reduces toughness; discontinuous granular form enhances strength, while continuous network form reduces strength." }, { "idx": 1381, "question": "Using dislocation theory, explain one of the reasons for work hardening in pure metal single crystals.", "answer": "Dislocation intersections occur, and the resulting jogs impede dislocation motion." }, { "idx": 1382, "question": "What is the phenomenon of work hardening in metals?", "answer": "The phenomenon where the applied flow stress continuously increases with the increase of strain during the plastic deformation of metallic materials is called work hardening. Alternatively, the phenomenon where the strength and hardness of metallic materials increase, while the plasticity and toughness decrease after cold plastic deformation, is called work hardening." }, { "idx": 1383, "question": "Using dislocation theory, what is the second reason for the work hardening of pure metal single crystals?", "answer": "Dislocation reactions occur, forming immobile dislocations that hinder dislocation motion" }, { "idx": 1384, "question": "Solid solution", "answer": "A solid solution is a solid phase formed by taking one component of an alloy as the solvent and other components as solutes, which has the same crystal structure as the solvent and slightly changed lattice constants." }, { "idx": 1385, "question": "Critical deformation degree", "answer": "Critical deformation degree is the minimum pre-cold deformation required for metal to undergo recrystallization at a given temperature." }, { "idx": 1386, "question": "What are the beneficial aspects of the work hardening characteristics of metals for the use of metal materials?", "answer": "As a means to improve the strength of metal materials; facilitating the plastic forming of metal materials; enabling metal parts to resist accidental overload." }, { "idx": 1387, "question": "Unit cell", "answer": "The unit cell is the most basic unit that constitutes a crystal lattice." }, { "idx": 1388, "question": "Calculate the coordination number of the cation in the compound MgO, given r(Mg2+)=0.078nm, r(O2-)=0.132nm", "answer": "6" }, { "idx": 1389, "question": "Using dislocation theory, what is the third reason for the work hardening of pure metal single crystals?", "answer": "Dislocation multiplication occurs, and the increase in dislocation density further enhances the resistance to dislocation motion" }, { "idx": 1390, "question": "Critical nucleus", "answer": "The critical nucleus is the smallest nucleus whose growth can decrease the Gibbs free energy of the system." }, { "idx": 1391, "question": "Unit dislocation", "answer": "A unit dislocation is a dislocation whose Burgers vector equals a lattice vector." }, { "idx": 1392, "question": "What are the adverse effects of the work hardening characteristics of metals on the use of metal materials?", "answer": "It makes the metal difficult to undergo further cold plastic deformation." }, { "idx": 1393, "question": "Calculate the coordination number of the cation in the compound Cr2O3, given r(Cr3+)=0.064nm, r(O2-)=0.132nm", "answer": "6" }, { "idx": 1394, "question": "Calculate the coordination number of the cation in the compound CaF2, given r(Ca2+)=0.106nm and r(F-)=0.133nm", "answer": "8" }, { "idx": 1395, "question": "Calculate the coordination number of the cation in the compound K2O, given r(K+)=0.132nm, r(O2-)=0.132nm", "answer": "12" }, { "idx": 1396, "question": "Determine whether the following statement is correct: After recovery annealing of cold-deformed metal, its mechanical properties can return to the state before deformation.", "answer": "Incorrect" }, { "idx": 1397, "question": "What basic conditions must be met for atomic diffusion in solids?", "answer": "Atomic diffusion in solids requires the presence of a chemical potential gradient." }, { "idx": 1398, "question": "According to the concentration distribution of the diffusing component, what are the basic types of diffusion?", "answer": "Self-diffusion and interdiffusion" }, { "idx": 1399, "question": "Determine whether the following statement is correct: Edge dislocations can undergo cross-slip.", "answer": "Incorrect" }, { "idx": 1400, "question": "According to the path of atomic diffusion, what are the basic types of diffusion?", "answer": "Bulk diffusion, grain boundary diffusion, dislocation diffusion, surface diffusion" }, { "idx": 1401, "question": "Determine whether the following statement is correct: The distinction between hot (deformation) working and cold (deformation) working of metals is based on the temperature of the deformation process.", "answer": "Incorrect" }, { "idx": 1402, "question": "Determine whether the following statement is correct: Recrystallization is a process of nucleation and core growth, therefore it is a phase transformation process.", "answer": "Incorrect" }, { "idx": 1403, "question": "Determine whether the following statement is correct: The growth of recrystallization nuclei is accompanied by the movement of grain boundaries, so the driving force is grain boundary energy.", "answer": "Incorrect" }, { "idx": 1404, "question": "According to the microscopic mechanism of diffusion, what are the basic types of diffusion?", "answer": "Interstitial diffusion and vacancy diffusion" }, { "idx": 1405, "question": "Determine whether the following statement is correct: Point defects in crystals are a type of thermodynamic equilibrium defect, while dislocations are not thermodynamic equilibrium defects.", "answer": "Correct" }, { "idx": 1406, "question": "According to the relationship between the diffusion direction of components and their concentration gradient direction, what are the basic types of diffusion?", "answer": "Downhill diffusion and uphill diffusion" }, { "idx": 1407, "question": "According to whether new phases are formed during diffusion, what are the basic types of diffusion?", "answer": "Single-phase diffusion and reactive diffusion" }, { "idx": 1408, "question": "In interstitial solid solutions, by what mechanism do solute atoms diffuse?", "answer": "In interstitial solid solutions, solute atoms diffuse by the interstitial mechanism." }, { "idx": 1409, "question": "In substitutional solid solutions, by what mechanism do solute atoms diffuse?", "answer": "In substitutional solid solutions, solute atoms diffuse by the vacancy mechanism." }, { "idx": 1410, "question": "What type of diffusion is the growth of uniform austenite grains?", "answer": "The growth of uniform austenite grains belongs to self-diffusion." }, { "idx": 1411, "question": "What type of diffusion does the homogenization of intracrystalline segregation during diffusion annealing belong to?", "answer": "The homogenization of intracrystalline segregation during diffusion annealing belongs to interdiffusion." }, { "idx": 1412, "question": "By what mechanism do metal materials conduct heat?", "answer": "Metal materials mainly conduct heat through free electrons" }, { "idx": 1413, "question": "Will thermoplastic with a lower glass transition temperature (Tg) exhibit 'work hardening' after deformation? Please briefly explain.", "answer": "Thermoplastics with a lower glass transition temperature will exhibit 'work hardening' after deformation. This is because as the degree of deformation increases, the molecular chains gradually align along the direction of the external force, leading to strain hardening." }, { "idx": 1414, "question": "What is the electron magnetic moment?", "answer": "The electron magnetic moment refers to the orbital magnetic moment generated by the electron's motion around the nucleus and the spin magnetic moment produced by the electron's spin." }, { "idx": 1415, "question": "Will ceramic materials exhibit 'work hardening' after deformation? Please briefly explain.", "answer": "Ceramic materials will not exhibit 'work hardening' after deformation, because ceramic materials cannot undergo plastic deformation." }, { "idx": 1416, "question": "By what mechanism do polymer materials conduct heat?", "answer": "Polymer materials mainly conduct heat through molecular conduction" }, { "idx": 1417, "question": "What conditions must be satisfied for spontaneous dislocation reactions?", "answer": "Geometric conditions and energy conditions." }, { "idx": 1418, "question": "What is atomic magnetic moment?", "answer": "Atomic magnetic moment refers to the total magnetic moment composed of the orbital magnetic moment and the spin magnetic moment of electrons." }, { "idx": 1419, "question": "The bonding type of a material determines the level of its elastic modulus. What type of bond predominates in oxide ceramic materials, how strong is the bonding, and thus how is their elastic modulus?", "answer": "Oxide ceramic materials are predominantly ionic bonded, with strong bonding, hence their elastic modulus is relatively high." }, { "idx": 1420, "question": "What is the difference in the conduction mechanisms between intrinsic semiconductors and doped semiconductors?", "answer": "In intrinsic semiconductors, the charge carriers participating in conduction are electrons in the conduction band and an equal number of holes in the valence band, with the Fermi level located at the center of the band gap. In doped semiconductors, the charge carriers participating in conduction are electrons in the conduction band and an unequal number of holes in the valence band, with the Fermi level not located at the center of the band gap—either shifting upward (as in $\\mathbf{n}$-type semiconductors) or downward (as in p-type semiconductors)." }, { "idx": 1421, "question": "The type of bonding in a material determines the level of its elastic modulus. What type of bond is present along the molecular chains of polymer materials, and what type of bond exists between the molecular chains, hence how is their elastic modulus?", "answer": "The bonds along the molecular chains of polymer materials are covalent bonds, and the bonds between the molecular chains are molecular bonds, hence their elastic modulus is the lowest." }, { "idx": 1422, "question": "What is the relationship between the magnetic properties of a substance and the filling of electrons outside the atomic nucleus?", "answer": "Substances with completely filled electron shells in atoms are diamagnetic; substances with unfilled electron shells in atoms are paramagnetic or ferromagnetic." }, { "idx": 1423, "question": "Write the names and reaction formulas of all possible three-phase equilibrium reactions in a binary phase diagram.", "answer": "Eutectic reaction: $\\\\operatorname{L}{\\\\overrightarrow{\\\\longrightarrow}}\\\\alpha+\\\\beta$ Eutectoid reaction: $\\\\gamma\\\\mathrm{=}\\\\mathrm{=}\\\\alpha+\\\\beta$ Monotectic reaction: $\\\\mathrm{L}_{1}\\\\longrightarrow\\\\mathrm{L}_{2}+\\\\updelta$ Syntectic reaction: $\\\\updelta\\\\rightleftharpoons\\\\mathbf{L}+\\\\upgamma$ Peritectic reaction: $\\\\mathbf{L}+\\\\alpha{\\\\stackrel{\\\\rightharpoonup}{\\\\longrightarrow}}\\\\beta$ Peritectoid reaction: $\\\\gamma+\\\\alpha{\\\\stackrel{\\\\textstyle}{\\\\longrightarrow}}\\\\beta$ Syntectic reaction: $\\\\mathrm{L}_{1}+\\\\mathrm{L}_{2}\\\\stackrel{\\\\rightharpoonup}{\\\\longrightarrow}\\\\delta$" }, { "idx": 1424, "question": "By what mechanism do ceramic materials conduct heat?", "answer": "Ceramic materials primarily conduct heat through phonons" }, { "idx": 1425, "question": "What is an extended dislocation? Please give an example.", "answer": "A dislocation configuration consisting of two partial dislocations and a stacking fault sandwiched between them. For example, in face-centered cubic crystals, $\\\\frac{a}{6}$ [121] $^+$ stacking fault $+{\\\\frac{a}{6}}$ [211]." }, { "idx": 1426, "question": "What is the intrinsic magnetic moment of a substance?", "answer": "The intrinsic magnetic moment of a substance refers to the sum of all atomic magnetic moments in the absence of an external magnetic field." }, { "idx": 1427, "question": "Which is the most closely packed plane in a face-centered cubic crystal?", "answer": "111" }, { "idx": 1428, "question": "What is the most close-packed direction in a face-centered cubic crystal?", "answer": "<110>" }, { "idx": 1429, "question": "The type of bonding in a material determines the level of its elastic modulus. What type of bond predominates in metallic materials, how strong is the bonding, and consequently, what is their elastic modulus like?", "answer": "Metallic materials are primarily bonded by metallic bonds, which are relatively weak, resulting in a lower elastic modulus." }, { "idx": 1430, "question": "Which is the most closely packed plane of a body-centered cubic crystal?", "answer": "110" }, { "idx": 1431, "question": "The commonly used method to refine grains in casting processes is (19)", "answer": "(19) Increasing the cooling rate" }, { "idx": 1432, "question": "What is the most densely packed direction in a body-centered cubic crystal?", "answer": "<111>" }, { "idx": 1433, "question": "What is the most closely packed direction in a hexagonal close-packed crystal?", "answer": "<1120>" }, { "idx": 1434, "question": "A metal with finer grains has (16)_ strength and hardness compared to the same metal with coarser grains", "answer": "(16) higher" }, { "idx": 1435, "question": "Ferrite is a solid solution of carbon in what type of iron?", "answer": "α" }, { "idx": 1436, "question": "Metals with finer grains have _ (17) plasticity and toughness than the same metals with coarser grains", "answer": "(17) better" }, { "idx": 1437, "question": "Which is the most closely packed plane in a hexagonal close-packed crystal?", "answer": "(0001)" }, { "idx": 1438, "question": "The commonly used method for refining grains in casting processes is (20)", "answer": "(20) Adding nucleating agents" }, { "idx": 1439, "question": "The commonly used method for refining grains in casting processes is (21)", "answer": "(21) stirring or vibration" }, { "idx": 1440, "question": "What is the solid solution method of carbon in ferrite?", "answer": "Interstitial" }, { "idx": 1441, "question": "A metal with finer grains exhibits this phenomenon compared to the same metal with coarser grains, which is called (18) strengthening", "answer": "(18) fine grain" }, { "idx": 1442, "question": "For an edge dislocation line, what is the relationship between its climb direction and the Burgers vector?", "answer": "Perpendicular" }, { "idx": 1443, "question": "For a screw dislocation line, what is the relationship between its Burgers vector and the dislocation line?", "answer": "Parallel" }, { "idx": 1444, "question": "What strengthening mechanism results in the change of ferrite properties?", "answer": "Solid solution" }, { "idx": 1445, "question": "There are five possible types of cementite in iron-carbon alloys, listed in order of formation temperature from high to low: (22) cementite, (23) cementite, (24) cementite, (25) cementite, and (26) cementite.", "answer": "(22) primary; (23) eutectic; (24) secondary; (25) eutectoid; (26) tertiary;" }, { "idx": 1446, "question": "For a screw dislocation line, what is the relationship between its cross-slip motion direction and the Burgers vector?", "answer": "Perpendicular" }, { "idx": 1447, "question": "For an edge dislocation line, what is the relationship between the direction of its slip motion and the Burgers vector?", "answer": "Parallel" }, { "idx": 1448, "question": "Compared to pure iron, how do the plasticity and toughness of ferrite change?", "answer": "Lower" }, { "idx": 1449, "question": "Compared to pure iron, how do the strength and hardness of ferrite change?", "answer": "Higher" }, { "idx": 1450, "question": "For a screw dislocation line, what is the relationship between its slip direction and the Burgers vector?", "answer": "Perpendicular" }, { "idx": 1451, "question": "For an edge dislocation line, what is the relationship between its Burgers vector and the dislocation line?", "answer": "Perpendicular" }, { "idx": 1452, "question": "For a mixed dislocation line, what is the relationship between its Burgers vector and the dislocation line?", "answer": "Neither perpendicular nor parallel" }, { "idx": 1453, "question": "After cold plastic deformation, the strength and hardness of metal (40)", "answer": "increase" }, { "idx": 1454, "question": "After cold plastic deformation of metal, its plasticity and toughness (41)", "answer": "decrease" }, { "idx": 1455, "question": "For a mixed dislocation line, what is the relationship between the direction of its slip motion and the dislocation line?", "answer": "Perpendicular" }, { "idx": 1456, "question": "After metal undergoes cold plastic deformation, this phenomenon is called (42) strengthening or (43)", "answer": "deformation; work hardening" }, { "idx": 1457, "question": "For metals that have undergone pre-cold plastic deformation, (44) annealing should be performed before further cold plastic deformation to improve their (45)", "answer": "recrystallization; plasticity and toughness" }, { "idx": 1458, "question": "For metals that have undergone pre-cold plastic deformation, recrystallization annealing should be performed before further cold plastic deformation, and the annealing temperature is (46)", "answer": "0.4Tm" }, { "idx": 1459, "question": "For cold-formed components, (47) annealing should be performed promptly after forming to remove (48) and prevent the components from (49) or (50) during use.", "answer": "stress relief; residual internal stress; deformation; cracking" }, { "idx": 1460, "question": "What effect do second-phase particles have on the plastic deformation of alloys? Use dislocation theory to explain the mechanism.", "answer": "Key points: They increase the strength of the alloy and the flow resistance during plastic deformation. This is because the resistance increases when dislocations cut through deformable second-phase particles or bypass non-deformable second-phase particles." }, { "idx": 1461, "question": "What is intrinsic semiconductor?", "answer": "A semiconductor with high purity and no doped impurities is called an intrinsic semiconductor." }, { "idx": 1462, "question": "What is constitutional supercooling?", "answer": "During the solidification process of an alloy, although the actual temperature distribution is constant, the liquid ahead of the solid/liquid interface becomes supercooled due to changes in solute distribution in the liquid. This supercooling, determined by both the liquid composition changes and the actual temperature distribution, is called constitutional supercooling." }, { "idx": 1463, "question": "The direction of crystal slip is the direction of dislocation line movement within the slip plane.", "answer": "False" }, { "idx": 1464, "question": "What is a doped semiconductor?", "answer": "A semiconductor formed by intentionally adding a small amount of impurity elements to an intrinsic semiconductor is called a doped semiconductor. If the doping element is from Group VA in the periodic table, it forms an n-type semiconductor; if the doping element is from Group IIIA, it forms a p-type semiconductor." }, { "idx": 1465, "question": "What is the difference in Fermi levels between intrinsic semiconductors and doped semiconductors?", "answer": "The Fermi level of an intrinsic semiconductor is located at the center of the band gap, while the Fermi level of a doped semiconductor is not at the center of the band gap. The Fermi level of an n-type semiconductor shifts upward relative to the center of the band gap, and the Fermi level of a p-type semiconductor shifts downward relative to the center of the band gap." }, { "idx": 1466, "question": "A pure edge dislocation loop can form in a crystal.", "answer": "True" }, { "idx": 1467, "question": "What is the difference between the carriers in intrinsic semiconductors and doped semiconductors?", "answer": "The carriers in intrinsic semiconductors are electrons in the conduction band and holes in the valence band, with an equal number of electrons and holes. In doped semiconductors, the carriers are also electrons in the conduction band and holes in the valence band, but the number of electrons and holes is not equal. In n-type semiconductors, the number of electrons is greater than the number of holes, while in p-type semiconductors, the number of holes is greater than the number of electrons." }, { "idx": 1468, "question": "A screw dislocation line can only undergo glide motion, not climb motion.", "answer": "Correct" }, { "idx": 1469, "question": "In a binary alloy phase diagram, the greater the distance between the liquidus and solidus lines, the worse the fluidity of the alloy.", "answer": "Correct" }, { "idx": 1470, "question": "The presence of vacancies always increases the free energy of a crystal.", "answer": "False" }, { "idx": 1471, "question": "The presence of dislocations always increases the free energy of a crystal.", "answer": "Correct" }, { "idx": 1472, "question": "The driving force for the growth of recrystallization nuclei is the reduction of deformation stored energy, while the driving force for the growth of recrystallized grains is the reduction of total grain boundary energy.", "answer": "Correct" }, { "idx": 1473, "question": "In essence, a twin is also a stacking fault", "answer": "Correct" }, { "idx": 1474, "question": "Under the condition of no solid-phase diffusion and complete liquid-phase mixing, constitutional supercooling cannot occur during the solidification of solid solutions.", "answer": "Correct" }, { "idx": 1475, "question": "In the vertical section of a ternary phase diagram, the lever rule cannot be used to determine the mass fractions of the three equilibrium phases.", "answer": "Correct" }, { "idx": 1476, "question": "At 727°C, the maximum carbon content in the equilibrium state of iron-carbon alloy is w_C=0.0218% for ferrite and w_C=0.77% for austenite. Where are the carbon atoms located in the ferrite and austenite crystals?", "answer": "The carbon atoms are located at the centers of the flattened octahedral interstitial sites in ferrite crystals and at the centers of the regular octahedral interstitial sites in austenite crystals." }, { "idx": 1477, "question": "Explain the characteristics of grain growth process during heating of cold-deformed metals", "answer": "Grain growth. It causes some property changes, such as decreases in strength, plasticity, and toughness. Along with grain growth, other structural changes also occur, such as recrystallization texture." }, { "idx": 1478, "question": "Using the constitutional undercooling theory, explain the relationship between the casting properties (fluidity, distribution characteristics of shrinkage porosity) of an alloy and the vertical distance between the liquidus and solidus lines in its phase diagram.", "answer": "The casting properties (fluidity, distribution characteristics of shrinkage porosity) of an alloy are related to the vertical distance between the liquidus and solidus lines in its phase diagram. The larger the vertical distance, the poorer the fluidity of the alloy, and the more likely shrinkage porosity is to be distributed dispersedly. The condition for constitutional undercooling is given by: $\\\\frac{-m C_{0}(1-k_{0})}{D k_{0}}$, where G is the actual temperature gradient in the liquid ahead of the liquid/solid interface, R is the velocity of the liquid/solid interface movement, $\\\\pmb{D}$ is the diffusion coefficient of solute atoms in the liquid phase, $\\\\mathbf{\\abla}m$ is the slope of the liquidus line in the alloy phase diagram, $C_{0}$ is the composition of the alloy, and $k_{0}$ is the equilibrium partition coefficient of the alloy. It can be proven that $\\\\frac{\\\\mathrm{-}m C_{0}~\\\\left(1-k_{0}\\\\right)}{k_{0}}$ in the above equation is the vertical distance between the liquidus and solidus lines for the $C_{0}$ alloy in its phase diagram. Clearly, the larger the vertical distance between the liquidus and solidus lines, the greater the tendency for constitutional undercooling in the alloy, and the more the liquid/solid interface tends to grow dendritically. This hinders the fluidity of the alloy melt and also causes dispersed shrinkage porosity to form in the interdendritic regions due to insufficient feeding." }, { "idx": 1479, "question": "For the equilibrium state of T12 steel (w_C=1.2%), write its phase constituents at room temperature and calculate the mass fraction of each phase constituent.", "answer": "Phase constituents: α+Fe3C. w_α=(6.69-1.2)/(6.69-0.0008)×100%≈82.07%. w_Fe3C=(1.2-0.0008)/(6.69-0.0008)×100%≈17.93%." }, { "idx": 1480, "question": "Explain the characteristics of the recovery process when cold-deformed metal is heated", "answer": "During the recovery process, the microstructure does not change, and the elongated grains in the deformed state are still maintained. The recovery process completely eliminates the macroscopic first-class stresses caused by deformation and removes most of the microscopic second-class stresses. Generally, mechanical properties change little during recovery, with hardness and strength slightly decreasing, plasticity slightly improving, and some physical properties undergoing significant changes—resistivity notably decreases, while density increases. The deformation-stored energy is partially released during the recovery stage." }, { "idx": 1481, "question": "Qualitatively compare the elastic modulus of ceramic materials, metal materials, and polymer materials", "answer": "Among the three types of materials, ceramic materials have the highest elastic modulus, metal materials have the next highest elastic modulus, and polymer materials have the lowest elastic modulus." }, { "idx": 1482, "question": "At 727°C, the maximum carbon content in the equilibrium iron-carbon alloy is w_C=0.0218% for ferrite, while the carbon content in austenite is w_C=0.77%. Explain why the carbon content differs so greatly between the two.", "answer": "Because the radius of the flat octahedral interstitial sites in the ferrite crystal is much smaller than that of the regular octahedral interstitial sites in the austenite crystal." }, { "idx": 1483, "question": "Explain the role of Cottrell atmosphere in strengthening metals", "answer": "Cottrell atmosphere: In BCC crystals (such as carbon steel), small-sized atoms like C and N preferentially distribute in the tensile stress region of edge dislocations, pinning the dislocations. To move the dislocations, they must be torn away from the pinning effect, requiring additional stress, thereby increasing the material's strength." }, { "idx": 1484, "question": "There is a face-centered cubic single crystal with a right-handed screw dislocation on the (11) plane, whose Burgers vector is a/2 [101], and another right-handed screw dislocation on the (111) plane, whose Burgers vector is a/2 [011]. These two dislocations meet at the intersection line of the two slip planes and form a new perfect dislocation. Explain what type of dislocation the newly formed perfect dislocation is, whether this dislocation can glide, and why?", "answer": "Since the dislocation line is the intersection line of the two slip planes, the dislocation line is [1 0 -1]. It can be seen that the dislocation line is neither parallel nor perpendicular to the Burgers vector, so the newly formed dislocation is a mixed-type dislocation. Given the dislocation line and Burgers vector of the newly formed dislocation, their cross product yields the normal vector of the slip plane for the new dislocation, which is [1 -1 1]. The slip plane of this dislocation is (1 -1 1). Because this slip plane is a close-packed plane in the face-centered cubic structure, the dislocation can glide." }, { "idx": 1485, "question": "From the perspective of bonding in materials, analyze the reasons for the differences in elastic modulus among ceramic materials, metal materials, and polymer materials", "answer": "The magnitude of a material's elastic modulus depends on the strength of the bonding within the material. Ceramic materials are bonded by strong ionic or covalent bonds, hence their elastic modulus is very large; metal materials are bonded by weaker metallic bonds, hence their elastic modulus is smaller; in polymer materials, the molecular chains are bonded by strong covalent bonds, but the chains themselves are bonded by very weak secondary bonds, hence their elastic modulus is very small." }, { "idx": 1486, "question": "Given that the recrystallization activation energy of a Cu-30%Zn alloy is 250 kJ/mol, and it takes 1 hour for this alloy to complete recrystallization at a constant temperature of 400°C, calculate how many hours it will take for this alloy to complete recrystallization at a constant temperature of 390°C.", "answer": "From the formula, we have $$\\\\begin{array}{l}{{\\\\displaystyle\\\\frac{t_{2}}{t_{1}}=\\\\mathrm{\\\\bar{exp}}\\\\Big[-\\\\frac{Q}{R}\\\\Big(\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{2}}\\\\Big)\\\\Big]}}\\\\ {{\\\\displaystyle=\\\\mathrm{exp}\\\\Big[-\\\\frac{250\\\\times10^{3}}{8.314}\\\\times\\\\Big(\\\\frac{1}{400+273}-\\\\frac{1}{390+273}\\\\Big)\\\\Big]}}\\\\ {{\\\\displaystyle=1.962}}\\\\end{array}$$ Therefore, $$t_{2}=t_{1}\\\\times1.962=1.962{\\\\mathrm{h}}$$" }, { "idx": 1487, "question": "Explain the role of Suzuki atmosphere in strengthening metals", "answer": "Suzuki atmosphere: In FCC crystals (such as stainless steel), alloying elements like Ni and α preferentially distribute in stacking fault regions, reducing the stacking fault energy and expanding the extended dislocation zone. To move these extended dislocations, additional stress is required, thereby increasing the material's strength." }, { "idx": 1488, "question": "For the equilibrium state of T12 steel (w_C=1.2%), write its room temperature microstructure constituents and calculate the mass fraction of each constituent.", "answer": "Microstructure constituents: P+Fe3CⅡ. w_P=(6.69-1.2)/(6.69-0.77)×100%≈92.74%. w_Fe3CⅡ=(1.2-0.77)/(6.69-0.77)×100%≈7.26%" }, { "idx": 1489, "question": "Which hazardous substances are prohibited by the EU RoHS Directive (now enacted as a regulation) that came into effect on July 1, 2006? What are their maximum allowable concentrations?", "answer": "Prohibited substances include lead (Pb, 0.1%), mercury (Hg, 0.1%), cadmium (Cd, 0.01%), hexavalent chromium (Cr+6, 0.1%), polybrominated biphenyls (PPB, 0.1%), and polybrominated diphenyl ethers (PBDE, 0.1%)." }, { "idx": 1490, "question": "There is a face-centered cubic single crystal with a right-handed screw dislocation on the (11) plane, whose Burgers vector is a/2 [101], and another right-handed screw dislocation on the (111) plane, whose Burgers vector is a/2 [011]. These two dislocations meet at the intersection line of the two slip planes and form a new perfect dislocation. If a tensile stress of 17.2MPa is applied along the [010] crystal direction, calculate the force per unit length acting on the newly formed perfect dislocation and indicate its direction (given the lattice constant a=0.2nm).", "answer": "cosφ=1/√3, cosλ=1/√2. According to Schmid's law, the resolved shear stress acting on the slip direction of the newly formed dislocation's slip plane is τ0=17.2×1/√3×1/√2 MPa=7.0MPa. Therefore, the force per unit length acting on the dislocation line is f=τb=(0.2×10^-9×7×10^6)/√2 N/m=10^-3 N/m. Its direction is perpendicular to the dislocation line direction [101], pointing towards the unslipped region." }, { "idx": 1491, "question": "What is aging?", "answer": "Aging refers to the process where a supersaturated solid solution is retained at room temperature or elevated temperature for a period of time, leading to the precipitation of a second phase from the matrix. The aging precipitation process is controlled by solute diffusion, and a series of metastable phases (transition phases) may form during the precipitation." }, { "idx": 1492, "question": "What are the main solder systems that can replace lead-containing solders?", "answer": "The main alternatives to lead-containing solders are: Sn-Ag-Cu, Sn-Zn, Sn-Bi, and Sn-In systems." }, { "idx": 1493, "question": "Explain the characteristics of the recrystallization process during heating of cold-deformed metals", "answer": "The microstructure undergoes changes, transforming from elongated grains due to cold deformation into new equiaxed grains. Mechanical properties change drastically, with strength and hardness sharply decreasing and plasticity increasing, returning to the state before deformation. The stored energy from deformation is fully released during recrystallization, eliminating the three types of stresses (lattice distortion) and reducing dislocation density." }, { "idx": 1494, "question": "A single crystal (BCC) with a rod axis [213] is stretched along the direction to undergo plastic deformation. Given that the slip plane of the single crystal is {110}, determine the initial slip system.", "answer": "The single crystal has a BCC structure, and [213] lies within the orientation triangle [001]~[1¯11]~[101]. Therefore, the initial slip system is (011)[111]." }, { "idx": 1495, "question": "A single crystal (BCC) rod is stretched along the rod axis [213] direction to undergo plastic deformation. Given that the slip plane of the single crystal is {110}, determine the double slip system.", "answer": "The double slip system is (011)[111]－(011)[111]." }, { "idx": 1496, "question": "What is the closest packed direction in a body-centered cubic crystal?", "answer": "<111>" }, { "idx": 1497, "question": "What is the most closely packed plane in a face-centered cubic crystal?", "answer": "{111}" }, { "idx": 1498, "question": "What is the most closely packed plane in a hexagonal close-packed crystal?", "answer": "(0001)" }, { "idx": 1499, "question": "What is the reason for age hardening in Al-4.5%Cu alloy?", "answer": "The reasons for age hardening are: first, when the precipitated metastable disk-shaped phase has a certain orientation relationship with the parent phase, it generates significant elastic strain in the matrix, which can markedly strengthen the alloy; second, when the alloy undergoes deformation, the interaction between dispersed particles and dislocations strengthens the alloy. If the precipitate particles are deformable, the work required to increase the surface energy of the particles when dislocations cut through them raises the resistance to dislocation motion, thereby strengthening the alloy. If the precipitate particles are strong and coherent with the matrix, the dislocation lines find it difficult to cut through the particles and will bypass them under applied stress, leaving dislocation loops. For the alloy to continue deforming, the stress exerted by the particles as the dislocation lines bypass them must be overcome, requiring further increase in the applied stress, meaning the alloy achieves age hardening." }, { "idx": 1500, "question": "What is the most close-packed direction in a face-centered cubic crystal?", "answer": "<110>" }, { "idx": 1501, "question": "What is the most closely packed plane of a body-centered cubic crystal?", "answer": "{110}" }, { "idx": 1502, "question": "A single crystal (BCC) rod is stretched along the rod axis [213] direction to undergo plastic deformation. Given that the slip plane of the single crystal is {110}, determine the shear strain γₒ at the onset of double slip.", "answer": "Using L = l + γ(l⋅n)b, let L = [u0w], then L = [2¯13] + 4γ[111]/√6. From this, it can be determined that u = 3, w = 4, γ = √6/4. Therefore, the crystal orientation is [304], and the shear strain is √6/4." }, { "idx": 1503, "question": "How do the strength and hardness of a metal with finer grains compare to the same metal with coarser grains", "answer": "Higher" }, { "idx": 1504, "question": "What is the most closely packed direction in a hexagonal close-packed crystal?", "answer": "<11\\overline{20}>" }, { "idx": 1505, "question": "How do the plasticity and toughness of a metal with finer grains compare to the same metal with coarser grains", "answer": "Better" }, { "idx": 1506, "question": "Compared with pure iron, how do the plasticity and toughness of ferrite change", "answer": "Lower" }, { "idx": 1507, "question": "What is the reason for the change in properties of ferrite compared to pure iron", "answer": "Solid solution" }, { "idx": 1508, "question": "What is the phenomenon called when a metal with finer grains exhibits higher strength and hardness but lower plasticity and toughness compared to the same metal with coarser grains?", "answer": "Grain refinement strengthening" }, { "idx": 1509, "question": "Ferrite is a solid solution of carbon in what－Fe", "answer": "α" }, { "idx": 1510, "question": "A single crystal (BCC) rod is stretched along the rod axis [213] direction to undergo plastic deformation. Given that the slip plane of the single crystal is {110}, determine the final orientation of the test rod (assuming the rod does not fracture before reaching a stable orientation).", "answer": "Let the stable orientation be [u′0w′]. To satisfy n = [000], it is required that [u′0w′] × ([111] ± [1¯11]) = [000], which implies u′ = w′. Therefore, the final stable orientation is [101]." }, { "idx": 1511, "question": "Compared with pure iron, how do the strength and hardness of ferrite change", "answer": "Higher" }, { "idx": 1512, "question": "A single crystal (BCC) rod is stretched along the rod axis [213] direction to undergo plastic deformation. Given that the slip plane of the single crystal is {110}, determine the rotation law and rotation axis during the slip process.", "answer": "During double slip, the specimen axis on one hand rotates towards [111], with the rotation axis n₁ = [304] × [111] = [413], and on the other hand also rotates towards [1¯11]; the rotation axis n₂ = [304] × [1¯11] = [41¯3]; the resultant rotation axis is [020] i.e., [010], so after double slip, point F moves along the edge [001]－[101]." }, { "idx": 1513, "question": "For an edge dislocation line, the direction of its slip motion is (18) to the Burgers vector", "answer": "(18) parallel" }, { "idx": 1514, "question": "What are the common methods for refining grains in casting processes (the second method)", "answer": "Adding nucleating agents, stirring or vibration" }, { "idx": 1515, "question": "What is the common method to refine grains in casting process (the first method)", "answer": "Increase the cooling rate" }, { "idx": 1516, "question": "For an edge dislocation line, its Burgers vector is __(17)__ to the dislocation line", "answer": "(17) perpendicular" }, { "idx": 1517, "question": "For an edge dislocation line, its climb direction is _（19）_ to the Burgers vector", "answer": "（19）perpendicular" }, { "idx": 1518, "question": "What type of solid solution is ferrite formed by carbon in α-Fe", "answer": "Interstitial" }, { "idx": 1519, "question": "For a screw dislocation line, its Burgers vector is _（20） to the dislocation line", "answer": "（20） parallel" }, { "idx": 1520, "question": "For a screw dislocation line, the direction of its slip motion is _（21）_ to the Burgers vector", "answer": "（21）perpendicular" }, { "idx": 1521, "question": "After cold plastic deformation, the strength and hardness of metal (23)—", "answer": "(23) increase" }, { "idx": 1522, "question": "After cold plastic deformation, the plasticity and toughness of metal (24)", "answer": "(24) decrease" }, { "idx": 1523, "question": "For cold-formed components, (29)__ annealing should be performed promptly after forming to remove (30)__ and prevent deformation or cracking during use.", "answer": "(29) stress relief; (30) residual internal stress" }, { "idx": 1524, "question": "After metal undergoes cold plastic deformation, this phenomenon is called (25) strengthening or (26) __", "answer": "(25) deformation; (26) work hardening" }, { "idx": 1525, "question": "For a screw dislocation line, the direction of its cross-slip motion is __(22)__ to the Burgers vector", "answer": "(22) perpendicular" }, { "idx": 1526, "question": "For metals that have undergone pre-cold plastic deformation, (27) annealing should be performed before further cold plastic deformation to improve their (28)", "answer": "(27) recrystallization; (28) plasticity and toughness" }, { "idx": 1527, "question": "Give examples of the basic strengthening forms of materials", "answer": "The basic strengthening forms of materials include: solid solution strengthening, strain hardening, precipitation strengthening and dispersion strengthening, grain refinement strengthening, etc." }, { "idx": 1528, "question": "Provide the Hall-Petch formula and explain the meaning and units of each physical quantity in the formula.", "answer": "The Hall-Petch formula describes the relationship between yield strength and grain size. The formula is σ_s=σ_0+K·d^(-1/2). In the formula, σ_0 (MPa) and K (MPa·m^(1/2)) are constants; d is the grain diameter (m); σ_s is the yield strength (MPa)." }, { "idx": 1529, "question": "Provide the grain boundary segregation formula and explain the meaning and units of each physical quantity in the formula", "answer": "The grain boundary segregation formula is C_gb=C_0·exp(-ΔE/kT)=C_0exp(-Q/RT). In the formula, C_gb and C_0 represent the grain boundary concentration and grain center concentration, respectively (both can use the same units, such as mass percentage concentration or volume concentration); ΔE is the interaction energy (eV); Q is the molar interaction energy (J), which reflects the change in grain boundary energy caused by segregation; R is the gas constant (J/K); k is the Boltzmann constant (eV/K); T is the absolute temperature (K)." }, { "idx": 1530, "question": "Explain the strengthening mechanism of solid solution strengthening", "answer": "Solid solution strengthening: For both substitutional atoms and interstitial atoms, under suitable conditions, atomic segregation may occur to form atmospheres. For substitutional lattices, when solute atoms are larger than solvent atoms, solute atoms tend to accumulate in the expanded region of edge dislocations; conversely, they accumulate in the compressed region. Interstitial atoms always tend to accumulate in the expanded region. This phenomenon of diffusion-driven accumulation near dislocations is called Cottrell atmosphere. The Cottrell atmosphere has a pinning effect on dislocations, thereby increasing strength." }, { "idx": 1531, "question": "Explain the strengthening mechanism of work hardening", "answer": "Work hardening: During the plastic deformation of cold-worked metals, a large number of dislocations are formed. Some of these dislocations become immobile dislocations, thereby increasing their resistance to mobile dislocations. This makes further deformation of the material difficult, resulting in work hardening or strain hardening." }, { "idx": 1532, "question": "Explain the strengthening mechanisms of precipitation strengthening and dispersion strengthening", "answer": "Precipitation strengthening and dispersion strengthening: A supersaturated solid solution undergoes decomposition upon temperature decrease or during prolonged holding (aging). The aging process is often complex; for example, in aluminum alloys, GP zones form first during aging, followed by the precipitation of transition phases (θ′′ and θ′), and finally the formation of thermodynamically stable equilibrium phases (θ). Fine precipitates dispersed in the matrix hinder dislocation motion, thereby producing a strengthening effect. This is known as 'precipitation strengthening' or 'age hardening'." }, { "idx": 1533, "question": "What is the thermodynamic condition for uphill diffusion to occur?", "answer": "The thermodynamic condition for uphill diffusion to occur is ∂²G/∂x²<0." }, { "idx": 1534, "question": "What is uphill diffusion?", "answer": "Uphill diffusion refers to the process where substances diffuse from regions of low concentration to regions of high concentration, resulting in an increased concentration gradient. Examples include the copper-rich clusters formed in the early stages of aging in aluminum-copper alloys and the solute-enriched regions formed during spinodal decomposition in certain alloy solid solutions. The true driving force for uphill diffusion is the chemical potential gradient, not the concentration gradient. Although diffusion leads to an increase in the concentration gradient, the chemical potential gradient actually decreases." }, { "idx": 1535, "question": "Recrystallization annealing", "answer": "Recrystallization annealing: For plastically deformed metals, during the reheating process, when the temperature exceeds the recrystallization temperature, new grains with low defect density are formed, restoring properties such as strength to their pre-deformation levels, while the phase structure remains unchanged." }, { "idx": 1536, "question": "Crystal", "answer": "Crystal: A solid composed of atoms, molecules, or ions arranged in a regular periodic pattern" }, { "idx": 1537, "question": "Taking Al-4.5% Cu alloy as an example, explain the aging process and the changes in its properties (hardness).", "answer": "After quenching, the aluminum alloy forms a supersaturated solid solution. Subsequent heating and holding cause the sequential precipitation of GP zones, θ'', θ', and θ phases within the solid solution. The formation of GP zones increases the material's hardness (the first aging peak). Prolonged aging leads to the dissolution of GP zones and a decrease in hardness. The formation of θ'' further increases the hardness (the second aging peak). When θ'' completely dissolves and transforms into θ' and θ' transforms into θ, the hardness begins to decline." }, { "idx": 1538, "question": "Taking the tensile curve of low-carbon steel as an example, use dislocation theory to explain the yield phenomenon", "answer": "The yield of low-carbon steel is due to the fact that carbon in low-carbon steel is an interstitial atom, which interacts with dislocations in ferrite to form solute atom atmospheres, known as Cottrell atmospheres. These atmospheres have a pinning effect on dislocations, and only under higher stress can dislocations break away from the pinning of solute atoms, manifesting as the upper yield point on the stress-strain curve. Once dislocations are unpinned and continue to glide, less stress is required, which appears as the lower yield point and the horizontal plateau on the stress-strain curve." }, { "idx": 1539, "question": "Please explain the effect of stacking fault energy level on the cross-slip of screw dislocations", "answer": "For metals with high stacking fault energy, during deformation, perfect dislocations are less likely to dissociate. When encountering obstacles, they can continue moving through cross-slip until they meet other dislocations and form tangles. In contrast, for metals with low stacking fault energy, their perfect dislocations easily dissociate into configurations consisting of two partial dislocations and a stacking fault, making cross-slip difficult and resulting in poor mobility of dislocation configurations." }, { "idx": 1540, "question": "Give the Hall-Petch formula and explain the meaning and units of each physical quantity in the formula", "answer": "The Hall-Petch formula describes the relationship between yield strength and grain size. The formula is: σ_s = σ_0 + K·d^(-1/2). In the formula, σ_0 (MPa) is a constant; K (MPa·m^(1/2)) is a constant; d (m) is the grain diameter; σ_s (MPa) is the yield strength." }, { "idx": 1541, "question": "Uphill diffusion", "answer": "Uphill diffusion: The diffusion of atoms from a low concentration to a high concentration area, where the driving force for diffusion is the chemical potential gradient" }, { "idx": 1542, "question": "Give the one-dimensional form of Fick's second law and explain the meaning and units of each physical quantity in the formula", "answer": "The one-dimensional form of Fick's second law is: ∂C/∂t = D·∂²C/∂x². In the formula, C (g/cm³) is the concentration; D (cm²/s) is the diffusion coefficient; t (s) is the time; x (cm) is the distance." }, { "idx": 1543, "question": "First-order phase transition", "answer": "First-order phase transition: A phase transition where the free energies of the old and new phases are equal before and after the transition, but their first-order partial derivatives are unequal." }, { "idx": 1544, "question": "Taking the tensile curve of low-carbon steel as an example, use dislocation theory to explain the phenomenon of work hardening", "answer": "When deformation continues, the stress increases again due to a significant rise in the number of dislocations, a phenomenon known as work hardening. This occurs because cold-deformed metals generate a large number of dislocations during plastic deformation, some of which become immobile dislocations. These immobile dislocations increase resistance to mobile dislocations, making further deformation of the material more difficult, resulting in work hardening or strain hardening." }, { "idx": 1545, "question": "5．Extended dislocation", "answer": "A complete dislocation dissociates into two partial dislocations and the stacking fault between them, and this dislocation configuration is called an extended dislocation." }, { "idx": 1546, "question": "Please explain the effect of stacking fault energy on the work hardening rate of metals", "answer": "For screw dislocations, cross-slip allows two screw dislocations of opposite signs (left and right) on the same slip plane to meet and annihilate each other, thereby reducing the dislocation multiplication rate. For metals with low stacking fault energy, cross-slip of dislocations is difficult, resulting in a high dislocation multiplication rate during deformation and an increased work hardening rate." }, { "idx": 1547, "question": "Using Al-4.5%Cu alloy as an example, explain the microstructural transformation during 130% aging", "answer": "After quenching, the aluminum alloy forms a supersaturated solid solution. Subsequent heating and holding will cause the sequential precipitation of GP zones, θ′′, θ′, and θ phases within the solid solution. Among these, α and θ are equilibrium phases; GP zones, θ′′, and θ′ are metastable phases." }, { "idx": 1548, "question": "Taking the Al-4.5%Cu alloy as an example, explain the hardness changes during the 130% aging process", "answer": "The formation of GP zones will increase the hardness of the material (the first aging peak). After prolonged aging, the GP zones dissolve, and the hardness decreases. The formation of θ′′ causes the hardness to continue increasing (the second aging peak). When θ′′ completely dissolves and transforms into θ′ and θ′ transforms into θ, the hardness begins to decrease." }, { "idx": 1549, "question": "1. Pearlitic transformation", "answer": "The phase transformation in which the austenite of Fe-C alloy simultaneously transforms into the eutectoid structure of ferrite and cementite below the eutectoid transformation temperature during cooling." }, { "idx": 1550, "question": "2. Intermediate phase", "answer": "A compound formed by two pure components with a structure different from theirs, whose composition lies between the two pure components." }, { "idx": 1551, "question": "4.Coordination number", "answer": "The number of nearest neighbor atoms around any lattice node atom in the crystal lattice." }, { "idx": 1552, "question": "3.Solid solution strengthening", "answer": "The presence of solute atoms in the solid solution causes lattice distortion of the solvent, thereby hindering dislocation motion and increasing its strength." }, { "idx": 1553, "question": "7. Up-hill diffusion", "answer": "The diffusion of atoms from a low concentration to a high concentration region, driven by the chemical potential gradient." }, { "idx": 1554, "question": "6. Eutectic transformation", "answer": "The transformation in a binary alloy where a single liquid phase simultaneously transforms into two different solid phases at a constant temperature." }, { "idx": 1555, "question": "8. Aging of aluminum alloys", "answer": "After high-temperature solution treatment, aluminum alloys are rapidly cooled to form a supersaturated solid solution, followed by the precipitation of metastable phases during subsequent heating and holding processes." }, { "idx": 1556, "question": "Based on the relationship between dislocation motion and crystal slip, analyze the relationship between the Burgers vector of a pure screw dislocation and the direction of the dislocation line", "answer": "The Burgers vector is parallel to the direction of the dislocation line" }, { "idx": 1557, "question": "10. First-order phase transition", "answer": "A phase transition where the free energies of the old and new phases are equal before and after the transition, but their first-order partial derivatives are unequal." }, { "idx": 1558, "question": "Give the Hall-Petch formula and explain the meaning and units of each physical quantity in the formula", "answer": "The Hall-Petch formula describes the relationship between yield strength and grain size: σₒ=σ₀+K·d^(-1/2). In the formula, σ₀ (MPa) and K (MPa·m^(1/2)) are constants; d is the grain diameter (m); σₒ is the yield strength (MPa)." }, { "idx": 1559, "question": "9. Recrystallization", "answer": "For plastically deformed metals, during the reheating process, when the temperature exceeds the recrystallization temperature, new grains with low defect density are formed, restoring their properties such as strength to the pre-deformation level, while the phase structure remains unchanged." }, { "idx": 1560, "question": "State Fick's second law (one-dimensional) and explain the meaning and units of each physical quantity in the formula", "answer": "Fick's second law (one-dimensional): ∂C/∂t=D(∂²C/∂x²). In the formula, C is the concentration (g/cm³); D is the diffusion coefficient (cm²/s); t is the time (s); x is the distance (cm)." }, { "idx": 1561, "question": "Provide the grain boundary segregation formula and explain the meaning and units of each physical quantity in the formula", "answer": "Grain boundary segregation formula: C_gb=C₀·exp(-ΔE/kT)=C₀exp(-Q/RT). In the formula, C_gb and C₀ represent the grain boundary concentration and grain center concentration, respectively, and both can use the same units, such as mass percentage concentration or volume concentration; ΔE is the interaction energy (eV); Q is the mole interaction energy (J); R is the gas constant (J/K); k is the Boltzmann constant (eV/K); T is the absolute temperature (K)." }, { "idx": 1562, "question": "Based on the relationship between dislocation motion and crystal slip, analyze the relationship between the Burgers vector of a pure screw dislocation and the direction of dislocation line motion", "answer": "The Burgers vector is perpendicular to the direction of dislocation line motion" }, { "idx": 1563, "question": "Based on the relationship between dislocation motion and crystal slip, analyze the relationship between the Burgers vector of a pure screw dislocation and the direction of crystal slip", "answer": "The Burgers vector is parallel to the direction of crystal slip" }, { "idx": 1564, "question": "Based on the relationship between dislocation motion and crystal slip, analyze the relationship between the Burgers vector of a pure edge dislocation and the direction of dislocation line motion", "answer": "The Burgers vector is parallel to the direction of dislocation line motion" }, { "idx": 1565, "question": "Based on the relationship between dislocation motion and crystal slip, analyze the relationship between the Burgers vector of a pure edge dislocation and the direction of the dislocation line", "answer": "The Burgers vector is perpendicular to the direction of the dislocation line" }, { "idx": 1566, "question": "Based on the relationship between dislocation motion and crystal slip, analyze the relationship between the Burgers vector of a pure edge dislocation and the direction of crystal slip", "answer": "The Burgers vector is parallel to the direction of crystal slip" }, { "idx": 1567, "question": "Describe the main characteristics of martensitic transformation.", "answer": "The fundamental characteristics of martensitic transformation are as follows (primarily the first two points): (1) Diffusionless nature. During martensitic transformation, atomic diffusion is not required, and there is no process of atomic mixing or remixing. The new phase M has exactly the same chemical composition as the parent phase A. (2) Shear nature. Specifically manifested as: $\\textcircled{1}$ Coordinated consistency of transformation: $\\mathbf{A}{\\longrightarrow}\\mathbf{M}$ $\\mathbf{FCC}{\\rightarrow}\\mathbf{BCC}$ ). Through the coordinated movement of atoms (shear), the crystal structure changes from FCC to BCC. The displacement distance of atoms is less than the interatomic spacing. $\\textcircled{2}$ Surface relief effect. On a polished surface, if martensitic transformation occurs, a surface relief effect will be produced during shear. This is due to the lattice deformation causing shape changes in the transformation region. $\\textcircled{3}$ Habit plane. M always precipitates on specific crystallographic planes of the parent phase, accompanied by the shear of the M phase transformation, generally parallel to this plane. This plane is shared by the matrix and the M phase and is called the habit plane. $\\textcircled{4}$ A definite crystallographic orientation relationship exists between the new phase and the parent phase. Two well-known orientation relationships (for steel M transformation) are the K-S relationship and the Nishiyama relationship. In practical materials, martensitic transformation usually deviates from these relationships by a few degrees. (3) Lattice distortion accompanies martensitic transformation. (4) Martensitic transformation has a start temperature $M_{\\mathrm{s}}$ and a finish temperature $M_{\\mathrm{f}}$ (or $M_{z}$ )." }, { "idx": 1568, "question": "1. Precipitation (secondary crystallization)", "answer": "The separation of another solid phase from a solid solution." }, { "idx": 1569, "question": "Briefly describe the recovery process of deformed metal during heating and the changes in its microstructure and properties", "answer": "When the temperature is below the recrystallization temperature, the main processes include the reduction in the concentration of point defects, the elimination of internal stresses, and the change in dislocation configuration, with no alteration in the optical microstructure; at this stage, mechanical properties such as strength, hardness, and plasticity remain largely unchanged, but electrical resistance decreases significantly." }, { "idx": 1570, "question": "Briefly describe the recrystallization process of deformed metal during heating and the changes in its microstructure and properties", "answer": "After reaching or exceeding the recrystallization temperature, new grains with low defect density will form within the original deformed grains, and the grains are basically equiaxed. At this time, mechanical properties such as strength and physical properties quickly recover to the levels before deformation." }, { "idx": 1571, "question": "2. Space group", "answer": "A symmetry group formed by the combination of all symmetry elements (including microscopic symmetry elements) in a crystal structure." }, { "idx": 1572, "question": "8. Dynamic recrystallization", "answer": "The phenomenon where deformation and recrystallization occur simultaneously above the recrystallization temperature." }, { "idx": 1573, "question": "Give examples of the basic strengthening mechanisms of materials", "answer": "The basic strengthening mechanisms of materials include: solid solution strengthening, work hardening, grain refinement strengthening, and dispersion strengthening. Specific examples: carbon forms a solid solution in ferrite, causing solid solution strengthening; the hardness of pure aluminum sheets increases after repeated processing, and the strength of cold-drawn steel wires significantly improves; modification treatment can refine the strength of cast aluminum alloys, mainly due to grain refinement; the carbides in granular pearlite play the role of dispersion strengthening, and additionally, the finely dispersed carbides in tempered martensite have a good strengthening effect." }, { "idx": 1574, "question": "5. Austenite", "answer": "An interstitial solid solution of carbon dissolved in γ-Fe." }, { "idx": 1575, "question": "4. Constitutional supercooling", "answer": "During crystallization, the phenomenon where a supercooled zone forms in the liquid phase near the solid-liquid interface due to the redistribution of components between the solid and liquid phases is called constitutional supercooling." }, { "idx": 1576, "question": "In NaCl-type crystals, what type of voids are entirely filled by Na+ ions?", "answer": "Octahedral" }, { "idx": 1577, "question": "6．Critical deformation", "answer": "The minimum pre-deformation required to cause recrystallization in a metal material when heated above the recrystallization temperature." }, { "idx": 1578, "question": "Explain in detail one of the strengthening mechanisms (using solid solution strengthening as an example)", "answer": "Dislocation theory explanation of solid solution strengthening: It is necessary to elaborate on the hindering effect of various strengthening factors on dislocation motion to improve material strength. For example, whether substitutional atoms or interstitial atoms, under suitable conditions, may undergo atomic segregation to form atmospheres. For substitutional lattices, when solute atoms are larger in diameter than solvent atoms, solute atoms tend to accumulate in the expanded region of edge dislocations; conversely, they accumulate in the compressed region. Interstitial atoms always tend to accumulate in the expanded region. This phenomenon of diffusion-induced enrichment near dislocations is called the Cottrell atmosphere. The Cottrell atmosphere has a pinning effect on dislocations, increasing the resistance to dislocation motion, thereby enhancing strength." }, { "idx": 1579, "question": "What type of void does the Cs+ ion occupy in CsCl crystal?", "answer": "Cubic" }, { "idx": 1580, "question": "9. Spinodal decomposition", "answer": "A transformation in which a solid solution decomposes into two solid solutions with the same structure as the parent phase but different compositions through uphill diffusion." }, { "idx": 1581, "question": "10. Habit plane", "answer": "During solid-state phase transformation, the new phase often forms along specific atomic planes of the parent phase. The parent phase crystal plane that is parallel to the main plane of the new phase is called the habit plane." }, { "idx": 1582, "question": "7. Deformation texture", "answer": "As the amount of plastic deformation increases, the phenomenon where a certain crystallographic orientation of different grains in a polycrystal tends to become consistent." }, { "idx": 1583, "question": "3. Dislocation intersection", "answer": "The phenomenon where dislocations moving on different slip planes meet and cut through each other during motion." }, { "idx": 1584, "question": "1. The macroscopic symmetry elements of crystals include (1), (2), (3), (4), and (5)", "answer": "(1) center of symmetry; (2) axis of symmetry; (3) plane of symmetry; (4) rotoinversion axis; (5) rotoreflection axis" }, { "idx": 1585, "question": "In fluorite, what type of voids are entirely occupied by F- ions?", "answer": "Tetrahedral" }, { "idx": 1586, "question": "The microscopic mechanisms of diffusion in crystalline solids include (11)", "answer": "(11) interstitial mechanism" }, { "idx": 1587, "question": "In the heterogeneous nucleation model, what value of the contact angle θ between the nucleus and the substrate plane cannot promote nucleation?", "answer": "π" }, { "idx": 1588, "question": "The microscopic mechanisms of diffusion in crystalline solids include (12)", "answer": "(12) interstitial mechanism" }, { "idx": 1589, "question": "The microscopic mechanisms of diffusion in crystalline solids include (13)", "answer": "(13) vacancy mechanism" }, { "idx": 1590, "question": "In the heterogeneous nucleation model, when the contact angle θ between the nucleus and the substrate plane is π/2, what fraction of the homogeneous nucleation energy is the nucleation energy?", "answer": "Half" }, { "idx": 1591, "question": "The microscopic mechanisms of diffusion in crystalline solids include (14)", "answer": "(14) exchange mechanism" }, { "idx": 1592, "question": "Small-angle grain boundaries are composed of dislocations, among which symmetric tilt grain boundaries are composed of what type of dislocations?", "answer": "Edge" }, { "idx": 1593, "question": "Small-angle grain boundaries are composed of dislocations, among which twist boundaries are composed of what type of dislocations?", "answer": "Screw" }, { "idx": 1594, "question": "1. What is aging treatment?", "answer": "The process of treating the precipitation of a supersaturated solid solution is called aging treatment." }, { "idx": 1595, "question": "6．Adsorption occurring on solid surfaces can be divided into two types: (17) and (18).", "answer": "(17) Physical adsorption; (18) Chemical adsorption" }, { "idx": 1596, "question": "How does temperature affect atomic diffusion in crystalline solids?", "answer": "The higher the temperature, the greater the diffusion coefficient and the faster the diffusion rate." }, { "idx": 1597, "question": "How does the third component affect atomic diffusion in crystalline solids?", "answer": "Depending on the nature of the added third component, some promote diffusion while others hinder it." }, { "idx": 1598, "question": "How do crystal structure and solid solution type affect atomic diffusion in crystalline solids?", "answer": "In crystal structures with smaller packing density, the diffusion activation energy is lower, making diffusion easier to occur; in crystal structures with lower symmetry, the anisotropy of the diffusion coefficient is more pronounced; in interstitial solid solutions, the diffusion activation energy is much smaller than in substitutional solid solutions, facilitating easier diffusion." }, { "idx": 1599, "question": "What is the difference in deformation mode between slip and twinning, the two mechanisms of metal plastic deformation?", "answer": "The slip process involves the relative sliding of one part of the crystal relative to another, while the twinning process involves the uniform shear of one part of the crystal relative to another." }, { "idx": 1600, "question": "Explain the reason for the shape memory effect produced by martensitic transformation", "answer": "The fundamental reasons are the diffusionless nature, coherent shear characteristics, and reversible transformation of martensitic transformation. During the cooling process of the parent phase, external stress induces martensitic transformation, utilizing the pseudoelasticity of martensitic transformation to produce macroscopic deformation. During the heating process, when the temperature exceeds the reverse transformation temperature of martensitic transformation, shape recovery occurs along with the reverse transformation of thermoelastic martensite, completing the shape memory process." }, { "idx": 1601, "question": "3. In practical applications, what treatment methods can be used to eliminate age hardening? Why?", "answer": "Age hardening can be eliminated through reversion treatment or re-solution treatment. This is because the second phase precipitated during aging will dissolve back into the solid solution during reversion treatment or re-solution treatment. When the precipitated phase has already become a stable phase, only solution treatment can be used." }, { "idx": 1602, "question": "How do crystal defects affect atomic diffusion in crystalline solids?", "answer": "The diffusion coefficient along grain boundaries is much greater than that of bulk diffusion; the activation energy for diffusion along dislocation pipes is smaller, thus dislocations accelerate diffusion." }, { "idx": 1603, "question": "What is the difference in continuity between the two metal plastic deformation mechanisms of slip and twinning?", "answer": "The slip process can proceed continuously while the twinning process cannot proceed continuously." }, { "idx": 1604, "question": "What is the difference in critical shear stress between the two metal plastic deformation mechanisms of slip and twinning?", "answer": "The critical shear stress required for twinning is much greater than that for slip." }, { "idx": 1605, "question": "2. Explain the reasons for strengthening through aging treatment.", "answer": "During the precipitation process of a supersaturated solid solution, the initially formed metastable precipitates maintain coherent or semi-coherent interfaces with the matrix phase. If the precipitate particles possess high strength, they will cause the gliding dislocations to bend and bypass the second-phase particles, leaving dislocation loops behind. This increases the dislocation line length, and both the second-phase particles and the dislocation loops enhance the resistance to subsequent dislocation motion, resulting in second-phase strengthening. If the precipitate particles can undergo deformation, new phase interfaces will form, disrupting the coherent (or semi-coherent) interfaces between the precipitates and the matrix phase. The slip planes will experience misfit, potentially destroying the ordered arrangement. In summary, macroscopic strengthening is achieved." }, { "idx": 1606, "question": "What is the difference between the effects of slip and twinning, these two metal plastic deformation mechanisms, on the crystal orientation relationship?", "answer": "The twinning process alters the crystal orientation relationship. The slip process does not change the crystal orientation relationship." }, { "idx": 1607, "question": "What are the causes and movement characteristics of Shockley partial dislocations in face-centered cubic crystals?", "answer": "In face-centered cubic crystals, Shockley partial dislocations are generated by non-uniform slip. They can be edge dislocations, screw dislocations, or mixed dislocations, and are capable of slipping." }, { "idx": 1608, "question": "7．The main resistances to solid-state phase transformation are (19) and (20)", "answer": "(19) interfacial energy; (20) elastic strain energy" }, { "idx": 1609, "question": "What is a perfect dislocation?", "answer": "A dislocation whose Burgers vector equals a lattice vector is called a perfect dislocation." }, { "idx": 1610, "question": "What are the differences in the roles of slip and twinning as two mechanisms of metal plastic deformation during plastic deformation?", "answer": "Slip is the primary mechanism of plastic deformation. When the slip system is unfavorable for slip deformation to occur, twinning can alter the orientation of the slip system relative to the external force, enabling further slip deformation to proceed." }, { "idx": 1611, "question": "What is the shape memory effect?", "answer": "When certain metal materials are deformed and then heated above a specific temperature, the deformed metal materials return to their pre-deformation shape. This phenomenon is called the shape memory effect." }, { "idx": 1612, "question": "What are the formation and movement characteristics of Frank partial dislocations in face-centered cubic crystals?", "answer": "Frank partial dislocations are caused by the removal or insertion of a close-packed plane, and their Burgers vector is perpendicular to the slip plane. Therefore, Frank partial dislocations cannot glide but can only climb." }, { "idx": 1613, "question": "Metallic bond", "answer": "The bonding force between metal atoms formed by the interaction between metal cations and free electrons is called metallic bond" }, { "idx": 1614, "question": "Interstitial solid solution", "answer": "When the solute atoms are relatively small, they can enter the interstitial sites of the solvent lattice, and the solid solution formed in this way is called an interstitial solid solution." }, { "idx": 1615, "question": "Schottky vacancy", "answer": "A vacancy formed by the migration of a displaced atom to the outer surface or internal interface (such as grain boundaries, etc.) is called a Schottky vacancy" }, { "idx": 1616, "question": "The _（3）__ of an atom or ion refers to the number of atoms directly adjacent to it in the crystal structure or the number of all hetero-ions combined with it.", "answer": "（3）coordination number" }, { "idx": 1617, "question": "The equilibrium concentration of point defects increases with the rise of __ (4).", "answer": "(4) temperature" }, { "idx": 1618, "question": "When titanium oxide is oxygen-deficient, the following reaction can occur: $\\\\mathrm{TiO}_{2}-\\\\frac{1}{2}0_{2}{\\\\rightarrow}\\\\mathrm{Ti^{\\\\prime}}_{\\\\mathrm{Ti}}+\\\\mathrm{V}_{0}^{\\\\ast}$. Please correctly write the defect equation and explain the meaning of each term.", "answer": "The defect equation is $2\\\\mathrm{TiO}_{2}-\\\\frac{1}{2}0_{{2}}{\\\\rightarrow}2\\\\mathrm{Ti^{\\\\prime}}_{\\\\mathrm{Ti}}+\\\\mathrm{V}_{0}^{\\\\ast}+30_{0}$. $\\\\mathrm{Ti^{\\\\prime}}_{\\\\mathrm{Ti}}$: Titanium dioxide loses oxygen, generating $\\\\mathbb{T}^{3+}$ occupying the $\\\\mathbf{Ti}^{4+}$ lattice site, with an effective charge of -1. $\\\\mathbf{V}_{0}^{*}$: Oxygen vacancy, with an effective charge of $^{+2}$. $0_{\\\\mathfrak{o}}$: Oxygen still occupies the oxygen lattice site." }, { "idx": 1619, "question": "Habit plane", "answer": "During solid-state phase transformation, the new phase often forms along specific atomic planes of the parent phase. The parent phase crystal plane that is parallel to the main plane of the new phase is called the habit plane." }, { "idx": 1620, "question": "Aging", "answer": "The subsequent precipitation process of solute atoms from a supersaturated solid solution at room temperature or above" }, { "idx": 1621, "question": "The process in which a single solid phase simultaneously precipitates two new solid phases with different compositions and crystal structures is called (8)__ transformation", "answer": "(8) eutectoid" }, { "idx": 1622, "question": "If different atomic planes in a crystal are represented by letters such as A, B, C, D, etc., the stacking sequence of atoms in a face-centered cubic crystal is _(1)_, and the stacking sequence of atoms in a hexagonal close-packed crystal is _(2)_.", "answer": "(1) ABCABCABC...; (2) ABABAB." }, { "idx": 1623, "question": "Deformation texture", "answer": "The phenomenon where a certain crystallographic orientation of different grains in a polycrystal tends to become consistent with increasing plastic deformation." }, { "idx": 1624, "question": "The formation of nuclei in a homogeneous liquid phase through conditions such as structural fluctuations and energy fluctuations is called (10) nucleation", "answer": "(10) homogeneous" }, { "idx": 1625, "question": "What is the driving force for nucleation in solid-state phase transformations?", "answer": "The difference in free energy between the new phase and the parent phase" }, { "idx": 1626, "question": "The imperfect dislocation caused by inserting or removing a close-packed plane is called _ (7) _ imperfect dislocation", "answer": "(7) Frank" }, { "idx": 1627, "question": "Given that the yield strength of industrial pure copper is $\\\\sigma{\\\\mathrm{s}}=70\\\\mathrm{MPa}$, with a grain size of $N{\\\\mathbf{A}}=18$ per $\\\\nearrow\\\\mathbf{mm}^{2}$; when $N{\\\\mathbf{A}}=4025$ per $/\\\\mathbf{m}\\\\mathbf{m}^{2}$, $\\\\pmb{\\\\sigma{\\\\mathrm{s}}=95\\\\mathrm{MPa}}$; calculate the yield strength ${\\\\pmb{\\\\sigma}}{\\\\mathfrak{s}}$ when $N{\\\\mathrm{A}}=260$ per $\\\\scriptstyle\\\\left/\\\\mathbf{mm}^{2}\\\\right.$.", "answer": "The relationship between grain size and yield strength satisfies the Hall-Petch formula, i.e., $$ \\\\sigma_{\\\\mathrm{s}}=\\\\sigma_{0}+K d^{-1/2} $$ Representing grain size by the diameter of an equal-area circle, thus $$ \\\\begin{array}{c}{{d=\\\\mathrm{\\\\Gamma}\\\\left({8}/{3\\\\pi}/{N_{\\\\mathrm{a}}}\\\\right)^{1/2}}}\\\\\\\\ {{d_{1}=\\\\mathrm{\\\\Gamma}\\\\left({8}/{3\\\\pi}/{18}\\\\right)^{1/2}=0.22\\\\mathrm{mm}}}\\\\\\\\ {{d_{2}=\\\\mathrm{\\\\Gamma}\\\\left({8}/{3\\\\pi}/{4025}\\\\right)^{1/2}=1.45\\\\times10^{-2}\\\\mathrm{mm}}}\\\\\\\\ {{d_{3}=\\\\mathrm{\\\\Gamma}\\\\left({8}/{3\\\\pi}/{260}\\\\right)^{1/2}=5.71\\\\times10^{-2}\\\\mathrm{mm}}}\\\\\\\\ {{\\\\sigma_{\\\\mathrm{s}1}=\\\\sigma_{0}+K d_{1}^{\\\\mathrm{\\\\Gamma}-1/2},\\\\sigma_{\\\\mathrm{s}2}=\\\\sigma_{0}+K d_{2}^{\\\\mathrm{\\\\Gamma}-1/2}}}\\\\end{array} $$ Substituting ${\\\\pmb\\\\sigma}_{\\\\mathrm{s}1}=70\\\\mathrm{MPa},{\\\\pmb\\\\sigma}_{\\\\mathrm{s}2}=95\\\\mathrm{MPa}$ to solve for $K=0.13\\\\mathrm{MPa\\\\cdot m^{1/2}},\\\\sigma_{0}=61.3\\\\mathrm{MPa}$, hence $\\\\sigma_{\\\\mathrm{s}}=78.3\\\\mathrm{MPa}$." }, { "idx": 1628, "question": "Negative temperature gradient", "answer": "In the liquid phase at the crystallization front, the distribution where the temperature gradually decreases from the solid-liquid interface toward the interior of the liquid phase is called a negative temperature gradient" }, { "idx": 1629, "question": "A dislocation whose Burgers vector equals a lattice vector is called __ (5)_", "answer": "(5) perfect dislocation" }, { "idx": 1630, "question": "In the equilibrium phase diagram of an alloy, to determine the proportions of the various phases present in the alloy at a certain temperature and composition, the __(9)__ rule can be applied using an isothermal tie line for calculation.", "answer": "(9) lever" }, { "idx": 1631, "question": "What are the main resistances to nucleation in solid-state phase transformations?", "answer": "Interfacial energy and elastic strain energy" }, { "idx": 1632, "question": "In face-centered cubic crystals, the partial dislocation caused by (6)__ is called the Shockley partial dislocation", "answer": "(6) non-uniform slip" }, { "idx": 1633, "question": "The dislocation that can produce cross-slip must be a _ (20) dislocation", "answer": "screw-type" }, { "idx": 1634, "question": "The adsorption occurring on the surface of solids can be divided into two types: (15) and (16).", "answer": "(15) Physical adsorption; (16) Chemical adsorption" }, { "idx": 1635, "question": "According to the different atomic arrangement structures at the interface, the phase interfaces in solids can be divided into _(17), (18), and (19) interfaces.", "answer": "(17) coherent; (18) incoherent; (19) semi-coherent" }, { "idx": 1636, "question": "If 6kg of austenite with a carbon content of w_c=0.45% is slowly cooled below 727%, what is the proeutectoid phase?", "answer": "If 6kg of austenite with a carbon content of w_c=0.45% is slowly cooled below 727%, the proeutectoid phase is proeutectoid ferrite." }, { "idx": 1637, "question": "Plastic deformation of metal polycrystals requires at least (14) independent slip systems to be activated", "answer": "(14) five" }, { "idx": 1638, "question": "What changes occur in the microstructure of plastically deformed metals?", "answer": "The strengthening factors of martensite include the interstitial solid solution strengthening effect of carbon atoms; the pinning effect of C atom clusters on dislocations; C atoms entering the flattened octahedral centers of the martensite crystal structure, causing asymmetric lattice expansion (forming a distorted dipole stress field), resulting in strong interstitial solid solution strengthening; and the strengthening effects of grain boundaries, dislocations, and twins." }, { "idx": 1639, "question": "What is the total amount of ferrite and cementite in kilograms in the final room-temperature equilibrium microstructure?", "answer": "According to the lever rule, the amounts of ferrite and cementite in the final room-temperature equilibrium microstructure are m_α=(6.70-0.45)/(6.70-0.022)×100%×6kg=5.64kg, m_Fe3C=(0.45-0.022)/(6.70-0.022)×100%×6kg=0.36kg, respectively." }, { "idx": 1640, "question": "What changes occur in the properties of plastically deformed metals?", "answer": "The strength and toughness of martensite are closely related to its carbon content, microstructure morphology, and substructure. In iron-carbon alloys, when the carbon content wc<0.3%, lath martensite forms with a dislocation substructure, exhibiting lower strength but good plasticity and toughness; when wc>1.0%, plate martensite forms with a twin substructure, showing high strength but poor plasticity and toughness; for 0.3%2)$ in the crystal structure, crystals are divided into lower (no higher-order axis), intermediate (one higher-order axis), and higher (more than one higher-order axis) crystal families." }, { "idx": 1654, "question": "Phase equilibrium", "answer": "The chemical potentials of each phase in the system are equal, and the chemical potentials of each component in each phase are equal." }, { "idx": 1655, "question": "Electronic compound", "answer": "Belongs to intermetallic compounds, where the electron concentration, defined as the ratio of the total number of valence electrons to the total number of atoms, is approximately 1.4 when the alloy reaches its maximum solubility." }, { "idx": 1656, "question": "What changes occur in the intragranular structure of plastically deformed metals?", "answer": "The strength and toughness of martensite are closely related to its carbon content, microstructure morphology, and substructure. In iron-carbon alloys, when the carbon content wc<0.3%, lath martensite forms with a dislocation substructure, exhibiting lower strength but good plasticity and toughness; when wc>1.0%, plate martensite forms with a twin substructure, showing high strength but poor plasticity and toughness; when 0.3% 0, the diffusion result causes the solute to ____; when the thermodynamic factor of diffusion coefficient < 0, the diffusion result causes the solute to ____. a. segregate b. concentration remains unchanged c. concentration tends to be uniform", "answer": "c a" }, { "idx": 1758, "question": "When n spheres form a cubic close packing, what is the number of octahedral voids?", "answer": "n" }, { "idx": 1759, "question": "What is the viscosity corresponding to the characteristic temperature Tf?", "answer": "(19)108dPa·s" }, { "idx": 1760, "question": "The crystal structure is related to its (10) (11) and (12).", "answer": "(10) chemical composition; (11) relative size of particles; (12) polarization properties" }, { "idx": 1761, "question": "The reason why a base-centered lattice cannot exist in the cubic system is _(9)", "answer": "If a base-centered lattice existed in the cubic system, the characteristic 4L3 symmetry of the cubic system would no longer exist, which does not conform to the symmetry characteristics of the cubic system" }, { "idx": 1762, "question": "When 6 mol% of MgO is added to the UO2 lattice to form a substitutional solid solution, write its solid solution formula", "answer": "U0.94Mg0.06O1.94" }, { "idx": 1763, "question": "In the wetting of solid-liquid interfaces, what is the second method to improve wetting?", "answer": "Increase the surface energy of the solid" }, { "idx": 1764, "question": "What is the viscosity corresponding to the characteristic temperature Tg?", "answer": "(18)1013dPa·s" }, { "idx": 1765, "question": "In the wetting of solid-liquid interfaces, what is the third method to improve wetting?", "answer": "Altering the surface roughness of the solid" }, { "idx": 1766, "question": "7. The reason why glass is isotropic is (15)", "answer": "The long-range disorder of the glass structure exhibits statistical homogeneity macroscopically" }, { "idx": 1767, "question": "In the wetting of solid-liquid interfaces, one of the methods to improve wetting is?", "answer": "Reduce the solid-liquid interfacial energy" }, { "idx": 1768, "question": "When 6 mol% of MgO is added to the UO2 lattice to form a substitutional solid solution, write the defect reaction equation.", "answer": "MgO→UO2Mg0′′+O0+V0∗" }, { "idx": 1769, "question": "What are the two characteristic temperatures on the curve of glass properties changing with temperature?", "answer": "(16) Tg; (17) Tf" }, { "idx": 1770, "question": "10. What type of solid solution can albite Na(AlSi3O8) and anorthite Ca(Al2Si2O8) form?", "answer": "Continuous substitutional solid solution" }, { "idx": 1771, "question": "11．The crystallographic characteristics of martensitic transformation are manifested as: _（26）_, etc.", "answer": "（26）the presence of habit plane" }, { "idx": 1772, "question": "11. The crystallographic characteristics of martensitic transformation are manifested as: _（27）_, etc.", "answer": "（27）strict orientation relationship" }, { "idx": 1773, "question": "11. The crystallographic characteristics of martensitic transformation are manifested as: _(28)_, etc.", "answer": "(28) maintaining coherent relationship" }, { "idx": 1774, "question": "10. What ions in plagioclase Na(AlSi3O8) and what ions in anorthite Ca(Al2Si2O8) can substitute for each other?", "answer": "Na+ and Si4+ in plagioclase can substitute for Ca2+ and Al3+ in anorthite" }, { "idx": 1775, "question": "Polymorphism", "answer": "Substances with the same chemical composition can form crystals with different structures under different thermodynamic conditions" }, { "idx": 1776, "question": "A green body with an initial particle size of 5μm is sintered for 2h to reach x/r=0.1. If sintering continues until x/r=0.2 (without considering grain growth), what is the required sintering time when material transport occurs via dissolution-precipitation?", "answer": "128h" }, { "idx": 1777, "question": "There is a green body with an initial particle size of 5μm. After sintering for 2h, x/r=0.1. If sintering proceeds until x/r=0.2 (without considering grain growth), what is the required sintering time when material transport is dominated by viscous flow?", "answer": "8h" }, { "idx": 1778, "question": "There is a green body with an initial particle size of 5μm. After sintering for 2h, x/r=0.1. If sintering continues until x/r=0.2 (without considering grain growth), what is the required sintering time when the material transport is dominated by evaporation-condensation?", "answer": "16h" }, { "idx": 1779, "question": "A green body has an initial particle size of 5μm. After sintering for 2h, x/r=0.1. If sintering proceeds until x/r=0.2 (without considering grain growth), what is the required sintering time when the process is driven by diffusion mass transfer?", "answer": "64h" }, { "idx": 1780, "question": "First-order phase transition", "answer": "First-order phase transition: During the phase transition, the chemical potentials of the two phases are equal, but the first-order partial derivatives of the chemical potential are not equal. A first-order phase transition involves latent heat and volume changes." }, { "idx": 1781, "question": "Component defect", "answer": "3. Component defect: In solid solutions with non-equivalent substitution, in order to maintain the electrical neutrality of the crystal, vacancies or interstitial ions are inevitably generated in the crystal. This type of crystal defect is called a component defect." }, { "idx": 1782, "question": "Degrees of freedom in phase diagrams", "answer": "Degrees of freedom in phase diagrams: In a phase equilibrium system, the independent variables that can be arbitrarily changed within a certain range without causing the disappearance of old phases or the formation of new phases are called degrees of freedom." }, { "idx": 1783, "question": "Inverse spinel structure", "answer": "Inverse spinel structure: It belongs to the cubic crystal system, where oxygen ions can be considered to be arranged in a cubic close-packed structure. Divalent cations A occupy octahedral sites, while trivalent cations B are half-filled in octahedral sites and half-filled in tetrahedral sites." }, { "idx": 1784, "question": "Electrokinetic potential of clay colloids", "answer": "Electrokinetic potential of clay colloids: When charged clay colloids are dispersed in water, a diffuse double layer forms at the interface between the colloidal particles and the liquid phase. The adsorption layer and the diffuse layer carry opposite charges, and the potential difference between them during relative movement is called the electrokinetic potential." }, { "idx": 1785, "question": "Pb has a face-centered cubic structure with an atomic radius of 0.1750 nm. Calculate the volume of its unit cell.", "answer": "V = a³ = (2√2r)³ = 0.1212 nm³" }, { "idx": 1786, "question": "The chemical composition of a glass is: 24mol% Na2O, 12mol% Al2O3, 64mol% SiO2. Calculate the four structural parameters Z, R, X, and Y of this glass.", "answer": "Converted to 6Na2O·3Al2O3·16SiO2; Z=4; R=2.17; Y=3.66; X=0.34." }, { "idx": 1787, "question": "The defect concentration of the non-stoichiometric compound TiO2-x is related to the nature and partial pressure of the surrounding atmosphere. When the oxygen partial pressure is increased, explain through calculation what changes will occur in the density of TiO2-x.", "answer": "As the value of x in TiO2-x increases, its density decreases." }, { "idx": 1788, "question": "In the manufacturing of Al2O3 ceramics, the particle size of the raw material is 2μm. After holding at the sintering temperature for 30min, the measured grain size is 10μm. What will be the grain size in μm after holding at the same sintering temperature for 2h?", "answer": "d1=20μm(D2=kt)" }, { "idx": 1789, "question": "The defect concentration of the non-stoichiometric compound Zn1+xO is related to the nature and partial pressure of the surrounding atmosphere. When the oxygen partial pressure is increased, explain through calculation what changes will occur in the density of Zn1+xO.", "answer": "ZnO↔Zni+2e'+1/2O2; Zn1+xO; the concentration of zinc interstitial ions is inversely proportional to the oxygen partial pressure. When the oxygen partial pressure increases, the x value in Zn1+xO decreases, resulting in a reduction of its density." }, { "idx": 1790, "question": "To inhibit grain growth, 0.2% MgO was added to the raw powder, and the temperature was held for 2 hours at the same sintering temperature. What is the grain size in μm?", "answer": "d2=16μm (D³=kt)" }, { "idx": 1791, "question": "Conditions necessary for mud peptization", "answer": "① The medium must be alkaline, ② Monovalent alkali metal cations must exchange with the higher-valent ions originally adsorbed on the clay, ③ Polymerization of anions." }, { "idx": 1792, "question": "Structurally compare the arrangement of silicon-oxygen frameworks in silicate crystals and silicate glasses", "answer": "In silicate crystals, the silicon-oxygen frameworks are arranged in an orderly manner according to certain symmetrical rules; in silicate glasses, the arrangement of silicon-oxygen frameworks is disordered." }, { "idx": 1793, "question": "Briefly describe the mechanism of slurry peptization", "answer": "Due to the difference in electrical charges on the edge and face surfaces of plate-like clay particles, face-face, edge-face, or edge-edge associations occur, forming a certain structure in the slurry that increases flow resistance. The peptization process of the slurry involves breaking down this internal structure, transforming edge-face and edge-edge associations into face-face arrangements." }, { "idx": 1794, "question": "Structurally compare the distribution characteristics of extra-framework metal ions in silicate crystals and silicate glasses", "answer": "In silicate crystals, the extra-framework metal ions occupy certain positions in the lattice; whereas in silicate glasses, the network-modifying ions are statistically distributed within the voids of the framework, balancing the negative charge of oxygen." }, { "idx": 1795, "question": "Compare the conditions for the substitution of extra-framework cations in silicate crystals and silicate glasses from a structural perspective", "answer": "In silicate crystals, isomorphous substitution can only occur when the radius of the foreign cation is similar to that of the cation in the crystal; in silicate glasses, extra-framework cations can undergo substitution regardless of their radius, as long as electrical neutrality is maintained." }, { "idx": 1796, "question": "Structurally compare the characteristics of oxide composition ratios in silicate crystals and silicate glasses", "answer": "In crystals, the original components (oxides) have simple fixed quantitative ratios between each other; whereas in glasses, oxides are mixed in almost any proportion." }, { "idx": 1797, "question": "What is extrinsic diffusion?", "answer": "Extrinsic diffusion is the migration phenomenon caused by vacancies created through doping with heterovalent impurity ions." }, { "idx": 1798, "question": "What are the diffusion coefficient, activation energy, and characteristics of intrinsic diffusion?", "answer": "The activation energy of intrinsic diffusion consists of two parts: vacancy formation energy and particle migration energy, with intrinsic diffusion dominating at high temperatures." }, { "idx": 1799, "question": "Briefly describe the difference between grain growth and secondary recrystallization", "answer": "Grain growth is the continuous increase in average grain size, with uniform grain size growth, and pores remaining at grain boundaries or grain boundary intersections; secondary recrystallization is the abnormal growth of individual grains, with pores being trapped inside the grains. Secondary recrystallization is also related to the particle size of the raw material." }, { "idx": 1800, "question": "What are the diffusion coefficient, activation energy, and characteristics of extrinsic diffusion?", "answer": "The activation energy of extrinsic diffusion only includes the migration energy of particles, and extrinsic diffusion dominates at low temperatures." }, { "idx": 1801, "question": "What is intrinsic diffusion?", "answer": "Intrinsic diffusion refers to the migration phenomenon caused by vacancies originating from the intrinsic thermal defects of the crystal." }, { "idx": 1802, "question": "Discuss the grain growth process based on the limiting grain size", "answer": "The grain growth process is briefly described." }, { "idx": 1803, "question": "Causes of secondary recrystallization", "answer": "Non-uniform particle size of raw materials, excessively high sintering temperature, too fast sintering rate." }, { "idx": 1804, "question": "Methods to prevent secondary recrystallization", "answer": "Control sintering temperature, sintering time, control the uniformity of raw material particle size, introduce sintering additives." }, { "idx": 1805, "question": "What are grain growth and secondary recrystallization?", "answer": "Grain growth is the process in which the average grain size of a strain-free material continuously increases during heat treatment without altering its distribution. Within the body, grain sizes grow uniformly, and during grain growth, pores remain at grain boundaries or their intersections. Secondary recrystallization is an abnormal growth process where a few large grains grow at the expense of fine grains, representing the abnormal growth of individual grains. During secondary recrystallization, pores become trapped inside the grains." }, { "idx": 1806, "question": "In the orthorhombic system, the possible types of space lattices are a.P, I, C b.P, 1, F c.P, C, F d.P, F, 1, C", "answer": "d" }, { "idx": 1807, "question": "In the spinel (MgAl2O4) structure, O2- forms a cubic close packing, and Mg2+ ions fill a. 1/2 of the tetrahedral voids b. 1/2 of the octahedral voids c. 1/8 of the tetrahedral voids d. 1/8 of the octahedral voids", "answer": "c" }, { "idx": 1808, "question": "Kaolinite belongs to the layered silicate structure, and its structural feature is a. two-layer type trioctahedral structure b. three-layer type trioctahedral structure c. two-layer type dioctahedral structure d. three-layer type dioctahedral structure", "answer": "c" }, { "idx": 1809, "question": "In the non-stoichiometric compound Cd1+xO, the type of non-stoichiometric structural defect present is a. anion vacancy b. cation vacancy c. anion interstitial d. cation interstitial", "answer": "d" }, { "idx": 1810, "question": "Among various layered silicate structures, the unit cell parameters that are similar are a.a0 and b0 b.a0 and c0 c.c0 and b0 d.c0", "answer": "a" }, { "idx": 1811, "question": "The melt is a mixture of . a. polymers with the same degree of polymerization and free alkali b. various polymers with different degrees of polymerization c. various oligomers d. various high polymers", "answer": "b" }, { "idx": 1812, "question": "The viscosity corresponding to the glass transition temperature Tg is a.108dPa·s b.1010dPa·s c.1011dPa·s d.1013dPa·s", "answer": "d" }, { "idx": 1813, "question": "Ionic crystals usually reduce their surface energy through the polarization deformation and rearrangement of surface ions. Among the following ionic crystals, the one with the smallest surface energy is a.CaF2 b.PbF2 c.PbI2 d.BaSO4 e.SrSO4", "answer": "c" }, { "idx": 1814, "question": "In the sintering process, the mass transfer method that only changes the pore shape without causing shrinkage of the green body is a. diffusion mass transfer b. flow mass transfer c. evaporation-condensation mass transfer d. grain boundary diffusion", "answer": "c" }, { "idx": 1815, "question": "Pauling's rules apply to (1)__ crystal structures.", "answer": "(1) ionic" }, { "idx": 1816, "question": "In the phase diagram of a ternary condensed system, if there are n boundary lines, the number of connection lines in this phase diagram must be a. (n+1) lines b. (n-1) lines c. n lines d. none", "answer": "c" }, { "idx": 1817, "question": "In the calculation of solid-state reactions, the Ginstling equation has a wider range of applicability than the Jander equation, and the biggest difference between the two is a. Different calculation methods b. The former is solved by the unstable diffusion equation c. Different geometric models", "answer": "c" }, { "idx": 1818, "question": "In the thermodynamic relation of diffusion coefficients, $\\\\left(1+\\\\frac{\\\\partial\\\\mathrm{ln}\\\\gamma_{i}}{\\\\partial\\\\mathrm{ln}N_{\\\\mathrm{i}}}\\\\right)$ is called the thermodynamic factor of the diffusion coefficient. In non-ideal mixing systems: when the thermodynamic factor of the diffusion coefficient > 0, the diffusion result causes the solute to ____; when the thermodynamic factor of the diffusion coefficient < 0, the diffusion result causes the solute to ____. a. segregate b. remain unchanged in concentration c. tend to uniform concentration", "answer": "c, a" }, { "idx": 1819, "question": "The essence of a crystal is (2)", "answer": "(2) Particles are arranged in a periodic repetition in three-dimensional space" }, { "idx": 1820, "question": "When selecting the unit parallelepiped of a space lattice, the primary principle is _ (3)", "answer": "(3) The symmetry of the unit parallelepiped should conform to the symmetry of the entire space lattice" }, { "idx": 1821, "question": "One of the conditions for clay slurry peptization is (6)", "answer": "The medium is alkaline" }, { "idx": 1822, "question": "同质多晶现象是指 (4)", "answer": "（4）chemical substances with the same composition form crystals with different structures under different thermodynamic conditions" }, { "idx": 1823, "question": "A crystal with space group Fm3m has a (5)_ structure.", "answer": "(5) face-centered cubic structure" }, { "idx": 1824, "question": "Glass has the following general properties: (12)", "answer": "Isotropic" }, { "idx": 1825, "question": "One of the conditions for the peptization of clay slurry is (8)", "answer": "Polymerization of anions" }, { "idx": 1826, "question": "What type of diffusion problem does Fick's first law apply to?", "answer": "Steady-state diffusion" }, { "idx": 1827, "question": "What type of diffusion problems is Fick's second law applicable to solve?", "answer": "Non-steady-state diffusion" }, { "idx": 1828, "question": "One of the conditions for the peptization of clay slurry is (7)", "answer": "Exchange of original cations with monovalent alkali metal cations" }, { "idx": 1829, "question": "The vacancy mechanism of diffusion refers to (11)", "answer": "(11) When vacancies exist in the lattice, the diffusion mode of mass points mediated by vacancy migration" }, { "idx": 1830, "question": "Glass has the following general properties: (13)", "answer": "Metastability" }, { "idx": 1831, "question": "Glass has the following general properties: (14)", "answer": "The transformation process is reversible and gradual" }, { "idx": 1832, "question": "What is the migration of particles caused by intrinsic diffusion?", "answer": "Vacancies generated by intrinsic thermal defects" }, { "idx": 1833, "question": "Glass has the following general properties: (15)", "answer": "Continuous change in properties during transformation" }, { "idx": 1834, "question": "What are the two components of the activation energy for intrinsic diffusion?", "answer": "Vacancy formation energy and particle migration energy" }, { "idx": 1835, "question": "What factors primarily determine the concentration of component defects?", "answer": "Doping amount" }, { "idx": 1836, "question": "As a result of what reason, 'compositional defects' are inevitably produced in the crystal structure?", "answer": "Doping with inequivalent ions" }, { "idx": 1837, "question": "What is the relationship between the diffusion coefficient of intrinsic diffusion and temperature?", "answer": "D=D0exp(-(ΔHf/2 + ΔHm)/RT)" }, { "idx": 1838, "question": "What factors does the concentration of component defects mainly depend on?", "answer": "Solid solubility" }, { "idx": 1839, "question": "What are the four main mass transfer mechanisms in sintering?", "answer": "(23) Evaporation-condensation mass transfer; (24) Diffusion mass transfer; (25) Flow mass transfer; (26) Dissolution-precipitation mass transfer" }, { "idx": 1840, "question": "What is the relationship between △L/L and sintering time in the mass transfer process?", "answer": "(29) △L/L∝t" }, { "idx": 1841, "question": "What is the relationship between △L/L and sintering time in the evaporation-condensation mass transfer process?", "answer": "(27) △L/L=0" }, { "idx": 1842, "question": "What is the relationship between △L/L and sintering time in the dissolution-precipitation mass transfer process?", "answer": "(30) △L/L∝t^(1/3)" }, { "idx": 1843, "question": "Frenkel defect", "answer": "Frenkel defect: When the lattice undergoes thermal vibration, some atoms with sufficient energy leave their equilibrium positions and squeeze into the interstitial sites of the lattice, forming interstitial atoms and leaving vacancies at the original normal lattice points." }, { "idx": 1844, "question": "What is the relationship between △L/L and sintering time in the diffusion mass transfer process?", "answer": "(28) △L/L∝t^(2/5)" }, { "idx": 1845, "question": "Homogeneous nucleation", "answer": "Homogeneous nucleation: refers to the probability of nuclei forming from a uniform single-phase melt being the same everywhere." }, { "idx": 1846, "question": "First-order phase transition", "answer": "First-order phase transition: During the phase transition, the chemical potentials of the two phases are equal, but the first-order partial derivatives of the chemical potential are not equal. A first-order phase transition involves latent heat and volume changes." }, { "idx": 1847, "question": "Connection rule", "answer": "Connection rule: In a ternary system, if the boundary (or its extension) between two adjacent phase regions (primary crystallization regions) intersects with the corresponding connecting line (or its extension), then this intersection point is the temperature maximum on that boundary." }, { "idx": 1848, "question": "Unit cell", "answer": "Unit cell: A unit cell is the smallest structural unit that fully reflects the characteristics of a crystal structure." }, { "idx": 1849, "question": "Sintering", "answer": "Sintering: A process in which one or more solid powders are formed and then heated to a certain temperature, causing them to shrink and become a dense, hard sintered body below the melting point temperature. Alternatively: Due to the mutual attraction of molecules (or atoms) in solids, heating causes the powder particles to bond, and through material migration and diffusion, the powder gains strength, leading to densification and recrystallization—this process is called sintering." }, { "idx": 1850, "question": "To inhibit grain growth, 0.1% grain growth inhibitor is added. If the holding time is also 2h, what is the grain size?", "answer": "d2=16μm(D3=kt)" }, { "idx": 1851, "question": "Calculate the four structural parameters Z, R, X, and Y of the glass Na2O·CaO·Al2O3·2SiO2", "answer": "1.Z=4; R=9/4=2.25; X=0.5; Y=3.5." }, { "idx": 1852, "question": "Write the solid solution formula corresponding to 20mol% YF3 doped into CaF2", "answer": "Ca0.8Y0.2F2.2, Ca0.7Y0.2F2" }, { "idx": 1853, "question": "From the perspective of crystal chemistry, discuss the rationality of the defect reaction equation for doping 20mol% YF3 into CaF2", "answer": "From the viewpoint of crystal chemistry, the substitution of Y3+ for Ca2+ will generate cation vacancies or interstitial fluoride ions to maintain electrical neutrality, both of these defect reactions are reasonable." }, { "idx": 1854, "question": "Try to write the defect reaction equation for 20mol% YF3 doped into CaF2", "answer": "2.YF3→CaF2YCa*+Fi'+2FF, 2YF3→CaF22YCa*+VCa''+6FF" }, { "idx": 1855, "question": "A certain powder compact has an average particle size of 2μm. After sintering to the highest temperature and holding for 0.5h, the measured grain size is 10μm. If the holding time is extended to 2h, what will the grain size be?", "answer": "d1=20μm(D2=kt)" }, { "idx": 1856, "question": "Give two examples of forming continuous substitutional solid solutions", "answer": "PbTiO3-PbZrO3, albite-orthoclase, MgO-NiO, etc." }, { "idx": 1857, "question": "For an AB-type compound crystal with a face-centered cubic structure, the molecular weight of the AB compound is 26, and its measured density is 2.6g/cm3. Based on this, calculate the unit cell parameter of the AB crystal.", "answer": "The number of formula units per unit cell Z=4, the number of AB molecules in 1mol is 2.6/26×N0=6.02×1022; the number of unit cells=6.02/4×1022=1.5×1022; 1.5×1022V=1.5×1022a3=1021nm3, thus a=0.41nm" }, { "idx": 1858, "question": "Briefly describe the conditions for forming continuous substitutional solid solutions", "answer": "The conditions for forming continuous substitutional solid solutions are: ionic size factor, the radius difference between the two substituting ions (r1-r2)/r1<15%; the two components must have exactly the same crystal structure type; continuous substitutional solid solutions can only form when the ionic charges are the same or the total charges of complex substitutions are equal; similar electronegativity and polarization properties." }, { "idx": 1859, "question": "In the face-centered cubic close-packed structure of the NaCl unit cell, how many octahedral voids are there?", "answer": "There are a total of 4 octahedral voids in the NaCl unit cell." }, { "idx": 1860, "question": "In the face-centered cubic close-packed structure of the NaCl unit cell, how many tetrahedral voids are there?", "answer": "There are a total of 8 tetrahedral voids in the NaCl unit cell." }, { "idx": 1861, "question": "What is martensitic transformation?", "answer": "Martensitic transformation refers to a phase transformation that occurs at an extremely rapid speed through the shear of a discrete volume of a crystal under the action of external stress. Martensitic transformation is one of the basic forms of solid-state phase transformations." }, { "idx": 1862, "question": "In the face-centered cubic close-packed structure of the NaCl unit cell, where are the tetrahedral void positions located?", "answer": "In the NaCl unit cell, the tetrahedral voids are located along the body diagonal direction of the unit cell. They are formed by the corner Cl- ions and the point Cl- ions in the three adjacent edges, constituting the tetrahedral voids, which are also the centers of the 8 small cubes. There are a total of 8 tetrahedral voids." }, { "idx": 1863, "question": "In the face-centered cubic close-packed structure of the NaCl unit cell, where are the octahedral void positions located?", "answer": "In the NaCl structure, Cl- forms a face-centered cubic close packing, and Na+ occupies all the octahedral voids. The positions of the octahedral voids include one at the center of the unit cell, and the midpoint of each edge is also an octahedral void position, totaling 12. The number of such voids belonging to this unit cell is 12 × 1/4 = 3. Therefore, there are a total of 4 octahedral voids in the NaCl unit cell." }, { "idx": 1864, "question": "Briefly describe the crystallographic characteristics of martensitic transformation.", "answer": "The crystallographic characteristics of martensitic transformation include: a strict orientation relationship between the new phase and the parent phase during the transformation, maintenance of a coherent relationship through shear, the presence of a habit plane that remains undistorted and unrotated before and after the transformation, and macroscopic continuity." }, { "idx": 1865, "question": "What is grain growth?", "answer": "Grain growth is the process in which the average grain size of a strain-free material continuously increases during heat treatment without altering its distribution. Within the bulk, grain sizes grow uniformly, and pores remain at grain boundaries or grain boundary junctions during grain growth." }, { "idx": 1866, "question": "What causes secondary recrystallization?", "answer": "The causes of secondary recrystallization: uneven particle size of the raw material, excessively high sintering temperature, and too fast sintering rate." }, { "idx": 1867, "question": "What are the methods to prevent secondary recrystallization?", "answer": "Methods to prevent secondary recrystallization: control sintering temperature, sintering time, control the uniformity of raw material particle size, introduce sintering additives." }, { "idx": 1868, "question": "What is secondary recrystallization?", "answer": "Secondary recrystallization is an abnormal growth process where a few large grains grow at the expense of fine grains, representing the abnormal growth of individual grains. During secondary recrystallization, pores are enclosed within the grains, and it is also related to the particle size of the raw material." }, { "idx": 1869, "question": "Among the seven crystal systems, the number of crystal systems belonging to the highest crystal category is A.1 B.2 C.3 D.4", "answer": "A" }, { "idx": 1870, "question": "The crystal structure with point group L6L27PC belongs to the crystal system. A. Cubic B. Hexagonal C. Tetragonal D. Orthorhombic", "answer": "B" }, { "idx": 1871, "question": "Kaolinite belongs to the silicate of . A. Chain structure B. Layered structure C. Framework structure D. Island structure", "answer": "B" }, { "idx": 1872, "question": "The crystal structure of diamond is , and the coordination number of carbon atoms is A. face-centered cubic lattice B. simple cubic lattice C. body-centered cubic lattice D. 3 E. 4 F. 6 G. 8", "answer": "A, E" }, { "idx": 1873, "question": "Most solid-phase reactions fall into A. Chemical reaction kinetics range B. Diffusion kinetics range C. Transition range", "answer": "B" }, { "idx": 1874, "question": "When water-based substances melt into a liquid state, their volume undergoes the phenomenon of . A. expansion B. contraction C. no change", "answer": "B" }, { "idx": 1875, "question": "At low temperatures, the diffusion that generally occurs in solid materials is A. Intrinsic diffusion B. Extrinsic diffusion C. Random diffusion", "answer": "B" }, { "idx": 1876, "question": "Among the following types of crystals, the order of forming interstitial solid solutions is A. zeolite > CaF2 > TiO2 > MgO B. MgO > TiO2 > CaF2 > zeolite C. CaF2 > TiO2 > MgO > zeolite D. TiO2 > MgO > CaF2 > zeolite", "answer": "A" }, { "idx": 1877, "question": "Substances that easily form glass often have bond types such as . A. Ionic bond B. Metallic bond C. Polar covalent bond D. Covalent bond", "answer": "C" }, { "idx": 1878, "question": "The viscosity corresponding to the glass transition temperature Tf is A.105dPa·s B.109dPa·s C.1011dPa·s D.1013dPa·s", "answer": "B" }, { "idx": 1879, "question": "When a small amount of CaO is doped into ThO2, write the possible defect reaction equation (7)", "answer": "7) CaO → CaTh'' + O0 + V0''" }, { "idx": 1880, "question": "Under the same system and the same degree of supercooling, the heterogeneous nucleation rate is always the homogeneous nucleation rate. A. Greater than or equal to B. Less than or equal to C. Equal to D. Not necessarily", "answer": "A" }, { "idx": 1881, "question": "How many space groups can appear in crystal structures?", "answer": "230" }, { "idx": 1882, "question": "When a small amount of CaO is doped into ThO2, write the corresponding solid solution formula (9)", "answer": "9) Th1-xCaxO2-x" }, { "idx": 1883, "question": "What is a space group?", "answer": "The collection of all symmetry elements in a crystal structure" }, { "idx": 1884, "question": "Fick's first law can be directly applied to solve the problem of . A．steady-state diffusion B．non-steady-state diffusion C．any diffusion", "answer": "A" }, { "idx": 1885, "question": "When a small amount of CaO is doped into ThO2, write the possible defect reaction equation (8)", "answer": "8) 2CaO → CaTh'' + Cai'' + 2O0" }, { "idx": 1886, "question": "When a small amount of CaO is doped into ThO2, write the corresponding solid solution formula (10)", "answer": "10) Th1-xCa2xO2" }, { "idx": 1887, "question": "4. Melt is a high-energy state of matter existing above the liquidus temperature, and during the cooling process of the melt, three different phase transition processes can occur: (11), (12), and (13).", "answer": "(11) Crystallization; (12) Vitrification; (13) Phase separation" }, { "idx": 1888, "question": "Crystals have basic properties such as (1) (2) (3) and (4).", "answer": "1) Symmetry; (2) Crystallographic homogeneity; (3) Anisotropy; (4) Self-confinement (or minimal internal energy property)" }, { "idx": 1889, "question": "5．According to the thermodynamic theory of diffusion, the driving force of the diffusion process is", "answer": "chemical potential gradient" }, { "idx": 1890, "question": "Wetting is an important behavior at the solid-liquid interface, and one of the methods to improve wetting is (16)", "answer": "Reduce the solid-liquid interfacial energy" }, { "idx": 1891, "question": "5. According to the thermodynamic theory of diffusion, the condition for reverse diffusion to occur is", "answer": "<0 or the thermodynamic factor of the diffusion coefficient is less than 0" }, { "idx": 1892, "question": "7. Martensitic transformation has the following characteristics: (19) etc.", "answer": "(19) Presence of habit plane" }, { "idx": 1893, "question": "Wetting is an important behavior at the solid-liquid interface, one of the methods to improve wetting is (18)", "answer": "Changing surface roughness" }, { "idx": 1894, "question": "7．Martensitic transformation has the following characteristics: (20) etc.", "answer": "(20) Orientation relationship" }, { "idx": 1895, "question": "Wetting is an important behavior at the solid-liquid interface, one of the methods to improve wetting is (17)", "answer": "Increase the surface energy of the solid (remove the adsorbed film on the solid surface)" }, { "idx": 1896, "question": "7. Martensitic transformation has the following characteristics: (21) etc.", "answer": "(21) Diffusionless" }, { "idx": 1897, "question": "8.The characteristic of reversible polymorphic transformation is: (23)", "answer": "(23) The polymorphic transformation temperature is lower than the melting points of both crystal forms" }, { "idx": 1898, "question": "7. Martensitic transformation has the following characteristics: (22) etc.", "answer": "(22) Fast speed (or no specific transformation temperature)" }, { "idx": 1899, "question": "What are the causes of secondary recrystallization?", "answer": "Non-uniform particle size of raw materials; excessively high sintering temperature; too fast sintering rate" }, { "idx": 1900, "question": "What is secondary recrystallization?", "answer": "During the middle and late stages of sintering, it is an abnormal growth process where a few large grains grow at the expense of fine grains." }, { "idx": 1901, "question": "What are the main mass transfer mechanisms in solid-state sintering?", "answer": "(28) Evaporation-condensation mass transfer; (29) Diffusion mass transfer" }, { "idx": 1902, "question": "What are the main mass transfer mechanisms in liquid phase sintering?", "answer": "(30) Flow mass transfer; (31) Dissolution-precipitation mass transfer" }, { "idx": 1903, "question": "What is the relationship between △L/L and sintering time in the evaporation-condensation mass transfer process?", "answer": "(32) △L/L=0" }, { "idx": 1904, "question": "What is the relationship between △L/L and sintering time in the dissolution-precipitation mass transfer process?", "answer": "(35) △L/L∝t^(1/3)" }, { "idx": 1905, "question": "Glass phase separation", "answer": "3. Glass phase separation: A homogeneous glass phase may, within certain temperature and composition ranges, separate into two mutually insoluble or partially soluble glass phases that coexist. This phenomenon is called glass phase separation." }, { "idx": 1906, "question": "Unit parallelepiped", "answer": "Unit parallelepiped: In a space lattice, the parallelepiped selected according to the selection principle is called the unit parallelepiped." }, { "idx": 1907, "question": "Tangent rule", "answer": "4. Tangent rule: Draw a tangent at a certain point on the boundary line and intersect it with the corresponding connecting line. If the intersection point lies on the connecting line, it indicates that the boundary at that point has a eutectic property; if the intersection point lies on the extension of the connecting line, it indicates that the boundary at that point has a peritectic property, and the crystalline phase farther from the intersection point is peritectically consumed." }, { "idx": 1908, "question": "Component defect", "answer": "2. Component defect: Due to the doping of unequal valence ions, in order to maintain the electrical neutrality of the crystal, defects such as vacancies or interstitial ions are inevitably generated in the crystal structure. This type of defect is called a 'component defect'." }, { "idx": 1909, "question": "What is the relationship between △L/L and sintering time in the diffusion mass transfer process?", "answer": "(33) △L/L∝t^(2/5)" }, { "idx": 1910, "question": "What is the relationship between △L/L in the mass transfer process and sintering time?", "answer": "(34) △L/L∝t" }, { "idx": 1911, "question": "Network former", "answer": "6. Network former: Substances with single bond energy ≥335kJ/mol that can independently form glass." }, { "idx": 1912, "question": "Normal spinel structure", "answer": "5. Normal spinel structure: It belongs to the cubic crystal system, in which oxygen ions can be regarded as arranged in cubic close packing, divalent cations A fill one-eighth of the tetrahedral voids, and trivalent cations B fill half of the octahedral voids." }, { "idx": 1913, "question": "Calculate the four structural parameters Z, R, X, and Y of the glass Na2O·CaO·Al2O3·3SiO2", "answer": "z=4 R=11/5=2.2; X=0.4; Y=3.6. The percentage of non-bridging oxygen =X/(X+Y/2)×100%=0.4/(0.4+1.8)×100%=18.2%" }, { "idx": 1914, "question": "Write the defect reaction equation for the formation of a solid solution of YF3 in CaF2 (atomic weights are Y88.9, Ca40.0, F19.0)", "answer": "2.YF3→CaF2YCa·+Fi′+2FF Ca0,8Y0,2F2,2 2YF3→CaF22YCa∗+V Ca′′+6FF Ca0,7Y0,2F2" }, { "idx": 1915, "question": "When 20 mol% YF3 is added to CaF2 to form a solid solution, the experimentally measured density of the solid solution is 3.64 g/cm3, and the lattice parameter at this time is a=0.55 nm. Calculate and determine the type of this solid solution. (The atomic weights are Y 88.9, Ca 40.0, F 19.0).", "answer": "The unit cell of the fluorite-type crystal structure contains 4 CaF2, ρ_vacancy=3.34 g/cm3; ρ_interstitial=3.65 g/cm3; ρ_interstitial is close to the measured ρ=3.65 g/cm3, so under these conditions, an interstitial solid solution is formed." }, { "idx": 1916, "question": "Explain the formation conditions and characteristics of structural defects in anion interstitial non-stoichiometric compounds", "answer": "Under oxidizing atmosphere, low valence state changes to high valence state, p-type semiconductor." }, { "idx": 1917, "question": "In the non-stoichiometric compound FexO, Fe3+/Fe2+=0.1. Determine the value of x in the non-stoichiometric compound FexO.", "answer": "According to the reaction formula Fe_xO → Fe2O3 → Fe0 → 2Fe_Fe^· + V_Fe^′′ + 3O_∘; Fe_1-3y^2+ Fe_2y^3+ O, let y be the mole fraction of Fe3+, then 2y/(1-3y) = 0.1. Solving gives y = 1/23 ≈ 0.043. The value of x is 1 - y = 1 - 0.043 ≈ 0.957." }, { "idx": 1918, "question": "In the non-stoichiometric compound FexO, Fe3+/Fe2+=0.1. Find the vacancy concentration in the non-stoichiometric compound FexO.", "answer": "According to the reaction formula Fe_xO → Fe2O3 → Fe0 → 2Fe_Fe^· + V_Fe^′′ + 3O_∘; Fe_1-3y^2+ Fe_2y^3+ O, let y be the mole fraction of Fe3+, then 2y/(1-3y) = 0.1. Solving gives y = 1/23 ≈ 0.043. The vacancy concentration is y/(1 + x), where x = 1 - y = 0.957, thus the vacancy concentration is 0.043 / (1 + 0.957) ≈ 2.2%." }, { "idx": 1919, "question": "Briefly describe the reasons for the formation of structural defects in non-stoichiometric compounds", "answer": "Reasons for formation: presence of variable valence elements, changes in the nature of the surrounding atmosphere." }, { "idx": 1920, "question": "Explain the formation conditions and characteristics of structural defects in anion vacancy-type non-stoichiometric compounds", "answer": "Under reducing atmosphere, high valence state changes to low valence state, n-type semiconductor." }, { "idx": 1921, "question": "Explain the formation conditions and characteristics of structural defects in cation interstitial non-stoichiometric compounds", "answer": "Under reducing atmosphere, high valence state changes to low valence state, n-type semiconductor." }, { "idx": 1922, "question": "Explain the formation conditions and characteristics of structural defects in cation vacancy-type non-stoichiometric compounds", "answer": "Under oxidizing atmosphere, low valence state changes to high valence state, p-type semiconductor." }, { "idx": 1923, "question": "Given a powder compact with an average particle size of 5μm, after sintering for 2 hours, the neck growth rate x/r=0.1. If grain growth is not considered, how much time is needed to sinter the compact to a neck growth rate x/r=0.2 via material transport by flow?", "answer": "The formula for material transport by flow: x/r = k r^(-1/2) t^(1/2). Substituting x/r=0.1, r=5μm, t=2h to solve for k: 0.1 = k (5)^(-1/2) (2)^(1/2), yielding k. Then substituting x/r=0.2, r=5μm, and k to solve for t: 0.2 = k (5)^(-1/2) t^(1/2), yielding t=8h. Material transport by flow requires 8h." }, { "idx": 1924, "question": "Please explain what a perfect dislocation is", "answer": "Perfect dislocation: The Burgers vector is an integer multiple of the lattice vector." }, { "idx": 1925, "question": "What is intrinsic diffusion?", "answer": "Intrinsic diffusion refers to the migration phenomenon caused by vacancies originating from the intrinsic thermal defects of the crystal. The activation energy of intrinsic diffusion consists of two parts: the vacancy formation energy and the particle migration energy. At high temperatures, intrinsic diffusion dominates." }, { "idx": 1926, "question": "Why do primitive, face-centered, and body-centered lattices exist in the cubic system, but not base-centered lattices?", "answer": "The characteristic of the cubic system is the presence of 4L3 axes, which exist in the primitive, face-centered, and body-centered lattices of the cubic system. If a base-centered lattice were to exist in the cubic system, the 4L3 axes would not be possible in such a lattice. Therefore, the base-centered lattice does not conform to the symmetry characteristics of the cubic system and cannot exist in it." }, { "idx": 1927, "question": "Please write the Burgers vector of the shortest unit dislocation in BCC crystals", "answer": "BCC: a/2<111>" }, { "idx": 1928, "question": "Given a powder compact with an average particle size of 5μm, after sintering for 2 hours, the neck growth ratio x/r=0.1. If grain growth is not considered, how much time is required to sinter the compact to a neck growth ratio x/r=0.2 through diffusion mass transport?", "answer": "Diffusion mass transport formula: x/r = k r^(-3/5) t^(1/5). Substituting x/r=0.1, r=5μm, t=2h to solve for k: 0.1 = k (5)^(-3/5) (2)^(1/5), yielding k. Then substituting x/r=0.2, r=5μm and k to solve for t: 0.2 = k (5)^(-3/5) t^(1/5), yielding t=64h. Diffusion mass transport requires 64h." }, { "idx": 1929, "question": "Please write the Burgers vector of the shortest unit dislocation in HCP crystals", "answer": "HCP: a/3<11\\\\bar{2}0>" }, { "idx": 1930, "question": "What is the difference between intrinsic diffusion and extrinsic diffusion?", "answer": "The activation energy of intrinsic diffusion consists of both vacancy formation energy and particle migration energy, while the activation energy of extrinsic diffusion only includes particle migration energy; intrinsic diffusion dominates at high temperatures, whereas extrinsic diffusion dominates at low temperatures." }, { "idx": 1931, "question": "Compare the similarities between solid-phase sintering and liquid-phase sintering", "answer": "The similarities between solid-phase sintering and liquid-phase sintering: the driving force for sintering is surface energy in both cases, and the sintering process consists of stages such as particle rearrangement, pore filling, and grain growth." }, { "idx": 1932, "question": "Please explain what is a partial dislocation", "answer": "Partial dislocation: The Burgers vector is not an integer multiple of the lattice vector." }, { "idx": 1933, "question": "How does the thermal motion of atoms affect diffusion?", "answer": "The enhancement of thermal motion will increase the jump distance, jump probability, and jump frequency of atoms, thereby increasing the diffusion coefficient." }, { "idx": 1934, "question": "What is extrinsic diffusion?", "answer": "Extrinsic diffusion is a migration phenomenon caused by vacancies generated from the doping of inequivalent impurity ions. The activation energy of extrinsic diffusion only includes the migration energy of mass points, and extrinsic diffusion dominates at low temperatures." }, { "idx": 1935, "question": "How to distinguish between hot deformation and cold deformation of metals?", "answer": "It is distinguished based on the relationship between the deformation temperature and the recrystallization temperature. Deformation above the recrystallization temperature is hot deformation, and vice versa is cold deformation." }, { "idx": 1936, "question": "Please write the Burgers vector of the shortest unit dislocation in an FCC crystal", "answer": "FCC: a/2<110>" }, { "idx": 1937, "question": "It is known that atomic radius is related to crystal structure. When the coordination number decreases, how does the atomic radius change? Why?", "answer": "The radius contracts. If the radius remains unchanged, a decrease in coordination number would cause an increase in crystal volume. To minimize volume change, the atomic radius will contract." }, { "idx": 1938, "question": "Discuss the conditions and characteristics of dissolution-precipitation mass transfer", "answer": "The conditions for dissolution-precipitation mass transfer are: a considerable amount of liquid phase, high solubility of the solid phase in the liquid phase, and the liquid phase can wet the solid phase; the characteristics are: dissolution at particle contact points and deposition on flat surfaces, dissolution of small grains and deposition on large grains, and the mass transfer process is also a grain growth process." }, { "idx": 1939, "question": "Homogeneous nucleation and heterogeneous nucleation have the same critical nucleus radius, and the critical nucleation work for heterogeneous nucleation is also equal to one-third of the surface energy. Why is heterogeneous nucleation easier than homogeneous nucleation?", "answer": "Because in heterogeneous nucleation, impurities or mold cavities act as part of the nucleus. That is to say, fewer atoms need to be mobilized." }, { "idx": 1940, "question": "For metals without phase transformation in the solid state, how to refine grain size through hot working without remelting?", "answer": "Perform hot working to induce dynamic recrystallization." }, { "idx": 1941, "question": "For metals without phase transformation in the solid state, can the grain size be refined without remelting?", "answer": "Yes." }, { "idx": 1942, "question": "In an FCC crystal, does the dislocation reaction a/2[10-1] + a/6[-121] → a/3[11-1] satisfy the energy condition?", "answer": "Energy condition: |a/2√2|^2 + |a/6√6|^2 = (a²/2 + a²/6) = 2a²/3 > a²/3, the energy condition is satisfied." }, { "idx": 1943, "question": "In an FCC crystal, does the dislocation reaction a/2[10-1] + a/6[-121] → a/3[11-1] satisfy the geometric conditions?", "answer": "Geometric conditions: b1 + b2 = (1/2 - 1/6)a + (0 + 2/6)b + (-1/2 + 1/6)c = 1/3a + 1/3b - 1/3c = a/3[11-1], satisfying the geometric conditions." }, { "idx": 1944, "question": "For metals without phase transformation in the solid state, if not remelted, how to refine grains through cold deformation and recrystallization?", "answer": "By performing significant cold deformation followed by recrystallization at an appropriate temperature to obtain fine grains." }, { "idx": 1945, "question": "What is texture?", "answer": "Texture is the phenomenon where crystal planes and orientations in a crystal tend to align uniformly." }, { "idx": 1946, "question": "Compare the differences between solid-phase sintering and liquid-phase sintering", "answer": "Differences: Due to the faster mass transfer rate by flow compared to diffusion, liquid-phase sintering has a higher densification rate and lower sintering temperature. Additionally, the rate of the liquid-phase sintering process is also related to factors such as the amount of liquid phase, its properties (viscosity, surface tension, etc.), the wetting condition between the liquid and solid phases, and the solubility of the solid phase in the liquid phase. The factors affecting liquid-phase sintering are more complex than those for solid-phase sintering." }, { "idx": 1947, "question": "What is equilibrium crystallization?", "answer": "Equilibrium crystallization refers to crystallization that occurs at a very slow rate, with sufficient diffusion in both the liquid and solid phases." }, { "idx": 1948, "question": "What is non-equilibrium crystallization?", "answer": "Non-equilibrium crystallization refers to crystallization that occurs under conditions of relatively fast crystallization rates and insufficient diffusion." }, { "idx": 1949, "question": "It is known that for a certain crystal at 500°C, 1 vacancy can form for every 10^10 atoms. What is the vacancy formation energy of this crystal? (Given: the constant A=0.0539 for this crystal, Boltzmann constant k=1.381×10^-23 J/K)", "answer": "$$ x\\\\begin{array}{c}{{c=A\\\\exp\\\\enspace\\\\big(\\\\enspace-\\\\frac{\\\\Delta E_{V}}{k T}\\\\big)}}\\\\ {{\\\\Delta E_{V}=-k T\\\\mathrm{ln}\\\\frac{c}{A}=\\\\mathrm{-\\\\enspace\\\\left[1.381\\\\times10^{-23}\\\\times\\\\enspace\\\\left(500+273\\\\right)\\\\right]\\\\enspace\\\\mathrm{ln}\\\\frac{10^{-10}}{0.0539}\\\\mathrm{\\\\scriptsize\\\\mathrm{~J}}}}}\\\\\\\\ {{=1.068\\\\times10^{-20}\\\\times17.8=1.9\\\\times10^{-19}\\\\mathrm{\\\\scriptsize{\\\\times10^{-19}}}}}\\\\\\\\ \\n $$" }, { "idx": 1950, "question": "What are the application conditions of the first law of diffusion?", "answer": "The application condition of the first law of diffusion is steady-state diffusion, that is, diffusion independent of time." }, { "idx": 1951, "question": "What types of textures are included?", "answer": "Textures include recrystallization texture and deformation texture. Among them, deformation texture further includes fiber texture and sheet texture." }, { "idx": 1952, "question": "What are the common phase structures in solids?", "answer": "The common phase structures in solids include: solid solution (element), compound, ceramic crystalline phase, amorphous phase, and molecular phase." }, { "idx": 1953, "question": "2. Please calculate the angle between (111) and (111) in a simple cubic crystal.", "answer": "$${{\\\\cos\\\\alpha=\\\\frac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{\\\\sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}\\\\sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}}}\\\\mathrm{\\\\textstyle=\\\\frac{1-1+1}{\\\\sqrt{3}\\\\times\\\\sqrt{3}}=\\\\frac{1}{3}}}}\\\\ {{\\\\alpha=70^{\\\\circ}32^{\\\\prime}\\\\quad\\\\qquad\\\\quad}}\\\\end{array} $$" }, { "idx": 1954, "question": "What is constitutional supercooling?", "answer": "During the solidification process, as the solid-liquid interface advances, solute atoms accumulate on the liquid side near the solid-liquid interface, causing a change in the melting point of the liquid phase. The resulting supercooling phenomenon is called constitutional supercooling." }, { "idx": 1955, "question": "What is a substitutional solid solution?", "answer": "An alloy phase where solute atoms replace solvent atoms while maintaining the solvent structure is called a substitutional solid solution." }, { "idx": 1956, "question": "For cases where the concentration gradient changes over time, can the first law of diffusion be applied?", "answer": "The first law of diffusion can also be applied to non-steady-state diffusion, but it must be modified." }, { "idx": 1957, "question": "What are the factors affecting the solubility of substitutional solid solutions?", "answer": "The influencing factors include: ①Atomic size; ②Crystal structure; ③Electronegativity; ④Electron concentration." }, { "idx": 1958, "question": "How does constitutional supercooling affect the growth morphology of solid solutions?", "answer": "In the absence of constitutional supercooling, the solid solution grows in a planar manner, forming equiaxed crystals; with a small degree of supercooling, cellular structures form; with a large degree of constitutional supercooling, dendritic crystals form." }, { "idx": 1959, "question": "What are the conditions for forming an infinite solid solution?", "answer": "The same crystal structure of the two components is a necessary condition for forming an infinite solid solution." }, { "idx": 1960, "question": "What is the difference in diffusion coefficients between substitutional diffusion and interstitial diffusion?", "answer": "The interstitial diffusion coefficient is independent of vacancy concentration, whereas the substitutional diffusion coefficient is related to vacancy concentration (can be expressed by a formula). Generally, the interstitial diffusion coefficient is greater than the substitutional diffusion coefficient." }, { "idx": 1961, "question": "An FCC crystal yields under a normal stress of 2MPa in the [123] direction. The activated slip system has been measured as (111)[101]. Determine the resolved shear stress τ that activates this slip system.", "answer": "$$ \\\\begin{array}{l}\\\\cos\\\\phi=\\\\frac{[\\\\overline{1}23]}{|[\\\\overline{1}23]|}\\\\cdot\\\\frac{[111]}{|[111]|}=\\\\frac{-1+2+3}{\\\\sqrt{14}\\\\sqrt{3}}=0.617\\\\\\\\ \\\\cos\\\\lambda=\\\\frac{[\\\\overline{1}23]}{|[\\\\overline{1}23]|}\\\\cdot\\\\frac{[\\\\overline{1}01]}{|[\\\\overline{1}01]|}=\\\\frac{1+0+3}{\\\\sqrt{14}\\\\sqrt{2}}=0.756\\\\end{array}$$ $$\\\\tau=2\\\\times0.617\\\\times0.756\\\\mathrm{MPa}=0.933\\\\mathrm{MPa}$$" }, { "idx": 1962, "question": "In a diffusion couple, if it is interstitial diffusion, will the Kirkendall effect occur? Why?", "answer": "It will not occur. Because interstitial diffusion considers the directional flow of interstitial atoms and does not account for substitutional diffusion." }, { "idx": 1963, "question": "At room temperature, a sheet of iron (with a melting point of 1538°C) is bent back and forth. What phenomenon will occur as the bending continues? Why?", "answer": "According to T_recrystallization = (0.35–0.45)T_melting, the processing of Fe at room temperature is cold working. Therefore, as bending proceeds, the sheet undergoes work hardening, and continued deformation leads to fracture of the iron sheet." }, { "idx": 1964, "question": "Please compare the similarities between binary eutectic transformation and peritectic transformation", "answer": "Similarities: isothermal and constant composition transformation; both appear as horizontal lines on the phase diagram." }, { "idx": 1965, "question": "When bending a tin plate (with a melting point of 232°C) back and forth at room temperature, what phenomenon will occur as the bending proceeds? Why?", "answer": "According to T_recrystallization=(0.35～0.45)Tm, the processing of Sn at room temperature is considered hot working. Therefore, as bending proceeds, dynamic recrystallization occurs in the Sn plate, allowing it to be bent for an extended period." }, { "idx": 1966, "question": "What is solid solution strengthening? Please briefly describe its strengthening mechanism.", "answer": "Solid solution strengthening is the phenomenon where solute atoms impede dislocation motion, thereby enhancing the strength of the alloy. The main mechanisms include: ① Cottrell atmosphere, where the elastic stress field of solute atoms hinders dislocation motion; ② Suzuki atmosphere, where solute atoms reduce the stacking fault energy of the matrix, causing dislocations to decompose into extended dislocations and impeding dislocation cross-slip or climb; ③ Electrical interaction, where charged solute atoms form electrostatic interactions with dislocations, hindering dislocation motion." }, { "idx": 1967, "question": "Please compare the differences between binary eutectic transformation and peritectic transformation", "answer": "Differences: eutectic is a decomposition-type reaction, peritectic is a synthesis-type reaction; the eutectic line is entirely a solidus line, while only part of the peritectic line is a solidus line; the eutectic triangle is above the horizontal line, the peritectic triangle is below the horizontal line." }, { "idx": 1968, "question": "Please write out the crystal planes included in the {110} and {100} crystal plane families in the cubic crystal system, respectively.", "answer": "{110}: (110) (101) (011) (110) (101) (011) {100}: (100) (010) (001) (001) (010) (100)" }, { "idx": 1969, "question": "Please compare the distortion energy of the two dislocations b1= a/2[111] and b2=a[100] in FCC crystals, which one is greater.", "answer": "∣b₁∣=a/2√(1+1+1)=√3/2a ∣b₂∣=a√(1+0+0)=a Therefore, the distortion energy of b1 is smaller." }, { "idx": 1970, "question": "When CN=6, the ionic radius of Na+ is 0.097nm, what is its radius when CN=4?", "answer": "0.088nm" }, { "idx": 1971, "question": "Nickel has a face-centered cubic structure with an atomic radius of r_Ni=0.1246 nm. Determine how many atoms are present in 1 mm^2 on the (100) plane of nickel.", "answer": "6.1×10^13 atoms/mm^2" }, { "idx": 1972, "question": "When CN=6, the ionic radius of Na+ is 0.097nm, what is its radius when CN=8?", "answer": "0.100nm" }, { "idx": 1973, "question": "In the <100> direction of copper (fcc, a=0.361nm), what is the linear density of atoms?", "answer": "The linear density of Cu atoms is 2.77×10^6 atoms/mm" }, { "idx": 1974, "question": "Write the equivalent crystal planes included in the crystal plane families {100}, {110}, {111}, {112} in cubic crystals.", "answer": "{100}=(100)+(010)+(001), totaling 3 equivalent planes. {110}=(110)+(1̅10)+(101)+(1̅01)+(011)+(01̅1), totaling 6 equivalent planes. {111}=(111)+(1̅11)+(11̅1)+(111̅), totaling 4 equivalent planes. {112}=(112)+(1̅12)+(11̅2)+(112̅)+(121)+(121)+(121)+(121)+(211)+(211)+(211)+(211), totaling 12 equivalent planes." }, { "idx": 1975, "question": "In the <100> direction of iron (bcc, a=0.286nm), what is the linear density of atoms?", "answer": "The linear density of Fe atoms is 3.50×10^6 atoms/mm" }, { "idx": 1976, "question": "Nickel has a face-centered cubic structure with an atomic radius of r_Ni=0.1246 nm. Determine how many atoms are present in 1 mm^2 on the (111) plane of nickel.", "answer": "1.86×10^13 atoms/mm^2" }, { "idx": 1977, "question": "Determine whether the following dislocation reaction can proceed: $a[100]\\\\rightarrow\\\\frac{a}{2}[101]+\\\\frac{a}{2}[10{\\\\overline{{1}}}]$", "answer": "No. Energy condition: $\\\\sum b_{\\\\mathbb{M}}^{2}=\\\\sum b_{E}^{2}=a^{2}$, the energy on both sides is equal." }, { "idx": 1978, "question": "The atomic packing density of magnesium, like all hcp metals, is 0.74. Calculate the volume of the unit cell of magnesium. Given the density of Mg ρMg=1.74 Mg/m³, relative atomic mass of 24.31, and atomic radius r=0.161 nm.", "answer": "The volume of the unit cell is Vuc=0.14 nm³ (or 1.4×10⁻²⁸ m³)." }, { "idx": 1979, "question": "Nickel has a face-centered cubic structure with an atomic radius of r_Ni=0.1246 nm. Determine the number of atoms per 1 mm^2 on the (110) plane of nickel.", "answer": "1.14×10^13 atoms/mm^2" }, { "idx": 1980, "question": "Determine whether the following dislocation reaction can occur: $\\\\frac{a}{3}[112]+\\\\frac{a}{2}[111]\\\\rightarrow\\\\frac{a}{6}[11\\\\overline{{{1}}}]$", "answer": "No. Geometric condition: $\\\\sum b_{\\\\mathbb{H}}=\\\\frac{a}{b}[557],\\\\sum b_{\\\\mathbb{H}}=\\\\frac{a}{b}[11\\\\overline{{1}}]$, cannot be satisfied." }, { "idx": 1981, "question": "The density of quartz (SiO2) is 2.65 Mg/m3. How many silicon atoms (and oxygen atoms) are there in 1 m3?", "answer": "5.29×10^28 silicon atoms /m3" }, { "idx": 1982, "question": "At $800^{\\circ}\\mathrm{C}$, one atom in $10^{10}$ atoms has sufficient energy to move within the solid, while at $900^{\\circ}\\mathrm{C}$, only one atom in $10^{9}$ atoms does. Calculate the activation energy (J/atom).", "answer": "$0.4\\times10^{-18}~\\mathrm{J}/$ atom" }, { "idx": 1983, "question": "If a face-centered cubic crystal has a unit dislocation with b=a/2[1̄01] and a partial dislocation with b=a/6[121̄], these two dislocations meet and undergo a dislocation reaction. Can this reaction proceed? Why?", "answer": "It can proceed. Because it satisfies both the geometric condition: ∑bH=∑bE=a/3[1̄11], and the energy condition: ∑bM²=2/3a²>∑bE²=1/3a²." }, { "idx": 1984, "question": "When the radii of silicon and oxygen in quartz (SiO2) are 0.038 nm and 0.114 nm respectively, what is the packing density (assuming the atoms are spherical)?", "answer": "0.33" }, { "idx": 1985, "question": "Determine whether the following dislocation reaction can proceed: $\\\\frac{a}{2}[10\\\\overline{{{1}}}]+\\\\frac{a}{6}[\\\\overline{{{1}}}21]\\\\rightarrow\\\\frac{a}{3}[11\\\\overline{{{1}}}]$", "answer": "Yes. Geometric condition: $\\\\sum b_{\\\\tt H}=\\\\sum b_{\\\\tt G}=\\\\frac{\\\\alpha}{3}[11\\\\overline{{1}}]$; energy condition: $\\\\sum b_{\\\\mathbb{H}}^{2}={\\\\frac{2}{3}}a^{2}>$ $\\\\sum b_{E}^{2}={\\\\frac{1}{3}}a^{2}.$" }, { "idx": 1986, "question": "Given the Burgers vector b=0.25 nm, if the misorientation angle θ of a symmetric tilt grain boundary is 10°, calculate the distance between dislocations at the grain boundary.", "answer": "When θ=10°, D=1.4 nm." }, { "idx": 1987, "question": "Determine whether the following dislocation reaction can proceed: $a[100]\\\\rightarrow\\\\frac{a}{2}[111]+\\\\frac{a}{2}[1\\\\overline{{{11}}}]_{\\\\circ}$.", "answer": "No. Energy condition: $\\\\sum b_{\\\\mathbb{H}}^{2}=a^{2}<\\\\sum b_{\\\\mathbb{H}}^{2}={\\\\frac{3}{2}}a^{2}$, meaning the energy increases after the reaction." }, { "idx": 1988, "question": "What conclusion can be drawn from the calculation results?", "answer": "When θ=10°, the distance between dislocations is only 5∼6 atomic spacings, indicating that the dislocation density is too high at this point. This suggests that the model is no longer applicable when the θ angle is large." }, { "idx": 1989, "question": "Calculate the shear stress required for slip to occur in a nickel crystal with this F-R dislocation source. Given that for Ni, G=7.9×10^10 Pa, a=0.350 nm.", "answer": "τ_Ni=1.95×10^7 Pa." }, { "idx": 1990, "question": "If the dislocation density in a certain crystal is known to be ρ=10^6~10^7 cm/cm^3, and the average length of F-R dislocation sources measured experimentally is 10^-4 cm, determine the number of F-R dislocation sources in the dislocation network.", "answer": "Assuming that the dislocation lines in the crystal are entangled and mutually pinned, the possible number of dislocation sources is n= ρ/L=10^10~10^11 per cm^3." }, { "idx": 1991, "question": "Given the Burgers vector b=0.25 nm, if the misorientation angle θ of the symmetric tilt grain boundary is 1°, calculate the distance between dislocations at the grain boundary.", "answer": "When θ=1°, D=14 nm." }, { "idx": 1992, "question": "If a piece of iron is heated to $850^{\\\\circ}\\\\mathrm{C}$ and then rapidly cooled to $20^{\\\\circ}\\\\mathrm{C}$, calculate how many times the number of vacancies should increase before and after the treatment (assuming the energy required to form one mole of vacancies in iron is 104600J)", "answer": "$0.616\\\\times10^{14}$ times" }, { "idx": 1993, "question": "Among the three common crystal structures of metals, which structure cannot serve as a space lattice?", "answer": "Hexagonal close-packed structure" }, { "idx": 1994, "question": "In the expression for calculating the force on dislocation motion, $f=\\tau b$, what does $\\tau$ refer to?", "answer": "The resolved shear stress of the external force in the slip direction on the slip plane." }, { "idx": 1995, "question": "If a face-centered cubic crystal has a unit dislocation with b=α/2[1̄01] and a partial dislocation with b=a/6[121̄], these two dislocations meet and undergo a dislocation reaction. Write the Burgers vector of the resulting dislocation and specify the type of the resulting dislocation.", "answer": "bŝ=a/3[1̄11]; this dislocation is a Frank partial dislocation." }, { "idx": 1996, "question": "What is the difference between a space lattice and a crystal lattice?", "answer": "A crystal lattice, also known as a crystal structure, refers to the specific arrangement of atoms; whereas a space lattice abstracts atoms as pure geometric points by ignoring their volume." }, { "idx": 1997, "question": "How are jogs generally formed on dislocation lines?", "answer": "Intersection of dislocations." }, { "idx": 1998, "question": "Can a dislocation loop be formed by inserting a columnar half-atom plane in a crystal?", "answer": "No. Because a dislocation loop can only be formed when the crystal inside the loop undergoes slip while the crystal outside does not." }, { "idx": 1999, "question": "When a dislocation is subjected to force, its motion direction is everywhere perpendicular to the dislocation line and can vary during movement. What should be the direction of the relative sliding of the crystal?", "answer": "It is always the direction of the Burgers vector." }, { "idx": 2000, "question": "The atomic radius is related to the crystal structure. How does the atomic radius change when the coordination number of the crystal structure decreases?", "answer": "The atomic radius contracts. This is because the atom tries to maintain the volume it occupies as unchanged or with minimal change [the volume occupied by the atom $V_{A}=$ the volume of the atom () + the interstitial volume]. When the coordination number of the crystal structure decreases, the interstitial volume increases. To maintain the balance of the above equation, the atomic radius must contract." }, { "idx": 2001, "question": "What is the interface with the lowest interfacial energy?", "answer": "Coherent interface." }, { "idx": 2002, "question": "Is the statement 'Small-angle grain boundaries are all formed by edge dislocations arranged into walls' correct?", "answer": "No. Twist boundaries are formed by crossed screw dislocations of the same sign." }, { "idx": 2003, "question": "A subgrain boundary is composed of $n$ edge dislocations with a misorientation of $0.057^{\\\\circ}$. Assuming there is no interaction between the dislocations before forming the subgrain boundary, by what factor does the distortion energy change after forming the subgrain boundary (given $R{=}10^{-4}$, $r_{0}=b=10^{-8}$; after forming the subgrain boundary, $$ R=D\\\\approx{\\\\frac{b}{\\\\theta}})?", "answer": "The distortion energy is 0.75 times the original value (indicating that the dislocation energy decreases after forming the subgrain boundary)." }, { "idx": 2004, "question": "What are the characteristics of atomic arrangement in ordered alloys?", "answer": "An ordered solid solution, where atoms of each component occupy their respective Bravais sublattices, known as sublattices. The entire solid solution forms a complex lattice composed of the sublattices of each component, also referred to as a superlattice or superstructure." }, { "idx": 2005, "question": "What is the relationship between the atomic arrangement and bonding in ordered alloys?", "answer": "This arrangement is related to the bonding energy (bond) between atoms. The greater the bonding energy, the less likely the atoms are to combine. If the bonding energy between dissimilar atoms is less than that between similar atoms, i.e., E_{A B}<(E_{A A}+E_{B B})/2, then the solute atoms will exhibit partially ordered or completely ordered arrangements." }, { "idx": 2006, "question": "Why do many ordered alloys become disordered at high temperatures?", "answer": "The driving force for ordering is the mixing energy parameter (=AB一 (ε^{m}=ε_{A B}-1/2(E_{A A}+E_{B B}))ε^{m}<0, while the resistance to ordering is the configurational entropy; increasing temperature enhances the contribution of the latter to the free energy (-T S), and beyond a certain critical temperature, the disordered solid solution becomes more stable, causing the ordered solid solution to disappear and transform into a disordered solid solution." }, { "idx": 2007, "question": "Analyze the type of solid solution formed by H in α-Fe and γ-Fe, their locations, and the solubility (mole fraction). The atomic radii of the elements are as follows: H: 0.046 nm, α-Fe: 0.124 nm, γ-Fe: 0.126 nm", "answer": "H forms an interstitial solid solution in α-Fe. Due to the significant difference in size factors, the solubility (mole fraction) is very small. The location of H in α-Fe is mostly at the octahedral interstitial centers. The solubility decreases rapidly with decreasing temperature. In γ-Fe, H also forms an interstitial solid solution, located at the octahedral interstitial centers, with slightly higher solubility than in α-Fe." }, { "idx": 2008, "question": "The known solid solubility limits (mole fractions) of elements such as Cd, Zn, Sn, and Sb in Ag are x_Cd=42.5×10^-2, x_Zn=20×10^-2, x_Sn=12×10^-2, x_Sb=7×10^-2, respectively, and their atomic diameters are 0.3042 nm, 0.314 nm, 0.316 nm, 0.3228 nm, while Ag is 0.2883 nm. Analyze the reasons for the differences in their solid solubility limits (mole fractions).", "answer": "When the atomic size factors are similar, the solid solubility limits (mole fractions) of the above elements in Ag are influenced by the atomic valence factor, i.e., the valence electron concentration e/a is an important factor determining the solid solubility limit (mole fraction). Their valences are 2, 3, 4, and 5, respectively, while Ag is 1." }, { "idx": 2009, "question": "Calculate the electron concentration at the solid solubility limit (mole fraction) for Cd, Zn, Sn, and Sb, given their valences are 2, 3, 4, and 5 respectively, Ag is 1, and the solid solubility limits (mole fractions) are x_Cd=42.5×10^-2, x_Zn=20×10^-2, x_Sn=12×10^-2, x_Sb=7×10^-2", "answer": "The electron concentration at the solid solubility limit can be calculated using the formula c=Z_A(1-x_B)+Z_Bx_B. Here, Z_A and Z_B are the numbers of valence electrons for components A and B respectively; x_B is the mole fraction of component B. The electron concentration for Cd=1×(1-0.425)+2×0.425=1.43; for Zn=1×(1-0.20)+3×0.20=1.42; for Sn=1×(1-0.12)+4×0.12=1.39; for Sb=1×(1-0.07)+5×0.07=1.31." }, { "idx": 2010, "question": "Analyze the type of solid solution formed by N in α-Fe and γ-Fe, their locations, and the solubility (mole fraction). The atomic radii of the elements are as follows: N: 0.071 nm, α-Fe: 0.124 nm, γ-Fe: 0.126 nm", "answer": "N forms an interstitial solid solution in α-Fe, with a maximum solubility (mole fraction) of about 0.1×10^-2 at 590°C, decreasing to 0.001×10^-2 at room temperature. In α-Fe, N is mostly located at the octahedral interstitial sites. In γ-Fe, N forms an interstitial solid solution, located at the octahedral interstitial sites, with a maximum solubility (mass fraction) of about 2.8×10^-2." }, { "idx": 2011, "question": "The intermetallic compound AlNi has a CsCl-type structure with a lattice constant of $a=0.288\\\\mathrm{~nm}$. Calculate its density (the relative atomic mass of Ni is 58.71, and the relative atomic mass of Al is 26.98).", "answer": "Density $\\\\rho=5.97~\\\\mathrm{g}/\\\\mathrm{cm}^{3}$." }, { "idx": 2012, "question": "Analyze the type of solid solution formed by C in α-Fe and γ-Fe, their locations, and the solubility (mole fraction). The atomic radii of the elements are as follows: C: 0.077 nm, α-Fe: 0.124 nm, γ-Fe: 0.126 nm", "answer": "In α-Fe, C forms an interstitial solid solution, with the maximum solubility (mole fraction) reaching 0.0218×10^-2 at 727°C and decreasing to 0.006×10^-2 at room temperature. The location of C in α-Fe is mostly at the octahedral interstitial centers. In γ-Fe, C forms an interstitial solid solution located at the octahedral interstitial centers, with the maximum solubility (mass fraction) being 2.11×10^-2." }, { "idx": 2013, "question": "Why can only the two components of substitutional solid solutions mutually dissolve infinitely, while interstitial solid solutions cannot?", "answer": "This is because when a solid solution forms, the dissolution of solute atoms causes lattice distortion in the solvent structure, thereby increasing the system's energy. The greater the difference in atomic size between the solute and solvent, the more severe the lattice distortion, leading to higher distortion energy, lower structural stability, and smaller solubility. Generally, the lattice distortion caused by solute atoms in interstitial solid solutions is more significant, so they cannot mutually dissolve infinitely but only have limited solubility." }, { "idx": 2014, "question": "The density of ZnS is $4.1~\\\\mathrm{Mg}/\\\\mathrm{m}^{3}$, calculate the distance between the centers of the two ions based on this", "answer": "The distance between the centers of the two ions is $0.234\\\\mathrm{~n}\\\\mathrm{m}$" }, { "idx": 2015, "question": "The maximum solid solubility (mole fraction) of carbon in γ-Fe is x_C=8.9×10^-2. Given that C atoms occupy octahedral interstitial sites, calculate the percentage of octahedral interstitial sites occupied by C atoms.", "answer": "The percentage of octahedral interstitial sites occupied by C atoms is 10.2%." }, { "idx": 2016, "question": "Calculate the distance between the center of a sodium ion and its nearest neighboring positive ion in NaCl (given the radii of Na+ and Cl− are 0.097nm and 0.181nm, respectively).", "answer": "0.393nm" }, { "idx": 2017, "question": "The maximum solid solubility (mole fraction) of nitrogen in γ-Fe is x_N=10.3×10^-2. Given that N atoms occupy octahedral interstitial sites, calculate the percentage of octahedral interstitial sites occupied by N atoms.", "answer": "The percentage of octahedral interstitial sites occupied by N atoms is 12.5%." }, { "idx": 2018, "question": "Calculate the distance between the center of a sodium ion and its nearest neighboring ion in NaCl (given the radii of Na+ and Cl− are 0.097nm and 0.181nm, respectively).", "answer": "0.278nm" }, { "idx": 2019, "question": "Calculate the distance between the center of a sodium ion and the next nearest Cl− ion in NaCl (given the radii of Na+ and Cl− are 0.097nm and 0.181nm, respectively).", "answer": "0.482nm" }, { "idx": 2020, "question": "Analyze the type of solid solution formed by B in α-Fe and γ-Fe, its location, and the solubility (mole fraction). The atomic radii of the elements are as follows: B: 0.091 nm, α-Fe: 0.124 nm, γ-Fe: 0.126 nm", "answer": "In α-Fe, due to the larger atomic size of B, it is more difficult to dissolve interstitially, sometimes partially dissolving substitutionally, forming interstitial or substitutional solid solutions, with relatively low solubility. In γ-Fe, B forms an interstitial solid solution, located at the center of octahedral interstitial sites, with slightly higher solubility than in α-Fe." }, { "idx": 2021, "question": "Calculate the distance between the center of a sodium ion and the third nearest Cl− ion in NaCl (given the radii of Na+ and Cl− are 0.097nm and 0.181nm, respectively).", "answer": "0.622nm" }, { "idx": 2022, "question": "A common polymer has C2H2Cl2 as its monomer, with an average molecular weight of 60000u (using the relative atomic masses of the elements as Ar(C)=12, Ar(H)=1, Ar(Cl)=35.5). What is its degree of polymerization?", "answer": "The degree of polymerization is n = 60000 / 97 = 620" }, { "idx": 2023, "question": "A solid solution contains xMgO=30/10−2 and xLiF=70/10−2. What are the mass fractions of Li+, Mg2+, F−, and O2−?", "answer": "wLi+=16/10−2, wMg2+=24/10−2, wF−=44/10−2, wO2−=16/10−2" }, { "idx": 2024, "question": "A ceramic insulator contains $1\\\\%$ (by volume) of pores after sintering, with the pores being cubes of side length $13.7\\\\mathrm{mm}$. If during the manufacturing process, the powder can be pressed to contain $24\\\\%$ pores, what should the size of the mold be?", "answer": "The size of the mold should be $l=15.0\\\\mathrm{~mm}$." }, { "idx": 2025, "question": "The theoretical strength of amorphous materials is calculated to be $G/6\\\\sim G/4$, where $G$ is the shear modulus. Given $\\\\nu=0.25$, estimate the theoretical strength of glass (an amorphous material) based on its elastic properties (given $E=70~000~\\\\mathrm{MPa}$).", "answer": "Therefore, the theoretical strength ranges between 0.4E and 0.4E, that is, $$4~900\\\\sim7~000~\\\\mathrm{MPa}$$" }, { "idx": 2026, "question": "Calculate the distance between the centers of sodium ions at the nearest identical positions in NaCl (given the radii of Na+ and Cl− are 0.097nm and 0.181nm, respectively).", "answer": "0.393nm" }, { "idx": 2027, "question": "An organic compound has the composition ${w_{\\mathrm{C}}=62.1/10^{-2}}$, ${w_{\\mathrm{H}}}=10.3/1\\\\time10^{-2}$, $\\\\varpi_{0}=27.6/10^{-2}$. Try to write the possible name of the compound.", "answer": "\\\\mathrm{{C}:\\\\mathrm{{H}:\\\\mathrm{{O}=\\\\frac{62.\\\\mathrm{{1}}}{12.\\\\mathrm{{,011}}}:\\\\frac{10.\\\\mathrm{{3}}}{1.\\\\mathrm{{007}\\\\mathrm{{97}}}}:\\\\frac{27.6}{15.\\\\mathrm{{9994}}}=5.2:10.2:1.7\\\\approx3:6:10.4:1.7\\\\approx3:6:1.74:\\\\mathrm{{8}.\\\\mathrm{{1}:\\\\mathrm{{7}}}}}}} Therefore, the possible compound is $\\\\mathrm{CH}_{3}\\\\mathrm{COCH}_{3}$ (acetone)." }, { "idx": 2028, "question": "Assuming the density of MgO is 3.6 g/cm3 and the density of LiF is 2.6 g/cm3, what is the density of the solid solution?", "answer": "The density of the solid solution ρ=2.9 g/cm3" }, { "idx": 2029, "question": "A common polymer has C2H2Cl2 as its monomer, with an average molecular weight of 60000u (using the relative atomic masses of the elements as Ar(C)=12, Ar(H)=1, Ar(Cl)=35.5). Calculate the mass of the monomer.", "answer": "The mass of the monomer is 12×2 + 1×2 + 35.5×2 = 97u/mol" }, { "idx": 2030, "question": "What are the types of silicate structures", "answer": "Silicates are divided into the following categories: (1) silicates containing finite silicon-oxygen groups; (2) chain silicates; (3) layer silicates; (4) framework silicates." }, { "idx": 2031, "question": "Why can rapid changes in external temperature cause many ceramic devices to crack or break?", "answer": "Because most ceramics are mainly composed of crystalline and glass phases, and the thermal expansion coefficients of these two phases differ significantly. When rapidly cooled from high temperatures, the different contractions of each phase generate internal stresses sufficient to cause the ceramic devices to crack or break." }, { "idx": 2032, "question": "A polymeric material contains polyvinyl chloride, with 900 monomers in one molecule. If each molecule can be stretched into a linear molecule, calculate the theoretically maximum strain that can be achieved for this polymer (assuming the bond length of each C-C bond is $0.154\\\\mathrm{nm}$).", "answer": "The theoretically maximum strain is $3380\\\\%$." }, { "idx": 2033, "question": "Describe the thermodynamic conditions of crystallization phase transition", "answer": "Analysis of the change in system free energy during crystallization phase transition shows that the thermodynamic condition for crystallization is ΔG<0. From the change in free energy per unit volume ΔGB=-LmΔT/Tm, it can be seen that only when ΔT>0 can ΔGg<0 be achieved. That is, only undercooling can make ΔG<0." }, { "idx": 2034, "question": "Describe the structural conditions of crystalline phase transformation", "answer": "The structural fluctuations present in the liquid serve as the foundation for nucleation during crystallization. Therefore, structural fluctuations are the essential structural condition required for the crystallization process." }, { "idx": 2035, "question": "Polyvinyl chloride eccC2H3Cl∂n is dissolved in an organic solvent, with its C-C bond length set at 0.154nm, and the number of bonds in the chain x=2n. For a molecule with a molecular mass of 28500g, what is its root mean square length?", "answer": "Root mean square length 4.65nm." }, { "idx": 2036, "question": "For a copolymer ABS with equal mass fractions of each component, what is the ratio of the monomers (A—acrylonitrile; B—butadiene; S—styrene)?", "answer": "The mole fractions of the monomers are \\n\\n$$\\nx_{\\\\#Z\\\\#}=20/10^{-2},\\\\quad x_{\\\\sf T\\\\lceil-w\\\\rceil}=40/10^{-2},\\\\quad x_{\\\\#\\\\mathbb{W}\\\\#\\\\mathbb{W}}=40/10^{-2}\\n$$" }, { "idx": 2037, "question": "Polyvinyl chloride C2H3Cln is dissolved in an organic solvent, with its C-C bond length set at 0.154nm, and the number of bonds in the chain x=2n. If the root mean square length is only half of that in (1), what is the molecular mass?", "answer": "The molecular mass M=7125g." }, { "idx": 2038, "question": "Given that the relative atomic mass of Cu is 63.5 and its density is 8.9 g/cm^3, find the number of atoms in the critical nucleus.", "answer": "n≈261" }, { "idx": 2039, "question": "If the maximum undercooling during the solidification of pure nickel is 0.18 times its melting point $(t_{\\\\mathrm{m}}=1453^{\\\\circ}\\\\mathrm{C})$, calculate the driving force for solidification. $(\\\\Delta H=-18075.\\\\mathrm{J/mol})$", "answer": "The driving force for solidification $\\\\Delta G=-3~253.5~\\\\mathrm{J/mol},$" }, { "idx": 2040, "question": "What are the main bonding types in ceramic materials?", "answer": "The main bonding types in ceramic materials are ionic bonds and covalent bonds." }, { "idx": 2041, "question": "Discuss the energy conditions of crystalline phase transformation", "answer": "From the critical nucleus formation work oS, it can be seen that when a critical nucleus is formed, 1/3 of the surface energy must still be provided by energy fluctuations in the liquid." }, { "idx": 2042, "question": "Describe the kinetic conditions of crystalline phase transformation", "answer": "The kinetic condition is that the temperature of the liquid at the liquid-solid interface front Tir_k have the potential to nucleate; while embryos with r=r_k may either disappear or grow stably. Therefore, an embryo with radius r_k is called a critical nucleus. Its physical meaning is that the short-range ordered atomic clusters emerging in the undercooled liquid can become nuclei and grow when their size r≥r_k." }, { "idx": 2049, "question": "What is the quantitative relationship between the critical nucleus radius r_k and the undercooling ΔT?", "answer": "The critical nucleus radius r_k, whose size is related to the undercooling, is given by r_k=(2σT_m)/(L_m)×1/ΔT" }, { "idx": 2050, "question": "What is dynamic undercooling?", "answer": "During crystal growth, a certain degree of undercooling in the liquid ahead of the liquid-solid interface is required to satisfy (dN/dt)F>(dN/dt)M. This undercooling is called dynamic undercooling (ΔTl=Tm−Ti), which is a necessary condition for crystal growth." }, { "idx": 2051, "question": "What are the characteristics of ingot structure?", "answer": "In the ingot structure, there are generally 3 crystal zones. (1) The outermost layer is the fine-grained zone. Its formation is due to the lower temperature of the mold wall and the larger undercooling of the liquid, resulting in a higher nucleation rate. (2) The middle layer is the columnar crystal zone. Its formation is mainly due to the increase in the temperature of the mold wall, where the growth rate of crystal nuclei is greater than the nucleation rate, and heat dissipation is more favorable in the direction perpendicular to the mold wall. In the fine-grained zone, grains with favorable orientations preferentially grow into columnar crystals. (3) The center is the equiaxed crystal zone. Its formation is due to the further increase in mold wall temperature and the further decrease in liquid undercooling, where the directional heat dissipation of the remaining liquid is no longer obvious, and it is in a state of uniform cooling. At the same time, unmelted impurities, broken dendrites, etc., tend to concentrate in the remaining liquid, all of which promote the formation of equiaxed crystals. It should be noted that not all ingot structures have 3 crystal zones. Due to different solidification conditions, an ingot may only have one type of crystal zone or only two types of crystal zones." }, { "idx": 2052, "question": "Point out the errors in the following concepts and correct them: (1) The so-called degree of undercooling refers to the difference between the temperature at which a plateau appears on the cooling curve during crystallization and the melting point; while the dynamic degree of undercooling refers to the difference between the actual temperature of the liquid phase during crystallization and the melting point.", "answer": "The difference between the actual crystallization temperature on the cooling curve and the melting point; the difference between the temperature of the liquid at the liquid-solid interface front and the melting point." }, { "idx": 2053, "question": "Briefly describe the growth mechanism of pure metal crystals", "answer": "The crystal growth mechanism refers to the microscopic growth mode of crystals, which is related to the liquid-solid interface structure. For substances with rough interfaces, approximately $50\\\\%$ of the atomic positions on the interface are vacant. These vacancies can accept atoms, allowing liquid atoms to individually occupy the vacancies and connect with the crystal. The interface advances perpendicularly along its normal direction, exhibiting continuous growth. For crystals with smooth interfaces, growth does not occur through the attachment of individual atoms. Instead, it proceeds via homogeneous nucleation, forming a two-dimensional nucleus one atomic layer thick on the crystallographic facet interface. This creates a step between the nucleus and the original interface. Individual atoms can then fill the step, allowing the two-dimensional nucleus to grow laterally. Once the layer is filled, a new two-dimensional nucleus forms on the new interface, repeating the process. If the smooth interface of a crystal has an exposed screw dislocation, the interface becomes a spiral surface, forming a step that never disappears. Atoms attach to the step, enabling crystal growth." }, { "idx": 2054, "question": "What is undercooling?", "answer": "The temperature difference between the actual crystallization temperature and the theoretical crystallization temperature is called undercooling (ΔT=Tm−Tn). It is required by the thermodynamic conditions of phase transformation. Only when ΔT>0 can the condition that the free energy of the solid phase is lower than that of the liquid phase be achieved. The free energy difference between the liquid and solid phases is the driving force for crystallization." }, { "idx": 2055, "question": "What is critical undercooling?", "answer": "In an undercooled liquid, the degree of undercooling at which an embryo can form with a radius equal to the critical nucleus radius is called the critical undercooling (ΔT∗). Clearly, when the actual undercooling ΔT<ΔT∗, even the largest embryo in the undercooled liquid is smaller than the critical nucleus radius, making nucleation difficult. Only when ΔT>ΔT∗ can homogeneous nucleation occur. Therefore, critical undercooling is required for nucleation." }, { "idx": 2056, "question": "Analyze the basic conditions for the formation of single crystals", "answer": "The basic condition for forming a single crystal is to ensure that only one nucleus is generated (or only one nucleus can grow) during the crystallization of liquid metal, which then grows into a single crystal." }, { "idx": 2057, "question": "Given the melting point of Cu tm=1083°C, latent heat of fusion Lm=1.88×10^3 J/cm^3, and specific surface energy σ=1.44×10^5 J/cm^2. Calculate the critical nucleus radius for homogeneous nucleation of Cu at 853°C.", "answer": "rk=9.03×10^-10 m" }, { "idx": 2058, "question": "Point out the errors in the following concepts and correct them: (2) During metal crystallization, atoms transition from a disordered arrangement in the liquid phase to an ordered arrangement in the solid phase, which reduces the entropy of the system, making it a spontaneous process.", "answer": "Reduces the free energy of the system." }, { "idx": 2059, "question": "Point out the errors in the following concept and correct them: (8) During the crystallization of certain castings, due to the faster cooling rate, the homogeneous nucleation rate N1 increases, and the heterogeneous nucleation rate N2 also increases. Therefore, the total nucleation rate is N=N1+N2.", "answer": "Then the total nucleation rate is N=N2." }, { "idx": 2060, "question": "Point out the errors in the following concepts and correct them: (4) The maximum structural fluctuations appearing in the liquid phase at any temperature are nuclei.", "answer": "Under a certain degree of undercooling (>ΔT*)." }, { "idx": 2061, "question": "To obtain metallic glass, why is it generally necessary to choose binary systems with a steep liquidus line and thus a low eutectic temperature?", "answer": "Metallic glass is obtained by ultra-rapid cooling methods, which suppress the liquid-solid crystallization process, resulting in an amorphous structure with exceptional properties. Glass is an undercooled liquid. This type of liquid has high viscosity and low atomic mobility, making crystallization difficult. For example, polymer materials (silicates, plastics, etc.) can achieve a glassy state under normal cooling conditions. Metals, however, are different. Due to the low viscosity of liquid metals, they rapidly crystallize when cooled below the liquidus line, thus requiring extremely high cooling rates (estimated >10^10 °C/s) to achieve a glassy state. To obtain metallic glass at lower cooling rates, the stability of the liquid must be increased, allowing it to exist over a wider temperature range. Experiments have shown that when the liquidus line is steep, resulting in a low eutectic temperature, the stability of the liquid is enhanced. Therefore, such binary systems (e.g., Fe-B, Fe-C, Fe-P, Fe-Si, etc.) are selected. To improve performance, other elements (e.g., Ni, Mo, Cr, Co, etc.) can be added. This type of metallic glass can be obtained at cooling rates of 10^5–10^6 °C/s." }, { "idx": 2062, "question": "Point out the errors in the following concept and correct them: (5) The so-called critical nucleus is the size of the embryo when the decrease in the system's free energy fully compensates for the increase in surface free energy.", "answer": "The decrease in the system's free energy should compensate for 2/3 of the surface free energy." }, { "idx": 2063, "question": "Derive the K.A. Jackson equation formula", "answer": "The so-called equilibrium structure of an interface refers to the most stable state of the interface under the condition of minimum interfacial energy. The essence of the problem is to analyze the relative change in interfacial free energy when the interface becomes rough. For this purpose, the following assumptions are made: (1) The equilibrium between the liquid and solid phases is under isothermal conditions; (2) The structures of the liquid and solid phases are identical near the interface; (3) Only configurational entropy is considered, while vibrational entropy is neglected. Let $N$ be the total number of atomic positions on the liquid-solid interface, with $n$ being the number of solid-phase atomic positions, and the occupation fraction $x=\\\\frac{n}{N}$; the vacancy fraction on the interface is $1-x$, and the number of vacancies is $N(1-x)$. The formation of vacancies causes changes in internal energy and structural entropy, leading to a corresponding change in surface Gibbs free energy: $$\\\\Delta G_{\\\\mathrm{s}}=\\\\Delta H-T\\\\Delta S=(\\\\Delta u+P\\\\Delta S)-T\\\\Delta S\\\\approx\\\\Delta u-T\\\\Delta S$$ The increase in internal energy due to the formation of $N(1-x)$ vacancies is determined by the product of the number of broken solid bonds and the bond energy of a pair of atoms. The change in internal energy is: $$\\\\Delta u=N\\\\xi L_{\\\\mathrm{~m~}}x(1-x)$$ Here, $\\\\boldsymbol{\\\\xi}$ is related to the crystal structure and is called the crystallographic factor. Next, the entropy change is calculated. From the definition of entropy change, we have: $$\\\\Delta S=k\\\\ln w=k\\\\ln\\\\frac{N!}{(N x)![N-(N x)]!}=k\\\\ln\\\\frac{N!}{(N x)![N(1-x)]!}$$ Using Stirling's approximation, when $N$ is large, we obtain: $$\\\\Delta S=-k N[x\\\\mathrm{ln}x+(1-x)\\\\mathrm{ln}(1-x)]$$ Finally, the total change in free energy on the liquid-solid interface is calculated as: $$\\\\Delta G_{\\\\mathrm{s}}=\\\\Delta u-T_{\\\\mathrm{m}}\\\\Delta S=N\\\\hat{\\\\xi}L_{\\\\mathrm{m}}x(1-x)+k T_{\\\\mathrm{m}}N[x\\\\mathrm{ln}x+(1-x)\\\\mathrm{ln}(1-x)]$$ Thus: $$\\\\frac{\\\\Delta G_{\\\\mathrm{S}}}{N k T_{\\\\mathrm{m}}}=\\\\frac{\\\\xi L_{\\\\mathrm{m}}}{k T_{\\\\mathrm{m}}}x(1-x)+x\\\\mathrm{ln}x+(1-x)\\\\mathrm{ln}(1-x)$$ $$\\\\alpha=\\\\frac{\\\\xi L_{\\\\mathrm{~m~}}}{k T_{\\\\mathrm{~m~}}}$$ Therefore: $$\\\\frac{\\\\Delta G_{\\\\mathrm{S}}}{N k T_{\\\\mathrm{m}}}=\\\\alpha x\\\\left(1-x\\\\right)+x{\\\\ln x+(1-x)\\\\ln(1-x)}$$" }, { "idx": 2064, "question": "Point out the errors in the following concepts and correct them: (6) In liquid metal, any embryo that emerges with a radius smaller than the critical nucleus radius cannot nucleate, but as long as there is sufficient energy fluctuation to provide the nucleation work, nucleation is still possible.", "answer": "Nucleation cannot occur, even if there is sufficient energy fluctuation to provide it, nucleation still cannot occur." }, { "idx": 2065, "question": "Point out the errors in the following concept and correct them: (7) Measuring the maximum undercooling during the crystallization of a pure metal casting, where the measured value is basically consistent with the calculated value using the formula ΔT=0.2Tm.", "answer": "Measure the effective undercooling during homogeneous nucleation of a pure metal." }, { "idx": 2066, "question": "Point out the errors in the following concepts and correct them: (3) At any temperature, the maximum structural fluctuations appearing in liquid metals are all embryos.", "answer": "In undercooled liquids, the maximum structural fluctuations appearing in liquid metals are all embryos." }, { "idx": 2067, "question": "Point out the errors in the following concepts and correct them: (14) During the growth of pure metals, regardless of whether the liquid-solid interface is rough or smooth, the liquid-phase atoms are connected one by one along the vertical direction of the solid-phase surface.", "answer": "If the liquid-solid interface is rough, its liquid-phase atoms." }, { "idx": 2068, "question": "Point out the errors in the following concepts and correct them: (10) From the calculation formula of heterogeneous nucleation work A_hetero = A_g(2-3cosθ+cos³θ), it can be seen that when the wetting angle θ=0°, the nucleation work for heterogeneous nucleation is the largest.", "answer": "The nucleation work for heterogeneous nucleation is the smallest." }, { "idx": 2069, "question": "Point out the errors in the following concept and correct them: (12) Heterogeneous nucleation is always easier than homogeneous nucleation because the former uses foreign particles as the crystallization core, unlike the latter which forms an interface and causes an increase in free energy.", "answer": "Because the former uses foreign particles as the substrate, the nucleation work is smaller." }, { "idx": 2070, "question": "Point out the errors in the following concept and correct them: (11) In order to produce a batch of sand castings with significant thickness variations and require uniform grain size, it can be satisfied by merely adding nucleating agents in the process.", "answer": "It can be satisfied by merely accelerating the cooling of thicker sections (such as adding chills) in the process." }, { "idx": 2071, "question": "Point out the errors in the following concepts and correct them: (13) When studying the process of refining grains in a certain metal, the main focus is to find nucleating agents with low melting points and lattice constants similar to those of the metal, as their nucleation catalytic efficiency is the highest.", "answer": "The main focus is to find those with high melting points, and." }, { "idx": 2072, "question": "Point out the errors in the following concept and correct them: (9) If 10,000 nucleation agents are added to an undercooled liquid, then 10,000 grains will form after crystallization.", "answer": "then tens of thousands of grains will form after crystallization." }, { "idx": 2073, "question": "Point out the errors in the following concepts and correct them: (18) When a nucleation agent is added to liquid pure metal, its growth morphology always appears dendritic.", "answer": "Its growth morphology will not change." }, { "idx": 2074, "question": "Point out the errors in the following concepts and correct them: (15) Regardless of the temperature distribution, common pure metals always grow with a dendritic interface.", "answer": "Only under negative temperature gradient conditions, common pure metals." }, { "idx": 2075, "question": "Point out the errors in the following concept and correct them: (17) It is impossible to observe the dendritic growth process of extremely pure metals, so the dendritic growth morphology is merely a speculation.", "answer": "The growth process can be observed through experimental methods, such as pouring off the remaining liquid of the crystallizing metal or performing overall quenching, so the dendritic growth morphology is not a speculation." }, { "idx": 2076, "question": "Point out the errors in the following concepts and correct them: (16) The microstructure morphology of a saturated aqueous solution of ammonium chloride and pure metal at the end of crystallization is the same, with the former exhibiting dendritic crystals and the latter also exhibiting dendritic crystals.", "answer": "The microstructure morphology at the end of crystallization is different; the former exhibits dendritic crystals (with water between the branches), while the latter exhibits individual (blocky) grains." }, { "idx": 2077, "question": "In the Al-Mg alloy, $x_{\\\\mathrm{Mg}}=0.05$, calculate the mass fraction of Mg ($w_{\\\\mathrm{Mg}}$) in the alloy (given that the relative atomic mass of Mg is 24.31 and Al is 26.98).", "answer": "$w_{\\\\mathrm{Mg}}=0.0453$." }, { "idx": 2078, "question": "Point out the errors in the following concepts and correct them: (21) Pure metal crystallization grows in a dendritic morphology or a planar morphology, which is unrelated to the melting entropy of the metal.", "answer": "Because it is also related to the structure of the liquid-solid interface (α=ξΔSm/k), i.e., it is related to the melting entropy of the metal." }, { "idx": 2079, "question": "Point out the errors in the following concepts and correct them: (20) From a macroscopic perspective, if the liquid-solid interface is straight, it is called a smooth interface structure; if it is zigzag like metal, it is called a rough interface structure.", "answer": "The straight one is called a rough interface structure; the zigzag one is called a smooth interface structure." }, { "idx": 2080, "question": "Point out the errors in the following concepts and correct them: (23) During metal crystallization, the dynamic undercooling required for crystal growth is sometimes greater than the critical undercooling required for nucleation.", "answer": "The dynamic undercooling is smaller than the critical undercooling required for nucleation." }, { "idx": 2081, "question": "Point out the errors in the following concepts and correct them: (22) During the crystallization of actual metals, the nucleation rate increases with the increase of undercooling, and after exceeding a certain maximum value, the opposite change occurs.", "answer": "increases, but due to the limited undercooling capacity of metals, it does not exceed a certain maximum value." }, { "idx": 2082, "question": "Should high carbon steel or low carbon steel be used to manufacture automobile fenders?", "answer": "High carbon steel. Because high carbon steel has high strength and can withstand greater impact force without deformation. In contrast, low carbon steel is softer and more prone to deformation under force." }, { "idx": 2083, "question": "At 800°C, which phases exist in Fe-0.002C steel?", "answer": "α phase, γ phase." }, { "idx": 2084, "question": "Calculate the relative amounts of secondary cementite, eutectic cementite, and eutectoid cementite in ledeburite", "answer": "In ledeburite, the relative amount of Fe_3C_I is 10.15%, the relative amount of Fe_3C_3tan is 41.21%, and the relative amount of eutectoid Fe:C is 3.9%." }, { "idx": 2085, "question": "Calculate the relative amounts of cementite and ferrite in the pearlite microstructure", "answer": "In pearlite, the relative amount of F is 9.38%, and the relative amount of Fe_3C_## is 1.22%." }, { "idx": 2086, "question": "In the Al-Cu phase diagram, given $K=0.16$, $m=3.2$. If the solidification rate of the casting is $R=3\\\\times10^{-4}~\\\\mathrm{cm/s}$, the temperature gradient is $G=30^{\\\\circ}\\\\mathrm{C}/\\\\mathrm{cm}$, and the diffusion coefficient is $D=3\\\\times10^{-5}~\\\\mathrm{cm}^{2}/\\\\mathrm{s}$, find the extreme value of $\\\\scriptstyle{\\\\mathcal{W}}_{\\\\mathrm{Cu}}$ in the alloy that maintains planar interface growth.", "answer": "$w_{\\\\mathrm{Cu}}^{\\\\mathrm{C}_{0}}=\\\\frac{G D}{R m}\\\\frac{K}{1-K}=0.1744$" }, { "idx": 2087, "question": "Point out the errors in the following concept and correct them: (19) When a pure metal crystallizes and grows vertically, its interface is sometimes smooth and sometimes rough, growing alternately.", "answer": "The interface is of the rough type." }, { "idx": 2088, "question": "Based on microstructural analysis, the volume of graphite in a gray cast iron accounts for 12%, and the volume of ferrite accounts for 88%. Determine the value of ωC (given that the density of graphite ρG=2.2 g/cm³, and the density of ferrite ρα=7.8 g/cm³).", "answer": "ωC=0.037." }, { "idx": 2089, "question": "Calculate the relative amounts of pearlite, secondary cementite, and ledeburite in the microstructure of an iron-carbon alloy with a carbon content of w_c=0.04 after metastable cooling to room temperature.", "answer": "In the microstructure, the relative amount of P is 10.6%, the relative amount of Fe_3C_∥ is 3.10%, and the relative amount of L_d′ is 86.3%." }, { "idx": 2090, "question": "At 800°C, what are the compositions of the α phase and γ phase in Fe-0.002C steel?", "answer": "α: wC=0.0001, wFe=0.9999; γ: wC=0.0046, wFe=0.9954." }, { "idx": 2091, "question": "How to obtain the mass and Cu content of solid α3 by heating solid α2 to melting and slowly cooling it to 920°C, then pouring off the liquid?", "answer": "Reheat α2 to melting, slowly cool to 920°C, pour off the liquid, leaving only α3, with a mass of 260g and w_Cu≈0.02." }, { "idx": 2092, "question": "How to obtain the mass and Cu content of solid α4 by heating solid α3 to melting and slowly cooling it to 935°C before pouring off the liquid?", "answer": "Heat α3 to melting again, slowly cool it to 935°C, pour off the liquid, leaving only α4, with a mass of 180g and a Cu content of w_Cu≈0.013." }, { "idx": 2093, "question": "Assuming the Cu content in α phase at 100°C can be considered zero, calculate how many copper atoms are present in each θ particle, given that the atomic radius of Al is 0.143 nm.", "answer": "Approximately 150 Cu atoms per θ particle." }, { "idx": 2094, "question": "How to obtain the mass and Cu content of solid α2 by heating solid α1 to melting, slowly cooling to 900°C, and pouring off the liquid?", "answer": "Heat the solid (α1) in (1) to melting, slowly cool to 900°C, pour off the liquid, and the remaining solid α2 has a weight of 390g with w_Cu≈0.03." }, { "idx": 2095, "question": "At 800°C, what are the fractions of α phase and γ phase in Fe-0.002C steel?", "answer": "nα=0.58, nγ=0.42." }, { "idx": 2096, "question": "A eutectic reaction in the Mg-Ni system is given by $$ \\\\operatorname{L}_{w_{\\\\mathrm{Ni}}=0.235}\\\\frac{570^{\\\\circ}\\\\mathrm{C}}{\\\\Longleftarrow}\\\\alpha_{(\\\\#\\\\mathbb{M}_{\\\\operatorname{g}})}+\\\\mathrm{Mg}_{2}\\\\mathrm{Ni}_{w_{\\\\mathrm{Ni}}=0.546} $$ Let $\\\\omega_{\\\\mathrm{Ni}}^{1}=C_{1}$ be a hypoeutectic alloy and $\\\\boldsymbol{w_{\\\\mathrm{Ni}}^{2}}=C_{2}$ be a hypereutectic alloy. The mass fraction of the proeutectic phase in these two alloys is equal, but the total amount of $\\\\upalpha$ in alloy $\\\\mathrm{C}_{1}$ is 2.5 times that in alloy $\\\\mathrm{C}_{2}$. Calculate the compositions of $\\\\mathrm{C}_{1}$ and $\\\\mathrm{C}_{2}$.", "answer": "The composition of alloy $C_{1}$ is $w_{\\\\mathrm{Mg}}=0.873$, $w_{\\\\mathrm{Ni}}=0.127$. The composition of alloy $\\\\mathrm{C}_{2}$ is $w_{\\\\mathrm{Mg}}=0.66,w_{\\\\mathrm{Ni}}=0.368$." }, { "idx": 2097, "question": "If there is a certain Cu-Ag alloy (w_Cu=0.075, w_Ag=0.925) weighing 1000g, how can the mass of solid α1 and its Cu content be obtained by heating it above 900°C to melt, then slowly cooling to 850°C and pouring off the liquid part?", "answer": "When 1000g of this alloy is heated above 900°C to melt, then slowly cooled to 850°C and the liquid part is poured off, the remaining solid α1 weighs 780g with a w_Cu≈0.055." }, { "idx": 2098, "question": "Under equilibrium cooling conditions, a carbon steel obtains a microstructure containing 50% pearlite and 50% ferrite. What is the mass fraction of carbon in this alloy?", "answer": "x=wC=0.385" }, { "idx": 2099, "question": "14. At 550°C, an aluminum-copper alloy solid solution with composition x_Cu=0.02 is quenched and then reheated to 100°C to precipitate θ phase (CuAl2). The θ phase forms many small particles dispersed in the alloy with an average particle spacing of 5.0 nm. Approximately how many particles are formed in 1 mm³ of the alloy?", "answer": "Approximately 8×10^24 particles/m³ (equivalent to 8×10^15 particles/mm³)." }, { "idx": 2100, "question": "A carbon steel, under equilibrium cooling conditions, obtains a microstructure consisting of 50% pearlite and 50% ferrite. Question: If this alloy is heated to 730°C, what microstructure will be obtained under equilibrium conditions?", "answer": "The microstructure will be F+A" }, { "idx": 2101, "question": "Under equilibrium cooling conditions, a carbon steel obtains a microstructure consisting of 50% pearlite and 50% ferrite. If it is heated to 850°C, what microstructure will be obtained?", "answer": "Entirely austenite (A) microstructure" }, { "idx": 2102, "question": "Point out the errors in the following concept and correct them: Repeatedly 'melting and solidifying' a solid solution alloy rod and using directional rapid solidification can effectively purify the metal.", "answer": "Repeatedly perform zone melting and use directional slow solidification." }, { "idx": 2103, "question": "Point out the errors in the following concept and correct them: Although the crystallization rate of solid solution alloys is very fast, at a certain moment during solidification, the chemical potentials of components A and B in the liquid and solid phases are equal.", "answer": "At the phase interface, the chemical potentials of components A and B in the liquid and solid phases are equal." }, { "idx": 2104, "question": "Point out the errors in the following concepts and correct them: In the solid solution alloy, whether during equilibrium or non-equilibrium crystallization, the liquid phase composition at the liquid-solid interface changes along the average liquid composition line; the solid phase composition changes along the average solid composition line.", "answer": "The liquid phase composition at the liquid-solid interface changes along the liquidus line; the solid phase composition changes along the solidus line." }, { "idx": 2105, "question": "Point out the errors in the following concepts and correct them: On the eutectic line, the lever rule can be used to calculate the relative amount of the eutectic. Since the eutectic line belongs to the three-phase region, the lever rule is not only applicable to the two-phase region but also to the three-phase region.", "answer": "However, the lever rule is only applicable to the two-phase region, so the relative amount of the eutectic is actually calculated in the two-phase region." }, { "idx": 2106, "question": "Point out the error in the following concept and correct it: In the directional solidification process of a solid solution alloy rod, the faster the liquid-solid interface advances, the more severe the macrosegregation in the rod.", "answer": "the less severe the macrosegregation in the rod." }, { "idx": 2107, "question": "Given the solid solubility equation of carbon in α-Fe in equilibrium with cementite as $$ w_{\\mathrm{C}}^{\\alpha}=2.55\\exp{\\frac{-11.3\\times10^{3}}{R T}} $$, assuming the solid solubility equation of carbon in austenite is similar to this equation, write this equation based on the Fe-Fe3C phase diagram.", "answer": "Let the solid solubility equation of C in γ be $$ w_{\\mathrm{c}}^{\\gamma}=A\\exp\\bigl(-\\frac{Q}{R T}\\bigr) $$ Taking the logarithm of both sides, we get $$ \\ln w_{\\mathrm{c}}^{\\gamma}=\\ln A-\\frac{Q}{R T} $$ From the Fe-Fe3C phase diagram, we have $\\ln0.77=\\ln A-{\\frac{Q}{R\\times1000}}$ $$ \\ln2.11=\\ln A-{\\frac{Q}{R\\times1~421}}\\quad. $$ Combining these two equations, we obtain $Q=28\\mathrm{kJ}$, $A=22,3$, thus $$ w_{\\mathrm{c}}^{\\mathrm{r}}=22.3\\exp\\Bigl(-\\frac{2.8\\times10^{3}}{R T}\\Bigr) $$" }, { "idx": 2108, "question": "Point out the incorrect concept and correct it: In solid solution grains, dendritic segregation exists, the composition of the main axis differs from that of the interdendritic regions, so the entire grain is not a single phase.", "answer": "Therefore, the entire grain is a single phase." }, { "idx": 2109, "question": "Point out the incorrect concept and correct it: In Ni-Cu alloy castings with uneven thickness, thin sections tend to form dendritic structures after crystallization, while thick sections tend to form cellular structures.", "answer": "After crystallization, thin sections tend to form cellular structures, while thick sections tend to form dendritic structures." }, { "idx": 2110, "question": "Point out the incorrect concept and correct it: Using circulating water to cool the metal mold is beneficial for obtaining a columnar crystal zone to improve the compactness of the casting.", "answer": "It is not beneficial for obtaining a columnar crystal zone." }, { "idx": 2111, "question": "Point out the error in the following concept and correct it: In the Fe-Fe3C system alloys, only hypereutectoid steels have secondary cementite in their equilibrium crystallization structure.", "answer": "Only when the iron-carbon alloy with a carbon mass fraction of 0.0077 < w_c < 0.043 undergoes equilibrium crystallization." }, { "idx": 2112, "question": "Point out the errors in the following concept and correct them: From the condition for constitutional supercooling <0, c=0. The left eutectoid steel rod is divided into volume elements with thickness d, so the carbon content per unit area is c0 d. Considering only the effect of this volume element, it is equivalent to an instantaneous plane source diffusion problem. After diffusion time t, the concentration at point P(x) at a distance ξ from this volume element is cξ=(c0 dξ)/(2√(πDt))exp(-ξ²/(4Dt)). Using the superposition principle, the total concentration at point P at time t should be the sum of contributions from all volume elements, i.e., c(x,t)=(c0)/(2√(πDt))∫_x^∞exp(-ξ²/(4Dt))dξ=(c0)/√π∫_(x/√π)^∞exp(-η²)dη. Here, η=ξ/2√Dt; erf(β)=2/√π∫_0^βe^(-η²)dη. erf(β) is the error function, and its values are given in Table 5-1. Note that erf(β)=∫_β^∞e^(-η²)dη=∫_0^∞e^(-η²)dη-∫_0^βe^(-η²)dη=√π/2[1-erf(β)]. Thus, c(x,t)=c0/2 erf(x²/(2√Dt)), as shown in Figure 5-7. When x<0, the concentration increases with |x|; when x>0, the concentration decreases with x; at x=0, when t>0, c=c0/2." }, { "idx": 2141, "question": "In a ternary system during diffusion, can a three-phase coexistence region appear within the diffusion layer? Why?", "answer": "A three-phase coexistence region cannot exist within the diffusion layer of a ternary system. The reasons are as follows: If three-phase equilibrium coexistence occurs in a ternary system, the compositions of the three phases are fixed, and the chemical potentials of the same component in different phases are equal, resulting in a zero chemical potential gradient, making diffusion impossible." }, { "idx": 2142, "question": "For NaCl crystals doped with a small amount of $\\\\mathrm{Cd}^{2+}$, the number of $\\\\mathrm{Na^{+}}$ vacancies related to Schottky defects at high temperatures is much higher than that related to $\\\\mathrm{Cd}^{2+}$, so intrinsic diffusion dominates. At low temperatures, the vacancies caused by the presence of $\\\\mathrm{Cd}^{2+}$ ions can accelerate the diffusion of $\\\\mathrm{Na^{+}}$ ions. Analyze how the transition point temperature in Figure 5-5 would shift if the concentration of $\\\\mathrm{Cd}^{2+}$ is reduced.", "answer": "The transition point shifts toward lower temperatures." }, { "idx": 2143, "question": "Point out the errors in the following concept: (1) If there is no diffusion flow in a solid, it means that atoms are not diffusing.", "answer": "Even if there is no macroscopic diffusion flow in a solid, diffusion still occurs due to the migration jumps of atomic thermal vibrations. Self-diffusion in pure substances is a typical example." }, { "idx": 2144, "question": "In a ternary system during diffusion, can a two-phase coexistence region appear within the diffusion layer? Why?", "answer": "A two-phase coexistence region can exist within the diffusion layer of a ternary system. The reasons are as follows: In a ternary system, when two phases coexist, due to the degree of freedom being 2, at a constant temperature, the composition of the coexisting phases can change. This disrupts the chemical potential balance of the same component atoms in the two phases, leading to diffusion." }, { "idx": 2145, "question": "Point out the error in the following concept: (2) Because the direction of each jump of solid atoms is random, the diffusion flux is zero under any circumstances.", "answer": "The direction of each jump of atoms is random. Only when the system is in thermal equilibrium, the probability of atoms jumping in any direction is equal. At this time, although there is atomic migration (i.e., diffusion), there is no macroscopic diffusion flow. If the system is in a non-equilibrium state, there must be a gradient of thermodynamic potential in the system (specifically expressed as concentration gradient, chemical potential gradient, strain energy gradient, etc.). The probability of atoms jumping in the direction of decreasing thermodynamic potential will be greater than that in the direction of increasing thermodynamic potential. As a result, macroscopic diffusion flow occurs." }, { "idx": 2146, "question": "Point out the error in the following concept: (3) The atomic arrangement at grain boundaries is disordered, and there are no vacancies, so atoms diffusing by the vacancy mechanism cannot diffuse at grain boundaries.", "answer": "The atomic arrangement at grain boundaries is disordered, similar to that in amorphous phases, and the atomic packing density is much lower than inside the grains, resulting in weaker constraints on atoms. The energy and vibration frequency ν of grain boundary atoms are significantly higher than those of intragranular atoms. Therefore, atoms at grain boundaries have higher mobility. The grain boundary diffusion coefficient is also significantly higher than the intragranular diffusion coefficient." }, { "idx": 2147, "question": "Point out the error in the following concept: (4) In an interstitial solid solution, the higher the solute concentration, the more interstitial sites are occupied by the solute, leaving fewer vacant interstitial sites for diffusion, i.e., the value of $_z$ decreases, leading to a reduction in the diffusion coefficient.", "answer": "In fact, this scenario is impossible. The solubility of solute atoms in an interstitial solid solution is very limited. Even in a supersaturated state, the number of solute atoms is several orders of magnitude smaller than the total number of interstitial sites in the crystal. Therefore, the interstitial sites around the solute atoms can be considered entirely vacant. That is, for a given crystal structure, 𝓏 is a constant." }, { "idx": 2148, "question": "For a pre-annealed metal polycrystal, the relationship between true stress σ_T and true strain ε_T in the uniform plastic deformation stage of the true stress-strain curve is given by σ_T=Kε_T^n, where K is the strength coefficient and n is the strain hardening exponent. If there are two metals A and B with roughly equal K values, but n_A=0.5 and n_B=0.2, which metal has higher hardening capability and why?", "answer": "Since σ_T=Kε_T^n, then dσ_T=n Kε_T^(n-1)dε_T, and θ=dσ_T/dε_T=n Kε_T^(n-1). From this equation, it can be seen that when ε_T<1 and 0n_B, metal B has a higher dislocation density at the same plastic strain." }, { "idx": 2152, "question": "For a face-centered cubic crystal, the movable slip system is (111)[110]. If the slip is caused by a pure edge dislocation, indicate the direction of the dislocation line.", "answer": "The dislocation line lies on the slip plane (111). Let the direction of the dislocation line be [uow], then u+v-w=0; the dislocation line is perpendicular to b, i.e., perpendicular to [110], so -u+v=0. From the above two equations, u:v:w=1:1:2, thus the direction of the dislocation line is [112]." }, { "idx": 2153, "question": "If the surface of a single crystal copper is exactly the {100} crystal plane, assuming the crystal can slip on various slip systems, discuss the possible morphology of slip lines (the orientation of slip lines and the angles between them) that may be observed on the surface.", "answer": "Copper crystal has a face-centered cubic lattice, and its slip systems are {111}<110>. If the surface of the single crystal copper is the {100} crystal plane, when plastic deformation occurs, the slip lines appearing on the crystal surface should be the intersection lines of {111} and {100}, which are <110>. That is, the slip lines observed on the crystal surface are either parallel to each other or intersect at a 90° angle." }, { "idx": 2154, "question": "For a face-centered cubic crystal with a movable slip system of (111)[110], indicate the direction of dislocation line movement during slip in the cases of pure edge dislocation and pure screw dislocation.", "answer": "In the case of pure edge dislocation, the direction of dislocation line movement is parallel to b; in the case of pure screw dislocation, the direction of dislocation line movement is perpendicular to b." }, { "idx": 2155, "question": "For a face-centered cubic crystal with a movable slip system of (111)[110], if the slip is caused by pure screw dislocations, indicate the direction of the dislocation line.", "answer": "The dislocation line lies on the slip plane and is parallel to b, so the direction of the dislocation line is [110]." }, { "idx": 2156, "question": "When a tensile force is applied along the [0001] direction of a hexagonal close-packed single crystal, what are the possible deformations and the primary mode of deformation?", "answer": "The slip plane of hexagonal close-packed metals is (0001), and the resolved shear stress on the slip plane in the [0001] direction is zero, so the single crystal cannot slip. During tension, the possible deformations of the single crystal are elastic deformation or subsequent brittle fracture." }, { "idx": 2157, "question": "A 20m long aluminum rod with a diameter of 14.0mm is drawn through a die with an aperture of 12.7mm. Calculate the cold working rate that this aluminum rod will undergo.", "answer": "The cold working rate (CW) can be expressed as the percentage reduction in cross-sectional area caused by plastic deformation, i.e., CW=[π×(14.0/2)^2-π×(12.7/2)^2]/[π×(14.0/2)^2]=18%." }, { "idx": 2158, "question": "The critical resolved shear stress of an aluminum single crystal at room temperature is $7.9\\\\times10^{5}~\\\\mathrm{Pa}$. If a tensile test is performed on an aluminum single crystal specimen at room temperature with the tensile axis in the [123] direction, and the possible activated slip system is (111)[101], determine the stress required to cause the specimen to yield.", "answer": "The aluminum crystal has a face-centered cubic lattice, and its slip system is $\\\\{111\\\\}<110>$. When performing a tensile test on an aluminum single crystal: $$ \\\\sigma_{s}=\\\\frac{\\\\tau_{\\\\mathrm{c}}}{\\\\cos\\\\varphi\\\\cos\\\\lambda} $$ Given $\\\\tau_{\\\\mathrm{c}}=7.9\\\\times10^{5}~\\\\mathrm{Pa}$, when the external force axis direction is [123], the possible activated slip system is (111)[101]. Thus, $\\\\varphi$ is the angle between [123] and the normal [111] of the (111) crystal plane, and $\\\\lambda$ is the angle between [123] and [101]. $$ \\\\mathrm{cos}\\\\varphi={\\\\frac{{\\\\overline{{1}}}+2+3}{\\\\sqrt{14}\\\\times\\\\sqrt{3}}}={\\\\frac{4}{\\\\sqrt{42}}} $$ $$ \\\\cos\\\\lambda={\\\\frac{1+0+3}{\\\\sqrt{14}\\\\times{\\\\sqrt{2}}}}={\\\\frac{2}{\\\\sqrt{7}}} $$ Therefore, $$ \\\\sigma_{\\\\mathrm{s}}=\\\\frac{7.9\\\\times10^{5}}{\\\\frac{4}{\\\\sqrt{42}}\\\\times\\\\frac{2}{\\\\sqrt{7}}}=1.69\\\\times10^{6}~\\\\mathrm{Pa} $$" }, { "idx": 2159, "question": "For a pre-annealed metal polycrystal, the relationship between true stress σ_T and true strain ε_T in the uniform plastic deformation stage of the true stress-strain curve is given by σ_T=Kε_T^n, where K is the strength coefficient and n is the strain hardening exponent. Derive the mathematical relationship between the strain hardening exponent n and the strain hardening rate θ=dσ_T/dε_T.", "answer": "The definition of the strain hardening rate θ is θ=dσ_T/dε_T. From σ_T=Kε_T^n, we obtain dσ_T/dε_T=n Kε_T^(n-1), that is, θ=n Kε_T^(n-1). Further derivation yields θ=n(σ_T/ε_T^n)ε_T^(n-1)=n(σ_T/ε_T)." }, { "idx": 2160, "question": "If the surface of a single crystal copper is the {111} crystal plane, assuming that the crystal can slip on various slip systems. Discuss the possible morphology of slip lines on the surface (the orientation of slip lines and the angles between them).", "answer": "When the outer surface of a copper single crystal is the {111} family of crystal planes, the slip lines appearing on the surface are <110>, which are either parallel to each other or intersect at an angle of 60°." }, { "idx": 2161, "question": "For a face-centered cubic crystal with a movable slip system of (111)[110], indicate the Burgers vector of the unit dislocation causing the slip (taking the lattice constant a=0.2nm).", "answer": "The Burgers vector of the unit dislocation causing the slip is b=a/2[110], which is the vector represented by the line connecting two adjacent atoms along the slip direction." }, { "idx": 2162, "question": "When a compressive force is applied along the [0001] direction of a hexagonal close-packed single crystal, what are the possibilities of deformation and the primary mode of deformation?", "answer": "The slip plane of hexagonal close-packed metals is (0001), and the resolved shear stress on the slip plane in the [0001] direction is zero, so the single crystal cannot slip. During compression, after elastic deformation, twinning may occur." }, { "idx": 2163, "question": "Calculate the volume fraction of the Fe3C phase in 40 steel, given the mass fraction of carbon ωC=0.004", "answer": "φFe3C=0.004/0.069=0.06" }, { "idx": 2164, "question": "The movable slip system of a certain face-centered cubic crystal is (111)[110]. Assuming a shear stress of magnitude 0.7MPa acts on this slip system, calculate the magnitude and direction of the force per unit length on a unit edge dislocation and a unit screw dislocation (take the lattice constant a=0.2nm).", "answer": "Under the applied shear stress τ, the magnitude of the force per unit length on the dislocation line is F=τb, and the direction is perpendicular to the dislocation line. Here, |b|=√((a/2)^2+(a/2)^2)=√2/2a, so F=τb=0.7×√2/2a=0.7×√2×0.2×10^-9/2=9.899×10^-11MN/m. The direction of F is perpendicular to the dislocation line; the direction of F is also perpendicular to the dislocation line." }, { "idx": 2165, "question": "Calculate the interparticle spacing λ, given the number of Fe3C particles per unit volume NV=1.43×1013", "answer": "λ=NV-1/3=(1.43×1013)-1/3=0.24×10-5m=2.4μm" }, { "idx": 2166, "question": "Calculate the number of Fe3C particles per unit volume NV, given the volume fraction of Fe3C phase φFe3C=0.06 and the spherical particle radius r=10×10-6m", "answer": "NV=φFe3C/(4/3πr3)=0.06/(4/3π×(10×10-6)3)≈1.43×1013" }, { "idx": 2167, "question": "Calculate the Burgers vector b of α-Fe, given the lattice constant of α-Fe a=0.28nm", "answer": "b=√3/2a=√3/2×0.28=0.25nm" }, { "idx": 2168, "question": "Calculate the shear strength τ of 40 steel, given the shear modulus of iron GFe=7.9×1010Pa, Burgers vector b=0.25nm, and particle spacing λ=2.4μm", "answer": "τ=Gb/λ=7.9×1010×2.5×10-10/2.4×10-5=8.23×105Pa" }, { "idx": 2169, "question": "Give the relationship between the lattice resistance to dislocation motion and the crystal structure.", "answer": "τ_p≈(2G/(1-ν))exp[-(2πa)/((1-ν)b)]≈(2G/(1-ν))exp(-2πw/b) where w is the dislocation width (w=a/(1-ν)), a is the interplanar spacing of the slip plane, b is the atomic spacing in the slip direction, and ν is the Poisson's ratio." }, { "idx": 2170, "question": "What is single slip? What are the characteristics of its slip lines?", "answer": "Single slip refers to the slip occurring on only one slip system. The slip lines appear as a series of parallel straight lines. This is because in single slip, only one set of slip systems is active, where all slip planes are parallel to each other and the slip directions are identical." }, { "idx": 2171, "question": "For annealed pure iron with a grain size of $N_{\\\\mathrm{A}}=16$ grains/mm², its yield strength $\\\\sigma_{\\\\textrm{s}}=100$ MPa; when $N_{\\\\mathrm{A}}=4~096$ grains/mm², $\\\\sigma_{\\\\mathrm{s}}=250\\\\mathrm{MPa}$. Calculate $\\\\sigma_{s}$ when $N_{\\\\mathrm{A}}=250$ grains/mm².", "answer": "Let the average grain diameter be $d$, and the number of grains per square millimeter be $N_{\\\\mathrm{A}}$. It can be proven that $$ d=\\\\sqrt{\\\\frac{8}{3\\\\pi N_{\\\\mathrm{A}}}} $$ Therefore, $$ d_{1}=\\\\sqrt{\\\\frac{8}{3\\\\pi\\\\times16}}=0.053\\\\mathrm{~mm} $$ $$ d_{2}=\\\\sqrt{\\\\frac{8}{3\\\\pi\\\\times4~096}}=2.072\\\\times10^{-4}~\\\\mathrm{mm} $$ $$ d_{3}=\\\\sqrt{\\\\frac{8}{3\\\\pi\\\\times250}}=3.395\\\\times10^{-3}\\\\mathrm{mm} $$ Substituting into the Hall-Petch formula, $$ \\\\begin{array}{l}{\\\\left\\\\{100=\\\\sigma_{0}+K(0.053)^{-1/2}\\\\right.} \\\\\\\\ {\\\\left.250=\\\\sigma_{0}+K(2.072\\\\times10^{-4})^{-1/2}\\\\right.}\\\\end{array} $$ Solving gives $$ \\\\sigma_{0}=90\\\\mathrm{MPa} $$ Therefore, when $N_{\\\\mathrm{A}}=250$ grains/mm² and $d=3.395\\\\times10^{-3}\\\\mathrm{mm}$, $$ \\\\sigma_{\\\\mathrm{s}}=90+2.303\\\\times(3.395\\\\times10^{-3})^{-1/2}=129.5~\\\\mathrm{MPa} $$" }, { "idx": 2172, "question": "Given that the critical resolved shear stress τc for the {111}[110] slip system in pure copper is 1MPa, what stress should be applied in the [001] direction to produce slip in the [101] direction on the (111) plane?", "answer": "For cubic crystals, the angle between two crystallographic directions [u1v1w1] and [u2v2w2] is given by cosφ=(u1u2+v1v2+w1w2)/(√(u1²+v1²+w1²)*√(u2²+v2²+w2²)). Therefore, the angle between the normal direction [111] of the slip plane (111) and the tensile axis [001] is cosφ=(1×0+1×0+1×1)/(√(1²+1²+1²)*√(0²+0²+1²))=1/√3=0.577. The angle between the slip direction [101] and the tensile axis [001] is cosλ=(1×0+0×0+1×1)/(√(1²+0²+1²)*√(0²+0²+1²))=1/√2=0.707. The applied stress σ=τc/(cosφcosλ)=1/(0.577×0.707)=2.45MPa." }, { "idx": 2173, "question": "When stretching a copper single crystal, if the direction of the tensile axis is [001], $\\\\sigma=10^{6}$ Pa. Calculate the force on the screw dislocation line with Burgers vector $b=$ $\\\\frac{a}{2}$ [101] on the (111) plane $(a_{\\\\mathrm{Cu}}=0.36\\\\mathrm{nm})$.", "answer": "Let the resolved shear stress of the applied tensile stress along the [101] direction on the (111) slip plane be $\\\\tau$, then $$ x \\\\tau=\\\\sigma{\\\\cos}\\\\varphi{\\\\cos}\\\\lambda\\n $$\\n where, $\\\\varphi$ is the angle between [001] and the normal [111] of the (111) plane, $\\\\lambda$ is the angle between [001] and the [101] direction. \\n\\nSo $$ \\\\tau=10^{6}\\\\times{\\\\frac{1}{{\\\\sqrt{1}}\\\\times{\\\\sqrt{3}}}}\\\\times{\\\\frac{1}{\\\\sqrt{2}}}=4.0825\\\\times10^{5}{\\\\mathrm{~Pa}}\\n $$\\n If the force on the screw dislocation line is $F_{\\\\mathrm{d}}$, then $$ F_{\\\\mathrm{d}}=\\\\tau b=4.0825\\\\times10^{5}\\\\times{\\\\frac{\\\\sqrt{2}}{2}}\\\\times0.36\\\\times10^{-9}=1.039\\\\times10^{-4}~{\\\\mathrm{N/m}}\\n $$" }, { "idx": 2174, "question": "What is cross-slip? What are the characteristics of its slip lines?", "answer": "Cross-slip refers to the simultaneous or alternating slip of two or more slip planes along a common slip direction. Their slip lines are usually zigzag or wavy. This is the result of screw dislocations repeatedly 'expanding' on different slip planes." }, { "idx": 2175, "question": "Give examples of the impact of Lüders bands on industrial production and methods to prevent them.", "answer": "Lüders bands can cause the surface of workpieces made from low-carbon thin steel sheets to become rough and uneven during stamping and forming. The solution, based on the strain aging principle, involves subjecting the steel sheet to a slight cold rolling process (e.g., 1%∼2% reduction) before stamping to eliminate the yield point, followed by stamping and forming. Alternatively, adding small amounts of Ti, Al, C, and N to the steel to form compounds can also eliminate the yield point." }, { "idx": 2176, "question": "How is texture formed?", "answer": "After cold working, the orientations of metal grains exhibit certain relationships. For instance, certain crystal planes or directions become parallel to each other and also parallel to an external reference direction of the component. Such an orientation distribution is called preferred orientation or simply texture. The formation of texture is not limited to cold working, but here it mainly refers to deformation texture. In terms of both orientation and properties, polycrystalline materials with texture lie between single crystals and completely randomly oriented polycrystals." }, { "idx": 2177, "question": "What is multiple slip? What are the characteristics of its slip lines?", "answer": "Multiple slip refers to the simultaneous or alternating slip of two or more different slip systems. Their slip lines are either parallel or intersect at a certain angle. This is because certain crystal structures have specific slip systems, and there are certain angles between the slip planes and slip directions of these slip systems." }, { "idx": 2178, "question": "For face-centered cubic crystals, it is generally required to have five independent slip systems for slip to occur. Is this conclusion correct? Please explain the reason and the conditions under which this conclusion applies.", "answer": "This conclusion is correct. Because generally, representing a deformation requires nine strain components, i.e., $$ \\\\mathfrak{\\\\varepsilon}_{i j}=\\\\left|\\\\begin{array}{l l l}{\\\\varepsilon_{x x}}&{\\\\mathfrak{\\\\varepsilon}_{x y}}&{\\\\mathfrak{\\\\varepsilon}_{x z}}\\\\\\\\{\\\\mathfrak{\\\\varepsilon}_{y y}}&{\\\\mathfrak{\\\\varepsilon}_{y x}}&{\\\\mathfrak{\\\\varepsilon}_{y z}}\\\\\\\\{\\\\mathfrak{\\\\varepsilon}_{z z}}&{\\\\mathfrak{\\\\varepsilon}_{z x}}&{\\\\mathfrak{\\\\varepsilon}_{z y}}\\\\end{array}\\\\right| $$ but $\\\\varepsilon_{x y}=\\\\varepsilon_{y x}, \\\\varepsilon_{y z}=\\\\varepsilon_{z y}, \\\\varepsilon_{z x}=\\\\varepsilon_{x z}$; thus, there are only six components left. Since the deformation is required to be uniform and continuous, the volume remains unchanged before and after deformation, i.e., $\\\\Delta V{=}_{\\\\epsilon_{x x}}+\\\\epsilon_{y y}+\\\\epsilon_{z z}{=}0$. With this constraint, there are only five independent strain components. Each independent strain component is generated by an independent slip system; therefore, five independent slip systems are needed to produce five independent strain components. When applying this conclusion, attention must be paid to the size of the crystal. The volume cannot be too small; it must be larger than the spacing between slip bands so that the plastic deformation can be considered uniform. However, the volume cannot be too large either; it must remain within the range of linear plastic deformation, such as not exceeding the size of a single grain." }, { "idx": 2179, "question": "Given that the critical resolved shear stress τc for the {111}[110] slip system in pure copper is 1MPa, what stress should be applied in the [001] direction to produce slip in the [110] direction on the (111) plane?", "answer": "Since the dot product of the slip direction [110] and the [001] direction is zero, it is known that the two crystallographic directions are perpendicular, cosλ=0, σ=∞. That is, when the applied stress direction is [001], slip will not occur in the [110] direction." }, { "idx": 2180, "question": "How is fibrous structure formed?", "answer": "After cold working of materials, in addition to transforming the disordered orientation of polycrystalline materials into materials with preferred orientation, it also causes deformation of insoluble impurities, second phases, and various defects in the material. Since grains, impurities, second phases, and defects are all elongated into fibrous shapes along the main deformation direction of the metal, it is called fibrous structure." }, { "idx": 2181, "question": "What is the difference between fibrous structure and texture?", "answer": "Fibrous structure is formed by grains, impurities, second phases, defects, etc., being elongated into fibrous shapes along the main deformation direction of the metal; whereas texture refers to the certain orientation relationship among grains after cold working, where certain crystal planes or directions are parallel to each other and all parallel to an external reference direction of the component." }, { "idx": 2182, "question": "Explain why crystal slip usually occurs on the most closely packed atomic planes and directions.", "answer": "As can be seen from the relation, the larger the value of a, the smaller the τ_p, so the slip plane should be the one with the largest interplanar spacing, i.e., the most closely packed atomic plane; the smaller the value of b, the smaller the τ_p, so the slip direction should be the most closely packed atomic direction." }, { "idx": 2183, "question": "Use dislocation theory to explain the yield phenomenon of low-carbon steel.", "answer": "The yield phenomenon of low-carbon steel can be explained by dislocation theory. Since low-carbon steel is an alloy based on ferrite, carbon (nitrogen) atoms in ferrite interact with dislocations and tend to aggregate in the tensile stress regions of dislocation lines to reduce the distortion energy of the system, forming Cottrell atmospheres that 'pin' the dislocations, thereby increasing σs. Once dislocations break free from the pinning of these atmospheres, they can continue to move under lower stress, leading to the appearance of the lower yield point on the stress-strain curve. For a specimen that has already yielded, if it is unloaded and immediately reloaded in tension, the yield point does not reappear because the dislocations have already escaped the pinning of the atmospheres. However, if the unloaded specimen is left for a long time or slightly heated before retesting, the solute atoms will diffuse back and re-aggregate around the dislocation lines to form atmospheres, causing the yield phenomenon to reappear." }, { "idx": 2184, "question": "What effect does fibrous structure have on the properties of metals?", "answer": "Generally speaking, fibrous structure makes the longitudinal (fiber direction) strength of metals higher than the transverse strength. This is because on the cross-section, the cross-sectional area of brittle, low-strength 'components' such as impurities, second phases, and defects is small, while on the longitudinal section, the cross-sectional area is large. When parts bear large loads or are subjected to impact and alternating loads, this anisotropy may pose significant risks." }, { "idx": 2185, "question": "Briefly analyze the essential similarities and differences in work hardening", "answer": "Work hardening is caused by dislocation pile-up, entanglement, and their interactions, which hinder further dislocation movement, with the flow stress σd=αGb√ρ" }, { "idx": 2186, "question": "Briefly analyze the essential similarities and differences of solid solution strengthening", "answer": "Solid solution strengthening results from the interaction between dislocations and solute atoms, namely the Cottrell atmosphere hindering dislocation motion." }, { "idx": 2187, "question": "What effect does texture have on the properties of metals?", "answer": "Due to texture-induced anisotropy in metals, it often causes inconvenience in metal processing. For example, cold-rolled magnesium sheets develop a (0001)<11\\overline{2}0> texture, making them prone to cracking during further processing; earing in deep-drawn metal cups; thermal cycling growth of metals, etc. However, in some cases, it also has beneficial aspects." }, { "idx": 2188, "question": "Briefly analyze the essential similarities and differences of grain refinement strengthening", "answer": "Grain refinement strengthening occurs because the atoms at grain boundaries are irregularly arranged, with many impurities and defects, resulting in higher energy that hinders dislocation movement, σs=σ0+Kd−1/2; moreover, when grains are fine, deformation is uniform, stress concentration is low, and cracks are less likely to initiate and propagate." }, { "idx": 2189, "question": "Briefly analyze the essential similarities and differences of dispersion strengthening", "answer": "Dispersion strengthening occurs because dislocations bypass or cut through second-phase particles, requiring additional energy (such as surface energy or misfit energy); meanwhile, the elastic stress field around the particles interacts with dislocations, hindering their movement." }, { "idx": 2190, "question": "When the bubble density in tungsten wire increases from 100/cm² to 400/cm², the tensile strength can approximately double, because bubbles can hinder dislocation motion. Analyze the mechanism by which bubbles impede dislocation motion.", "answer": "The mechanism by which bubbles hinder dislocation motion is that when dislocations pass through bubbles, they cut the bubbles, increasing the area of the bubble-metal interface. This requires additional external shear stress to perform work, thereby enhancing the strength of tungsten metal." }, { "idx": 2191, "question": "Given the Burgers vector of a dislocation as b, the bubble radius as r, and the specific interfacial energy between the bubble and metal as σ, determine the increment of shear stress Δτ.", "answer": "The increased interfacial area between the bubble and metal after the dislocation cuts the bubble is A=2rb. The increment of interfacial energy is 2rbσ. If the increment of shear stress for a dislocation cutting one bubble is Δτ', then the work done by the stress is Δτ'b. Therefore, 2rbσ=Δτ'b, Δτ'=2rσ. When the bubble density is n, the total increment of shear stress Δτ=nΔτ'=2nrσ. It can be seen that the increment of shear stress is proportional to the bubble density." }, { "idx": 2192, "question": "Why is the actual tensile strength of ceramics lower than the theoretical yield strength?", "answer": "This is due to the unavoidable microscopic voids present during the sintering of ceramic powders. During cooling or thermal cycling, thermal stresses generate microcracks, and surface cracks caused by corrosion result in ceramics, unlike metals, having inherent microcracks. At the crack tip, severe stress concentration occurs. According to elastic mechanics estimates, the maximum stress at the crack tip has already reached the theoretical fracture strength or theoretical yield strength (because ceramic crystals have very few mobile dislocations, and dislocation motion is extremely difficult, so once the yield strength is reached, fracture occurs). Conversely, one can calculate the nominal stress at which the crystal fractures when the maximum stress at the crack tip equals the theoretical yield strength, and this value is extremely close to the actual measured tensile strength." }, { "idx": 2193, "question": "This shows, what effect does recovery have on recrystallization?", "answer": "After polygonization, the dislocation energy decreases, reducing the stored energy and thereby diminishing the driving force for subsequent recrystallization." }, { "idx": 2194, "question": "Given that the elastic modulus of sintered alumina is $370\\\\mathrm{GPa}$ when its porosity is $5\\\\%$, if the elastic modulus of another sintered alumina is $270~\\\\mathrm{GPa}$, calculate its porosity.", "answer": "The relationship between the elastic modulus $E$ of ceramic materials and their pore volume fraction $\\\\varphi$ can be expressed as $$ E=E_{\\\\circ}(1-1.9\\\\varphi+0.9\\\\varphi^{2}) $$ where $E_{0}$ is the elastic modulus of the material without pores. Given that when $\\\\varphi=0.05$, $E=370~\\\\mathrm{GPa}$, thus $$ E_{\\\\circ}={\\\\frac{E}{1-1.9{\\\\varphi}+0.9{\\\\varphi}^{2}}}={\\\\frac{370}{1-1.9\\\\times0.05+0.9\\\\times(0.05)^{2}}}=407.8{\\\\mathrm{~GPa}} $$ When $E=270$ GPa, $$ 270=407.8(1-1.9\\\\varphi+0.9\\\\varphi^{2}) $$ which simplifies to $$ 0.9\\\\varphi^{2}-1.9\\\\varphi+0.338=0 $$ Therefore, $$ \\\\varphi=0.196=19.6\\\\% $$" }, { "idx": 2195, "question": "Why is the compressive strength of ceramics always higher than the tensile strength?", "answer": "The compressive strength of ceramics is generally about 15 times the tensile strength. This is because under tension, when a crack reaches the critical size, it becomes unstable and propagates rapidly, leading to fracture; whereas under compression, cracks either close or propagate steadily and slowly, turning parallel to the compression axis. In other words, under tension, the tensile strength of ceramics is determined by the maximum crack size in the crystal, while the compressive strength is determined by the average size of the cracks." }, { "idx": 2196, "question": "A subgrain boundary is composed of several edge dislocations with a misorientation of 0.057 degrees. Assuming there is no interaction between the dislocations before polygonization, by what factor does the distortion energy change after the formation of the subgrain?", "answer": "The energy per unit length of a dislocation line is W_I = (G b^2) / (4π(1-ν)) * ln(R/r_0), where r_0 is the radius of the dislocation core region and R is the radius of the maximum range of the dislocation stress field. If we take r_0 ≈ b = 10^-8 cm and R ≈ 10^-4 cm, then before polygonization, W_D = (G b^2) / (4π(1-ν)) * ln(10^4). After polygonization, R = D = b / θ = 10^-8 / 10^-3 = 10^-5 cm, and W_31* = (G b^2) / (4π(1-ν)) * ln(10^3). Thus, W_I* / W_I1 = ln(10^3) / ln(10^4) = 0.75." }, { "idx": 2197, "question": "Given that the recovery activation energy of a zinc single crystal is $20000\\\\mathrm{J/mol}$, removing $2\\\\%$ of the work hardening at $50^{\\\\circ}\\\\mathrm{C}$ requires $13\\\\textup{d}$; how much should the temperature be increased to remove the same amount of work hardening in $5\\\\textrm{min}$?", "answer": "When cold-plastically deformed metal undergoes recovery, if the recovery amount $R$ (in this case, the removed portion of work hardening) is constant, the relationship between the required recovery time $t$ and the recovery temperature $T$ can be expressed as $$ \\\\ln t=a+{\\\\frac{Q}{R}}{\\\\frac{1}{T}}$$ where $a$ is a constant; $\\\\boldsymbol{Q}$ is the recovery activation energy. Accordingly, we have $$ \\\\frac{t_{1}}{t_{2}}=\\\\exp\\\\biggl[-\\\\frac{Q}{R}\\\\Bigl(\\\\frac{1}{T_{2}}-\\\\frac{1}{T_{1}}\\\\Bigr)\\\\biggr]$$ From the problem statement, $$ T_{2}=-50^{\\\\circ}C=223\\\\mathrm{~K~}$$ $$ t_{2}=13~\\\\mathrm{d}=18~500~\\\\mathrm{min},\\\\quad Q=20~000~\\\\mathrm{J/mol}$$ When $t_{1}=5\\\\mathrm{min}$ is required, then $$ {\\\\frac{5}{18500}}{=}\\\\exp\\\\Bigl[-{\\\\frac{20000}{2}}\\\\Bigl({\\\\frac{1}{223}}-{\\\\frac{1}{T_{1}}}\\\\Bigr)\\\\Bigr]$$ $$ {\\\\frac{1}{T_{1}}}={\\\\frac{1}{223}}-{\\\\frac{\\\\ln3700}{10000}}$$ Therefore, the recovery temperature $$ T_{1}=273\\\\mathrm{~K~}$$" }, { "idx": 2198, "question": "The strength of OFHC copper (oxygen-free high-conductivity copper) can be increased by more than 2 times after cold drawing deformation. If the safety factor for allowable stress is taken as 2, calculate the service life of an OFHC copper component working at $130^{\\\\circ}\\\\mathrm{C}$. (Given: $A=10^{12}1/\\\\operatorname*{min}, \\\\frac{Q}{R}=$ $1.5\\\\times10^{4}~\\\\mathrm{K}, t_{0.5}$ is the time required for 50% recrystallization)", "answer": "Since OFHC copper operates at $130^{\\\\circ}C$, with a strength design safety factor of 2, for cold-worked and strengthened materials, only 50% recrystallization is allowed, i.e., $$ \\\\begin{array}{r}{\\\\frac{1}{t_{0.5}}=A\\\\mathrm{exp}\\\\left(-\\\\frac{Q}{R T}\\\\right)} \\\\ {\\\\mathrm{lg}(A t_{0.5})=\\\\frac{Q}{R T}\\\\mathrm{lg}\\\\mathrm{~e~}}\\\\end{array}$$ $A=10^{12}~1/\\\\operatorname*{min}{},\\\\frac{Q}{R}=1.5\\\\times10^{4}~\\\\mathrm{K}{},T=130^{\\\\circ}\\\\mathrm{C}=403~\\\\mathrm{K}$. Therefore, $$ t_{0.5}=14497\\\\mathrm{min}=242\\\\mathrm{h}$$ Thus, the service life of OFHC copper under these working conditions is $242\\\\mathrm{~h~}$ \\\\lg(10^{12}t_{0.5})={\\\\frac{1.5\\\\times10^{4}}{403}}\\\\times0.4342$$" }, { "idx": 2199, "question": "What is the difference between crazing and cracking?", "answer": "Crazing is different from cracking. The two open surfaces of a crack are completely empty, whereas the surfaces of a craze consist of highly oriented fiber bundles and voids, which still retain some strength. The formation of crazes is due to localized yielding and cold drawing of the material under tensile stress." }, { "idx": 2200, "question": "The melting point of iron is known to be 1,538℃, estimate the minimum recrystallization temperature of iron", "answer": "The minimum recrystallization temperature of iron is T_r=0.4×(1,538+273)=723 K" }, { "idx": 2201, "question": "Based on the minimum recrystallization temperature of iron, 723 K, select the recrystallization annealing temperature for iron", "answer": "The recrystallization annealing temperature for iron T=823∼923 K" }, { "idx": 2202, "question": "Given that the melting point of copper is 1,083℃, estimate the minimum recrystallization temperature of copper", "answer": "The minimum recrystallization temperature of copper is T_r=0.4×(1,083+273)=542 K" }, { "idx": 2203, "question": "Based on the minimum recrystallization temperature of copper being 542 K, determine the recrystallization annealing temperature for copper", "answer": "The recrystallization annealing temperature for copper T=643∼743 K" }, { "idx": 2204, "question": "Pure zirconium requires 40 h and 1 h of isothermal annealing at $553^{\\\\circ}\\\\mathrm{C}$ and $627^{\\\\circ}\\\\mathrm{C}$ respectively to complete recrystallization. Calculate the recrystallization activation energy of this material.", "answer": "From Equation (7-6) in Section 7.3: $$Therefore$$ Q=R\\\\ln\\\\frac{t_{1}}{t_{2}}\\\\bigg/\\\\Big(\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{2}}\\\\Big)$$ Substituting the known values, we get $$ Q={\\\\frac{8.31\\\\ln{\\\\frac{40}{1}}}{{\\\\frac{1}{553+273}}-{\\\\frac{1}{627+273}}}}=3.08\\\\times10^{5}~\\\\mathrm{J/mol}$$ \\\\ln{\\\\frac{t_{1}}{t_{2}}}=\\\\frac{Q}{R}\\\\Big(\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{2}}\\\\Big)" }, { "idx": 2205, "question": "For metals and alloys that do not undergo phase transformations in the solid state, can their grain size be changed without remelting? What methods can be used to alter it?", "answer": "Yes. It can be achieved through cold deformation followed by recrystallization annealing." }, { "idx": 2206, "question": "A cold-drawn steel wire rope was used to hoist a large workpiece into a furnace and was heated together with the workpiece to $1,000^{\\circ}\\mathrm{C}$. Upon completion of heating, when the workpiece was lifted out, the steel wire rope broke. Analyze the reason for this.", "answer": "The processing of the cold-drawn steel wire rope is a cold-working process. Due to work hardening, the strength and hardness of the steel wire increase, thereby enhancing its load-bearing capacity. When heated, if the temperature exceeds its recrystallization temperature, the steel wire rope undergoes recrystallization, leading to a reduction in strength and hardness. Once the external load exceeds its load-bearing capacity, it will break." }, { "idx": 2207, "question": "There are three types of ingots: pure Ti, Al, and Pb. It is known that the melting point of Ti is 1672°C, with a hexagonal close-packed structure below 883°C and a face-centered cubic structure above 883°C; the melting point of Al is 660°C, with a face-centered cubic structure; the melting point of Pb is 328°C, with a face-centered cubic structure. Determine the order of their rolling difficulty at room temperature (20°C).", "answer": "At room temperature, Pb is the easiest to roll, followed by Al, and Ti is the most difficult." }, { "idx": 2208, "question": "What measures should be taken for A1 and Ti to enable them to be rolled into thin sheets?", "answer": "For A1 and Ti, intermediate annealing (i.e., recrystallization annealing) should be employed." }, { "idx": 2209, "question": "Given that brass containing ${w_{\\\\mathrm{Zn}}}=0.30$ requires $^{1\\\\textrm{h}}$ to complete recrystallization at a constant temperature of $400^{\\\\circ}\\\\mathrm{C}$, and $2\\\\textrm{h}$ at $390^{\\\\circ}\\\\mathrm{C}$, calculate the time required to complete recrystallization at a constant temperature of $420^{\\\\circ}\\\\mathrm{C}$.", "answer": "The rate of recrystallization is given by $Q$ is the activation energy for recrystallization) Let $t$ be the time required to complete recrystallization, then$$ $$ \\\\begin{array}{r}{\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{2}}=\\\\ln\\\\frac{t_{2}}{t_{1}}}\\\\\\\\ {\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{3}}=\\\\ln\\\\frac{t_{3}}{t_{1}}}\\\\end{array}$$ Substituting $T_{1}=673~\\\\mathrm{K},t_{1}=1~\\\\mathrm{h};T_{2}=663~\\\\mathrm{K},t_{2}=2~\\\\mathrm{h};T_{3}=693~\\\\mathrm{K}$ into the above equations, we obtain$$ t_{3}\\\\approx0.26~\\\\mathrm{h}$$ That is, completing recrystallization at a constant temperature of $420^{\\\\circ}\\\\mathrm{C}$ requires approximately $0,26\\\\mathrm{~h~}$ \\\\begin{array}{c}{{V_{\\\\parallel\\\\parallel}t=1}}\\\\\\\\ {{A\\\\mathrm{e}^{\\\\frac{-Q}{R T_{1}}}t_{1}=A\\\\mathrm{e}^{\\\\frac{-Q}{R T_{2}}}t_{2}=A\\\\mathrm{e}^{\\\\frac{-Q}{R T_{3}}}t_{3}}}\\\\\\\\ {{-\\\\frac{Q}{R}\\\\Big(\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{2}}\\\\Big)=\\\\ln\\\\frac{t_{2}}{t_{1}}}}\\\\\\\\ {{-\\\\frac{Q}{R}\\\\Big(\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{3}}\\\\Big)=\\\\ln\\\\frac{t_{3}}{t_{1}}}}\\\\end{array}" }, { "idx": 2210, "question": "Design an experimental method to determine the variation of recrystallization nucleation rate $\\dot{N}$ with time at a certain temperature (T).", "answer": "The recrystallization nucleation rate $\\dot{N}$ can be determined using the metallographic method. Specific steps: Heat a batch of specimens deformed with a large strain to a certain temperature (T) and hold, then quench one specimen at regular time intervals t. Observe the prepared metallographic samples under a microscope and count the number of recrystallization nuclei N. After obtaining a set of data (several points), plot an N-t graph. The slope at each point on the N-t curve represents the recrystallization nucleation rate $\\dot{N}$ of the material at temperature T for different holding times." }, { "idx": 2211, "question": "Design an experimental method to determine the variation of growth linear velocity G with time at a certain temperature (T).", "answer": "The growth linear velocity G can be determined using the metallographic method. Specific steps: Perform metallographic observation on a set of quenched samples from the nucleation rate experiment, measure the linear dimension D of the largest nucleus in each sample (representing different holding times), plot a D-t graph, and the slope at each point on the D-t curve represents the growth linear velocity G at temperature T for different holding times." }, { "idx": 2212, "question": "When cold-drawn copper wires are used as overhead conductors (requiring certain strength), what final heat treatment process should be adopted?", "answer": "Stress relief annealing (low-temperature annealing) should be adopted." }, { "idx": 2213, "question": "Can metal castings refine grain size through recrystallization annealing?", "answer": "Recrystallization annealing must be applied to materials that have undergone cold plastic deformation processing, with the purpose of improving the microstructure and properties of the material after cold deformation. The temperature of recrystallization annealing is relatively low, generally below the critical point. If recrystallization annealing is applied to castings, their microstructure will not undergo phase transformation, nor will there be a driving force for the formation of new nuclei (such as stored energy from cold deformation, etc.), so new grains will not form, and grain refinement cannot be achieved." }, { "idx": 2214, "question": "There are three types of ingots: pure Ti, Al, and Pb. It is known that the melting point of Ti is 1672°C, with a hexagonal close-packed structure below 883°C and a face-centered cubic structure above 883°C; the melting point of Al is 660°C, with a face-centered cubic structure; the melting point of Pb is 328°C, with a face-centered cubic structure. Determine whether they can be continuously rolled?", "answer": "Only lead can be continuously rolled (because rolling at 20°C already constitutes hot deformation)." }, { "idx": 2215, "question": "Point out the unreasonable aspects of the above process and formulate a reasonable grain refinement process", "answer": "Under the condition of 80% deformation, annealing at 150°C for 1h can refine its grains." }, { "idx": 2216, "question": "To refine the grain of a pure aluminum part, it was cold deformed by 5% and then annealed at 650°C for 1 hour, but the structure coarsened instead. Analyze the reason for this.", "answer": "Since the deformation of the aluminum part was under the critical deformation degree, only a few recrystallization nuclei could form during annealing, resulting in extremely coarse grains." }, { "idx": 2217, "question": "Explain the meanings of N0, G0, Qg, Qn and their influencing factors.", "answer": "N0 and G0 are constants in the Arrhenius equation; Qn is the activation energy for recrystallization nucleation; Qg is the activation energy for recrystallized grain growth. Qn and Qg are mainly influenced by deformation amount, metal composition, metal purity, and original grain size. When the deformation amount exceeds 5%, Qn and Qg are approximately equal. For high-purity metals, the value of Qg is roughly equivalent to the activation energy for grain boundary self-diffusion." }, { "idx": 2218, "question": "When cold-drawn copper wire is used as lamp filament wire (requiring good toughness), what final heat treatment process should be adopted?", "answer": "Recrystallization annealing (high-temperature annealing) should be adopted." }, { "idx": 2219, "question": "To refine the grain size of a pure aluminum part, the cold deformation was increased to 80%, followed by annealing at 650°C for 1 hour, but coarse grains were still obtained. Analyze the reason.", "answer": "The reason is the inappropriate selection of recrystallization annealing temperature (the temperature was too high). According to the estimation of T_rec ≈ 0.4T_melt, where T_melt = 100°C, the recrystallization temperature should not exceed 200°C. Since annealing was performed at 650°C for 1 hour, the grains remained coarse." }, { "idx": 2220, "question": "Compare the differences in dislocation motion between stress relief annealing and dynamic recovery processes.", "answer": "In the stress relief annealing process, dislocations rearrange through climb and glide, transitioning from a high-energy state to a low-energy state; in the dynamic recovery process, it is through cross-slip of screw dislocations and climb of edge dislocations that opposite-sign dislocations cancel each other out, maintaining a dynamic balance between dislocation multiplication rate and dislocation annihilation rate." }, { "idx": 2221, "question": "A low-carbon steel part has developed banded structure, how to eliminate or improve the banded structure through heat treatment methods?", "answer": "For materials that have developed banded structure, heating in the single-phase region and normalizing treatment can eliminate or improve it." }, { "idx": 2222, "question": "How to distinguish dynamic and static recovery, and dynamic and static recrystallization from the microstructure", "answer": "During static recovery, clear subgrain boundaries can be observed, while static recrystallization forms equiaxed grains. In contrast, dynamic recovery forms cellular substructures, and dynamic recrystallization produces dislocation-tangled cells within equiaxed grains, which are finer than those formed by static recrystallization." }, { "idx": 2223, "question": "A low-carbon steel part requires isotropy, but a relatively obvious banded structure forms after hot working. How can the formation of banded structure be mitigated or eliminated by controlling the hot working process?", "answer": "First, avoid deformation in the two-phase region; second, reduce the content of impurity elements; third, use high-temperature diffusion annealing to eliminate element segregation." }, { "idx": 2224, "question": "Assuming the recrystallization temperature is defined as the temperature at which 95% recrystallization is completed within 1h, according to the Arrhenius equations Ṅ=N0exp(−Qn/RT) and G=G0exp(−Qg/RT), it can be inferred that the recrystallization temperature will be a function of G and Ṅ. Determine the functional relationship between the recrystallization temperature and G0, N0, Qg, Qn.", "answer": "According to the J-M equation, if the temperature at which 95% recrystallization is completed within 1h is defined as TF, then 0.95=1−exp(−π/3 ṄG3t0^4). Therefore, t0=(2.86/ṄG3)^1/4. Substituting the Arrhenius equations, we obtain N0G0^3exp(−(Qn+3Qg)/RTF)=k. Rearranging gives TF=(Qn+3Qg)/(R ln(N0G0^3/k))=k′(Qn+3Qg). This equation represents the functional relationship between TF and N0, G0, Qn, Qg." }, { "idx": 2225, "question": "Determine whether the following statement is correct. (1) Appropriate recrystallization annealing can refine the grain size of metal castings.", "answer": "Incorrect. Only for metals that have undergone cold deformation (with significant deformation) can the grain size be refined through appropriate recrystallization annealing." }, { "idx": 2226, "question": "Why do metallic materials exhibit better mechanical properties after hot working compared to the as-cast state?", "answer": "During hot working, metallic materials undergo dynamic deformation, dynamic recovery, and recrystallization processes. The columnar crystal zones and coarse equiaxed crystal zones disappear and are replaced by finer equiaxed grains. Many dispersed shrinkage pores and microcracks in the original ingot vanish due to mechanical welding effects, and microsegregation is also reduced to some extent through compression and diffusion. Consequently, the material's density and mechanical properties (especially plasticity and toughness) are improved." }, { "idx": 2227, "question": "Determine whether the following view is correct. (7) Recovery, recrystallization, and grain growth are all processes of nucleation and growth, and their driving force is the stored energy.", "answer": "Incorrect. Only the recrystallization process is a nucleation and growth process, and its driving force is the stored energy." }, { "idx": 2228, "question": "Determine whether the following statement is correct. (3) Polygonization causes the scattered dislocations to gather together to form dislocation walls, and the superposition of dislocation stress fields increases the lattice distortion.", "answer": "Incorrect. During the polygonization process, the vacancy concentration decreases and dislocations recombine, causing dislocations of opposite signs to cancel each other out, reducing dislocation density and thereby alleviating lattice distortion." }, { "idx": 2229, "question": "Determine whether the following statement is correct. (4) For any metal that undergoes cold deformation followed by recrystallization annealing, the grain size can be refined.", "answer": "Incorrect. If the metal is deformed at the critical deformation degree, the grain size will coarsen after recrystallization annealing." }, { "idx": 2230, "question": "The tungsten filament in a light bulb operates at very high temperatures, leading to significant grain growth. When large grains spanning the filament form, the filament becomes brittle under certain conditions and may fracture due to stresses caused by thermal expansion during heating and cooling. Find a method to extend the lifespan of the tungsten filament.", "answer": "Dispersed, particulate second phases (such as $\\\\mathrm{ThO}_{2}$) can be introduced into the tungsten filament to restrict grain growth. If the volume fraction of $\\\\mathrm{ThO}_{2}$ is $\\\\varphi$ and the radius is $r$, the limiting grain size $R = \\\\frac{4r}{3\\\\varphi(1+\\\\cos\\\\alpha)}$ (where $\\\\alpha$ is the contact angle). By selecting appropriate values of $\\\\varphi$ and $r$ to minimize $R$, grain growth can be effectively halted. Since grain refinement significantly reduces the brittleness of the filament, it becomes less prone to fracture, thereby effectively extending its lifespan." }, { "idx": 2231, "question": "Determine whether the following view is correct. (5) The recrystallization temperature of a certain aluminum alloy is 320°C, indicating that this alloy can only undergo recovery below 320°C, while recrystallization must occur above 320°C.", "answer": "Incorrect. Recrystallization is not a phase transformation. Therefore, it can occur over a wide temperature range." }, { "idx": 2232, "question": "Determine whether the following statement is correct. (2) Dynamic recrystallization only occurs under hot deformation conditions, therefore, metals deformed at room temperature will not undergo dynamic recrystallization.", "answer": "Incorrect. The recrystallization temperature of some metals is below room temperature, so deformation at room temperature is also considered hot deformation, and dynamic recrystallization can occur." }, { "idx": 2233, "question": "Determine whether the following statement is correct. (6) The melting point of 20# steel is lower than that of pure iron, so its recrystallization temperature is also lower than that of pure iron.", "answer": "Incorrect. The presence of trace solute atoms (wC=0.002 in 20# steel) hinders the recrystallization of the metal, thereby increasing its recrystallization temperature." }, { "idx": 2234, "question": "Determine whether the following view is correct. (8) The greater the deformation of the metal, the more likely the recrystallization mechanism of grain boundary bulging nucleation will occur.", "answer": "Incorrect. When the cold deformation degree of the metal is small, uneven deformation is more likely to occur in adjacent grains, that is, the dislocation density differs, making the recrystallization mechanism of grain boundary bulging nucleation more likely to occur." }, { "idx": 2235, "question": "Determine whether the following view is correct. (9) Normal grain growth is the consumption of small grains by large grains, while abnormal growth is the consumption of large grains by small grains.", "answer": "Incorrect. Normal grain growth is uniform growth driven by interface curvature, whereas abnormal growth is non-uniform growth where large grains consume small grains." }, { "idx": 2236, "question": "Determine whether the following statement is correct. (10) The second-phase particles in an alloy generally can hinder recrystallization but promote grain growth.", "answer": "Incorrect. The second-phase particles in an alloy generally can hinder recrystallization and also impede grain growth." }, { "idx": 2237, "question": "Determine whether the following statement is correct. (13) Recrystallization is a nucleation-growth process, so it is also a phase transformation process.", "answer": "Incorrect. Although recrystallization is a nucleation-growth process, the crystal lattice type does not change, so it is not a phase transformation process." }, { "idx": 2238, "question": "Determine whether the following statement is correct. (11) Recrystallization texture is the deformation texture retained during the recrystallization process.", "answer": "Incorrect. Recrystallization texture is the texture formed during the recrystallization (primary, secondary) process of cold-deformed metals. It is formed on the basis of deformation texture, with two scenarios: one is retaining the original deformation texture, and the other is the disappearance of the original deformation texture, replaced by a new recrystallization texture." }, { "idx": 2239, "question": "In Fe-Si steel (with $\\\\mathrm{{\\\\tau}}\\\\mathrm{{{w}}}_{\\\\mathrm{{Si}}}$ being 0.03), the measured diameter of $\\\\mathrm{MnS}$ particles is $0.4\\\\mu\\\\mathrm{m}$, and the number of particles per $1~\\\\mathrm{m}\\\\mathrm{m}^{2}$ is $2\\\\times10^{5}$. Calculate the effect of $\\\\mathrm{MnS}$ on the austenite grain growth during normal heat treatment of this steel (i.e., calculate the austenite grain size).", "answer": "Let the number of $\\\\mathrm{MnS}$ particles per unit volume be $N_{\\\\mathrm{V}}(1/\\\\mathrm{m}\\\\mathrm{m}^{3})$. Given the number of MnS particles per unit area $N_{\\\\mathrm{A}}=$ $2\\\\times10^{5}~1/\\\\mathrm{m}\\\\mathrm{m}^{2}$ and the particle diameter $d=0.4~\\\\mu\\\\mathrm{m}$. According to the principles of quantitative metallography, $$ N_{\\\\mathrm{A}}=d N_{\\\\mathrm{v}} $$ The volume fraction of MnS is $$ \\\\varphi={\\\\frac{1}{6}}\\\\pi d^{3}N_{\\\\mathrm{v}}={\\\\frac{1}{6}}\\\\pi d^{2}N_{\\\\mathrm{A}}= $$ $$ \\\\frac{1}{6}\\\\pi\\\\times(0.4\\\\times10^{-3})^{2}\\\\times2\\\\times10^{5}=0.0167 $$ Therefore, during the heating of this steel, due to the effect of $\\\\mathrm{MnS}$ particles, the limiting size for austenite grain growth is $$ \\\\overline{{\\\\cal D}}_{\\\\mathrm{lim}}=\\\\frac{4r}{3\\\\varphi}=\\\\frac{4\\\\times0.2}{3\\\\times0.016~7}=16~\\\\mu\\\\mathrm{m} $$" }, { "idx": 2240, "question": "If σ_α/β=200×10^-3 J/m^2, what is the total interfacial energy per cubic meter of the alloy? What is the total interfacial energy per mole of the alloy (V_m=10^-5 m^3/mol)?", "answer": "The total interfacial energy per cubic meter of the alloy is 1.2×10^8×200×10^-3=240×10^5 J/m^3. The interfacial energy per mole of the alloy is γ=1.2×10^8×200×10^-3×10^-5=240 J/mol" }, { "idx": 2241, "question": "If the interfacial energy is the same as (3), how much driving force for phase transformation remains in the alloy?", "answer": "420.8-240=180.8 J/mol. That is, after overcoming the interfacial energy during the phase transformation, there remains a driving force of 180.8 J/mol." }, { "idx": 2242, "question": "Determine whether the following view is correct. (12) When the deformation is large and relatively uniform, the grains after recrystallization tend to undergo normal growth, otherwise, abnormal growth is more likely to occur.", "answer": "Incorrect. Normal grain growth is the process of uniform grain growth that occurs during continued heating or holding after recrystallization is completed; whereas abnormal grain growth is the process of non-uniform grain growth that occurs under certain conditions (i.e., stable grains after recrystallization, the presence of a few grains favorable for growth, and high-temperature heating) following normal grain growth." }, { "idx": 2243, "question": "The carbon in iron, c0=wC=0.0001, E=-0.5 eV, calculate the critical temperature for solute segregation.", "answer": "c=c0exp(-E/kT), when complete solute segregation occurs, c=wx=1, T=T∥s, 1=c0exp(-E/kT), then T∥⊥=-E/kln(1/c0). Substituting c0=wC=0.0001, E=-0.5 eV=-1.602×10^-19×0.5 J, T∥∥=(1.602×10^-19×0.5)/(1.381×10^-23×ln(1/0.0001))=629 K." }, { "idx": 2244, "question": "In the regular melt α, the total driving force ΔG for the precipitation of β can be approximately expressed as ΔG=RT[x_o ln(x_o/x_e)+(1-x_o)ln((1-x_o)/(1-x_e))]-2Ω(x_o-x_e)^2. Given T=600K, x_o=0.1, x_e=0.02, Ω=0, use this expression to estimate the total driving force for α→α′+β.", "answer": "Given T=600K, x_o=0.1, x_e=0.02, R=8.31 J/(mol·K). ΔG=RT[x_o ln(x_o/x_e)+(1-x_o)ln((1-x_o)/(1-x_e))]=8.31×600×[0.1 ln(0.1/0.02)+(1-0.1)ln((1-0.1)/(1-0.02))]=420.8 J/mol" }, { "idx": 2245, "question": "Using classical nucleation theory, calculate the shape factor η of a new phase when a cubic nucleus is formed by $n$ atoms during a solid-state phase transformation.", "answer": "According to classical nucleation theory, during a solid-state phase transformation, the change in the system's free energy is given by $$ \\\\Delta G=n\\\\Delta G_{V}+S\\\\gamma+n E_{s}$$ where $\\\\Delta G_{V}$ is the free energy difference per atom between the old and new phases; S is the surface area of the nucleus; $\\\\gamma$ is the average surface energy; $E_{s}$ is the strain energy per atom in the nucleus; $n$ is the number of atoms in the nucleus. Assuming the density of the new phase is $\\\\rho$ and the relative atomic mass is $M,$ the volume occupied per gram-atom of the new phase material is $M/\\\\rho;$ the volume occupied per atom of the new phase is $M/(\\\\rho N_{0});$ the volume of a nucleus with $n$ atoms is $n M/(\\\\rho N_{0})$. If a cubic nucleus is formed, its edge length is $(n M/(\\\\rho N_{\\\\circ}))^{1/3}$, and the surface area of the nucleus is $6(n M/(\\\\rho N_{\\\\circ}))^{2/3}$. Thus, the shape factor $$\\\\eta=(M/(\\\\rho N_{\\\\circ}))^{2/3}$$ and the free energy change is $$ \\\\Delta G=n\\\\Delta G_{V}+6(M/(\\\\rho N_{\\\\L_{0}}))^{2/3}n^{2/3}\\\\gamma+n E_{s}$$." }, { "idx": 2246, "question": "If an alloy has β-phase precipitates with a spacing of 50 nm after heat treatment, calculate the total area of α/β interfaces per cubic meter (assuming the precipitates are cubes).", "answer": "The number of β precipitates is n_β=1/(50×10^-9)^3=8×10^21. The surface area of each β is S_β=6×(50×10^-9)^2=1.5×10^-14 m^2. The total interface area in 1 m^3 is S_A=8×10^21×1.5×10^-14=1.2×10^8 m^2." }, { "idx": 2247, "question": "During solid-state phase transformation, assuming the new phase nucleus is spherical and the volume free energy change per atom ΔGv=200ΔT/Tc (J/cm³), the critical transformation temperature Tc=1000K, the strain energy Es=4 J/cm³, the coherent interface energy γ_coherent=40×10⁻⁷ J/cm², and the incoherent interface energy γ_incoherent=400×10⁻⁷ J/cm², calculate ΔT when ΔG_coherent* = ΔG_incoherent*.", "answer": "Let ΔG_coherent* = ΔG_incoherent*, then (40×10⁻⁷)³ / (200×ΔT/1000 - 4)² = (400×10⁻⁷)³ / (200×ΔT/1000)². Solving gives ΔT ≈ 21K." }, { "idx": 2248, "question": "For zinc in copper, c0=wZn=0.0001, E=-0.12 eV, calculate the critical temperature for solute segregation.", "answer": "c=c0exp(-E/kT). When complete solute segregation occurs, c=wx=1, T=T∥s, 1=c0exp(-E/kT), then T∥⊥=-E/kln(1/c0). Substituting c0=wZn=0.0001, E=-0.12 eV=-1.602×10^-19×0.12 J, T∥∥=(1.602×10^-19×0.12)/(1.381×10^-23×ln(1/0.0001))=151 K." }, { "idx": 2249, "question": "Analyze the main reasons for dislocations promoting nucleation, focusing on the case where the new phase nucleates on the dislocation line", "answer": "When the new phase nucleates on the dislocation line, the dislocation disappears at the nucleation site, and the released elastic strain energy reduces the nucleation work, thereby promoting nucleation." }, { "idx": 2250, "question": "During solid-state phase transformation, assuming the new phase nucleus is spherical and the volume free energy change per atom ΔGv=200ΔT/Tc(J/cm³), the critical transformation temperature Tc=1000K, strain energy Es=4 J/cm³, coherent interface energy γ_coherent=40×10⁻⁷ J/cm², incoherent interface energy γ_incoherent=400×10⁻⁷ J/cm², calculate the ratio of critical nucleation energy ΔG_coherent*/ΔG_incoherent* when ΔT=50K.", "answer": "ΔG_coherent* = (16πγ_coherent³) / [3(ΔGv - Es)²] = (16π(40×10⁻⁷)³) / [3(200×50/1000 - 4)²]\\nΔG_incoherent* = (16πγ_incoherent³) / (3ΔGv²) = (16π(400×10⁻⁷)³) / [3(200×50/1000)²]\\nΔG_coherent*/ΔG_incoherent* = [(200×50/1000)² × (40×10⁻⁷)³] / [(200×50/1000 - 4)² × (400×10⁻⁷)³] = 0.01" }, { "idx": 2251, "question": "During the solid-state phase transformation, assuming the nucleation rate $\\\\dot{N}$ and growth rate $G$ are constants, the volume fraction of the new phase formed after time $t$ can be obtained using the Johnson-Mehl equation, i.e., $$ \\\\begin{array}{r}{\\\\varphi=1-\\\\exp\\\\left(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\right)}\\\\end{array}$$ Given the nucleation rate $\\\\dot{N}=1~000/(\\\\mathrm{cm}^{3}\\\\bullet\\\\mathrm{s}),G=3\\\\times10^{5}~\\\\mathrm{cm/s},$ calculate the time required to achieve a $50\\\\%$ transformation.", "answer": "$$ \\\\varphi=1-\\\\exp\\\\Bigl(-\\\\textstyle{\\\\frac{\\\\pi}{3}}\\\\dot{N}G^{3}t^{4}\\\\Bigr)$$ $$ 50\\\\%=1-\\\\exp\\\\Bigl(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\Bigr)$$ $$ 0.6931={\\\\frac{\\\\pi}{3}}\\\\times1000\\\\times(3\\\\times10^{-5})^{3}t^{4}$$ $$ t^{4}=2.45\\\\times10^{10}$$ $$ t=395\\\\mathrm{~s~}$$" }, { "idx": 2252, "question": "Analyze the resistance to solid-state phase transformation.", "answer": "The resistance to nucleation during solid-state phase transformation arises from the interfacial energy $E_{\\\\gamma}$ introduced by the formation of an interface between the new-phase nucleus and the matrix, as well as the volumetric strain energy (i.e., elastic energy) $E_{\\\\mathrm{e}}$. The interfacial energy $E_{\\\\gamma}$ consists of two parts: one is the chemical energy caused by changes in the strength and quantity of like and unlike bonds when forming the new-phase interface in the parent phase, referred to as the chemical term of the interfacial energy; the other is the interfacial strain energy caused by the mismatch of interfacial atoms and the resulting strain in atomic spacing, referred to as the geometric term of the interfacial energy. The strain energy $E_{\\\\mathrm{e}}$ arises because the formation of the new phase in the parent phase, due to their differing specific volumes, induces volumetric strain. This volumetric strain is typically accommodated by elastic strain between the new phase and the parent phase, resulting in volumetric strain energy. Overall, as the size of the new-phase nucleus increases and the new phase grows, $(E_{\\\\gamma}+E_{e})$ will increase. Of course, $E_{\\\\gamma}$ and $E_{\\\\mathrm{e}}$ can also adjust to each other through factors such as the precipitation location, particle shape, and interface state of the new phase to minimize $(E_{\\\\gamma}+E_{e})$. When the parent phase is liquid, there is no issue of volumetric strain energy, and the interfacial energy of a solid phase is much higher than that of a liquid-solid interface. In comparison, the resistance to solid-state phase transformation is greater." }, { "idx": 2253, "question": "Analyze the driving force for particle coarsening when spherical second-phase particles precipitate from the solid solution.", "answer": "Assume spherical β particles with radius r precipitate in the α parent phase, with volume V and β/α interfacial area S. The free energy can be expressed as $$ where G_V and E_s are the chemical free energy and elastic strain energy per unit volume of the new phase, respectively, and γ is the specific interfacial energy. The chemical potential of a certain component, such as the solute component, can be expressed as G = V(G_V + E_s) + Sγ $$ where Ω is the molar volume, i.e., the volume of the new phase corresponding to one mole of solute atoms. From the above two equations, we obtain $$ μ = Ω(G_V + E_s) + Ω(∂S/∂V)γ $$ where ∂S/∂V is the increase in surface area per unit volume increase. For spherical particles, we have Thus, $$ ∂S/∂V = d(4πr²)/d(4/3πr³) = 2/r $$ $$ μ = Ω(G_V + E_s) + 2Ωγ/r $$ Clearly, the chemical potential of solute atoms in spherical particles depends on the particle radius r. The smaller r is, the higher μ becomes, making such particles less stable. Assume two spherical β particles with radii r₁ and r₂ (r₁ > r₂) precipitate in the parent phase and are adjacent to each other. The difference in their chemical potentials is $$ Δμ = μ₂ - μ₁ = 2Ωγ(1/r₂ - 1/r₁) $$ This is the driving force for solute atoms to diffuse from smaller particles to larger ones, leading to particle coarsening." }, { "idx": 2254, "question": "Analyze the main reasons why dislocations promote nucleation, focusing on the scenario where dislocations do not disappear and adhere to the new phase interface", "answer": "The dislocations do not disappear and instead adhere to the new phase interface, becoming part of the dislocations in the semi-coherent interface, compensating for the mismatch. This reduces the energy, thereby decreasing the energy required for nucleus formation and promoting nucleation." }, { "idx": 2255, "question": "Analyze the main reasons for dislocations promoting nucleation in the case where the composition of the new phase differs from that of the parent phase", "answer": "When the composition of the new phase differs from that of the parent phase, the segregation of solute atoms along dislocation lines (forming Cottrell atmospheres) facilitates the precipitation of the new phase and also promotes nucleation." }, { "idx": 2256, "question": "During the solid-state phase transformation, assuming the nucleation rate $\\\\dot{N}$ and growth rate $G$ of the new phase are constants, the volume fraction of the new phase formed after time $t$ can be obtained using the Johnson-Mehl equation, i.e., $$ \\\\begin{array}{r}{\\\\varphi=1-\\\\exp\\\\left(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\right)}\\\\end{array}$$ Given the nucleation rate $\\\\dot{N}=1~000/(\\\\mathrm{cm}^{3}\\\\bullet\\\\mathrm{s}),G=3\\\\times10^{5}~\\\\mathrm{cm/s},$ calculate the maximum transformation rate during the phase transformation.", "answer": "$$ \\\\frac{\\\\mathrm{d}\\\\varphi}{\\\\mathrm{d}t}=\\\\left(\\\\frac{4}{3}\\\\pi\\\\dot{N}G^{3}t^{3}\\\\right)\\\\exp\\\\left(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\right)$$ Substitute $t_{\\\\mathrm{max}}=403$ into the above equation to calculate the maximum transformation rate." }, { "idx": 2257, "question": "What is the main difference between precipitation decomposition and spinodal decomposition in the formation of precipitated phases?", "answer": "The main difference between the two in the formation of precipitated phases lies in the nucleation driving force and the compositional change of the new phase. During precipitation transformation, the formation of the new phase requires significant concentration fluctuations, and the composition of the new phase changes abruptly compared to the parent phase, resulting in interfacial energy. This necessitates a larger nucleation driving force to overcome the interfacial energy, meaning a greater degree of undercooling is required. In contrast, spinodal decomposition does not involve a nucleation process or abrupt compositional changes; any small concentration fluctuation can lead to the formation and growth of the new phase." }, { "idx": 2258, "question": "The following equation represents the change in the system's free energy caused by the formation of a crystal embryo containing $n$ atoms. $$ \\\\Delta G=-\\\\:b n\\\\left(\\\\Delta G_{V}-E_{\\\\mathrm{s}}\\\\right)+a n^{2/3}\\\\gamma_{\\\\upalpha/\\\\upbeta} $$ where $\\\\Delta G_{V}$ is the free energy change per unit volume of the crystal embryo; $\\\\gamma_{\\\\mathfrak{a}/\\\\mathfrak{g}}$ is the interfacial energy; $E_{\\\\mathrm{s}}$ is the strain energy; $a, b$ are coefficients whose values are determined by the shape of the crystal embryo. \\n\\nFind the values of $a$ and $b$ when the crystal embryo is spherical. If $\\\\Delta G_{V}, \\\\gamma_{\\\\scriptscriptstyle\\\\alpha/\\\\beta}, E_{s}$ are all constants, derive the nucleation energy $\\\\Delta G^{*}$ for a spherical nucleus.", "answer": "If precipitation occurs simultaneously throughout the parent phase, and the composition of the parent phase changes continuously with the formation of the new phase, but the grain morphology and orientation remain unchanged, it is called continuous precipitation. \\n\\nIn contrast to continuous precipitation, when precipitation occurs, the solid solution within a certain surrounding range immediately changes from a supersaturated state to a saturated state, forming a distinct interface with the original composition of the parent phase. After nucleation at grain boundaries, it grows into the grain interior in an alternating lamellar distribution. Not only does the composition change abruptly at the interface, but the orientation also changes. This is discontinuous precipitation. The main difference lies in the length of the diffusion path. In the former, the diffusion field extends over a considerable distance, while in the latter, the diffusion distance is on the order of the lamellar spacing (generally less than $1~\\\\mu\\\\mathrm{m}$). \\n\\nDiscontinuous precipitation has the following characteristics: \\n\\n(1) At the interface between the precipitate and the matrix, the composition is discontinuous; the interfaces between the precipitate and the matrix are all high-angle incoherent interfaces, indicating that the crystal orientation is also discontinuous. (2) Cellular precipitates usually nucleate at the grain boundaries of the matrix $({\\\\alpha^{\\\\prime}})$ and always grow into one of the adjacent grains of the $\\\\alpha^{'}$ phase. (3) During the growth of cellular precipitates, the redistribution of solute atoms is achieved through interfacial diffusion between the precipitate and the parent phase, with the diffusion distance typically less than $1~\\\\mu\\\\mathrm{m}$." }, { "idx": 2259, "question": "In the process of solid-state phase transformation, assuming the nucleation rate $\\\\dot{N}$ and growth rate $G$ are constants, the volume fraction of the new phase formed after time $t$ can be obtained using the Johnson-Mehl equation, i.e., $$ \\\\begin{array}{r}{\\\\varphi=1-\\\\exp\\\\left(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\right)}\\\\end{array}$$ Given the nucleation rate $\\\\dot{N}=1~000/(\\\\mathrm{cm}^{3}\\\\bullet\\\\mathrm{s}),G=3\\\\times10^{5}~\\\\mathrm{cm/s},$ calculate the time at which the phase transformation rate is the fastest.", "answer": "$$ \\\\varphi=1-\\\\exp\\\\Bigl(-\\\\textstyle{\\\\frac{\\\\pi}{3}}\\\\dot{N}G^{3}t^{4}\\\\Bigr)$$ $$ \\\\frac{\\\\mathrm{d}\\\\varphi}{\\\\mathrm{d}t}=\\\\left(\\\\frac{4}{3}\\\\pi\\\\dot{N}G^{3}t^{3}\\\\right)\\\\exp\\\\left(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\right)$$ $$ \\\\frac{\\\\mathrm{d}^{2}\\\\varphi}{\\\\mathrm{d}t^{2}}=-\\\\left(\\\\frac{4}{3}\\\\pi\\\\dot{N}G^{3}t^{3}\\\\right)^{2}\\\\exp\\\\left(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\right)+\\\\left(\\\\frac{12}{3}\\\\pi\\\\dot{N}G^{3}t^{2}\\\\right)\\\\exp\\\\left(-\\\\frac{\\\\pi}{3}\\\\dot{N}G^{3}t^{4}\\\\right)$$ ${\\\\frac{\\\\mathrm{d}^{2}\\\\varphi}{\\\\mathrm{d}t^{2}}}=0$ , i.e.,$$ -\\\\left({\\\\frac{4}{3}}\\\\pi{\\\\dot{N}}G^{3}t^{3}\\\\right)^{2}+\\\\left({\\\\frac{12}{3}}\\\\pi{\\\\dot{N}}G^{3}t^{2}\\\\right)=0$$ $$ t_{\\\\mathrm{max}}=\\\\left(\\\\frac{9}{4\\\\pi\\\\dot{N}G^{3}}\\\\right)^{1/4}=\\\\left[\\\\frac{9}{4\\\\times3.14\\\\times1000\\\\times(3\\\\times10^{-5})^{3}}\\\\right]^{\\\\frac{1}{4}}=403$$" }, { "idx": 2260, "question": "What is the difference between continuous and discontinuous precipitation?", "answer": "If precipitation occurs simultaneously throughout the parent phase, and the composition of the parent phase changes continuously with the formation of the new phase, but the grain morphology and orientation remain unchanged, it is called continuous precipitation. In contrast to continuous precipitation, when precipitation occurs, the solid solution within a certain range around it immediately changes from a supersaturated state to a saturated state, forming a distinct interface with the original composition of the parent phase. After nucleation at grain boundaries, it grows into the grain interior with an alternating lamellar distribution. Not only does the composition change abruptly at the interface, but the orientation also changes, which is called discontinuous precipitation. The main difference lies in the length of the diffusion path. In the former, the diffusion field extends over a considerable distance, while in the latter, the diffusion distance is only on the order of the lamellar spacing (generally less than 1 μm)." }, { "idx": 2261, "question": "Describe the precipitation sequence of Al-Cu alloy", "answer": "The precipitation sequence of Al-Cu alloy is: GP zones → θ′′ metastable phase → θ′ metastable phase → θ phase (equilibrium phase)" }, { "idx": 2262, "question": "What are the main characteristics of discontinuous precipitation?", "answer": "Discontinuous precipitation has the following characteristics: (1) At the interface between the precipitate and the matrix, the composition is discontinuous; the interfaces between the precipitate and the matrix are all high-angle non-coherent interfaces, indicating that the crystal orientation is also discontinuous. (2) Cellular precipitates usually nucleate at the grain boundaries of the matrix (α') and always grow into one of the adjacent grains of the α' phase. (3) When cellular precipitates grow, the partitioning of solute atoms is achieved through their interfacial diffusion between the precipitate phase and the parent phase, with the diffusion distance typically less than 1 μm." }, { "idx": 2263, "question": "What is the essence of aging?", "answer": "The essence of aging is the process of separating a new phase from a supersaturated solid solution, usually caused by temperature changes. The microstructure after aging consists of a matrix and precipitates. The matrix has the same crystal structure as the parent phase but differs in composition and lattice constants, while the precipitates can have different crystal structures and compositions from the parent phase. Due to variations in the properties, size, shape, and distribution of the precipitates in the microstructure, the performance of the alloy can change significantly." }, { "idx": 2264, "question": "For a φ5mm carbon steel sample with a carbon mass fraction of w_c=0.003, after heating and quenching at 860°C, the sample is tempered. How will the microstructure change during the tempering process?", "answer": "For carbon steel with w_c=0.003, when tempered below 200°C, the changes in microstructure morphology are minor, and the hardness change is also small. However, the tendency for carbon atoms to segregate near dislocation lines increases. When the tempering temperature exceeds 250°C, cementite precipitates between laths or along dislocation lines, leading to a decrease in strength and plasticity. When the tempering temperature reaches 300~400°C, lamellar or rod-shaped cementite precipitates, causing a significant reduction in hardness and strength, while plasticity begins to increase. During tempering at 400~700°C, the carbides undergo aggregation, growth, and spheroidization, and the α phase undergoes recovery and recrystallization. At this stage, hardness and strength gradually decrease, while plasticity gradually increases." }, { "idx": 2265, "question": "If a new phase in a solid-state phase transformation precipitates from the parent phase as spherical particles, and the change in free energy per unit volume is $10^{8}\\\\mathrm{~J~}\\\\cdot\\\\mathrm{~m~}^{-3}$, the specific surface energy is $1~\\\\mathrm{J}\\\\cdot\\\\mathrm{m}^{-2}$, and the strain energy is negligible, calculate the diameter of the new phase particles when the surface energy is $1\\\\%$ of the volume free energy.", "answer": "Diameter $2r=6\\\\times10^{-6}\\\\mathrm{~m~}$." }, { "idx": 2266, "question": "Describe the characteristics of diffusionless phase transformation.", "answer": "Diffusionless phase transformation has the following characteristics: (1) There is a shape change caused by uniform shear, resulting in a change in the crystal's external form. (2) Since the phase transformation process is diffusionless, the new phase and the parent phase have the same chemical composition. (3) There is a certain crystallographic orientation relationship between the parent phase and the new phase. (4) The phase boundary moves extremely fast, approaching the speed of sound." }, { "idx": 2267, "question": "Describe the basic characteristics of the possible precipitates in Al-Cu alloys", "answer": "GP zones are disc-shaped, with a thickness of 0.3~0.6nm and a diameter of about 8nm, formed on the {100} planes of the parent phase. Their lattice is the same as that of the matrix α (fcc) and is fully coherent with the α phase. The θ′′ transition phase is disc-shaped, with a thickness of 2nm and a diameter of 30~40nm, formed on the {100} planes of the parent phase. It has a tetragonal lattice with lattice constants a=b=0.404nm, c≈0.78nm, and is fully coherent with the matrix. However, due to the difference in lattice constants along the Z-axis, there is about 4% mismatch, creating an elastic coherency strain field around θ′′. The θ′ transition phase also forms on the {100} planes of the matrix, with a tetragonal structure and lattice constants a=b=0.404nm, c=0.58nm, and its nominal composition is CuAl2. Due to the large mismatch along the Z-axis, it can only maintain partial coherence with the matrix. The θ phase has a tetragonal structure with lattice constants a=b=0.607nm, c=0.487nm. This equilibrium precipitate phase is completely incoherent with the matrix." }, { "idx": 2268, "question": "Why do transition phases appear during the precipitation process?", "answer": "Transition phases appear during the precipitation process because the system requires a series of intermediate transition phases to gradually reduce free energy during the transformation from a supersaturated solid solution to the equilibrium phase. These transition phases, with their crystal structures and compositions intermediate between GP zones and the equilibrium phase, can reduce nucleation work and strain energy, making the phase transformation process easier to proceed." }, { "idx": 2269, "question": "For carbon steel samples with carbon mass fractions of w_c=0.003 and w_C=0.012 and a diameter of φ5mm, both heated at 860°C and then quenched, explain the microstructure morphology, fine structure, and composition obtained after quenching.", "answer": "Heated at 860°C, both steels are in the single-phase region (refer to the Fe-Fe3C phase diagram), and after quenching, both form martensite. The carbon steel with w_C=0.012 contains a certain amount of retained austenite. For the carbon steel with w_c=0.003, the martensite composition is w_c=0.003, with a lath-like morphology and a fine structure of dislocations. For the carbon steel with w_c=0.012, the martensite composition is w_c=0.012, with an acicular morphology and a fine structure of twins." }, { "idx": 2270, "question": "Referring to the composite material in the previous question. Given that the yield strength of steel is 280 MPa and the yield strength of copper is 140 MPa. What is the maximum tensile load the composite material can withstand without undergoing plastic deformation?", "answer": "The maximum tensile force is F⊥=Fcu+Fst=(140×10^6 N/m^2)×(2.4×10^−6 m^2)+(260×10^6 N/m^2)×(0.8×10^−6 m^2)=540 N" }, { "idx": 2271, "question": "For a φ5mm carbon steel sample with a carbon mass fraction of w_c=0.012, after quenching at 860°C and then tempering, how will the microstructure change during the tempering process?", "answer": "For carbon steel with w_c=0.012, when tempered below 100°C, carbon atoms form carbon-rich regions; during tempering at 100~200°C, a large number of fine carbides precipitate, resulting in a slight increase in hardness; during tempering at 200~300°C, retained austenite transforms into tempered martensite (or bainite), causing an increase in hardness, but at the same time, the hardness of martensite decreases, so overall, the hardness change is minimal; when tempered above 300°C, carbides continue to precipitate, followed by carbide growth and spheroidization, while the α phase undergoes recovery and recrystallization, leading to a decrease in hardness and an increase in toughness." }, { "idx": 2272, "question": "Referring to the composite material in the previous question. Given that the yield strength of steel is 280 MPa and the yield strength of copper is 140 MPa. If the composite material is subjected to tension, which metal will yield first?", "answer": "Since the stress ratio is 1.86, when the stress on steel is 140 MPa, the stress on copper is also 140 MPa. Therefore, copper will yield first." }, { "idx": 2273, "question": "A steel wire with a diameter of 0.89 mm has σ_ss=980 MPa and σ_sb=1,130 MPa. There is also an aluminum alloy with σ_as=255 MPa and σ_ab=400 MPa. Their densities are known to be ρ_s=7.85 g/cm³ and ρ_a=2.7 g/cm³, respectively. If the aluminum wire is to bear the same maximum load without deformation, how much heavier or lighter is it compared to the steel wire (expressed as a percentage)?", "answer": "σ_ss=φ_max/A_s=980, σ_as=φ_max/A_a=255. From these two equations, we know A_s/A_a=255/980. A_a=3.84A_s. m_a/m_s=(3.84×2.7)/(1×7.85)=1.32. 1.32-1=0.32=32%. This aluminum wire is 32% heavier than the steel wire." }, { "idx": 2274, "question": "A steel wire with a diameter of 0.89 mm has σ_ss=980 MPa and σ_sb=1,130 MPa. Another aluminum alloy has σ_as=255 MPa and σ_ab=400 MPa. Their densities are known to be ρ_s=7.85 g/cm³ and ρ_a=2.7 g/cm³, respectively. What if they do not break?", "answer": "σ_sb=φ_max/A_s=1,130, σ_ab=φ_max/A_a=400. From the above two equations, A_s/A_a=400/1,130. A_a=2.825A_s. m_a/m_s=(2.825×2.7)/(1×7.85)=0.97. 0.97-1=-0.03=-3%. That is, the aluminum wire is 3% lighter than the steel wire." }, { "idx": 2275, "question": "Referring to the composite material in the previous question. Given that the elastic modulus of steel is 205 GPa and the elastic modulus of copper is 110 GPa. What is the elastic modulus of this composite material?", "answer": "E=(φE)st+(φE)cu=0.25×(205000 MPa)+0.75×(110000 MPa)=130000 MPa" }, { "idx": 2276, "question": "Please derive the rule of mixtures for the longitudinal elastic modulus of a fiberglass composite fishing rod (composed of longitudinally aligned glass fibers and polyester resin as the bonding fiber).", "answer": "The derivation assumes that the two constituent materials have the same Poisson's ratio, and the transverse strain is negligible. Under longitudinal loading, the strains of the two components must be equal, i.e., $$ {\\\\varepsilon}_{\\\\mathfrak{g}1}={\\\\varepsilon}_{\\\\mathfrak{m}}={\\\\varepsilon}_{\\\\mathfrak{p}\\\\mathfrak{r}}$$ $$ {\\\\frac{F/f_{\\\\mathrm{^{gl}}}}{E_{\\\\mathrm{gl}}}}={\\\\frac{F/A}{E_{\\\\mathrm{m}}}}={\\\\frac{F/f_{\\\\mathrm{pr}}}{E_{\\\\mathrm{pr}}}}$$ where $E_{\\\\mathrm{m}}$ is the elastic modulus of the mixture; $_{\\\\mathrm{g1}}$ represents glass fiber; pr represents resin fiber; $A$ is the total cross-sectional area (can be considered as 1), and $f$ is the cross-sectional area (or volume) fraction of different materials. Then, $$ \\\\begin{array}{c}{F_{\\\\mathrm{g1}}=\\\\frac{f_{\\\\mathrm{g1}}E_{\\\\mathrm{g1}}F}{E_{\\\\mathrm{m}}}}\\\\\\\\ {F_{\\\\mathrm{pr}}=\\\\frac{f_{\\\\mathrm{pr}}E_{\\\\mathrm{pr}}F}{E_{\\\\mathrm{m}}}}\\\\end{array}$$ Since $F=F_{\\\\mathrm{gl}}+F_{\\\\mathrm{pr}}$, it follows that $$ F=\\\\frac{(f_{\\\\mathrm{gl}}E_{\\\\mathrm{gl}}+f_{\\\\mathrm{pr}}+E_{\\\\mathrm{pr}})F}{E_{\\\\mathrm{m}}}$$ $$ E_{\\\\mathrm{m}}=f_{\\\\mathrm{g1}}E_{\\\\mathrm{g1}}+f_{\\\\mathrm{pr}}E_{\\\\mathrm{pr}}$." }, { "idx": 2277, "question": "From the perspective of wettability, should β-SiC whiskers or Al2O3 short fibers be selected as the reinforcement for aluminum matrix composites? It is known that the wettability of molten aluminum on β-SiC is higher than that on Al2O3.", "answer": "The wettability of molten aluminum on β-SiC is higher than that on Al2O3. From the perspective of wettability, β-SiC whiskers should be selected as the reinforcement." }, { "idx": 2278, "question": "Calculate the critical length Lc of β-SiC whiskers, given a diameter of 0.5μm, tensile strength of 70000MPa, and interfacial shear strength τy approximated by the matrix yield strength of 76MPa.", "answer": "For β-SiC whiskers, the critical length Lc = (0.5×10^-6 × 70000) / (2 × 38) = 460.5×10^-6 m = 460.5μm. Since the actual length of 70μm is less than Lc, the fibers do not achieve the maximum strengthening effect." }, { "idx": 2279, "question": "Calculate the critical length Lc of Al2O3 short fibers, given a diameter of 10μm, tensile strength of 2275MPa, and interfacial shear strength τy approximated by the matrix yield strength of 76MPa.", "answer": "For Al2O3 short fibers, the critical length Lc = (10×10^-6 × 2275) / (2 × 38) = 299.3×10^-6 m = 299.3μm. Since the actual length of 4mm is greater than Lc, the fibers can achieve maximum strengthening effect." }, { "idx": 2280, "question": "A steel wire with a diameter of 0.89 mm has σ_ss=980 MPa and σ_sb=1,130 MPa. There is also an aluminum alloy with σ_as=255 MPa and σ_ab=400 MPa. Their densities are ρ_s=7.85 g/cm³ and ρ_a=2.7 g/cm³, respectively. If an aluminum wire is to bear a load of 40 kg and have the same elastic deformation as the steel wire, by what percentage will this aluminum wire be heavier or lighter than the steel wire?", "answer": "Given the steel's σ_ss=980 MPa, σ_sb=1,130 MPa, E_ss=205,000 MPa; the aluminum's σ_as=255 MPa, σ_ab=400 MPa, E_as=70,000 MPa, 40 kg=392 N, σ_s=F/A_s=392/A_ss, σ_a=392/A_a. σ_s/E_s=σ_a/E_a, then 392/(205 A_s)=392/(70 A_a), so A_a=2.93 A_s. m_a/m_s=(A_a ρ_a)/(A_s ρ_s)=(2.93×2.7)/(1×7.85)=1.008. 1.008-1=0.008=0.8%. This aluminum wire will be 0.8% heavier than the steel wire." }, { "idx": 2281, "question": "Predict the strength σLu of an Al2O3 short fiber reinforced composite at a volume fraction φf=0.1, given the fiber stress σf=2190MPa and the matrix yield strength σy=76MPa.", "answer": "For Al2O3 short fiber composites, the strength σLu = σf × φf + σy × (1 - φf) = 2190 × 0.1 + 76 × 0.9 = 386MPa." }, { "idx": 2282, "question": "Calculate the critical volume fraction φtr of β-SiC whisker reinforced composite, given the matrix tensile strength σm=186MPa, yield strength σy=76MPa, and average fiber stress σf=5320MPa.", "answer": "For β-SiC whiskers, the critical volume fraction φtr = (σm - σy) / (σf - σy) = (186 - 76) / (5320 - 76) = 0.021." }, { "idx": 2283, "question": "Calculate the fiber stress σf of Al2O3 short fibers when the length exceeds the critical length, given the fiber tensile strength σfa=2275MPa, critical length Lc=299.3μm, and actual length L=4mm.", "answer": "For Al2O3 short fibers, the fiber stress σf = (1 - Lc/(2L)) × σfa = (1 - 299.3×10^-6/(2 × 4000×10^-6)) × 2275 = 2190MPa." }, { "idx": 2284, "question": "Predict the strength σLu of a β-SiC whisker-reinforced composite with a volume fraction φf=0.1, given the average fiber stress σf=5320MPa and the matrix yield strength σy=76MPa.", "answer": "For the β-SiC whisker composite, the strength σLu = σf × φf + σy × (1 - φf) = 5320 × 0.1 + 76 × 0.9 = 700MPa." }, { "idx": 2285, "question": "Calculate the critical volume fraction φtr of Al2O3 short fiber reinforced composite, given the matrix tensile strength σm=186MPa, yield strength σy=76MPa, and fiber stress σf=2190MPa.", "answer": "For Al2O3 short fibers, the critical volume fraction φtr = (σm - σy) / (σf - σy) = (186 - 76) / (2190 - 76) = 0.052." }, { "idx": 2286, "question": "A steel wire (with a diameter of $1~\\\\mathrm{mm}$) is coated with a layer of copper (total diameter of $2~\\\\mathrm{mm}$). What is the thermal expansion coefficient of this composite material? The elastic modulus of steel is $E_{\\\\mathrm{st}}{=}205~\\\\mathrm{GPa}$, and that of copper is $E_{\\\\mathrm{cu}}{=}110~\\\\mathrm{GPa}$; their expansion coefficients are $\\\\alpha_{\\\\mathrm{st}}=1.1\\\\times10^{-6}/\\\\mathsf{\\\\Pi}^{\\\\circ}\\\\mathsf{C}, \\\\alpha_{\\\\mathrm{Cu}}=17\\\\times10^{-6}/\\\\mathsf{\\\\Pi}^{\\\\circ}\\\\mathsf{C}$ respectively.", "answer": "Under no-load conditions, the composite material satisfies $(\\\\Delta l/L)_{\\\\mathrm{st}}=(\\\\Delta l/L)_{\\\\mathrm{cu}}$, and the forces $F_{\\\\mathrm{Cu}}=-F_{\\\\mathrm{st}}$. If $\\\\Delta t= 1^{\\\\circ}\\\\mathrm{C}$, then $$ A_{\\\\mathrm{Cu}}=\\\\pi(\\\\frac{d}{2})^{2}=(\\\\pi/4)(0.002\\\\mathrm{m})^{2}-0.8\\\\times10^{-6}\\\\mathrm{m}^{2}=2.4\\\\times10^{-6}\\\\mathrm{m}^{2}$$ $$ (\\\\Delta l/L)_{\\\\mathrm{st}}=(\\\\Delta l/L)_{\\\\mathrm{cu}}$$ $$ \\\\begin{array}{r}{\\\\left[\\\\alpha\\\\Delta t+(F/A)/E\\\\right]_{\\\\ast}=\\\\left[\\\\alpha\\\\Delta t+(F/A)/E\\\\right]_{\\\\mathrm{cu}}}\\\\end{array}$$ $$ 1\\\\times10^{-6})\\\\times1+{\\\\frac{F_{\\\\mathrm{st}}/(0.8\\\\times10^{-6}{\\\\mathrm{m}}^{2})}{205\\\\times10^{9}{\\\\mathrm{N/m}}^{2}}}=(17\\\\times10^{-6})\\\\times1+{\\\\frac{-F_{\\\\mathrm{st}}/(2.4\\\\times10^{-6}{\\\\mathrm{m}}^{2})}{110\\\\times10^{9}{\\\\mathrm{N/m}}^{2}}}$$ When $\\\\Delta t=1~^{\\\\circ}C$, then $$ (\\\\Delta l/L)_{\\\\mathrm{cu}}=(17\\\\times10^{-6}/^{9}\\\\mathbb{C}\\\\times1^{9}\\\\mathbb{C})+\\\\frac{=0.61~\\\\mathrm{N}/(2.4\\\\times10^{-6}~\\\\mathrm{m}^{2})}{110\\\\times10^{9}~\\\\mathrm{N/m}^{2}}=15\\\\times10^{-6}$$ Thus, the thermal expansion coefficient of the composite material is $$ \\\\overline{{\\\\alpha}}=15\\\\times10^{-6}/^{\\\\circ}\\\\mathrm{C}$$ $A_{\\\\mathrm{st}}=\\\\pi(\\\\frac{d}{2})^{2}=(\\\\pi/4)(0.001~\\\\mathrm{m})^{2}=0.8\\\\times10^{-6}~\\\\mathrm{m}^{2}$" }, { "idx": 2287, "question": "Calculate the average stress σf of β-SiC whiskers when the critical length is not reached, given the matrix yield strength σy=76MPa, fiber length L=70μm, and diameter df=0.5μm.", "answer": "For β-SiC whiskers, the average stress σf = (τy × L) / df = (σy × L) / (2 × df) = (76 × 70×10^-6) / (2 × 0.5×10^-6) = 5320MPa." }, { "idx": 2288, "question": "A contact device is made of silver-tungsten composite material. Its production process involves first creating a porous tungsten powder metallurgy blank, and then infiltrating pure silver into the pores. Before silver infiltration, the density of the tungsten compact is 14.5 g/cm³. Calculate the mass fraction of silver in the blank after infiltration. The densities of pure tungsten and pure silver are 19.3 g/cm³ and 10.49 g/cm³, respectively. Assume the tungsten blank is very thin and entirely composed of open pores.", "answer": "The mass fraction of silver after infiltration is wAg = (0.25 × 10.49) / (0.25 × 10.49 + 0.75 × 19.3) = 15.4%" }, { "idx": 2289, "question": "A contact device is made of silver-tungsten composite material. Its production process involves first creating a porous tungsten powder metallurgy blank, then infiltrating pure silver into the pores. Before silver infiltration, the density of the tungsten compact is 14.5 g/cm³. Calculate the volume fraction of the pores. The density of pure tungsten is known to be 19.3 g/cm³. Assume the tungsten blank is very thin and all pores are open.", "answer": "The pore density ρ_pore is obviously zero. Therefore, φ_pore = 1 - φ_tungsten = 1 - (ρc / ρ_tungsten) = 1 - (14.5 / 19.3) = 0.25" }, { "idx": 2290, "question": "In randomly oriented short fiber composites, what is the relationship between the value of $\\\\frac{L_{\\\\mathrm{c}}}{L}$ and the number of pulled-out fibers and fractured fibers on the composite fracture surface? Please explain.", "answer": "If the fiber strength variability is not significant, for a single ideal straight short fiber, the probability of fracture occurring at any location is equal. In the composite, for a segment of the composite with the same fiber length, the probability of fracture occurring at any location is also equal. The probability that the fracture occurs at a distance less than $L_{\\\\mathrm{c}}/2$ from the fiber end should be $L_{\\\\mathrm{c}}/L$ (where $L$ is the fiber length). When the fracture occurs at a distance less than $L_{\\\\mathrm{c}}/2$ from the fiber end, the fiber will not fracture but will instead be pulled out from the matrix. This probability is the same for any fiber. Therefore, $L_{\\\\mathrm{c}}/L$ represents the probability of fiber pull-out on the fracture surface of the short fiber composite, and $1-L_{\\\\mathrm{c}}/L$ represents the probability of fiber fracture on the fracture surface, meaning the number of fractured fibers is the total number of fibers on the fracture surface multiplied by $(1-L_{\\\\mathrm{c}}/L)$." }, { "idx": 2291, "question": "What is the role and mechanism of particle (granular) reinforcement?", "answer": "Particle reinforcement: Hard particles with a diameter of 1~50 μm are added to the matrix. The particles can bear part of the load, but the matrix bears the main load. The particles restrict the deformation of the matrix through mechanical constraint. Properly sized and uniformly distributed particles can effectively strengthen the material." }, { "idx": 2292, "question": "What role does the matrix play in composite materials?", "answer": "The matrix primarily serves to fix and adhere the reinforcement, transferring the applied load to the reinforcement through the interface while also bearing a small portion of the load itself. The matrix acts like a separator, keeping the reinforcements apart. When some reinforcements are damaged or fractured, cracks are prevented from propagating from one reinforcement to another. During the processing and use of composite materials, the matrix also protects the reinforcements from chemical interactions with the environment and physical damage." }, { "idx": 2293, "question": "Briefly describe the classification of different reinforcement particle sizes in composite materials and their strengthening mechanisms.", "answer": "From the perspective of structural units and scales in composite materials, reinforcement particles with sizes of 1~50 μm are called particle-reinforced composites, those with sizes of 0.01~0.1 μm are called dispersion-strengthened composites, and those at submicron to nanometer scales are called fine composites, each with distinct strengthening mechanisms." }, { "idx": 2294, "question": "What role does the interface play in composite materials?", "answer": "The interface serves to coordinate the deformation of the matrix and the reinforcement. Through the interface, the stress from the matrix can be transferred to the reinforcement. The matrix and reinforcement are bonded via the interface, but the bonding strength must be appropriate—neither too strong nor too weak. Excessive bonding strength can reduce the toughness of the composite material, while insufficient bonding strength fails to transfer stress effectively, making the interface prone to cracking." }, { "idx": 2295, "question": "In a glass fiber reinforced nylon composite, the volume fraction of E-glass fiber is 0.3. The elastic moduli of E-glass and nylon are 72.4 GPa and 2.76 GPa, respectively. Calculate the fraction of the load borne by the fibers in the composite.", "answer": "If the interface bonding is good, the strain of the composite under force should be equal to that of the matrix and the fibers, i.e., $$ \\\\varepsilon_{\\\\mathrm{c}}={\\\\frac{\\\\sigma_{\\\\mathrm{c}}}{E_{\\\\mathrm{c}}}}=\\\\varepsilon_{\\\\mathrm{m}}={\\\\frac{\\\\sigma_{\\\\mathrm{m}}}{E_{\\\\mathrm{m}}}}=\\\\varepsilon_{\\\\mathrm{f}}={\\\\frac{\\\\sigma_{\\\\mathrm{f}}}{E_{\\\\mathrm{f}}}} $$ Thus, $$ \\\\frac{\\\\sigma_{\\\\mathrm{f}}}{\\\\sigma_{\\\\mathrm{m}}}=\\\\frac{E_{\\\\mathrm{f}}}{E_{\\\\mathrm{m}}}=\\\\varepsilon_{\\\\mathrm{f}}=\\\\frac{72.4~\\\\mathrm{GPa}}{2.76~\\\\mathrm{GPa}}=26.25 $$ The fraction of the load borne by the fibers is $$ \\\\varphi=\\\\frac{\\\\sigma_{\\\\mathrm{f}}}{\\\\sigma_{\\\\mathrm{f}}+\\\\sigma_{\\\\mathrm{m}}}=\\\\frac{1}{1+\\\\frac{\\\\sigma_{\\\\mathrm{m}}}{\\\\sigma_{\\\\mathrm{f}}}}=\\\\frac{1}{1+\\\\frac{1}{26.25}}=0.96 $$" }, { "idx": 2296, "question": "What is the role and mechanism of dispersion strengthening?", "answer": "Dispersion strengthening: mainly applied to metal matrices, by adding hard particles such as Al2O3, TiC, SiC, etc., with particle sizes around 0.01~0.1μm. These particles dispersed in the metal or alloy can effectively hinder the movement of dislocations, providing significant strengthening effects, but the matrix remains the primary load-bearing component." }, { "idx": 2297, "question": "Given a hard alloy containing WC, TiC, TaC, and Co with mass fractions of 0.25, 0.15, 0.05, and 0.05, respectively. Their densities are $15.77\\\\mathrm{g/cm^{3}}, 4.94\\\\mathrm{g/cm^{3}}, 14.5\\\\mathrm{g/cm^{3}}, 8.9\\\\mathrm{g/cm^{3}}$. Calculate the density of this composite material.", "answer": "The density can be calculated using $\\\\rho_{\\\\mathrm{c}}=\\\\rho_{\\\\mathrm{wc}}\\\\varphi_{\\\\mathrm{wc}}+\\\\rho_{\\\\mathrm{Tic}}\\\\varphi_{\\\\mathrm{Tic}}+\\\\rho_{\\\\mathrm{Tac}}\\\\varphi_{\\\\mathrm{TaC}}+\\\\rho_{\\\\mathrm{co}}\\\\varphi_{\\\\mathrm{co}}$. Convert the mass fractions to volume fractions, i.e., $$ $$ \\\\begin{array}{l}{\\\\displaystyle{\\\\varphi_{\\\\mathrm{\\\\tiny~\\\\mathrm{{TC}}}}=\\\\frac{\\\\frac{15}{4.94}}{8.70}=0.349}}\\\\ {~}\\\\ {\\\\displaystyle{\\\\varphi_{\\\\mathrm{\\\\tiny~\\\\mathrm{{Tac}}}}=\\\\frac{\\\\frac{5}{14.5}}{8.70}=0.040}}\\\\ {~}\\\\ {\\\\displaystyle{\\\\varphi_{\\\\mathrm{\\\\tiny~{co}}}=\\\\frac{\\\\frac{5}{8.90}}{8.70}=0.064}}\\\\end{array}$$ $$ \\\\rho\\\\mathrm{c}=\\\\sum\\\\rho_{i}\\\\varphi_{i}=11.5~\\\\mathrm{g/cm}^{3}$$ \\\\varphi\\\\mathrm{wc}={\\\\frac{\\\\frac{25}{15.77}}{{\\\\frac{25}{15.77}}+{\\\\frac{15}{4.94}}+{\\\\frac{5}{14.5}}+{\\\\frac{5}{8.9}}}}={\\\\frac{4.76}{8.70}}=0.547$$" }, { "idx": 2298, "question": "What role does the reinforcement play in composite materials?", "answer": "From the perspective that the reinforcement in structural composites mainly bears the load, it is generally required that the reinforcement has high strength and high modulus. The volume fraction of the reinforcement and its bonding performance with the matrix greatly influence the properties of the composite material. The combined action of the reinforcement, matrix, and interface can alter the toughness, fatigue resistance, creep resistance, impact resistance, and other properties of the composite material." }, { "idx": 2299, "question": "Briefly describe the relationship between the size of composite material specimens and their strength.", "answer": "The larger the composite material specimen, the higher the probability of defects, and the lower the strength." }, { "idx": 2300, "question": "Briefly describe in which aspects the size effect of fiber-reinforced composites is manifested.", "answer": "The fiber failure probability F(σ)=1-exp(-α lσ^β) and the average fiber strength σ̄=(αl)^(-1/β)𝒯(1+1/β) are both related to the fiber length l. The performance of fiber-reinforced composites is not only related to the fiber length but also to the fiber aspect ratio l/d, as well as the thickness of the composite plate. These are all manifestations of the size effect in composites." }, { "idx": 2301, "question": "What is the role and mechanism of fiber reinforcement?", "answer": "Fiber reinforcement: 1) Continuous fiber reinforcement can be explained by the rule of mixtures, where the load and modulus are primarily determined by the fibers. Since the strength and modulus of the fibers are much higher than those of the matrix and exceed the critical fiber volume fraction, they provide reinforcement, and the interfacial bonding should be moderate. 2) In short fiber and whisker reinforced composites, the fiber length should be greater than the critical length, or the aspect ratio should exceed the critical value. The fibers are the main contributors to strength and modulus, as their strength and modulus are much higher than those of the matrix, and the interfacial bonding should be moderate." }, { "idx": 2302, "question": "What are the requirements for forming a composite, and can any two materials form a composite material after combination? What are the requirements for interfacial bonding strength?", "answer": "To form a composite material, the two materials must establish a certain bonding strength at the interface. The interfacial bonding strength can generally be divided into physical bonding strength and chemical bonding strength." }, { "idx": 2303, "question": "Explain the significance of the critical volume fraction", "answer": "The significance of the critical volume fraction: when the fiber volume fraction is greater than the critical volume fraction, the strength of the composite material is higher than that of the matrix. When the fiber volume fraction is less than the critical volume fraction, the strength of the composite material is lower than that of the matrix, and no reinforcement effect is achieved." }, { "idx": 2304, "question": "Compare the critical volume fractions of continuous fiber and short fiber reinforced composites", "answer": "Continuous fiber: φ_Cr=(σ_mu−σ_m∗)/(σ_fu−σ_m), where σ_mu and σ_fu are the tensile strengths of the matrix and fiber, respectively; σ_m∗ is the stress borne by the matrix when the fiber reaches its fracture strain. Short fiber: φ_cr=(σ_mu−σ_m∗)/(σ_f−σ_m∗), where σ_f is the average stress of the fiber. Since σ_f is less than σ_fu, the critical volume fraction of short fibers is greater than that of continuous fibers." }, { "idx": 2305, "question": "Explain why in short-fiber composites with $LL_c" }, { "idx": 2312, "question": "What are the requirements for compounding? Can any two materials form a composite material after compounding? What are the requirements for dissolution and wetting bonding?", "answer": "For dissolution and wetting bonding, the matrix must wet the reinforcement, and mutual diffusion and dissolution should occur to form a bond. For reaction bonding, the matrix and reinforcement should react to form favorable interfacial products, whose thickness must be controlled below a critical value." }, { "idx": 2313, "question": "If unidirectional continuous fibers are unevenly distributed but all well-aligned in parallel, will it affect the elastic modulus? Please explain.", "answer": "No effect. The modulus of the composite material $\\\\boldsymbol{E}_{\\\\mathrm{{L}}}=\\\\varphi_{\\\\mathrm{{f}}}\\\\boldsymbol{E}_{\\\\mathrm{{f}}}+\\\\varphi_{\\\\mathrm{{m}}}\\\\boldsymbol{E}_{\\\\mathrm{{m}}}$ varies linearly with the fiber volume percentage. Areas with dense fiber arrangement have a higher elastic modulus, while areas with sparse fiber arrangement have a lower elastic modulus. However, the linear relationship is additive, and the overall average modulus still equals the elastic modulus of a composite material with uniformly arranged fibers of the same volume fraction, meaning there is no effect." }, { "idx": 2314, "question": "In what aspects should matrix modification be considered to improve interface bonding?", "answer": "Matrix modification" }, { "idx": 2315, "question": "When unidirectional continuous composites are subjected to longitudinal stress and the fibers fracture, what length will the fibers break into, and analyze why this type of fracture occurs.", "answer": "The fibers will break into short segments with lengths of $L_{\\\\mathrm{c}}$ or $2L_{\\\\mathrm{c}}$. According to the shear-lag theory analysis, when the short fiber length is $L_{\\\\mathrm{c}}$, the maximum normal stress in the fiber can reach the fiber fracture stress." }, { "idx": 2316, "question": "How to select reasonable composite processes and conditions for improving interface bonding?", "answer": "Select reasonable composite processes and conditions" }, { "idx": 2317, "question": "Derive the relational expression for φ_min in continuous fiber reinforced composites", "answer": "For continuous fiber reinforcement, from σ_Lu=σ_fuφ_f+σ_m*(1-φ_f)=σ_mu(1-φ_f), we can obtain φ_min=φ_f=(σ_mu-σ_m*)/(σ_fu+σ_mu-σ_m*)" }, { "idx": 2318, "question": "What aspects should be considered for fiber surface modification to improve interfacial bonding?", "answer": "Fiber surface modification" }, { "idx": 2319, "question": "In a Kevlar fiber-epoxy resin composite, the fiber volume fraction is 0.3, the density of epoxy resin is 1.25 g/cm³, and the density of Kevlar fiber is 1.44 g/cm³. Calculate the density of this composite material.", "answer": "The density is ρ=1.31 g/cm³. Calculation process: ρ_L=1.25×0.7+1.44×0.3=1.3078 g/cm³." }, { "idx": 2320, "question": "Nickel is alloyed with Th having ${\\\\boldsymbol{w}}_{\\\\mathrm{Th}}$ of 0.01 to form an alloy, which is then made into powder, pressed into the desired shape, and sintered into the final product, during which Th is completely oxidized. Calculate the volume fraction of $\\\\mathrm{ThO}_{2}$ produced in this Th-Ni material. Given the density of $\\\\mathrm{ThO}_{2}$ is $9.86~\\\\mathrm{g/cm^{3}}$, Ni is $8.98~\\\\mathrm{g/cm}^{3}$, and Th is $11.72~\\\\mathrm{g/cm}^{3}$.", "answer": "The volume fraction of $\\\\mathrm{ThO}_{2}$ is $0.0084 (0.84\\\\%)$." }, { "idx": 2321, "question": "In a Kevlar fiber-epoxy resin composite, the fiber volume fraction is 0.3, the elastic modulus of the epoxy resin is 31 GPa, and the elastic modulus of the Kevlar fiber is 124 GPa. Calculate the modulus parallel to the fiber direction.", "answer": "The modulus in the fiber direction is 58.9 GPa. Calculation process: E_T=0.3×124+0.7×31=58.9 GPa." }, { "idx": 2322, "question": "In a Kevlar fiber-epoxy resin composite, the fiber volume fraction is 0.3, the elastic modulus of the epoxy resin is 31 GPa, and the elastic modulus of the Kevlar fiber is 124 GPa. Calculate the modulus perpendicular to the fiber direction.", "answer": "The modulus perpendicular to the fiber direction is 40 GPa. Calculation process: 1/E_T=0.3/124+0.7/31=0.025, E_T=40 GPa." }, { "idx": 2323, "question": "If manufacturing a continuous aluminum fiber cable with epoxy resin as the matrix and a volume fraction of 0.30, predict the electrical conductivity of the cable. The electrical conductivity of aluminum is $3.8\\\\times10^{7}~\\\\mathrm{S/m}$, and the electrical conductivity of epoxy resin is $10^{-11}~\\\\mathrm{S/m}$.", "answer": "$1.14\\\\times10^{5}~\\\\mathrm{{S/m}}$." }, { "idx": 2324, "question": "Why are most covalent solids low-density materials?", "answer": "Most covalent solids have low packing density because the interatomic bonds are constrained by bond saturation and directionality." }, { "idx": 2325, "question": "According to solidification theory, how does adding a nucleating agent refine the grains of a casting?", "answer": "After adding a nucleating agent, it can promote heterogeneous nucleation in the undercooled liquid. This not only increases the substrates required for heterogeneous nucleation but also reduces the critical nucleus radius, both of which will increase the number of nuclei, thereby refining the grains." }, { "idx": 2326, "question": "In an aluminum specimen, the dislocation density within the grains was measured to be 2×10^12 m^−2. Assuming all dislocations are edge dislocations and are entirely concentrated at subgrain boundaries (i.e., low-angle grain boundaries), with their Burgers vector b=a/2⟨110⟩. If the cross-sections of the subgrain boundaries are all regular hexagons, and the tilt angle between subgrains is 2°, calculate the dislocation spacing at the subgrain boundary (given the lattice constant of aluminum a=0.404 nm).", "answer": "The dislocation spacing D is calculated by the formula D = b / θ, where b is the Burgers vector and θ is the tilt angle. The Burgers vector b = (√2/2) × 0.404 nm = 0.286 nm. The tilt angle θ = 2° = 2 × (3.14/180) rad = 0.0349 rad. Therefore, D = 0.286 nm / 0.0349 = 8.188 nm." }, { "idx": 2327, "question": "In an aluminum specimen, the dislocation density within the grains was measured to be 2×10^12 m^−2. Assuming all dislocations are edge dislocations and are entirely concentrated at subgrain boundaries (i.e., low-angle grain boundaries), with their Burgers vector b=a/2⟨110⟩. If the cross-sections of the subgrain boundaries are all regular hexagons, and the tilt angle between subgrains is 2°, calculate the number of subgrains per square meter (given the lattice constant of aluminum a=0.404 nm).", "answer": "The number of dislocations in each regular hexagon N = 3l / D = 3 × 7.05×10^−5 m / 8.188×10^−10 m = 0.26×10^7. The number of subgrains per square meter n = ρ / N = 2×10^12 / 0.26×10^7 = 7.69×10^7 m^−2." }, { "idx": 2328, "question": "In an aluminum specimen, the dislocation density within the grain interior was measured to be 2×10^12 m^−2. Assuming all dislocations are edge dislocations and are entirely concentrated on subgrain boundaries (i.e., low-angle grain boundaries), with their Burgers vector b=a/2⟨110⟩. If the cross-sections of the subgrain boundaries are all regular hexagons, and the tilt angle between subgrains is 2°, determine the side length of each regular hexagon (given the lattice constant of aluminum a=0.404 nm).", "answer": "Let the side length of the regular hexagon be l, the area of each regular hexagon S = (3√3/2) l^2. The dislocation density ρ = 2×10^12 m^−2, the number of dislocations N = ρ × S. The dislocation spacing D = 8.188 nm, N = 3l / D. Combining the equations yields l = 3 / (8.188×10^−10 × 2×10^12 × 1.5√3) = 7.05×10^−5 m." }, { "idx": 2329, "question": "Why are metals high-density materials?", "answer": "The bonding in metal crystals has neither directionality nor saturation requirements, so atoms are packed as closely as possible." }, { "idx": 2330, "question": "Both silver and aluminum have face-centered cubic lattices, and their atomic radii are very close, $r_{\\\\text{Ag}}=0.288\\\\mathrm{~nm}$ $r_{\\\\text{Al}}=0.286\\\\mathrm{nm}$, but they cannot form infinite solid solutions in the solid state. Explain the reason. (Northwestern Polytechnical University postgraduate entrance exam question)", "answer": "For substitutional solid solutions, the solute and solvent having the same crystal structure type and similar atomic radii are necessary conditions for forming infinite solid solutions. However, the $15\\\\%$ rule indicates that when $|\\\\delta|>15\\\\%$, the solid solubility (mole fraction) becomes very small, showing that when the size factor is within a favorable range for forming infinite solid solutions, its importance becomes secondary, meaning the extent of solid solubility depends on other factors. Here, the valence factor is crucial. Because the valence of $\\\\text{Ag}$ is 1, while that of Al is 3, i.e., the solid solubility of a high-valence element as a solute in a low-valence element is greater than that of a low-valence element in a high-valence element. Therefore, silver and aluminum cannot form infinite solid solutions in the solid state." }, { "idx": 2331, "question": "According to solidification theory, how does increasing the undercooling ΔT refine the grain size of castings?", "answer": "Increase the undercooling ΔT. As ΔT increases, both N and Vs increase, but the growth rate of N is greater than that of Vs. Therefore, the value of N/Vs increases, meaning z increases." }, { "idx": 2332, "question": "According to solidification theory, how does vibration crystallization refine the grain size of castings?", "answer": "Vibration crystallization, on one hand, provides the energy required for nucleation, and on the other hand, can break the growing crystals to form more crystallization nuclei, thereby refining the grains." }, { "idx": 2333, "question": "Why are ionic solids medium-density materials?", "answer": "The structural unit of an ionic crystal is an electrically neutral molecule composed of a set of positive and negative ions. The stacking of ions in ionic crystals must consider the role of electrical charge. In actual ionic crystals, the radii of positive and negative ions differ significantly. Generally, the negative ions are stacked in a cubic or hexagonal close-packed arrangement, while the smaller positive ions fill the gaps in this close-packed structure. This filling method of positive ions can uniformly separate the negative ions, increasing the distance between them so they no longer contact each other. Such a stacking structure can achieve higher packing density than covalently bonded crystals while satisfying the requirement of alternating arrangement of oppositely charged ions." }, { "idx": 2334, "question": "What type of solid solution is formed when hydrogen atoms dissolve in aluminum?", "answer": "The crystal structure of aluminum is fcc. From the appendix, the radius of the hydrogen atom is found to be rH=0.046nm, and the radius of the aluminum atom is rAl=0.143nm. The radius ratio rH/rAl=0.046/0.143=0.32." }, { "idx": 2335, "question": "Given that the recrystallization activation energy of a Cu-Zn alloy (w_Zn=0.30) is 250 kJ/mol, and this alloy takes 1 hour to complete recrystallization at a constant temperature of 400°C, how many hours does it take for this alloy to complete recrystallization at a constant temperature of 390°C? (Shanghai Jiao Tong University postgraduate entrance exam question)", "answer": "Given T_1=400°C=673 K, t_1=1 h, T_2=390°C=663 K, Q=2.5×10^5 J/mol. Let the time required to complete recrystallization at 390°C be t_2, then from A e^(-Q/(R T_1 )) t_1=A e^(-Q/(R T_2 )) t_2, we get t_1/t_2=e^[-Q/R (1/T_2 -1/T_1 )]=e^[-2.5×10^5/8.31×(1/663-1/673)]=0.509. Therefore, t_2=1.96 h." }, { "idx": 2336, "question": "Assuming the lattice constant of the alloy solid solution changes linearly with the increase of Zn atoms, and given the atomic radii of Cu and Zn are r_Cu=0.128nm and r_Zn=0.133nm respectively, find the average atomic radius r of the alloy with x_Zn=3%.", "answer": "The average atomic radius r is r = r_Cu + (r_Zn - r_Cu) × 3% = 0.128 + (0.133 - 0.128) × 0.03 nm = 0.1282 nm" }, { "idx": 2337, "question": "Given the diffusion constant of carbon in γ-Fe, D₀ = 2.0 × 10⁻⁵ m²·s⁻¹, and the activation energy for diffusion, Q = 1.4 × 10⁵ J/mol (R = 8.31 J/(mol·K)). Under the condition of a carbon potential Cₚ = 1.1% C, carburization of 20# steel is performed at 880°C. To achieve the same effect as carburizing at 927°C for 5 hours, how long should the carburization time be? (Huazhong University of Science and Technology postgraduate entrance exam question)", "answer": "T₁ = 927°C = 1200 K. From D = D₀ exp(−Q/RT), we have D₁ = D₁₂₀₀ = 2.0 × 10⁻⁵ exp(−1.4 × 10⁵ / (8.31 × 1200)) = 1.6 × 10⁻¹¹ m²·s⁻¹. T₂ = 880°C = 1153 K. D₂ = D₁₁₅₃ = 2.0 × 10⁻⁵ exp(−1.4 × 10⁵ / (8.31 × 1153)) = 0.902 × 10⁻¹¹ m²·s⁻¹. Therefore, tₓ = (1.6 × 10⁻¹¹ × 5) / (0.902 × 10⁻¹¹) = 9 h." }, { "idx": 2338, "question": "Calculate the packing density of a crystal structure with a coordination number of 6 for a monoatomic substance", "answer": "When the coordination number of a monoatomic substance is 6, it forms a simple cubic structure, where the atomic radius r1=a1×7. The packing density η=(1×(4/3)πr3)/a3=(4πa3)/(3×8×a3)=0.523" }, { "idx": 2339, "question": "Where should hydrogen atoms be located in the aluminum lattice? Why?", "answer": "The octahedral interstice is larger than the tetrahedral interstice, so hydrogen atoms reside in the octahedral interstice." }, { "idx": 2340, "question": "Calculate the ratio of the atomic radius when the coordination number of the elemental atom is 6 to that when the coordination number is 12.", "answer": "Since the molar volumes are equal for coordination numbers 6 and 12, according to the packing density relation: η12Vmol=(4NAπr123)/3, η6Vmol=(4NAπr63)/3. Dividing the two equations gives: r6/r12=(η6/η12)1/3=(0.52/0.74)1/3=0.89" }, { "idx": 2341, "question": "The measured lattice constant a=0.3795nm and density of 14.213g/cm^3 for the Cu-40%Au solid solution. Calculate and explain what type of solid solution this alloy is. The relative atomic mass of Au is found to be 196.97 from the appendix.", "answer": "The average weight per atom A in the Cu-40%Au solid solution is (63.55×0.6+196.97×0.4)/(6.0238×10^23)g=1.941×10^-22g. The number of atoms per unit cell n is calculated as a^3×ρ/A=(0.3795×10^-7)^3×14.213/1.941×10^-22=4.002. Both Cu and Au have a face-centered cubic structure, with 4 atoms per unit cell. The calculation yields 4.002 atoms per unit cell, where the decimal is due to calculation and measurement errors, thus this is a substitutional solid solution." }, { "idx": 2342, "question": "The density of copper is 8.96g/cm^3, calculate the lattice constant and atomic radius of copper. The relative atomic mass of copper is found to be 63.55 from the appendix, and the Avogadro constant is 6.0238×10^23.", "answer": "The mass of each Cu atom is A_Cu=63.55/(6.0238×10^23)g=10.55×10^-23g. Cu has an fcc structure, with each unit cell containing 4 atoms. Let the lattice constant of the unit cell be a, the mass of one unit cell is the mass of 4 Cu atoms, so a^3×ρ=4×A_Cu, i.e., a=(4×A_Cu/ρ)^1/3=(4×10.55×10^-23/8.96)^1/3cm=0.3611×10^-7cm=0.3611nm. The relationship between the atomic radius r and the lattice constant a in an fcc structure is r=a×√2/4, so the atomic radius of Cu is r_Cu=0.3611×√2/4nm=0.1277nm." }, { "idx": 2343, "question": "Given that the alloy is a face-centered cubic solid solution with an average atomic radius r=0.1282nm, find the average lattice constant a", "answer": "The average lattice constant a is a = 4r / √2 = 4 × 0.1282 / √2 nm = 0.3625 nm" }, { "idx": 2344, "question": "Given the relative atomic masses of Cu and Zn are 63.55 and 65.38 respectively, the Avogadro constant is 6.0238×10^23, and x_Zn=3%, find the average mass of component atoms A in the alloy solid solution.", "answer": "The average mass of component atoms A is A = (63.55 × 0.97 + 65.38 × 0.03) / (6.0238 × 10^23) g = 10.56 × 10^-23 g" }, { "idx": 2345, "question": "Given the lattice constant a=0.3625nm of a face-centered cubic solid solution and the average atomic mass A=10.56×10^-23g of the component atoms, calculate the theoretical density ρ of the alloy.", "answer": "The theoretical density ρ is ρ = 4A / a^3 = 4 × 10.56 × 10^-23 / (0.3625 × 10^-7)^3 g·cm^-3 = 8.87 g·cm^-3" }, { "idx": 2346, "question": "Based on the above calculation results, explain the relationship between the atomic radius and volume change when pure iron transforms from bcc to fcc structure at 17°C.", "answer": "From the calculated results above, it can be seen that if the atomic radius remains unchanged before and after the transformation, the volume change after the transformation would be very large, which is inconsistent with the actual measured results and also contradicts the nature of metallic bonds. Therefore, for the same metal, the atomic radius of different structures should change to minimize the volume change as much as possible." }, { "idx": 2347, "question": "Pure iron transforms from bcc structure to fcc structure at 17°C, with a volume reduction of 1.06%. Calculate the relative change in atomic radius.", "answer": "The relative change in atomic radius is (r_f - r_b)/r_b = 1 - 1/0.9758 = -2.47%." }, { "idx": 2348, "question": "Assuming the atomic radius remains unchanged before and after the transformation, calculate the volume change of pure iron when it transforms from bcc to fcc structure at 17°C.", "answer": "Assuming the atomic radius remains unchanged before and after the transformation, the volume change after the transformation is (a_f^3 - 2a_b^3)/(2a_b^3) = ((4r_f/√2)^3 - 2(4r_b/√3)^3)/(2(4r_b/√3)^3) = ((4/√2)^3 - 2(4/√3)^3)/(2(4/√3)^3) = -8.196%." }, { "idx": 2349, "question": "Calculate the theoretical density of CsCl", "answer": "The experimentally measured density is 3.99 g/cm³." }, { "idx": 2350, "question": "Pure iron transforms from bcc structure to fcc structure at 17°C, with a volume reduction of 1.06%. Calculate the atomic radius of the bcc structure based on the atomic radius of the fcc structure.", "answer": "Let the lattice constant of the bcc structure be a_b, and the lattice constant of the fcc structure be a_f. When transforming from bcc to fcc, the volume reduces by 1.06%: since each bcc unit cell contains 2 atoms, and the fcc unit cell contains 4 atoms, 2 bcc unit cells transform into 1 fcc unit cell. Therefore, (a_f^3 - 2a_b^3)/(2a_b^3) = -1.06/100, solving for a_f = (2×(100 - 1.06)/100)^(1/3)×a_b = 1.255a_b. The atomic radius of the bcc structure r_b = √3a_b/4, and the atomic radius of the fcc structure r_f = √2a_f/4. Substituting a_f = 1.255a_b, we get r_b = (√3a_b)/4 = (√3a_f)/(4×1.255) = (√3×4r_f)/(4×1.255×√2) = 0.9758r_f." }, { "idx": 2351, "question": "Iron carbide Fe3C is an orthorhombic intermetallic compound with lattice constants of a=0.4514nm, b=0.5080nm, c=0.6734nm, and a density of 7.65g/cm³. Determine the number of iron and carbon atoms in the unit cell.", "answer": "From the appendix, the atomic masses of Fe and C are found to be 55.85g/mol and 12g/mol, respectively. The mass of each Fe and C atom is A_Fe = [55.85/(6.023×10^23)]g = 9.273×10^-23g, A_C = [12/(6.023×10^23)]g = 1.992×10^-23g. The atomic ratio of Fe to C is 3:1. Assuming there are n C atoms in the unit cell, there should be 3n Fe atoms. The density of Fe3C is ρ_Fe3C = (n × A_Fe + 3n × A_C) / V_unit cell. Therefore, 7.65 = [(9.273 × 3n + 1.992n) × 10^-23] / [0.4514 × 10^-7 × 0.5080 × 10^-7 × 0.6734 × 10^-7] = [29.811 × 10^-23 × n] / [1.544 × 10^-22]. Solving for n gives n = [7.65 × 1.544 × 10^-22] / [29.811 × 10^-23] = 3.97. Thus, there are 4 C atoms and 3 × 4 = 12 Fe atoms in the unit cell." }, { "idx": 2352, "question": "Predict the crystal structure of CsCl", "answer": "From the appendix, the electronegativities of Cs and Cl are found to be 0.82 and 3.16, respectively: the ionic radii are r(Cs+)=0.165 nm, r(Cl-)=0.181 nm. The electronegativity difference ΔEN=3.16-0.82=2.34, the ionic bond character is 75%, the covalent character is 25%, primarily ionic bonding. The ratio of anion radius to cation radius r(Cs+)/r(Cl-)=0.165/0.181=0.912. From the data given in Table 3-5 in the text, the cation should occupy the interstitial site at the center of a cube formed by the anions, corresponding to CN(Cl-)=8; the ratio of the number of anions to cations is 1:1, so CN(Cs+)=8. This crystal structure should be a simple cubic structure, with Cl- at the cube corners and Cs+ at the cube center." }, { "idx": 2353, "question": "Brass $(77,77)$ has a B2 structure, with a $z_{17}$ to $C11$ atomic ratio of 46:54. At $450-20=20$, if $900\\\\div\\\\cdots$ of the ($112$ 1/2, 1/2) positions are occupied by copper atoms, what percentage of the $(0,0,0)$ positions are occupied by copper atoms?", "answer": "The stoichiometric composition of $C112.77$ is $5(\\\\angle17)=500^{\\\\circ}$. In the stoichiometric alloy, the ($112$, $117$, $112$) positions are entirely occupied by copper atoms, and the $(0,0,0)$ positions are entirely occupied by zinc atoms. Currently, the ratio of $217$ to $Cu$ atoms is $46:54$, and at $450-40$, $90\\\\%$ of the ($112$, 1/2, 1/2) positions are occupied by copper atoms. Thus, in the alloy, $\\\\sum11$ atoms have $0.5\\\\times0.0=4500$ located at the (1/2, 1/2, 1/2) positions. The remaining $0.54-0.45=0.00$ are located at the (0, 0, 0) positions, meaning $0.09/0.5=180\\\\dot{0}$ of the (0, 0, 0) positions are occupied by $c_{11}$ atoms. This can also be verified in the opposite direction. From the calculation results above, it shows that in the alloy, $z u$ atoms have $0.5-0.45=500$ located at the ($112$ 1/2, 1/2) positions and $0.5\\\\times(1-1).18)=410.0$ located at the (0, 0, 0) positions. Therefore, the content of $z_{1}$ in the alloy should be $0.41+0.05=4600$, which matches the given alloy composition." }, { "idx": 2354, "question": "(2) Calculate the theoretical density of CsCl.", "answer": "First, calculate the lattice constant from the ionic radii. Since the crystal structure is simple cubic, the diagonal length of the cube should be the sum of the diameters of the negative and positive ions, i.e., equal to $2(0.181+0.165)\\\\mathrm{nm}=0.692$ nm. The diagonal length equals $\\\\sqrt{3}a$, so $a=0.692/\\\\sqrt{3}=0.3995$ nm. From the appendix, the relative atomic masses of Cl and Cs are $35.45$ and $132.91$, respectively. The mass of each Cl and Cs atom is $35.45/6.023\\\\times10^{23}=5.885\\\\times10^{-23}$ g and $132.91/6.023\\\\times10^{23}=2.207\\\\times10^{-22}$ g. A unit cell contains one Cl ion and one Cs ion, so the theoretical density of CsCl is $\\\\rho_{\\\\text{(CsCl)}}=\\\\frac{5.885\\\\times10^{-23}+2.207\\\\times10^{-22}}{(0.3995\\\\times10^{-7})^{3}}\\\\text{g/cm}^{3}=4.38\\\\text{g/cm}^{3}$. The calculated theoretical density is slightly higher than the experimentally measured density, possibly because the estimated lattice constant is smaller than the actual value." }, { "idx": 2355, "question": "Calculate the packing density of the CsCl crystal structure (assuming ions are in contact along the body diagonal).", "answer": "The CsCl crystal has a simple cubic structure. The radii of the Cs⁺ and Cl⁻ ions are r_Cs⁺ = 0.165 nm and r_Cl⁻ = 0.181 nm, respectively. Assuming the ions are in contact along the body diagonal, the diagonal length d = 2(r_Cs⁺ + r_Cl⁻) = 2(0.165 + 0.181) nm = 0.692 nm. The lattice constant a = d/√3 = 0.692 nm/√3 = 0.3995 nm. A unit cell contains one Cs⁺ ion and one Cl⁻ ion, so the packing density η_CsCl of the CsCl crystal structure is η_CsCl = [4π(r_Cs⁺³ + r_Cl⁻³)/3] / a³ = [4π(0.165³ + 0.181³)] / (3 × 0.3995³) = 0.685." }, { "idx": 2356, "question": "Why do commercial oxides such as windows and beverage glass require the addition of CaO to SiO2?", "answer": "CaO cannot form a network in glass. Calcium ions have only one functionality, and Ca2+ has only two functionalities. When they are added to the SiO2 network and connected with oxygen ions, they inevitably break a primary bond of the network, reducing the density of primary bonds in the network and thereby lowering the glass transition temperature. For general wide-mouth bottles and window glass, which are used at low temperatures, they do not require an excessively high glass transition temperature Tg. A low Tg makes the product easier to handle and reduces costs." }, { "idx": 2357, "question": "Is the glass transition temperature of ordinary window glass above or below room temperature?", "answer": "The glass transition temperature of ordinary window glass should be above room temperature. If it were below room temperature, the glass would remain in a liquid state at room temperature, albeit with high viscosity and some fluidity, meaning the atoms would still be delocalized. Consequently, it could not maintain a fixed shape at room temperature." }, { "idx": 2358, "question": "What is the role of SiO2 in commercial oxide glasses?", "answer": "SiO2 oxide satisfies Zachariasen's rules and can form a large-scale three-dimensional glass network, acting as the network former in glass." }, { "idx": 2359, "question": "What is the relationship between functionality and the structural morphology of polymers?", "answer": "If the functionality of a polymer monomer equals 2, the monomer can only form long molecular chains without branches, resulting in a linear structure. If the functionality equals 3 or is greater than 3, during polymerization, one functionality of a trifunctional monomer provides covalent cross-linking with another molecular chain, forming a network, which is a three-dimensional structure. Such a polymer is essentially a single molecule with an extremely large molecular weight." }, { "idx": 2360, "question": "Why are the atomic masses of Si and Al very close (28.09 and 26.98, respectively), while the densities of SiO2 and Al2O3 differ significantly (2.55 g/cm³ and 3.95 g/cm³, respectively)? Explain this difference using crystal structure and Pauling's rules.", "answer": "The ionic radii of silicon, aluminum, and oxygen are r_Si4+=0.039 nm, r_Al3+=0.057 nm, and r_O2-=0.132 nm, respectively. The radius ratio of silicon to oxygen is 0.295, and that of aluminum to oxygen is 0.431. Therefore, silicon ions occupy the tetrahedral interstices of oxygen ions, while aluminum ions occupy the octahedral interstices of oxygen ions. Si is located at the center of an oxygen-coordinated tetrahedron with a coordination number (CN) of 4. Si is tetravalent, so the electrostatic bond strength S=4/4=1. O2- is divalent, i.e., Z-=2. According to Pauling's second rule, since Z-=∑S_i=2, the summation index i is 2, meaning each O2- at the tetrahedron vertex is shared by two tetrahedra. Thus, silicon-oxygen tetrahedra share all vertices, forming a three-dimensional framework structure of silicates. Because one oxygen ion is connected to only two silicon ions, such a low coordination number prevents SiO2 from achieving close packing, resulting in a generally open silicate structure. Al is located at the center of an oxygen-coordinated octahedron with a coordination number (CN) of 6. Al is trivalent, so the electrostatic bond strength S=3/6=0.5. O2- is divalent, i.e., Z-=2. According to Pauling's second rule, since Z-=∑S_i=2, the summation index i is 4, meaning each O2- at the octahedron vertex is shared by four octahedra. Because one oxygen ion is connected to four aluminum ions, close packing is possible. Oxygen ions are arranged in a hexagonal close-packed structure, with aluminum ions occupying the octahedral interstices. Since the number of atoms and octahedral interstices in a hexagonal close-packed structure are equal, 1/3 of the octahedral interstices remain unfilled. Because SiO2 has a relatively open structure and Al2O3 has a relatively close-packed structure, despite the very similar atomic masses of Si and Al, the densities of SiO2 and Al2O3 differ significantly." }, { "idx": 2361, "question": "How to compare the density difference between SiO2 and Al2O3 by calculating the packing density?", "answer": "The relative atomic mass of SiO2 is 16. For SiO2, the mass of each SiO2 molecule is A_SiO2=(28.09+2×16)/(6.023×10^23) g=9.977×10^-23 g. The number of SiO2 molecules per cm³ is n_SiO2=ρ_SiO2/A_SiO2=2.55/9.977×10^-23=2.556×10^22 per cm³. The number of Si4+ and O2- ions per cm³ are n_Si4+=2.556×10^22 per cm³ and n_O2-=5.112×10^22 per cm³, respectively. The packing density of SiO2 is η_SiO2=(n_Si4+×4πr_Si4+^3/3 + n_O2-×4πr_O2-^3/3)/1=(4π/3)(2.556×0.039^3 + 5.112×0.132^3)×10^22×10^-21=0.577. For Al2O3, the mass of each Al2O3 molecule is A_Al2O3=(2×26.98+3×16)/(6.023×10^23) g=1.693×10^-22 g. The number of Al2O3 molecules per cm³ is n_Al2O3=ρ_Al2O3/A_Al2O3=3.95/1.693×10^-22=2.330×10^22 per cm³. The number of Al3+ and O2- ions per cm³ are n_Al3+=4.660×10^22 per cm³ and n_O2-=6.990×10^22 per cm³, respectively. The packing density of Al2O3 is η_Al2O3=(n_Al3+×4πr_Al3+^3/3 + n_O2-×4πr_O2-^3/3)/1=(4π/3)(4.660×0.057^3 + 6.990×0.132^3)×10^22×10^-21=0.714. Since η_Al2O3=0.714 is greater than η_SiO2=0.577, the density of Al2O3 is greater than that of SiO2." }, { "idx": 2362, "question": "Is the glass transition temperature of rubber band polymers above or below room temperature?", "answer": "The glass transition temperature of rubber band polymers is below room temperature. This is because rubber band polymers exhibit significant elasticity at room temperature; if they were in a glassy state at room temperature, they would not display elasticity but rather brittleness." }, { "idx": 2363, "question": "Why do commercial oxides such as windows and beverage glass require the addition of Na2O to SiO2?", "answer": "Na2O cannot form a network in glass. Sodium ions have only one functionality, Na+ has only one functionality. When they are added to the SiO2 network and connected with oxygen ions, they inevitably break a primary bond of the network, reducing the density of primary bonds in the network and thereby lowering the glass transition temperature. For general wide-mouth bottles and window glass, which are used at low temperatures, they do not require an excessively high glass transition temperature Tg. A low Tg makes the products easier to handle and reduces costs." }, { "idx": 2364, "question": "The copolymer contains PVC (polyvinyl chloride — $T_{2}H_{2}C-1-$) and PVA (polyvinyl alcohol —CHO—). The mass fraction of PVC is $1\\\\Psi_{\\\\mathrm{W}_{\\\\mathrm{E}}}=\\\\mathrm{B},5\\\\Psi_{\\\\mathrm{W}}$, and the mass fraction of PVA is $14^{\\\\circ}=15^{\\\\circ}$. Determine the mole fractions of PVC and PVA.", "answer": "First, calculate the molar masses of PVC and PVA. The relative atomic mass of C is 12.0I, $\\\\mathsf{H}$ is 1.008, Cl is 35.45, and O is $16.00$. The molar mass of PVC is $(2\\\\times12+3\\\\times1+35.45)\\\\underline{{{\\\\mathrm{g}}}}/110|=62.45\\\\underline{{{\\\\mathrm{g}}}}/1101$. The molar mass of PVA is $(4\\\\times12+6\\\\times1+2\\\\times16)\\\\mathrm{g/mol}=86\\\\mathrm{g/mol}$. In $100\\\\underline{{\\\\underline{{\\\\mathbf{I}}}}}$ of the polymer, there are $85/62,45=1,36\\\\mathrm{PVC}$ moles and $15:80=0.174$ PVA moles. Therefore, the mole fraction of PVC in the polymer, Ac, and the mole fraction of PVC, $f^{\\\\prime}(x)$, are: $$ f_{\\\\mathrm{vc}}={\\\\frac{1.36}{1.36+0.174}}=0.887\\\\qquadf_{\\\\mathrm{vA}}={\\\\frac{0.174}{1.36+0.174}}=0.113$$" }, { "idx": 2365, "question": "A lightly cross-linked rubber band is stretched several times its original length and returns to its original length after the force is removed. If the stretched rubber band is placed below its glass transition temperature, what will happen to its length after the force is removed? Why?", "answer": "When the stretched rubber band is placed below its glass transition temperature, it transitions into a glassy state. The glass transition causes the atoms to change from delocalized to localized (solidification), so it cannot return to its original length after the external force is removed." }, { "idx": 2366, "question": "In the copolymer of PVC vinyl chloride $(C_{2}H_{3}C)$ and PVA vinyl alcohol $(C_{4}H_{5}C)_{2}$, the ratio of vinyl chloride to vinyl alcohol is 10:1. If the average molar mass of the copolymer molecular chain is $\\lvert5000\\rvert\\lvert\\bar{\\mathbf{g}}/17\\rvert\\lvert\\bar{\\mathbf{U}}\\rvert$, find its degree of polymerization (DP).", "answer": "First, calculate the molar mass of the repeating unit. In question 49, the molar masses of PVC and PVA were determined to be $\\bar{6}\\bar{2}.45\\bar{5}\\bar{5}^{\\prime}\\bar{1}710\\bar{1}$ and $86\\\\mathrm{g/mol}$, respectively. In this copolymer, the ratio of PVC to PVA is 10:1, so the molar mass of the repeating unit is $111$ $$\\\\begin{array}{l l}{{}}&{{M=\\\\displaystyle\\\\frac{10}{11}\\\\mathrm{PVC}\\\\mathrm{\\\\Ddot{H}^{\\\\sharp}J}\\\\frac{\\\\mathrm{s}\\\\mathrm{s}}{\\\\mathrm{1}}\\\\mathrm{s}^{\\\\prime}\\\\mathrm{H}\\\\overline{{{\\\\mathcal{T}}}}^{\\\\sharp}\\\\mathrm{IW}\\\\mathrm{E}\\\\mathrm{H}+\\\\displaystyle\\\\frac{1}{11}\\\\mathrm{PVA}\\\\mathrm{\\\\Ddot{H}^{\\\\sharp}}}}\\\\ {{}}&{{=\\\\displaystyle[\\\\frac{10}{11}\\\\mathrm{6}2.45+\\\\frac{1}{11}86]\\\\mathrm{g}/\\\\mathrm{mol}=\\\\mathrm{64.6}\\\\mathrm{g}/\\\\mathrm{mol}}}\\\\end{array}$$ The degree of polymerization (DP) of the copolymer is: DP = molar mass of the copolymer / molar mass of the repeating unit = 16000g/mol / 64.6g/mol = 248" }, { "idx": 2367, "question": "After washing and drying clothes containing artificial fibers, the clothes shrink. Please explain.", "answer": "The fiber polymers in clothes made of artificial fibers are stretched and have a certain crystalline state. When the clothes are washed, the stretched state is removed, causing the polymer chains to return to a curled state. This is because the value of the curled state is greater than that of the stretched state, resulting in the clothes shrinking." }, { "idx": 2368, "question": "A lightly cross-linked rubber band is stretched several times its original length. After removing the applied force, the rubber band returns to its original length. If the stretched rubber band is placed above its glass transition temperature, what will happen to the length of the rubber after the force is removed? Why?", "answer": "When the stretched rubber band is placed above its glass transition temperature, if the polymer does not crystallize during stretching, the molecular segments between cross-linking points are elongated. After removing the external force, the molecules return to their original relaxed network state, increasing configurational entropy and reducing free energy. If the polymer crystallizes after stretching and generates heat, the stretched state can be maintained if the generated heat is dissipated. Only if heat is provided to melt the crystals will the elongation recover." }, { "idx": 2369, "question": "Calculate the equilibrium vacancy concentration in a Cu crystal at |000|.", "answer": "The equilibrium vacancy concentration in metals is x_s = exp(ΔS_f / k_B) * exp(-ΔH_f / (k_B * T))." }, { "idx": 2370, "question": "A transparent PET film remains generally transparent when stretched slowly at room temperature. However, if stretched at $130^{\\circ}$, it becomes opaque. Why?", "answer": "The glass transition temperature of PET (polyethylene terephthalate) is $10-C$. Stretching at $130^{\\circ}$, which is above the glass transition temperature, allows molecular chains to flow more easily. External stretching can promote polymer crystallization, and crystallization reduces transparency, with opacity increasing as crystallinity rises. However, slow stretching at room temperature (below the glass transition temperature) makes crystallization difficult for the polymer, so it retains its transparency." }, { "idx": 2371, "question": "According to the $F E=F e=C$ phase diagram, calculate the relative amounts of phases in an iron-carbon alloy at room temperature under equilibrium conditions, where the mass fraction of carbon is $0.14$.", "answer": "At room temperature, the equilibrium phases of the iron-carbon alloy are $\\alpha$-Fe (the mass fraction of carbon is $0.008$) and $Fe_3C$ (the mass fraction of carbon is $6.67$). Therefore, for an alloy with a mass fraction of carbon of $0.14$ at room temperature under equilibrium conditions, the relative amounts (mass fractions) of the $\\alpha$-Fe phase and the $Fe_3C$ phase are: $$ A^{\\alpha}=\\frac{6.67-0.14}{6.67-0.008}=98.62\\% \\qquad A^{Fe_3C}=1-98.62\\%=1.38\\%$$" }, { "idx": 2372, "question": "According to the $FE=Fe=C$ phase diagram, calculate the relative amount of eutectoid (pearlite) in an iron-carbon alloy with a carbon mass fraction of $0.14$ at room temperature under equilibrium conditions.", "answer": "The microstructure of an alloy with a carbon mass fraction of $0.14$ at room temperature under equilibrium conditions consists of $\\\\alpha$-Fe and eutectoid (pearlite). The microstructure can be approximately considered the same as after the eutectoid transformation. At the eutectoid temperature, the carbon composition of $\\\\alpha$-Fe is $0.022$, and the carbon composition of the eutectoid structure is $0.76$. Therefore, the relative amount of eutectoid in the microstructure of an alloy with a carbon mass fraction of $0.14$ at room temperature is: $$ A^{P}=\\\\frac{0.14-0.022}{0.76-0.022}=10.58\\\\%$$" }, { "idx": 2373, "question": "According to the $FE=Fe=C$ phase diagram, calculate the percentage of cementite precipitated from all primary phases relative to the total system (the entire system) for an iron-carbon alloy with a carbon mass fraction of $3.6$ at the eutectoid temperature.", "answer": "For an alloy with a carbon mass fraction of $3.6$ just after solidification, the relative amount of all phases (including primary phases and phases in the eutectic) is: $$ A^{\\\\Psi}=\\\\frac{6.67-3.6}{6.67-2.14}=67.77\\\\%$$ During the cooling process, the relative amount of $Fe_3C$ precipitated from the primary phases relative to the total system is: $$ A^{tFe_3C}=67.77\\\\% \\\\times 23.35\\\\%=15.82\\\\%$$" }, { "idx": 2374, "question": "Assuming an ionic crystal has monovalent cations and anions, with a cation vacancy formation energy of 2000, an anion vacancy formation energy of 101-1'tra01, and a cation interstitial formation energy of 30k=1,17701, calculate the relative concentration of Schottky defects.", "answer": "The formation energy of a Schottky defect is the sum of the cation vacancy formation energy and the anion vacancy formation energy, which is (20+40) kJ/mol = 60 kJ/mol. The relative concentration of Schottky defects is exp(-60×10^3/RT). For example, at room temperature (300K), the above ratio equals 0.018." }, { "idx": 2375, "question": "According to the $F E=F e=C$ phase diagram, calculate the relative amounts of phases in an iron-carbon alloy with a carbon mass fraction of $3.6$ under equilibrium conditions at room temperature.", "answer": "At room temperature, the equilibrium phases of the iron-carbon alloy are $\\alpha$-Fe (carbon mass fraction is $0.008$) and $Fe_3C$ (carbon mass fraction is $6.67$). Therefore, for an alloy with a carbon mass fraction of $3.6$ under equilibrium conditions at room temperature, the relative amount (mass fraction) of the $\\alpha$-Fe phase and the relative amount of the $Fe_3C$ phase are: $$ A^{\\alpha}=\\frac{6.67-3.6}{6.67-0.008}=46.08\\% \\qquad A^{Fe_3C}=1-46.08\\%=53.92\\%$$" }, { "idx": 2376, "question": "Assuming an ionic crystal has monovalent cations and anions, with a cation vacancy formation energy of 2000, anion vacancy formation energy of 101-1'tra01, and cation interstitial formation energy of 30k=1,17701, calculate the relative concentration of Frenkel defects.", "answer": "The formation energy of Frenkel defects is the sum of the cation vacancy formation energy and the cation interstitial formation energy, which is (20+30) kJ/mol = 50 kJ/mol. The relative concentration of Frenkel defects is exp(-50×10^3/RT)." }, { "idx": 2377, "question": "According to the $FE=Fe=C$ phase diagram, calculate the percentage of the final transformed eutectoid in the eutectic structure relative to the total system for an iron-carbon alloy with a carbon mass fraction of $3.6$.", "answer": "The percentage of phases in the eutectic structure is: $$ A_{6}^{7}=\\\\frac{6.67-4.3}{6.67-2.14}=52.3\\\\%$$ Therefore, the percentage of the final transformed eutectoid in the eutectic structure is: $$ A_{G}^{P}=67.6\\\\% \\\\times 52.3\\\\% \\\\times (1-23.35\\\\%)=27.1\\\\%$$" }, { "idx": 2378, "question": "According to the $FE=Fe=C$ phase diagram, calculate the relative amount of eutectoid (pearlite) in an iron-carbon alloy with a carbon mass fraction of $1.20$ at room temperature under equilibrium conditions.", "answer": "The microstructure of the alloy with a carbon mass fraction of $1.20$ at room temperature under equilibrium conditions consists of $Fe_3C$ and eutectoid (pearlite). The relative amount of eutectoid in the microstructure at room temperature is: $$ A^{P}=\\\\frac{6.67-1.2}{6.67-0.76}=92.55\\\\%$$" }, { "idx": 2379, "question": "Can all parts of a dislocation loop be screw dislocations? Why?", "answer": "The Burgers vector of a screw dislocation is parallel to the dislocation line. A single dislocation has only one Burgers vector, and a dislocation loop cannot be parallel to one direction everywhere. Therefore, a dislocation loop cannot have all its parts as screw dislocations." }, { "idx": 2380, "question": "According to the $FE=Fe=C$ phase diagram, calculate the relative amounts of the primary phase (austenite) and eutectic in an iron-carbon alloy with a carbon mass fraction of $3.6$ just after solidification is completed.", "answer": "Just after solidification is completed, the carbon compositions of the primary phase and eutectic structure are $2.14$ and $4.3$ respectively. Therefore, the relative amounts of the primary phase and eutectic just after solidification are: $$ A^{\\\\gamma}=\\\\frac{4.3-3.6}{4.3-2.14}=32.4\\\\% \\\\qquad A^{0}=1-32.4\\\\%=67.6\\\\%$$" }, { "idx": 2381, "question": "MgO dissolves into Al2O3 to form a solid solution. Assuming Mg2+ replaces Al3+, write the reaction equation using Kroger-Vink notation.", "answer": "2MgO forms 2MgAl' + V..O + 2OOX through Al2O3" }, { "idx": 2382, "question": "MgO dissolves into Al2O3 to form a solid solution, assuming the formation of cation interstitials. Write the reaction equation using Kroger-Vink notation.", "answer": "3MgO forms 3Mg••i + 2V•Al + 3OO× through Al2O3" }, { "idx": 2383, "question": "Assuming a monovalent cation and anion in an ionic crystal, the cation vacancy formation energy is 2000, the anion vacancy formation energy is 101-1'tra01, the cation interstitial formation energy is 30k=1,17701, calculate the concentration ratio of Schottky defects to Frenkel defects.", "answer": "The concentration ratio of Schottky defects to Frenkel defects is exp(-60×10^3/RT)/exp(-50×10^3/RT)=exp(-10×10^3/RT). For example, at room temperature (300K), the above ratio equals 0.018." }, { "idx": 2384, "question": "At which temperature is the equilibrium vacancy concentration 10 times that at 1000K?", "answer": "The ratio of vacancy concentrations at two different temperatures T1 and T2 is x_r(T2)/x_r(T1) = exp[-ΔH_f / k_B * (1/T2 - 1/T1)]. Let T1 = 1000K, and the vacancy concentration at T2 be 10 times that at T1, then ln10 = -ΔH_f / k_B * (1/T2 - 1/1000). 1/T2 = (1/1000 - k_B * ln10 / ΔH_f) K^-1. From textbook Table 6-1, ΔH_f for Cu is 1.22 eV, and k_B = 8.61×10^-5 eV·K^-1. Substituting these values into the equation gives 1/T2 = (1/1000 - 8.61×10^-5 * ln10 / 1.22) K^-1 = 8.375×10^-4 K^-1. Therefore, T2 = 1144 K. That is, the vacancy concentration at 1144 K is 10 times that at 1000 K." }, { "idx": 2385, "question": "Can all parts of a dislocation loop be edge dislocations? Why?", "answer": "The Burgers vector of an edge dislocation is perpendicular to the dislocation line. If the Burgers vector is perpendicular to the plane of the dislocation loop, then every part of the dislocation loop is an edge dislocation. The slip plane of such a dislocation is the prismatic plane formed by the dislocation loop and the direction of the Burgers vector. This type of dislocation is also called a prismatic dislocation." }, { "idx": 2386, "question": "Explain how to use the method of measuring density to determine which reaction is correct when MgO dissolves into Al2O3 to form a solid solution", "answer": "By experimentally measuring the crystal lattice constants, calculate the theoretical densities under two defect scenarios and compare them with the measured density. First, calculate the number of ions in the unit cell for each type of defect based on the solid solution concentration, then use the lattice constants obtained from X-ray experiments to calculate the densities for both cases. The defect mode corresponding to the theoretical density closest to the measured density is the correct one." }, { "idx": 2387, "question": "In a simple cubic crystal, the Burgers vector of dislocation (2) is $b^{(2)}=a[0,10]$, and the tangent direction of the dislocation line is $l^{(2)}=[001]$. Identify the type of this dislocation and the slip plane of the dislocation. If the slip plane is not unique, explain the constraints on the slip plane.", "answer": "For dislocation (2), since the Burgers vector is perpendicular to the dislocation line, it is an edge dislocation. The slip plane is the plane that contains both the Burgers vector and the dislocation line, so the slip plane of dislocation (2) should be the (100) plane." }, { "idx": 2388, "question": "The vacancy formation energy is 75 kJ/mol, and the crystal is quenched from 1000K to room temperature (approximately 300K). The Burgers vector h of the edge dislocation is about 0.25 nm. What is the osmotic force experienced by the dislocation?", "answer": "When there is an unbalanced vacancy concentration, the chemical force per unit length on an edge dislocation is Fs=(kBT/h²)ln(x/x̄), where kB is the Boltzmann constant, T is the temperature, h is the Burgers vector, x is the actual vacancy concentration, and x̄ is the equilibrium vacancy concentration. The equilibrium vacancy concentration at different temperatures is x=exp(-Gf/kBT). Therefore, the vacancy concentrations at 1000K and 300K are exp(-Gf/1000kB) and exp(-Gf/300kB), respectively. The normal stress σs on the edge dislocation when the crystal is quenched from 1000K to 300K is σs=(kB300/h³)(Gf/R)(1/300-1/1000)=(300/(0.25×10⁻⁹)³)×(75000/6.02×10²³)(1/300-1/1000)Pa=5.43×10⁹Pa." }, { "idx": 2389, "question": "The vacancy formation energy is 75 kJ/mol, the crystal is quenched from 1000 K to room temperature (approximately 300 K), and the Burgers vector h of the edge dislocation is about 0.25 nm. Estimate whether the dislocation can climb.", "answer": "The climb stress σs acting on the dislocation is 5.43×10⁹ Pa, which is close to the theoretical shear strength of general metals, so the dislocation will climb." }, { "idx": 2390, "question": "According to the $FE=Fe=C$ phase diagram, calculate the relative amounts of phases in an iron-carbon alloy at room temperature under equilibrium conditions, where the mass fraction of carbon is $1.20$.", "answer": "At room temperature, the equilibrium phases of the iron-carbon alloy are $\\alpha$-Fe (with a mass fraction of carbon of $0.008$) and $Fe_3C$ (with a mass fraction of carbon of $6.67$). Therefore, for an alloy with a mass fraction of carbon of $1.20$, the relative amounts (mass fractions) of the $\\alpha$-Fe phase and the $Fe_3C$ phase at room temperature under equilibrium conditions are: $$ A^{\\alpha}=\\frac{6.67-1.2}{6.67-0.008}=82.11\\% \\qquad A^{Fe_3C}=1-82.11\\%=17.89\\%$$" }, { "idx": 2391, "question": "In a simple cubic crystal, the Burgers vector of dislocation (1) is $b^{(1)}=a[0,10]$ and the tangent direction of the dislocation line is $l^{(1)}=[010]$. Identify the type of this dislocation and the slip plane of the dislocation. If the slip plane is not unique, explain the constraints on the slip plane.", "answer": "For dislocation (1), the Burgers vector is parallel and opposite to the dislocation line, so it is a left-handed screw dislocation. In principle, all planes passing through the dislocation line are slip planes. However, due to the lattice resistance encountered during dislocation motion, only the close-packed planes have the least lattice resistance and thus are the actual slip planes for dislocations. The close-packed planes in a simple cubic lattice are of the {100} type, so the slip planes for dislocation (1) should be the (100) and (001) planes." }, { "idx": 2392, "question": "A face-centered cubic single crystal undergoes tensile deformation with the tensile axis along [001]. Determine the force on a dislocation with Burgers vector $b{=}a[\\overline{{\\textbf{\\vert}}}0\\mathbf{\\vert}]/2$ and line direction $t$ parallel to [121] in the slip direction. The lattice constant is given as $=0.35\\\\pi m$.", "answer": "The force $F$ per unit length of the dislocation line in the slip plane is the product of the resolved shear stress $=5$ in the slip direction of the applied stress field and the Burgers vector $h$: $F_{n}=n+$. Under uniaxial tension (stress $(\\overline{o})$), $\\\\tau=\\\\sigma\\\\mathrm{cos}\\\\lambda\\\\cos\\\\varphi$, where $\\\\lambda$ is the angle between the tensile axis and the Burgers vector, and $11\\\\pi$ is the angle between the tensile axis and the normal to the slip plane. Since $b{=}a[\\overline{{\\textbf{l}}}()]\\\\backslash2$ and $t$ is parallel to [121], the slip plane is the plane containing both the Burgers vector and the dislocation line, hence the slip plane is (111). Therefore, 2 is the angle between [001]-[101], and $42$ is the angle between [001]-[111]. Using the crystallographic angle formula for cubic systems, $$\\\\cos\\\\lambda={\\\\frac{1}{\\\\sqrt{1}\\\\sqrt{1+1}}}={\\\\frac{1}{\\\\sqrt{2}}}\\\\qquad\\\\cos\\\\varphi={\\\\frac{1}{\\\\sqrt{1}\\\\sqrt{1+1+1}}}={\\\\frac{1}{\\\\sqrt{3}}}$$ Thus, $=\\\\sigma/\\\\sqrt{6}=0.408\\\\sigma$, and the magnitude of $11$ is $\\\\alpha\\\\sqrt{2}/2=0.36\\\\times10^{-9}\\\\times\\\\sqrt{2}/2=2.55\\\\times10^{-10}\\\\mathrm{m}$. Finally, $$ F_{\\\\mathrm{g}}=\\\\tau b=0.408\\\\times2.55\\\\times10^{-10}\\\\sigma\\\\mathrm{N}/\\\\mathrm{m}=1.04\\\\times10^{-10}\\\\sigma\\\\mathrm{N}/\\\\mathrm{m}$$ where the unit of $\\\\frac{1}{a^{2}}$ is $\\\\mathsf{P}_{\\\\perp}$." }, { "idx": 2393, "question": "A face-centered cubic single crystal undergoes tensile deformation with the tensile axis along [001]. Determine the force on a dislocation with Burgers vector $b{=}a[\\overline{1}01]/2$ and line direction $t$ parallel to [121] in the climb direction. The lattice constant is given as $0.35\\\\pi m$.", "answer": "The force per unit length of the dislocation line in the climb direction, $F_{*}$, is the product of the normal stress $0.5$ on the half-plane of the edge dislocation due to the external stress field and the Burgers vector $h$: $F_{1}=-\\\\sigma_{1}h$. Since $b$ is perpendicular to the dislocation line, the dislocation in question is an edge dislocation. The normal vector to its half-plane is $b$, which is [101], so $\\\\sigma_{0}=\\\\sigma\\\\mathrm{COS}^{2}\\\\bar{\\\\varphi}^{\\\\prime}=\\\\sigma_{0}^{12}$. The climb force acting per unit length of the dislocation line is $$F_{\\\\mathrm{c}}=\\\\frac{\\\\sigma}{2}2.55\\\\times10^{-10}\\\\sigma\\\\mathrm{N}/\\\\mathrm{m}=1.275\\\\times10^{-10}\\\\sigma\\\\mathrm{N}/\\\\mathrm{m}$$ where $\\\\sigma$ is in units of $P F_{A}$." }, { "idx": 2394, "question": "In a crystal, there is a pair of parallel edge dislocations on the slip plane. How large should their spacing be to prevent movement due to their interaction? Assume the slip resistance (shear stress) of the dislocation is $\\\\square A^{\\\\prime\\\\prime}\\\\approx10^{5}Pa$, $1=0.3$, $G=5\\\\times10^{10}Pa$ (express the answer in terms of b).", "answer": "Two parallel edge dislocations (denoted as A and B), with dislocation A located on the $\\\\underline{{\\\\underline{{\\\\pi}}}}$ axis of the coordinate frame. The force per unit length exerted by dislocation A on parallel dislocation B at position $(x,y)$ in the slip direction is F. Since both dislocations are on the same slip plane, the force is $F_{s}^{\\\\mathrm{A\\\\toB}}=\\\\frac{G b^{2}}{2\\\\pi(1-\\\\nu)}\\\\frac{1}{x}$ G²1, where x is the distance between the two dislocations. When this force equals or exceeds the resistance $\\\\cdot E_{1}b$ that the dislocation must overcome to slip, the two dislocations can slide. Therefore, when $$\\\\leq\\\\frac{\\\\vec{\\\\left(r\\\\right)}^{h}}{\\\\underline{{\\\\vec{2}\\\\pi(1-\\\\nu^{\\\\prime})}}}\\\\frac{1}{\\\\tau_{\\\\perp}}$$, the two dislocations will slide. That is, $$ \\\\displaystyle{x\\\\leq\\\\frac{G b}{2\\\\pi(1-\\\\nu)}\\\\frac{1}{\\\\tau_{\\\\mathrm{pe}}}}{=\\\\frac{5\\\\times10^{10}b}{2\\\\pi(1-0.3)}\\\\frac{1}{9.8\\\\times10^{5}}}{=1.16\\\\times10^{4}b}$$. If the two dislocations have the same sign, they will move toward each other until the distance between them is less than the calculated $^\\\\texttt{I}$ and then remain stationary. If the two dislocations have opposite signs, they will attract and annihilate each other when the distance between them is less than the calculated $x$. The two dislocations will remain stationary only if the distance between them is greater than $\\\\gamma^{-}$. The force in the climb direction is 0, so no climb will occur regardless of the distance between the two dislocations." }, { "idx": 2395, "question": "If the resistance (shear stress) that dislocation slip needs to overcome is 9.8×10^5 Pa for copper, the shear modulus G of copper is 4×10^10 Pa, and the lattice constant is 0.55 nm. How thick is the low-dislocation-density layer on the surface of copper? It is known that copper has a face-centered cubic structure, and the Burgers vector length is 0.36/√2 nm = 0.255 nm.", "answer": "Due to the effect of the surface image force, dislocations near the surface are subjected to the image force F_im. When the image force is greater than or equal to the dislocation slip resistance, the dislocation slips out of the surface, reducing the dislocation density on the surface. Taking a screw dislocation as an example, the image force F_im per unit length parallel to the surface is F_im = τ_in b = G b^2 / (4π d), where d is the distance of the dislocation from the surface. When F_im equals the dislocation slip resistance, the corresponding d is the thickness of the low-dislocation-density layer on the surface. Therefore, d ≤ G b / (4π τ_⊥). The thickness of the low-dislocation-density layer for copper is d = (4×10^10 × 0.255×10^-9) / (4π × 9.8×10^5) m = 8.28×10^-7 m." }, { "idx": 2396, "question": "Assuming the resistance (shear stress) that dislocation slip needs to overcome is 1.5×10^8 Pa for 30-51=Fe alloy, the shear modulus G for 3% Si-Fe alloy is 3.5×10^11 Pa, and the lattice constant is 0.28 nm. How thick is the low-dislocation-density layer near the surface of the 3% Si-Fe alloy? It is known that iron-silicon alloy has a body-centered cubic structure, and the Burgers vector length is 0.28√3/2 nm = 0.242 nm.", "answer": "Due to the effect of surface image force, dislocations near the surface are subjected to the image force F_im. When the image force is greater than or equal to the resistance to dislocation slip, the dislocation slips out of the surface, reducing the dislocation density near the surface. Taking a screw dislocation as an example, the image force F_im per unit length parallel to the surface is F_im = τ_in b = G b^2 / (4π d), where d is the distance of the dislocation from the surface. When F_im equals the resistance to dislocation slip, d represents the thickness of the low-dislocation-density layer near the surface. Therefore, d ≤ G b / (4π τ_⊥). The thickness of the low-dislocation-density layer for the iron-silicon alloy is d = (3.5×10^11 × 0.242×10^-9) / (4π × 1.5×10^8) m = 4.88×10^-8 m." }, { "idx": 2397, "question": "A simple cubic crystal has a screw dislocation with $b=[001]$ on the (100) plane. An edge dislocation with $b=[010]$ on the (001) plane intersects with it. After the intersection, do kinks or jogs form on the two dislocations?", "answer": "After the intersection of the two dislocations, a small segment of dislocation with the same magnitude and direction as the Burgers vector of the other dislocation is left on each dislocation. If this small segment lies on the slip plane of the original dislocation, it is a kink; otherwise, it is a jog. Let dislocation A be the screw dislocation with $b=[001]$ on the (100) plane, and dislocation B be the edge dislocation with $b=[010]$ on the (001) plane. After the intersection of dislocation A and dislocation B, dislocation A produces a small segment of dislocation in the [010] direction. The slip plane of dislocation A is (100), and [010]·[100]=0, meaning the small segment lies on the slip plane of dislocation A, so it is a kink. On dislocation B, a small segment of dislocation in the [001] direction is produced. The slip plane of dislocation B is (001), and [001]·[001]=1, meaning the small segment does not lie on the slip plane of dislocation B, so it is a jog." }, { "idx": 2398, "question": "In a face-centered cubic crystal, there are two dislocations with their Burgers vectors $b$ and the planes on which they are gliding as follows: Dislocation A has $b^{[i]}=a[0]1]12$ and glides on the (111) plane; Dislocation B has $b^{(h)}=a[10\\overline{{1}}]/2$ and glides on the (111) plane. If the two dislocations are parallel, what are the tangent vectors of these two dislocations?", "answer": "Since the (111) and (111) planes intersect, when the dislocations on these two planes are parallel to each other, the direction of the dislocation line can only be the line of intersection of the two planes. Using the zone law, it is easy to determine that the line of intersection of the (111) and (111) planes is [110], so the tangent vectors of the two dislocation lines are [110]." }, { "idx": 2399, "question": "A simple cubic crystal has a screw dislocation with $b=[001]$ on the (100) plane. There is a screw dislocation with $b=[100]$ on the (001) plane intersecting with it. After the intersection, do kinks or jogs form on the two dislocations?", "answer": "After the intersection of the two dislocations, a small segment of dislocation with the same magnitude and direction as the Burgers vector of the other dislocation is left on each dislocation. If this small segment lies on the slip plane of the original dislocation, it is a kink; otherwise, it is a jog. Let dislocation A be the screw dislocation with $b=[001]$ on the (100) plane, and dislocation C be the screw dislocation with $b=[100]$ on the (001) plane. After the intersection of dislocation A and dislocation C, dislocation A produces a small segment of dislocation in the [100] direction. One of the slip planes of dislocation A is (100), and [100]·[100]=1, meaning the small segment does not lie on this slip plane of dislocation A. However, (010) is also a slip plane of dislocation A, and [100]·[010]=0, so it is a kink. On dislocation C, a small segment of dislocation in the [001] direction is produced. The slip planes of dislocation C are (001) and (010), and [001]·[001]=1 while [001]·[010]=0, meaning the small segment lies on one of the slip planes of dislocation C, so it is a kink." }, { "idx": 2400, "question": "In a Ni single crystal, there is a dislocation with Burgers vector b=a[0]12 on the (11T) plane. The lattice constant a=0.35 nm. What is the length of the dislocation's Burgers vector?", "answer": "Solution: The Ni crystal has an fcc structure, and the Burgers vector b is of the [110]/2 type. Therefore, the length of the Burgers vector is b=a√2/2=0.35√2/2=0.254 nm" }, { "idx": 2401, "question": "In a face-centered cubic crystal, there are two dislocations with their Burgers vectors $b$ and the planes they are gliding on as follows: Dislocation A has $b^{[i]}=a[0]1]12$ and glides on the (111) plane; Dislocation B has $b^{(h)}=a[10\\\\overline{{1}}]/2$ and glides on the (111) plane. Can they react when they meet during gliding? What is the Burgers vector of the resulting dislocation after the reaction? What is the glide plane?", "answer": "If the two dislocations meet and react, the reaction equation is $${\\\\frac{a}{2}}[011]+{\\\\frac{a}{2}}[10{\\\\overline{{1}}}]\\\\rightarrow{\\\\frac{a}{2}}[110]$$. The sum of the squares of the Burgers vectors of the two dislocations before the reaction is $a^{2}/2+a^{2}/2=a^{2}$, and the square of the Burgers vector of the resulting dislocation after the reaction is $=2-17$. According to the Frank criterion, this reaction reduces energy, so the reaction can proceed. Since the direction of the resulting dislocation line is [110] and the Burgers vector direction is [110], its glide plane is (001)." }, { "idx": 2402, "question": "In a face-centered cubic crystal, there are two dislocations with their Burgers vectors $b$ and the planes on which they are gliding as follows: Dislocation A has $b^{[i]}=a[0]1]12$ and glides on the (111) plane; Dislocation B has $b^{(h)}=a[10\\overline{{1}}]/2$ and glides on the (111) plane. Does the generation of such dislocations make them easier or more difficult to glide, and why?", "answer": "The glide plane of the generated dislocation is not the original glide planes of the two dislocations, the (111) plane and the (111) plane, and its glide plane (001) is not the easiest glide plane for the fcc structure, so this dislocation is difficult to glide." }, { "idx": 2403, "question": "The density of a crystal with NaCl-type structure is 3.55g/cm³, what should its Burgers vector be?", "answer": "In NaCl-type structures, the common dislocation Burgers vector is of the <110>/2 type." }, { "idx": 2404, "question": "The shear modulus of Ni is G=75 GPa, and the intrinsic stacking fault energy is γ=150 mJ/m². Calculate the equilibrium width of dislocation extension.", "answer": "Since the perfect dislocation before decomposition is an edge dislocation, the equilibrium width d₀ of the extended dislocation is d₀=Gb²/(8πγ)×(2+ν)/(1−ν)=75×10⁹×(0.147×10⁻⁹)²/(8π×150×10⁻³)×(2+1/3)/(1−1/3) m=1.505×10⁻⁹ m" }, { "idx": 2405, "question": "Calculate the length of the Burgers vector for a NaCl-type structure crystal, given the lattice constant a=0.4151 nm.", "answer": "The Burgers vector b=a√2/2=0.4151×√2/2 nm=0.2935 nm." }, { "idx": 2406, "question": "If the dislocation is an edge dislocation, can it decompose into an extended dislocation? Write the decomposition reaction equation.", "answer": "As long as the stacking fault energy is sufficiently low, any type of perfect dislocation on the {111} plane can decompose into an extended dislocation. If the given perfect dislocation decomposes into an extended dislocation, the reaction equation is a/2[101]→a/6[112]+a/6[21̅1]. This is because the decomposition into an extended dislocation involves the perfect dislocation splitting into two Shockley partial dislocations, whose Burgers vectors are of the a/6<112> type, with a Burgers vector length of b=a√6/6=0.35√6/6 nm=0.147 nm. Additionally, the slip planes of the dislocations before and after decomposition remain the original slip plane. Therefore, when writing the reaction equation, two points must be ensured: (1) the Burgers vectors of the two Shockley partial dislocations must lie on the (111) plane; (2) the Burgers vector must be conserved in the decomposition equation, meaning the Burgers vector of the dislocation before decomposition equals the sum of the Burgers vectors of the two dislocations after decomposition." }, { "idx": 2407, "question": "Verify the correctness of the calculation: The radius of Mg²⁺ ion is 0.078 nm, the radius of O²⁻ ion is 0.132 nm, and the lattice constant of NaCl-type structure is the sum of the diameters of one cation and one anion.", "answer": "The lattice constant a=2(0.078+0.132) nm=0.420 nm, which is basically consistent with the result of 0.4151 nm calculated based on density." }, { "idx": 2408, "question": "After polishing and etching lithium fluoride polycrystals, etch pits arranged in straight lines with a spacing of 10μm were observed. After applying shear stress, grain boundary dislocations moved perpendicular to the grain boundary plane. Explain the structure of the grain boundary.", "answer": "The crystal structure of lithium fluoride is a √3-1<1 type structure, and its slip plane is {0}11}. The etch pits reveal the outcrops of dislocations, and the straight-line arrangement of the etch pits indicates that this line is a set of parallel dislocation outcrops, which is a small-angle grain boundary with a dislocation spacing of 10μm. Since the grain boundary dislocations all move perpendicular to the boundary, it indicates that only a set of like-signed dislocations form the grain boundary, which can only be edge dislocations, and the slip plane of the dislocations is perpendicular to the grain boundary. Assuming the dislocation slip plane is (110), then the grain boundary plane is (110), and the Burgers vector of the dislocations can only be [1¯1¯1]¯1¯2. Therefore, this is a symmetric tilt grain boundary." }, { "idx": 2409, "question": "Using the etch pit method, the distance between small-angle grain boundary etch pits was measured to be $[5,8][11]$. X-ray diffraction measured the misorientation between the two grains as 30 ($1\\\\because=0.000280$). What is the magnitude of the Burgers vector of the dislocations in the grain boundary?", "answer": "Assuming the relationship between the dislocation spacing $n$, the Burgers vector $b$, and the misorientation in the grain boundary can be approximately estimated by the following formula: $$ D={\\\\frac{\\\\tilde{b}}{\\\\theta}} $$. Based on the given data, the Burgers vector $b$ is calculated as $$b=D\\\\pmb{\\\\theta}=6.87\\\\times10^{3}\\\\times30\\\\times0.0002\\\\&(\\\\pi/180)\\\\mathrm{nm}=1.00\\\\mathrm{nm}$$." }, { "idx": 2410, "question": "Given the density of a crystal with NaCl-type structure is 3.55 g/cm³, calculate its unit cell lattice constant. A unit cell contains 4 cations and 4 anions, the relative atomic mass of Mg is 24.31, and that of O is 16.", "answer": "According to the density formula ρ=4(24.31+16)/(a³×6.022×10²³), solving for the lattice constant a=[4(24.31+16)/(3.55×6.022×10²³)]^(1/3) cm=4.151×10⁻⁸ cm=0.4151 nm." }, { "idx": 2411, "question": "Based on the calculation results at 700°C and 550°C, find the average bonding energy between sulfur atoms and 12 iron atoms (express the answer in kJ/mol and eV).", "answer": "Averaging over these two temperatures yields -ΔG=69 kJ/mol=0.71 eV. If it is approximately assumed that the distortion energy of sulfur atoms at grain boundaries is negligible, then this value can be considered as the bonding energy between sulfur atoms and Fe atoms." }, { "idx": 2412, "question": "At 700°C (973K), the enrichment rate of sulfur at the grain boundaries of iron is 7170. Estimate the bonding energy between sulfur atoms and iron (answer in kJ/mol and eV).", "answer": "Since the concentrations at the grain boundaries and within the grains are both much smaller than 1, the simplified formula for concentration segregation is used: -ΔG=8.314×973×ln7170=71.8 kJ/mol =71.8×10^3/(6.022×10^23×1.603×10^-19) eV=0.74 eV" }, { "idx": 2413, "question": "At 550°C (823K), the enrichment rate of sulfur at the grain boundaries of γ-Fe is 15700. Estimate the bonding energy between sulfur atoms and iron (provide the answer in kJ/mol and eV).", "answer": "Since the concentrations at both the grain boundaries and within the grains are much smaller than 1, the simplified formula for concentration segregation is used: -ΔG=8.314×823×ln15700=66.1 kJ/mol =66.1×10^3/(6.022×10^23×1.603×10^-19) eV=0.68 eV" }, { "idx": 2414, "question": "A Cu (a=0.3615nm) thin film is deposited on an Au (a=0.4079nm) substrate, and their mismatch is relaxed by misfit dislocations at the interface. If the interface is the (100) plane, determine the magnitude of the Burgers vector of the misfit dislocations and the dislocation spacing.", "answer": "If the (100) plane is taken as the interface, the two phases remain parallel to each other along the [110] direction at the interface, and their mismatch δ is still 0.121, so the spacing of the misfit dislocations is also 2.256nm. Since the grain boundary dislocations relax the mismatch along the [110] direction, the Burgers vector of the dislocations is [110]/2, and thus the dislocation line is along the [110] direction. There are two [110] directions on the (100) plane, so there are two sets of misfit dislocations on the interface, perpendicular to the [110] directions and mutually forming a 90° angle." }, { "idx": 2415, "question": "If the Burgers vector magnitude of a grain boundary dislocation is 0.253nm, what is the misorientation between the two sides of the grain boundary?", "answer": "The misorientation between the two grains at the grain boundary is θ = b/D = (0.253 × 10^-9)/(10 × 10^-6) × (180/π) = 1.45 × 10^-3 degrees." }, { "idx": 2416, "question": "A Cu (a=0.3615nm) thin film is deposited on an Au (a=0.4079nm) substrate, and their mismatch is relaxed by misfit dislocations at the interface. Assuming the interface is the (111) plane, find the magnitude of the Burgers vector of the misfit dislocations and the dislocation spacing.", "answer": "Both Au and Cu have an fcc crystal structure, and the (111) plane is commonly used as the interface, with the [110] directions aligned parallel. The atomic spacing along the [110] direction is a×√2/2. For Cu, it is 0.3615nm×√2/2=0.2556nm; for Au, it is 0.4079nm×√2/2=0.2884nm. The mismatch along the [111] direction is δ=(0.2884-0.2556)/((0.2556+0.2884)/2)=0.121. To accommodate the mismatch with grain boundary dislocations, the basis vector length of the coincident site lattice is Dδ=b/δ=((bCu+bAu)/2)/(bAu-bCu)=((0.2556+0.2884)/2)/(0.2884-0.2556)=2.256nm. This means a misfit dislocation occurs every 2.256nm. Since the grain boundary dislocations relax the mismatch along the [110] direction, the Burgers vector of the dislocations is [110]/2, and the dislocation line is along the [112] direction. There are three [110] directions on the (111) plane, so there are three sets of misfit dislocations on the interface, each perpendicular to a [110] direction and mutually forming 120° angles along the [112] directions." }, { "idx": 2417, "question": "Consider a two-dimensional rectangular crystal with side lengths $L_{1}$ and $L_{\\\\pm}$ $L_{1}$ and $L_{\\\\perp}$. The interfacial energies for the two sides are $71$ and p, respectively. If the interfacial area remains constant ($L_{1}L_{1}=$ constant), prove that the equilibrium shape satisfies the relation $L_{1}/L_{2}=m g/y_{2}$.", "answer": "The total surface energy of this crystal is $E=2(L_{1}y+L_{2}y_{2})$. At equilibrium shape, the following relation holds: $$ \\\\mathrm{d}E=2(L_{1}\\\\mathrm{d}\\\\gamma_{1}+\\\\gamma_{1}\\\\mathrm{d}L_{1}+L_{2}\\\\mathrm{d}\\\\gamma_{2}+\\\\gamma_{2}\\\\mathrm{d}L_{2})=0 $$ Since $72$ and $79$ are independent of length, the equation simplifies to $$ \\\\gamma_{1}\\\\mathrm{d}L_{1}+\\\\gamma_{1}\\\\mathrm{d}L_{2}=0 $$ Given that the area $[1L]=$ constant, i.e., $\\\\angle1\\\\mathrm{d}L_{2}+L_{2}\\\\mathrm{d}L_{1}=0$, substituting this relation into the above equation yields $L_{1}:L_{2}=m:21$. Q.E.D." }, { "idx": 2418, "question": "Two parallel low-angle tilt boundaries have misorientation angles of 6 and degrees, respectively. Is it possible for these two boundaries to merge into a single boundary with a misorientation angle of $\\\\theta_{1}+\\\\theta_{2}$?", "answer": "The energy of a low-angle grain boundary is $$ \\\\gamma_{_\\\\mathrm{\\\\scriptsize{ab}}}=\\\\gamma_{_0}\\\\theta(\\\\mathcal{A}^{\\\\mathrm{\\\\scriptsize{r}}}-\\\\ln\\\\theta) $$ where $M^{\\\\prime}$ is a constant. The energy before the merging of the two boundaries is $$ \\\\gamma_{\\\\bar{0}}\\\\theta_{\\\\bar{1}}(A^{\\\\prime}-\\\\ln\\\\theta_{1})+\\\\gamma_{\\\\bar{0}}\\\\theta_{\\\\bar{z}}(A^{\\\\prime}-\\\\ln\\\\theta_{\\\\bar{z}}) $$ The energy after merging is: $$ \\\\gamma_{\\\\mathrm{o}}(\\\\theta_{\\\\mathrm{i}}+\\\\theta_{\\\\mathrm{z}})[A^{\\\\prime}-\\\\mathrm{ln}(\\\\theta_{\\\\mathrm{i}}+\\\\theta_{\\\\mathrm{z}})] $$ The energy difference $\\\\bar{A}E$ before and after merging is: $$ \\\\begin{array}{r l}&{\\\\Delta E=\\\\gamma_{\\\\mathrm{0}}(\\\\theta_{\\\\mathrm{1}}+\\\\theta_{\\\\mathrm{2}})[A^{\\\\prime}-\\\\ln(\\\\theta_{\\\\mathrm{1}}+\\\\theta_{\\\\mathrm{2}})]-\\\\gamma_{\\\\mathrm{0}}\\\\theta_{\\\\mathrm{1}}(A^{\\\\prime}-\\\\ln\\\\theta_{\\\\mathrm{1}})-\\\\gamma_{\\\\mathrm{0}}\\\\theta_{\\\\mathrm{2}}(A^{\\\\prime}-\\\\ln\\\\theta_{\\\\mathrm{2}})} &{\\\\qquad=\\\\gamma_{\\\\mathrm{0}}A^{\\\\prime}(\\\\theta_{\\\\mathrm{1}}+\\\\theta_{\\\\mathrm{2}}-\\\\theta_{\\\\mathrm{1}}-\\\\theta_{\\\\mathrm{2}})-\\\\gamma_{\\\\mathrm{0}}[(\\\\theta_{\\\\mathrm{1}}+\\\\theta_{\\\\mathrm{2}})\\\\ln(\\\\theta_{\\\\mathrm{1}}+\\\\theta_{\\\\mathrm{2}})-\\\\theta_{\\\\mathrm{1}}\\\\ln\\\\theta_{\\\\mathrm{1}}-\\\\theta_{\\\\mathrm{2}}\\\\ln\\\\theta_{\\\\mathrm{2}}]} &{\\\\qquad=\\\\gamma_{\\\\mathrm{0}}[\\\\theta_{\\\\mathrm{1}}\\\\ln\\\\frac{\\\\theta_{\\\\mathrm{1}}}{\\\\theta_{\\\\mathrm{1}}+\\\\theta_{\\\\mathrm{2}}}+\\\\theta_{\\\\mathrm{2}}\\\\ln\\\\frac{\\\\theta_{\\\\mathrm{2}}}{\\\\theta_{\\\\mathrm{1}}+\\\\theta_{\\\\mathrm{2}}}]}\\\\end{array} $$ Because $y=\\\\frac{1}{2}$ $C A$ and $C H_{2}$ are both positive, $\\\\left|\\\\Pi\\\\right|\\\\left|\\\\hat{H}_{l}\\\\right|\\\\left|\\\\hat{H}_{1}+\\\\hat{H}_{2}\\\\right|\\\\right|$ and $\\\\|\\\\bar{\\\\boldsymbol{\\\\Pi}}\\\\|\\\\hat{\\\\cal{Q}}^{\\\\prime}(\\\\hat{\\\\cal{Q}}_{1}+\\\\hat{\\\\cal{Q}}_{2})\\\\|$ are both less than 0. This process reduces energy, so it can proceed. In fact, during the recovery process of deformed materials, a large number of such processes occur, namely the coarsening of subgrains." }, { "idx": 2419, "question": "Which side does the marker wire in the diffusion couple move towards?", "answer": "Because D_Zn > D_Cu, the marker plane moves towards the Cu side." }, { "idx": 2420, "question": "A diffusion couple is formed by pure copper and a Zn=10° alloy with w(Zn)=3000. A marker wire is inserted at the bonding interface. After annealing at 7851 for 56 days, the marker wire is found to have moved 0.0105πm, with x(Zn)=200 at the marker plane, a concentration gradient (∂x_Zn/∂x)=-0.089 mm^-1, and ∫_{0}^{0.27} x dN_Zn=0.016 mm. Calculate the interdiffusion coefficient D̃ for the alloy with concentration x(Zn)=22.74°.", "answer": "The diffusion duration is 56 days, i.e., t=56d=56×24×3600s=4838400s. Based on the given data, the interdiffusion coefficient D̃ is: D̃ = (-1)/(2t (dC/dx)) ∫_0^∞ x dC = (-0.016 mm)/(-2×4838400s×0.089 mm^-1) = 1.869×10^-8 mm^2 s^-1." }, { "idx": 2421, "question": "To remove hydrogen from the system through palladium membrane diffusion, it is known that the diffusion coefficient of hydrogen in palladium at $500^{\\\\circ}C$ is $1.0\\\\times10^{-8}177^{-15}$. The hydrogen concentrations on both surfaces of a palladium plate are maintained at $2.4 kg/m^{3}$ and $0.61-2^{7}m^{3}$. Under steady-state diffusion, calculate the amount of hydrogen passing through a palladium plate with an area of $0.277^{2}$ and a thickness of $5\\\\pi\\\\pi\\\\pi$ per hour (expressed in $k=1$).", "answer": "Since this is a steady-state thin-film diffusion problem, and the diffusion coefficient does not vary with concentration, the concentration distribution within the film is linear. According to the problem statement, substituting the corresponding values, the concentration gradient within the film is calculated as: $$\\\\frac{\\\\mathrm{d}C}{\\\\mathrm{d}x}=\\\\frac{0.6-2.4}{5\\\\times10^{-3}}kg/m^{4}=-360\\\\mathrm{kg/m^{-4}}$$ The one-dimensional steady-state problem can be solved using the first law of diffusion: $$J=-D{\\\\frac{\\\\mathrm{d}C}{\\\\mathrm{d}x}}$$ The amount of hydrogen passing through the palladium plate with an area of $0.2171^{3}$ per hour is: $$Q=J\\\\times A\\\\times t=-D{\\\\frac{\\\\mathrm{d}C}{\\\\mathrm{d}x}}.A t=1.0\\\\times10^{-8}\\\\times360\\\\times0.2\\\\times3600\\\\mathrm{kg}=2.592\\\\times10^{-3}\\\\mathrm{kg}$$" }, { "idx": 2422, "question": "Assuming that at 1000°C, atoms diffuse through an Al2O3 crystal a distance x in 3h, estimate the time required for Al atoms to diffuse through the same Al2O3 crystal a distance of 9x.", "answer": "Since the diffusion distance is proportional to the square root of time, when the diffusion distance is 9x, the required time is t2 = (9x)^2 / x^2 * 3h = 243h." }, { "idx": 2423, "question": "A $C_{5}^{5}$ Fe plate with a thickness of $1.2\\\\pi m m$. One side is carburizing and the other side is decarburizing atmosphere, with surface concentrations of $\\\\mathrm{Pr}(C)=0.01200$ and $1+1(1-)=0.007500$, respectively. At $750-10$, when steady-state diffusion is achieved in the plate, the measured diffusion flux is $1.4\\\\times10^{-8}k g/m^{2}\\\\cdot5$. Find the diffusion coefficient of carbon at this temperature. $\\\\mathrm{{.}_{W}(C)=0.8\\\\%\\\\equiv60k g/m^{3})}$", "answer": "First, convert the mass concentration to volume concentration. Since $\\\\begin{array}{r}{w(\\\\mathbf{C})=0,8\\\\%=60\\\\mathrm{kg/m^{3}}}\\\\end{array}$, the volume concentrations of $\\\\mathbb{C}_{1}{=}\\\\mathrm{w}(\\\\mathbb{C})=0,012\\\\%$ and $\\\\mathrm{C_{2}}{=}{}_{\\\\mathrm{W}}(\\\\mathrm{C}){=}0.00750/\\\\$ are: $$ C_{1}=\\\\frac{0.012}{0.8}\\\\times60=0.9\\\\mathrm{kg\\\\cdotm^{-3}}\\\\quad C_{2}=\\\\frac{0.0075}{0.8}\\\\times60=0.5625\\\\mathrm{kg\\\\cdotm^{-3}}$$ This is a one-dimensional steady-state diffusion problem, and the diffusion coefficient does not vary with concentration, so the concentration distribution in the plate is linear. Based on the given data, the concentration gradient in the plate can be calculated: $$ \\\\frac{\\\\mathrm{d}C}{\\\\mathrm{d}x}=\\\\frac{0.5625-0.9}{1.2\\\\times10^{-3}}\\\\mathrm{kg/m^{4}}=-281.25\\\\mathrm{kg/m^{4}}$$ According to the one-dimensional steady-state diffusion equation, the diffusion coefficient is $$ {\\\\cal D}=-J{\\\\frac{\\\\mathrm{d}\\\\mathrm{r}}{\\\\mathrm{d}C}}=1.4\\\\times10^{\\\\mathrm{~8}}\\\\times{\\\\frac{1}{281.25}}\\\\mathrm{m}^{2}/\\\\mathrm{s}=4.98\\\\times10^{-11}\\\\mathrm{m}^{2}/\\\\mathrm{s}$$" }, { "idx": 2424, "question": "A diffusion couple is composed of pure copper and a Zn=10° alloy with w(Zn)=3000. A marker wire is inserted at the bonding interface. After annealing at 7851 for 56 days, the marker wire is found to have moved 0.0105πm. The marker plane has x(Zn)=200, the concentration gradient (∂x_Zn/∂x)=-0.089 mm^-1, and ∫_{0}^{0.27} x dN_Zn=0.016 mm. Determine the intrinsic diffusion coefficients of Zn and Cu in the alloy with x(Zn)=22.74°.", "answer": "The velocity of the marker plane movement v = marker displacement / (2×time) = 0.0105 mm / (2×4838400s) = 1.085×10^-9 mm/s. According to the relations v = (D_Cu - D_Zn) (dx_Zn/dy) and D̃ = (x_Zn D_Cu + x_Cu D_Zn), at the marker plane where x_Zn=0.22, the simultaneous equations are: (D_Cu - D_Zn)×(-0.089 mm^-1) = 1.085×10^-9 mm/s and (0.22 D_Cu + 0.78 D_Zn) = 1.869×10^-8 mm^2 s^-1. Solving the simultaneous equations yields: D_Zn = 2.137×10^-8 mm^2 s^-1, D_Cu = 0.918×10^-8 mm^2 s^-1." }, { "idx": 2425, "question": "Calculate the diffusion rate of AI through Al2O3 at a temperature 10 times that at 1000°C (i.e., find D(T)=10D(1000°C)). The activation energy Q for AI diffusion in Al2O3 crystal is known to be 477 kJ/mol.", "answer": "According to the relationship between diffusion coefficient and temperature D = D0 exp(-Q/RT), let T1=1000°C=1273K, and find the temperature T2 where D(T2)=10D(T1). From 10 = exp(-Q/R (1/T2 - 1/T1)), we get 1/T2 = 1/T1 - (R ln10)/Q = 1/1273K - (8.314*ln10)/477000 = 7.454×10^-4 K^-1. Therefore, T2=1341.6K, and the temperature increases by 68.6K." }, { "idx": 2426, "question": "The turbine blade is oxidized as metal atoms reach the metal surface through oxide diffusion and react with oxygen to form oxides. This material forms a $B H//H T$ thick oxide layer after $50\\\\textcircled{<}107$. What is the thickness of the oxide layer after 100h at the same temperature?", "answer": "This process involves two sub-processes: diffusion of metal atoms in the oxide layer and the reaction of metal atoms with oxygen. Because diffusion is slow, the entire process is controlled by diffusion. The diffusion distance $l$ and diffusion time $^ Ḋ t Ḍ$ follow a square root relationship, i.e., $l=11$ after $5-50=50$ holding time $\\\\lfloor0\\\\rfloor_{\\\\lfloor1\\\\rfloor}$, forming $3[1171(l)$. Then, the oxide layer thickness $l-$ after $\\\\lvert[\\\\rvert]\\\\rvert\\\\lvert\\\\chi\\\\rvert/\\\\lvert\\\\chi\\\\rvert)$ is $$ I_{\\\\mathrm{{z}}}=I_{\\\\mathrm{{|\\\\sqrt{\\\\frac{t_{\\\\mathrm{z}}}{t_{\\\\mathrm{l}}}}=8\\\\sqrt{\\\\frac{100}{10}}\\\\mu\\\\mathrm{{m}=25.3\\\\upmu5}}}}$$" }, { "idx": 2427, "question": "At what temperature can the contribution of grain boundary diffusion be neglected?", "answer": "From the data, it can be seen that at 1200K, the difference between Dapp and D1 is already very small, and the special contribution of grain boundary diffusion can be neglected." }, { "idx": 2428, "question": "If the diameter of polyester (PET) carpet fibers is doubled to 100 microns, how much time is required for water, dye, and dye bags to penetrate to the center of the fiber, respectively?", "answer": "The penetration depth is 50 microns. The time required for water to penetrate to the center of the fiber is t=(50×10^-6)^2/(2×1.0×10^-12)=1250 s; the time required for dye to penetrate to the center of the fiber is t=(50×10^-6)^2/(2×1.0×10^-13)=12500 s; the time required for dye bags to penetrate to the center of the fiber is t=(50×10^-6)^2/(2×1.0×10^-14)=125000 s." }, { "idx": 2429, "question": "In polycrystalline ceramics, does diffusion predominantly occur along grain boundaries or through the lattice?", "answer": "Grain boundaries are fast paths for diffusion, so in polycrystals, if the temperature is not too high, grain boundary diffusion predominates. However, at very high temperatures, the difference between grain boundary diffusion and lattice diffusion diminishes." }, { "idx": 2430, "question": "Both Ge and Cu exhibit self-diffusion via a vacancy mechanism. When diffusion occurs at comparable temperatures, which one has a larger self-diffusion coefficient? Why?", "answer": "Referring to Appendix B of the materials, the melting points of Ge and $[\\\\frac{\\\\pi}{2}]$ are $\\\\textcircled{1}17.57.757$ and $108.1.575\\\\times$, respectively. At first glance, it might seem that Cu's self-diffusion coefficient is lower than Ge's at comparable temperatures. However, $111$ crystal is metallically bonded, while Ge crystal is covalently bonded. Covalent bonds are much stronger than metallic bonds. Therefore, in reality, $\\\\sum11$'s self-diffusion coefficient is higher than Ge's." }, { "idx": 2431, "question": "Is the diffusion in oxide ceramics a vacancy exchange mechanism or a rotary exchange mechanism?", "answer": "The bonding in oxide ceramics is ionic bonding, and the diffusion mechanism in ionic crystals is primarily the vacancy exchange mechanism. The activation energy for the rotary exchange mechanism is too high and often disrupts the ionic bonding, so it cannot be the diffusion mechanism." }, { "idx": 2432, "question": "What is texture?", "answer": "The orientation distribution of a polyoriented body deviating from a random distribution is called texture." }, { "idx": 2433, "question": "The diameter of polyester (PET) carpet fiber is 50 micrometers, immersed in a dye bath containing water, dye, and dye bags at boiling temperature. The diffusion coefficients of water, dye, and dye bags are 1.0×10^-12 m²/s, 1.0×10^-13 m²/s, and 1.0×10^-14 m²/s, respectively. Estimate the time required for water, dye, and dye bags to penetrate into the center of the fiber.", "answer": "Using the simple one-dimensional diffusion distance-time relationship l=√(2Dt) for approximate estimation, the penetration depth is 25 micrometers. The time required for water to penetrate into the fiber center is t=(25×10^-6)^2/(2×1.0×10^-12)=312.5 s; the time required for dye to penetrate into the fiber center is t=(25×10^-6)^2/(2×1.0×10^-13)=3125 s; the time required for dye bags to penetrate into the fiber center is t=(25×10^-6)^2/(2×1.0×10^-14)=31250 s." }, { "idx": 2434, "question": "Given the volume diffusion coefficient of silver D1=7.2×10^-5 exp(-190 kJ/mol / (R T)) m²·s^-1 and the grain boundary diffusion coefficient Db=1.4×10^-5 exp(-90 kJ/mol / (R T)) m²·s^-1, with a grain size of 7×10^-5 m and a grain boundary thickness of 5×10^-10 m, calculate the effective diffusion coefficient Dapp at temperatures 577°C, 727°C, and 0.776 Tm.", "answer": "The apparent diffusion coefficient of a polycrystal containing grain boundaries is expressed as Dapp=f Db + (1-f) D1, where f is the fraction of defects (grain boundaries). Assuming the grains are cubes with side length a equivalent to the grain diameter, each face of the cube is shared by adjacent cubes, so there are only 3 grain boundaries per cube, hence f≈3δ/d=7.5×10^-5. The calculation results are as follows: Temperature K: 800, 1000, 1200; D (m²·s^-1): 2.82×10^-17, 8.5×10^-15, 3.86×10^-13; Db (m²·s^-1): 1.86×10^-11, 2.78×10^-10, 1.69×10^-9; Dapp (m²·s^-1): 1.42×10^-15, 2.94×10^-14, 5.13×10^-13; Dapp/D: 50.47, 3.44, 1.33." }, { "idx": 2435, "question": "How is texture formed in crystals?", "answer": "A crystal is a three-dimensionally periodic ordered arrangement, which has a certain orientation relative to a reference coordinate. The orientations of the grains in a polycrystal have an orientation distribution relative to the reference coordinate. If the orientation distribution of the polycrystal deviates from a random distribution, texture is formed." }, { "idx": 2436, "question": "The critical resolved shear stress of aluminum is $2.40\\\\approx10^{5}$ Pa. When the tensile axis is [001], what is the tensile stress required to cause yielding?", "answer": "The relationship between tensile stress $\\\\sqrt{h}$ and the resolved shear stress on the slip system is =coscos, where and are the angles between the tensile direction and the slip direction and the normal to the slip plane, respectively. The crystal structure of aluminum is fcc, and the slip system is $11117=110.3$. When the tensile axis is [001], since it is perpendicular to [110] and [110], the resolved shear stress on the slip systems composed of them is 0, and they will not activate. The geometric relationship between the [001] axis and its slip systems is equivalent, and their resolved shear stresses are equal. Taking the (111)[011] slip system as an example, the tensile stress required to cause yielding is calculated. The cosine of the angle $\\\\varphi$ between [001] and the slip plane normal [111], cos$\\\\varphi$, and the cosine of the angle $\\\\lambda$ between [001] and the slip direction [011], cos$\\\\lambda$, are respectively $$ \\\\cos\\\\varphi={\\\\frac{1}{\\\\sqrt{1}{\\\\sqrt{3}}}}={\\\\frac{1}{\\\\sqrt{3}}}\\\\cos\\\\lambda={\\\\frac{1}{\\\\sqrt{1}{\\\\sqrt{2}}}}={\\\\frac{1}{\\\\sqrt{2}}}$$ When the critical resolved shear stress $\\\\frac{\\\\sqrt{1}}{2}=\\\\frac{1}{2}$ is $2.40\\\\times0^{-}10^{5}Pa$, the corresponding stress $\\\\alpha$ is the yield stress: $$ \\\\sigma_{\\\\circ}={\\\\frac{\\\\tau_{\\\\circ}}{\\\\cos\\\\varphi\\\\cos\\\\lambda}}=2.4\\\\times10^{5}\\\\times{\\\\sqrt{3}}\\\\times{\\\\sqrt{2}}\\\\mathbf{Pa}=5.89\\\\times10^{5}\\\\mathbf{Pa}$$" }, { "idx": 2437, "question": "For a face-centered cubic crystal stretched along the [131] axis, determine the resolved shear stress of the slip system (111)[10-1]. The tensile stress is 6.9×10^5 Pa", "answer": "For the slip system (111)[10-1], the cosine of the angle between [131] and the slip plane normal [111] is cosφ=5/(√3×√11)=0.870; the cosine of the angle between [131] and the slip direction [10-1] is cosλ=(1×1 + 3×0 + 1×(-1))/(√11×√(1+0+1))=0/(√11×√2)=0. The resolved shear stress τ=6.9×10^5×0.870×0 Pa=0 Pa" }, { "idx": 2438, "question": "Mg2+ diffuses in MgO. What concentration of trivalent cation impurities needs to be added near the melting point of 2800°C to ensure that Mg2+ mainly diffuses via the non-intrinsic vacancy mechanism?", "answer": "When the concentration of non-intrinsic vacancies exceeds that of intrinsic vacancies, the non-intrinsic vacancy mechanism becomes the dominant diffusion mechanism. In MgO crystals, the formation enthalpy of Schottky defects is ΔH = 6 eV (as shown in Table 6-2 of the textbook). Thus, the equilibrium concentration of Schottky vacancies at 2800°C is: xv ≈ exp(-ΔHc / 2kBT) = exp(-6 / (2 × 8.62 × 10^-5 × 3073)) = 1.206 × 10^-5. If the non-intrinsic vacancy concentration exceeds this equilibrium intrinsic vacancy concentration, diffusion will primarily occur via the non-intrinsic vacancy mechanism. Now, by adding trivalent cations, every two trivalent cations replacing two Mg2+ ions will generate one cation vacancy. To achieve a non-intrinsic cation vacancy concentration of 1.206 × 10^-5, the required trivalent cation concentration x is: x = 2 × 1.206 × 10^-5 = 2.412 × 10^-5. Therefore, at 2800°C, for Mg2+ in MgO to mainly diffuse via the non-intrinsic vacancy mechanism, the trivalent cation impurity concentration must exceed 2.412 × 10^-5." }, { "idx": 2439, "question": "For a face-centered cubic crystal stretched along the [131] axis, determine the resolved shear stress on the slip system (111)[0-11]. The tensile stress is 6.9×10^5 Pa.", "answer": "According to the relationship between tensile stress and resolved shear stress on the slip system, τ=σcosλcosφ. For the slip system (111)[0-11], the cosine of the angle between [131] and the slip plane normal [111], cosφ, and the cosine of the angle between [131] and the slip direction [0-11], cosλ, are: cosφ=(1×1 + 3×1 + 1×1)/(√(1+1+1)×√(1+9+1))=5/(√3×√11)=0.870; cosλ=(1×0 + 3×(-1) + 1×1)/(√(1+9+1)×√(0+1+1))=-2/(√11×√2)=-0.426. The resolved shear stress τ=6.9×10^5×0.870×(-0.426) Pa=-2.56×10^5 Pa." }, { "idx": 2440, "question": "For a face-centered cubic crystal stretched along the [131] axis, determine the resolved shear stress of the slip system (111)[1-10]. The tensile stress is 6.9×10^5 Pa.", "answer": "For the slip system (111)[1-10], the cosine of the angle between [131] and the slip plane normal [111] is cosφ=5/(√3×√11)=0.870; the cosine of the angle between [131] and the slip direction [1-10] is cosλ=(1×1 + 3×(-1) + 1×0)/(√11×√(1+1+0))=-2/(√11×√2)=-0.426. The resolved shear stress τ=6.9×10^5×0.870×(-0.426) Pa=-2.56×10^5 Pa." }, { "idx": 2441, "question": "What are the similarities between crystal texture and polymer texture in a broad sense?", "answer": "In a broad sense, the textures of the two are similar." }, { "idx": 2442, "question": "How is texture formed in polymers?", "answer": "Under the action of tensile force on polymers, the molecular chains in the amorphous regions of the polymer become ordered in one-dimensional or two-dimensional directions. In the crystalline regions (folded lamellae), the layers rotate to orient along the tensile axis, meaning the direction of the molecular chains and the orientation of the crystalline blocks are not randomly distributed, thus forming the polymer texture." }, { "idx": 2443, "question": "For a ceramic oxide, in which temperature ranges does the vacancy diffusion mechanism occur as intrinsic or extrinsic?", "answer": "If the ceramic oxide is a solid solution, the incorporation of solutes with different valences will generate extrinsic point defects. The concentration of intrinsic point defects changes drastically with temperature, while the concentration of extrinsic point defects does not vary with temperature. Therefore, at low temperatures, the concentration of extrinsic vacancies is higher than that of intrinsic vacancies, and this difference becomes more pronounced as the temperature decreases. Thus, in the low-temperature range, the extrinsic vacancy diffusion mechanism dominates." }, { "idx": 2444, "question": "What changes occur in polymer texture when heated?", "answer": "Heating can cause oriented polymers to disorient." }, { "idx": 2445, "question": "What are the differences in the formation mechanisms between crystal texture and polymer texture?", "answer": "Polycrystalline texture is formed by the deformation and rotation of grains under force, while polymer texture is formed by the ordered arrangement and distribution of molecular chains under force, along with the rotational orientation of crystalline blocks in the amorphous region." }, { "idx": 2446, "question": "The possible slip planes for a body-centered cubic crystal are {110}, {112}, and {123}. If the slip direction is [1-11], what are the specific slip systems?", "answer": "The slip direction of a slip system must lie on its slip plane. According to the crystallographic zone law, when the slip direction is [1-1 1], for the {110} slip plane, the possible slip planes are (110), (011), and (10-1). For the {112} slip plane, the possible slip planes are (121), (21-1), and (-112). For the {123} slip plane, the possible slip planes are (132), (231), (32-1), (-123), (-213), and (31-2)." }, { "idx": 2447, "question": "When using X-ray detection, how does the diffraction of polycrystalline texture appear?", "answer": "The diffraction of polycrystalline texture appears as the diffraction rings of polycrystals becoming discontinuous rings, and in extreme cases, turning into diffraction spots like those of single crystals." }, { "idx": 2448, "question": "What is the critical curvature radius of particles that substantially affects the solubility of Fe3C in α-Fe? Given the interfacial energy of Fe3C is 0.71 J/m², and the molar volume of Fe is 23.4×10^-6 m³/mol.", "answer": "When the particle curvature radius is large (e.g., 1000nm), the solubility already approaches the equilibrium concentration. When the curvature radius exceeds this size, it no longer has a substantial effect on solubility." }, { "idx": 2449, "question": "The equilibrium solubility of Fe3C in α-Fe measured by internal friction method is given by C=0.736exp(-4850/T), where T is the temperature. Determine the equilibrium solubility of Fe3C in α-Fe at 627°C (i.e., 900K).", "answer": "C∞=0.736exp(-4850/T)=0.736exp(-4850/900)=3.36×10^-3" }, { "idx": 2450, "question": "When using X-ray detection, how does the diffraction of polymer texture appear?", "answer": "The diffuse rings of polymer texture consist of diffuse peaks (rather than the discrete reflections observed in crystals)." }, { "idx": 2451, "question": "At 627°C (i.e., 900K), what is the solubility of Fe3C particles with a radius of 100nm in α-Fe? The interfacial energy of Fe3C is known to be 0.71 J/m², and the molar volume of Fe is 23.4×10^-6 m³/mol.", "answer": "Cr=C∞(1+(2×23.4×10^-6×0.71)/(8.314×900×100×10^-9))=3.36×10^-3(1+4.440×10^-9/100×10^-9)=3.509×10^-3" }, { "idx": 2452, "question": "At 627°C (i.e., 900K), what is the solubility of Fe3C particles with a radius of 1000nm in α-Fe? The interfacial energy of Fe3C is known to be 0.71 J/m², and the molar volume of Fe is 23.4×10^-6 m³/mol.", "answer": "Cr=C∞(1+(2×23.4×10^-6×0.71)/(8.314×900×1000×10^-9))=3.36×10^-3(1+4.440×10^-9/1000×10^-9)=3.375×10^-3" }, { "idx": 2453, "question": "For the allotropic transformation of a pure metal $(x\\\\rightarrow\\\\beta)$, at a certain degree of undercooling, the volume Gibbs free energy difference between the two phases is $7\\\\times10^{8} kJ/m^{3}$, and the interfacial energy is $0.5 J/m^{2}$. If the strain energy of nucleation is neglected, calculate the critical nucleus size and the critical nucleation work for forming a cubic nucleus.", "answer": "For a cubic nucleus, the critical edge length $a^{*}=-\\\\frac{4\\\\gamma}{\\\\Delta G_{V}}=\\\\frac{4\\\\times0.6}{7\\\\times10^{8}}m=3.428\\\\times10^{-9}m$. The critical nucleation work is $\\\\Delta G^{*}=\\\\frac{1}{3}\\\\gamma A^{*}=\\\\frac{1}{3}\\\\times6\\\\times(3.428\\\\times10^{-9})^{2}\\\\times0.6J=1.44\\\\times10^{-17}J$." }, { "idx": 2454, "question": "At 627°C (i.e., 900K), what is the solubility of Fe3C particles with a radius of 10nm in α-Fe? The interfacial energy of Fe3C is known to be 0.71 J/m², and the molar volume of Fe is 23.4×10^-6 m³/mol.", "answer": "Cr=C∞(1+(2×23.4×10^-6×0.71)/(8.314×900×10×10^-9))=3.36×10^-3(1+4.440×10^-9/10×10^-9)=4.952×10^-3" }, { "idx": 2455, "question": "For the allotropic transformation of pure metal $(x\\\\rightarrow\\\\beta)$, at a certain degree of undercooling, the volume Gibbs free energy difference between the two phases is $7\\\\times10^{8} kJ/m^{3}$, and the interfacial energy is $0.5 J/m^{2}$. If the strain energy of nucleation is neglected, calculate the critical nucleus size and the critical nucleation work for forming a spherical nucleus.", "answer": "For a spherical nucleus, the critical nucleus radius $r^{*}=-\\\\frac{2\\\\gamma}{\\\\Delta G_{V}}=\\\\frac{2\\\\times0.6}{7\\\\times10^{8}}m=1.71\\\\times10^{-9}m$. The critical nucleation work is $\\\\Delta G^{*}=\\\\gamma A^{*}/3=0.6\\\\times4\\\\pi\\\\times(1.71\\\\times10^{-9})^{2}/3J=7.35\\\\times10^{-18}J$." }, { "idx": 2456, "question": "For the allotropic transformation of pure metal $(x\\\\rightarrow\\\\beta)$, at a certain degree of undercooling, the volume Gibbs free energy difference between the two phases is $7\\\\times10^{8} kJ/m^{3}$, and the interfacial energy is $0.5 J/m^{2}$. If the strain energy of nucleation is neglected, calculate the critical nucleus size and the critical nucleation energy for the formation of a disk-shaped nucleus with a diameter $(D)$ to thickness $(1)$ ratio $(D//)$ of 20.", "answer": "For the disk-shaped nucleus, the critical nucleus diameter $D^{*}=-\\\\frac{88}{3}\\\\frac{\\\\gamma}{\\\\Delta G_{V}}=\\\\frac{88\\\\times0.6}{3\\\\times7\\\\times10^{8}}m=2.514\\\\times10^{-8}m$. The critical nucleation energy is $\\\\Delta G^{*}=[-\\\\frac{\\\\pi}{80}\\\\times(2.514\\\\times10^{-8})^{3}\\\\times7\\\\times10^{8}+\\\\frac{11\\\\pi}{20}\\\\times(2.514\\\\times10^{-8})^{2}\\\\times0.6]J=2.184\\\\times10^{-16}J$." }, { "idx": 2457, "question": "Assuming the shear moduli G of the parent phase and the precipitate phase are the same, and the parent phase is an isotropic continuous medium. If a coherent nucleus forms, derive the expression for the size at which the spherical nucleus loses coherency during growth.", "answer": "For a spherical nucleus, the sum of the misfit strain energy and interfacial energy before the loss of coherency is (4/3)πr^3×4Gδ^2+4πr^2γ_ci0. After losing coherency, there is no misfit strain energy, and the interfacial energy becomes 4πr^2γ_m. Let the critical size for losing coherency be r*. At this size, the energies of the coherent and incoherent states should be equal, hence (4/3)π(r*)^3×4Gδ^2+4π(r*)^2γ_ci0=4π(r*)^2γ_m. Solving this gives r*=3(γ_m−γ_ci0)/(4Gδ^2)." }, { "idx": 2458, "question": "The equilibrium melting point of nickel is 172515, the solid phase ρ̄=0.6cm³/1770l, the liquid/solid interface energy y=2.25×10⁻⁵·50⁻⁷. If the radius of the spherical particle is 1cm, how much is the melting point reduced? Assume the molar melting ΔHₘ=[Γ̄Γ̄ḠḠJ†m0]", "answer": "The melting point Tᵣ of a crystal with a curvature radius of 1cm is Tᵣ = Tₘ - (2γVₛTₘ)/(rΔHₘ). Substituting the data gives Tᵣ = 1728K - (2×6.6×2.25×10⁻⁵×1728)/(1×18066)K ≈ 1727.99K, ΔT = 0.01K." }, { "idx": 2459, "question": "The equilibrium melting point of nickel is 172515, the solid phase ρ̄=0.6cm³/1770l, the liquid/solid interface energy y=2.25×10⁻⁵·50⁻⁷. If the spherical particle radius is 1um, how much does the melting point decrease? Assume the molar melting ΔHₘ=[Γ̄Γ̄ḠḠJ†m0]", "answer": "The melting point Tᵣ of a crystal with a curvature radius of 1um is Tᵣ = Tₘ - (2γVₛTₘ)/(rΔHₘ). Substituting the data gives Tᵣ = 1728K - (2×6.6×2.25×10⁻⁵×1728)/(10⁻⁴×18066)K ≈ 1727.72K, ΔT = 0.28K." }, { "idx": 2460, "question": "Estimate the number of atomic clusters containing 60 atoms in 1cm³ of copper at its melting point temperature. The atomic volume of liquid copper is 1.6×10⁻²⁹m³, the surface energy is 0.177J/m², and the melting point is 1356K.", "answer": "According to nᵢ = n exp(-ΔG / k_B T), first calculate the energy ΔG₆₀ of a 60-atom cluster. The radius of the cluster r = (3×60×1.6×10⁻²⁹ / 4π)^(1/3), and the surface area A₆₀ = 4πr² = 4π(3×60×1.6×10⁻²⁹ / 4π)^(2/3). ΔG₆₀ = A₆₀γ = 4π(3×60×1.6×10⁻²⁹ / 4π)^(2/3)×0.177 = 8.33×10⁻¹⁹J. The number of atoms per cm³ n = 1 / (1.6×10⁻²³) = 6.25×10²² cm⁻³. n₆₀ = 6.25×10²² exp(-8.33×10⁻¹⁹ / (1.38×10⁻²³×1356)) = 2.91×10³ cm⁻³." }, { "idx": 2461, "question": "What is the ratio of nucleation rates between intragranular coherent disc-shaped nucleation and grain boundary incoherent double spherical cap-shaped nucleation? Given the grain diameter Π=10^-3 m, temperature T=1000 K, and Boltzmann constant k=1.38×10^-23 J/K.", "answer": "The calculation process for the nucleation rate ratio I2/I1 is as follows: I2/I1 = (d/L)exp[-(ΔGb* - ΔG*)/kT] = (10^-9/10^-3)exp[-(2.615×10^-18 - 1.26×10^-18)/(1.38×10^-23×1000)] = 2.27×10^-49. The result shows that the nucleation rate for intragranular coherent nucleation is higher than that for grain boundary incoherent nucleation." }, { "idx": 2462, "question": "AI-Mg substitutional solid solution, estimate the misfit strain energy generated by solute atoms $1\\\\cdot11$, expressed in $\\\\|\\\\cdot\\\\|\\\\Pi\\\\mathbb{O}\\\\|^{-1}$ and $E V_{1}$ per atom. State the assumptions used in your estimation. The atomic radius of AI is $0.14317171$, shear modulus $G=2.5\\\\times10^{10}P a$, and Mg's $\\\\therefore\\\\frac{1}{2}$ axis length is $0.32\\\\mathrm{nm}$.", "answer": "Assumption $\\\\boxed{1}$ the matrix is an isotropic continuous medium, $121$ solute atoms are far apart, and the stress field of each solute atom does not overlap. The misfit strain energy generated by solute atoms $N/E$ can be estimated using the following formula: \\n\\n$$ \\n\\\\Delta G_{x}=4G\\\\delta^{2}F^{\\\\prime} \\n$$ \\n\\nwhere $0.51$ lattice misfit is \\n\\n$$ \\n\\\\delta=\\\\frac{r_{\\\\mathrm{{Mg}}}-r_{\\\\mathrm{{Al}}}}{r_{\\\\mathrm{{Al}}}}=\\\\frac{0.16-0.143}{0.143}=0.1189 \\n$$ \\n\\n$\\\\int\\\\limits_{a}^{b}$ is the atomic volume of AI: \\n\\n$$ \\n\\\\begin{array}{l}{{{\\\\displaystyle{\\\\cal V}=\\\\frac{4}{3}\\\\pi v_{\\\\mathrm{Al}}^{3}=\\\\frac{4}{3}\\\\pi(0.143\\\\times10^{-9})^{3}=1.225\\\\times10^{-29}m^{3}}}}\\\\ {{{\\\\mathrm{}}}}\\\\ {{{\\\\Delta G_{\\\\mathrm{sl}}}=4G\\\\delta^{2}V=4\\\\times2.5\\\\times10^{10}\\\\times(0.1189)^{2}\\\\times1.225\\\\times10^{-29}J/a t=1.73\\\\times10^{-20}J/a t}}\\\\end{array} \\n$$ \\n\\nTherefore, \\n\\nsince $\\\\mathsf{I e V}{=}1.602\\\\times10^{-19}J$, \\n\\n$$ \\n\\\\Delta G_{*}=1.73\\\\times10^{-20}/1.602\\\\times10^{-19}{\\\\mathrm{~eV/at}}=0.108{\\\\mathrm{~eV/at}} \\n$$ \\n\\nMultiply the value expressed in $\\\\int\\\\limits_{0}^{1}\\\\frac{2}{c}d t$ by Avogadro's number $N_{10}$ to obtain the value in $\\\\vert11770-5\\\\vert$ units: \\n\\n$$ \\n\\\\Delta G_{_{x t}}=1.73\\\\times10^{-20}\\\\times6.025\\\\times10^{23}\\\\mathrm{J/mol}=10.4\\\\times10^{3}\\\\mathrm{J/mol} \\n$$" }, { "idx": 2463, "question": "Estimate the number of clusters containing 10 atoms in 1cm³ of copper at its melting point temperature. The atomic volume of liquid copper is 1.6×10⁻²⁹m³, the surface energy is 0.177J/m², and the melting point is 1356K.", "answer": "According to nᵢ = n exp(-ΔG / k_B T), first calculate the energy ΔG₁₀ of a 10-atom cluster. The radius of the cluster r = (3×10×1.6×10⁻²⁹ / 4π)^(1/3), and the surface area A₁₀ = 4πr² = 4π(3×10×1.6×10⁻²⁹ / 4π)^(2/3). ΔG₁₀ = A₁₀γ = 4π(3×10×1.6×10⁻²⁹ / 4π)^(2/3)×0.177 = 2.52×10⁻¹⁹J. The number of atoms per cm³ n = 1 / (1.6×10⁻²³) = 6.25×10²² cm⁻³. n₁₀ = 6.25×10²² exp(-2.52×10⁻¹⁹ / (1.38×10⁻²³×1356)) = 8.85×10¹⁰ cm⁻³." }, { "idx": 2464, "question": "Given that the shear moduli G of the parent phase and the precipitate phase are the same, and the parent phase is an isotropic continuous medium. If a coherent nucleus forms, derive the expression for the size at which the disc-shaped nucleus loses coherence during growth.", "answer": "For a disc-shaped nucleus, let the radius be r and the thickness be t. Before coherence is lost, the sum of the misfit strain energy and the interfacial energy is πr^2t×4Gδ^2+2πr^2γ_ci0+2πrtγ_in. After coherence is lost, there is no misfit strain energy, and the disc interface changes from coherent to incoherent, so the energy becomes 2πr^2γ_in+2πrtγ_in. At the critical size, the energies of coherent and incoherent states should be equal, i.e., πr^2t*×4Gδ^2+2πr^2γ_ci0+2πrt*γ_in=2πr^2γ_in+2πrt*γ_in. Solving gives t*=(γ_in−γ_ci0)/(2Gδ^2)." }, { "idx": 2465, "question": "For heterogeneous nucleation at grain boundaries (double spherical cap shape, with all interfaces being incoherent), what is the critical nucleus formation energy ΔGb*? Given the contact angle θ=50°, nucleation driving force ΔG1=5×10^8 J/m^3, and incoherent interface energy γin=0.5 J/m^2.", "answer": "The calculation process for the critical nucleus formation energy ΔGb* is as follows: First, calculate the shape factor f(θ) = 0.5×(2-3cos50°+cos^3 50°) = 0.3125. The critical nucleus radius for homogeneous nucleation (spherical shape) rin* = 2γin/ΔG1 = 2×0.5/(5×10^8) m = 2×10^-9 m. The critical nucleus formation energy for homogeneous nucleation ΔGin* = (16π/3)(γin^3)/(ΔG1^2) = (16π/3)(0.5^3)/(5×10^8)^2 J = 8.37×10^-18 J. The critical nucleus formation energy for heterogeneous nucleation ΔGb* = f(θ)ΔGin* = 8.37×10^-18×0.3125 J = 2.615×10^-18 J." }, { "idx": 2466, "question": "What is the critical nucleation energy ΔG* when nucleation occurs as coherent disc-shaped particles within grains? Given the ratio of disc diameter to thickness D/N=10, the nucleation driving force ΔG1=5×10^8 J/m^3, the coherent interfacial energy γco=0.05×(1/17)^2, and the incoherent interfacial energy γin=0.5 J/m^2.", "answer": "The calculation process for the critical nucleation energy ΔG* is as follows: The relationship between the disc radius r and thickness t is t = r/5. The disc volume equals πr^2t = πr^3/5. The disc top surface area equals πr^2, and the disc side surface area equals 2πrt = 2πr^2/5. The energy required for nucleation ΔG is ΔG = (πr^3/5)ΔG1 + 2πr^2γco + (2πr^2/5)γin. Differentiating with respect to r and setting it to zero, the critical size r* = (4×5γco + 4γin)/(3ΔG1) = (4×5×0.05 + 4×0.5)/(3×5×10^8) m = 2×10^-9 m. Substituting r* back into the ΔG equation, we obtain ΔG* = [-(π(2×10^-9)^3)/5 ×5×10^8 + 2π×(2×10^-9)^2×0.05 + (2π(2×10^-9)^2)/5×0.5] J = 1.26×10^-18 J." }, { "idx": 2467, "question": "What is the interdiffusion coefficient D for the A-B binary system at 550K with molar fractions x_A=0.6 and x_B=0.4, given D_B^AB=9×10^-12 cm²/s, D_A^AB=2×10^-12 cm²/s, and d²G/dx_B²=-95.325?", "answer": "The interdiffusion coefficient D is calculated as D = (x_A * D_B^AB + x_B * D_A^AB) * (x_A * x_B / (R * T)) * (d²G/dx_B²) = (0.6 * 2×10^-16 + 0.4 * 9×10^-16) * (0.6 * 0.4 / (8.314 * 550)) * (-95.325) m²/s = -2.4×10^-18 m²/s." }, { "idx": 2468, "question": "What are the similarities between polymer crystallization and metal crystallization in terms of the crystallization process?", "answer": "Polymer crystallization and metal crystallization follow the general phase transition rules: 1 The crystallization process consists of nucleation and growth processes: (1) 1/2 Heterogeneous nucleation is dominant: (1-1) The kinetic equation has the same form as the Avrami equation: 4 Large undercooling results in small grain size. 5 The melting point of small particles is lower than that of large particles, exhibiting a Gibbs-Thomson-like effect." }, { "idx": 2469, "question": "For a silicon-containing low-alloy steel ingot with dendritic segregation and a dendrite arm spacing of 500μm, diffusion annealing is performed at 1200°C. To reduce the segregation amplitude to 10% of its original value, how long should the holding time be for carbon? Given that the diffusion coefficient of carbon in austenite at 1200°C is 2.23×10^(-6)cm^2/s.", "answer": "When dendritic segregation exists, the solute composition is distributed with the dendrite arm spacing as the period, where the period l is 500μm=0.05cm. According to the trigonometric series solution of the diffusion equation, the concentration amplitude decays with a decay factor, i.e., C=C¯+C0sin(πx/l)exp(-π^2Dt/l^2). If the amplitude decays to 1/10 of its original value, then exp(-π^2Dt/l^2)=0.1. Therefore, the required time for carbon is: t=l^2/(π^2D)ln0.1=-(0.025)^2/(π^2×2.23×10^(-6))ln0.1s=65.3s." }, { "idx": 2470, "question": "What cooling rate v is required to avoid spinodal decomposition at 550K when cooling from 850K to 550K, given the required time t=173.6 s?", "answer": "The cooling rate v must be greater than (850K - 550K) / 173.6 s = 1.44 K/s." }, { "idx": 2471, "question": "What are the main differences in crystallization ability between polymer crystallization and metal crystallization?", "answer": "Only thermoplastic polymers can crystallize. These polymers have a linear structure, and the crystallization process involves the arrangement of molecular chains from disorder to order, with no fixed melting temperature. Due to factors such as the long molecular chains, complex structure, and high melt viscosity of polymers, most polymer crystallizations are much slower than metal crystallizations and cannot fully crystallize, typically achieving a crystallinity of around 50%. The degree of crystallinity is closely related to the molecular structure of the polymer, and the more complex the structure, the more difficult it is to crystallize, often leading to the formation of a glassy state. Therefore, the ability of a polymer to form a crystalline state has an inverse relationship with its ability to form a glassy state." }, { "idx": 2472, "question": "What is the time t required for spinodal decomposition to occur over a diffusion distance of λ_m/2=50 nm (with λ_m=100 nm) at 550K, given the interdiffusion coefficient D=-2.4×10^-18 m²/s?", "answer": "The time t required is calculated as t = (λ_m/2)² / (6D) = (50×10^-9)² / (6 * 2.4×10^-18) s = 173.6 s." }, { "idx": 2473, "question": "A low-alloy steel ingot containing silicon exhibits dendritic segregation with a dendrite arm spacing of 500μm. After diffusion annealing at 1200°C, the segregation amplitude is reduced to 10% of its original value. How long should the holding time be for the silicon element? Assume the diffusion coefficient of silicon in austenite at 1200°C is 7.03×10^(-11)cm^2/s.", "answer": "When dendritic segregation exists, the solute composition is distributed with the dendrite arm spacing as the period, where the period l is 500μm=0.05cm. According to the trigonometric series solution of the diffusion equation, the concentration amplitude decays with a decay factor, i.e., C=C¯+C0sin(πx/l)exp(-π^2Dt/l^2). If the amplitude decays to 1/10 of its original value, then exp(-π^2Dt/l^2)=0.1. Therefore, the required time for the Si element is: t=l^2/(π^2D)ln0.1=-(0.025)^2/(π^2×7.03×10^(-11))ln0.1s=2.07×10^6s. Since Si in Fe is a substitutional solute atom while C is an interstitial solute atom, the diffusion rate of Si atoms is much slower than that of C, so the homogenization time for Si is several orders of magnitude slower than that for C." }, { "idx": 2474, "question": "The equilibrium melting point of nickel is 172515, the solid phase ρ̄=0.6cm³/1770l, the liquid/solid interface energy y=2.25×10⁻⁵·50⁻⁷. If the radius of spherical particles is 0.01μm, how much is the melting point reduced? Assume the molar melting ΔHₘ=[Γ̄Γ̄ḠḠJ†m0]", "answer": "The melting point Tᵣ of a crystal with a curvature radius of 0.01μm is Tᵣ = Tₘ - (2γVₛTₘ)/(rΔHₘ). Substituting the data gives Tᵣ = 1728K - (2×6.6×2.25×10⁻⁵×1728)/(10⁻⁶×18066)K ≈ 1699.6K, ΔT = 28.4K." }, { "idx": 2475, "question": "If two such subgrain boundaries merge to form one new subgrain boundary, by how much does the misorientation increase?", "answer": "If two such subgrain boundaries merge to form one new subgrain boundary, the dislocation density at the grain boundary doubles. Since the misorientation is proportional to the dislocation density, the misorientation becomes 2 × 10^-3 radians." }, { "idx": 2476, "question": "If the stored energy is mainly contributed by dislocations, establish the relationship between dislocation density and critical nucleus size.", "answer": "The energy of a dislocation per unit length is approximately Gb^2. Let the dislocation density be ρ, then the dislocation energy per unit volume is ρGb^2. If the stored energy E is mainly contributed by dislocations, the relationship between the critical nucleus radius and dislocation density is r* = 4γ/(ρGb^2)." }, { "idx": 2477, "question": "How much energy is released before and after the merging?", "answer": "The grain boundary energy per unit area after merging is 2θ E0(A - ln2θ), the energy of the two grain boundaries before merging is 2θ E0(A - lnθ), and the relative value of released energy after merging ΔE is ΔE = (2θ E0(A - lnθ) - 2θ E0(A - ln2θ))/(2θ E0(A - lnθ)) = (ln2θ - lnθ)/(0.5 - lnθ) = (ln(2 × 10^-3) - ln10^-3)/(0.5 - ln10^-3) = 9.35%." }, { "idx": 2478, "question": "A copper block subjected to large cold deformation has a stored energy of 2×10^5 J·m^-3, and the typical value of high-angle boundary energy is 0.5 J·m^-2. According to the classical homogeneous nucleation theory, what is the critical nucleus size for recrystallization?", "answer": "According to classical nucleation theory, the critical nucleus radius r* = 2γ/ΔG, where γ is the boundary energy and ΔG is the stored energy per unit volume. Substituting the boundary energy and stored energy, we obtain r* = (2×0.5)/(2×10^5) m = 5×10^-7 m." }, { "idx": 2479, "question": "What are the basic characteristics of the Bravais lattice?", "answer": "It has periodicity and symmetry, and each node is an equivalent point." }, { "idx": 2480, "question": "If a nucleus with a radius of 2×10^-9 m is to be formed, what should the dislocation density at the nucleation site be? (G=4×10^10 Pa, b=0.25×10^-9 m)", "answer": "If a nucleus with a radius of 2×10^-9 m is to be formed, the required dislocation density ρ is: ρ = 4γ/(r*Gb^2) = (4×0.5)/(2×10^-9×4×10^10×(0.25×10^-9)^2) m^-2 = 4×10^17 m^-2." }, { "idx": 2481, "question": "Demonstrate why there are no fewer than 14 Bravais lattices", "answer": "For any one of the 14 lattices, it is impossible to find a method of connecting nodes to form a new unit cell while preserving the symmetry." }, { "idx": 2482, "question": "Comment on the possibility of classical nucleation. It is known that the dislocation density of heavily worked metallic materials is approximately 10^15 m^-2.", "answer": "The dislocation density of heavily worked metallic materials is approximately 10^15 m^-2, making the formation of a critical nucleus of such size essentially impossible. The critical nucleus radius for classical nucleation discussed above is 5×10^-7 m, clearly indicating that nucleation cannot occur via the classical nucleation mechanism." }, { "idx": 2483, "question": "The subgrain boundaries formed after polygonization contain π edge dislocations, and the misorientation between subgrains is 10^-3 radians. Assuming there is no interaction between dislocations before polygonization, how much energy is released after polygonization (expressed as a percentage)? (The grain boundary energy E⊥-E0θ(A-1)=β, where A is set to 0.5)", "answer": "Before polygonization, the dislocations have no interaction, meaning their energy is the same as when they exist independently. Assuming the dislocations are edge dislocations, the total energy E1 of these π dislocations when they exist separately is πEe, where Ee is the energy per unit length of an edge dislocation: Ee = (μb^2)/(4π(1-ν)) ln(R/r0) ≈ 10θ E0, where E0 = μb^2/(4π(1-ν)), and the approximation ln(R/r0) = 10 is used. After polygonization, the energy Eb of the grain boundary formed by these π dislocations per unit area is Eb = E0θ(0.5 - lnθ). The relative energy released per unit area of the grain boundary before and after polygonization compared to E1 is ΔE/E1 = (E1 - Eb)/E1 = (10θ E0 - E0θ(0.5 - lnθ))/(10θ E0) = (10 - 0.5 + ln10^-3)/10 = 25.99%." }, { "idx": 2484, "question": "What is the lattice constant?", "answer": "The lengths a, b, c of the three adjacent edges in a unit cell and the angles α, β, γ between these edges determine the size and shape of the unit cell. These six parameters are called the lattice constants." }, { "idx": 2485, "question": "Demonstrate why there are no more than 14 Bravais lattices", "answer": "If each crystal system includes four types of lattices: simple, face-centered, body-centered, and base-centered, the seven crystal systems would yield 28 Bravais lattices. However, some of these 28 can be connected to form one of the 14 lattices without changing the symmetry. For example, a body-centered monoclinic lattice can be connected to form a base-centered monoclinic lattice, so it is not a new lattice type." }, { "idx": 2486, "question": "Write the specific indices of all equivalent crystal planes in the {123} plane family of the cubic crystal system", "answer": "{123}=(123)+(123̄)+(1̄23)+(12̄3)+(132)+(13̄2)+(1̄32)+(132̄)+(213)+(21̄3)+(2̄13)+(213̄)+(231)+(231̄)+(2̄31)+(23̄1)+(312)+(31̄2)+(3̄12)+(312̄)+(321)+(321̄)+(3̄21)+(32̄1)" }, { "idx": 2487, "question": "Using BCC, FCC, and hexagonal lattices as examples, explain the similarities and differences between unit cells and primitive cells.", "answer": "Both unit cells and primitive cells can reflect the periodicity of the lattice, meaning that infinite stacking of unit cells or primitive cells can reconstruct the entire lattice. However, unit cells are required to reflect the symmetry of the lattice, and the smallest volume unit under this premise is the unit cell; whereas primitive cells only require the smallest volume, and the primitive cells of Bravais lattices contain only one lattice point. For example: in a BCC unit cell, the number of lattice points is 2, while in a primitive cell it is 1; in an FCC unit cell, the number of lattice points is 4, while in a primitive cell it is 1; in a hexagonal lattice unit cell, the number of lattice points is 3, while in a primitive cell it is 1." }, { "idx": 2488, "question": "Calculate the length of the [10 0] crystal direction in a hexagonal crystal (in units of lattice constants a and c)", "answer": "The formula for the length of the [10 0] crystal direction in a hexagonal crystal: L = a" }, { "idx": 2489, "question": "Write all the equivalent crystallographic direction indices in the <123> direction family of the cubic crystal system", "answer": "<123>=[123]+[1̄23]+[12̄3]+[123̄]+[132]+[1̄32]+[13̄2]+[132̄]+[213]+[2̄13]+[21̄3]+[213̄]+[231]+[2̄31]+[23̄1]+[231̄]+[312]+[3̄12]+[31̄2]+[312̄]+[321]+[3̄21]+[32̄1]+[321̄]" }, { "idx": 2490, "question": "Calculate the length of the [11 0] crystal direction in a hexagonal crystal (in units of lattice constants a and c)", "answer": "The formula for the length of the [11 0] crystal direction in a hexagonal crystal: L = a√3" }, { "idx": 2491, "question": "Calculate the length of the [0001] direction in a hexagonal crystal (in units of lattice constants a and c)", "answer": "Formula for the length of the [0001] direction in a hexagonal crystal: L = c" }, { "idx": 2492, "question": "Calculate the lengths L (in units of the unit cell edge a) of low-index crystallographic directions with indices not exceeding 3 in cubic crystals.", "answer": "The length of a crystallographic direction is given by L=a·sqrt(u²+v²+w²), yielding:\\n<100>: 1\\n<110>: √2\\n<111>: √3\\n<200>: 2\\n<210>: √5\\n<211>: √6\\n<220>: 2√2\\n<221>: 3\\n<300>: 3\\n<310>: √10\\n<311>: √11\\n<222>: 2√3\\n<320>: √13\\n<321>: √14\\n<322>: √17\\n<330>: 3√2\\n<331>: √19\\n<332>: √22\\n<333>: 3√3" }, { "idx": 2493, "question": "How many lattice constants are there in each crystal system?", "answer": "
Crystal SystemRelationships among a, b, c, α, β, γNumber of Lattice Constants
Triclinica≠b≠c, α≠β≠γ≠90°6 (a, b, c, α, β, γ)
Monoclinica≠b≠c, α=γ=90°≠β or α=β=90°≠γ4 (a, b, c, γ or a, b, c, β)
Orthorhombica≠b≠c, α=β=γ=90°3 (a, b, c)
Tetragonala=b≠c, α=β=γ=90°2 (a, c)
Cubica=b=c, α=β=γ=90°1 (a)
Hexagonala=b≠c, α=β=90°, γ=120°2 (a, c)
Rhombohedrala=b=c, α=β=γ≠90°2 (a, α)
" }, { "idx": 2494, "question": "Calculate the interplanar spacings (in units of the unit cell edge length a) for low-index planes with indices not exceeding 3 in a cubic crystal.", "answer": "The interplanar spacing is given by d=a/sqrt(h²+k²+l²), yielding:\\n{100}: 1\\n{110}: √2/2\\n{111}: √3/3\\n{200}: 1/2\\n{210}: √5/5\\n{211}: √6/6\\n{220}: √2/4\\n{221}: 1/3\\n{300}: 1/3\\n{310}: √10/10\\n{311}: √11/11\\n{222}: √3/6\\n{320}: √13/13\\n{321}: √14/14\\n{322}: √17/17\\n{330}: √2/6\\n{331}: √19/19\\n{332}: √22/22\\n{333}: √3/9" }, { "idx": 2495, "question": "Based on the stacking characteristics of FCC and HCP crystals, demonstrate that the sizes of octahedral and tetrahedral interstices in these two types of crystals must be the same.", "answer": "Examining the (111) close-packed planes of FCC crystals and the (0001) close-packed planes of HCP crystals, it is found that the atomic arrangements of the two are identical. Further studying the adjacent two layers of close-packed planes reveals no difference in the way the layers fit together. In fact, only when examining three adjacent layers can the distinction between FCC and HCP be observed. However, both octahedral and tetrahedral interstices are only related to two layers of close-packed atoms. Therefore, for these two types of interstices, the microscopic environments provided by FCC and HCP are exactly the same, and their sizes must also be identical." }, { "idx": 2496, "question": "Calculate the length of the [10 1] crystal direction in a hexagonal crystal (in units of lattice constants a and c)", "answer": "The formula for the length of the [10 1] crystal direction in a hexagonal crystal: L = √(a² + c²)" }, { "idx": 2497, "question": "Calculate the angles between low-index crystallographic directions with indices not exceeding 3 in cubic crystals (presented in a list)", "answer": "Use the crystallographic direction angle formula cosθ=(u1u2+v1v2+w1w2)/sqrt((u1²+v1²+w1²)*(u2²+v2²+w2²)) to calculate. The angle between two crystallographic direction families may have multiple values depending on the selected directions. The specific calculation results need to list all crystallographic direction combinations with indices not exceeding 3 and their corresponding angles." }, { "idx": 2498, "question": "Compare the obtained results with the previous question", "answer": "The obtained results are exactly the same as the previous question, only the crystal plane indices {hkl} are replaced with [uvw]. Superficially, this is because the formula for the angle between crystal directions is identical to that for the angle between crystal planes. Upon deeper analysis, it is found that the crystal direction [uvw] is the normal direction of the crystal plane (hkl), forming a perpendicular relationship. Therefore, the angle between two crystal planes is always equal to the angle between crystal directions with the same indices." }, { "idx": 2499, "question": "Derive the four-index zone equation from the three-index zone equation of the hexagonal crystal system.", "answer": "The three-index zone equation for the hexagonal crystal system is $H\\\\sqcup+K V+L W=0$; the plane (H K L) is converted to four-index $(\\\\textsf{h k i}\\\\textsf{l})$, with $\\\\forall=h,\\\\quad\\\\forall=k,\\\\quad\\\\mathsf{L}=1$; the direction [U $\\\\pmb{\\\\upnu}$ W] is converted to four-index $[u\\\\textbf{v t}\\\\pmb{w}]$, with $\\\\Psi=2{\\\\mathsf{u}}+{\\\\mathsf{v}},\\\\quad\\\\forall{\\\\mathsf{{\\\\mathbf{\\\\bar{\\\\alpha}}}}}2{\\\\mathsf{v}}+{\\\\mathsf{u}},\\\\quad\\\\forall{\\\\mathsf{I}}={\\\\mathsf{w}};$; substituting into the zone equation, we get ${\\\\sf h}\\\\left(2{\\\\sf u}+{\\\\sf v}\\\\right)+{\\\\sf k}\\\\left(2{\\\\sf v}+{\\\\sf u}\\\\right)+{\\\\sf I}{\\\\sf w}=0;$; substituting $\\\\dag=-(h+k)$, $\\\\mathtt{t=-\\\\tau(u+v)}$ into the above equation, we obtain ${\\\\mathsf{h}}{\\\\mathsf{u}}+{\\\\mathsf{k}}{\\\\mathsf{v}}+{\\\\mathsf{i}}{\\\\mathsf{t}}+{\\\\mathsf{I}}{\\\\mathsf{w}}=0{\\\\mathsf{c}}$." }, { "idx": 2500, "question": "What is the bonding type of graphite?", "answer": "Within the same layer of graphite, the bonding is covalent, while between adjacent layers, the bonding is van der Waals forces." }, { "idx": 2501, "question": "What is the crystal structure of diamond?", "answer": "The crystal structure of diamond is FCC with tetrahedral interstices, where carbon atoms are located at the bonding points of the FCC lattice and four non-adjacent tetrahedral interstice positions." }, { "idx": 2502, "question": "What is the bonding type in diamond?", "answer": "In diamond, carbon atoms are bonded by covalent bonds." }, { "idx": 2503, "question": "What are the performance characteristics of diamond?", "answer": "Diamond has high hardness and a dense structure." }, { "idx": 2504, "question": "What is the reason for the difference in property changes between short-period elements and long-period elements?", "answer": "This is because the number of subshell electrons in the long-period transition elements also affects the properties of the elements." }, { "idx": 2505, "question": "What are the performance characteristics of graphite?", "answer": "Graphite has a loose structure and certain electrical conductivity, and is commonly used as a lubricant." }, { "idx": 2506, "question": "Derive the conversion formula between the lattice constant $a_{R}$ in rhombohedral axes and the lattice constant $\\pmb{a}_{\\sf H}$ in hexagonal axes for a rhombohedral crystal.", "answer": "Under $\\pmb{a}_{\\sf H}$, bH, $\\pmb{\\mathbb{G H}}$, $a_{R}=1/s$ [1 1], so the lattice constant $a_{R}=\\mathsf{L}$ $\\begin{array}{r l}&{=\\mathsf{a_{H}}\\bullet\\mathsf{s q r t}(\\mathsf{U}^{2}+\\mathsf{V}^{2}+\\mathsf{W}^{2}\\mathsf{c_{H}}^{2}/\\mathsf{a_{H}}^{2}-\\mathsf{U V})}\\ &{=\\V_{3}\\vee(3\\mathsf{a_{H}}^{2}+\\mathsf{c_{H}}^{2}),}\\end{array}$ Since ${\\pmb{\\upalpha}}_{\\aleph}$ is the angle between the crystallographic directions $\\%$ [1 1] and $\\%$ [121], the lattice constant $\\texttt{a}_{\\texttt{R}}$ $\\begin{array}{r l}{=\\mathsf{a r c o s}(c_{\\mathsf{G}_{\\mathsf{H}}}{}^{2}/\\mathsf{a}_{\\mathsf{H}}{}^{2}{}-3/2)/(3+\\mathsf{c}_{\\mathsf{H}}{}^{2}/\\mathsf{a}_{\\mathsf{H}}{}^{2})}\\ {=\\mathsf{a r c o s}(}&{(2\\mathsf{c}_{\\mathsf{H}}{}^{2}{}-3\\mathsf{a}_{\\mathsf{H}}{}^{2})/(6\\mathsf{a}_{\\mathsf{H}}{}^{2}+2\\mathsf{c}_{\\mathsf{H}}{}^{2}))\\circ}\\end{array}$ ) gives a H = aR·sqrt (2(1-cosα)); $c_{H}=a_{R}\\cdot\\mathsf{s q r t}(3(1+2\\cos\\theta)\\circ$." }, { "idx": 2507, "question": "Why do the properties of elements change periodically with atomic number?", "answer": "Because the properties of elements are mainly determined by the number of outer valence electrons, and the number of valence electrons changes periodically with atomic number, thus reflecting the periodic changes in the properties of elements." }, { "idx": 2508, "question": "What are the differences in the property changes between short-period elements and long-period elements?", "answer": "The properties of long-period elements change more continuously and gradually, while the properties of short-period elements differ more significantly." }, { "idx": 2509, "question": "Given the lattice constants of $\\mathsf{\\Omega}\\mathsf{a}-\\mathsf{A l}\\mathsf{\\Omega}_{2}\\mathsf{0}_{3}$ (rhombohedral crystal) as $\\mathtt{a}_{\\mathsf{R}}={\\mathsf{5}}.$ .12$\\mathring{\\mathbf{A}}$ 、 $a_{\\scriptscriptstyle\\textsf{R}}=55^{\\circ}17^{\\circ}$ ’, find its lattice constants $\\pmb{a}_{\\sf H}$ and $\\mathbf{GHo}$ in the hexagonal axes.", "answer": "Using the formula from the previous question and substituting the values of aR and $\\texttt{a}_{\\texttt{R}}$, we obtain $a_{\\scriptscriptstyle\\ H}={\\bf\\nabla}4.75\\mathrm{~\\AA~}$ 、 $\\pmb{\\upalpha}_{\\mathsf{H}}=12.97\\mathrm{~\\AA~}$." }, { "idx": 2510, "question": "What is the crystal structure of graphite?", "answer": "The crystal structure of graphite is a simple hexagonal lattice, with carbon atoms located at the lattice points." }, { "idx": 2511, "question": "Derive the conversion formula between the lattice constant $\\\\mathfrak{a}_{R}$ in rhombohedral axes and the lattice constant $\\\\pmb{\\\\complement H}$ in hexagonal axes for a rhombohedral crystal.", "answer": "In $\\\\pmb{a}_{\\\\sf H}$, $b_H$, $\\\\pmb{\\\\mathbb{G H}}$, $a_{R}=1/s$ [1 1], so the lattice constant $a_{R}=\\\\mathsf{L}$ $\\\\begin{array}{r l}&{=\\\\mathsf{a_{H}}\\\\bullet\\\\mathsf{s q r t}(\\\\mathsf{U}^{2}+\\\\mathsf{V}^{2}+\\\\mathsf{W}^{2}\\\\mathsf{c_{H}}^{2}/\\\\mathsf{a_{H}}^{2}-\\\\mathsf{U V})}\\\\ &{=\\\\V_{3}\\\\vee(3\\\\mathsf{a_{H}}^{2}+\\\\mathsf{c_{H}}^{2}),}\\\\end{array}$ Also, since ${\\\\pmb{\\\\upalpha}}_{\\\\aleph}$ is the angle between the crystallographic directions $\\\\%$ [1 1] and $\\\\%$ [121], the lattice constant $\\\\texttt{a}_{\\\\texttt{R}}$ $\\\\begin{array}{r l}{=\\\\mathsf{a r c o s}(c_{\\\\mathsf{G}_{\\\\mathsf{H}}}{}^{2}/\\\\mathsf{a}_{\\\\mathsf{H}}{}^{2}{}-3/2)/(3+\\\\mathsf{c}_{\\\\mathsf{H}}{}^{2}/\\\\mathsf{a}_{\\\\mathsf{H}}{}^{2})}\\\\ {=\\\\mathsf{a r c o s}(}&{(2\\\\mathsf{c}_{\\\\mathsf{H}}{}^{2}{}-3\\\\mathsf{a}_{\\\\mathsf{H}}{}^{2})/(6\\\\mathsf{a}_{\\\\mathsf{H}}{}^{2}+2\\\\mathsf{c}_{\\\\mathsf{H}}{}^{2}))\\\\circ}\\\\end{array}$ ) gives $a_H = a_R\\\\cdot\\\\sqrt{2(1-\\\\cos\\\\alpha)}$; $c_{H}=a_{R}\\\\cdot\\\\mathsf{s q r t}(3(1+2\\\\cos\\\\theta)\\\\circ$." }, { "idx": 2512, "question": "Calculate the angles between low-index crystal planes with indices not exceeding 3 in cubic crystals (presented in a list). Why are the angles independent of the lattice constant?", "answer": "Using the crystal plane angle formula $\\\\cos\\\\phi=(h_{1}h_{2}+k_{1}k_{2}+1,1,2)/\\\\{(h_{1}^{2}+k_{1}^{2}+1,^{2})\\\\ast(h_{2}^{2}+k_{2}^{2}+1,^{2}))$ to calculate. There may be multiple angles between two crystal plane families depending on the selected planes, only one is listed below, others are not discussed here.
cos in[100}[110}[111}[210}[211}[221}{310}
[100}1√2/2√3/32√5/5√ 6/32/33√10/10
[110}1√6/33√10/10√3/22√2/32√5/5
[111}1√15/52√2/35 √3/92√30/15
[210}1√30/62√5/57 √2/10
[211}17 √6/187 √15/30
[221}14√10/15
[310}1
The remaining results are omitted." }, { "idx": 2513, "question": "Discuss the significance of atomic radius in covalent crystals and its influencing factors, with examples", "answer": "For covalent crystals, the atomic radius is defined as half the distance between the nearest neighboring atomic nuclei of the same element in the crystal. The type of bonding between atoms in covalent crystals—whether single, double, or triple bonds—will affect the atomic radius, so the largest single-bond atomic radius r(1) is generally used." }, { "idx": 2514, "question": "Discuss the significance of atomic radius in metal crystals and its influencing factors, and provide examples for illustration", "answer": "For metal crystals, the atomic radius is defined as half the distance between the nuclei of nearest-neighbor atoms in the crystal of the same element. In metal crystals, the coordination number affects the atomic radius. For example, α-Fe (CN=8) has an atomic radius 3% smaller than that of γ-Fe (CN=12). Generally, the atomic radius for CN=12 is used." }, { "idx": 2515, "question": "Discuss the significance of atomic radius in molecular crystals of nonmetals and its influencing factors, with examples provided.", "answer": "For molecular crystals of nonmetals, there exist two atomic radii: one is the covalent radius, and the other is the van der Waals atomic radius (half the distance between adjacent molecules). For example, in the chlorine molecular crystal, the two radii are 0.099nm and 0.180nm, respectively." }, { "idx": 2516, "question": "Explain the term: component", "answer": "Each element (metal, non-metal) that constitutes an alloy." }, { "idx": 2517, "question": "Explain the term: alloy", "answer": "A material formed by the chemical bonding of a metal with one or more other elements." }, { "idx": 2518, "question": "Explain the term: sublattice", "answer": "The respective Bravais lattices occupied by the constituent atoms in an ordered solid solution." }, { "idx": 2519, "question": "Explain the term: intermetallic compound", "answer": "A compound formed between metals." }, { "idx": 2520, "question": "Explain the term: microstructure", "answer": "Under certain external conditions, the collective of several different phases that constitute an alloy of a certain composition." }, { "idx": 2521, "question": "Discuss the significance of atomic radius in ionic crystals and its influencing factors, with examples provided.", "answer": "For ionic crystals, the sizes of cations and anions are represented by ionic radii r+ and r-, respectively. Assuming that the same ion has the same radius in different ionic crystals, the ionic radius can be roughly determined. However, the ionic radius is only an approximate concept, as electrons cannot completely detach from the cation. Therefore, many ionic bonds have more or less covalent character. When this feature is particularly prominent, the significance of the ionic radius becomes less precise." }, { "idx": 2522, "question": "What is the relationship between the atomic arrangement and bonding in ordered alloys?", "answer": "This arrangement is caused by metallic bonds between atoms, where valence electrons collectively arrange the atoms in a regular pattern." }, { "idx": 2523, "question": "Explain the term: solid solution", "answer": "The atoms of the solute and the solvent occupy a common Bravais lattice, and this lattice type is the same as that of the solvent; the content of the components can vary within a certain range without causing a change in the lattice type. Metal or non-metal compounds possessing these two properties are called solid solutions." }, { "idx": 2524, "question": "Explain the term: phase", "answer": "A portion or region within an alloy that has the same (or continuously varying) composition, structure, and properties." }, { "idx": 2525, "question": "Explain the term: electronegativity", "answer": "A parameter indicating the ability of an element to attract electrons when forming compounds or solid solutions with other elements." }, { "idx": 2526, "question": "Explain the term: electron concentration", "answer": "The average number of valence electrons per atom in an alloy." }, { "idx": 2527, "question": "Explain the term: superstructure (superlattice)", "answer": "A complex lattice composed of the sublattices of each component in an ordered solid solution." }, { "idx": 2528, "question": "What are the characteristics of atomic arrangement in ordered alloys?", "answer": "In ordered alloys, atoms of each component occupy their respective Bravais lattices, and the entire alloy forms a superlattice composed of these sublattices." }, { "idx": 2529, "question": "Why do many ordered alloys become disordered at high temperatures?", "answer": "At high temperatures, the increased thermal motion of atoms causes them to break free from their original lattice positions, resulting in disordered atomic arrangements." }, { "idx": 2530, "question": "How to theoretically determine the order-disorder transition temperature (Curie temperature)?", "answer": "Theoretically, the transition temperature of ordered alloys can be determined by the relationship between the strength of metallic bonds and the magnitude of the average molecular free energy." }, { "idx": 2531, "question": "Briefly describe the influence of electronegativity difference on solid solubility according to the Hume-Rothery rules", "answer": "If the electronegativity difference between alloy components is large, the solid solubility is extremely small." }, { "idx": 2532, "question": "Briefly describe the practical significance of the Hume-Rothery rules", "answer": "Although the Hume-Rothery rules are only negative rules, qualitative or semi-quantitative rules, and the latter three are limited to specific situations, they summarize some patterns of alloy solid solubility, helping to predict the extent of solid solubility. Thus, they are highly useful for determining the properties and heat treatment behavior of alloys." }, { "idx": 2533, "question": "Briefly describe the law of solid solubility of specific group elements in the Hume-Rothery rules", "answer": "The solid solubility of solute elements from groups 11B to VB in solvent elements of group IB is the same (θ/a=1.36), regardless of the specific element types." }, { "idx": 2534, "question": "Briefly describe the influence of atomic valence on solid solubility according to the Hume-Rothery rules", "answer": "The solid solubility of two elements is related to their atomic valence. The solubility of a high-valence element in a low-valence element is greater than that of a low-valence element in a high-valence element." }, { "idx": 2535, "question": "Briefly describe the influence of atomic radius difference on solid solubility according to the Hume-Rothery rules", "answer": "If the difference in atomic radii of the elements forming the alloy exceeds 14%~15%, the solid solubility is extremely limited." }, { "idx": 2536, "question": "Using the Darken-Gurry diagram to analyze elements that may have relatively high solid solubility in Mg", "answer": "Possible elements include Cd, Nb, Ti, Ce, Hf, Zr, Am, P, Sc, and lanthanides." }, { "idx": 2537, "question": "What is the relationship between the mechanical properties of a solid solution and those of its pure components? Please provide a qualitative explanation.", "answer": "The strength and hardness of a solid solution are often higher than those of its components, while the plasticity is lower. This is because: (1) For interstitial solid solutions, solute atoms tend to preferentially distribute along dislocation lines, forming interstitial atom 'atmospheres' that firmly pin the dislocations, thereby strengthening the material; (2) For substitutional solid solutions, solute atoms are usually uniformly distributed within the lattice, causing lattice distortion, which increases the resistance to dislocation motion, though this strengthening effect is relatively smaller." }, { "idx": 2538, "question": "What is Vegard's law?", "answer": "Experiments have found that when two isomorphous salts form a continuous solid solution, the lattice constant of the solid solution has a linear relationship with the composition, meaning the lattice constant is proportional to the concentration of any component. This is Vegard's law." }, { "idx": 2539, "question": "Briefly describe the influence of crystal structure on solubility according to the Hume-Rothery rules", "answer": "Only when the two components have the same crystal structure can they form an infinite (or continuous) solid solution." }, { "idx": 2540, "question": "Why do real solid solutions often not conform to Vegard's law?", "answer": "Because Vegard's law reflects the influence of composition on the structure of alloy phases, but the structure of alloy phases is affected not only by composition but also by other factors (such as electron concentration, electronegativity, etc.), which cause real solid solutions to deviate from Vegard's law." }, { "idx": 2541, "question": "What is the relationship between the physical properties of a solid solution and those of the pure components? Please provide a qualitative explanation.", "answer": "The electrical, thermal, magnetic, and other physical properties of a solid solution also change continuously with composition, but generally not in a linear relationship. This is because solute atoms usually disrupt the original physical properties of the solvent. However, when the alloy is in an ordered state, the physical properties undergo abrupt changes, exhibiting superior physical performance." }, { "idx": 2542, "question": "Discuss the general rules of oxide structures.", "answer": "An important characteristic of oxide structures is the close packing of oxygen ions. In most simple oxide structures, oxygen ions are arranged in face-centered cubic, hexagonal close-packed, or approximately close-packed simple cubic configurations, while the cations occupy octahedral interstitial sites, tetrahedral interstitial sites, or the body center of the simple cubic lattice." }, { "idx": 2543, "question": "Analyze the crystal structure of rutile using Pauling's rules", "answer": "For rutile: (1) The cation-anion radius ratio is 0.48. According to Table 2-8 on page 104 of the textbook, it can be seen that the anion polyhedron is an octahedron, and the cation coordination number is 6; (2) Z+=4, Z-=2, so CN-=CN+•Z-/Z+=6/2=3." }, { "idx": 2544, "question": "Summarize and compare various intermetallic compounds such as valence compounds, electron compounds, TCP phases, and interstitial phases (interstitial compounds) in terms of the main factors determining their structures and theoretical foundations.", "answer": "
Valence compoundsElectron compoundsTCP phasesInterstitial phases
Electronegativity, electron shell theoryElectron concentration, electron theoryComponent atomic radius ratio, topologyComponent atomic radius ratio, spatial geometry
" }, { "idx": 2545, "question": "Describe Pauling's rules regarding the structure of ionic compounds", "answer": "(1) A coordination polyhedron of anions is formed around the cation, with the distance between the cation and anion determined by the sum of their ionic radii, and the coordination number determined by the ratio of the cation to anion radii; (2) The number of valence electrons provided by the cation equals the number of valence electrons received by the anion, hence Z+/CN+=Z-/CN-; (3) In a coordination structure, when coordination polyhedra share edges, especially faces, their stability decreases, and this effect is more pronounced for cations with higher valence and lower coordination numbers; (4) In crystals containing more than one type of cation, the coordination polyhedra of anions around cations with higher valence and smaller coordination numbers tend to connect by sharing vertices; (5) The types of coordination polyhedra in a crystal tend to be minimized." }, { "idx": 2546, "question": "Summarize and compare the performance characteristics of various intermetallic compounds such as valence compounds, electron compounds, TCP phases, and interstitial phases (interstitial compounds)", "answer": "
Valence compoundsElectron compoundsTCP phasesInterstitial phases
Non-metallic or semiconducting propertiesDistinct metallic characteristicsHard and brittle phases, Cr3Si-type structure alloys mostly exhibit superconducting propertiesWide mutual solid solubility range, distinct metallic properties, very high melting points, extremely high hardness and brittleness
" }, { "idx": 2547, "question": "Summarize and compare various intermetallic compounds such as valence compounds, electron compounds, TCP phases, and interstitial phases (interstitial compounds) in terms of their constituent elements.", "answer": "
Valence compoundsElectron compoundsTCP phasesInterstitial phases
Metal and metal, metal and metalloidElements, Group VII (ironSmall metal elementsGroup metal elements and atomic radii
" }, { "idx": 2548, "question": "Summarize and compare various intermetallic compounds such as valence compounds, electron compounds, TCP phases, and interstitial phases (interstitial compounds) in terms of their structural characteristics", "answer": "
Valence compoundsElectron compoundsTCP phasesInterstitial phases
Transfer or sharing to form stable 8-electronStructure mainly determined by electron concentrationStacked in a certain order by close-packed tetrahedronsFCC or CPH structure, metalloid atoms located in tetrahedral
" }, { "idx": 2549, "question": "What are the main types of silicate structures?", "answer": "The main types of silicate structures can be classified according to the spatial arrangement of silicon-oxygen tetrahedra into: island, chain, sheet, and framework." }, { "idx": 2550, "question": "Given that the outer surface of a copper single crystal sample is (111), analyze the possible angles between slip lines that may appear on this outer surface when the crystal undergoes slip at room temperature.", "answer": "The possible slip planes are the {111} family of crystal planes, and their intersections with the (111) plane may be [ 10], [0 1], [10 ], so the slip lines are either parallel or at a 60-degree angle to each other." }, { "idx": 2551, "question": "Summarize and compare various intermetallic compounds such as valence compounds, electron compounds, TCP phases, and interstitial phases (interstitial compounds) in terms of bonding types.", "answer": "
Valence compoundsElectron compoundsTCP phasesInterstitial phases
Ionic or covalent bondsPrimarily metallic bondsMetallic bondsMixed type: metallic bonds between metal atoms, and covalent bonds between metal and metalloid atoms
" }, { "idx": 2552, "question": "If tension is applied along the [2 3] direction of an aluminum single crystal, determine the initial slip system.", "answer": "The aluminum single crystal has an FCC structure, and [2 3] lies within the orientation triangle [001]―[1 1]―[101]. Therefore, the initial slip system is (111)[0 1]." }, { "idx": 2553, "question": "Write all possible slip systems for an FCC crystal at room temperature (specific crystallographic plane and direction indices are required).", "answer": "There are a total of 12 possible slip systems: (111)[10 ], (111)[01 ], (111)[1 0], ( 11)[110], ( 11)[0 1], ( 11)[101], (1 1)[110], (1 1)[10 ], (1 1)[011], (11 )[011], (11 )[101], (11 )[1 0]." }, { "idx": 2554, "question": "Summarize and compare various intermetallic compounds such as valence compounds, electron compounds, TCP phases, and interstitial phases (interstitial compounds) from typical examples.", "answer": "
Valence compoundsElectron compoundsTCP phasesInterstitial phases
MgSe, Pt2P, Mg2Si, MnS, MgS, MnAsCuZn, CusZn8, CuZn3MgCu2, MgZn2, MgNi2 (Laves phases), Fe-Cr alloys (phases), Cr3SiFe4N, Fe2N, NaH, TiH2 (simple); Fe3C, Cr23C6, Fe4W2C (complex)
" }, { "idx": 2555, "question": "If tension is applied along the [2 3] direction of an aluminum single crystal, determine the rotation rule and the rotation axis.", "answer": "The specimen axis turns toward [0 1], and the rotation axis is [2 1 3]×[0 1 1]=[2 1 2], i.e., [1 ]." }, { "idx": 2556, "question": "If tension is applied along the [110] direction of an FCC crystal, please write down the possible activated slip systems.", "answer": "There are four possible activated slip systems, which are (111)[101], (111)[011], (111)[110], (111)[011]." }, { "idx": 2557, "question": "Using computer verification, it is determined that the image rule for slip systems applies to both FCC crystals and BCC crystals with {110}<111> slip systems. (Hint: For any given external force direction, use a computer to calculate the orientation factors for all equivalent slip systems.)", "answer": "μ=cosλcosφ, calculate μ for all equivalent slip systems, and it can be found that μ_max must correspond to the slip system selected by the image rule." }, { "idx": 2558, "question": "If tension is applied along the [2 3] direction of an aluminum single crystal, determine the double slip system.", "answer": "The double slip system is (111)[0 1]－( 1)[101]." }, { "idx": 2559, "question": "Given that the outer surface of a copper single crystal sample is (001), analyze the possible angles between slip lines that may appear on this outer surface when the crystal undergoes slip at room temperature.", "answer": "The possible slip planes are the {111} family of crystal planes, and their intersection lines with the (001) plane can only be [110] and [1 0], so the slip lines are either parallel or perpendicular to each other." }, { "idx": 2560, "question": "If a single crystal aluminum rod with a diameter of $5mm$ starts to slip when a tensile force of 40N is applied along the rod axis [123], find the critical resolved shear stress of aluminum during slipping.", "answer": "Single crystal aluminum has an FCC structure, and the slip system is {111}<110>. Using the imaging rule, the slip plane and slip direction are determined to be (11)[101]. The angles between them and the axis are respectively: $$ \\begin{array}{r l}&{\\mathtt{cos}\\Phi=\\left[123\\right]\\bullet[\\overline{{1}}11]/(\\left|\\left[123\\right]\\right|\\left[\\overline{{1}}11\\right]|)=4/\\surd42;}\\ &{\\mathtt{cos}\\mathtt{\\lambda}=\\left[123\\right]\\bullet[101]/(\\left|\\left[123\\right]\\right|\\left[101\\right]|)=2/\\surd7;}\\end{array} $$ Therefore, the critical resolved shear stress τc = $\\mathsf{\\bar{cos}\\lambda cos\\Phi/A_{0}=\\lambda\\cdots=0.95MPa}.$" }, { "idx": 2561, "question": "What are the basic characteristics of silicate structures?", "answer": "The basic characteristics of silicate structures include: (1) The fundamental structural unit of silicates is the [SiO4] tetrahedron, with silicon atoms located in the interstices of the oxygen tetrahedron; (2) Each oxygen can be shared by at most two [SiO4] tetrahedra; (3) [SiO4] tetrahedra can exist isolated in the structure or be interconnected by sharing vertices; (4) The Si-O-Si bond forms a bent line." }, { "idx": 2562, "question": "If a single crystal of aluminum is stretched along the [2 3] direction, determine the final orientation (stable orientation) of the crystal.", "answer": "From (5), it can be seen that the final orientation of the crystal is [1 2]." }, { "idx": 2563, "question": "If tension is applied along the [2 3] direction of an aluminum single crystal, determine the crystal orientation and shear strain at the onset of double slip.", "answer": "Using L=ds/+dt dt∨(/⋅↗)db, let L=[u u w], then L=[2 1 3]+4Y[0 1 1]/√6, from which it can be determined that u=2, v=4, γ=√6/4. Therefore, the crystal orientation is [2 4], i.e., [1 2], and the shear strain is √6/4." }, { "idx": 2564, "question": "Niobium single crystal has a BCC structure, with [2 3] located in the orientation triangle [001]―[1 1]―[101]. What is the initial slip system?", "answer": "The initial slip system is (0 1)[111]." }, { "idx": 2565, "question": "Change the tension in the previous question to compression, determine the initial slip system", "answer": "[2 3] is located in the orientation triangle [001]―[1 1]―[101], so the initial slip system is (111)[0 1]" }, { "idx": 2566, "question": "Change the tension in the previous question to compression, determine the double slip system", "answer": "The double slip system is (111)[0 1]－(1 1)[011]" }, { "idx": 2567, "question": "If tension is applied along the [2 3] direction of an aluminum single crystal, determine the rotation law and rotation axis of the crystal during double slip.", "answer": "During double slip, the specimen axis rotates towards [0 1], with the rotation axis n1=[1 1 2]×[0 1 1]=[1 1 1], and simultaneously rotates towards [101], with the rotation axis n2=[1 1 2]×[101]=[1 1 1]. The resultant rotation axis is [000], so the crystal no longer rotates." }, { "idx": 2568, "question": "The sample axis is rotated to [111], what is the rotation axis?", "answer": "The rotation axis is [2 1 3]×[111]=[4 1 3]." }, { "idx": 2569, "question": "Change the tension in the previous question to compression, determine the specimen axis rotation direction and rotation axis", "answer": "The specimen axis rotation direction is [111], and the rotation axis is $[2\\bar{1}3]\\times[111]=[\\bar{4}13]$" }, { "idx": 2570, "question": "Change the tension in the previous question to compression, and calculate the crystal orientation and shear strain", "answer": "Using $A=a-\\textsf{v}(a\\cdot b)n,$, let $\\textsf{A}=\\textsf{[u0w]}$, we get $\\mathsf{A}=[2\\bar{1}3]-4\\lor[111]/\\surd6$. From this, it can be seen that $\\scriptstyle{\\boldsymbol{\\mathsf{u}}}=3$, $v=4$, $\\gamma=-\\sqrt{6/4}$, so the crystal orientation is [304], and the shear strain is - $\\surd6/4$" }, { "idx": 2571, "question": "Change the tension in the previous question to compression, analyze the rotation of the specimen axis and the resultant rotation axis during double slip.", "answer": "During double slip, the specimen axis rotates towards [111], with the rotation axis $\\mathsf{n}_{1}=[304]\\times[111]=[\\bar{4}13]$, and simultaneously rotates towards [1 1], with the rotation axis $n_{2}=[304]\\times[1\\overline{{{1}}}1]=[41\\overline{{{3}}}]$. The resultant rotation axis is [020], i.e., [010], so after double slip, point F moves along the edge [001]－[101]." }, { "idx": 2572, "question": "What is a double slip system?", "answer": "The double slip system is (0 1)[111]－(011)[1 1]." }, { "idx": 2573, "question": "Change the tension in the previous question to compression, and determine the stable orientation", "answer": "Let the stable orientation be $[\\\\boldsymbol{\\\\mathbf{u}}^{},~\\\\boldsymbol{\\\\mathbf{0}}\\\\boldsymbol{\\\\mathbf{w}}^{}~]$. To make $\\\\boldsymbol{\\\\mathsf{n}}=[000]$, it is required that $\\\\begin{array}{r l}{[\\\\mathbf{u}^{\\\\prime}}&{{}0\\\\mathbf{w}^{\\\\prime}]\\\\times([111]+[1\\\\bar{1}1])}\\\\end{array}$ $=[000]$, i.e., $\\\\begin{array}{r l}{\\\\upmu^{\\\\prime}}&{{}=\\\\upmu^{\\\\prime}}\\\\end{array}$. Therefore, the stable orientation is [101]" }, { "idx": 2574, "question": "Using L=ds/+ds†(′⋅η)b, let L=[u 0 w], find the crystal orientation and shear strain.", "answer": "L=[2 1 3]+4∨[111]/√6, from which it can be seen that u=3, v=4, γ=√6/4, so the crystal orientation is [304], and the shear strain is √6/4." }, { "idx": 2575, "question": "Assuming the stable orientation is [u′, 0 w′], what should the stable orientation be to make n=[000]?", "answer": "To make n=[000], it is required that [u′ 0 w′]×([111]+[1 1 1])=[000], i.e., u′=u′’, thus the stable orientation is [101]." }, { "idx": 2576, "question": "During double slip, the specimen axis turns towards [111] and [1 1]. What is the resultant rotation axis? Along which direction does point F move after double slip?", "answer": "During double slip, the specimen axis turns towards [111] with a rotation axis n1=[304]×[111]=[4 1 3], and simultaneously turns towards [1 1] with a rotation axis n2=[304]×[1 1 1]=[4 1 3]. The resultant rotation axis is [020], i.e., [010]. Therefore, after double slip, point F moves along the edge [001]－[101]." }, { "idx": 2577, "question": "Practice has shown that highly cold-rolled magnesium sheets tend to crack during deep drawing. Analyze the reasons for this.", "answer": "After cold rolling, magnesium sheets develop a (0001)<11 0> texture. When stress is applied parallel or perpendicular to the sheet surface, the orientation factor is zero, resulting in almost no plasticity, making further processing prone to cracking." }, { "idx": 2578, "question": "It is known that the critical resolved shear stress required for twinning in magnesium single crystals is several times larger than that for slip. When a magnesium single crystal is stretched along the [0001] direction, what is the deformation mode of the crystal?", "answer": "The slip systems of magnesium single crystals are (0001)<11-20> and {10-10}<11-20>, and the possible slip directions are all perpendicular to [0001], so slip does not occur in this case; c/a=1.62<√3, so [0001] is in the obtuse angle region of K1 and K2, and twinning will increase during deformation. Therefore, twinning occurs when stretched along the [0001] direction, and twinning changes the crystal orientation, which may further lead to slip." }, { "idx": 2579, "question": "Can inserting an additional columnar half-atom plane in a crystal form a dislocation loop? Why?", "answer": "A dislocation loop cannot be formed. Assuming a dislocation loop could form, it would consist entirely of edge dislocations. According to l⊥b, the Burgers vector $\\pmb{b}$ at each point of the loop should be along the radial direction, meaning the $\\pmb{b}$ at different points on the loop would vary. This contradicts the fact that a single dislocation line has only one $\\pmb{b}$." }, { "idx": 2580, "question": "Highly cold-rolled aluminum sheets will form a well-developed {001}<100> texture (cube texture) after high-temperature annealing. If such an aluminum sheet is deep-drawn into a cup, how many ears will be produced?", "answer": "Eight ears will be produced during deep drawing." }, { "idx": 2581, "question": "What is the fundamental reason for the formation of deformation texture (or processing texture)?", "answer": "The fundamental reason for the formation of deformation texture is that during processing, each grain slips along certain slip planes and rotates according to specific rules, causing the slip direction to align with the principal strain direction or the slip plane to align with the compression plane. Therefore, when the deformation is sufficiently large, the slip directions or slip planes of a large number of grains will become parallel to the tensile direction or compression plane, thus forming the texture." }, { "idx": 2582, "question": "What is texture (or preferred orientation)?", "answer": "After cold working, the orientations of metal grains exhibit a certain relationship, and such a distribution of orientations is called preferred orientation, i.e., texture." }, { "idx": 2583, "question": "Highly cold-rolled aluminum sheets form a well-developed {001}<100> texture (cube texture) after high-temperature annealing. If such an aluminum sheet is deep-drawn into a cup, where will the ears appear?", "answer": "Four ears appear in the [010], [0 0], [100], [ 00] directions, while four small ears may also form in the <110>, <1 0>, < 0>, < 10> directions." }, { "idx": 2584, "question": "In a simple cubic crystal, pure bending of the (010) plane around the [001] axis will form what type of dislocation (specify the direction of the dislocation line and the Burgers vector).", "answer": "Edge type, dislocation line direction=[001], Burgers vector=a[010]" }, { "idx": 2585, "question": "For a simple cubic crystal, pure bending of the (110) plane around the [001] axis will form what type of dislocations (specify the direction of the dislocation line and the Burgers vector).", "answer": "Edge type, dislocation line direction=[001], Burgers vector=a[100] or a[010]" }, { "idx": 2586, "question": "FCC crystal, pure bending occurs on the (110) plane around the [001] axis, what kind of dislocations will form (specify the direction of the dislocation line and the Burgers vector).", "answer": "Edge type, dislocation line direction=[001], Burgers vector=a[110]/2" }, { "idx": 2587, "question": "When a simple cubic crystal is twisted by an angle θ around the [001] axis, what kind of dislocation will form (specify the direction of the dislocation line and the Burgers vector)?", "answer": "Screw type, dislocation line direction=[001], Burgers vector=a[001]" }, { "idx": 2588, "question": "It is known that the critical resolved shear stress required for twinning in magnesium single crystals is several times larger than that for slip. When compressing along the [0001] direction of a Mg single crystal, what is the deformation mode of the crystal?", "answer": "The slip systems of magnesium single crystals are (0001)<11-20> and {10-10}<11-20>, and the possible slip directions are all perpendicular to [0001], so slip does not occur in this case; c/a=1.62<√3, so [0001] is in the obtuse angle region of K1 and K2, and it will elongate during twinning. When compressed along the [0001] direction, neither slip nor twinning can occur, and the crystal exhibits strong brittleness." }, { "idx": 2589, "question": "When the crystal around an edge dislocation contains excess interstitial atoms, how will the dislocation climb?", "answer": "Negative climb" }, { "idx": 2590, "question": "When the crystal around an edge dislocation contains an excess of vacancies, how will the dislocation climb?", "answer": "Positive climb" }, { "idx": 2591, "question": "When the crystal around an edge dislocation contains vacancies below the equilibrium concentration, how will the dislocation climb?", "answer": "Negative climb" }, { "idx": 2592, "question": "A screw dislocation with $b=a/2$ [10] moves on the (111) plane. If it encounters an obstacle during motion and undergoes cross-slip, please identify the cross-slip system.", "answer": "∵ All planes containing the direction of the screw dislocation are slip planes, ∴ For the FCC crystal slip plane (111), only (111) and (11 ) contain $I=b={\\\\sf a}/2$ [10]. ∴ If cross-slip occurs, it must be from the (111) plane to the (11 ) plane." }, { "idx": 2593, "question": "A closed dislocation loop lies on the slip plane of a prism with a square cross-section. The two sides of the square are along the x and y axes, and the Burgers vector is along the z-axis. If the dislocation loop can only glide, determine the equilibrium shape of the dislocation loop and the critical stress for initiation under the stress distribution condition τxz = τyz = τ = const. (Assume the line tension is approximately constant.)", "answer": "f = (σ ⋅ b) × υ = (u - v)τbk; the initiation stress τp = 2G exp(-2πw/b)/(1-υ)." }, { "idx": 2594, "question": "If the applied stress is uniformly distributed, find the net force acting on any dislocation loop.", "answer": "In general, the formula for the force on a dislocation is ${\\mathsf{d}}F=({\\textsf{\\sigma}}\\cdot{\\textsf{\\em b}})\\times{\\mathsf{d}}/,$. Since the applied stress is uniformly distributed, $\\pmb{\\upsigma}$ and $\\pmb{b}$ are both constants. Therefore, the net force on any dislocation loop is $\\oint\\mathbf{d}F=\\oint{\\bigl(}\\upsigma\\bullet b{\\bigr)}\\times\\mathbf{d}/={\\bigl(}\\upsigma\\bullet b{\\bigr)}\\oint\\mathbf{d}/={\\bigl(}\\upsigma$ • $b)\\times0=0.$." }, { "idx": 2595, "question": "When the crystal around an edge dislocation contains interstitial atoms below the equilibrium concentration, how will the dislocation climb?", "answer": "Positive climb" }, { "idx": 2596, "question": "Derive the elastic energy formula for a mixed dislocation.", "answer": "$\\\\begin{array}{r l}&{\\\\mathsf{E}=\\\\mathsf{E}_{\\\\mathsf{o}1}\\\\left(\\\\mathcal{V}\\\\right)+\\\\mathsf{E}_{\\\\mathsf{e}1}\\\\left(\\\\frac{\\\\mathsf{i}\\\\mathcal{R}}{\\\\mathsf{s}\\\\mathsf{t}_{\\\\mathrm{A}}}\\\\right)=\\\\mathsf{G b}^{2}|\\\\mathsf{s}\\\\mathsf{i n}^{2}\\\\mathsf{a}\\\\star|\\\\mathsf{n}\\\\left(\\\\mathsf{R}/\\\\mathsf{r}_{0}\\\\right)/\\\\left[4\\\\pi\\\\left(1-\\\\upsilon\\\\right)\\\\right]+\\\\mathsf{G b}^{2}|\\\\mathsf{c o s}^{2}\\\\mathsf{a}\\\\star|\\\\mathsf{n}\\\\left(\\\\mathsf{R}/\\\\mathsf{r}_{0}\\\\right)}\\\\ &{\\\\mathrm{~\\\\ensuremath~{\\\\left/~\\\\left(4\\\\pi\\\\right)~\\\\tau~=~\\\\mathsf{Gb}^2|\\\\tau~\\\\left(1-\\\\upsilon\\\\cos^2\\\\theta\\\\right)~\\\\cdot~|\\\\mathsf{n}\\\\left(\\\\mathsf{R}/\\\\mathsf{r}_{0}\\\\right)~\\\\middle/~\\\\left[4\\\\pi\\\\left(1-\\\\upsilon\\\\right)\\\\right]~\\\\right.~}}\\\\end{array}$" }, { "idx": 2597, "question": "A closed dislocation loop lies on the slip plane of a prism with a square cross-section. The two sides of the square are along the x and y axes, and the Burgers vector is along the z-axis. If the dislocation loop can only glide, determine the equilibrium shape of the dislocation loop and the critical stress for initiation under the stress distribution condition τyz = τ = const, τxz = 0. (Assume the line tension is approximately constant.)", "answer": "f = (σ ⋅ b) × υ = -uτbk; the initiation stress τp = 2G exp(-2πw/b)/(1-υ)." }, { "idx": 2598, "question": "What kind of a pair of dislocations is equivalent to a row of vacancies (or a row of interstitial atoms)?", "answer": "When the half-atomic planes of a positive edge dislocation and a negative edge dislocation lie on the same plane, leaving out one atomic site in between will form a row of vacancies, while overlapping one atomic site will form a row of interstitial atoms." }, { "idx": 2599, "question": "On the (111) plane of a copper single crystal, there is a right-handed screw dislocation with $b=a/2$ [10 ], where $a=0.36\\\\mathsf{n m}$. Now, a tensile stress of $10^{6}\\\\mathsf{P a}$ is applied along the [001] direction. Find the force acting on the screw dislocation.", "answer": "Using the Peach-Koehler formula, we obtain \\n\\n$$\\n{f}=(\\\\sigma\\\\cdot b)\\\\times\\\\upsilon\\n$$\\n\\n$$\\n\\\\begin{array}{r l}&{\\\\langle\\\\left[\\\\begin{array}{l}{\\\\bar{\\\\mathbb{O}}\\\\bar{\\\\mathbb{O}}\\\\bar{\\\\mathbb{O}}}\\\\ {\\\\bar{\\\\mathbb{O}}\\\\bar{\\\\mathbb{O}}\\\\bar{\\\\mathbb{O}}}\\\\end{array}\\\\right]\\\\bullet\\\\frac{\\\\bar{\\\\mathbb{a}}}{2}\\\\left[\\\\begin{array}{l}{1}\\\\ {\\\\bar{\\\\mathbb{O}}}\\\\ {-1}\\\\end{array}\\\\right]\\\\rangle\\\\times\\\\frac{\\\\sqrt{2}}{2}\\\\left[\\\\begin{array}{l}{1}\\\\ {\\\\bar{\\\\mathbb{O}}}\\\\ {-1}\\\\end{array}\\\\right]=\\\\frac{\\\\rho}{(-\\\\sqrt{2/4})\\\\mathsf{a}\\\\upsigma_{z}j},}\\\\end{array}\\n$$\\n\\nThus, $\\\\pounds=1.27\\\\times10^{-4}$ N/m, and the direction is along the negative $\\\\pmb{\\\\ y}$-axis." }, { "idx": 2600, "question": "Estimate the equilibrium width of an extended dislocation in Al. Given the lattice constant of Al a = 0.404 nm, shear modulus G = 3×10^6 N/cm^2, and stacking fault energy γ_I = 166×10^-6 J/m^2.", "answer": "Al has a face-centered cubic structure, and the equilibrium width of an extended dislocation d_0 = G a^2 / (24πγ_I). Substituting the data: d_0 = (3×10^6 N/cm^2) × (0.404×10^-9 m)^2 / (24π × 166×10^-6 J/m^2) = 3.91×10^-7 m." }, { "idx": 2601, "question": "Which four quantum numbers can determine the spatial position and energy of an electron in an atom?", "answer": "Principal quantum number n, orbital angular momentum quantum number li, magnetic quantum number mi, and spin angular momentum quantum number si" }, { "idx": 2602, "question": "Estimate the equilibrium width of an extended dislocation in Cu. Given: lattice constant of Cu a = 0.361 nm, shear modulus G = 5×10^6 N/cm^2, stacking fault energy γ_I = 45×10^-6 J/m^2.", "answer": "Cu has a face-centered cubic structure, and the equilibrium width of an extended dislocation d_0 = G a^2 / (24πγ_I). Substituting the data: d_0 = (5×10^6 N/cm^2) × (0.361×10^-9 m)^2 / (24π × 45×10^-6 J/m^2) = 1.92×10^-6 m." }, { "idx": 2603, "question": "In multi-electron atoms, what principles should the arrangement of extranuclear electrons follow?", "answer": "The principle of lowest energy, Pauli exclusion principle, Hund's rule." }, { "idx": 2604, "question": "In the periodic table, what are the differences in atomic structure of elements in the same period from left to right?", "answer": "From left to right, the nuclear charge increases sequentially, and the atomic radius gradually decreases." }, { "idx": 2605, "question": "In the periodic table, how do the properties of elements in the same main group change from top to bottom?", "answer": "From top to bottom, the ionization energy decreases, the ability to lose electrons increases, the ability to gain electrons decreases, the metallic character increases, and the non-metallic character decreases." }, { "idx": 2606, "question": "In the periodic table, what are the common characteristics of atomic structures for elements in the same main group?", "answer": "Elements in the same main group have the same number of outer electrons." }, { "idx": 2607, "question": "Why is the relative atomic mass of an element not always a whole number?", "answer": "Because different isotopes contain varying numbers of neutrons (while having the same number of protons), elements with multiple isotopes in different abundances have a relative atomic mass that is not a whole number." }, { "idx": 2608, "question": "Estimate the equilibrium width of extended dislocations in stainless steel. Given the lattice constant of stainless steel a = 0.356 nm, shear modulus G = 10×10^6 N/cm^2, and stacking fault energy γ_I = 15×10^-6 J/m^2.", "answer": "Stainless steel has a face-centered cubic structure, and the equilibrium width of extended dislocations d_0 = G a^2 / (24πγ_I). Substituting the data: d_0 = (10×10^6 N/cm^2) × (0.356×10^-9 m)^2 / (24π × 15×10^-6 J/m^2) = 1.12×10^-5 m." }, { "idx": 2609, "question": "In the periodic table, what are the differences in atomic structure from top to bottom for elements in the same main group?", "answer": "From top to bottom, the number of electron shells increases, and the atomic radius becomes larger." }, { "idx": 2610, "question": "In the periodic table, what common characteristics do the atomic structures of elements in the same period have?", "answer": "Elements in the same period have the same number of electron shells outside the atomic nucleus." }, { "idx": 2611, "question": "What are isotopes?", "answer": "Substances that occupy the same position in the periodic table, although their masses are different, yet their chemical properties are the same, are called isotopes." }, { "idx": 2612, "question": "In the periodic table, how do the properties of elements change from left to right within the same period?", "answer": "From left to right, ionization energy increases, the ability to lose electrons decreases, the ability to gain electrons increases, metallic character weakens, and non-metallic character strengthens." }, { "idx": 2613, "question": "The atomic number of chromium is 24, and it has a total of 4 isotopes: 4.31% of Cr atoms contain 26 neutrons, 83.76% contain 28 neutrons, 9.55% contain 29 neutrons, and 2.38% contain 30 neutrons. Calculate the relative atomic mass of chromium.", "answer": "$A{\\tau}=0.0431\\times(24+26)+0.8376\\times(24+28)+0.0955\\times(24+29)$ $+0.0238\\times(24+30)=52.057$ 1-6 $A" }, { "idx": 2614, "question": "The atomic number of tin is 50, and except for the 4f subshell, all other inner electron subshells are filled. Determine the number of valence electrons in tin from the perspective of atomic structure.", "answer": "1s²2s²2p63s²3p63d104s²4p64d105s²5p²; the number of valence electrons in tin is 4." }, { "idx": 2615, "question": "Given that the atomic number of an element is 32, based on the knowledge of atomic electronic structure, write its electron configuration.", "answer": "1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2" }, { "idx": 2616, "question": "The atomic number of copper is 29, its relative atomic mass is 63.54, and it has two isotopes, Cu63 and Cu65. Calculate the percentage content of the two copper isotopes.", "answer": "$A{\\\\mathrm{r}}=63.54=63x+65\\\\times(1-x);\\\\quad x={\\\\frac{65-63.54}{2}}=0.73\\\\rightarrow73\\\\mathcal{V}_{0}\\\\left(\\\\mathrm{Cu}^{\\\\mathrm{63}}\\\\right);$ $1-x=0.27\\\\substack{\\\\rightarrow27\\\\%(\\\\mathrm{Cu}^{65})}$" }, { "idx": 2617, "question": "What are the characteristics of covalent bonds?", "answer": "Covalent bonds: shared electron pairs, with saturation and directionality." }, { "idx": 2618, "question": "What are the characteristics of physical bonds (van der Waals forces)?", "answer": "Physical bonds: secondary bonds, also known as van der Waals forces." }, { "idx": 2619, "question": "Given that the atomic number of an element is 32, based on its electron configuration, indicate which period it belongs to.", "answer": "The fourth period" }, { "idx": 2620, "question": "The atomic number of platinum is 78, it has only 9 electrons in the 5d subshell and no electrons in the 5f subshell. How many electrons are there in the 6s subshell of Pt?", "answer": "$\\\\mathrm{ls^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}4f^{14}5s^{2}5p^{6}5d^{9}6s^{1}};$ $2+8+18+32+17=77;\\\\quad78-77=1$" }, { "idx": 2621, "question": "How many types of bonds exist between atoms?", "answer": "There are 5 types of bonds between atoms: metallic bond, ionic bond, covalent bond, physical bond (van der Waals force), hydrogen bond." }, { "idx": 2622, "question": "Given that the atomic number of an element is 32, based on its electron configuration, identify which group it belongs to.", "answer": "Group IVA" }, { "idx": 2623, "question": "What are the characteristics of ionic bonds?", "answer": "Ionic bonds: the bonding units are ions rather than atoms, with no directionality or saturation." }, { "idx": 2624, "question": "Given that the atomic number of an element is 32, determine the strength of its metallic character based on its electron configuration.", "answer": "Metalloid Ge" }, { "idx": 2625, "question": "What are the characteristics of hydrogen bonds?", "answer": "Hydrogen bond: intermolecular force, hydrogen bridge, with saturation." }, { "idx": 2626, "question": "Given that the relative atomic mass of Si is 28.09, if there are 5×10^10 electrons capable of free movement in 100g of Si, calculate the proportion of freely moving electrons to the total number of valence electrons.", "answer": "Number of atoms = (100 / 28.09) × 6.023 × 10^23 = 2.144 × 10^24; Number of valence electrons = 4 × number of atoms = 4 × 2.144 × 10^24 = 8.576 × 10^24; Proportion = (5 × 10^10) / (8.576 × 10^24) = 5.830 × 10^-15" }, { "idx": 2627, "question": "What does the short-range structure (primary structure) of polymer chain include?", "answer": "Chemical structure, atomic arrangement in the molecular chain, bonding sequence of structural units, branching, cross-linking, etc." }, { "idx": 2628, "question": "The density of Al2O3 is 3.8g/cm3, calculate how many atoms are present in 1 mm3?", "answer": "The relative molecular mass of Al2O3 M=26.98×2+16×3=101.96, the number of atoms in 1mm3 is 1×3.8×10-3/101.96×6.023×1023×5=1.12×1020" }, { "idx": 2629, "question": "Although the relative molecular mass of HF is relatively low, please explain why the boiling temperature of HF (19.4 ℃) is higher than that of HCl (-85 ℃)?", "answer": "This is because the intermolecular force in HF is hydrogen bonding, while that in HCl is van der Waals force. The bond energy of hydrogen bonding is higher than that of van der Waals force, so the boiling point of HF is higher than that of HCl." }, { "idx": 2630, "question": "What are the characteristics of metallic bonds?", "answer": "Metallic bonds: electron delocalization, no saturation, no directionality." }, { "idx": 2631, "question": "Given that the relative atomic mass of Si is 28.09, if there are 5×10^10 electrons capable of free movement in 100g of Si, calculate the proportion of covalent bonds that must be broken.", "answer": "Number of covalent bonds = number of atoms = 2.144 × 10^24; Number of covalent bonds to be broken = (5 × 10^10) / 2 = 2.5 × 10^10; Proportion = (2.5 × 10^10) / (2.144 × 10^24) = 1.166 × 10^-14" }, { "idx": 2632, "question": "What does the long-range structure (secondary structure) of polymer chain structures include?", "answer": "Relative molecular mass and its distribution, chain flexibility and conformation" }, { "idx": 2633, "question": "The density of Al2O3 is 3.8g/cm3, calculate how many atoms are contained in 1g?", "answer": "The relative molecular mass of Al2O3 M=26.98×2+16×3=101.96, the number of atoms contained in 1g is 1/101.96×6.023×1023×5=2.95×1022" }, { "idx": 2634, "question": "The percentage of ionic character in the bonding between elements A and B can be approximately expressed by the following formula: IC(%)=[1-e^(-0.25(x_A-x_B)^2)]×100, where x_A and x_B are the electronegativity values of elements A and B, respectively. Given that the electronegativities of In and Sb are 1.7 and 1.9, respectively, calculate the IC(%) for InSb.", "answer": "For InSb: IC(%)=[1-e^(-0.25)(1.9-1.7)^2]×100=1.0%" }, { "idx": 2635, "question": "Explain the characteristics of thermoplastic polymer materials from the perspective of polymer chain structure", "answer": "Thermoplastic polymers have linear and branched polymer chain structures, which soften upon heating and can be repeatedly processed and reshaped" }, { "idx": 2636, "question": "The percentage of ionic character in the bonding between elements A and B can be approximately calculated using the following formula: IC(%)=[1-e^(-0.25(x_A-x_B)^2)]×100, where x_A and x_B are the electronegativity values of elements A and B, respectively. Given that the electronegativities of Ti and O are 1.5 and 3.5, respectively, calculate the IC(%) for TiO2.", "answer": "For TiO2: IC(%)=[1-e^(-0.25)(3.5-1.5)^2]×100=63.2%" }, { "idx": 2637, "question": "The chemical behavior of S sometimes resembles that of a hexavalent element, while other times it resembles a tetravalent element. Please explain the reason for this behavior of S.", "answer": "The outermost electrons of S are 3s²3p⁴. When S combines with H to form H₂S, it accepts 2 electrons, and since S has 6 valence electrons in its outer shell, it behaves as a hexavalent element. When S combines with O to form SO₂, it provides 4 electrons, thus behaving as a tetravalent element." }, { "idx": 2638, "question": "Explain the characteristics of thermosetting polymer materials from the perspective of polymer chain structure", "answer": "Thermosetting polymers have a three-dimensional (cross-linked network) polymer chain structure, are insoluble in any solvent, and cannot be melted. Once set, their shape cannot be altered and they cannot be recycled." }, { "idx": 2639, "question": "Determine whether [-110] lies on the (111) plane", "answer": "Using the zone law: h u + k v + l w = 0, here 1×(-1) + 1×1 + 1×0 = 0, thus [-110] lies on the (111) plane." }, { "idx": 2640, "question": "Calculate the number-average relative molecular mass $\\overline{M}_{n}$ of this polymer material, given the molecular weight range, average molecular weight M, and molecular number fraction.", "answer": "The number-average relative molecular mass $\\overline{M}_{n} = \\sum x_{i}M_{i} = 375 + 2000 + 3850 + 6075 + 5500 + 2600 + 750 = 21150$." }, { "idx": 2641, "question": "Calculate the linear density in the [-110] direction", "answer": "The formula for linear density is K = (4 × r) / l, where r = (√2 × a) / 4, l = √2 × a. Substituting gives K = (4 × (√2 × a / 4)) / (√2 × a) = 1. Therefore, the linear density in the [-110] direction is 1." }, { "idx": 2642, "question": "Label the coordinates of all points on the (111) plane in a face-centered cubic unit cell", "answer": "The coordinates of all points on the (111) plane in a face-centered cubic unit cell include: (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), (1,1,1)." }, { "idx": 2643, "question": "Calculate the weight-average relative molecular mass $\\overline{M}_{w}$ of this polymer material, given the average molecular weight M and mass fraction.", "answer": "The weight-average relative molecular mass $\\overline{M}_{w} = \\sum w_{i}M_{i} = 150 + 1250 + 3150 + 6525 + 7150 + 4225 + 750 = 23200." }, { "idx": 2644, "question": "There is a copolymer ABS (A—acrylonitrile, B—butadiene, S—styrene), with the same mass fraction for each monomer. Calculate the mole fraction of each monomer.", "answer": "Acrylonitrile (—C2H3CN—) monomer has a relative molecular mass of 53; butadiene (—C2H3C2H3—) monomer has a relative molecular mass of 54; styrene (—C2H3C6H5—) monomer has a relative molecular mass of 104. Assuming each weighs 1 g, then acrylonitrile has 1/53 mol, butadiene has 1/54 mol, and styrene has 1/104 mol. Therefore, the mole fractions of the monomers are: x_acrylonitrile = (1/53) / (1/53 + 1/54 + 1/104) = 40.1%; x_butadiene = (1/54) / (1/53 + 1/54 + 1/104) = 39.4%; x_styrene = (1/104) / (1/53 + 1/54 + 1/104) = 20.5%." }, { "idx": 2645, "question": "An organic compound consists of $\\\\mathbf{\\\\pi}\\\\mathbf{\\\\pi}(\\\\mathbf{C})$ at $62.1\\\\%$, $w(\\\\mathrm{H})$ at $\\\\mathtt{10.3\\\\%}$, and $\\\\boldsymbol{\\\\mathbf{\\\\hat{\\\\rho}}}\\\\mathbf{\\\\hat{\\\\rho}}\\\\mathbf{\\\\hat{\\\\rho}}$ at $27.6\\\\%$. Try to deduce the name of the compound.", "answer": "The mole fractions of each component are $$ x_{\\\\mathrm{C}}=\\\\frac{62.1}{12.011}=5.2$$ $$ x_{{\\\\scriptscriptstyle\\\\mathrm{H}}}=\\\\frac{10,3}{1.008}{=}10,2$$ $$ x_{0}=\\\\frac{27\\\\cdot6}{16}=1.7$$ $$ \\\\begin{array}{r l}{\\\\mathrm{~C~}^{:}\\\\mathrm{~H~}^{:}\\\\mathrm{~O}=5.2:10.2:1.7\\\\approx3:6:1\\\\quad}&{{}\\\\mathrm{~H~}\\\\stackrel{\\\\mathrm{~O~}\\\\mathrm{~H~}}{\\\\longrightarrow}\\\\mathrm{C}-\\\\mathrm{C}-\\\\mathrm{H}}\\\\\\\\ {\\\\mathrm{~H~}^{:}\\\\mathrm{~H~}\\\\stackrel{\\\\mathrm{~O~}\\\\mathrm{~H~}}{\\\\longrightarrow}\\\\mathrm{~H~}}&{{}\\\\mathrm{~H~}\\\\stackrel{\\\\mathrm{~H~}}{\\\\longrightarrow}\\\\mathrm{~H~}}\\\\end{array}$$ Therefore, the possible compound is $\\\\mathrm{CH}_{3}\\\\mathrm{COCH}_{3}$ (acetone)." }, { "idx": 2646, "question": "Calculate the number-average degree of polymerization $n_{\\\\eta}$ of this polymer material, given the number-average relative molecular mass $\\\\overline{M}_{n}$ and the mass of each repeating unit.", "answer": "The mass of each repeating unit $\\\\overline{m} = 2 \\\\times 12.01 + 3 \\\\times 1.008 + 35.45 = 62.50$. The number-average degree of polymerization $n_{\\\\eta} = \\\\frac{\\\\overline{M}_{n}}{\\\\overline{m}} = \\\\frac{21150}{62.50} = 338$." }, { "idx": 2647, "question": "High-density polyethylene can be chlorinated by replacing hydrogen atoms in the structural units with chlorine atoms. If 8% of the hydrogen atoms in polyethylene are replaced by chlorine, calculate the mass fraction of chlorine that needs to be added.", "answer": "From the previous question, it is known that the structural unit of polyethylene contains 2 C atoms and 4 H atoms. If 8% of the H atoms are replaced by Cl atoms, the mass fraction of Cl that needs to be added is $$ \\\\begin{array}{r}{\\\\overbrace{2\\\\times A_{c}+4\\\\times(0.08\\\\times A_{\\\\mathrm{G}}+0.92\\\\times A_{\\\\mathrm{H}})}^{4\\\\times0.08\\\\times A_{\\\\mathrm{G}}}=\\\\overbrace{2\\\\times12.01+4\\\\times(0.08\\\\times35.45+0.921\\\\times1.001\\\\times1.001\\\\times1.001\\\\times1.001\\\\times1.01\\\\times1.01\\\\times1.01\\\\times1.01\\\\times1.01\\\\times1.01\\\\times1.01}^{4\\\\times0.08\\\\times35.45}}\\\\\\\\ {=0,290=29.0\\\\%\\\\end{array}$$" }, { "idx": 2648, "question": "It is known that pure titanium at 20°C is α-Ti (hcp structure), with lattice constants a=0.2951nm and c=0.4679nm. Calculate the interplanar spacing of the (112) plane.", "answer": "For the hcp structure, when h+2k=3n (n=0,1,2,3...), and l is an odd number, there is an additional plane. d_(112)=1/√[(4/3)((h²+hk+k²)/a²)+(l/c)²]=1/√[(4/3)((1²+1×1+1²)/0.2951²)+(2/0.4679)²]=0.1248nm" }, { "idx": 2649, "question": "It is known that pure titanium at 20°C is α-Ti (hcp structure) with lattice constants a=0.2951nm and c=0.4679nm. Calculate the interplanar spacing of the (001) plane.", "answer": "For the (001) plane of hcp structure, d_(001)=1/2×1/√(l/c)²=1/2×1/√(1/0.4679)²=0.2339nm" }, { "idx": 2650, "question": "It is known that pure titanium at 900°C is β-Ti (bcc structure) with a lattice constant a=0.3307nm. Calculate the interplanar spacing of the (112) plane.", "answer": "For the bcc structure, there is an additional plane when h+k+l=odd number. d_(112)=a/√(h²+k²+l²)=0.3307/√(1²+1²+2²)=0.135nm" }, { "idx": 2651, "question": "Given that pure titanium at 900°C is β-Ti (bcc structure) with a lattice constant a=0.3307nm, calculate the (001) interplanar spacing.", "answer": "For the (001) plane of bcc structure, d_(001)=1/2×a/√1²=1/2×0.3307/1=0.1653nm" }, { "idx": 2652, "question": "Indicate the crystal plane with the largest interplanar spacing in a face-centered cubic crystal", "answer": "From the above calculation results, it is known that the (111) crystal plane, which has the densest atomic arrangement, has the largest interplanar spacing." }, { "idx": 2653, "question": "Calculate the interplanar spacing and planar density of the (100) plane in a face-centered cubic crystal", "answer": "In a face-centered cubic crystal, when (h k l) are not all odd or all even, there are additional planes. d_(100)=1/2*a/sqrt(1^2+0+0)=0.5a; K_100=((1/4×4+1)πr^2)/a^2=(2πr^2)/((4/sqrt(2))r)^2=0.785" }, { "idx": 2654, "question": "Using the method of analytic geometry, how to determine the crystallographic direction indices of the intersection line between two crystal planes in a cubic crystal system", "answer": "Let there be two non-parallel crystal planes (h1,k1,l1) and (h2,k2,l2) in a cubic crystal system, and their intersection line is [u v w]. According to geometric relations, this crystallographic direction should lie on both crystal planes simultaneously, hence the following system of equations can be obtained: h1u+k1v+l1w=0, h2u+k2v+l2w=0. Solving the above system of equations yields u:v:w=(k1l2-l1k2):(l1h2-h1l2):(h1k2-h1h2)." }, { "idx": 2655, "question": "Using the method of analytic geometry, how to determine the crystallographic plane indices defined by two crystallographic directions in a cubic crystal system", "answer": "Assume there are two non-parallel crystallographic directions [u1,v1,w1] and [u2,v2,w2] in the crystal. The crystallographic plane indices defined by them are (hkl). According to the zone law, we have the equation system: u1h+v1k+w1l=0, u2h+v2k+w2l=0. Solving the above equations yields h:k:l=(v1w2-w1v2):(w1u2-u1w2):(u1v2-v1u2)." }, { "idx": 2656, "question": "Using the method of analytic geometry, how to determine the angle θ between two crystallographic directions in a cubic crystal system", "answer": "Let the two crystallographic directions in the cubic crystal system be [u1,v1,w1] and [u2,v2,w2]. From the scalar product of vectors, we know: [u1,v1,w1]·[u2,v2,w2]=|[u1,v1,w1]|·|[u2,v2,w2]|·cosθ. Therefore, the angle θ between the two crystallographic directions can be obtained from its cosine value: cosθ=(u1u2+v1v2+w1w2)/(√(u1²+v1²+w1²)·√(u2²+v2²+w2²)), θ=arccos(cosθ)." }, { "idx": 2657, "question": "Calculate the interplanar spacing and planar density of the (110) plane in a face-centered cubic crystal.", "answer": "In a face-centered cubic crystal, when (h k l) are not all odd or all even, there is an additional plane. d_(110)=1/2*a/sqrt(1^2+1^2+0)=0.354a; K_110=((1/4×4+1/2×2)πr^2)/(sqrt(2)*a^2)=(2πr^2)/(sqrt(2)(4/sqrt(2))r)^2=0.555" }, { "idx": 2658, "question": "Using the method of analytic geometry, how to determine the angle θ between two crystal planes in a cubic crystal system", "answer": "Let there be two crystal planes (h1,k1,l1) and (h2,k2,l2) in the cubic crystal system. The angle θ between them is the angle between their respective normals [h1,k1,l1] and [h2,k2,l2]. Therefore, cosθ=(h1h2+k1k2+l1l2)/(√(h1²+k1²+l1²)·√(h2²+k2²+l2²)), and the angle θ between the two crystal planes is θ=arccos(cosθ)." }, { "idx": 2659, "question": "Calculate the interplanar spacing and planar density of the (111) plane in a face-centered cubic crystal.", "answer": "In a face-centered cubic crystal, when (h k l) are not all odd or all even, there are additional planes. d_(111)=a/sqrt(1^2+1^2+1^2)=0.577a; K_111=((1/6×3+1/2×3)πr^2)/(sqrt(3)/4*(sqrt(2)a)^2)=(2πr^2)/(sqrt(3)/4(sqrt(2)*4/sqrt(2))r)^2=0.907" }, { "idx": 2660, "question": "Using Cu's ka (λ=0.1542 nm), the first line of Cr's X-ray diffraction spectrum was measured at 2θ=64.6°. If the lattice constant a of (bcc) Cr is 0.2885 nm, find the Miller indices corresponding to these spectral lines.", "answer": "According to the formula d_hkl=a/sqrt(h^2+k^2+l^2)=λ/(2sinθ), if 2θ=64.6°, then h^2+k^2+l^2={(2sin(64.6°/2)×0.2885)/0.1542}^2=3.9976≈4. Therefore, this plane is (200), or (020) or (002)." }, { "idx": 2661, "question": "The crystal structure of Ni is face-centered cubic, and its atomic radius is r=0.1243nm. Calculate the lattice constant of Ni.", "answer": "a=4r/sqrt(2)=4×0.1243/sqrt(2)=0.3516(nm)" }, { "idx": 2662, "question": "Using Cu's ka (λ=0.1542 nm), the first line of Cr's X-ray diffraction spectrum was measured at 2θ=44.4°. If the lattice constant a of (bcc) Cr is 0.2885 nm, find the Miller indices corresponding to these spectral lines.", "answer": "According to the formula d_hkl=a/sqrt(h^2+k^2+l^2)=λ/(2sinθ), if 2θ=44.4°, then sqrt(h^2+k^2+l^2)=(2sin(44.4°/2)×0.2885)/0.1542=1.4138, h^2+k^2+l^2=1.999≈2. Therefore, the plane is (110), or (1-10), or (101), or (10-1), or (011), or (01-1)." }, { "idx": 2663, "question": "Using Cu's ka (λ=0.1542 nm), the first line of Cr's X-ray diffraction spectrum was measured at 2θ=81.8°. If the lattice constant a of (bcc)Cr is 0.2885 nm, find the Miller indices corresponding to these spectral lines.", "answer": "According to the formula d_hkl=a/sqrt(h^2+k^2+l^2)=λ/(2sinθ), if 2θ=81.8°, then h^2+k^2+l^2={(2sin(81.8°/2)×0.2885)/0.1542}^2=6.0023≈6. Therefore, the plane is (112), or (1-12) or (1-1-2) or (11-2) or (121) or (12-1) or (1-21) or (1-2-1) or (211) or (21-1) or (2-11) or (2-1-1)." }, { "idx": 2664, "question": "The crystal structure of Ni is face-centered cubic, with an atomic radius of r=0.1243nm and a lattice constant of a=0.3516nm. Calculate the density of Ni.", "answer": "ρ=4Ar/(a³×NA)=4×58.69/((3.516×10^-8)^3×6.023×10^23)=8.967(g/cm³)" }, { "idx": 2665, "question": "One of the allotropes of Mn has a cubic structure with a lattice constant a of 0.632 nm, an atomic radius r of 0.122 nm, and 20 atoms in the unit cell. What is its packing density?", "answer": "The packing density K = (20 * (4/3) * π * r³) / a³. Substituting the data: K = (20 * (4/3) * π * (0.122)^3) / (0.632)^3 = 0.466." }, { "idx": 2666, "question": "The lattice constant of Cr is $\\mathbf{a=0.2884nm}$, and the density is $\\rho{=}7.19\\mathrm{g}/\\mathrm{cm}^{3}$. Determine the crystal structure of $\\mathrm{Cr}$ at this time.", "answer": "p= $\\rho{=}\\frac{n A_{\\tau}}{a\\times N_{\\mathrm{A}}}\\Rightarrow n=\\frac{\\rho a^{3}N_{\\mathrm{A}}}{A_{\\tau}}{=}\\frac{7.19\\times(2.884\\times10^{-8})^{3}\\times6.023\\times10^{23}}{52.0}{=}1.9977\\approx2$, hence it is a bcc structure." }, { "idx": 2667, "question": "In indium with a tetragonal structure, the relative atomic mass A_r=114.82, atomic radius r=0.1625 nm, lattice constants a=0.3252 nm, c=0.4946 nm, and density ρ=7.286 g/cm³. What is the packing fraction of In?", "answer": "K=(2×(4/3)πr³)/(a²c)=(2×(4/3)π(0.1625)^3)/((0.3252)^2×0.4946)=0.6873." }, { "idx": 2668, "question": "In indium with a tetragonal structure, the relative atomic mass A_r=114.82, atomic radius r=0.1625 nm, lattice constants a=0.3252 nm, c=0.4946 nm, and density ρ=7.286 g/cm³. How many atoms are there in the unit cell of In?", "answer": "n=(ρa²c×N_A)/A_r=(7.286×(3.252×10^-8)^2×(4.946×10^-8)×6.023×10^23)/114.82=1.9991≈2. Therefore, there are 2 atoms in the unit cell of In." }, { "idx": 2669, "question": "One of the allotropes of Mn has a cubic structure with a lattice constant a of 0.632nm, ρ of 7.26g/cm³, and Ar of 54.94. How many atoms are there in the Mn unit cell?", "answer": "According to the formula ρ = (n * Ar) / (a³ * NA), it can be derived that n = (ρ * a³ * NA) / Ar. Substituting the data: n = (7.26 * (6.32 * 10^-8)^3 * 6.023 * 10^23) / 54.94 = 20.091 ≈ 20. Therefore, there are 20 atoms in each unit cell." }, { "idx": 2670, "question": "The crystal structure of Mo is body-centered cubic, with a lattice constant $a{=}0.3147{\\mathrm{~nm}}$. Determine the atomic radius $\\\\scriptstyle{\\\\mathcal{r}}_{\\\\circ}$ of Mo.", "answer": "$$ \\\\alpha{=}\\\\frac{4r}{\\\\sqrt{3}}\\\\Rightarrow r{=}\\\\frac{\\\\sqrt{3}}{4}\\\\alpha{=}\\\\frac{\\\\sqrt{3}}{4}\\\\times0.3147{=}0.1363(\\\\mathrm{nm})$$" }, { "idx": 2671, "question": "Compared with question 1, explain the reasons for the difference.", "answer": "Reasons for the difference: different crystal structures and different atomic radii; when the coordination number of atoms in the crystal structure decreases, the atomic radius contracts." }, { "idx": 2672, "question": "If the atomic fraction of Zn dissolved in a Cu crystal is 10%, what is the maximum additional atomic fraction of Sn that can be dissolved?", "answer": "1.36=(1(100-10-x3)+2×10+4x3)/100, solving gives x3=8.67, meaning a maximum of 8.67% Sn can be additionally dissolved." }, { "idx": 2673, "question": "What is the maximum atomic fraction of Zn or Sn that can be dissolved in Cu-Zn and Cu-Sn solid solutions?", "answer": "The limiting electron concentration for Cu-based solid solutions is 1.36. For the Cu-Zn solid solution, 1.36=(1(100-x1)+2x1)/100→x1=36, so the maximum solubility is 36% Zn; for the Cu-Sn solid solution, 1.36=(1(100-x2)+4x2)/100→x2=12, so the maximum solubility is 12% Sn." }, { "idx": 2674, "question": "According to X-ray diffraction measurements, at 912°C, the lattice parameter a of α-Fe is 0.2892 nm, and that of γ-Fe is 0.3633 nm. Calculate the volume expansion when γ-Fe transforms into α-Fe.", "answer": "fcc: r = sqrt(2)/4 * a = sqrt(2)/4 * 0.3633 = 0.1284 nm; bcc: r = sqrt(3)/4 * a = sqrt(3)/4 * 0.2892 = 0.1251 nm; ΔV = (1/2 * (0.2892)^3 - 1/4 * (0.3633)^3) / (1/4 * (0.3633)^3) = 0.87%" }, { "idx": 2675, "question": "Cementite (Fe3C) is an interstitial compound with an orthorhombic crystal structure. Its lattice constants are a=0.4514nm, b=0.508nm, c=0.6734nm, and its density ρ=7.66g/cm³. Determine the number of Fe and C atoms per unit cell of Fe3C.", "answer": "Assume the number of C atoms in the Fe3C unit cell is x, then the number of Fe atoms is 3x. ρ = (x Ar(C) + 3 × Ar(Fe)) / (a b c × NA), 7.66 = (x × 12.011 + 3x × 55.85) / (4.514 × 5.08 × 6.743 × 10^-24 × 6.023 × 10^23). x = (7.66 × 4.514 × 5.08 × 6.734 × 0.602) / (12.011 + 3 × 55.85) = 3.968 ≈ 4. 3x = 12. Therefore, in the Fe3C compound, each unit cell contains 4 C atoms and 12 Fe atoms." }, { "idx": 2676, "question": "According to the hard sphere model of crystals, if the diameter of the spheres remains unchanged, calculate the volume expansion when Fe transforms from fcc to bcc.", "answer": "a_fcc = 4 / sqrt(2) * r ⇒ V_fcc = a_fcc^3 = 64 / (2 * sqrt(2)) * r^3；a_bcc = 4 / sqrt(3) * r ⇒ V_bcc = a_bcc^3 = 64 / (3 * sqrt(3)) * r^3；ΔV = (1/2 * 64 / (3 * sqrt(3)) * r^3 - 1/4 * 64 / (2 * sqrt(2)) * r^3) / (1/4 * 64 / (2 * sqrt(2)) * r^3) = 9%" }, { "idx": 2677, "question": "What is an interstitial phase? Explain its characteristics from the perspective of crystal structure.", "answer": "Interstitial phases are intermediate phases dominated by atomic size factors. They are formed when nonmetal elements with smaller atomic radii occupy the interstitial sites of the lattice, yet the lattice of the interstitial phase differs from that of any of their constituent elements. Their composition can fluctuate within a certain range. However, the constituent elements generally have a specific atomic ratio, which can be represented by a chemical formula. When rB/rA < 0.59, interstitial phases usually form, characterized by a simple crystal structure, extremely high melting points, and hardness." }, { "idx": 2678, "question": "Given that the ionic radii of K⁺ and Cl⁻ are 0.133nm and 0.181nm respectively, and KCl has a CsCl-type structure, calculate its density ρ.", "answer": "ρ = (Aₜ(K) + Aₜ(Cl)) / [(2(rₖ⁺ + rCl⁻) / √3)³ × Nₐ] = (39.102 + 35.453) / [(2 × (0.133 + 0.181) / √3)³ × 6.023 × 10²³ × 10⁻²⁴] = 2.597 g/cm³" }, { "idx": 2679, "question": "MgO has a NaCl-type structure. The ionic radius of Mg2+ is 0.078nm, and the ionic radius of O2- is 0.132nm. Calculate the density (ρ) of MgO.", "answer": "ρ = 4[Ar(Mg) + Ar(O)] / (2rMg + 2rO)^3 × NA = 4 × 24.31 + 4 × 16.00 / 8 × (0.78 + 1.32)^3 × 10^-24 × 6.023 × 10^23 = 3.613 g/cm^3" }, { "idx": 2680, "question": "A solid solution contains x(MgO) at 30% and x(LiF) at 70%. Calculate the mass fractions of Li1+, Mg2+, F1-, and O2-. The relative atomic masses of Mg, O, Li, and F are 24.31, 16.00, 6.94, and 19.00, respectively.", "answer": "w(Li+) = (0.7 × 6.94) / [0.3 × (24.31 + 16) + 0.7 × (6.94 + 19)] = 16%\\nw(Mg2+) = (0.3 × 24.31) / [0.3 × (24.31 + 16) + 0.7 × (6.94 + 19)] = 24%\\nw(F-) = (0.7 × 19) / [0.3 × (24.31 + 16) + 0.7 × (6.94 + 19)] = 44%\\nw(O2-) = (0.3 × 16) / [0.3 × (24.31 + 16) + 0.7 × (6.94 + 19)] = 16%" }, { "idx": 2681, "question": "MgO has a NaCl-type structure. The ionic radius of Mg2+ is 0.078nm, and the ionic radius of O2- is 0.132nm. Calculate the packing fraction (K) of MgO.", "answer": "K = [4 × (4/3)πrMg^3 + 4 × (4/3)πrO^3] / (2rMg + 2rO)^3 = [16/3π × (0.78^3 + 1.32^3)] / [8 × (0.78 + 1.32)^3] = 0.627" }, { "idx": 2682, "question": "What is an interstitial solid solution? Explain its characteristics from the perspective of crystal structure.", "answer": "An interstitial solid solution is formed when solute atoms are distributed in the interstitial sites of the solvent lattice. The solute atoms that form interstitial solid solutions are usually non-metallic elements with atomic radii smaller than 0.1nm, such as H, B, C, N, O, etc. The interstitial solid solution retains the crystal structure of the parent phase (solvent), and its composition can fluctuate within a certain solubility limit, but cannot be expressed by a molecular formula." }, { "idx": 2683, "question": "A solid solution contains x(MgO) at 30% and x(LiF) at 70%. If the density of MgO is 3.6 g/cm3 and the density of LiF is 2.6 g/cm3, what is the density of this solid solution?", "answer": "The density of the solid solution ρ = 0.3 × 3.6 + 0.7 × 2.6 = 2.9 g/cm3" }, { "idx": 2684, "question": "What are interstitial compounds? Explain their characteristics from the perspective of crystal structure.", "answer": "Interstitial compounds belong to intermediate phases dominated by atomic size factors. They are formed when non-metal elements with smaller atomic radii occupy the interstitial sites of the crystal lattice, yet the lattice of interstitial compounds differs from that of any of their constituent elements. Their composition can vary within a certain range. However, the constituent elements generally maintain a specific atomic ratio, which can be represented by a chemical formula. When rB/rA≥0.59, interstitial compounds form, exhibiting a complex crystal structure." }, { "idx": 2685, "question": "Given the ionic radii of K⁺ and Cl⁻ are 0.133nm and 0.181nm respectively, and KCl has a CsCl-type structure, calculate its packing fraction κ.", "answer": "κ = [(4/3)πrₖ⁺³ + (4/3)πrCl⁻³] / [2(rₖ⁺ + rCl⁻) / √3]³ = [(4/3)π(0.133³ + 0.181³)] / [2(0.133 + 0.181) / √3]³ = 0.728" }, { "idx": 2686, "question": "In a ZrO2 solid solution, for every 6 Zr4+ ions, 1 Ce3+ ion is added to form a cubic lattice of ZrO2. If these cations form an fcc structure and the O2- ions occupy the tetrahedral interstitial sites, calculate how many O2- ions are needed for 100 cations.", "answer": "The total charge of 100 cations = (100/7) × (6×4 + 1×2) = 371.4, so 371.4/2 = 185.7 O2- ions are required to balance this charge." }, { "idx": 2687, "question": "What is glass?", "answer": "The so-called glass refers to an amorphous solid that has a glass transition temperature. The difference between glass and other amorphous states lies in the presence or absence of a glass transition temperature. The glassy state also refers to amorphous metals and alloys (amorphous metal), which are essentially supercooled liquid metals." }, { "idx": 2688, "question": "In the ZrO2 solid solution, cations form an fcc structure, and O2- ions are located at tetrahedral interstitial sites. Given that 100 cations form 25 unit cells, and each unit cell has a total of 8 tetrahedral interstitial sites, calculate the percentage of tetrahedral interstitial sites occupied by O2- ions.", "answer": "Since the fcc structure contains 4 atoms per unit cell, 100 cations form 25 unit cells. Each unit cell has a total of 8 tetrahedral interstitial sites, so the percentage of tetrahedral interstitial sites occupied by O2- ions is 185.7/(25×8)=92.9%." }, { "idx": 2689, "question": "Given that the number-average relative molecular weight of linear polytetrafluoroethylene is 5×10^5, its C-C bond length is 0.154 nm, and the bond angle θ is 109°, calculate its root-mean-square length.", "answer": "The root-mean-square length r=d√N=0.154×√(1.0×10^4)=15.4 nm." }, { "idx": 2690, "question": "Polypropylene is polymerized from propylene, with the chemical formula C3H6. Crystalline polypropylene belongs to the monoclinic crystal system, with lattice constants α = 0.665 nm, b = 2.096 nm, c = 0.65 nm, α = γ = 90°, β = 99.3°, and a density ρ = 0.91 g/cm³. Calculate the number of C and H atoms in the unit cell of crystalline polypropylene.", "answer": "Assuming there are x C atoms in the polypropylene unit cell, then there are 2x H atoms. The volume of the unit cell is $$ V = a b c \\\\sin \\\\beta $$ $$ \\\\rho = \\\\frac{x \\\\cdot 12.011 + 2x \\\\cdot 1.008}{(6.65 \\\\times 10^{-8})(20.96 \\\\times 10^{-8})(6.5 \\\\times 10^{-8}) \\\\sin 99.3 \\\\times 6.023 \\\\times 10^{23}} = 0.91 $$ $$ x = \\\\frac{0.91 \\\\times 6.65 \\\\times 2.096 \\\\times 6.5 \\\\times \\\\sin 99.3 \\\\times 6.023}{12.011 + 2 \\\\times 1.008} = 34.936 \\\\approx 35 $$ $$ 2x = 2 \\\\times 35 = 70 $$ Therefore, the C3H6 unit cell contains 35 C atoms and 70 H atoms." }, { "idx": 2691, "question": "The ionic radii of $\\\\mathrm{Al}^{3+}$ and $\\\\mathrm{O}^{2-}$ are $0.051\\\\mathrm{nm}$ and $0.132\\\\mathrm{nm}$ respectively. Determine the coordination number of $\\\\mathbf{Al}_{2}\\\\mathbf{O}_{3}$.", "answer": "The ratio of the two ionic radii is $\\\\frac{0.051}{0.132}{=}0.386$. The coordination number (CN) of an ionic crystal depends on the ratio of the cation to anion radii. Referring to the table, when $\\\\frac{r_{+}}{r_{-}}$ is between $0.225\\\\sim0.414$, the CN is 4. The shape of the anion polyhedron is tetrahedral." }, { "idx": 2692, "question": "Diamond is a crystalline form of carbon with a lattice constant α = 0.357 nm. When it transforms into the graphite (ρ = 2.25 g/cm³) structure, calculate the change in volume.", "answer": "The crystal structure of diamond is a complex face-centered cubic structure, with each unit cell containing a total of 8 carbon atoms. The density of diamond is $$ ρ = (8 × 12) / ((0.357 × 10⁻⁷)³ × 6.023 × 10²³) = 3.503 (g/cm³) $$. For 1 g of carbon, when it is in the diamond structure, its volume is $$ V₁ = 1 / 3.503 = 0.285 (cm³) $$. When it is in the graphite structure, its volume is $$ V₂ = 1 / 2.25 = 0.444 (cm³) $$. Therefore, the volume expansion when transforming from diamond to graphite is = (V₂ - V₁) / V₁ = (0.444 - 0.285) / 0.285 ≈ 55.8%." }, { "idx": 2693, "question": "Given that the number-average relative molecular weight of linear polytetrafluoroethylene is 5×10^5, the C-C bond length is 0.154 nm, and the bond angle θ is 109°, calculate its total chain length L.", "answer": "For a linear polymer, the total chain length L depends on the bond length d between atoms, the number of bonds N, and the angle θ between adjacent bonds, i.e., L=Ndsin(θ/2). For polytetrafluoroethylene, each chain unit has 2 C atoms and 4 F atoms. First, calculate its degree of polymerization πn=M/m=5×10^5/(2×12.01+4×19.00)=5×10^3. Each chain unit has 2 C atoms, so each chain unit has two C-C main bonds. Therefore, the total number of bonds in this polymer is N=2πn=2×5×10^3=1.0×10^4. If each C-C bond length d=0.154 nm and bond angle θ=109°, then L=Ndsin(θ/2)=1.0×10^4×0.154×sin(109°/2)=1253.738 nm." }, { "idx": 2694, "question": "What are the main differences between amorphous and crystalline substances in terms of internal atomic arrangement and properties?", "answer": "From the perspective of internal atomic arrangement, the fundamental characteristic of a crystalline structure is that atoms are arranged periodically in three-dimensional space, exhibiting long-range order, whereas the atomic arrangement in amorphous materials lacks long-range order. In terms of properties, crystals have fixed melting points and anisotropy, while amorphous materials do not have fixed melting points and are isotropic." }, { "idx": 2695, "question": "Calculate the theoretical density of α-Fe (given a=0.286nm for α-Fe, Ar(Fe)=55.85, NA=6.023×10²³)", "answer": "ρtheoretical=2×Ar(Fe)/(a³×NA)=2×55.85/((2.86×10⁻⁸)³×6.023×10²³)=7.9276(g/cm³)" }, { "idx": 2696, "question": "Calculate the packing density of the diamond structure.", "answer": "Diamond is the most typical covalent crystal, entirely bonded by covalent bonds. Its crystal structure belongs to a complex fcc structure, where each C atom (d=0.1544 nm) has 4 equidistant nearest neighbors, conforming to the 8-N rule. The distance between the nearest neighbors is equivalent to the bond length. Based on the crystal structure of diamond, it can be determined that: Therefore, $$ a={\\frac{4\\times0.1544}{\\sqrt{3}}}=0.3566({\\mathrm{nm}})$$ $$ {\\cal K}=\\frac{8\\times\\frac43\\pi r^{3}}{a^{3}}=\\frac{8\\times\\frac43\\pi\\left(\\frac{0.1544}{2}\\right)^{3}}{(0.3566)^{3}}=0.34$$" }, { "idx": 2697, "question": "Calculate the theoretical packing density of α-Fe (given rFe=0.1241nm, a=0.286nm)", "answer": "Ktheoretical=8/3×π×rFe³/a³=8/3×π×(0.1241)³/(0.286)³=0.6844" }, { "idx": 2698, "question": "If in fcc Cu, 1 out of every 500 atoms is missing, and its lattice constant is 0.3615 nm, calculate the density of Cu.", "answer": "$$ \\\\rho{=}\\\\frac{4\\\\times\\\\left(1-\\\\frac{1}{500}\\\\right)63.54}{(3.615\\\\times10^{-8})^{3}\\\\times6.023\\\\times10^{23}}{=}8.915({\\\\mathrm{g/cm}}^{3})$$" }, { "idx": 2699, "question": "There is a glass containing soda, with a mass fraction of SiO2 at 80% and Na2O at 20%. Calculate the fraction of non-bridging O atoms.", "answer": "According to the problem, assume there is 100g of glass, containing 80g of SiO2 and 20g of Na2O. First, calculate their molar fractions: The amount of SiO2 is 80/(28.09+2×16.00)=1.331(mol) ⇒ x_SiO2=80.47%. The amount of Na2O is 20/(2×22.99+16.00)=0.323(mol) ⇒ x_Na2O=19.53%. Now, using 100mol as the basis, then: 80.47SiO2 = 80.47Si + 160.94O 19.53Na2O = 39.06Na + 19.53O Since each Na+ produces one non-bridging oxygen ion, there are 39.06 non-bridging oxygen atoms, while the bridging oxygen atoms are: (160.94 + 19.53) - 39.06 = 141.41 The fraction of non-bridging oxygen atoms is then: 39.06/180.47 = 0.216" }, { "idx": 2700, "question": "Given the vibration frequency of atoms around a vacancy in Cu as 1×10¹³ s⁻¹, ΔEv as 0.15×10⁻¹⁸ J, and exp(ΔSm/k) approximately 1, calculate the migration frequency of vacancies at room temperature 27°C (300K).", "answer": "ν=ν₀zexp(−ΔEv/kT)exp(ΔSm/k), ν₃₀₀=1×10¹³×12×exp(−0.15×10⁻¹⁸/(1.38×10⁻²³×300))×1=2.207×10⁻² s⁻¹" }, { "idx": 2701, "question": "Given that the vibration frequency of atoms around a vacancy in Cu is 1×10¹³ s⁻¹, ΔEv is 0.15×10⁻¹⁸ J, and exp(ΔSm/k) is approximately 1, calculate the migration frequency of vacancies at 700K.", "answer": "ν=ν₀zexp(−ΔEv/kT)exp(ΔSm/k), ν₇₀₀=1×10¹³×12×exp(−0.15×10⁻¹⁸/(1.38×10⁻²³×700))×1=2.165×10⁷ s⁻¹" }, { "idx": 2702, "question": "Calculate the actual packing density of α-Fe (given that for every 200 iron atoms there is 1 H atom, rFe=0.1241nm, rH=0.036nm, a=0.286nm)", "answer": "K actual=8/3×π×(rFe³+rH³/200)/a³=8/3×π×((0.1241)³+(0.036)³/200)/(0.286)³=0.6845" }, { "idx": 2703, "question": "Calculate the actual density of α-Fe (given that for every 200 iron atoms there is 1 H atom, Ar(H)=1.008, a=0.286nm, NA=6.023×10²³)", "answer": "ρ_actual=2×(Ar(Fe)+1/200×Ar(H))/(a³×NA)=2×(55.85+1.008/200)/((2.86×10⁻⁸)³×6.023×10²³)=7.9283(g/cm³)" }, { "idx": 2704, "question": "To dissolve MgF2 into LiF, what type of vacancies (anion or cation) need to be introduced into LiF?", "answer": "To dissolve MgF2 into LiF by replacing Mg2+ with Li+, anion vacancies must be introduced to balance the charge and maintain the original MgF2 structure." }, { "idx": 2705, "question": "In a diffusion experiment of a certain crystal, it was found that at 500°C, 1 in 10^10 atoms had sufficient activation energy to jump out of their equilibrium positions and enter interstitial positions; at 600°C, this ratio increased to 10^9. Calculate the activation energy required for this jump.", "answer": "The thermal activation process can generally be described by the famous Arrhenius equation. Let E be the energy required to form an interstitial atom, so the ratio of the number of atoms n with energy exceeding the average energy to the total number of atoms N is C = n/N = A e^(-E/kT), where A is the proportionality constant, k is the Boltzmann constant, and T is the absolute temperature. Taking the logarithm of both sides, we have ln C = ln A - E/kT. Solving the simultaneous equations gives ln 10^(-10) = ln A - E/(1.38×10^(-23)×773) and ln 10^(-9) = ln A - E/(1.38×10^(-23)×873). Thus, ln A = -2.92, E = 2.14×10^(-19) (J)." }, { "idx": 2706, "question": "The density of $\\\\mathbf{M}_{\\\\mathbf{g}}\\\\mathbf{O}$ is $3.58\\\\:\\\\mathbf{g}/\\\\mathrm{cm}^{3}$, and its lattice constant is $0.42\\\\mathfrak{n m}$. Determine the number of Schottky defects per unit cell of $\\\\mathbf{M}_{\\\\mathbf{g}}\\\\mathrm{O}$.", "answer": "Let the number of Schottky defects per unit cell be x. $$ \\\\rho{=}\\\\frac{(4-x)\\\\times\\\\left[A_{\\\\mathrm{r}}(\\\\mathrm{M}g)+A_{\\\\mathrm{r}}(\\\\mathrm{O})\\\\right]}{a^{3}N_{\\\\mathrm{A}}}$$ $$ x=4-{\\\\frac{\\\\rho\\\\alpha^{3}N_{\\\\mathrm{{A}}}}{A_{\\\\mathrm{{r}}}(\\\\mathrm{Mg})+A_{\\\\mathrm{{r}}}(\\\\mathrm{O})}}=4-{\\\\frac{3.58\\\\times(4.2\\\\times10^{-8})^{3}\\\\times6.023\\\\times10^{23}}{24.31+16.00}}=0.0369$$" }, { "idx": 2707, "question": "If LiF is dissolved in MgF2, what type of vacancies (anion or cation) must be introduced into MgF2?", "answer": "To dissolve LiF into MgF2, where Mg2+ replaces Li+, cation vacancies must be introduced because the valence charges of the replaced ion and the newly added ion must be equal." }, { "idx": 2708, "question": "At 700°C, what is the proportion of atoms with sufficient energy?", "answer": "According to the Arrhenius equation, ln C = ln A - E/kT. Given ln A = -2.92, E = 2.14×10^(-19) J, k = 1.38×10^(-23) J/K, and T = 973 K. Substituting these values yields ln C = -2.92 - (2.14×10^(-19))/(1.38×10^(-23)×973). The calculation gives C = n/N = 6×10^(-9)." }, { "idx": 2709, "question": "Given that Al has an fcc crystal structure with a lattice constant α=0.405 nm, and the vacancy concentration at 550°C is 2×10-6, calculate the average spacing between these vacancies if they are uniformly distributed in the crystal.", "answer": "The number of lattice points in $1~\\\\mu\\\\mathrm m^{3}$ volume of Al is $$ N={\\\\frac{1}{a^{3}}}\\\\times4\\\\Rightarrow{\\\\frac{1}{(0.405\\\\times10^{-6})^{3}}}=6.021\\\\times10^{10}$$ Therefore, the number of vacancies in $1~\\\\mu\\\\mathrm m^{3}$ volume is $$ n_{V}=\\\\mathrm{CN}=6.021\\\\times10^{10}\\\\times2\\\\times10^{-6}=1.204\\\\times10^{5}$$ Assuming the vacancies are uniformly distributed in the crystal, the average spacing between them is $$ L=\\\\sqrt[3]{\\\\frac{1}{n_{v}}}=\\\\sqrt[3]{\\\\frac{1}{1.204\\\\times10^{5}}}=0.02025(\\\\mu\\\\mathrm{m})=20.25\\\\mathrm{nm}$$" }, { "idx": 2710, "question": "Given that W at 20℃ has one vacancy per 10^23 unit cells, and each unit cell contains 2 W atoms, calculate the vacancy concentration C at 20℃.", "answer": "C_20 = 1 / (2 × 10^23) = 5 × 10^-24." }, { "idx": 2711, "question": "The activation energy required to form a vacancy in a certain crystal is 0.32×10^-18 J. At 800°C, there is one vacancy per 1×10^4 atoms. At what temperature will there be one vacancy per 10^3 atoms?", "answer": "According to the Arrhenius equation: $$ \\\\ln{\\\\frac{n}{N}}=\\\\ln A-{\\\\frac{E}{{k}\\\\ T}}$$ Substituting the known conditions into the equation: $$ \\\\ln10^{-4}=\\\\ln A-\\\\frac{0.32\\\\times10^{-18}}{1.38\\\\times10^{-23}\\\\times1073}$$ We obtain: $$ \\\\ln A=12.4$$ And: $$ {\\\\bf{In}}10^{-3}=12.4-\\\\frac{0.32\\\\times10^{-18}}{1.38\\\\times10^{-23}\\\\times\\\\mathit{T}}.$$ Therefore: $$ \\\\operatorname{T}=1201\\\\operatorname{K}=928^{\\\\circ}{\\\\mathbb{C}}$$" }, { "idx": 2712, "question": "When the temperature increases from 20℃ to 1020℃, the lattice constant expands by (4 × 10^-4)%, and the density decreases by 0.012%. Assuming the total mass remains unchanged, calculate the total volume change rate ΔV/V.", "answer": "(1 + ΔV/V) × (1 - 0.00012) = 1 → ΔV/V ≈ 0.012%." }, { "idx": 2713, "question": "For a cubic crystal with side length L, derive the volume change rate ΔV0/V0 due to thermal expansion when the temperature increases from T1 to T2.", "answer": "ΔV0/V0 = (L + ΔL)^3 - L^3 / L^3 = 3 × (ΔL/L)." }, { "idx": 2714, "question": "The vacancy formation energy (E_v) and interstitial atom formation energy (E_i) of Al are 0.76 eV and 3.0 eV, respectively. Calculate the ratio of the equilibrium concentration of vacancies to the equilibrium concentration of interstitial atoms in Al at 500°C.", "answer": "At 500°C: C_V/C_i = e^((1/(8.617×10^-5×773))(3.0-0.76)) = e^33.63 = 4.026×10^14" }, { "idx": 2715, "question": "Solve the equations to find the vacancy formation energy Ev and entropy Sv.", "answer": "Ev = 1.45 eV, Sv = 3.3 × 10^-4 eV." }, { "idx": 2716, "question": "Assuming the vacancy concentration at T1 is negligible compared to T2, calculate the equilibrium vacancy concentration Cv at 1020℃.", "answer": "Cv = ΔV0/V0 - ΔVa/Va = 3 × (ΔL/L - Δa/a) = (0.012% - 3 × 4 × 10^-4%) = 1 × 10^-4." }, { "idx": 2717, "question": "Using the vacancy concentrations at 20℃ and 1020℃, set up the equations to solve for the vacancy formation energy Ev and entropy Sv.", "answer": "5 × 10^-24 = exp(Sv/k) × exp[-Ev/(293k)], 1 × 10^-4 = exp(Sv/k) × exp[-Ev/(1293k)]." }, { "idx": 2718, "question": "When the temperature decreases from 600°C to 300°C, the equilibrium vacancy concentration in $\\mathtt{Ge}$ crystal decreases by 6 orders of magnitude. Calculate the vacancy formation energy in the Ge crystal.", "answer": "$$ \\\\begin{array}{r l}&{\\\\frac{C_{\\\\tau_{1}}}{C_{\\\\tau_{2}}}=\\\\frac{1}{10^{6}}=\\\\frac{A\\\\exp\\\\left(-\\\\frac{E_{\\\\mathrm{v}}}{k T_{1}}\\\\right)}{A\\\\exp\\\\left(-\\\\frac{E_{\\\\mathrm{v}}}{k T_{2}}\\\\right)}=\\\\mathrm{e}^{\\\\frac{\\\\varepsilon_{\\\\mathrm{v}}}{k}\\\\left(\\\\frac{1}{T_{2}}-\\\\frac{1}{T_{1}}\\\\right)}-\\\\ln10^{6}}\\\\ &{\\\\qquad=\\\\frac{E_{\\\\mathrm{v}}}{\\\\frac{1}{k}\\\\left(\\\\frac{1}{T_{2}}-\\\\frac{1}{T_{1}}\\\\right)}}\\\\ &{E_{\\\\mathrm{v}}=\\\\frac{-\\\\ln10^{6}}{\\\\frac{1}{873}-\\\\frac{1}{573}}=\\\\frac{-13,8\\\\times8,617\\\\times10^{-5}}{1.145\\\\times10^{-3}-1.745\\\\times10^{-3}}=1.98\\\\mathrm{(eV)}}\\\\end{array}$$" }, { "idx": 2719, "question": "Derive the volume change rate ΔVa/Va due to lattice constant expansion when the temperature increases from T1 to T2.", "answer": "ΔVa/Va = (a + Δa)^3 - a^3 / a^3 = 3 × (Δa/a)." }, { "idx": 2720, "question": "If the positive direction of a dislocation line is defined as the original opposite direction, will the Burgers vector of this dislocation change?", "answer": "According to the method of determining the Burgers vector of a dislocation by the Burgers circuit, the Burgers vector of this dislocation will reverse." }, { "idx": 2721, "question": "If the positive direction of a dislocation line is defined as the original opposite direction, does the type and nature of the dislocation change?", "answer": "The type and nature of this dislocation do not change." }, { "idx": 2722, "question": "The vacancy formation energy (E_v) and interstitial atom formation energy (E_i) of Al are 0.76 eV and 3.0 eV, respectively. Calculate the ratio of the equilibrium concentration of vacancies to the equilibrium concentration of interstitial atoms in Al at room temperature (20℃).", "answer": "At 20℃: C_V/C_i = e^((1/(8.617×10^-5×293))(3.0-0.76)) = e^88.72 = 3.395×10^38" }, { "idx": 2723, "question": "Assume there is an edge dislocation with Burgers vector b in the [0-10] direction gliding along the (100) plane. If another edge dislocation with Burgers vector in the [010] direction moves through the (001) plane, will this dislocation form a kink or a jog when passing through the aforementioned dislocation?", "answer": "Kink" }, { "idx": 2724, "question": "Assume there is an edge dislocation with a b in the [0-10] direction gliding along the (100) crystal plane. If a screw dislocation with a b direction of [100] gliding on the (001) crystal plane passes through the aforementioned dislocation, will it form a kink or a jog?", "answer": "Jog" }, { "idx": 2725, "question": "There are two left-handed screw dislocation lines, each with an energy of $E_{1}$. When they approach infinitely close, what is the total energy?", "answer": "Since the strain energy of a dislocation is proportional to $b^{2}$, and the energies of like-signed screw dislocations are the same, their Burgers vectors must be identical. If they approach infinitely close and merge into a new dislocation with a Burgers vector of 2b, the total energy should be $4E_{1}$. However, in reality, this dislocation reaction cannot proceed because the energy increases after merging. Moreover, like dislocations repel each other, and the repulsive force between two like-signed dislocations will prevent them from approaching infinitely close." }, { "idx": 2726, "question": "Are the types of dislocations the same at each point on a dislocation loop?", "answer": "Judging by the angle between the dislocation line and the Burgers vector, if the Burgers vector of a dislocation loop is perpendicular to the dislocation line at each point on the loop, then the nature of the dislocations is the same at each point on the loop, all being edge dislocations; however, if the Burgers vector of the dislocation loop is parallel to the plane where the dislocation line lies, then some are pure edge dislocations, some are pure screw dislocations, and others are mixed dislocations; when the Burgers vector intersects the dislocation loop line at a certain angle, although each point on this dislocation loop is a mixed dislocation, the edge and screw components at each point are different." }, { "idx": 2727, "question": "Indicate the nature types of the three dislocations in the dislocation reaction $$ \\frac{a}{2}[10\\overline{{{1}}}] + \\frac{a}{6}[\\overline{{{1}}}21] \\rightarrow \\frac{a}{3}[11\\overline{{{1}}}] $$?", "answer": "Referring to the Thompson tetrahedron, this dislocation reaction corresponds to $\\begin{array}{r l r l r l r l r l}{{\\bf C}{\\bf A}}&{{}}&{+}&{{}{\\bf a}{\\bf C}}&{}&{{}\\searrow}&{}&{{\\bf a}{\\bf A}}\\end{array}$" }, { "idx": 2728, "question": "Calculate the dislocation density ρ when the shear strength is $42\\mathrm{MPa}$.", "answer": "The dislocation density ρ can be calculated using the formula ρ = (τ̄/αGb)^2, where α is a constant (typically taken as 0.5). Substituting τ̄ = 42×10^6 Pa, G = 50×10^9 Pa, and b = 3.12×10^-10 m, we obtain ρ = (42×10^6 / (0.5×50×10^9×3.12×10^-10))^2 ≈ 7.3×10^12 m^-2." }, { "idx": 2729, "question": "Given that the lattice constant of a $\\\\mathrm{\\\\hat{Cu}}$ crystal is $a=0.35\\\\mathrm{nm}$, the shear modulus $G=4\\\\times10^{4}MPa$, and there is a dislocation with Burgers vector $\\\\pmb{\\\\mathscr{b}}==\\\\frac{a}{2}[\\\\overline{{1}}01]$, where the dislocation line direction is [-101], calculate the strain energy of this dislocation.", "answer": "Based on the relationship between the Burgers vector and the dislocation line, it is known that this dislocation is a screw dislocation, and its strain energy is $$ \\\\begin{array}{r}{E=\\\\frac{G b^{2}}{4\\\\pi}\\\\mathrm{ln}\\\\frac{{R}}{r_{\\\\mathrm{0}}}}\\\\end{array}$$ $r_{0}\\\\approx b-{\\\\frac{\\\\sqrt{2}}{2}}a=0.707\\\\times0.35\\\\times10^{-6}=2.475\\\\times10^{-10}({\\\\mathrm{m}})$ $R{\\\\approx}1\\\\times10~^{\\\\circ}\\\\mathrm{m}$ $E={\\\\frac{4\\\\times10^{10}\\\\times(2.475\\\\times10^{-16})^{2}}{4\\\\pi}}\\\\ln{\\\\frac{10^{-2}}{2.475\\\\times10^{-10}}}=3.415\\\\times10^{-9}(\\\\mathrm{N}\\\\cdot\\\\mathrm{m}/\\\\mathrm{m})$" }, { "idx": 2730, "question": "In fcc, can the dislocation reaction $$ \\\\frac{a}{2}[10\\\\overline{{{1}}}] + \\\\frac{a}{6}[\\\\overline{{{1}}}21] \\\\rightarrow \\\\frac{a}{3}[11\\\\overline{{{1}}}] $$ proceed?", "answer": "Dislocation reaction geometric condition: $b_{1}+b_{2}=\\\\Big(\\\\frac{1}{2}-\\\\frac{1}{6}\\\\Big)a+\\\\frac{2}{6}b+\\\\Big(-\\\\frac{1}{2}+\\\\frac{1}{6}\\\\Big)c=\\\\frac{1}{3}a+\\\\frac{1}{3}b-\\\\frac{1}{3}c=\\\\frac{a}{3}[11\\\\bar{1}]$ Energy condition: $\\\\left|{\\\\frac{a}{2}}{\\\\surd2}\\\\right|^{2}+\\\\left|{\\\\frac{a}{6}}{\\\\surd6}\\\\right|^{2}=\\\\Bigl({\\\\frac{a^{2}}{2}}+{\\\\frac{a^{2}}{6}}\\\\Bigr)>\\\\frac{a^{2}}{3}$ Therefore $\\\\frac{a}{2}[10\\\\overline{{{1}}}]+\\\\frac{a}{6}[\\\\overline{{{1}}}21]{\\\\rightarrow}\\\\frac{a}{3}[11\\\\overline{{{1}}}]$ The dislocation reaction can proceed." }, { "idx": 2731, "question": "Given the stacking fault energy Y=0.01 J/m², shear modulus G=7 × 10^10 Pa, lattice constant α=0.3 nm, Poisson's ratio v=0.3 for an fcc crystal, determine the equilibrium separation distance between the two partial dislocations $\\frac{a}{6}[11\\vec{2}]$ and $\\frac{a}{6}[2\\overline{{{1}}}\\overline{{{1}}}]$.", "answer": "$d_{\\mathrm{s}}\\approx\\frac{G b^{\\mathrm{2}}}{24\\pi\\gamma}$ $b{=}{\\frac{a}{n}}{\\sqrt{1^{2}+1^{2}+2^{2}}}{=}{\\frac{\\sqrt{6}}{6}}a,$ $$ d_{*}\\approx\\frac{7\\times10^{10}\\times\\frac{1}{6}\\times(0.3\\times10^{-9})^{2}}{24\\times3.1416\\times0.01}=1.3926\\times10^{-9}(\\mathrm{m})$" }, { "idx": 2732, "question": "The dislocation spacing of a Ni crystal is ${\\\\bf2000n m}$. Assuming each dislocation is caused by an additional (110) atomic plane, calculate the $\\\\pmb\\\\theta$ angle of the small-angle grain boundary.", "answer": "The interplanar spacing of (110) is $d_{110}=\\\\frac{1}{2}\\\\times\\\\frac{\\\\alpha}{\\\\sqrt{1^{2}+1^{2}+0}}=\\\\stackrel{0.35238}{2\\\\sqrt{2}}=0.1246(\\\\mathrm{nm})$. $$ \\\\sin\\\\frac{\\\\theta}{2}=\\\\frac{d_{110}}{\\\\cal{I}}=\\\\frac{0.1246}{2000}=6.23\\\\times10^{-5},$$ $$ \\\\theta{=0.003569527^{\\\\circ}}$$" }, { "idx": 2733, "question": "Can the new dislocation generated by the dislocation reaction $$ \\\\frac{a}{2}[10\\\\overline{{{1}}}] + \\\\frac{a}{6}[\\\\overline{{{1}}}21] \\\\rightarrow \\\\frac{a}{3}[11\\\\overline{{{1}}}] $$ move on the slip plane?", "answer": "The dislocation line of the new dislocation $\\\\frac{a}{3}[11\\\\overline{{{1}}}]$ is located at the intersection of (111) and (111) planes on the (001) plane, and it is a pure edge dislocation. Since the (001) plane is a non-close-packed plane in fcc, it cannot move and is a fixed dislocation." }, { "idx": 2734, "question": "In an aluminum specimen, the dislocation density within the grains was measured to be 5×10^9/cm². Assuming all dislocations are concentrated on the subgrain boundaries; each subgrain has a regular hexagonal cross-section. The misorientation angle between subgrains is 5°, and if all dislocations are edge dislocations with b = a/2 [101], the magnitude of the Burgers vector is equal to 2×10^-10 m. Calculate the dislocation spacing on the subgrain boundary.", "answer": "D = b / (2 sin(θ/2)) ≈ b / θ = (2×10^-10) / 0.087 = 23×10^-10 m" }, { "idx": 2735, "question": "If this stress determines the shear strength of the material, to achieve a strength value of $\\\\frac{G}{100}$, and given that $G=50\\\\mathrm{GPa}$, $a=0.36\\\\mathrm{nm}$, what should the value of $D$ be?", "answer": "According to $\\\\overline{\\\\tau} = \\\\frac{G}{100}$, substituting $\\\\overline{\\\\tau} = \\\\frac{Gb}{D}$ yields $D = 100b$. The Burgers vector $b = \\\\frac{a\\\\sqrt{3}}{2} = \\\\frac{0.36\\\\times10^{-9}\\\\times\\\\sqrt{3}}{2} \\\\approx 3.12\\\\times10^{-10}\\\\mathrm{m}$, therefore $D = 100\\\\times3.12\\\\times10^{-10} = 3.12\\\\times10^{-8}\\\\mathrm{m}$." }, { "idx": 2736, "question": "The lattice constant of a copper single crystal is $a=0.36\\\\mathfrak{n m}$. When the copper single crystal sample is subjected to tensile deformation at a constant strain rate, the true strain of the sample is $6\\\\%$ after $\\\\mathtt{3s}$. If the average velocity of dislocation motion is $4\\\\times10^{-3}~\\\\mathrm{cm/s}$, calculate the average dislocation density in the crystal.", "answer": "① The two nodes in the dislocation network and the dislocation segments between them can act as F-R sources, and the shear stress required for dislocation multiplication is the minimum shear stress required to activate the F-R source; $$ D=\\\\frac{G b}{\\\\tau}$$ ② $$ D=\\\\frac{G b}{\\\\tau}=\\\\frac{G b}{\\\\frac{G}{100}}=100b=100~\\\\frac{\\\\sqrt{2}}{2}a=25.5\\\\left(\\\\mathrm{nm}\\\\right)$$ ③ For a three-dimensional dislocation network $$ \\\\rho={\\\\frac{2}{D^{2}}}=2\\\\left({\\\\frac{\\\\tau}{G b}}\\\\right)^{2}=2\\\\left({\\\\frac{42\\\\times10^{6}}{50\\\\times10^{8}\\\\times2.55\\\\times10^{-8}}}\\\\right)^{2}=2.17\\\\times10^{9}{\\\\mathrm{cm}}^{-2}$$" }, { "idx": 2737, "question": "The average distance between nodes of the entangled three-dimensional dislocation network in a copper single crystal is $D$. Calculate the stress $\\overline{\\tau}$ required for dislocation multiplication.", "answer": "The stress $\\overline{\\tau}$ required for dislocation multiplication can be calculated by the formula $\\overline{\\tau} = \\frac{Gb}{D}$, where $G$ is the shear modulus and $b$ is the Burgers vector of the dislocation." }, { "idx": 2738, "question": "In an aluminum sample, the dislocation density inside the grains is measured to be 5×10^9/cm². Assuming all dislocations are concentrated on the subgrain boundaries; each subgrain has a regular hexagonal cross-section. The misorientation angle between subgrains is 5°, and if all dislocations are edge dislocations, with b = a/2 [101], where the magnitude of the Burgers vector equals 2×10^-10 m, calculate the average size of the subgrains.", "answer": "The area of a regular hexagon S = (3/2)√3 a², and the total perimeter is 6a. The number of subgrains per unit area n = 1/S. The dislocation density ρ = 5×10^13 = (1/S) × 6a × (1/D) × (1/2). Solving gives a = 1×10^-5 m." }, { "idx": 2739, "question": "If an extra (111) plane is embedded, causing a small-angle grain boundary tilted by 1° in α-Fe, calculate the average distance between the dislocations.", "answer": "The lattice constant of α-Fe crystal $$ \\\\alpha=\\\\frac{4r_{\\\\mathrm{Fe}}}{\\\\sqrt{3}}=\\\\frac{4\\\\times0.1241}{\\\\sqrt{3}}=0.2866(\\\\mathrm{nm})$$ $$x d_{111}={\\\\frac{a}{\\\\sqrt{{{1}}^{2}+{{1}}^{2}+{{1}}^{2}}}}={\\\\frac{0.2866}{\\\\sqrt{3}}}=0.16547({\\\\mathrm{nm}})$$ $$ \\\\sin{\\\\frac{\\\\theta}{2}}={\\\\frac{d_{\\\\mathrm{{111}}}}{\\\\it{l}}}$$ $$ \\\\begin{array}{r}{\\\\lambda=\\\\frac{d_{\\\\mathrm{{n1}}}}{\\\\sin\\\\frac{\\\\theta}{2}}=\\\\frac{0.16547}{\\\\sin\\\\left(\\\\frac{1}{2}\\\\right)^{\\\\circ}}=18.9615(\\\\mathrm{nm})}\\\\end{array}$$" }, { "idx": 2740, "question": "To achieve homogeneous nucleation at 2045K, by how much must the atmospheric pressure be increased? Given the volume change during solidification ΔV= -0.26 cm^3/mol (1J=9.87×10^6 cm^3·Pa)", "answer": "For homogeneous nucleation to occur at 1726K, an undercooling of 319°C is required. To achieve this, the pressure must be increased to raise the solidification temperature of pure nickel from 1726K to 2045K: dP/dT=ΔH/TΔV. Integrating this equation: ∫1.013×10^5^P dP=∫1726^2045 ΔH/TΔV dT P-1.013×10^5=ΔH/ΔV ln2045/1726=18075/0.26×9.87×10^5×ln2045/1726=116366×10^5(Pa). Thus, when P=116366×10^5-1.013×10^5=116365×10^5(Pa), homogeneous nucleation can occur at 2045K." }, { "idx": 2741, "question": "It is known that liquid pure nickel undergoes homogeneous nucleation at an undercooling of 319°C under 1.013×10^5 Pa (1 atm). Given the critical nucleus radius as 1 nm, the melting point of pure nickel as 1726 K, the heat of fusion Lm = 18075 J/mol, and the molar volume V = 6.6 cm^3/mol, calculate the liquid-solid interfacial energy and the critical nucleation work for pure nickel.", "answer": "Since r* = 2σ/ΔGv = 2σTm/LmΔT = 2σTm V/ΔHmΔT, and for solidification, ΔGv = LmΔT/Tm, thus σ = r*ΔGv/2V = r*ΔHmΔT/2VTm = 1×10^-7 × 18075 × 319 / 2 × 1726 × 6.6 = 2.53×10^-5 (J/cm^2) = 0.253 (J/m^2). ΔG* = 16πσ^3Tm^2Vs^2 / 3ΔHm^2ΔT^2 = 16 × 3.14 × (2.53×10^-5)^3 × 1726^2 × 6.6^2 / 3 × 18075^2 × 319^2 = 1.06×10^-18 (J)." }, { "idx": 2742, "question": "Explain why polymers crystallized at lower temperatures have a wider melting range, and vice versa.", "answer": "When polymers crystallize at lower temperatures, the mobility of molecular chains is poor, resulting in less perfect crystals with a greater variation in perfection. Crystals with more defects will melt at lower temperatures, while those with fewer defects will melt at higher temperatures, leading to a wider melting range. Conversely, when polymers crystallize at higher temperatures, the mobility of molecular chains is stronger, resulting in more perfect crystals with less variation in perfection. Therefore, the melting range is narrower." }, { "idx": 2743, "question": "Calculate the axial ratio c/a of an ideal close-packed hexagonal crystal structure", "answer": "c/a=√(8/3)=1.633" }, { "idx": 2744, "question": "Indicate the close-packed plane of an ideal hexagonal close-packed crystal structure", "answer": "The close-packed plane is {0001}" }, { "idx": 2745, "question": "Calculate the change in the melting point of tin when the pressure increases to $500\\\\times10^{5}\\\\mathbf{Pa}$. It is known that at $\\\\mathfrak{10}^{5}\\\\mathbf{Pa}$, the melting point of tin is $505~\\\\mathrm{K}$, the heat of fusion is $7196\\\\mathrm{J/mol}$, the molar mass is $118.8\\\\times10^{-3}~\\\\mathrm{kg/mol}$, the density of solid tin is $\\\\small7.30\\\\times10^{3}~\\\\mathrm{kg/m^{3}}$, and the volume change during melting is $+2.7\\\\%$.", "answer": "The molar volume of tin $$ V_{\\\\mathrm{m}}={\\\\mathrm{molar mass}}/\\\\mathrm{density}={\\\\frac{118.7\\\\times10^{-3}}{7.30\\\\times10^{3}}}=1.626\\\\times10^{-5}(\\\\mathrm{m}^{3}/\\\\mathrm{mol})$$ $$ \\\\begin{array}{r}{\\\\Delta V_{\\\\mathrm{m}}=0.027\\\\times1.626\\\\times10^{-5}}\\\\\\\\ {=4.39\\\\times10^{-7}\\\\left(\\\\mathrm{m}^{3}/\\\\mathrm{mol}\\\\right)}\\\\end{array}$$ Assuming $\\\\Delta V_{\\\\mathfrak{m}}$ and $\\\\Delta H_{\\\\pmb{\\\\upmu}}$ remain constant within the considered temperature range, and $\\\\Delta T{\\\\ll}T$ $$ \\\\begin{array}{c}{{\\\\frac{\\\\Delta\\\\phi}{\\\\Delta T}=\\\\frac{\\\\Delta H_{\\\\mathrm{m}}}{T\\\\cdot\\\\Delta V_{\\\\mathrm{m}}}=\\\\frac{7196}{505\\\\times4.39\\\\times10^{-7}}}}\\\\\\\\ {{=3.25\\\\times10^{7}(\\\\mathrm{N}\\\\cdot\\\\mathrm{m}^{-2}\\\\cdot\\\\mathrm{K}^{\\\\mathrm{\\\\Delta_{\\\\parallel}}})}}\\\\end{array}$$ Then $$ \\\\Delta T={\\\\frac{(500-1)\\\\times10^{5}}{3.25\\\\times10^{7}}}=1.54(\\\\mathrm{K})$$" }, { "idx": 2746, "question": "Determine the solidification velocity $R$ required to maintain a planar liquid-solid interface during directional solidification of an 8% B binary alloy. Given the temperature gradient $G=225^{\\\\circ}C$/cm, the diffusion coefficient of component B $D=2\\\\times10^{-4}$ cm²/s, the equilibrium partition coefficient $\\\\pmb{k}_{0}=0.3$, and the slope of the binary alloy liquidus line $m=0.142^{\\\\circ}C$/%, which means the temperature decreases by $0.142^{\\\\circ}C$ for every 1% increase in B solute concentration.", "answer": "nan" }, { "idx": 2747, "question": "What is the experimental method for determining the critical resolved shear stress of a crystal?", "answer": "Experimental method: 1) Select an appropriate orientation in the single crystal to ensure that the initial slip of the crystal is single slip. 2) Determine the orientation of the crystal's tensile direction to obtain the orientation factor. 3) Use τc=σscosφcosλ and the σs from the tensile curve, along with the orientation factor, to calculate the τs, the type of crystal, purity, test temperature, strain rate, and other factors affecting the resolved shear stress τc value." }, { "idx": 2748, "question": "In a copper crystal, when the a/2[101] dislocation on the (111) plane reacts with the a/2[011] dislocation on the (11-1) plane, write the dislocation reaction equation and determine the direction of the reaction.", "answer": "[101]+[011]a2/2+a2/2>a2/2 , the reaction proceeds to the right" }, { "idx": 2749, "question": "The $c/a$ ratio for an ideal hexagonal close-packed metal is ", "answer": "B" }, { "idx": 2750, "question": "Indicate the close-packed directions of an ideal hexagonal close-packed crystal structure", "answer": "The close-packed directions are (11-20)" }, { "idx": 2751, "question": "In a copper crystal, when the a/2[101] dislocation on the (111) plane reacts with the a/2[011] dislocation on the (11-1) plane, describe the nature of the new dislocation.", "answer": "The new dislocation b=a/2[110] is a unit dislocation in the face-centered cubic structure; the direction of the dislocation line is along the intersection line [110] of the two crystal planes (111) and (111), so the dislocation is of edge type; the slip plane is (001), which is not a close-packed plane, hence it is a sessile dislocation." }, { "idx": 2752, "question": "Cesium chloride (CsCl) has an ordered body-centered cubic structure, which belongs to (A) body-centered cubic lattice (B) face-centered cubic lattice (C) simple cubic lattice", "answer": "C" }, { "idx": 2753, "question": "The [001] direction in a cubic crystal is", "answer": "B" }, { "idx": 2754, "question": "In the hexagonal crystal system, the interplanar spacing of (11-20) plane is (1010) plane spacing.", "answer": "A" }, { "idx": 2755, "question": "In a crystal, the defect formed by creating a vacancy while simultaneously generating an interstitial atom is called", "answer": "B" }, { "idx": 2756, "question": "Fick's first law describes the characteristics of steady-state diffusion, where the concentration does not vary with . \\n\\n(A) distance (B) time (C) temperature", "answer": "B" }, { "idx": 2757, "question": "The ordered structure formation temperature of any alloy is _____ the disordered structure formation temperature. (A) Lower than (B) Higher than (C) May be lower or higher than", "answer": "A" }, { "idx": 2758, "question": "The twinning plane of a face-centered cubic crystal is", "answer": "C" }, { "idx": 2759, "question": "There is a dislocation loop on the crystal slip plane, and a shear stress $\\\\boldsymbol{\\\\tau}$ is applied in the direction of its Burgers vector. What is the minimum radius required for the dislocation loop to remain stable in the crystal?", "answer": "The force acting on the ds arc segment of the dislocation line: $\\\\pmb{\\\\tau}\\\\cdot\\\\pmb{b}\\\\cdot\\\\mathbf{d}\\\\mathscr{s}.$ At the same time, the line tension on the dislocation line: $T{\\\\approx}\\\\alpha\\\\cdot G\\\\cdot b^{2}$ , where the horizontal component: $2(\\\\alpha G b^{2})\\\\sin\\\\frac{\\\\mathrm{d}\\\\theta}{2}$ Thus, $$ \\\\mathrm{d}s=r\\\\mathrm{d}{\\\\theta},\\\\sin\\\\frac{\\\\mathrm{d}\\\\theta}{2}\\\\approx\\\\frac{\\\\mathrm{d}\\\\theta}{2}$$ When the two forces reach equilibrium, they are equal, i.e., Therefore, $$ \\\\begin{array}{c}{{\\\\tau\\\\bullet b\\\\bullet\\\\mathrm{d}s=2(a G b^{2}){\\\\frac{\\\\mathrm{d}\\\\theta}{2}}}}\\\\\\\\ {{\\\\tau\\\\bullet b\\\\bullet r_{c}\\\\bullet\\\\mathrm{d}\\\\theta=a G b^{2}\\\\mathrm{d}\\\\theta}}\\\\\\\\ {{r_{\\\\bullet}={\\\\frac{a G b}{\\\\tau}}}}\\\\end{array}$$" }, { "idx": 2760, "question": "In a body-centered cubic structure, can a dislocation with Burgers vector $a[100]$ decompose into $\\frac{a}{2}[111]+\\frac{a}{2}[1\\overline{1}\\overline{1}]$? (A) No (B) Yes (C) Possible", "answer": "A" }, { "idx": 2761, "question": "In substitutional solid solutions, the general mode of atomic diffusion is", "answer": "C" }, { "idx": 2762, "question": "The condition for an alloy to develop constitutional supercooling during solidification is . (where $T_{\\\\ L}$ is the solidification start temperature of the alloy with composition $C_{\\\\mathrm{L}}$)", "answer": "A" }, { "idx": 2763, "question": "The driving force for atomic diffusion is", "answer": "B" }, { "idx": 2764, "question": "When a critical nucleus forms, the reduction in volume free energy can only compensate for the surface energy by ", "answer": "B" }, { "idx": 2765, "question": "In the Kirkendall effect, the main reason for marker drift is the diffusion couple's ", "answer": "C" }, { "idx": 2766, "question": "What is the critical resolved shear stress for crystal slip?", "answer": "The minimum resolved shear stress required for alternate slip in the slip systems of a crystal is called the critical resolved shear stress." }, { "idx": 2767, "question": "The difference between cast iron and carbon steel lies in the presence of", "answer": "A" }, { "idx": 2768, "question": "The effective distribution coefficient $K_{i}$ represents the degree of mixing in the liquid phase, and its value range is . (where $K_{\\circ}$ is the equilibrium distribution coefficient)", "answer": "B" }, { "idx": 2769, "question": "In a binary alloy phase diagram, the lever rule for calculating the relative amounts of two phases can only be applied in", "answer": "B" }, { "idx": 2770, "question": "According to the vertical section of a ternary phase diagram, one can", "answer": "B" }, { "idx": 2771, "question": "In a ternary phase diagram, the isothermal section of a three-phase region is always a connected triangle, with its vertices touching", "answer": "A" }, { "idx": 2772, "question": "In the ternary composition triangle, alloys whose compositions lie on the have equal contents of the two components represented by the other two vertices. (A) The perpendicular bisector passing through the vertex of the triangle (B) Any straight line passing through the vertex of the triangle (C) A straight line passing through the vertex of the triangle and forming a $45^{\\circ}$ angle with the opposite side", "answer": "A" }, { "idx": 2773, "question": "How many equivalent slip systems are there when a face-centered cubic metal single crystal is stretched along [001]? And specifically write out the indices of each slip system.", "answer": "When an fcc-structured crystal is stretched along the [001] axis, there are a total of 8 equivalent slip systems, which are: (111)[011], (111)[101], (111)[011], (111)[101], (1 11)[011], (1 11)[101], (111)[101], (111)[011]." }, { "idx": 2774, "question": "How many equivalent slip systems are there when a face-centered cubic metal single crystal is stretched along [111]? Also, specifically write the indices of each slip system.", "answer": "When stretched along the [111] direction, there are 6 equivalent slip systems, which are: (111)[011], (111)[110], (111)[011], (111)[101], (111)[101], (111)[110]." }, { "idx": 2775, "question": "Carburizing steel in a carbon-rich environment can harden the surface of the steel. It is known that during this carburizing heat treatment at 1000°C, the carbon content decreases from x=5% to x=4% at a distance of 1∼2mm from the surface of the steel. First, the content of solute carbon atoms should be converted from atomic fraction to volume fraction, so the number of atoms per unit volume of the solvent iron must be determined. The density of γ-Fe at 1000°C is known to be =7.63g/cm³.", "answer": "ρ=7.63×(6.023×10^23)/55.85=8.23×10^22 at/cm³" }, { "idx": 2776, "question": "Given that the recrystallization activation energy of a $\\mathtt{C u}{\\sim}30\\%\\mathtt{Z n}$ alloy is $250\\mathrm{kJ/mol}$, and it takes 1 hour for this alloy to complete recrystallization at a constant temperature of $400^{\\circ}C$, calculate how many hours it will take for this alloy to complete recrystallization at a constant temperature of $390\\mathrm{\\textperthousand}$.", "answer": "Solution: Using the formula $$ \\frac{t_{2}}{t_{1}}{=}\\ e^{-\\frac{Q}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)},$$ thus $$ \\begin{array}{r l}&{\\frac{t_{2}}{t_{1}}{=}\\exp\\Big({-}\\frac{Q}{R}\\Big(\\frac{1}{T_{1}}{-}\\frac{1}{T_{2}}\\Big)\\Big)}\\ &{\\quad\\quad{=}\\exp\\Big({-}\\frac{250\\times10^{3}}{8.314}\\Big(\\frac{1}{400+273}{-}\\frac{1}{390+273}\\Big)\\Big)}\\ &{\\quad{=}1.962}\\end{array}$$ Therefore, it will take 1.962 hours." }, { "idx": 2777, "question": "Based on the diffusion coefficients at 1223K and 1136K, calculate the activation energy Q for interstitial atom diffusion in face-centered cubic metals. The diffusion coefficient at 1223K is 4.34×10⁻⁸m²/s, and at 1136K it is 1.78×10⁻⁸m²/s.", "answer": "According to Arrhenius' law, D = D₀exp(-Q/RT). For two temperatures, we have: D₁/D₂ = exp[(-Q/R)(1/T₁ - 1/T₂)]. Substituting the data: 4.34×10⁻⁸ / 1.78×10⁻⁸ = exp[(-Q/8.314)(1/1223 - 1/1136)]. Solving gives: Q ≈ 1.2×10⁵J/mol." }, { "idx": 2778, "question": "At a temperature of 1223K, calculate the diffusion coefficient of interstitial atoms in a face-centered cubic metal. The thickness of the metal film is 0.25mm, the end face area is 1000mm², the rate of interstitial atoms passing through the film is 0.0025g/s, and the solubility of interstitial atoms in the film is 14.4kg/m³.", "answer": "According to Fick's first law, the formula for calculating the diffusion coefficient D is: D = -J * Δx / Δρ. Here, J = v / A = 0.0025g/s / 1000mm² = 0.0025×10⁻³kg/s / 10⁻³m² = 0.0025kg/(m²·s). Δx = 0.25mm = 0.25×10⁻³m. Δρ = 0 - 14.4kg/m³ = -14.4kg/m³. Therefore, D = -0.0025kg/(m²·s) * 0.25×10⁻³m / (-14.4kg/m³) ≈ 4.34×10⁻⁸m²/s." }, { "idx": 2779, "question": "According to Fick's first law, calculate the influx J (atoms/m²s) of carbon atoms entering the steel in the near-surface region.", "answer": "J=−D(Δρ/Δx)=−(2.98×10⁻¹¹)×(−8.23×10^29)=2.45×10^19 at/(m²s)" }, { "idx": 2780, "question": "The homogeneous nucleation rate of pure metals can be expressed by the following equation: $$ \\\\dot{N}=A\\\\exp\\\\Bigl(-\\\\frac{\\\\Delta G^{*}}{k T}\\\\Bigr)\\\\exp\\\\Bigl(-\\\\frac{Q}{k T}\\\\Bigr)$$ where $A\\\\approx10^{35}\\\\mathfrak{s e x p}\\\\left\\\\{-\\\\frac{Q}{k T}\\\\right\\\\}\\\\approx10^{-2}\\\\mathfrak{s}$, $\\\\Delta G^{*}$ is the critical nucleation work; $\\\\pmb{k}$ is the Boltzmann constant, with a value of $1.38\\\\times10^{-23}\\\\mathrm{J}/\\\\mathrm{K}$. \\n\\n$\\\\textcircled{1}$ Assuming undercooling $\\\\Delta T$ is $20\\\\mathrm{\\\\mathcal{C}}$ and ${\\\\bf 200}\\\\%$, interfacial energy $\\\\sigma{=}2\\\\times10^{-5}~\\\\mathrm{J/cm^{2}}$, heat of fusion $\\\\Delta H_{\\\\mathrm{m}}=$ $\\\\mathrm{12600J/mol}$, melting point $\\\\mathcal{T}_{\\\\mathbf{m}}{=}1000\\\\:\\\\mathrm{K}$, molar volume $V{=}6~\\\\mathrm{cm}^{3}/\\\\mathrm{mol}$, calculate the homogeneous nucleation rate $\\\\dot{N}$. \\n\\n$\\\\textcircled{2}$ If it is heterogeneous nucleation, with the contact angle between the nucleus and impurity $\\\\pmb{\\\\theta}\\\\mathrm{=}60^{\\\\circ}$, how does $\\\\dot{N}$ change? What is $\\\\Delta T$? \\n\\n$\\\\textcircled{3}$ Derive the relationship between $\\\\neq^{*}$ and $\\\\mathbf{\\\\Delta}\\\\pmb{\\\\Delta}^{T}$, and calculate $\\\\begin{array}{l}{\\\\underbrace{\\\\Delta\\\\mathcal{T}}_{\\\\mathcal{T}_{\\\\infty}}}\\\\end{array}$ when $r^{*}=1{\\\\mathrm{~nm}}$.", "answer": "$$ \\\\begin{array}{l}{{\\\\dot{N}=A\\\\exp\\\\Bigl(-\\\\frac{\\\\Delta G^{*}}{{\\\\dot{k}}T}\\\\Bigr)\\\\exp\\\\Bigl\\\\{-\\\\frac{Q}{k t}\\\\Bigr\\\\}}}\\\\ {{{}}}\\\\ {{=10^{35}\\\\times10^{-2}\\\\times\\\\exp\\\\Bigl(-\\\\frac{16\\\\pi\\\\sigma^{3}}{3\\\\Delta G_{v}^{2}k T}\\\\Bigr)}}\\\\ {{{}}}\\\\ {{=10^{33}\\\\times\\\\exp\\\\Bigl(-\\\\frac{16\\\\pi\\\\sigma^{3}T_{\\\\mathrm{a}}^{2}V^{2}}{3k T\\\\Delta{\\\\dot{H}}^{2}\\\\Delta{\\\\dot{T}}^{2}}\\\\Bigr)}}\\\\end{array}$$ $\\\\textcircled{1}$ \\n\\nWhen $\\\\Delta T=20\\\\mathrm{~C~}$: $$ {\\\\dot{N}}{=}10^{33}\\\\exp\\\\Bigl[{-\\\\frac{16\\\\times3.14\\\\times(2\\\\times10^{-5})^{3}\\\\times1000^{2}\\\\times6^{2}}{3\\\\times1.38\\\\times10^{-23}\\\\times800\\\\times12600^{2}\\\\times200^{2}}}\\\\Bigr]$$ $\\\\textcircled{2}$ When $\\\\theta=60^{\\\\circ}$: The heterogeneous nucleation free energy $$ \\\\Delta G_{\\\\mathrm{in}}^{*}=\\\\Delta G^{*}\\\\ \\\\Bigl(\\\\frac{2-3\\\\mathrm{cos}60^{\\\\circ}+\\\\mathrm{cos}^{3}60^{\\\\circ}}{4}\\\\Bigr)=0.156\\\\Delta G^{*}.$$ When $\\\\Delta T{=}20\\\\mathrm{~C~}$, $\\\\dot{N}=10^{33}\\\\exp(-0.156\\\\times5615,8)=0$. When $\\\\Delta{\\\\cal T}{=}200\\\\Upsilon$, $\\\\dot{N}{=}10^{33}\\\\exp(-0.156\\\\times68.79){=}2.2\\\\times10^{28}\\\\left({\\\\mathrm{cm}}^{-3}{\\\\mathrm{s}}^{-1}\\\\right)$. Assuming undercooling is $\\\\Delta T,T{=}T_{\\\\infty}{-}\\\\Delta T$, according to the given conditions: $$ 1=10^{33}\\\\exp\\\\Bigl(-\\\\frac{16\\\\times3.14\\\\times200^{3}\\\\times1000^{2}\\\\times6^{2}}{3\\\\times1.38\\\\times10^{-16}\\\\times(12600\\\\times10^{7})^{2}(1000-\\\\Delta T)\\\\Delta T^{2}}\\\\times0.156\\\\Bigr)$$ or $$ 10^{-33}=\\\\exp\\\\Bigl(-\\\\frac{3.43\\\\times10^{8}}{(1000-\\\\Delta T)\\\\Delta T^{2}}\\\\Bigr)$$ Taking the logarithm of both sides: $$ 75.98={\\\\frac{3.43\\\\times10^{8}}{(1000-\\\\Delta T)\\\\Delta T^{2}}}$$ $$ (1000-\\\\Delta T)\\\\Delta T^{\\\\circ}=4.51\\\\times10^{6}$$ Thus: $$ \\\\Delta T\\\\approx70^{\\\\circ}C$$ $\\\\textcircled{3}$ $$ r^{\\\\bullet}=\\\\frac{2\\\\sigma}{\\\\Delta G}=\\\\frac{2\\\\sigma T_{\\\\mathrm{m}}V}{\\\\Delta H\\\\Delta T}$$ $$ {\\\\frac{\\\\Delta\\\\mathrm{T}}{\\\\mathrm{T}_{\\\\mathrm{m}}}}={\\\\frac{2\\\\sigma V}{\\\\Delta H r^{\\\\ast}}}$$ When $r^{*}=1\\\\mathrm{nm}$: $$ \\\\frac{\\\\Delta T}{T_{\\\\mathrm{m}}}=\\\\frac{2\\\\times200\\\\times6}{12600\\\\times10^{7}\\\\times1\\\\times10^{-7}}=0.19$$" }, { "idx": 2781, "question": "Calculate the diffusion coefficient D of carbon in γ-Fe. Given the diffusion coefficient of carbon in γ-Fe D₀=2.0×10⁻⁵ m²/s, activation energy Q=142 kJ/mol, and temperature T=1000℃.", "answer": "D=D₀e^(−Q/RT)=2.0×10⁻⁵×exp(−142000/(8.314×1273))=2.98×10⁻¹¹ m²/s" }, { "idx": 2782, "question": "Approximately consider (number of carbon atoms + number of iron atoms) ≈ number of iron atoms, calculate the rate of change of carbon atom number Δρ/Δx at a distance of 1∼2mm from the steel surface.", "answer": "Δρ/Δx=(5%−4%)×(8.23×10^22)/(1−2)×10^6×10^3=−8.23×10^29 at/m⁴" }, { "idx": 2783, "question": "Given a diffusion couple composed of pure chromium and pure iron, after 1 hour of diffusion, the Matano plane moved by 1.52×10^(-3) cm. It is known that the ratio of the square of the distance moved by the Matano plane to the diffusion time is a constant. Determine the moving speed of the Matano plane.", "answer": "According to the given conditions: x²/t = k, the moving speed of the Matano plane vₐ = dx/dt = k/(2x) = x/(2t) = (1.52×10^(-3))/(2×3600) cm/s." }, { "idx": 2784, "question": "At a temperature of 1136K, calculate the diffusion coefficient of interstitial atoms in a face-centered cubic metal. It is known that the metal film thickness is 0.25mm, the end face area is 1000mm², the rate of interstitial atoms passing through the film is 0.0014g/s, and the solubility of interstitial atoms in the film is 19.6kg/m³.", "answer": "According to Fick's first law, the formula for calculating the diffusion coefficient D is: D = -J * Δx / Δρ. Here, J = v / A = 0.0014g/s / 1000mm² = 0.0014×10⁻³kg/s / 10⁻³m² = 0.0014kg/(m²·s). Δx = 0.25mm = 0.25×10⁻³m. Δρ = 0 - 19.6kg/m³ = -19.6kg/m³. Therefore, D = -0.0014kg/(m²·s) * 0.25×10⁻³m / (-19.6kg/m³) ≈ 1.78×10⁻⁸m²/s." }, { "idx": 2785, "question": "There is a silicon single crystal wafer with a thickness of 0.5 mm. On one end face, every 10^7 silicon atoms contain two atoms, while the other end face has an increased concentration of gallium after treatment. Determine how many gallium atoms per 10^7 silicon atoms are required on this face to achieve a concentration gradient of 2×10^26 atoms/m³·m. The lattice constant of silicon is 0.5407 nm.", "answer": "Silicon has a diamond structure, with 8 atoms per unit cell. Therefore, the volume corresponding to 10^7 atoms is: V = (10^7 / 8) × a₀³ = (10^7 / 8) × (0.5407 × 10^-9)^3 = 1.976 × 10^-22 m³. Let x be the number of gallium atoms per 10^7 silicon atoms on this end face. Then: Δρ/Δx = [(x / V) - (2 / V)] / (0.5 × 10^-3) = 2 × 10^26. x = 2 + 0.5 × 10^-3 × 2 × 10^26 × 1.976 × 10^-22 = 21.76 ≈ 22. Thus, this end face must contain 22 gallium atoms per 10^7 silicon atoms." }, { "idx": 2786, "question": "A carbon steel with w(C)=0.1% is carburized at 930°C, reaching a carbon concentration of 0.45% at a depth of 0.05cm. For all times t>0, the carburizing atmosphere maintains a surface composition of 1%. Assuming D=0.2×10^(-5) exp(-140000/RT)(m²/s), calculate the carburizing time.", "answer": "From Fick's second law, the solution is w=w_s-(w_s-w_0)erf(x/(2√(Dt))). Substituting the data gives (1%-0.45%)/(1%-0.1%)=erf(0.05/(2√(Dt))), i.e., 0.61=erf(0.05/(2√(Dt))). Looking up the table yields 0.05/(2√(Dt))=0.61, and D=0.2×exp(-140000/(8.314×1203))=1.67×10^(-7)(cm²/s). Therefore, the carburizing time t≈1.0×10^3(s)." }, { "idx": 2787, "question": "If w(C)=0.3% is specified as the measure of carburization layer thickness, how many times is the carburization layer thickness after 10h at 930°C compared to that after 10h at 870°C?", "answer": "x₉₃₀/x₈₇₀=√(D₉₃₀t₉₃₀)/√(D₈₇₀t₈₇₀). Since t₉₃₀=t₈₇₀, D₉₃₀=1.67×10^(-7)(cm²/s), D₈₇₀=0.2×exp(-140000/(8.314×1143))=8.0×10^(-8)(cm²/s). Therefore, x₉₃₀/x₈₇₀=√(D₉₃₀)/√(D₈₇₀)=√(1.6×10^(-7)/7.9×10^(-8))=1.45." }, { "idx": 2788, "question": "Given the initial carbon concentration of plain carbon steel w0=0.85%, after heating to 900°C and holding in air for 1h, the outer carbon concentration drops to zero. If the required carbon concentration in the outer layer of the part is 0.8%, how much depth should be turned off the surface? (D=1.1×10−7cm2/s)", "answer": "From w=w0erf(β), we get 0.80=0.85erf(x/(2√Dt)) → erf(x/(2√Dt))=0.94. Checking the error function table gives x/(2√Dt)=1.33. Calculating √Dt=√(1.1×10−7×3600)=0.0199 cm. Solving for x gives x=1.33×2×0.0199=0.053 cm." }, { "idx": 2789, "question": "Derive the solution to the decarburization diffusion equation, assuming that at t>0, ρ=0 at x=0. The initial conditions are known as t=0, x≥0, ρ=ρ0; boundary conditions are t>0, x=0, ρ=0; x=∞, ρ=ρ0.", "answer": "From Fick's second law, the general solution is obtained (assuming D is independent of ρ): ρ=A1∫0βexp(−β2)dβ+A2. Initial condition t=0, x≥0, ρ=ρ0, β=x/(2√Dt). Boundary condition t>0, x=0, ρ=0 → 0=A1∫0βexp(−β2)dβ+A2 → A2=0. Combined equation: ρ0=A1√π/2+A2 → A1=2ρ0/√π. Substituting into the general solution: ρ=2ρ0/√π∫0βexp(−β2)dβ=ρ0erf(β). Dividing by the alloy density gives w=w0erf(β)." }, { "idx": 2790, "question": "If the depth of the diffusion layer is doubled, how much time is required?", "answer": "From the relation x=A√(Dt), we have x₁=A√(Dt₁) and x₂=A√(Dt₂). Dividing the two equations gives (x₂²)/(x₁²)=(Dt₂)/(Dt₁). When the temperature is the same, D₁=D₂, thus t₂=(x₂²)/(x₁²)×t₁=(0.1²)/(0.05²)×1.0×10^4=4.0×10^4 s." }, { "idx": 2791, "question": "Calculate the root mean square displacement of the total migration of carbon atoms $\\\\sqrt{\\\\overline{R_{n}^{2}}}$, given the jump frequency of carbon atoms as $\\\\Gamma=1.7\\\\times10^{9}/\\\\mathrm{s}$, the jump distance as $2.53\\\\times10^{-10}\\\\mathrm{m}$, and the time as $4\\\\mathrm{h}$.", "answer": "$\\\\sqrt{\\\\overline{R_{n}^{2}}}=\\\\sqrt{n}\\\\cdot r=\\\\sqrt{1.7\\\\times10^{9}\\\\times4\\\\times3600}\\\\times2.53\\\\times10^{-10}=1.25\\\\times10^{-3}\\\\mathrm{m}\\\\approx1.3\\\\mathrm{mm}$" }, { "idx": 2792, "question": "Given a diffusion couple composed of pure chromium and pure iron, it is known that when the mole fraction x_Cr=0.478, ∂x/∂z=126/cm, and the interdiffusion coefficient D=1.43×10^(-8) m²/s, find the intrinsic diffusion coefficients D_Cr and D_Fe of chromium and iron.", "answer": "According to the Kirkendall effect, the marker velocity vₐ = (D_Cr - D_Fe)×126. The interdiffusion coefficient D = x_Fe D_Cr + x_Cr D_Fe → 0.478 D_Fe + (1-0.478) D_Cr = 1.43×10^(-8) m²/s. By solving these two equations together with the marker velocity vₐ at the Matano plane, we obtain D_Cr = 2.23×10^(-9) cm²/s and D_Fe = 0.56×10^(-9) cm²/s." }, { "idx": 2793, "question": "For pure iron carburized at $950\\\\mathrm{\\\\textperthousand}$, it is desired to achieve a carbon content of $\\\\mathfrak{w}{1}\\\\left(\\\\mathbb{C}\\\\right)=0,9\\\\%$ at a depth of $0.1\\\\mathrm{mm}$. Assuming the surface carbon content remains at $w{2}\\\\left(\\\\mathsf{C}\\\\right)=1.20\\\\%$, and the diffusion coefficient $D_{\\\\mathtt{Y F e}}{=}10^{{-}10}\\\\ensuremath{\\\\mathbf{m}}^{2}/\\\\mathbf{s},$, calculate the minimum carburization time required to meet this requirement.", "answer": "$$\\\\begin{array}{c}{\\\\frac x{2\\\\sqrt{D t}}=0,8163}}\\\\ {{{\\\\mathrm x}=2\\\\times0.8163\\\\times\\\\sqrt{1.1\\\\times10^{-1}\\\\times3600}}}\\\\ {{{\\\\mathrm x}=0.032\\\\mathrm{(cm)}}}\\\\ {{{\\\\mathrm w}_{2}-{\\\\mathrm w}_{1}=\\\\mathrm{erf}\\\\left({\\\\frac{x}{2\\\\sqrt{D t}}}\\\\right)}}\\\\ {{{\\\\mathrm x}_{2}-{\\\\mathrm w}_{0}=\\\\mathrm{erf}\\\\left({\\\\frac{0.1\\\\times10^{-3}}{2\\\\sqrt{10^{-10}\\\\times t}}}\\\\right)}}\\\\end{array}$$ $$\\\\operatorname{erf}{\\\\Big(}{\\\\frac{5}{\\\\sqrt{t}}}{\\\\Big)}=0.25$$ Look up the table to find $$\\\\frac{5}{\\\\sqrt{t}}\\\\approx0.2763,$$ hence $$t\\\\approx327(\\\\mathrm{s})$$" }, { "idx": 2794, "question": "There are two diffusion reactions with activation energies of Q_1=83.7 kJ/mol and Q_2=251 kJ/mol, respectively. Observe the effect of increasing the temperature from 25°C to 600°C on the diffusion with an activation energy of Q_2=251 kJ/mol, and comment on the results.", "answer": "From D=D_0 exp(-Q/RT), we get D_873/D_298=exp[-251000/(8.314)×(298-873)/(873×298)]=9.5×10^28. For the temperature increase from 298K to 873K, the diffusion rate D increases by 9.5×10^28 times, showing that the higher the activation energy, the more sensitive the diffusion rate is to temperature." }, { "idx": 2795, "question": "Determine whether the formula D=D0exp(-Q/RT) is applicable to the diffusion coefficient data of carbon in α-Ti; if applicable, calculate the diffusion constant D0 and activation energy Q", "answer": "By plotting D against temperature (converted to absolute temperature), it was found that lgD has a linear relationship with 1/T, thus satisfying the formula D=D0exp(-Q/RT). The slope is Q/(2.3R), and the calculated Q=2.3×R×0.92×10^4=175.9 kJ/mol. For T=1009 K (736℃), D736=2×10^-13 m^2/s. Substituting into the formula gives lgD0=lgD+Q/(2.3R)×1/T=lg(2×10^-13)+175900/(2.3×8.314×1009)≈-3.58, therefore D0=2.62×10^-4 m^2/s." }, { "idx": 2796, "question": "Calculate the diffusion activation energy Q1 of polycrystalline silver at 700°C (based on the given data).", "answer": "From the equation -10.72-(-12) = -Q_1/R (1.10×10^-3 -1.30×10^-3) lge, the solution gives Q_1 = 122.4 kJ" }, { "idx": 2797, "question": "Calculate the diffusion coefficient of carbon in α-Ti at 500℃", "answer": "Using the formula lgD500=lgD0-Q/(2.3×R)×1/T=lg(2.62×10^-4)-175.9×10^3/(2.3×8.314×773)=-3.58-11.90=-15.48, thus D500≈3.31×10^-16 m^2/s." }, { "idx": 2798, "question": "Calculate the total migration distance $S$ of carbon atoms, given the carbon atom jump frequency as $\\Gamma=1.7\\times10^{9}/\\mathrm{s}$, jump distance as $2.53\\times10^{-10}\\mathrm{m}$, and time as $4\\mathrm{h}$.", "answer": "$4\\mathrm{h}=4\\times3600\\mathrm{s}$, $S=\\Gamma \\cdot t \\cdot r=1.7\\times10^{9}\\times4\\times3600\\times2.53\\times10^{-10}=6193\\mathrm{m}$" }, { "idx": 2799, "question": "Calculate the total migration distance S of carbon atoms at 20%, given the transition frequency r=2.1×10^-9/s, transition step length of 2.53×10^-10m, and time of 4h.", "answer": "S=Γ·t·r=2.1×10^-9×4×3600×2.53×10^-10=7.65×10^-15m" }, { "idx": 2800, "question": "Indicate approximately in which temperature range grain boundary diffusion dominates.", "answer": "In the low-temperature region (when 1/T is large), the lnD value of grain boundary diffusion is higher than that of bulk diffusion, and grain boundary diffusion dominates at this time. As the temperature increases (1/T decreases), the two curves will intersect, and in the high-temperature region above the intersection point, bulk diffusion will dominate." }, { "idx": 2801, "question": "Calculate the root mean square displacement $\\sqrt{\\overline{R_{n}^{2}}}$ of carbon atoms at $20\\%$, given the transition frequency is $r=2.1\\times10^{-9}/\\mathrm{s}$, the transition step length is $2.53\\times10^{-10}\\mathrm{m}$, and the time is $4\\mathrm{h}$.", "answer": "$\\sqrt{\\overline{R_{n}^{2}}}=\\sqrt{n}\\cdot r=\\sqrt{2.1\\times10^{-9}\\times4\\times3600}\\times2.53\\times10^{-10}=1.39\\times10^{-12}\\mathrm{m}$" }, { "idx": 2802, "question": "Calculate the diffusion activation energy Q2 for single-crystal silver (based on the given data).", "answer": "From equation 8-14 = -Q_2/R lge (0.8×10^-3 -1.39×10^-3), we obtain Q_2 = 194.5 kJ" }, { "idx": 2803, "question": "For grain boundary diffusion and bulk diffusion, assuming the diffusion activation energy Q_AF≈1/2 Q_AF, plot the curve of InD versus the reciprocal of temperature 1/T.", "answer": "According to the Arrhenius equation D = D_0 exp(-Q/RT), taking the logarithm gives lnD = lnD_0 - Q/(RT). For grain boundary diffusion (Q_gb) and bulk diffusion (Q_bulk), when Q_gb ≈ 1/2 Q_bulk, both lnD vs 1/T curves are straight lines, with the grain boundary diffusion line having a smaller slope." }, { "idx": 2804, "question": "Introducing high-valence W6+ into NiO will generate vacancies of which ions?", "answer": "Vacancies of cations (Ni) will be generated. (Principle of electroneutrality)" }, { "idx": 2805, "question": "There are two diffusion reactions with activation energies of Q_1=83.7 kJ/mol and Q_2=251 kJ/mol, respectively. Observe the effect of increasing the temperature from 25°C to 600°C on the diffusion with activation energy Q_1=83.7 kJ/mol, and comment on the results.", "answer": "From D=D_0 exp(-Q/RT), we obtain D_873/D_298=exp[-83700/(8.314)×(298-873)/(873×298)]=4.6×10^9. For the temperature increase from 298K to 873K, the diffusion rate D increases by 4.6×10^9 times, demonstrating the significant impact of temperature on the diffusion rate." }, { "idx": 2806, "question": "In NiO, introducing high-valence W6+, how many vacancies will each W6+ generate?", "answer": "Each W6+ introduction generates 2 Ni2+ vacancies." }, { "idx": 2807, "question": "Explain why the diffusion activation energy of polycrystals is lower than that of single crystals.", "answer": "Polycrystals have grain boundaries, and the 'short-circuit' diffusion effect of grain boundaries increases the diffusion rate, thus resulting in a lower diffusion activation energy. The diffusion in single crystals is purely bulk diffusion, which has a higher activation energy." }, { "idx": 2808, "question": "Compare the oxidation resistance of NiO and W-doped NiO (i.e., NiO-WO3), which one is better?", "answer": "Due to the introduction of W, the vacancy concentration increases, making it easier for oxygen in the air and Ni2+ ions in the oxide to migrate in and out of the surface, thereby increasing the oxidation rate. As a result, the oxidation resistance is reduced." }, { "idx": 2809, "question": "The yield strength of the Mg alloy is 180MPa, and E is 45GPa. Under the maximum load, what is the elongation per mm of this magnesium plate?", "answer": "ε=σ/E=180×10^6/45×10^9=0.004" }, { "idx": 2810, "question": "Explain the reason for the difference in diffusion coefficients between Al and O in Al2O3.", "answer": "Because in Al2O3, the ionic radius of the cation Al is smaller than that of the anion O, the activation energy for Al diffusion in Al2O3 is lower than that for O diffusion in Al2O3. Therefore, the diffusion coefficient of the former is greater than that of the latter." }, { "idx": 2811, "question": "Given the diffusion coefficient of O in Al2O3, D0(O) = 0.19 m^2/s, and the activation energy Q = 636 kJ/mol, calculate its diffusion coefficient D at a temperature of 2000 K.", "answer": "D = D0 * exp(-Q / (R * T)) = 0.19 * exp(-636000 / (8.314 * 2000)) = 4.7×10^-18 m^2/s" }, { "idx": 2812, "question": "Given the diffusion coefficient of Al in Al2O3, D0(Al) = 2.8×10^-3 m^2/s, and the activation energy Q = 477 kJ/mol, calculate its diffusion coefficient D at a temperature of 2000 K.", "answer": "D = D0 * exp(-Q / (R * T)) = 2.8×10^-3 * exp(-477000 / (8.314 * 2000)) = 9.7×10^-16 m^2/s" }, { "idx": 2813, "question": "The yield strength of Mg alloy is 180MPa, E is 45GPa, find the maximum load that will not cause plastic deformation of a 10mm×2mm Mg plate.", "answer": "F=σA=180×10^6×10×2×10^-6=3600(N)" }, { "idx": 2814, "question": "A nickel-based alloy sample with a cross-section of 10mm×10mm and a length of 40mm was subjected to a tensile test, and the results are shown in the table below. Calculate its yield strength σ0.2.", "answer": "σ0.2 can be obtained from the stress-strain curve of the tensile test, which is 1000 MPa." }, { "idx": 2815, "question": "A Gu–30%Zn brass plate is cold rolled by 25%, reducing its thickness to 1 cm. Subsequently, the plate is further reduced to a thickness of 0.6 cm. Calculate the total cold deformation degree.", "answer": "Cold deformation degree = (A0 - AF) / A0 × 100%, 25% = (h w - 1 w) / (h w) × 100%, h = 4/3 cm. Total deformation degree = (4/3 w - 0.6 w) / (4/3 w) × 100% = 55%." }, { "idx": 2816, "question": "Analyze the effect of polymer molecular weight on the viscous flow temperature", "answer": "When the molecular weight is smaller, the internal frictional resistance between molecular chains is reduced, making the relative movement of molecular chains easier, thus lowering the viscous flow temperature." }, { "idx": 2817, "question": "What are the changes in properties of a Gu–30%Zn brass sheet after cold rolling?", "answer": "After cold rolling, the strength and hardness of the brass sheet increase, while plasticity and toughness decrease, which is the phenomenon of work hardening." }, { "idx": 2818, "question": "There is an aluminum wire with a length of 5 m and a diameter of 3 mm. Given that the elastic modulus of aluminum is 70 GPa, find the total length of the wire under a tensile force of 200 N.", "answer": "Within the elastic range, stress and strain obey Hooke's law ${\\pmb\\sigma}{=}{\\pmb E}{\\pmb\\varepsilon}$, and $\\mathsf{e}=\\frac{\\boldsymbol{\\ell}-\\boldsymbol{l}_{0}}{\\boldsymbol{l}_{0}}{=}\\frac{\\frac{\\boldsymbol{F}}{A}}{E}$. Therefore, $$l=l_{0}+\\frac{F}{E A}l_{\\circ}=l_{0}\\left(1+\\frac{F}{E A}\\right)=5\\left[1+\\frac{200}{70\\times10^{3}\\times\\frac{\\pi}{4}(3\\times10^{-3})^{2}}\\right]$$$$=5.00202(\\mathrm{m})=5002.02(\\mathrm{mm})$$" }, { "idx": 2819, "question": "There is a nickel-based alloy specimen with a cross-section of 10mm×10mm and a length of 40mm. The tensile test results are shown in the table below. Calculate its tensile strength σb.", "answer": "σb = 127600 / (10 × 10 × 10^-6) = 1.276 GPa" }, { "idx": 2820, "question": "It is known that the porosity of sintered $\\\\mathrm{Al}_{2}\\\\mathrm{O}_{3}$ is $5\\\\%$, and its $E=370\\\\mathrm{GPa}$. If another sintered $\\\\mathrm{Al}_{2}\\\\mathrm{O}_{3}$ has $E=270\\\\mathrm{GPa}$, find its porosity.", "answer": "The relationship between the $\\\\pmb{E}$ of ceramic materials and their pore volume fraction $\\\\varphi$ can be expressed by the following equation: $$E=E_{\\\\circ}(1-1.9\\\\varphi+0.9\\\\varphi^{2})$$where $E_{0}$ is the elastic modulus of the material without pores. Substituting the known conditions into the above equation, we obtain $$E_{\\\\circ}={\\\\frac{E}{1-1.9{\\\\varphi}+0.9{\\\\varphi}^{2}}}={\\\\frac{370\\\\times10^{\\\\vartheta}}{1-1.9\\\\times0.05+0.9\\\\times(0.05)^{2}}}=407.8({\\\\mathrm{GPa}})$$$$270\\\\times10^{9}=407.8\\\\times10^{9}(1-1.9\\\\varphi_{1}+0.9\\\\varphi_{1})$$Thus, $\\\\varphi_{1}=19.61\\\\%$" }, { "idx": 2821, "question": "A nickel-based alloy specimen with a cross-section of 10mm×10mm and a length of 40mm was subjected to a tensile test, and the results are shown in the table below. Calculate its elastic modulus E.", "answer": "E = (86200 / (10 × 10 × 10^-6)) / ((40.2 - 40) / 40) = 172.4 GPa" }, { "idx": 2822, "question": "A nickel-based alloy specimen with a cross-section of 10mm×10mm and a length of 40mm was subjected to a tensile test, and the results are shown in the table below. Calculate its elongation δ.", "answer": "δ = (50.2 - 40) / 40 = 25.5%" }, { "idx": 2823, "question": "Analyze the influence of molecular chain flexibility on the viscous flow temperature of polymers", "answer": "From the relationship between chain segments and the energy barrier difference (potential barrier) Lp=l exp{Δε/kT}, it can be seen that the better the molecular chain flexibility, the lower the potential barrier (Δε) for internal rotation, and the shorter the flow unit segments. According to the segmental motion mechanism of polymer flow, flexible molecules require smaller free volume space for flow, thus viscous flow can occur at relatively lower temperatures." }, { "idx": 2824, "question": "Given that the glass transition temperature of polyethylene is $T_{\\\\mathrm{g}}=-68~\\\\mathrm{^\\\\circ C}$, that of polyoxymethylene is $\\\\mathcal{T}_{\\\\varepsilon}=-83^{\\\\circ}\\\\mathrm{C}$, and that of polydimethylsiloxane is $T_{\\\\mathrm{{s}}}=-128^{\\\\circ}\\\\mathrm{{C}}$, analyze the general relationship between the flexibility of polymer chains and their $T_{\\\\mathfrak{s}}$.", "answer": "The repeating unit structure of polyethylene is: —CH2—CH2—; the repeating unit structure of polyoxymethylene is: —$\\\\mathbf{CH_{3}}$—O—, and the repeating unit structure of polydimethylsiloxane is: —Si$\\\\mathbf{CH_{3}}$$\\\\mathbf{CH_{3}}$—O—. Since the internal rotation of the Si—O bond is easier than that of the C—O bond, and the internal rotation of the C—O bond is easier than that of the C—C bond, the easier the internal rotation, the better the flexibility of the molecular chain. It can thus be concluded that as flexibility increases, the temperature $T_{g}$ decreases." }, { "idx": 2825, "question": "An aluminum rod with a length of 20m and a diameter of 14mm is drawn through a die with an aperture of 12.7mm. Calculate the length of the aluminum rod after drawing.", "answer": "During the deformation process, the total volume remains unchanged. Let the length after drawing be L, then π(14.0/2)^2×20×10^3 = π(12.7/2)^2×L×10^3, L=24.3m" }, { "idx": 2826, "question": "A 14mm diameter aluminum rod is drawn through a die with an aperture of 12.7mm. Calculate the cold working rate that this aluminum rod will undergo.", "answer": "The cold working rate is the reduction in cross-sectional area: φ=[π(14.0/2)^2-π(12.7/2)^2]/π(14.0/2)^2=18%" }, { "idx": 2827, "question": "Determine the engineering strain εe and true strain εT when the length increases from L to 1.1L, and explain which one better reflects the true deformation characteristics", "answer": "εe = (1.1 - 1)L / L = 10%; εT = ln(1.1L / L) = 9.5%" }, { "idx": 2828, "question": "Determine the engineering strain εe and true strain εT when compressed from h to 0.9h, and explain which one better reflects the true deformation characteristics", "answer": "εe = (0.9 - 1)h / h = -10%; εT = ln(0.9h / h) = -10.5%" }, { "idx": 2829, "question": "Determine the engineering strain εe and true strain εT when elongating from L to 2L, and explain which one better reflects the true deformation characteristics", "answer": "εe = (2 - 1)L / L = 100%; εT = ln(2L / L) = 69.3%" }, { "idx": 2830, "question": "Determine the engineering strain εe and true strain εT when compressed from h to 0.5h, and explain which one better reflects the true deformation characteristics", "answer": "εe = (0.5 - 1)h / h = -50%; εT = ln(0.5h / h) = -69.3%" }, { "idx": 2831, "question": "Compare the differences between engineering strain and true strain under tension and compression, and explain which one better reflects the true deformation characteristics", "answer": "εT ≠ εe, the larger the deformation, the greater the difference between εT and εe. Comparing the cases of elongation to 2L and compression to 0.5h, the absolute values of true strain are equal, while the absolute values of engineering strain are not equal. Therefore, true strain better reflects the true deformation characteristics." }, { "idx": 2832, "question": "For a pre-annealed metallic polycrystal, the true stress-strain curve in the plastic region can be approximately expressed as ∇σ_T= kε_T^n, where k and n are empirical constants; known as the strength coefficient and strain hardening exponent, respectively. Derive the mathematical relationship between the strain hardening exponent n and the strain hardening rate (θ=dσ_T/dε_T).", "answer": "θ=dσ_T/dε_T=n kε_T^(n-1). Substituting σ_T=kε_T^n, we obtain θ=nσ_T/ε_T." }, { "idx": 2833, "question": "For a pre-annealed metal polycrystal, the true stress-strain curve in the plastic region can be approximately expressed as ∇σ_T= kε_T^n, where k and n are empirical constants, referred to as the strength coefficient and strain hardening exponent, respectively. If there are two materials, A and B, with roughly equal k values, but n_A=0.5 and n_B=0.2, which material has higher hardening capability and why?", "answer": "For σ_T=kε_T^n, dσ_T=n kε_T^(n-1)dε_T, so dσ_T/dε_T=n kε_T^(n-1). When ε_T<1, if 0. Along which direction of tension (or compression) will slip not occur?", "answer": "Based on the characteristics of the slip systems in the magnesium oxide structure, slip will not occur only when tension (or compression) is applied along a direction perpendicular to all (110) planes. From the standard projection diagram of the cubic crystal system (001), it can be seen that there is no pole that is $90^\\mathfrak{o}$ away from all (110) poles. Therefore, for magnesium oxide, there is no direction of tension (or compression) that will not cause slip." }, { "idx": 2838, "question": "A cross-slip system consists of one slip direction and two crystallographic planes containing this slip direction, such as the (101) [111] (110) in bcc crystals. Write down three other cross-slip systems of the same type for bcc crystals.", "answer": "From the standard projection diagram of the cubic crystal system (001), it can be found that the other three cross-slip systems of the same type for bcc crystals are: (101) [111] (110), (011) [111] (110), (110) [111] (101)." }, { "idx": 2839, "question": "For a bcc crystal with a critical resolved shear stress of 60MPa on the (110)[111] slip system, how much stress must be applied in the [001] direction to initiate slip?", "answer": "Vector dot product:\\n[001] direction:\\na⋅b=|a|⋅|b|cosθ⇒cosθ=(a⋅b)/(|a|⋅|b|)=(a1b1+a2b2+a3b3)/(√(a1²+a2²+a3²)⋅√(b1²+b2²+b3²))\\ncosλ=1/(1×√3)=1/√3,\\ncosφ=0/(1×√2)=0\\nσ=τt/(cosλcosφ)=60/((1/√3)×0)=∞,\\nTherefore, slip cannot be initiated in this direction no matter how much stress is applied." }, { "idx": 2840, "question": "For a bcc crystal with a critical resolved shear stress of 60MPa on the (110)[111] slip system, how much stress must be applied in the [010] direction to initiate slip?", "answer": "[010] direction: cosλ=1/(1×√3)=1/√3, cosφ=|−1/(1×√2)|=1/√2 σ=τc/(cosλcosφ)=60/((1/√3)×(1/√2))=146.97(MPa)" }, { "idx": 2841, "question": "For a Cu single crystal, the stress σ corresponding to 1% plastic deformation is 40 MPa. From the standard projection diagram of the cubic crystal system (001), it is found that when the tensile axis is [111], the activated slip systems are (111)[011] and another 5 equivalent slip systems. The orientation factor for activating any one of these slip systems can be calculated as cosφcosλ=1/3×2/√6=2/3√6. Determine the dislocation density after 1% plastic deformation. Given τ0=700 kPa, G=42×10^3 MPa, b=0.256 nm, α=0.4.", "answer": "τ=σcosφcosλ=40×2/3√6=10.80 MPa. From τ=τ0+αGb√ρ, the dislocation density can be calculated as ρ=((τ−τ0)/αGb)^2=((10.89×10^3−700)/(0.4×42×10^6×0.256×10^−9))^2=5.61×10^8 cm−2" }, { "idx": 2842, "question": "The critical resolved shear stress of a single crystal at room temperature is $\\\\tau_{\\\\mathfrak{c}}=7.9\\\\times10^{5}~\\\\mathrm{Pa}$. If a tensile test is performed on an aluminum single crystal sample at room temperature with the tensile axis in the [123] direction, calculate the stress required to cause the sample to yield.", "answer": "Al has an fcc crystal structure, and its slip system is $\\\\{111\\\\}\\\\langle110\\\\rangle$. When the external force axis is [123], according to the standard projection diagram of the cubic crystal system, the first activated slip system is (111)[101]. Therefore, $\\\\phi$ is the angle between [123] and the normal [111] of the (111) crystal plane, and $\\\\lambda$ is the angle between [123] and [101]. Thus, $$\\\\cos\\\\phi{=}\\\\frac{-1+2+3}{\\\\sqrt{14}\\\\times\\\\sqrt{3}}=\\\\frac{4}{\\\\sqrt{42}}$$ $$\\\\cos\\\\lambda{=}\\\\frac{1+0+3}{\\\\sqrt{14}\\\\times\\\\sqrt{2}}=\\\\frac{2}{\\\\sqrt{7}}$$ $$\\\\sigma_{\\\\mathrm{{s}}}=\\\\frac{\\\\tau_{C}}{\\\\cos{\\\\phi}\\\\cos\\\\lambda}=\\\\frac{7.9\\\\times10^{5}}{\\\\frac{4}{\\\\sqrt{42}}\\\\times\\\\frac{2}{\\\\sqrt{7}}}=1.69({\\\\mathrm{MP}}a)$$" }, { "idx": 2843, "question": "For a bcc crystal with a critical resolved shear stress of ${60}\\\\mathbf{M}\\\\mathbf{Pa}$ on the (110)[111] slip system, how much stress must be applied in the [001] and [010] directions to initiate slip?", "answer": "$A C$ and $A^{\\\\prime}C^{\\\\prime}$ are the distances between two adjacent slip planes in the crystal before and after stretching, respectively. Since the distance between slip planes remains unchanged before and after stretching, i.e., $A C{=}A^{\\\\prime}C^{\\\\prime}$, we have \\n$$\\n\\\\epsilon={\\\\frac{A^{\\\\prime}B^{\\\\prime}-A B}{A B}}={\\\\frac{{\\\\frac{A^{\\\\prime}C^{\\\\prime}}{\\\\sin30^{\\\\circ}}}-{\\\\frac{A C}{\\\\sin45^{\\\\circ}}}}{\\\\frac{A C}{\\\\sin45^{\\\\circ}}}}\\n$$\\n\\n$$\\n={\\\\frac{2-{\\\\sqrt{2}}}{\\\\sqrt{2}}}=41.4\\\\%\\n$$" }, { "idx": 2844, "question": "When a single crystal sample of Mg is subjected to tensile testing, the three slip directions form angles of $38^{\\\\circ}, 45^{\\\\circ}, 85^{\\\\circ}$ with the tensile axis, respectively, while the normal to the basal plane forms an angle of $60^{\\\\circ}$ with the tensile axis. If plastic deformation is first observed at a tensile stress of $2.05\\\\,\\\\mathrm{MPa}$, what is the critical resolved shear stress of $\\\\mathrm{Mg}$?", "answer": "The slip plane of $\\\\mathrm{Mg}$ is the (0001) plane (basal plane). According to the resolved shear stress formula $\\\\tau = \\\\sigma \\\\cos\\\\lambda \\\\cos\\\\phi$, when $\\\\phi$ is fixed at $60^{\\\\circ}$, the smaller $\\\\lambda$ is, the larger $\\\\tau$ becomes. Therefore, under tensile stress, the crystal slips along the slip direction that forms a $38^{\\\\circ}$ angle with the tensile axis, resulting in plastic deformation. Thus, the critical resolved shear stress of $\\\\mathrm{Mg}$ is $$\\\\tau_{\\\\mathrm{c}} = \\\\sigma_{\\\\mathrm{s}} \\\\cos\\\\lambda \\\\cos\\\\phi = 2.05 \\\\times \\\\cos60^{\\\\circ} \\\\times \\\\cos38^{\\\\circ}$$ $$= 2.05 \\\\times 0.5 \\\\times 0.788 = 0.8077\\\\,(\\\\mathrm{MPa})$$." }, { "idx": 2845, "question": "For FCC and BCC metals during plastic deformation, the relationship between flow stress and dislocation density ρ is given by τ=τ0+αGb√ρ, where τ0 is the stress required for dislocation motion without interference from other dislocations, i.e., the shear stress without work hardening, G is the shear modulus, b is the Burgers vector of the dislocation, and α is a material-dependent constant, α=0.3∼0.5. If a Cu single crystal has τ0=700 kPa, initial dislocation density ρ0=10^5 cm−2, G=42×10^3 MPa, b=0.256 nm, α=0.4, what is the critical resolved shear stress?", "answer": "τ=τ0+αGb√ρ0=700+0.4×42×10^6×0.256×10^−9×√10^9=836 kPa" }, { "idx": 2846, "question": "Indicate the easy slip plane and easy slip direction of Cu crystal, and calculate the slip plane spacing, atomic spacing in the slip direction, and lattice resistance. (Given G_Cu=48.3GPa, ν=0.3)", "answer": "Cu has an fcc structure, with the easy slip plane being {111} and the easy slip direction being <110>. The slip plane spacing d_(111)=a/√3, and the atomic spacing in the slip direction b=√2/2a. The lattice resistance τ_PN=(2×48.3×10^9)/(1-0.3)×exp[-2π(a/√3)/((1-0.3)(√2/2)a)]=90.45MPa." }, { "idx": 2847, "question": "Calculate the volume fraction φ_Fe3C of the Fe3C phase in 40 steel, given the mass fraction of carbon w_c=0.004.", "answer": "φ_Fe3C = 0.004 / 0.0667 = 0.06" }, { "idx": 2848, "question": "Indicate the easy slip plane and easy slip direction of α-Fe crystal, and calculate the slip plane spacing, atomic spacing in the slip direction, and lattice resistance. (Given G_α-Fe=81.6GPa, ν=0.3)", "answer": "α-Fe has a bcc structure, with the slip plane being {110} and the easy slip direction being <111>. The slip plane spacing d_(110)=a/√2, and the atomic spacing in the slip direction b=√3/2a. The lattice resistance τ_PN=(2×81.6×10^9)/(1-0.3)×exp[-2π(a/√2)/((1-0.3)(√3/2)a)]=152.8MPa." }, { "idx": 2849, "question": "Calculate the shear strength τ of 40 steel, given that the shear modulus of Fe is G=7.9×10^4 MPa, the lattice constant of α-Fe is a=0.28 nm, and the average spacing of Fe3C particles is λ=41.2 μm.", "answer": "τ = (G * b) / λ = (G * (√3/2) * a) / λ = (7.9×10^4 * (√3/2) * 0.28×10^-3) / 41.2 = 0.465 MPa" }, { "idx": 2850, "question": "A single crystal of A1 is made into a tensile specimen (with a cross-sectional area of 9 mm²) for room-temperature tensile testing. The tensile axis forms an angle of $36.7^{\\\\circ}$ with [001], $19.1^{\\\\circ}$ with [011], and $22.2^{\\\\circ}$ with [111]. The load at the onset of yielding is 20.4 N. Determine the resolved shear stress on the primary slip system.", "answer": "From the known tensile axis direction, the primary slip system can be determined as (111)[101] using the standard projection diagram of the cubic crystal system (001). Let the stress axis direction be $[u\\\\tau\\\\mathbf{w}]$. From the given conditions, we have $$\\\\cos36.7^{\\\\circ}=\\\\frac{w}{\\\\sqrt{u^{2}+v^{2}+w^{2}}}$$$$\\\\cos19.1^{\\\\circ}=\\\\frac{v+w}{\\\\sqrt{2}\\\\sqrt{u^{2}+v^{2}+w^{2}}}$$$$\\\\cos22.2^{\\\\circ}={\\\\frac{u+v+w}{{\\\\sqrt{3}}{\\\\sqrt{{u}^{2}+{v}^{2}+{w}^{2}}}}}$$Let $u^{2}+v^{2}+w^{2}=1$, then solving gives $\\\\scriptstyle\\\\pmb{u}=0.26$, $v{=}0.54$, ${\\\\mathfrak{w}}{=}0.80$. Thus, $\\\\cos\\\\lambda={\\\\frac{0.26+0.80}{\\\\sqrt{2}}}=0.75$ $$\\\\cos\\\\phi=\\\\frac{-0.26+0.54+0.80}{\\\\sqrt{3}}=0.62$$$$\\\\tau=\\\\sigma\\\\cos\\\\lambda\\\\cos\\\\phi=\\\\frac{20.4}{9\\\\times10^{-6}}\\\\times0.75\\\\times0.62=1.01(\\\\mathrm{MPa})$$" }, { "idx": 2851, "question": "Calculate the average spacing λ of Fe3C particles, given the number of Fe3C particles per unit volume N_v=1.43×10^13 1/m^3.", "answer": "λ = (1 / N_v)^(1/3) = (1 / 1.43×10^13)^(1/3) = 4.12×10^-5 m = 41.2 μm" }, { "idx": 2852, "question": "Given that a straight dislocation line in an alloy is hindered by second-phase particles with a spacing of $\\\\pmb{\\\\lambda}$ during its motion, prove that the shear stress required for the dislocation to continue moving via the bypass mechanism is: 2T-Bln(2), where T is the line tension; b is the Burgers vector; $_{!G}$ is the shear modulus; $\\\\pmb{\\\\gamma}_{0}$ is the radius of the second-phase particle; and $B$ is a constant.", "answer": "Strengthening effect of non-deformable particles: When a moving dislocation encounters non-deformable particles, it will be blocked, causing the dislocation line to bend around them. Since the dislocation has line tension $\\\\pmb{T}$, bending the dislocation line requires overcoming the effect of its line tension. The shear stress required for the dislocation line to bypass particles with spacing $\\\\lambda$ is The line tension of a dislocation is analogous to the surface tension of a liquid and can be represented by the energy per unit length of the dislocation. The energy per unit length of the dislocation $T=E=\\\\frac{G b^{2}}{4\\\\pi k}\\\\ln\\\\frac{\\\\bar{R}}{r_{0}}$ Gn. Substituting this into the above equation, then" }, { "idx": 2853, "question": "For an Al2O3 specimen with a circular cross-section in a three-point bending test, the cross-sectional radius r=3.5mm, the span between the two supports is 50mm, and it fractures under a load of 950N. Calculate the fracture strength σfs of the material.", "answer": "σfs = (Ff * L) / (π * r^3) = (950 * 50 * 10^-3) / (π * (3.5 * 10^-3)^3) = 352.6 MPa" }, { "idx": 2854, "question": "For an Al2O3 specimen with a square cross-section of side length 12mm in a three-point bending test, the distance between the two supports is 40mm. Given the material's fracture strength σfs=352.6 MPa, find the load at fracture Ff.", "answer": "Ff = (2 * σfs * b^3) / (3 * L) = (2 * 352.6 * 10^6 * (12 * 10^-3)^3) / (3 * 40 * 10^-3) = 10154.9 N" }, { "idx": 2855, "question": "Calculate the number of Fe3C particles per unit volume N_v, given the volume fraction of Fe3C phase φ_Fe3C=0.06 and the radius of spherical cementite particles r=10×10^-6 m.", "answer": "N_v = φ_Fe3C / (4/3 * π * r^3) = 0.06 / (4/3 * π * (10×10^-6)^3) ≈ 1.43×10^13 (1/m^3)" }, { "idx": 2856, "question": "For many polymer materials, their tensile strength σi is a function of the number-average relative molecular mass Mn̅: the formula is given by σi = σ0 - A / Mn̅, where σ0 is the tensile strength at infinite molecular weight, and A is a constant. Given two types of poly(methyl methacrylate) with number-average relative molecular masses of 4×10^4 and 6×10^4, the corresponding tensile strengths are 107 MPa and 170 MPa, respectively. Determine the tensile strength σb when the number-average relative molecular mass is 3×10^4.", "answer": "The equations are: 107 = σ0 - A / (4×10^4), 170 = σ0 - A / (6×10^4). Solving these gives σ0 = 296 MPa and A = 7.56×10^6. Therefore, σb = σ0 - A / Mn̅ = 296 - 7.56×10^6 / (3×10^4) = 44 MPa." }, { "idx": 2857, "question": "The recovery activation energy of iron is $88.9\\\\mathrm{kJ/mol}$. If cold-deformed iron is subjected to recovery treatment at $400^{\\\\circ}\\\\mathrm{C}$ to retain 60% of its work hardening, it takes $160\\\\mathfrak{min}$. How much time is needed to achieve the same effect with recovery treatment at $450\\\\mathrm{~C~}$?", "answer": "$$\\\\frac{t_{1}}{t_{2}}=\\\\mathrm{e}^{-{\\\\frac{Q}{R}(\\\\frac{1}{T_{2}}-\\\\frac{1}{T_{1}})}} , t_{2}={\\\\frac{t_{1}}{\\\\mathrm{e}^{-{\\\\frac{\\\\mathrm{{Q}}}{\\\\scriptscriptstyle{R}}}\\\\big({\\\\frac{1}{T_{2}}}-{\\\\frac{1}{T_{1}}}\\\\big)}}}}={\\\\frac{160}{\\\\mathrm{e}^{-{\\\\frac{80.5}{831}}\\\\big({\\\\frac{1}{723}}-{\\\\frac{1}{673}}\\\\big)}}}=59(\\\\operatorname*{min})$$" }, { "idx": 2858, "question": "Briefly describe the characteristics of plastic deformation in ceramic materials (crystalline).", "answer": "In general, compared to metallic materials and high molecular materials, ceramic materials appear hard and brittle, which is determined by the type of bonding between their atoms. The atoms in ceramic materials are usually bonded by ionic or covalent bonds. In covalently bonded ceramics, atoms are bonded through shared electron pairs, exhibiting directionality and saturation, with relatively high bond energy. During plastic deformation, the movement of dislocations inevitably disrupts the covalent bonds between atoms, resulting in significant lattice resistance (Peierls-Nabarro force). Therefore, covalently bonded ceramics exhibit hard and brittle characteristics. For ionically bonded ceramic materials, there are two scenarios: single crystals (such as NaCl, FeO, etc.) can undergo considerable plastic deformation under compressive stress at room temperature. However, polycrystalline ionically bonded ceramics are often brittle and prone to crack formation at grain boundaries. This is because ionic crystals require alternating positive and negative ion arrangements. Under external forces, when a dislocation moves by one atomic spacing, the strong Coulombic repulsion between like-charged ions makes dislocation movement along directions perpendicular or parallel to the ionic bonds extremely difficult. However, if the dislocation moves along a $45^{\\circ}$ direction rather than horizontally, adjacent crystal planes remain mutually attracted by Coulombic forces during the slip process, resulting in relatively good plasticity. Nevertheless, during the deformation of polycrystalline ceramics, adjacent grains must deform in a coordinated and mutually constrained manner. Due to the limited number of slip systems in ceramics, this is difficult to achieve, leading to cracking at grain boundaries and ultimately brittle fracture. On the other hand, during the heating and cooling processes of sintered ceramic materials, the presence of thermal stresses often leads to the formation of microcracks. Additionally, factors such as corrosion can cause surface cracks. Therefore, inherent cracks are always present to some extent in ceramic materials. Under external forces, severe stress concentration occurs at crack tips. According to elastic mechanics estimates, the maximum stress at a crack tip can reach the theoretical fracture strength. Moreover, since ceramic crystals have few mobile dislocations and dislocation movement is difficult, brittle fracture often occurs once the yield strength is reached. Of course, under tensile or compressive conditions, the mechanical properties of ceramic materials also differ significantly, with compressive strength typically being higher than tensile strength." }, { "idx": 2859, "question": "The tensile strength of brittle materials can be expressed by the following formula: \\n\\n$$\\n\\\\sigma_{\\\\mathrm{m}}=2\\\\sigma_{\\\\mathrm{0}}\\\\bigg(\\\\frac{\\\\ell}{r}\\\\bigg)^{\\\\frac{1}{2}}\\n$$\\n\\nWhere, $\\\\sigma_{0}$ is the nominal applied tensile stress; $l$ is the length of the surface crack or half the length of an internal crack; $r$ is the radius of curvature at the crack tip; and $\\\\sigma_{\\\\infty}$ is actually the maximum stress caused by stress concentration at the crack tip. Now assume that the critical length of the surface crack in $\\\\mathrm{Al}_{2}\\\\mathrm{O}_{3}$ ceramic is $l = 2\\\\times10^{-3}\\\\mathrm{mm}$, its theoretical fracture strength is $\\\\frac{E}{10}$, where $E$ is the elastic modulus of the material and is $393\\\\mathrm{GPa}$. Calculate: When a tensile stress of $275\\\\mathrm{MPa}$ is applied to the $\\\\mathrm{Al}_{2}\\\\mathrm{O}_{3}$ ceramic specimen, what is the critical radius of curvature $r_{\\\\mathfrak{c}}$ at the crack tip that causes fracture?", "answer": "When the tensile stress reaches the fracture strength of the material ($\\\\frac{E}{10}$), $\\\\mathrm{Al}_{2}\\\\mathrm{O}_{3}$ fractures, therefore:\\n\\n$$\\n\\\\frac{E}{10}=2\\\\sigma_{\\\\circ}\\\\left(\\\\frac{\\\\lambda}{r}\\\\right)^{\\\\frac{1}{2}}\\\\Rightarrow r=\\\\frac{400\\\\ l\\\\sigma_{0}^{2}}{E^{2}}\\n$$\\n\\n$$\\nr_{\\\\mathrm{c}}={\\\\frac{400l\\\\sigma_{0}^{2}}{E^{2}}}={\\\\frac{400\\\\times2\\\\times10^{-3}\\\\times(275)^{2}}{(393\\\\times10^{3})^{2}}}=3.9\\\\times10^{-7}(\\\\mathrm{mm})=0.39(\\\\mathrm{nm})\\n$$" }, { "idx": 2860, "question": "Given that the yield strengths of pure iron with average grain diameters of $\\\\boldsymbol{1}\\\\mathfrak{mm}$ and $0.0625\\\\mathbf{mm}$ are 112.7MPa and 196$\\\\mathbf{MPa}$ respectively, what is the yield strength of pure iron with an average grain diameter of $0.0196\\\\mathrm{mm}$?", "answer": "$$\\\\tau={\\\\frac{G b}{\\\\lambda}}={\\\\frac{G{\\\\frac{\\\\sqrt{3}}{2}}a}{\\\\lambda}}={\\\\frac{7.9\\\\times10^{19}\\\\times{\\\\frac{\\\\sqrt{3}}{2}}\\\\times0.28\\\\times10^{-9}}{41.2\\\\times10^{-6}}}=0.465(\\\\mathrm{MPa})$$\\n\\n$$\\\\left\\\\{\\\\begin{array}{l}{{112,7=\\\\sigma_{\\\\circ}+k(1\\\\times10^{\\\\circ})^{-\\\\frac{1}{2}},}}\\\\ {{}}\\\\ {{196=\\\\sigma_{\\\\circ}+k(0.0625\\\\times10^{\\\\circ})^{-\\\\frac{1}{2}}}}\\\\end{array}\\\\right.,$$\\n\\nThe solution is\\n\\n$$\\\\begin{array}{l}{{\\\\pmb{\\\\mathscr{s}}}_{\\\\mathscr{0}}=84,935(\\\\mathrm{MPa})}\\\\ {\\\\backslash_{\\\\pmb{\\\\mathscr{k}}}=0.878}\\\\end{array}$$\\n\\nTherefore, $\\\\sigma_{\\\\ast}=84.935+0.878(0.0196\\\\times10^{-3})^{-\\\\frac{2}{2}}=283.255(\\\\mathrm{MPa})$" }, { "idx": 2861, "question": "Explain the phenomenon where the cross-sectional area of the neck remains essentially unchanged during uniaxial stretching of polymers.", "answer": "Many polymers often exhibit instability in uniform deformation during plastic deformation. For example, when a polymer sample is subjected to a uniaxial tensile test, the stress initially increases linearly with strain, and the specimen is uniformly elongated. After the yield point, the strain at a certain part of the specimen suddenly increases faster than the overall strain, causing the originally uniform cross-section to become non-uniform, forming one or several necks. As deformation continues, the necking zone expands and extends along the length of the specimen until the entire specimen's cross-section becomes uniformly thinner. During this deformation process, the stress remains almost constant. This is because, after exceeding the yield strength, the specimen undergoes plastic deformation, and work hardening occurs at the necking region. XRD analysis proves that, whether in the amorphous or crystalline state, the macromolecules in the polymer gradually align along the direction of the external force as deformation increases. Due to the directionality of bonds (mainly covalent bonds), strain hardening occurs after alignment." }, { "idx": 2862, "question": "Explain how pre-deformation degree, original grain size, and metal purity affect the recrystallization temperature.", "answer": "From the above analysis, it can be seen that increasing the pre-deformation degree and refining the original grain size will cause the $\\boldsymbol{T}_{\\mathtt{R}}$ to decrease. The influence of impurities on $T_{\\mathbf{R}}$ is dual. If the presence of impurities increases the distortion energy and this factor dominates, then metals with lower purity will have a lower $T_{\\tt R}$. Conversely, if the presence of impurities slows down interface migration and this factor dominates, then metals with higher purity will have a lower $T_{\\tt R}$. Different impurity atoms have different effects on $T_{\\mathbf{R}}$. Generally, the presence of a small amount of impurity atoms will hinder the recrystallization of the metal, thereby causing $T_{8}$ to rise, and the extent of the increase varies depending on the type of impurity." }, { "idx": 2863, "question": "Given that the yield strength of industrial pure copper is ${\\\\pmb{\\\\sigma{s}}}=70~{\\\\bf M}{\\\\bf P}{\\\\bf a}$, its grain size is $N{\\\\mathrm{A}}=18$ per $'\\\\mathbf{m}\\\\mathbf{m}^{2}$, and when $N{\\\\mathbf{A}}=$ $\\\\bf{4025}$ per $\\\\scriptstyle{\\\\prime}{\\\\mathbf{m}\\\\mathbf{m}^{2}}$, ${\\\\pmb{\\\\sigma}}{\\\\mathsf{S}}={\\\\pmb{95}}\\\\mathbf{M}\\\\mathbf{P}{\\\\bf a}$. Calculate ${\\\\pmb\\\\sigma_{\\\\mathfrak{s}}}$ when $N{\\\\mathbf{A}}=260$ per $\\\\scriptstyle{\\\\left/{\\\\mathfrak{m m}}^{2}\\\\right.}$.", "answer": "Let the average grain diameter be $^{d}$, and the number of grains per $\\\\scriptstyle\\\\mathtt{m m}^{2}$ be $N_{\\\\Lambda}$. It can be proven that: Thus $$d=\\\\sqrt{\\\\frac{8}{3\\\\pi N_{\\\\Lambda}}}$$$$d_{1}=\\\\sqrt{\\\\frac{8}{3\\\\pi\\\\times\\\\mathrm{i}8}}=0.217(\\\\mathrm{mm})$$$$d_{2}=\\\\sqrt{\\\\frac{8}{3\\\\pi\\\\times4025}}=1.452\\\\times10^{-2}(\\\\mathrm{mm})$$$$d_{3}=\\\\sqrt{\\\\frac{8}{3\\\\pi\\\\times260}}=5.714\\\\times10^{-2}(\\\\mathrm{mm})$$Substituting into the Hall-Petch formula: Solving $$\\\\left\\\\{\\\\begin{array}{l l}{{70=\\\\sigma_{0}+k(0,217\\\\times10^{-3})^{-\\\\frac{1}{2}}}}\\\\ {{95=\\\\sigma_{0}+k(1,452\\\\times10^{-5})^{-\\\\frac{1}{2}}}}\\\\end{array}\\\\right.$$$$\\\\begin{array}{r}{\\\\sigma_{0}=61,3\\\\mathrm{MPa},\\\\quad k=0,1285}\\\\end{array}$$$$\\\\sigma_{5}=61,3+0.1285\\\\times(5,714\\\\times10^{-5})^{-\\\\frac{1}{2}}=78.3(\\\\mathrm{MPa})$$" }, { "idx": 2864, "question": "There is currently a $\\\\phi{\\\\ell\\\\mathrm{mm}}$ aluminum wire that needs to be finally processed into $\\\\phi{\\\\mathrm{0.5mm}}$ aluminum material. However, to ensure product quality, the cold working amount of this aluminum material cannot exceed $85\\\\%$. How to formulate a reasonable processing procedure?", "answer": "Cold working amount $=\\\\frac{\\\\Delta A}{A}=\\\\frac{A_{0}-A_{1}}{A_{0}}=\\\\frac{\\\\frac{\\\\pi}{4}d_{0}^{2}-\\\\frac{\\\\pi}{4}\\\\times d_{1}^{2}}{\\\\frac{\\\\pi}{4}\\\\times d_{0}^{2}}=1-\\\\left(\\\\frac{d_{1}}{d_{0}}\\\\right)^{2}=85\\\\%$\\n$$\\nd_{1}=\\\\sqrt{1-0.85}\\\\times6=2.324(\\\\mathrm{mm}),\\\\quad d_{2}=\\\\sqrt{0.15}\\\\times2.324=0.9(\\\\mathrm{mm}),\\n$$\\n\\n$$\\nd_{3}=\\\\sqrt{0.15}\\\\times0.9=0.348(\\\\mathrm{mm})\\n$$\\n\\nTherefore, the $\\\\phi{\\\\mathrm{6mm}}$ aluminum wire can first be cold drawn to $\\\\phi{\\\\mathrm{2.324mm}}$, followed by recrystallization annealing to eliminate work hardening. Then, it is cold drawn to $\\\\phi_{\\\\mathrm{0.9mm}}$, followed by another recrystallization annealing, and finally cold drawn to $\\\\phi_{\\\\mathrm{0.5mm}}$." }, { "idx": 2865, "question": "After cold working, the dislocation density of $\\\\mathbf{A}\\\\mathbf{g}$ is $10^{12}/\\\\mathfrak{c m}^{2}$. Assuming that the recrystallization nucleus moves from a high-angle grain boundary into the deformed matrix, find the minimum radius of curvature for the bulging grain boundary $\\\\scriptstyle\\\\left(\\\\mathbf{A}\\\\mathbf{g}_{:}G=30\\\\mathrm{GPa},b=0.3\\\\mathrm{nm},\\\\gamma=0.4\\\\mathrm{J}/\\\\mathrm{m}^{2}\\\\right)$.", "answer": "The driving force $F$ for the movement of the $\\\\mathbf{Ag}$ recrystallization nucleus from a high-angle grain boundary into the deformed matrix is the stored energy from cold working, $F= G b^{2}\\\\left(\\\\rho_{1}-\\\\rho_{0}\\\\right)$. Since $\\\\rho_{1}\\\\gg\\\\rho_{0}$, $\\\\scriptstyle{F\\\\approx G b^{2}\\\\rho_{1}}$. The bulged grain boundary is subjected to a force $f$ directed toward its center of curvature. When the radius of curvature of the bulge is $R$, $f= \\\\frac{2\\\\gamma}{R},f$ $\\\\scriptstyle{\\\\dot{F}}$ $R_{\\\\mathrm{min}}$ $f$ $F$ $$F{=}f,$$$$G\\\\delta^{2}\\\\rho_{\\\\mathrm{f}}=\\\\frac{2\\\\gamma}{R_{\\\\mathrm{rein}}}$$$$R_{\\\\mathrm{aia}}={\\\\frac{2\\\\gamma}{G b^{2}\\\\rho_{1}}}={\\\\frac{2\\\\times0.4}{30\\\\times10^{8}\\\\times(3\\\\times10^{-10})^{2}\\\\times10^{16}}}=2.9\\\\times10^{-8}({\\\\mathrm{m}})=29({\\\\mathrm{nm}})$$" }, { "idx": 2866, "question": "A factory used a cold-drawn steel wire rope to lift a large steel component into a heat treatment furnace. Due to a momentary oversight, the wire rope was not removed but was heated together with the component to $860~\\\\mathrm{{^\\\\circC}}$. When the holding time was reached and the furnace door was opened to lift out the component, the wire rope broke. Analyze the cause.", "answer": "The cold-drawn steel wire rope is made by twisting cold-drawn steel wires with large deformation. The cold work hardening during the processing significantly improves the strength and hardness of the steel wires, enabling them to bear heavy loads. However, when heated to ${\\\\tt860^{\\\\circ}C}$, the temperature far exceeds the recrystallization temperature of the wire rope, leading to recovery and recrystallization phenomena. The effect of work hardening completely disappears, and the strength and hardness are greatly reduced. When used for lifting again, once the load exceeds its bearing capacity, it inevitably leads to the fracture of the wire rope." }, { "idx": 2867, "question": "When moving dislocations are pinned, their average spacing is $\\scriptstyle{\\ell=\\rho^{-{\\frac{1}{2}}}(\\rho}$ is the dislocation density). It is given that a $\\mathtt{Cu}$ single crystal has been strain-hardened to the extent that the resolved shear stress acting on the crystal is 14MPa. Given $G=40\\mathrm{GPa},b=0,256$ $\\scriptstyle{\\mathtt{n m}}$, calculate the dislocation density of the $\\mathtt{Cu}$ single crystal.", "answer": "After moving dislocations are pinned, a dislocation segment of length $\\boldsymbol{\\ell}$ can act as a dislocation source, and the resolved shear stress required to activate this dislocation source is $$\\tau_{\\mathrm{c}}=\\frac{G b}{l},$$$$14\\times10^{6}=\\frac{40\\times10^{9}\\times0.256\\times10^{-9}}{\\rho^{-\\frac{1}{2}}},$$$$\\rho{=1.869\\times10^{12}(\\mathrm{m}^{-2})}$$" }, { "idx": 2868, "question": "Based on the equation $t_{0.95}=\\\\left[\\\\frac{2.85}{\\\\dot{N}G^{3}}\\\\right]^{\\\\frac{1}{4}}$, derive the functional relationship between the recrystallization temperature $T_{R}$ and $G_{0},N_{0},Q_{g}$ and $Q_{n}$.", "answer": "$t_{0.95}=\\\\Big[\\\\frac{2.85}{\\\\tilde{N}\\\\bar{G}^{3}}\\\\Big]^{\\\\frac{1}{4}}=\\\\Big[\\\\frac{2.85}{N_{0}G_{0}^{3}}\\\\Big]^{\\\\frac{1}{4}}\\\\:\\\\mathrm{exp}\\\\Big(\\\\frac{Q_{n}+3Q_{\\\\ell}}{4k T}\\\\Big)$. Substituting $T=T_{\\\\parallel}\\\\circ\\\\mathcal{t}_{0,85}=1$, we get $$\\\\exp\\\\Big(\\\\frac{Q_{n}+3Q_{s}}{4k T_{\\\\mathrm{R}}}\\\\Big)=\\\\Big[\\\\frac{2,85}{N_{0}G_{0}^{3}}\\\\Big]^{-\\\\frac{1}{4}},\\\\frac{Q_{n}+3Q_{\\\\mathrm{g}}}{4k T_{\\\\mathrm{R}}}=-\\\\frac{1}{4}\\\\ln\\\\frac{2.85}{N_{0}G_{0}^{3}},$$ Therefore, $T_{\\\\mathbb{R}}=-\\\\frac{Q_{n}+3Q_{g}}{k\\\\ln\\\\Bigl(\\\\frac{2.85}{N_{0}G_{0}^{3}}\\\\Bigr)}$" }, { "idx": 2869, "question": "Given that the Tm of Fe is 1538°C, estimate the minimum recrystallization temperature of Fe.", "answer": "According to the empirical formula, the recrystallization temperature Tr≈0.4Tm. Therefore, the minimum recrystallization temperature of Fe is Tr=0.4×(1538+273)=724.4(K)=451.4(°C). In production, to improve efficiency, the actual recrystallization annealing temperature in factories is usually selected as Tr+(100~200)(°C)." }, { "idx": 2870, "question": "Given that the Tm of Cu is 1083°C, estimate the minimum recrystallization temperature of Cu.", "answer": "According to the empirical formula, the recrystallization temperature Tr≈0.4Tm. Therefore, the minimum recrystallization temperature of Cu is Tr=0.4×(1083+273)=542.4(K)=269.4(°C). In production, to improve efficiency, the actual recrystallization annealing temperature in factories is usually set at Tr+(100~200)(°C)." }, { "idx": 2871, "question": "Given a $1~\\\\mathrm{cm}^{3}$ brass sample annealed at $700\\\\textcircled{1}$ with an initial grain diameter of $2.16\\\\times10^{-3}~\\\\mathrm{cm}$, the grain boundary energy of brass is $0.5J/\\\\mathfrak{m}^{2}$. A calorimeter measured a total heat release of 0.035J after holding for $2\\\\mathtt{h}$. Determine the grain size after holding for $2\\\\mathtt{h}$.", "answer": "Assuming the heat released after holding for $2\\\\textrm{h}$ is due to grain growth, where the reduction in total grain boundary area releases energy. From quantitative metallography, the relationship between the interfacial area per unit volume $S_{V}$ and the grain diameter $d$ on the cross-section is $S_{v}={\\\\frac{2}{d}}$. Therefore, $$\\\\scriptstyle Q={\\\\Big(}{\\\\frac{2}{d_{1}}}-{\\\\frac{2}{d_{2}}}{\\\\Big)}\\\\gamma,\\\\quad{\\\\frac{1}{d_{2}}}={\\\\frac{1}{d_{1}}}-{\\\\frac{Q}{2\\\\gamma}}$$Substituting the corresponding data, we obtain $$d_{z}=8.9\\\\times10^{3}(\\\\mathrm{cm)}$$" }, { "idx": 2872, "question": "Industrial pure aluminum was rolled into strips with large deformation at room temperature, and the measured room-temperature mechanical properties were those of the cold-worked state. It was found from the table that the recrystallization temperature of industrial pure aluminum is $T_{\\\\mathbb{F}}=150~\\\\mathrm{\\\\textdegree}$. However, when the aforementioned industrial pure aluminum strip was heated to $100^{\\\\circ}C$, held for 16 days, and then cooled to room temperature before measuring its strength again, a significant decrease in strength was observed. Please explain the reason.", "answer": "The recrystallization temperature of industrial pure aluminum $T_{\\\\mathbb{R}}=150^{\\\\circ}C$ found in the table refers to the temperature at which recrystallization is completed after annealing for $^\\\\textrm{\\\\scriptsize1h}$. In reality, besides the annealing temperature, the holding time also affects the recrystallization process. For metal materials subjected to large cold deformation, even when annealed at $T{<}T_{\\\\mathbb{R}}$, recrystallization can still occur if the holding time is sufficient. Two methods can be used to determine this: $\\\\textcircled{1}$ metallographic examination; $\\\\textcircled{2}$ substituting the known $T_{1},t_{1},t_{2},Q$ into the formula $\\\\frac{t_{1}}{t_{2}}=\\\\mathrm{e}^{-\\\\frac{Q}{R}(\\\\frac{1}{T_{1}}-\\\\frac{1}{T_{2}})}$ to solve for $\\\\pmb{T_{2}}$, and comparing it with $100\\\\mathrm{\\\\mathcal{C}}$ to determine whether recrystallization has occurred." }, { "idx": 2873, "question": "Given that H70 brass [w(Zn)=30%] requires 1 hour to complete recrystallization at a constant temperature of 400°C, and 2 hours at 390°C, calculate the time required to complete recrystallization at a constant temperature of 420°C.", "answer": "Recrystallization is a thermally activated process, so the recrystallization rate; v_R = A exp(-Q/RT), and the recrystallization rate is inversely proportional to the time t required to achieve a certain volume fraction, i.e., v_R ∝ 1/t. 1/t = A' exp(-Q/RT) When recrystallization of the same degree occurs at two different constant temperatures, t1/t2 = e^(-Q/R (1/T2 - 1/T1)) Taking the natural logarithm on both sides: ln(t1/t2) = -Q/R (1/T2 - 1/T1). Similarly, ln(t1/t3) = -Q/R (1/T3 - 1/T1). Thus, (ln(t1/t2)) / (ln(t1/t3)) = (1/T2 - 1/T1) / (1/T3 - 1/T1) Substituting the corresponding data, we obtain t3 = 0.26 h." }, { "idx": 2874, "question": "Explain how the following factors affect $G_{0},N_{0},Q_{8}$, and $Q_{n}$: degree of pre-deformation; original grain size; metal purity.", "answer": "The driving force for primary recrystallization is the distortion energy of the crystal after deformation. The transition of the crystal from the high-energy state after distortion to the low-energy state of annealing is a spontaneous trend. However, this change in energy state requires atoms to overcome a potential barrier $\\\\Delta E$, the height of which depends on the lattice distortion energy after deformation. When the distortion energy is high, $\\\\Delta E$ decreases, and the nucleation and growth activation energies $\\\\mathbf{Q}_{\\\\mathfrak{n}}, \\\\mathbf{Q}_{\\\\mathfrak{g}}$ both decrease, thereby accelerating the recrystallization rate. Therefore, all factors affecting the distortion energy after deformation will influence $\\\\mathbf{Q}_{\\\\mathbf{n}}, \\\\mathbf{Q}_{\\\\mathbf{\\\\tilde{g}}}$, and $T_{\\\\mathsf{R}}$. From the above analysis, within a certain range of deformation, the higher the degree of pre-deformation and the finer the original grain size, the greater the distortion energy after deformation, and the lower $Q_{\\\\mathfrak{n}}, Q_{\\\\mathfrak{g}}$ will be. The effect of metal purity on $\\\\mathbf{Q}_{\\\\mathrm{n}}, \\\\mathbf{Q}_{\\\\mathrm{g}}$ can be considered from two aspects. On one hand, impurities increase the distortion energy, reducing $Q_{\\\\mathrm{n}}$ and $\\\\mathbf{Q}_{\\\\mathrm{g}}$. On the other hand, impurities also hinder interface migration, increasing $\\\\mathbf{\\\\hat{Q}}_{\\\\mathrm{n}}, \\\\mathbf{\\\\hat{Q}}_{\\\\mathrm{g}}$. These two opposing effects coexist, and the dominant one determines the outcome. $\\\\aleph_{\\\\mathfrak{g}}\\\\mathrm{G}_{\\\\mathfrak{o}}$ is only related to the intrinsic nature of the metal and is not significantly affected by the degree of pre-deformation, original grain size, or metal purity." }, { "idx": 2875, "question": "In a metal with a dislocation density of $\\\\mathrm{10^{12}/cm^{2}}$ after cold deformation, there exist second-phase particles that do not aggregate or grow upon heating, with a volume fraction $\\\\varphi{=}1\\\\%$ and a radius of $1\\\\mu\\\\mathfrak{m}$. Can the presence of these second-phase particles completely prevent recrystallization of this metal upon heating? (Given $G{=}10^{5}\\\\mathrm{MPa}$, $\\\\pmb{b=0.3\\\\mathrm{nm}}$, and interfacial energy ${\\\\pmb\\\\sigma}{=}0,5\\\\mathrm{J}/\\\\uppi^{2}$.)", "answer": "The driving force for recrystallization $$F=G b^{2}\\\\left(\\\\rho-\\\\rho_{\\\\circ}\\\\right)\\\\approx G b^{2}\\\\rho=10^{11}\\\\times(3\\\\times10^{-10})^{2}\\\\times10^{16}=9\\\\times10^{7}({\\\\mathrm{N/m}}^{2})$$The resistance to recrystallization $$f={\\\\frac{3\\\\varphi}{2r}}{\\\\sigma}={\\\\frac{3}{2}}\\\\times{\\\\frac{0.01}{{\\\\bar{1}}\\\\times}}{\\\\frac{0.01}{10^{-6}}}\\\\times0.5=7.5\\\\times10^{3}({\\\\mathrm{N/m^{2}}})$$$F{\\\\gg}f$, so the presence of these second-phase particles cannot completely prevent recrystallization." }, { "idx": 2876, "question": "For an Fe-3%Si alloy containing MnS particles with a radius of 0.05μm and a volume fraction of 0.01, during the annealing process below 850°C, when the average grain diameter of the matrix reaches 6μm, normal grain growth stops. Analyze the reason for this phenomenon.", "answer": "For silicon steel sheets containing MnS particles with a radius of 0.05μm and a volume fraction of 0.01 during recrystallization, the limiting grain diameter is calculated as: D_lim = (4r)/(3φ) = (4/3) × (0.05/0.01) = 6.67μm. It is precisely due to the presence of these dispersed particles that when the silicon steel sheet is annealed below 850°C, normal grain growth stops once the average grain diameter of the matrix reaches 6μm." }, { "idx": 2877, "question": "Briefly describe the driving forces of primary recrystallization and secondary recrystallization", "answer": "The driving force of primary recrystallization is the elastic distortion energy of the matrix, while the driving force of secondary recrystallization comes from the reduction of interfacial energy." }, { "idx": 2878, "question": "W has a very high melting point (Tm=3410°C) and is often chosen as the heating element for incandescent light bulbs. However, when large grains spanning the filament exist, the filament becomes brittle and may fracture under the thermal shock of frequent switching. Please introduce a method to extend the filament's lifespan.", "answer": "Factors affecting the normal growth of recrystallized grains, besides temperature, include the presence of finely dispersed second-phase particles, which play a significant role in grain boundary migration. For example, ThO2 second-phase particles can be formed in the tungsten filament to hinder grain growth during high-temperature operation. If the volume fraction of ThO2 particles is φ and the particle radius is r, the limiting grain size is given by: Dlim=4r/(3φ(1+cosα)), where α is the contact angle. Therefore, by selecting appropriate φ and r, Dlim can be minimized. Grain refinement can enhance strength while maintaining a high level of toughness, effectively extending the filament's service life." }, { "idx": 2879, "question": "In engineering, it is often considered that the grain size of steel does not grow when heated to 760°C, but will significantly grow at 870°C. If the original grain diameter of the steel is 0.05 mm, the empirical formula for grain growth is D^(1/n)−D0^(1/n)=c t, where D is the grain diameter after growth; D0 is the original grain diameter; c is the proportionality constant; t is the holding time. Given that at 760°C, n=0.1, c=6×10^(−16), find the grain diameter of steel with a carbon content of 0.8% after holding at 760°C for 1 hour.", "answer": "At 760°C: D^(1/n)=D0^(1/n)+c t=(0.05)^10+6×10^(−16)×60=13.37×10^(−14). D=0.0516(mm). Therefore, the grain has essentially not grown." }, { "idx": 2880, "question": "How to distinguish between cold and hot working?", "answer": "The recrystallization temperature is the dividing line between cold and hot working." }, { "idx": 2881, "question": "In practice, it is often considered that the grain size of steel will significantly grow when heated to 870°C. If the original grain diameter of the steel is 0.05 mm, the empirical formula for grain growth is D^(1/n)−D0^(1/n)=c t, where D is the grown grain diameter; D0 is the original grain diameter; c is the proportional constant; t is the holding time. Given that at 870°C, n=0.2, c=2×10^(−8), find the grain diameter of steel with 0.8% carbon content after holding at 870°C for 1 hour.", "answer": "At 870°C: D^5=(0.05)^5+2×10^(−8)×60=1.513×10^(−6) D=0.0686(mm). Compared to the original grain diameter, it has significantly grown (approximately 37%)." }, { "idx": 2882, "question": "Assuming we add Na2O with w(Na2O)=10% to SiO2, please calculate the ratio of oxygen to silicon.", "answer": "There are 90at% SiO2 and 10at% Na2O, so O:Si = (0.9×2 + 0.1) : 0.9 = 2.111. Since O:Si = 2.111 < 2.5, there is a good tendency for glass formation." }, { "idx": 2883, "question": "In a eutectic reaction of the Mg-Ni system, let C1 be the hypoeutectic alloy and C2 be the hypereutectic alloy. The mass fractions of the proeutectic phases in these two alloys are equal. Find the relationship between C1 and C2.", "answer": "According to the lever rule, the mass fractions of the proeutectic phases are: α_pro=(23.5-C1)/23.5 β_pro=(C2-23.5)/(54.6-23.5) Given that α_pro=β_pro, combining the above equations yields: C2=54.6-1.323C1" }, { "idx": 2884, "question": "What are the main differences in the microstructure between dynamic recrystallization and static recrystallization?", "answer": "Although the microstructure after dynamic recrystallization also consists of equiaxed grains, the grain boundaries are serrated, and the grains contain subgrains divided by dislocation tangles. This differs from the grains produced by static recrystallization, which have a very low dislocation density. Therefore, the strength and hardness of dynamically recrystallized structures with the same grain size are higher than those of statically recrystallized ones. The grain size after dynamic recrystallization is proportional to the flow stress. Additionally, the lower the strain rate and the higher the deformation temperature, the larger and more complete the grains after dynamic recrystallization will be." }, { "idx": 2885, "question": "What is a copolymer?", "answer": "A polymer formed by the polymerization of two or more monomers is called a copolymer." }, { "idx": 2886, "question": "In a eutectic reaction of the Mg-Ni system, let C1 be the hypoeutectic alloy and C2 be the hypereutectic alloy. The total amount of α in alloy C1 is 2.5 times that in alloy C2. Determine the compositions of C1 and C2.", "answer": "Let the total amount of α in C1 be α1, then: α1=(54.6-C1)/54.6 Let the total amount of α in C2 be α2, then: α2=(54.6-C2)/54.6 According to the given condition, α1=2.5α2, that is: (54.6-C1)/54.6=2.5*(54.6-C2)/54.6 Substituting C2=54.6-1.323C1 into the above equation, we can solve for: C1=w(Ni)=12.7% C2=w(Ni)=37.8%" }, { "idx": 2887, "question": "If O:Si ≈ 2.5 is the criterion for glass-forming tendency, what is the maximum amount of Na2O that can form a glass?", "answer": "Let the maximum content of Na2O be x, and SiO2 = 1 - x. According to (x + 2(1 - x)) / (1 - x) ≤ 2.5, we get (2 - x) / (1 - x) ≤ 2.5. Further derivation yields 1 + 1 / (1 - x) ≤ 2.5; 1 / (1 - x) ≤ 1.5; 1 - x ≥ 1 / 1.5; x ≤ 1 / 3. Therefore, the maximum amount of Na2O that can form a glass is 1/3." }, { "idx": 2888, "question": "Based on the shown CaO·ZrO2 phase diagram, write down all the isothermal three-phase transformations.", "answer": "There are three isothermal three-phase transformations in the shown ZrO2-σ·CaO phase diagram: peritectic reaction: L+T-ZrO3→C-ZrO2; eutectic reaction: L→(C-ZrO2+ZrCaO3); eutectoid reaction: T-ZrO2→M-ZrO2+C-ZrO2; where L represents liquid phase, T represents tetragonal, C represents cubic, M represents monoclinic." }, { "idx": 2889, "question": "What is a homopolymer?", "answer": "A polymer formed by the polymerization of a single monomer is called a homopolymer." }, { "idx": 2890, "question": "Using the data from the previous question, assuming the alloy composition is Al-0.5%Cu and there is no convection in the liquid, calculate the temperature gradient required to maintain a planar solid-liquid interface.", "answer": "G≥(m w0 R(1-k0))/(D k0)=(320×0.005×3×10^-4×(1-0.16))/(3×10^-5×0.16)=84℃/cm" }, { "idx": 2891, "question": "Using the data from the previous question, assuming the alloy composition is Al-2%Cu and there is no liquid convection, calculate the interface temperature at the beginning of solidification", "answer": "T=T0-m(w0/k0)=660.37-320×(0.02/0.16)=620.37℃" }, { "idx": 2892, "question": "Using the data from the previous question, assuming the alloy composition is Al-2%Cu and there is no liquid convection, calculate the temperature gradient required to maintain a planar solid-liquid interface.", "answer": "G≥(m w0 R(1-k0))/(D k0)=(320×0.02×3×10^-4×(1-0.16))/(3×10^-5×0.16)=336℃/cm" }, { "idx": 2893, "question": "What is homoaddition polymerization?", "answer": "A polymerization reaction involving a single monomer is called homoaddition polymerization, abbreviated as homoaddition; the resulting polymer has the same composition as its monomer." }, { "idx": 2894, "question": "What is co-condensation?", "answer": "The polycondensation reaction involving two or more monomers is called co-condensation polymerization, abbreviated as co-condensation, and the resulting polymer composition differs from the monomers." }, { "idx": 2895, "question": "Briefly describe the advantages of polymer alloying", "answer": "Through alloying, polymers can overcome certain performance weaknesses of single-component polymers (homopolymers). For example, polypropylene (PP) is prone to brittle fracture at low temperatures, but blending with butadiene rubber (BR) can significantly improve the toughness of polypropylene. It can also broaden the applications of polymers. For instance, blending polyethylene (PE) with different densities can produce foam plastics with various properties." }, { "idx": 2896, "question": "Calculate the relative amounts (in mol%) of monoclinic ZrO2 solid solution (Monoclinic ZrO2 SS) and cubic ZrO2 solid solution (Cubic ZrO2 SS) at room temperature for a CaO–ZrO2 ceramic with w(CaO)=4%. Assume that the solubilities of monoclinic ZrO2 solid solution and cubic ZrO2 solid solution at room temperature are 2mol% CaO and 15mol% CaO, respectively.", "answer": "Using the conversion formula between mole fraction and mass fraction, the mole fraction corresponding to w(Ca)=4% can be calculated: xA = (wA/ArA)/(wA/ArA + wB/ArB) = (4/(40+16))/(4/(40+16) + 96/(91+16×2)) ≈ 0.08; thus, 4wt% CaO = 8mol% CaO. Moreover, it can be observed from the figure that the solubility limit changes little below 900°C, yielding: monoclinic phase % = (xcub - x)/(xcub - xmono) × 100% = (15 - 8)/(15 - 2) × 100% = 53.8%; cubic phase % = (x - xmono)/(xcub - xmono) × 100% = (8 - 2)/(15 - 2) × 100% = 46.2%." }, { "idx": 2897, "question": "Using the data from the previous question, assuming the alloy composition is Al-0.5%Cu and there is no liquid convection, calculate the interface temperature at the beginning of solidification", "answer": "T=T0-m(w0/k0)=660.37-320×(0.005/0.16)=650.37℃" }, { "idx": 2898, "question": "What is a quasicrystal?", "answer": "Quasicrystals lack translational symmetry but are a type of solid atomic aggregation state that exhibits periodic order similar to crystalline materials. In three-dimensional space, they possess not only 5-fold symmetry axes but also 8, 10, or 12-fold symmetry axes, with diffraction patterns showing non-crystallographic symmetry. Most quasicrystalline phases are metastable and can only be obtained through rapid solidification methods. It is well known that periodic tiling of a plane can be achieved with equilateral triangles, squares, or regular hexagons, whereas regular pentagons cannot tile the plane without overlaps or gaps." }, { "idx": 2899, "question": "Briefly describe the characteristics of nanomaterials from the perspective of internal microstructure.", "answer": "Nanomaterials refer to materials that have at least one dimension in the three-dimensional space at the nanoscale or are composed of them as basic units. According to dimensionality, the basic units of nanomaterials can be divided into three categories: (1) Zero-dimensional, meaning all three spatial dimensions are at the nanoscale, such as nanopowder materials; (2) One-dimensional, meaning two dimensions in space are at the nanoscale, such as nanowires, nanorods, nanotubes, etc.; (3) Two-dimensional, meaning one dimension in three-dimensional space is at the nanoscale, such as ultrathin films, multilayer films, and superlattices. Due to the ultrafine size of nanoparticles, they are on the same order of magnitude as the wavelength of light waves, neutron wavelength, mean free path, etc., resulting in quantum size effects, small size effects, surface effects, macroscopic quantum tunneling effects, and when the volume fraction exceeds $50\\%$, the influence of grain boundary structure causes nanomaterials to exhibit unique mechanical, physical, and chemical properties." }, { "idx": 2900, "question": "A ternary alloy with mass fractions of 40% A, 30% B, and 30% C forms a three-phase equilibrium at the eutectic temperature. The compositions of the three phases are as follows: liquid phase (50% A, 40% B, 10% C), α phase (85% A, 10% B, 5% C), and β phase (10% A, 20% B, 70% C). Calculate the fractions of the liquid phase, α phase, and β phase.", "answer": "First, draw a concentration triangle and mark the compositions of each phase. The alloy composition point is: L% = (57 - 30) / (57 - 10) × 100% = 57.4%; α% = (40 - 35) / (85 - 35) × 100% = 10%; β% = 100% - 57.4% - 10% = 32.6%." }, { "idx": 2901, "question": "Briefly describe the methods of polymer alloying", "answer": "The preparation methods of polymer alloys can be divided into physical methods and chemical methods. Physical blending methods include dry powder blending, melt blending, and latex blending, with melt blending being the most commonly used. Chemical blending methods mainly include copolymerization-blending and interpenetrating polymer network methods." }, { "idx": 2902, "question": "Why can stretching improve the crystallinity of crystalline polymers?", "answer": "From $\\Delta G{=}\\Delta H{-}T\\Delta S$, it is known that the free energy $\\Delta G$ must be less than zero for the crystallization process to proceed spontaneously. When a substance transitions from an amorphous state to a crystalline state, the arrangement of molecules changes from disordered to ordered, which always results in a decrease, i.e., $\\Delta S{<}0$. In this case, $-T\\Delta S{>}0$, while $\\Delta H{<}0$ (exothermic). To ensure $\\Delta G{<}0$, it is necessary that $|\\Delta H|>T|\\Delta S|$. For some polymers transitioning from the amorphous phase to the crystalline phase, $\\vert\\Delta S\\vert$ is large, while the thermal effect of crystallization $\\Delta H$ is small. To satisfy $\\mid\\Delta H\\mid>T\\mid\\Delta S\\mid$, there are only two approaches: lowering $T$ or reducing $\\{\\Delta S\\}$. However, excessively lowering the temperature makes molecular movement difficult, potentially leading to a glassy state instead of crystallization. To reduce $\\mid\\Delta S\\mid$, stretching the polymer before crystallization can be employed, which imparts a certain degree of order to the polymer chains in the amorphous phase. This reduces the corresponding $|\\Delta S|$ during crystallization, enabling the process. Therefore, for crystalline polymers, stretching is beneficial for increasing crystallinity. For example: natural rubber takes decades to crystallize at room temperature, but when stretched, it crystallizes in just a few seconds." }, { "idx": 2903, "question": "The crystallization activation energy of amorphous alloys can be determined using the Ozawa plotting method, which utilizes the linear relationship between the crystallization temperature $\\scriptstyle{\\mathcal{T}}_{\\mathbf{z}}$ and the heating rate $^{a}$ measured under different continuous heating conditions. The $\\pmb{T_{x}}$ values for the pre-crystallization phase $\\alpha{\\boldsymbol{-}}\\mathbf{\\overline{{F}}}\\mathbf{e}$ of the amorphous FenBsSis alloy have been measured as shown in the table below. Calculate the activation energy.", "answer": "Based on the data in the table, plot the graph (see Figure 42), which approximates a straight line. Using the least squares method, the equations of the straight lines are fitted as follows: \\n$$\\n\\\\ln{\\\\frac{T_{x}}{a}}={\\\\frac{46}{T_{x}}}-53\\n$$\\n$$\\n\\\\ln{\\\\frac{T_{x}}{a}}={\\\\frac{49}{T_{x}}}-57\\n$$\\nFrom the slope of the straight lines, the activation energy for the precipitation stage of the ${\\\\mathfrak{a}}{\\\\mathfrak{F}}{\\\\mathfrak{e}}$ pre-crystallization phase is determined to be $(382\\\\sim407)\\\\mathrm{kJ/mol}_{\\\\circ}$." }, { "idx": 2904, "question": "Explain why crystal structures do not possess 5-fold or higher than 6-fold symmetry axes?", "answer": "5-fold or higher than 6-fold symmetry axes cannot satisfy the condition of identical surroundings around lattice points, lack translational symmetry, and thus cannot achieve a regularly periodic arrangement of crystal structures." }, { "idx": 2905, "question": "How to describe the quasicrystalline structure?", "answer": "The quasicrystalline structure cannot be represented by a unit cell like crystals, meaning it cannot achieve periodicity through translation operations. Currently, the more commonly used model is the tiling method to characterize quasicrystalline structures. For example, a 5-fold symmetric quasicrystalline structure can be constructed using two types of rhombuses with equal side lengths and angles of 36 and 144 degrees (narrow), and 72 and 108 degrees (wide), following specific matching rules." }, { "idx": 2906, "question": "Briefly describe the influencing factors of the glass transition temperature of polymers.", "answer": "There are many factors affecting the glass transition temperature, usually including (1) chain flexibility; (2) the influence of intermolecular forces; (3) the effect of copolymerization; (4) the influence of plasticizers, etc." }, { "idx": 2907, "question": "Briefly describe the method of determining phase boundaries in polymer phase diagrams using scattered light intensity.", "answer": "For multiphase polymer blends, when the size of the dispersed phase is comparable to the wavelength of visible light (i.e., several hundred nanometers), the scattered light intensity method can be used for measurement. When visible light passes through such materials, strong light scattering occurs, resulting in turbidity. In a single-phase system, there is no abrupt change in scattered light intensity. Therefore, for a binary polymer with composition w, if it is single-phase at low temperatures and undergoes a phase transition to two phases upon heating to a certain temperature ${\\\\bf\\\\cal I}_{1}$, the curve of scattered light intensity versus temperature will show an abrupt change. The temperature at this abrupt change point is often referred to as the 'cloud point,' which is the phase transition temperature. Similarly, the corresponding $T_{2}$ for ${\\\\mathfrak{w}}_{2}$ and $T_{3}$ for $w_{3}$ can be measured. By plotting the cloud points of blends with different compositions against their compositions, the phase boundaries can be obtained." }, { "idx": 2908, "question": "What is the glass transition temperature of a polymer?", "answer": "Amorphous linear polymers can be divided into three states based on their mechanical properties at different temperatures: glassy state, high elastic state, and viscous flow state. At lower temperatures, the thermal motion of molecules is limited, preventing not only the movement of entire macromolecular chains but also the movement of chain segments or even individual chain units, causing the entire macromolecule to lose flexibility. In this state, the polymer resembles ordinary silicate glass in a supercooled liquid state, hence it is called the glassy state. The highest temperature (Tg) at which a polymer exhibits the glassy state is called the glass transition temperature, which is the temperature at which the polymer transitions from the high elastic state to the glassy state. During the glass transition, in addition to discontinuous and significant changes in mechanical properties such as the elastic modulus (E) of the polymer, other properties like the expansion coefficient, heat capacity, and dielectric constant also undergo notable changes. Therefore, the glass transition is not a thermodynamic phase transition but rather a state transition under non-equilibrium conditions, which can be regarded as a volume relaxation process. The glass transition temperature is an extremely important property of polymeric materials and serves as the dividing line between plastics and rubbers." }, { "idx": 2909, "question": "Due to the incompleteness of crystallization, crystalline and amorphous regions always coexist in crystalline polymers. The volume fraction crystallinity and density of two crystalline polytetrafluoroethylene samples have been measured as φ₁=51.3%, φ₂=74.2% and ρ₁=2.144 g/cm³, ρ₂=2.215 g/cm³, respectively. Calculate the densities of completely crystalline and completely amorphous polytetrafluoroethylene.", "answer": "The density of a crystalline polymer ρ = φρ_c + (1-φ)ρ_a, where ρ_c and ρ_a are the densities of the crystalline and amorphous parts of the polymer, respectively, and φ is the volume fraction of the crystalline part. Solving the simultaneous equations: 2.144 = 0.513ρ_c + (1-0.513)ρ_a, 2.215 = 0.742ρ_c + (1-0.742)ρ_a, we obtain ρ_c = 2.296 g/cm³ and ρ_a = 1.984 g/cm³." }, { "idx": 2910, "question": "At the same temperature, when the α phase (85% A, 10% B, 5% C) and β phase (10% A, 20% B, 70% C) each account for 50%, estimate the composition of the alloy.", "answer": "Let the alloy composition be x, which must lie on the line connecting the α-β phase composition points. According to the lever rule: (α/β) = (x - 10%) / (85% - x) = 1; solving gives x = 47.5% A; then, from the concentration triangle, B = 14.5% and C = 38% are obtained." }, { "idx": 2911, "question": "For an Al-Cu alloy with an atomic fraction of Cu at 4.6%, after solution treatment at 550°C, the α phase contains x(Cu)=2%. When reheated to 100°C and held for a period, the precipitated θ phase spreads throughout the entire alloy volume. The θ phase has an fcc structure with r=0.143nm, and the average spacing between θ particles is 5nm. If the Cu atoms in the α phase can be neglected after θ precipitation, how many Cu atoms does each θ particle contain?", "answer": "The fcc structure has 4 atoms per unit cell, a=4r/√2=4×0.143/√2=0.404nm. Since x(Cu)=2%, the number of Cu atoms per cm³=0.02×4/(4.04×10⁻⁸)³=1.213×10²¹ atoms/cm³. Therefore, the number of Cu atoms per θ phase particle=1.213×10²¹/8×10¹⁸=151.6 atoms/particle" }, { "idx": 2912, "question": "The concentration fluctuation equation for spinodal decomposition is given by $C-C_{0}=\\\\mathtt{e}^{R(\\\\lambda)t}\\\\cos{\\\\frac{2\\\\pi}{\\\\lambda}{Z}}$. Find the critical wavelength, where $R(\\\\lambda)=-M\\\\frac{4\\\\pi^{2}}{\\\\lambda}\\\\biggl[\\\\vert G^{\\\\prime}+2\\\\eta Y+\\\\frac{8\\\\pi^{2}K}{\\\\lambda^{2}}\\\\biggr]$; M is the interdiffusion mobility; $\\\\eta$ is the mismatch due to concentration gradient; $Y=\\\\frac{E}{(1-\\\\nu)}$ ($E$ is the elastic modulus, $\\\\nu$ is the Poisson's ratio); $K$ is a constant; $\\\\lambda$ is the wavelength; $Z$ is the distance; $t$ is the time; $\\\\vert G^{\\\\prime}={\\\\frac{\\\\partial^{2}\\\\dot{G_{s}}}{\\\\partial x^{2}}}$ ($G_{s}$ is the free energy of the solid solution, $_x$ represents the composition of the solid solution).", "answer": "When $R\\\\langle\\\\lambda\\\\rangle=0$, \\n\\n$$\\nC-C_{\\\\mathrm{0}}=\\\\cos\\\\frac{2\\\\pi}{\\\\lambda}Z\\n$$\\n\\nAt this point, the composition fluctuation does not change with time, meaning spinodal decomposition does not occur. Spinodal decomposition can only occur when $R\\\\langle\\\\lambda\\\\rangle>0$, i.e., the $\\\\lambda$ value when $R(\\\\lambda)=0$ is the critical wavelength $\\\\lambda_{c}$. Therefore, \\n\\n$$\\nG^{\\\\prime\\\\prime}+2\\\\eta Y+\\\\frac{8\\\\pi^{2}K}{\\\\lambda_{\\\\odot}^{2}}=0\\n$$\\n\\n$$\\n\\\\lambda_{\\\\mathrm{{c}}}=\\\\Bigl[\\\\frac{-8\\\\pi^{2}K}{\\\\vec{G}^{\\\\eta}+2\\\\eta Y}\\\\Bigr]^{\\\\frac{3}{2}}\\n$$" }, { "idx": 2913, "question": "After quenching 1.2% steel to obtain martensite and a small amount of retained austenite structure, what changes will occur when heated to 300°C and held for 2h?", "answer": "The retained austenite decomposes and transforms into α plus fine carbides, and the martensite also decomposes into α plus fine carbides, with the original martensite morphology becoming less distinct." }, { "idx": 2914, "question": "An Al-Cu alloy with an atomic fraction of Cu at 4.6% undergoes solution treatment at 550°C, resulting in the α phase containing x(Cu)=2%. After reheating to 100°C and holding for a period, θ phase precipitates uniformly throughout the alloy volume. The θ phase has an fcc structure with r=0.143 nm, and the average spacing between θ particles is 5 nm. Calculate the number of θ phase particles per cm³ in the alloy.", "answer": "Assuming each θ phase particle occupies a volume of 5 nm³, the number of θ particles is 1/(5×10⁻⁷)³=8×10¹⁸ particles/cm³." }, { "idx": 2915, "question": "An Al-Cu alloy with an atomic fraction of 2% Cu is rapidly cooled from 520°C to 27°C and held for 3 hours, forming G.P. zones with an average spacing of 1.5×10^-6 cm. Given that the diffusion coefficient of Cu in Al at 27°C is D=2.3×10^-25 cm²/s, and assuming the process is diffusion-controlled, estimate the quenched vacancy concentration of the alloy.", "answer": "According to the vacancy concentration formula Cv=C0 exp(-Qv/RT), given the vacancy formation energy Qv=82.811 kJ/mol, temperature T=793 K, and R=8.314 J/(mol·K), the quenched vacancy concentration is Cv=2.3 exp(-82811/(8.314×793))=8.069×10^-6." }, { "idx": 2916, "question": "During solid-state phase transformation, assuming the volume free energy change per atom is ΔGB=200ΔT/Tc, in units of J/cm³, the critical transformation temperature Tc=1000K, strain energy ε=4J/cm³, coherent interface energy σ_coherent=4.0×10⁻⁶ J/cm², and incoherent interface energy σ_incoherent=4.0×10⁻⁵ J/cm², calculate the ratio of critical nucleation work ΔG_coherent* to ΔG_incoherent* when ΔT=50°C.", "answer": "The nucleation work for a spherical nucleus is ΔG_coherent*=16πσ³/3(ΔGB−ε)², and if the interface is incoherent, the strain energy can be neglected, with the nucleation work being ΔG_incoherent*=16πσ³/3ΔGB². The ratio ΔG_coherent*/ΔG_incoherent*=ΔGB²σ_coherent³/(ΔGB−ε)²σ_incoherent³ =[(200×50/1000)²×(4.0×10⁻⁶)³]/[(200×50/1000−4)²×(4.0×10⁻⁵)³]=2.777×10⁻³." }, { "idx": 2917, "question": "For a quenched alloy, precipitation begins to form from the supersaturated solid solution after aging at $15^{\\\\circ}\\\\mathrm{C}$ for 1h. If aging is performed at $100^{\\\\circ}\\\\mathrm{C}$, precipitation starts after $1\\\\mathfrak{min}$. To prevent precipitation within 1 day after quenching, at what temperature should the alloy be maintained? (Hint: Apply the Arrhenius rate equation)", "answer": "Rate $=A\\\\exp\\\\Bigl\\\\{-\\\\frac{Q}{R T}\\\\Bigr\\\\}$, so $\\\\ln t=A+\\\\frac{Q}{R T}$. Substituting the data, we obtain $Q{=}4.24\\\\times10^{4}~\\\\mathrm{J/mol}$. Substituting again, we get $\\\\pmb{T=243K(-30^{\\\\circ}C)}$" }, { "idx": 2918, "question": "After quenching 1.2% steel to obtain martensite and a small amount of retained austenite structure, what changes will occur when heated to 180°C and held for 2 hours?", "answer": "Fine carbides begin to precipitate from the martensite laths, which are easily etched and appear dark." }, { "idx": 2919, "question": "Under the given conditions, prove the following relationship holds: ΔFπr²=(1/6)Gϕ̂²πh_max[8r−9h_max], where ΔF is the free energy difference between austenite and martensite.", "answer": "The equation −ΔFπr²+(4/3)Gϕ̂²πrh_max−(3/2)Gϕ̂²πh_max²=0 is derived from the equilibrium condition dΔF_t/dh=0. Multiplying both sides by πr² and rearranging yields ΔFπr²=(1/6)Gϕ̂²πh_max[8r−9h_max]." }, { "idx": 2920, "question": "In the face-centered cubic (fcc) structure of aluminum crystal, the coordination number of each aluminum atom on the (111) plane is (A)12 (B) 6 (C) 4 (D) 3", "answer": "B" }, { "idx": 2921, "question": "Given the density of fully crystalline polytetrafluoroethylene ρ_c = 2.296 g/cm³ and the density of completely amorphous polytetrafluoroethylene ρ_a = 1.984 g/cm³, calculate the crystallinity of a polytetrafluoroethylene sample with a density of 2.26 g/cm³.", "answer": "Crystallinity φ = (ρ - ρ_a)/(ρ_c - ρ_a) = (2.26 - 1.984)/(2.296 - 1.984) = 0.276/0.312 = 0.885, thus φ = 88.5%." }, { "idx": 2922, "question": "In cubic crystals, the (110) and (211) planes belong to the same zone axis.", "answer": "D" }, { "idx": 2923, "question": "An Al-Cu alloy with an atomic fraction of 2% Cu is rapidly cooled from 520°C to 27°C and held for 3 hours, forming G.P. zones with an average spacing of 1.5×10^-6 cm. Given that the diffusion coefficient of Cu in Al at 27°C is D=2.3×10^-25 cm²/s, and assuming the process is diffusion-controlled, estimate the vacancy formation energy of this alloy.", "answer": "After quenching the Al-Cu alloy, the diffusion coefficient of Cu in the alloy is D=x²/t=(1.5×10^-6)²/(3×3600)=2.08×10^-16 cm²/s. This diffusion coefficient is larger than the normal diffusion coefficient, with an increase factor of 2.08×10^-16/2.3×10^-25=9.04×10^8. Since Cu is a substitutional solute in Al, if diffusion occurs via the vacancy mechanism, the increase factor can be attributed entirely to the contribution of quenched supersaturated vacancies. According to the vacancy concentration Cv=C0 exp(-Qv/RT), then Cv(525°C)/Cv(27°C)=exp[-Qv/R (300-793)/(793×300)]=9.04×10^8. Solving gives Qv=82.811 kJ/mol." }, { "idx": 2924, "question": "In polyethylene polymer materials, the C一H chemical bond belongs to (A) metallic bond (B) ionic bond (C) covalent bond (D) hydrogen bond", "answer": "C" }, { "idx": 2925, "question": "After quenching, 1.2% steel obtains a martensite and a small amount of retained austenite structure. What changes will occur when heated to 680°C and held for 2 hours?", "answer": "Carbides are distributed in granular form within the ferrite matrix, resulting in a granular pearlite structure." }, { "idx": 2926, "question": "When a pure metal undergoes an allotropic transformation during cooling from high temperature to room temperature with volume expansion, the atomic coordination number of the low-temperature phase is ___ that of the high-temperature phase. (A) Lower (B) Higher (C) The same", "answer": "A" }, { "idx": 2927, "question": "The packing density of a simple cubic crystal is (A) $100\\\\%$ (B) $65\\\\%$ (C) $52\\\\%$ (D) $58\\\\%$", "answer": "C" }, { "idx": 2928, "question": "During solid-state phase transformation, given the volume free energy change per atom as ΔGB=200ΔT/Tc, in units of J/cm³, the critical transformation temperature Tc=1000K, strain energy ε=4 J/cm³, coherent interface energy σ_coherent=4.0×10⁻⁶ J/cm², and incoherent interface energy σ_incoherent=4.0×10⁻⁵ J/cm², calculate ΔT when ΔG_coherent*=ΔG_incoherent*.", "answer": "ΔG_coherent*=ΔG_incoherent*\\n(4.0×10⁻⁶)³/(200×ΔT/1000−4)²=(4.0×10⁻⁵)³/(200×ΔT/1000)²\\nSolving gives ΔT=20,653°C" }, { "idx": 2929, "question": "In substitutional solid solution alloys, the diffusion of solute atoms is achieved through . (A) Atomic exchange mechanism (B) Interstitial diffusion mechanism (C) Vacancy mechanism", "answer": "C" }, { "idx": 2930, "question": "The flexibility of polymer materials mainly depends on the mobility of . (A) Main chain segments (B) Side groups (C) Functional groups or atoms within side groups", "answer": "A" }, { "idx": 2931, "question": "Two parallel screw dislocations, when their Burgers vectors are in the same direction, the interaction force between them is (A) zero (B) repulsive (C) attractive", "answer": "B" }, { "idx": 2932, "question": "In ionic crystals, if Schottky defects are formed in a local region, the concentration of cation vacancies in this region is equal to . \\n\\n(A) Anion vacancy concentration (B) Interstitial anion concentration (C) Interstitial cation concentration", "answer": "A" }, { "idx": 2933, "question": "Among the three crystal structures of fcc, bcc, and hcp materials, the one most prone to forming twins during plastic deformation is (A) fcc (B) bcc (C) hcp", "answer": "C" }, { "idx": 2934, "question": "For a lenticular martensite plate with thickness h and radius r, the volume can be approximated as πr²h, and the volume of the surrounding strain field can be taken as (4/3)πr³−πr²h. The strain energy per unit volume in the strain field can be expressed as (Gϕ̂²h²)/(2r²) (where G is the shear modulus and ϕ̂ is the shear angle). Assuming the diameter of the plate remains unchanged during martensite growth, explain that when the plate thickens, the thickness cannot exceed a maximum value h_max due to the limitation imposed by strain energy.", "answer": "If the semi-coherent interface energy is negligible, the change in the system's free energy during martensite plate growth is ΔF_t=−ΔFπr²h+Gϕ̂²h²((4/3)πr³−πr²h)/(2r²). Given that the diameter remains unchanged, setting dΔF_t/dh=0 yields −ΔFπr²+(4/3)Gϕ̂²πrh_0−(3/2)Gϕ̂²πh_0²=0. Simplifying this gives ΔFr²=(1/6)Gϕ̂²h_0[8r−9h_0], where h_0 is the h_max." }, { "idx": 2935, "question": "The dislocation that cannot undergo climb motion is (A) Shockley partial dislocation (B) Frank partial dislocation (C) edge perfect dislocation", "answer": "A" }, { "idx": 2936, "question": "When a deformed material is reheated and undergoes recovery and recrystallization phenomena, the significant decrease in point defect concentration occurs during (A) recovery stage (B) recrystallization stage (C) grain growth stage", "answer": "A" }, { "idx": 2937, "question": "A sample composed of three components is analyzed by X-ray diffraction (XRD) in air to study its phase transformation with temperature changes. The maximum number of coexisting phases that can be recorded is .", "answer": "C" }, { "idx": 2938, "question": "After welding A and A-B alloy, the Kirkendall effect occurs, and the interface is measured to move towards the A sample. Then (A) The diffusion rate of component A is greater than that of component B (B) The opposite of (A) (C) The diffusion rates of components A and B are the same", "answer": "A" }, { "idx": 2939, "question": "The Peierls-Nabarro force for dislocation slip is smaller.", "answer": "A" }, { "idx": 2940, "question": "The fundamental reason for diffusion to occur in materials is (A) temperature change (B) concentration gradient (C) chemical potential gradient", "answer": "C" }, { "idx": 2941, "question": "The relationship between the probability of annealing twin formation and the stacking fault energy of the crystal is (A) unrelated, only dependent on the annealing temperature and time (B) crystals with low stacking fault energy have a higher probability of forming annealing twins (C) crystals with high stacking fault energy have a higher probability of forming annealing twins", "answer": "B" }, { "idx": 2942, "question": "During the nucleation stage of solidification, only when the radius of the nucleus equals or exceeds the critical size can it become a crystallization core. When the formed nucleus has a radius equal to the critical size, the free energy change of the system is", "answer": "A" }, { "idx": 2943, "question": "A factory used 9Mn2V steel to make plastic molds, requiring a hardness of 58~63 HRC. After oil quenching at 790°C and tempering at 200~220°C, brittle fractures often occurred during use. Later, the process was changed to heating at 790°C followed by isothermal treatment in a nitrite salt bath at 260~280°C for 4 hours and then air cooling. Although the hardness decreased to 50 HRC, the service life was significantly improved. Please analyze the reason.", "answer": "After quenching and low-temperature tempering, the 9Mn2V steel mainly obtains a tempered structure of plate martensite. Since the substructure of plate martensite consists of twins and microcracks exist during its formation, the brittleness is relatively high. Isothermal quenching in nitrite salt produces lower bainite, whose matrix ferrite has a substructure of high-density dislocations and no microcracks, resulting in significantly reduced brittleness." }, { "idx": 2944, "question": "In a binary system composed of A and B, when α and β phases are in equilibrium, the relationship between composition (x) and free energy (G) for the two components is (A) $G^{\\\\alpha}=G^{\\\\beta}$ (B) $\\\\mathrm{d}G^{\\\\circ}/\\\\mathrm{d}x=\\\\mathrm{d}G^{\\\\beta}/\\\\mathrm{d}x$ (C) $G_{\\\\tt A}=G_{\\\\tt B}$", "answer": "B" }, { "idx": 2945, "question": "The relationship between the growth velocity (vg) of a certain crystal during solidification and the dynamic undercooling (ΔTx) at the liquid-solid interface front is vg proportional to ΔTx^2. This crystal belongs to which growth mechanism? (A) Continuous growth (B) Growth via screw dislocations (C) Two-dimensional nucleation", "answer": "B" }, { "idx": 2946, "question": "When most of the latent heat of crystallization can be dissipated through the liquid phase during ingot solidification, the main solid microstructure is (A) dendritic crystal (B) columnar crystal (C) spherulite", "answer": "A" }, { "idx": 2947, "question": "The dislocation that can undergo cross-slip must be (A) edge dislocation (B) screw dislocation (C) mixed dislocation", "answer": "B" }, { "idx": 2948, "question": "In ionic crystals, the diffusion rate of cations is (A) faster than that of anions (B) slower than that of anions (C) neither A nor B is correct", "answer": "A" }, { "idx": 2949, "question": "In an fcc crystal, there is a perfect dislocation with a Burgers vector of 1/2[011] on the (111) plane. When it decomposes into an extended dislocation, the leading dislocation is 1/6[121]. Determine the possible dislocation decomposition reaction and evaluate the possibility of decomposition using the structural condition and energy condition.", "answer": "The possible dislocation reaction on the (111) plane is 1/2[011](b_{4})→1/6[\\\\overline{1}21](b_{5})+1/6[112](b_{6}). Energy condition: b_{4}^{2}=1/2, b_{5}^{2}+b_{6}^{2}=1/3. b_{4}^{2}>b_{5}^{2}+b_{6}^{2}, decomposition is feasible." }, { "idx": 2950, "question": "In a binary system composed of A and B, when α and β phases are in equilibrium, the relationship between the composition (x) and free energy (G) of the two components is (A) $G^{\\\\alpha}=G^{\\\\beta}$ (B) $\\\\mathrm{d}G^{\\\\circ}/\\\\mathrm{d}x=\\\\mathrm{d}G^{\\\\beta}/\\\\mathrm{d}x$ (C) $G_{\\\\tt A}=G_{\\\\tt B}$", "answer": "B" }, { "idx": 2951, "question": "When two leading dislocations 1/6[2\\overline{1}\\overline{1}] and 1/6[\\overline{1}21] move on their respective slip planes and meet, a new dislocation reaction occurs. Determine the possible dislocation reaction.", "answer": "When b_{2} and b_{5} meet and react: b_{2}+b_{5}→1/6[110], forming a stair-rod dislocation." }, { "idx": 2952, "question": "When the crystal growth mechanism of Ge (germanium) follows the two-dimensional nucleation model, if the nuclei formed at the liquid-solid interface are cylindrical with each nucleus height h = 0.25 nm, calculate the critical nucleus diameter d*. (Given: melting point Tm = 1231 K, heat of fusion = 750000 kJ/m³, surface energy per unit area = 5.5×10⁻² J/m², undercooling during solidification ΔT = 0.01 Tm)", "answer": "ΔG = π(d/2)²hΔGv + πdhσ\\n\\n∂ΔG/∂d = 0 yields\\nπ(d/2)hΔGv + πhσ = 0\\nd* = -2σ/ΔGv\\nSince ΔGv = -LvΔT/Tm\\nThus, d* = 2σTm/(LvΔT)\\nSubstituting the values gives d* = 14.7 nm" }, { "idx": 2953, "question": "According to the relationship d²=kt and k=17.6μm²/min, find the grain diameter d after annealing for 90 minutes.", "answer": "d = √(17.6μm²/min × 90min) = √1584μm² = 39.8μm" }, { "idx": 2954, "question": "In an fcc crystal, there is a perfect dislocation with a Burgers vector of 1/2[110] on the (111) plane. When it decomposes into an extended dislocation, the leading dislocation is 1/6[211]. Determine the possible dislocation decomposition reaction and evaluate the possibility of decomposition using structural and energy conditions.", "answer": "The dislocation reaction on the (111) plane is 1/2 [1\\\\overline{1}0] (b_{1})→1/6[2\\\\overline{1}\\\\overline{1}](b_{2})+1/6[1\\\\overline{2}1](b_{3}). Energy condition: b_{1}^{2}=1/2, b_{2}^{2}+b_{3}^{2}=1/3. Since b_{1}^{2}>b_{2}^{2}+b_{3}^{2}, the decomposition is feasible." }, { "idx": 2955, "question": "Cesium chloride (CsCl) has an ordered body-centered cubic structure, which belongs to (A) body-centered cubic lattice (B) face-centered cubic lattice (C) simple cubic lattice", "answer": "C" }, { "idx": 2956, "question": "According to the relationship d²=kt and k=17.6μm²/min, find the grain diameter d after annealing for 60 minutes.", "answer": "d = √(17.6μm²/min × 60min) = √1056μm² = 32.5μm" }, { "idx": 2957, "question": "The main reason why it is difficult to form a single-phase structure in polymer alloys is that the alloy has (A) smaller mixing entropy (B) opposite to (A) (C) smaller mixing heat", "answer": "A" }, { "idx": 2958, "question": "Carbon atoms take 10 hours to diffuse into the surface of pure iron material at ${800}^{\\\\circ}\\\\mathrm{C}$ to a depth of $0.1\\\\mathfrak{c m}$. Calculate the time required to achieve the same carbon depth at $900^{\\\\circ}\\\\mathrm{C}$. (The diffusion activation energy of carbon atoms in fcc iron is $137520\\\\mathrm{J/mol}$.)", "answer": "Since $D{=}D_{0}\\\\exp(-Q/R T)$, substituting the values gives $D_{800}/D_{500}{=}0.27$. The diffusion depth $x^{2}/(4D t)=$ constant. Substituting the values gives $t_{900}=2.7\\\\mathrm{hr}$." }, { "idx": 2959, "question": "Explain the phenomenon of melting with increasing temperature in polymer crystals.", "answer": "Crystalline polymer materials exhibit a relatively wide melting temperature range (melting range) during the melting process. The reasons are as follows: (1) The crystallization rate of polymers is slow, and the usual cooling rate cannot ensure sufficient diffusion of chain segments in the polymer to form well-developed crystals; (2) When the temperature increases, less perfect crystals melt first at lower temperatures due to their thin lamellae and high defect content, while more perfect crystals melt within a higher temperature range. This results in the phenomenon of melting with increasing temperature; (3) Slower cooling rates during crystallization can narrow the melting range." }, { "idx": 2960, "question": "The [001] direction in a cubic crystal is", "answer": "B" }, { "idx": 2961, "question": "In the hexagonal crystal system, the interplanar spacing of $(11\\\\overline{2}0)$ is $(10\\\\overline{1}0)$ interplanar spacing.", "answer": "A" }, { "idx": 2962, "question": "In the seven polymorphic forms of $\\\\mathrm{{SiO}}_{2}$, there are two types of transformations: one is displacive transformation, and the other is reconstructive transformation. The activation energy required for displacive transformation is _____ that for reconstructive transformation. (A) greater than (B) less than (C) equal to", "answer": "B" }, { "idx": 2963, "question": "According to the relationship between grain diameter and annealing time d²=kt, given that the grain diameter is 23μm after annealing for 30 minutes, find the value of the constant k.", "answer": "k = (23μm)² / 30min = 529μm² / 30min = 17.6μm²/min" }, { "idx": 2964, "question": "The $c/a$ ratio for an ideal hexagonal close-packed metal is ", "answer": "B" }, { "idx": 2965, "question": "According to the Hall-Petch equation σ=σ₀+kd⁻¹/², given that when annealed for 30 minutes σ=112MPa, d=23μm, and when annealed for 60 minutes σ=103MPa, d=32.5μm, find the values of the constants σ₀ and k.", "answer": "Set up the system of equations: 112MPa = σ₀ + k×(23μm)⁻¹/² 103MPa = σ₀ + k×(32.5μm)⁻¹/² The solutions are: σ₀ = 55.3MPa, k = 272MPa·μm¹/²" }, { "idx": 2966, "question": "The twinning plane of a face-centered cubic crystal is", "answer": "C" }, { "idx": 2967, "question": "In a body-centered cubic structure with a lattice constant of $\\pmb{\\alpha}$, can a dislocation with a Burgers vector of a[100] decompose into $\\frac{a}{2}[111]+\\frac{a}{2}[1\\overline{{{1}}}\\overline{{{1}}}]$? (A) No (B) Yes (C) Possibly", "answer": "A" }, { "idx": 2968, "question": "In $\\\\mathrm{Ti}\\\\Omega_{2}$, when a portion of $\\\\mathrm{^{\\\\circ}Ti^{\\\\circ}}$ is reduced to $\\\\mathrm{Ti^{3+}}$, to balance the charge, there appears \\n\\n(A) oxygen ion vacancy (B) titanium ion vacancy (C) cation vacancy", "answer": "A" }, { "idx": 2969, "question": "Fick's first law describes the characteristics of steady-state diffusion, where the concentration does not vary with . \\n\\n(A) distance (B) time (C) temperature", "answer": "B" }, { "idx": 2970, "question": "From the symmetry of the repeating structural unit (mer) of polymers, the one that is most prone to crystallization is", "answer": "B" }, { "idx": 2971, "question": "According to the Hall-Petch formula σ=σ₀+kd⁻¹/², given σ₀=55.3MPa, k=272MPa·μm¹/², d=39.8μm, find the yield strength σ after annealing for 90 minutes.", "answer": "σ = 55.3MPa + 272MPa·μm¹/² × (39.8μm)⁻¹/² = 55.3MPa + 272MPa·μm¹/² × 0.1585μm⁻¹/² = 55.3MPa + 43.1MPa = 98.4MPa" }, { "idx": 2972, "question": "In the Kirkendall effect, the main reason for marker shift in a diffusion couple is", "answer": "C" }, { "idx": 2973, "question": "When forming a critical nucleus, the decrease in volume free energy can only compensate for the surface energy by", "answer": "B" }, { "idx": 2974, "question": "In crystals, the defect formed by creating vacancies while simultaneously generating interstitial atoms is called", "answer": "B" }, { "idx": 2975, "question": "In substitutional solid solutions, the mode of atomic diffusion is generally", "answer": "C" }, { "idx": 2976, "question": "The difference between cast iron and carbon steel lies in the presence or absence of (A) ledeburite (B) pearlite (C) ferrite", "answer": "A" }, { "idx": 2977, "question": "In binary phase diagrams, the lever rule for calculating the relative amounts of two phases can only be used in", "answer": "B" }, { "idx": 2978, "question": "Network polymers cannot exhibit", "answer": "C" }, { "idx": 2979, "question": "In the ionic compound $\\\\mathbf{MgO}$, the cation most likely to replace $\\\\mathbf{Mg}^{2+}$ in the compound (given the radii (nm) of each cation: (${\\\\bf Mg}^{2+}$) 0.066, ($\\\\mathbb{C}a^{2+}$) 0.099, ($\\\\mathrm{Li^{+}}$) 0.066, ($\\\\mathbf{Fe}^{\\\\mathbf{2+}}$) 0.074) is", "answer": "C" }, { "idx": 2980, "question": "In the concentration equilateral triangle of a ternary phase diagram, for alloys whose composition lies on , the content of the two components represented by the other two vertices is equal. (A) The perpendicular bisector passing through the vertex of the triangle (B) Any straight line passing through the vertex of the triangle (C) A straight line passing through the vertex of the triangle and forming a $45^{\\\\circ}$ angle with the opposite side", "answer": "A" }, { "idx": 2981, "question": "According to the Clausius-Clapeyron equation, as pressure increases, the temperature at which Y-Fe transforms into $\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}\\\\mathsf{\\\\Gamma}$", "answer": "B" }, { "idx": 2982, "question": "Given erf(0.71)=0.68, find the value of x/(2√(Dt)).", "answer": "x/(2√(Dt)) = 0.71." }, { "idx": 2983, "question": "Calculate the diffusion coefficient D of carbon at 920°C, given the activation energy for diffusion is 133984 J/mol and D0=0.23 cm²/s.", "answer": "D = D0 × exp(-Q/(RT)) = 0.23 × exp(-133984/(8.314 × (273 + 920))) = 3.12 × 10^-7 cm²/s." }, { "idx": 2984, "question": "In polymer materials, what is the bond between molecules (secondary bond)?", "answer": "Molecular bond" }, { "idx": 2985, "question": "Polymer materials are", "answer": "Materials with polymer compounds as the main component" }, { "idx": 2986, "question": "Calculate the carburizing time t, given x=0.2 cm, D=3.12 × 10^-7 cm²/s.", "answer": "t = (x/(2 × 0.71))² / D = (0.2/(2 × 0.71))² / (3.12 × 10^-7) = 17.7 hours." }, { "idx": 2987, "question": "What are the two synthesis methods of polymer materials", "answer": "Addition polymerization and condensation polymerization" }, { "idx": 2988, "question": "Using the carburization equation to calculate the value of erf(x/(2√(Dt))), given the surface carbon content ρs=1.2%, initial carbon content ρ0=0.1%, and target carbon content ρ(x,t)=0.45%.", "answer": "erf(x/(2√(Dt))) = (ρs - ρ(x,t))/(ρs - ρ0) = (1.2 - 0.45)/(1.2 - 0.1) = 0.682." }, { "idx": 2989, "question": "Into what categories can polymer materials be classified by application", "answer": "Plastics, rubber, fibers, adhesives, coatings" }, { "idx": 2990, "question": "In polymer materials, what is the atomic bonding between molecules (primary valence force)?", "answer": "Covalent bond" }, { "idx": 2991, "question": "Thermosetting plastics mainly have (5) molecular chains.", "answer": "(5) three-dimensional network" }, { "idx": 2992, "question": "Due to the very long molecular chains, how is the relationship between secondary valence forces and primary valence forces often?", "answer": "Greater than" }, { "idx": 2993, "question": "The geometric shapes of macromolecular chains are (1), (2), (3).", "answer": "(1) Linear; (2) Branched; (3) Network" }, { "idx": 2994, "question": "Thermoplastics mainly have (4) molecular chains.", "answer": "(4) linear" }, { "idx": 2995, "question": "The basic motion units corresponding to the three mechanical states of linear amorphous polymers are (4), (5), (6)", "answer": "(4) chain segments or side groups; (5) chain segments; (6) macromolecular chains" }, { "idx": 2996, "question": "The three mechanical states of linear amorphous polymers are (1), (2), (3)", "answer": "(1) Glassy state; (2) High elastic state; (3) Viscous flow state" }, { "idx": 2997, "question": "The aging of polymer materials involves structural changes such as (1) and (2).", "answer": "(1) cross-linking; (2) scission" }, { "idx": 2998, "question": "The three mechanical states corresponding to linear amorphous polymers are the usage states of (7), (8), and (9)", "answer": "(7) plastic; (8) rubber; (9) flow resin" }, { "idx": 2999, "question": "What are the mechanical states of high polymers with larger molecular weight that are not completely crystalline?", "answer": "(1) Glassy state; (2) High elastic state; (3) Leathery state; (4) Viscous flow state" }, { "idx": 3000, "question": "What are the usage states corresponding to these mechanical states?", "answer": "(5)Plastic; (6)Rubber; (7)Flow resin" }, { "idx": 3001, "question": "What usually breaks first when a polymer material fractures under force?", "answer": "Covalent bonds" }, { "idx": 3002, "question": "The five major engineering plastics are (1), (2), (3), (4), (5).", "answer": "(1)Polyoxymethylene; (2)Polyamide; (3)Polycarbonate; (4)ABS; (5)Polysulfone" }, { "idx": 3003, "question": "Rubber exhibits (1) physical state at -70℃.", "answer": "(1)glass transition" }, { "idx": 3004, "question": "When a factory used a nylon rope sling that had been in stock for two years, a fracture accident occurred even though the load-bearing capacity was far greater than the lifting stress. The reason for the fracture was (1).", "answer": "(1)aging" }, { "idx": 3005, "question": "The mechanical properties of polymers are mainly determined by their degree of polymerization, crystallinity, and intermolecular forces.", "answer": "√" }, { "idx": 3006, "question": "Linear polymers with high molecular weight exhibit glassy (or crystalline) state, high elastic state, and viscous flow state. Three-dimensional polymers with high crosslinking density do not possess high elasticity or viscous flow state.", "answer": "√" }, { "idx": 3007, "question": "ABS plastic is an engineering material with excellent comprehensive properties.", "answer": "√" }, { "idx": 3008, "question": "Engineering polymers can be divided into five major types based on performance: (1), for example: (2) belongs to the (3) category, for example (4) belongs to the (5) category, for example (6) belongs to the (7) category, for example (8) belongs to the (9) category, for example (10) belongs to the (11) category.", "answer": "(1)plastics, rubber, fibers, adhesives, coatings; (2)polyamide; (3)plastics; (4)styrene-butadiene rubber; (5)rubber; (6)polyester fiber; (7)fibers; (8)epoxy adhesive; (9)adhesives; (10)phenolic resin coating; (11)coatings" }, { "idx": 3009, "question": "Silicate glass, mica, and asbestos belong to which category of compounds? Options: A. Carbon-chain organic polymers B. Heterochain organic polymers C. Elemental organic compounds D. Inorganic polymers", "answer": "D" }, { "idx": 3010, "question": "To which category of compounds do silicone resins and silicone rubbers belong? Options: A. Carbon-chain organic polymers B. Heterochain organic polymers C. Element-organic compounds D. Inorganic polymers", "answer": "C" }, { "idx": 3011, "question": "To which category of compounds do nylon and polysulfone belong? Options: A. Carbon-chain organic polymers B. Heterochain organic polymers C. Elemental organic compounds D. Inorganic polymers", "answer": "B" }, { "idx": 3012, "question": "To which category of compounds do Teflon (polytetrafluoroethylene) and Plexiglas (polymethyl methacrylate) belong? Options: A. Carbon-chain organic polymers B. Heterochain organic polymers C. Elemental organic compounds D. Inorganic polymers", "answer": "A" }, { "idx": 3013, "question": "The form of polymer with the lowest expansion coefficient is (1). A. Linear B. Branched C. Crosslinked", "answer": "(1)C" }, { "idx": 3014, "question": "The usage state of synthetic fibers is (). A. Crystalline state B. Glassy state C. High elastic state D. Viscous flow state", "answer": "B" }, { "idx": 3015, "question": "The elasticity of a polymer is related to which temperature? A. $T_{\\\\mathrm{m}}$ B. $T_{_{8}}$ C. $T_{t}$ D.Ta", "answer": "C" }, { "idx": 3016, "question": "The usage state of plastic is (). A. Crystalline state B. Glassy state C. High elastic state D. Viscous flow state", "answer": "B" }, { "idx": 3017, "question": "The close-packed {111} planes of the face-centered cubic (fcc) structure are stacked in the sequence ABCABC..., while the close-packed {0001} planes of the hexagonal close-packed (hcp) structure are stacked in the sequence ABABAB.... Explain how and by introducing what type of dislocations the fcc structure can be entirely transformed into the hcp structure.", "answer": "Accordingly, if the C-layer atoms are moved to the E position (or E2, E positions, all referring to projection positions), the arrangement of the C atomic layer can be changed to that of the A atomic layer. If the B atomic layer is correspondingly moved at this time, it will change to the arrangement of the C atomic layer, and the A-layer atoms will change to the arrangement of the B layer. Therefore, introducing a partial dislocation of $\\frac{a}{6}(11\\overline{2})$ (or $\\frac{a}{6}(1\\overline{2}1)$, or $\\frac{a}{6}(211)$) into the second layer of the ABCABC... stacking and allowing it to sweep through the third layer and subsequent layers can change the atomic arrangement to ABABCABC.... Then, introducing the same dislocation into the fourth layer and performing the same operation can change the atomic arrangement to ABABABCABC..., and so on, resulting in the ABABAB... stacking sequence. In summary, by introducing partial dislocations of $\\frac{a}{6}(11\\overline{2})$ (or $\\frac{a}{6}(1\\overline{2}1)$, or $\\frac{a}{6}(211)$) into every second layer (111) plane (where n = 1, 2, 3,...) of the fcc structure's ABCABC... stacking, i.e., introducing a partial dislocation every other layer in the (111) plane, the stacking sequence can be changed to ABABAB..., thereby transforming it into the hcp stacking structure." }, { "idx": 3018, "question": "When a polymer material is subjected to force, the elasticity achieved by the elongation of bond length is (1). A. Ordinary elasticity B. High elasticity C. Viscoelasticity D. Forced elasticity", "answer": "C" }, { "idx": 3019, "question": "The one that is easier to obtain a crystalline structure is (1). B. Branched molecules A. Linear molecules C. Three-dimensional molecules", "answer": "(1)A" }, { "idx": 3020, "question": "If crystalline regions exist in polymer materials, then their melting point is (1). A. Fixed B. A temperature softening range C. Above the glass transition temperature D. Above the flow temperature", "answer": "(1)C" }, { "idx": 3021, "question": "The plasticity of a polymer is related to which temperature? A. $T_{\\mathrm{m}}$ B. $T_{_{8}}$ C. $T_{t}$ D.Ta", "answer": "D" }, { "idx": 3022, "question": "Comparison of mechanical properties, the (1) of polymers is better than that of metal materials. A. Stiffness B. Strength C. Impact strength (toughness) D. Specific strength", "answer": "(1)D" }, { "idx": 3023, "question": "When a polymer material is subjected to force, the elasticity achieved by the movement of chain segments is (2). A. Ordinary elasticity B. High elasticity C. Viscoelasticity D. Forced elasticity", "answer": "B" }, { "idx": 3024, "question": "To improve the strength of polymers, how many 'alloying' methods are there?", "answer": "Physical modification: mainly involves adding fillers to alter the physical and mechanical properties of polymers. For example, after treatment, asbestos used as a filler for polypropylene can increase tensile strength by 60% and flexural strength by 100%. Chemical modification: through copolymerization, block, grafting, blending, and compounding methods, polymers can acquire new properties." }, { "idx": 3025, "question": "What is the aging of polymers?", "answer": "The phenomenon where polymer materials gradually degrade in performance and eventually lose their use value due to external factors such as oxygen, light, heat, mechanical force, water vapor, and microorganisms during long-term storage and use is called aging. The fundamental cause of aging is the cross-linking and scission of polymer molecular chains under the influence of external factors." }, { "idx": 3026, "question": "Explain the role of crosslinking and how it alters the structure and properties of polymers.", "answer": "The so-called crosslinking reaction refers to the process in which polymers, under external factors, transform from a linear structure to a three-dimensional network structure, leading to increased strength, brittleness, and improved chemical stability. The crosslinking reaction makes polymer materials harder, more brittle, and even prone to cracking." }, { "idx": 3027, "question": "What are the steps involved in the production process of ceramic materials?", "answer": "(4) Preparation of raw materials; (5) Forming of the green body; (6) Sintering of the product" }, { "idx": 3028, "question": "How to prevent the aging of polymers?", "answer": "Currently, there are three measures taken to prevent aging.\\n① Modify the structure of the polymer. For example, chlorinating polyvinyl chloride can improve its thermal stability.\\n② Add anti-aging agents. Incorporating organic compounds such as salicylate esters, benzophenones, and carbon black into polymers can prevent photo-oxidation.\\n③ Surface treatment. Coating the surface of polymer materials with metals (such as silver, copper, nickel) or spraying anti-aging coatings (such as paint, paraffin) as protective layers isolates the material from air, light, water, and other aging-inducing media to prevent aging." }, { "idx": 3029, "question": "What are the advantages and disadvantages of parts made entirely of plastic?", "answer": "Advantages: Good chemical stability, excellent corrosion resistance, high specific strength, good friction and wear properties, good insulation. Disadvantages: Low strength and elastic modulus, low hardness, cold flow phenomenon, poor heat resistance, large expansion coefficient." }, { "idx": 3030, "question": "To improve the toughness of polymers, how many 'alloying' methods are there?", "answer": "Chemical modification; methods such as copolymerization, block, grafting, blending, and compounding are used to endow polymers with new properties. For example, ABS plastic is a type of plastic with excellent comprehensive performance obtained through ternary copolymerization." }, { "idx": 3031, "question": "What are the three categories of ceramic materials?", "answer": "(1)Glass; (2)Ceramics; (3)Glass-ceramics" }, { "idx": 3032, "question": "When designing plastic parts, compared to metals, list four limiting factors.", "answer": "Low strength, low hardness, low elastic modulus, cold flow phenomenon, poor heat resistance, large expansion coefficient." }, { "idx": 3033, "question": "The Tg of the glass phase in ceramics is (5).", "answer": "(5)Glass transition temperature" }, { "idx": 3034, "question": "The Tf of the glass phase in ceramics is (6).", "answer": "(6) viscous flow temperature" }, { "idx": 3035, "question": "Rubber is an excellent damping material and friction material due to its outstanding (1). A. High elasticity B. Viscoelasticity C. Plastic D. Anti-friction", "answer": "(1)A" }, { "idx": 3036, "question": "The roles of the glass phase in ceramics are (1), (2), (3), and (4).", "answer": "(1) bonding dispersed crystalline phases; (2) reducing sintering temperature; (3) inhibiting grain growth; (4) filling pores" }, { "idx": 3037, "question": "The basic raw materials of traditional ceramics are (1), (2), and (3), and its structure is composed of (4), (5), and (6).", "answer": "(1) clay; (2) quartz; (3) feldspar; (4) crystalline phase; (5) glass phase; (6) gas phase" }, { "idx": 3038, "question": "The structure of glass is an irregular network formed by silicon-oxygen tetrahedra in space.", "answer": "√" }, { "idx": 3039, "question": "What type of alloy is YT30?", "answer": "Tungsten-cobalt-titanium alloy" }, { "idx": 3040, "question": "Which compounds can be used to prepare high-temperature ceramics?", "answer": "(1) Oxides; (2) Carbides; (3) Nitrides; (4) Borides; (5) Silicides" }, { "idx": 3041, "question": "What are the main components of YT30 alloy?", "answer": "WC, TiC, and Co" }, { "idx": 3042, "question": "What can YT30 alloy be used to make?", "answer": "Cutting tool edges" }, { "idx": 3043, "question": "Oxide ceramics have a close-packed structure and exhibit high melting points and chemical stability due to strong ionic bonds.", "answer": "√" }, { "idx": 3044, "question": "What are the main bonding types of these compounds?", "answer": "(6) Covalent bond; (7) Ionic bond" }, { "idx": 3045, "question": "Ceramic materials can be used as high-temperature materials and also as wear-resistant materials.", "answer": "√" }, { "idx": 3046, "question": "What type of chemical bond is primarily found in nitrides? A. Metallic bond B. Covalent bond C. Molecular bond D. Ionic bond", "answer": "B" }, { "idx": 3047, "question": "Ceramic materials can be used as cutting tool materials, and can also be used as thermal insulation materials.", "answer": "√" }, { "idx": 3048, "question": "What type of chemical bond is primarily found in oxides? A. Metallic bond B. Covalent bond C. Molecular bond D. Ionic bond", "answer": "D" }, { "idx": 3049, "question": "Ceramic materials have low tensile strength but high compressive strength.", "answer": "√" }, { "idx": 3050, "question": "The porosity of special ceramics is (4). A.5%~10% B.<5% C.<0.5% D.>10%", "answer": "(4)C" }, { "idx": 3051, "question": "The porosity of cermet is (1). A.5%~10% B.<5% C.<0.5% D.>10%", "answer": "(1)D" }, { "idx": 3052, "question": "The porosity of ordinary ceramics is (2). A.5%~10% B.<5% C.<0.5% D.>10%", "answer": "(2)A" }, { "idx": 3053, "question": "The porosity of thermal insulation material is (3). A.5%~10% B.<5% C.<0.5% D.>10%", "answer": "(3)D" }, { "idx": 3054, "question": "What are the main types of special ceramics? Options: A. Cement B. Alumina C. Silicon carbide D. Boron nitride E. Refractory materials F. Household ceramics G. Silicon nitride H. Glass", "answer": "B,C,D,E,G" }, { "idx": 3055, "question": "What can alumina ceramic be used for? Options: A. Grinding wheel B. Blade C. Cutting tool D. Abrasive E. Crucible", "answer": "C,E" }, { "idx": 3056, "question": "What can SiC ceramics be used for? Options: A. Grinding wheel B. Blade C. Cutting tool D. Abrasive E. Crucible", "answer": "B" }, { "idx": 3057, "question": "What can silicon nitride ceramics be used for? Options: A. Grinding wheel B. Blade C. Cutting tool D. Abrasive E. Crucible", "answer": "A,D" }, { "idx": 3058, "question": "What are included in traditional ceramics? Options: A. Cement B. Alumina C. Silicon carbide D. Boron nitride E. Refractory materials F. Household ceramics G. Silicon nitride H. Glass", "answer": "A,F,H" }, { "idx": 3059, "question": "What material is suitable for a thermocouple sheath? Options: A.Polyvinyl chloride B.2Cr13 C.High-temperature ceramics D.Manganese brass", "answer": "C" }, { "idx": 3060, "question": "What type of chemical bond is primarily found in carbides? A. Metallic bond B. Covalent bond C. Molecular bond D. Ionic bond", "answer": "B" }, { "idx": 3061, "question": "What material is suitable for the handle of a test pen? Options: A.Polyvinyl chloride B.2Cr13 C.High-temperature ceramic D.Manganese brass", "answer": "A" }, { "idx": 3062, "question": "What material is suitable for steam turbine blades? Options: A.Polyvinyl chloride B.2Cr13 C.High-temperature ceramics D.Manganese brass", "answer": "B" }, { "idx": 3063, "question": "In which fields can ceramic materials be applied?", "answer": "Applied in insulation, wear resistance, corrosion resistance, and high-temperature resistant parts." }, { "idx": 3064, "question": "What is special ceramics?", "answer": "To improve the performance of ordinary ceramics, it was found that the impurities brought by natural raw materials were quite unfavorable. Therefore, high-purity artificially synthesized raw materials were adopted, and the forming and sintering processes of ordinary ceramics were followed to produce new types of ceramics. These ceramics are called special ceramics, such as oxide ceramics and piezoelectric ceramics." }, { "idx": 3065, "question": "What are the similarities and differences in composition between traditional ceramics and special ceramics?", "answer": "Similarities and differences in composition: a) Traditional ceramics are made from clay, feldspar, and quartz as raw materials. b) Special ceramics: They are formed and sintered using traditional processes with artificially synthesized raw materials (various compounds with no or fewer impurities, such as oxides, nitrides, carbides). Their compositional characteristic is fewer impurities compared to traditional ceramics, and since artificially synthesized powders are used, the composition can be adjusted." }, { "idx": 3066, "question": "Why are ceramic materials brittle?", "answer": "The brittleness of ceramic materials is influenced by both microstructural factors and macroscopic organization. From a microstructural perspective, the bonding in ceramic materials is ionic. If relative displacement occurs in ionic crystals, the electrical balance is lost, leading to the destruction of the ionic bonds, which is why materials bonded by ionic bonds are brittle. From a macroscopic organizational standpoint, ceramic materials contain a large number of pores (5%~10%), which reduce the load-bearing area when the ceramic is under stress, with pores being particularly prone to stress concentration. In summary, ceramic materials are brittle." }, { "idx": 3067, "question": "What are the characteristics of ceramic materials?", "answer": "Main performance characteristics: high hardness, high wear resistance, high elastic modulus, relatively high compressive strength, excellent high-temperature strength, good insulation, excellent high-temperature oxidation resistance, and good corrosion resistance. The disadvantages are high brittleness and low impact toughness. Solving the problem of ceramic toughening is the key to its application in the mechanical industry." }, { "idx": 3068, "question": "What is traditionally referred to as 'ceramics'?", "answer": "Ordinary ceramics are inorganic polycrystalline solid materials obtained by processing natural silicate minerals (such as clay, feldspar, and quartz) through raw material processing—forming—sintering. Therefore, this type of ceramics is also called silicate ceramics." }, { "idx": 3069, "question": "Why is the tensile strength of ceramics often much lower than the theoretical strength?", "answer": "Due to the splitting effect of pores and stress concentration during tension, the tensile strength of ceramics is lower than the theoretical strength." }, { "idx": 3070, "question": "What is reaction sintering?", "answer": "Reaction sintering is a process in which ceramic powders of various compounds (such as Si, Si-SiN4 powder, etc.) are pressed into shape and then subjected to special chemical treatment during sintering to obtain ceramics." }, { "idx": 3071, "question": "What is wood composed of?", "answer": "(1) Cellulose; (2) Lignin" }, { "idx": 3072, "question": "What is hot pressing sintering?", "answer": "Hot pressing sintering is a method that uses ceramic powders of various compounds as raw materials, adds a small amount of additives, loads them into graphite molds, and sinters and forms under high temperature and high pressure." }, { "idx": 3073, "question": "What is gray cast iron composed of?", "answer": "(3) steel matrix; (4) graphite" }, { "idx": 3074, "question": "What are the advantages and disadvantages of reaction sintering?", "answer": "The porosity of reaction-sintered ceramics is as high as 20%~30%, so their strength is not as good as that of hot-pressed sintered ceramics. However, reaction-sintered ceramics can often be machined during chemical processing, making them suitable for producing heat-resistant, wear-resistant, corrosion-resistant insulating products with complex shapes and high dimensional accuracy." }, { "idx": 3075, "question": "What are the advantages and disadvantages of hot pressing sintering?", "answer": "Hot pressing sintered ceramics are limited by the mold shape and can only process simple-shaped wear-resistant and high-temperature-resistant products (e.g., cutting tools). However, hot pressing sintered ceramics have higher strength, are dense, and have extremely low porosity." }, { "idx": 3076, "question": "In fiber composites, what should the length of carbon fiber be", "answer": "Tall" }, { "idx": 3077, "question": "In fiber-reinforced composites, the fibers with relatively good performance are mainly (1), (2), (3), (4).", "answer": "(1)glass fiber; (2)carbon fiber; (3)boron fiber; (4)silicon carbide fiber" }, { "idx": 3078, "question": "In fiber composite materials, how should the diameter of carbon fibers be", "answer": "Small" }, { "idx": 3079, "question": "In fiber composites, what should be the range of carbon fiber volume content", "answer": "Higher" }, { "idx": 3080, "question": "When designing fiber-reinforced composites, for a matrix with lower toughness, the coefficient of expansion of the fiber can be what? A. Slightly lower B. Very different C. Slightly higher D. The same", "answer": "C" }, { "idx": 3081, "question": "Tungsten-cobalt carbide is a composite material of (3) and (4)", "answer": "(3)WC; (4)Co" }, { "idx": 3082, "question": "In fiber-reinforced composites, the smaller the fiber diameter, the better the fiber reinforcement effect.", "answer": "√" }, { "idx": 3083, "question": "What materials can be used to manufacture the vehicle body itself? Options: A. Carbon fiber resin composite material B. Thermosetting fiberglass C. Boron fiber resin composite material", "answer": "B" }, { "idx": 3084, "question": "Fiberglass is a composite material of (1) and (2)", "answer": "(1) Resin; (2) Glass fiber" }, { "idx": 3085, "question": "The reinforcing effect is best when the diameter of the fine particle phase in fine-grained composites is (1). A. $<0.01~\\\\mu\\\\mathrm{m}$ B.0.01\\\\~0.1-pm $\\\\mathrm{C.}>0.1\\\\mu m$", "answer": "(1)B" }, { "idx": 3086, "question": "In order to achieve high strength in composite materials, the elastic modulus of the fibers must be very high.", "answer": "√" }, { "idx": 3087, "question": "When designing fiber composites, for a matrix with good plasticity, how can the coefficient of expansion of carbon fiber be? A. Slightly lower B. Very different C. Slightly higher D. The same", "answer": "A" }, { "idx": 3088, "question": "What materials can be used to manufacture rocket supports? Options: A. Carbon fiber resin composite B. Thermosetting fiberglass C. Boron fiber resin composite", "answer": "C" }, { "idx": 3089, "question": "What are the commonly used reinforcing fibers?", "answer": "Glass fiber, carbon fiber, boron fiber, aramid fiber, silicon carbide fiber" }, { "idx": 3090, "question": "What types can composites be divided into according to the category of matrix materials?", "answer": "Non-metal matrix composites and metal matrix composites." }, { "idx": 3091, "question": "What materials can be used to manufacture helicopter propeller blades? Options: A. Carbon fiber resin composite B. Thermosetting fiberglass C. Boron fiber resin composite", "answer": "A" }, { "idx": 3092, "question": "What is the reinforcement mechanism of fiber-reinforced composites?", "answer": "Fiber-reinforced composites are formed by combining high-strength, high-modulus continuous (long) fibers or discontinuous (short) fibers with a matrix (resin, metal, ceramic, etc.). When the composite material is subjected to force, the high-strength, high-modulus reinforcing fibers bear most of the load, while the matrix primarily acts as a medium to transfer and disperse the load." }, { "idx": 3093, "question": "What types can composites be classified into based on the types of reinforcing materials?", "answer": "Fiber-reinforced composites, particle-reinforced composites, and laminated composites." }, { "idx": 3094, "question": "What are the factors affecting the widespread application of composite materials?", "answer": "Composite materials refer to multiphase materials artificially synthesized from two or more materials with different properties through various processing methods. Composites not only retain the optimal characteristics of their constituent materials but also exhibit new properties after combination. For example, the fracture energy of glass fibers is only 7.5×10^-2 J, while common resins have a fracture energy of about 2.26×10^-2 J. However, the composite material composed of glass fibers and thermosetting resins, known as thermosetting fiberglass, achieves a fracture energy as high as 17.6 J, with strength significantly higher than that of resins and brittleness much lower than that of glass fibers. It is evident that 'compositing' has become an important means to improve material performance. Therefore, composite materials are attracting increasing attention, and the development and application of new composite materials are becoming more widespread." }, { "idx": 3095, "question": "What are the performance characteristics of carbon fiber?", "answer": "Low density, high strength and modulus. It has good high and low temperature performance, high chemical stability; small thermal expansion coefficient, high thermal conductivity, good electrical conductivity and self-lubrication. Its disadvantages are high brittleness and susceptibility to oxidation." }, { "idx": 3096, "question": "What are the performance characteristics of fiberglass?", "answer": "The density is 2.4~2.7g/cm³, similar to aluminum. The tensile strength is dozens of times higher than that of bulk glass and even higher than that of bulk high-strength alloy steel. The elastic modulus is 5~8 times higher than that of other artificial fibers, and the elongation is lower than that of other organic fibers. It has relatively high heat resistance, good corrosion resistance, and excellent chemical stability to other solvents. It is easy to produce and inexpensive." }, { "idx": 3097, "question": "Through what means can the performance of composite materials be further improved and their application scope expanded?", "answer": "Composite materials refer to multiphase materials artificially synthesized from two or more materials with different properties through various process methods. Composite materials not only retain the optimal characteristics of their constituent materials but also possess new properties resulting from their combination. For example, the fracture energy of glass fiber is only 7.5×10^-2J, while that of commonly used resin is about 2.26×10^-2J. However, the composite material composed of glass fiber and thermosetting resin, known as thermosetting fiberglass, has a fracture energy as high as 17.6J, with its strength significantly higher than that of resin and its brittleness much lower than that of glass fiber. It is evident that 'compositing' has become an important means to improve material performance. Therefore, composite materials are attracting increasing attention, and the development and application of new composite materials are becoming more widespread." }, { "idx": 3098, "question": "What are the performance characteristics of boron fibers?", "answer": "Boron fibers have a high melting point, high strength, high elastic modulus, and good oxidation resistance and corrosion resistance. Their disadvantages include high density, large diameter, complex production process, high cost, and expensive price." }, { "idx": 3099, "question": "What are the performance characteristics of silicon carbide fibers?", "answer": "High melting point, high strength, high modulus ceramic fibers, mainly used to reinforce metals and ceramics. The outstanding feature is excellent high-temperature strength." }, { "idx": 3100, "question": "What are the performance characteristics of aramid fiber?", "answer": "Its most notable characteristics are high specific strength and specific modulus. It has low density, good toughness, better heat resistance than glass fiber, excellent fatigue resistance, corrosion resistance, insulation properties, and processability, and is inexpensive." }, { "idx": 3101, "question": "What is the reinforcement mechanism of particle-reinforced composites?", "answer": "The reinforcement mechanism of particle-reinforced composites is divided into two types: 1. Dispersion-strengthened composites are materials formed by dispersing and uniformly distributing particles (less than 0.1 micrometers) of one or several materials within the matrix material. The reinforcement mechanism of such composites is: under external forces, the matrix of the composite primarily bears the load, while the uniformly dispersed reinforcing particles hinder the movement of dislocations causing collective plastic deformation (e.g., the bypass mechanism in metal matrices) or molecular chain movement (in polymer matrices). The strengthening effect is related to particle diameter and volume fraction; the smaller the particle size and the higher the volume fraction, the better the strengthening effect. Typically, particle diameters range from 0.01 to 0.1 micrometers, and volume fractions range from 1% to 15%. 2. Particle-reinforced composites are materials formed by bonding metal oxides, carbides, or nitrides, which have good heat resistance and high hardness but poor impact resistance, using metals or polymer binders. These materials combine the advantages of high hardness and heat resistance of ceramics with the benefits of low brittleness and good impact resistance, demonstrating outstanding composite effects. Since the reinforcing particles are relatively large (greater than 1 micrometer), they do not significantly hinder dislocation slip (in metal matrices) or molecular chain movement (in polymer matrices), so the strengthening effect is not significant. Particle-reinforced composites are primarily used not to enhance strength but to improve wear resistance or overall mechanical properties." }, { "idx": 3102, "question": "Elastic alloys can be divided into two major categories: (1) and (2).", "answer": "(1) High elastic alloys; (2) Constant elastic alloys" }, { "idx": 3103, "question": "What is the development direction of materials science?", "answer": "In the 21st century, materials science will inevitably develop towards high functionality, ultra-high performance, complexity (compositing and complicating), refinement, ecological environmentalization, and intelligence, based on the rapid advancement of science and technology, thereby making greater contributions to the material civilization of human society." }, { "idx": 3104, "question": "Commonly used expansion materials are divided into three categories: (1), (2), (3).", "answer": "(1) Low expansion materials; (2) Controlled expansion materials; (3) High expansion materials" }, { "idx": 3105, "question": "High elastic alloys have (1). A. high elastic limit and high elastic modulus B. high elastic limit and low elastic modulus C. low elastic limit and high elastic modulus D. low elastic limit and low elastic modulus", "answer": "(1)B" }, { "idx": 3106, "question": "Below the critical temperature Tc, superconductors possess complete (1) what property?", "answer": "Electrical conductivity" }, { "idx": 3107, "question": "Below the critical temperature Tc, superconductors exhibit complete (2) what property?", "answer": "Diamagnetism" }, { "idx": 3108, "question": "Shape memory alloys utilize the characteristics of (1) and (2) of the material to achieve shape recovery.", "answer": "(1) superelasticity; (2) shape memory effect" }, { "idx": 3109, "question": "Resistance materials can be divided into (1), (2), and (3) according to their characteristics and applications.", "answer": "(1) Precision resistance materials; (2) Film resistance materials; (3) Electric heating materials" }, { "idx": 3110, "question": "Hydrogen storage alloys are alloys that form hydrogen-containing solid solutions by absorbing hydrogen, and under certain conditions, these alloys decompose to release hydrogen", "answer": "√" }, { "idx": 3111, "question": "The temperature coefficient of resistance (1) of resistive materials. A. The larger the better B. The smaller the better C. No requirement on size", "answer": "(1)C" }, { "idx": 3112, "question": "Shape memory alloy components undergo plastic deformation in the (1) state and recover their original shape after (2). A. Martensite Heating transforms into parent phase B. Parent phase Cooling transforms into martensite C. Martensite Staying for several days D. Parent phase Heating transforms into new parent phase", "answer": "(1)A" }, { "idx": 3113, "question": "The active layer and passive layer of a thermostatic bimetal strip are respectively (1). A. Constant expansion alloy and high expansion alloy B. High expansion alloy and constant expansion alloy C. Constant expansion alloy and low expansion alloy D. High expansion alloy and low expansion alloy", "answer": "(1)D" }, { "idx": 3114, "question": "What characteristics do high-elastic alloys have?", "answer": "These alloys require a high elastic limit σe and a low elastic modulus E, meaning a high σe/E ratio, which minimizes the elastic after-effect of components and ensures stable operation. Additionally, they must possess high fatigue strength and good workability." }, { "idx": 3115, "question": "What are the characteristics of future materials science research?", "answer": "Currently, due to the increasing demand for new materials, there is a desire to enhance theoretical predictability and reduce blind experimentation in materials development as much as possible. Objectively, the in-depth development of modern fundamental sciences such as physics and chemistry has provided many new principles and concepts. More importantly, advancements in computer information processing technology, as well as various material preparation and characterization techniques, have led to some new features in materials development and design. 1) In terms of the microscopic structure design of materials, the focus will shift from the microstructural level (-1μm) to the molecular and atomic levels (1~10nm) and the electronic level (0.1~1nm) (developing micrometer and nanometer materials). 2) The conceptual idea of mixing organic, inorganic, and metallic materials at the atomic and molecular levels to form so-called 'hybrid' (Hybrid) materials, exploring new pathways for synthetic materials. 3) In the development of new materials, based on databases and knowledge repositories, computers are used to predict the properties of new materials and simulate the relationship between the microscopic structure and properties of new materials. 4) In-depth research into the production processes of materials under various conditions, employing new thinking and adopting new technologies to develop new materials, such as the design of semiconductor superlattice materials, exemplified by 'bandgap engineering' or 'atomic engineering.' This involves controlling the electronic structure of materials to create multilayer heterostructure periodic materials composed of alternating ultrathin layers of different semiconductors, thereby greatly advancing the development of semiconductor lasers. 5) Selecting key objectives and organizing multidisciplinary efforts to jointly design certain new materials, such as the concept and practice of 'functionally graded' materials (FGM) proposed according to the requirements of aerospace thermal protection materials." }, { "idx": 3116, "question": "What are the categories of high elastic alloys?", "answer": "High elastic alloys can be divided into steel, copper alloys, nickel-based and cobalt-based alloys by composition." }, { "idx": 3117, "question": "What are the characteristics of constant expansion alloys?", "answer": "Materials with a certain coefficient of expansion within a specific temperature range are called constant expansion materials." }, { "idx": 3118, "question": "What is resistive material?", "answer": "Materials used in physical components that serve a resistive function in circuits are called resistive materials." }, { "idx": 3119, "question": "What is the main use of controlled expansion alloys?", "answer": "This type of material is used in the electric vacuum industry to seal with glass, ceramics, etc., requiring its expansion coefficient to match that of the sealed material." }, { "idx": 3120, "question": "What are the characteristics of resistance materials?", "answer": "The general requirements are high and stable resistance values, a small temperature coefficient of resistance, and sufficient mechanical strength, while also requiring a small thermoelectric potential with copper, good corrosion resistance, and ease of machining and welding." }, { "idx": 3121, "question": "What is the critical temperature Tc of a superconductor?", "answer": "The temperature at which the resistance abruptly drops to zero is called the superconducting transition temperature or critical temperature Tc." }, { "idx": 3122, "question": "What is a superconductor?", "answer": "Generally, the DC resistivity of metals decreases as temperature drops, and near absolute zero, the resistivity no longer continues to decline but approaches a finite value. However, the DC resistivity of certain conductors suddenly drops to zero at a specific low temperature, a phenomenon known as zero resistance or superconductivity. Objects exhibiting such superconducting properties are typically referred to as superconductors." }, { "idx": 3123, "question": "What are the main uses of superconductors?", "answer": "Superconductors can not only be used in superconducting generators, but also prove effective in applications such as superconducting motors, superconducting power transmission, superconducting energy storage, maglev trains, magnetohydrodynamic power generation, and nuclear fusion." }, { "idx": 3124, "question": "Explain the basic concept and terminology: Superstructure", "answer": "Superstructure: The lattice constant of an ordered solid solution differs from that of a disordered solid solution, resulting in additional diffraction lines on the X-ray diffraction pattern, known as superstructure lines. Therefore, an ordered solid solution is also referred to as a superstructure or superlattice." }, { "idx": 3125, "question": "Explain the basic concept and terminology: uphill diffusion", "answer": "Uphill diffusion: The diffusion of atoms from a region of low concentration to a region of high concentration is called uphill diffusion, and the driving force for uphill diffusion is the chemical potential gradient." }, { "idx": 3126, "question": "What is the shape memory mechanism of shape memory alloys?", "answer": "After heating and quenching, shape memory alloy materials obtain thermoelastic martensite. The interface between this martensite and the parent phase has good coherency, so the martensite phase continuously shrinks and grows during heating and cooling. When the temperature exceeds the As point, the martensite undergoes plastic transformation into the parent phase. If pressure is applied in the martensitic state, the martensite lattice orientation changes, resulting in deformation. This deformed martensite undergoes reversible phase transformation upon heating, reverting to the parent phase state, causing the entire lattice to restore its original form, thereby eliminating the deformation." }, { "idx": 3127, "question": "Explain the basic concept and terminology: symmetric tilt boundary", "answer": "Symmetric tilt boundary: Formed by a series of edge dislocations with parallel Burgers vectors and the same sign arranged vertically. The two sides of the boundary are symmetric, and the misorientation between the two grains is very small. It is the simplest type of low-angle grain boundary." }, { "idx": 3128, "question": "What is the shape memory mechanism of shape memory polymers?", "answer": "The shape memory mechanism of polymers is the radiation crosslinking reaction that occurs when polymers are exposed to high-energy radiation. When the temperature exceeds the melting point and enters the high elastic state region, the crystals melt, allowing the polymer's shape to be arbitrarily altered by applying external force. If the temperature is then cooled below the crystalline melting point, the polymer chains become 'frozen' due to recrystallization, fixing the shape. Once the temperature rises above the melting point again (or above the glass transition temperature for polymers like polyvinyl chloride), the polymer returns to its original shape, demonstrating the shape memory effect." }, { "idx": 3129, "question": "Explain the basic concept and terminology: dendritic segregation", "answer": "Dendritic segregation: During non-equilibrium solidification, the initially solidified solid solution contains more high-melting-point components, while the later solidified solid solution contains more low-melting-point components. This results in non-uniform chemical composition within the grain, known as intracrystalline segregation. Since solid solutions often grow in a dendritic manner, it is also called dendritic segregation." }, { "idx": 3130, "question": "Explain the basic concept and terminology: Twinning", "answer": "Twinning: One of the important forms of cold plastic deformation, occurring as a uniform shear within the crystal, always along certain crystallographic planes and directions. After deformation, the deformed part and the undeformed part exhibit a mirror-symmetric orientation relationship, hence this type of plastic deformation is called twinning." }, { "idx": 3131, "question": "Explain the basic concept and terminology: Modification treatment", "answer": "Modification treatment: Before casting metal or alloy, adding certain solid-phase substances that can promote heterogeneous nucleation or hinder the growth of crystal nuclei into the molten metal to refine the grains of the casting is called modification treatment." }, { "idx": 3132, "question": "Explain the basic concept and terminology: superplasticity", "answer": "Superplasticity: Under specific conditions, metallic materials can achieve exceptionally large elongation rates when stretched, sometimes even reaching 1000%. This property is called superplasticity. During superplastic deformation, the strain rate sensitivity index m is very high, with m≈0.5, whereas for ordinary metallic materials it is only 0.01~0.04." }, { "idx": 3133, "question": "Explain the basic concept and terminology: Dynamic recovery", "answer": "Dynamic recovery: During hot deformation of high stacking fault energy metallic materials, due to the narrow width of extended dislocations, they are prone to constriction and cross-slip. Therefore, high-temperature recovery occurs simultaneously during hot deformation, hence it is called dynamic recovery. It is the primary or sole softening mechanism for high stacking fault energy metallic materials during hot deformation." }, { "idx": 3134, "question": "What is constitutional supercooling?", "answer": "During the solidification process of a liquid alloy, the supercooling determined jointly by the actual temperature distribution and compositional changes is called constitutional supercooling." }, { "idx": 3135, "question": "Compare the roles of twinning and slip in the plastic deformation process", "answer": "The direct contribution of twinning to plastic deformation is not as significant as that of slip. However, since twinning alters the crystal orientation, it can enable slip systems originally in hard orientations to rotate into soft orientations and participate in slip. This is particularly important for hexagonal close-packed metals with fewer slip systems." }, { "idx": 3136, "question": "What is secondary recrystallization (abnormal growth)?", "answer": "After recrystallization is completed, continued heating or holding may lead to discontinuous grain growth, where the growth of most grains is inhibited, while a few grains grow rapidly, which is called abnormal growth, also known as secondary recrystallization." }, { "idx": 3137, "question": "Briefly explain the differences between twinning and slip in the plastic deformation process", "answer": "Twinning causes a uniform shear in a portion of the crystal, while slip is concentrated on the slip plane; twinning changes the orientation of a portion of the crystal, whereas slip does not alter the crystal orientation; the twinning elements are usually different from the slip systems; the critical shear stress for twinning is much higher than that for slip, the stress-strain curve for twinning is serrated, while that for slip is smooth; the deformation rate of twinning is far higher than that of slip." }, { "idx": 3138, "question": "Determine whether the dislocation reaction $\\frac{1}{2}[111]\\rightarrow\\frac{1}{8}[110]+\\frac{1}{4}[112]+\\frac{1}{8}[110]=0.$ 110] can proceed, and why?", "answer": "$$ \\Sigma b_{\\mathbb{H}}=\\Sigma b_{\\mathbb{E}}=\\frac{1}{2}[111]$$Thus, this dislocation reaction satisfies the geometric condition$$ x\\Sigma b_{j j}^{2}=\\frac{3}{4}>\\Sigma b_{j j}^{2}=\\frac{1+1}{64}+\\frac{1+1+4}{16}+\\frac{1+1}{64}=\\frac{7}{16} $$Therefore, it also satisfies the energy condition, so this dislocation reaction can proceed spontaneously." }, { "idx": 3139, "question": "It is known that the stacking fault energy of a certain stainless steel is very low, γ=0.013 J/m². It is cold-pressed with a reduction of 8%, followed by recrystallization annealing. Briefly describe its recrystallization nucleation mechanism.", "answer": "At a reduction of 8%, the nucleation mechanism is the bow-out nucleation mechanism. Due to the small deformation amount and non-uniform deformation, the dislocation density between adjacent grains varies significantly. A small segment of the remaining original grain boundary will suddenly bow out towards the side with higher dislocation density. The swept small region releases all stored energy, and this area can become the recrystallization nucleus." }, { "idx": 3140, "question": "What are the conditions for the occurrence of secondary recrystallization (abnormal growth)?", "answer": "The conditions for its occurrence are the presence of recrystallization texture, second-phase particles, or surface thermal grooves." }, { "idx": 3141, "question": "What is the effect of constitutional supercooling on the growth morphology of solid solutions?", "answer": "When the constitutional supercooling zone is large, the alloy grows in a dendritic manner; when the constitutional supercooling zone is small, the alloy grows in a cellular manner; when there is no constitutional supercooling zone, the alloy grows in a planar manner." }, { "idx": 3142, "question": "γ-Fe has a lattice constant α=0.3633nm slightly above 910 ℃, find the atomic radius of γ-Fe", "answer": "The atomic radius of γ-Fe r=√2/4×a=√2/4×0.3633=0.1284nm" }, { "idx": 3143, "question": "For α-Fe slightly below 910°C, the lattice constant α=0.2892nm, find the atomic radius of α-Fe", "answer": "The atomic radius of α-Fe r=√3/4×a=√3/4×0.2892=0.1252nm" }, { "idx": 3144, "question": "It is known that the stacking fault energy of a certain stainless steel is very low, γ=0.013 J/m². After cold pressing deformation with reductions of 8% and 60%, recrystallization annealing is performed. Explain the difference in grain size after recrystallization annealing between the two cases. Why?", "answer": "After recrystallization annealing, with a reduction of 8%, the deformation is near the critical deformation level. The small deformation amount results in low stored energy and a small driving force for recrystallization. Although recrystallization can occur, the nucleation rate is low, leading to very coarse grains. With a reduction of 60%, the stored energy is high, the driving force for recrystallization is large, and the nucleation rate is high, resulting in fine grains after recrystallization." }, { "idx": 3145, "question": "The volume change rate during the γ-Fe→α-Fe transformation (given the lattice constant of γ-Fe α=0.3633nm, and the lattice constant of α-Fe α=0.2892nm)", "answer": "The volume change rate ΔV=(Vα-Vγ)/Vγ=(0.2892³×1/2-0.3633³×1/4)/(0.3633³×1/4)≈0.886" }, { "idx": 3146, "question": "In has a tetragonal structure with lattice constants a=0.3252 nm, c=0.4946 nm, atomic radius r=0.1625 nm, and number of atoms per unit cell n=2. Calculate the packing density.", "answer": "The packing density K=(n×v)/V=(2×(4/3)π(0.1625)³)/(0.3252²×0.4946)≈0.687." }, { "idx": 3147, "question": "It is known that the stacking fault energy of a certain stainless steel is very low, γ=0.013 J/m². It is cold-pressed and deformed with a reduction of 60%, followed by recrystallization annealing. Briefly describe its recrystallization nucleation mechanism.", "answer": "When the reduction is 60%, the nucleation mechanism is the subgrain coalescence mechanism. The stacking fault energy γ of this material is very low, and the width of extended dislocations d is very small. Extended dislocations are prone to constriction and cross-slip, making it easy to achieve the coalescence of small subgrains through polygonization, the movement of 'Y' nodes, and the rotation of adjacent subgrains. As subgrains coalesce, their size increases, and the misorientation with surrounding subgrains also becomes larger, eventually forming high-angle grain boundaries with high mobility. After the high-angle grain boundaries sweep through, they leave behind strain-free crystals, forming recrystallization nuclei." }, { "idx": 3148, "question": "The lattice constant of a copper crystal is $a=0.362\\mathrm{nm},$ the density is $\\rho=8.98~\\mathrm{g/cm}^{3},$ and the relative atomic mass is 63.55. Determine the crystal structure of copper.", "answer": "From the physical meaning of density $\\rho=\\frac{6.023\\times10^{23}}{(0.362\\times10^{-7})^{3}}=8.98(g/\\mathrm{cm}^{3})$ n. Solving gives $n=3.997{\\approx}4.$ A cubic unit cell with 4 atoms corresponds to a face-centered cubic structure." }, { "idx": 3149, "question": "Given the density of amorphous polyethylene ρa=0.854 g/cm³, and the density of typical commercial low-density polyethylene ρ=0.920 g/cm³, calculate the volume fraction of the crystalline region φc", "answer": "The volume fraction of the crystalline region φc = (ρ - ρa) / (ρc - ρa) = (0.92 - 0.854) / (1.01 - 0.854) ≈ 42.3%" }, { "idx": 3150, "question": "Using analytical geometry to determine the angle between two crystal planes $h_{1}k_{1}l_{1}$ and $h_{2}k_{2}l_{2}$ in a cubic crystal", "answer": "Let the two crystal planes in the cubic system be $h_{1}k_{1}l_{1}$ and $h_{2}k_{2}l_{2}$, and the angle between them be $\\phi$, then $$ \\cos\\phi={\\frac{h_{1}h_{2}+k_{1}k_{2}+l_{1}l_{2}}{\\sqrt{h_{1}^{2}+k_{1}^{2}+l_{1}^{2}}\\cdot\\sqrt{h_{2}^{2}+k_{2}^{2}+l_{2}^{2}}}} $$ Therefore, $$ \\phi=\\operatorname{arccos}(\\cos\\phi) $$" }, { "idx": 3151, "question": "In has a tetragonal structure, with a relative atomic mass Ar=114.82, lattice constants a=0.3252 nm, c=0.4946 nm, atomic radius r=0.1625 nm, density ρ=7.286 g/cm³. Determine the number of atoms per unit cell of In.", "answer": "From the density expression ρ=(n×114.82)/((0.3252×10⁻⁷)²×0.4946×10⁻⁷×6.023×10²³)≈7.286, the number of atoms per unit cell is obtained as n≈1.999, n=2, which should be a body-centered tetragonal structure." }, { "idx": 3152, "question": "Using analytical geometry to determine the angle between two crystallographic directions [$u_{1}v_{1}w_{1}$] and [$u_{2}v_{2}w_{2}$] in a cubic crystal", "answer": "Let the two crystallographic directions in the cubic system be [$u_{1}v_{1}w_{1}$] and [$u_{2}v_{2}w_{2}$], and the angle between them be θ, then $$ \\\\cos\\\\theta={\\\\frac{\\\\left[u_{1}v_{1}w_{1}\\\\right]\\\\cdot\\\\left[u_{2}v_{2}w_{2}\\\\right]}{|\\\\left[u_{1}v_{1}w_{1}\\\\right]|\\\\cdot|\\\\left[u_{2}v_{2}w_{2}\\\\right]|}}={\\\\frac{u_{1}u_{2}+v_{1}v_{2}+w_{1}w_{2}}{{\\\\sqrt{u_{1}^{2}+v_{1}^{2}+w_{1}^{2}}}\\\\cdot{\\\\sqrt{u_{2}^{2}+v_{2}^{2}+w_{2}^{2}}}}} $$ Therefore $$ \\\\theta=\\\\operatorname{arccos}(\\\\cos\\\\theta) $$" }, { "idx": 3153, "question": "In stable ZrO2 material, cations form an fcc structure, and anions occupy tetrahedral interstitial sites. If 20 mol% CaO is added, calculate how many anions are needed for 100 cations?", "answer": "Since the amount of CaO added to ZrO2 is 20 mol%, the total charge number for 100 cations is 20×2 + 80×4 = 360. To maintain electrical neutrality, the required number of O2− anions is 360 ÷ 2 = 180." }, { "idx": 3154, "question": "In stable ZrO2 material, cations form an fcc structure, and anions occupy tetrahedral interstitial sites. If 20 mol% CaO is added, calculate the percentage of occupied tetrahedral interstitial sites.", "answer": "Zr4+ and Ca2+ cations occupy the face-centered cubic lattice sites. 100 cations can form 25 unit cells, with a total of 25×8=200 tetrahedral interstitial sites. Therefore, the percentage of occupied tetrahedral interstitial sites is 180÷200=90%." }, { "idx": 3155, "question": "Determine the crystallographic direction in the diamond structure defined by the points (0,0,0) and $(\\\\frac{3}{4},\\\\frac{3}{4},\\\\frac{1}{4})$, and find the crystallographic plane perpendicular to this direction.", "answer": "Subtract the coordinates of the two points, then reduce them to a set of coprime integers. The direction indices are [331]. Since it is a cubic system, the crystallographic plane with the same indices is perpendicular to the direction, i.e., (331) is perpendicular to [331]." }, { "idx": 3156, "question": "It is known that polyethylene belongs to the body-centered orthorhombic structure, with lattice constants a=0.740nm, b=0.493nm, c=0.253nm, and two molecular chains passing through a unit cell. Calculate the density ρc of fully crystalline polyethylene.", "answer": "Polyethylene (C2H4) is produced through addition polymerization, with its repeating unit being CH2-CH2. Considering one repeating unit as a lattice point, it can be determined that each unit cell contains two repeating units. Therefore, the density ρc of fully crystalline polyethylene can be calculated as ρc = 2(12×2 + 1×4) / (6.023×10^23) / (0.740×0.493×0.253)×10^-21 ≈ 1.01 g/cm³." }, { "idx": 3157, "question": "The density of $\\\\mathrm{CaF}_{2}$ is $\\\\rho=3.18~\\\\mathrm{g/cm}^{3}$, the relative atomic mass of $\\\\mathrm{Ca}$ is $A_{\\\\mathrm{r}}=40.08$, and the relative atomic mass of $\\\\mathrm{F}$ is $A_{\\\\mathrm{r}}=19.00$. Find the lattice constant $a$.", "answer": "From the structure of $\\\\mathrm{CaF}_{2}$, it is known that the unit cell contains 4 $\\\\mathrm{Ca}^{2+}$ ions and 8 $\\\\mathrm{F^{-}}$ ions. Therefore, the relationship between its density and the lattice constant is as follows: $$ x\\\\rho=\\\\frac{4\\\\times\\\\frac{40.08+19\\\\times2}{6.023\\\\times10^{23}}}{a^{3}}\\\\approx3.18 $$ Solving the above equation gives $a=0.546(\\\\mathrm{nm})$." }, { "idx": 3158, "question": "Given the density of amorphous polyethylene ρa=0.854 g/cm³, the density of typical commercial low-density polyethylene ρ=0.920 g/cm³, and the density of fully crystalline polyethylene ρc=1.01 g/cm³, calculate the mass fraction of the crystalline region wc", "answer": "The mass fraction of the crystalline region wc = ρc(ρ - ρa) / ρ(ρc - ρa) = 1.01×(0.92 - 0.854) / 0.92×(1.01 - 0.854) ≈ 46.6%" }, { "idx": 3159, "question": "What type of alloy phase does CuZn belong to, and what are its structural characteristics?", "answer": "CuZn is an electron compound, c13=3.12, with a body-centered cubic structure." }, { "idx": 3160, "question": "What type does the alloy phase Fe3C belong to, and what are its structural characteristics?", "answer": "Fe3C is a complex lattice interstitial compound with an orthorhombic crystal system. Its unit cell contains 12 iron atoms and 4 carbon atoms. The iron atoms are arranged in a close-packed manner, forming octahedrons, with carbon atoms located in the octahedral interstitial sites. The corner atoms of the octahedrons are shared by two adjacent octahedrons, and there is a certain orientation between the octahedrons." }, { "idx": 3161, "question": "What type of alloy phase is VC, and what are its structural characteristics?", "answer": "VC is an interstitial phase, where V occupies the lattice points of a face-centered cubic lattice, and C occupies all the octahedral interstitial sites. It belongs to the face-centered cubic lattice and has a NaCl-type structure." }, { "idx": 3162, "question": "What type of alloy phase is ZrFe2, and what are its structural characteristics?", "answer": "ZrFe2 is a Laves phase among the topologically close-packed phases, with a complex cubic structure. The smaller iron atoms form small tetrahedrons, and the vertices of these tetrahedrons are interconnected to form a network. The larger Zr atoms are located in the gaps between the small tetrahedrons formed by Fe atoms and themselves form a diamond structure. This structure only contains tetrahedral gaps, so its packing density is higher than that of the face-centered cubic structure formed by equal-sized steel balls." }, { "idx": 3163, "question": "Calculate the number of atoms N in 1 cm³ of copper, given that the density of copper is 8.9 g/cm³, the relative atomic mass is 63.5, and Avogadro's number is 6.02×10²³.", "answer": "N = (6.02×10²³ × 8.9) / 63.5 ≈ 0.84×10²³" }, { "idx": 3164, "question": "Calculate the number of vacancies n in 1 cm³ of copper at 1000°C, given the average vacancy concentration C≈6.27×10⁻⁵ and the number of atoms N≈0.84×10²³.", "answer": "n = N × C = 0.84×10²³ × 6.27×10⁻⁵ ≈ 5.27×10¹⁸" }, { "idx": 3165, "question": "The vacancy formation energy of copper is 1.7×10⁻¹⁹ J. Calculate the average vacancy concentration C at 1000°C, given the Boltzmann constant K=1.38×10⁻²³ J/K.", "answer": "The average vacancy concentration C = e^(-Ev/KT) = exp{- (1.7×10⁻¹⁹) / (1.38×10⁻²³ × 1273)} ≈ 6.27×10⁻⁵" }, { "idx": 3166, "question": "To dissolve MgF2 into LiF, what type of vacancies should be introduced into LiF?", "answer": "Dissolving MgF2 into LiF, i.e., replacing Li+ with Mg2+, generates cation vacancies." }, { "idx": 3167, "question": "If LiF is dissolved in MgF2, what type of vacancies (anion or cation) must be introduced into MgF2?", "answer": "To maintain electrical neutrality, when a small amount of LiF is dissolved in MgF2, i.e., Li+ replaces Mg2+, anion vacancies are generated." }, { "idx": 3168, "question": "Given that Cr has a body-centered cubic structure, with a lattice constant a=0.2885 nm, density ρ=7.10 g/cm³, and relative atomic mass A=51.996, calculate the number of vacancies in Cr per 10⁶ lattice points.", "answer": "The fraction of vacancies is x. From the definition of density, xρ=(2×(51.996)/(6.023×10²³)×(1−x))/(2.885³×10⁻²⁴)=7.10. Solving gives x=1−(2.885³×10⁻²⁴×7.10×6.023×10²³)/(2×51.996)≈0.0126. Therefore, the number of vacancies per 10⁶ lattice points is 10⁶×0.0126=12,600 (vacancies)." }, { "idx": 3169, "question": "The temperature increases from $T_{1}$ to $T_{2}$. The relative change in lattice parameter measured by X-ray diffraction method is $\\frac{\\Delta a}{a}=($4\\times10^{6}$）%. For a cube with side length L, the measured value is =0.004%. Find the vacancy concentration at temperature T2.", "answer": "$$ xC_{v}=\\frac{\\Delta N}{N}=3\\Bigl[\\frac{\\Delta L}{L}-\\frac{\\Delta a}{a}\\Bigr]=3(0.004\\%-0.0004\\%)=1.08\\times10^{-4} $$" }, { "idx": 3170, "question": "Two edge dislocations with opposite signs on the same slip plane have Burgers vectors of b. If they approach infinitely close, what is the total energy?", "answer": "When two edge dislocations with opposite signs approach infinitely close, they annihilate each other, and the total energy becomes zero." }, { "idx": 3171, "question": "Two positive edge dislocations on the same slip plane, each with a Burgers vector b, when they approach infinitely close, what is the total energy?", "answer": "When two positive edge dislocations approach infinitely close, it is equivalent to the energy of a single large dislocation with a Burgers vector 2b, W=α⋅G(2b)²=4αGb²" }, { "idx": 3172, "question": "Two positive edge dislocations on the same slip plane, with Burgers vector b and separated by a distance L, when L is much larger than the magnitude of the Burgers vector, what is the total energy?", "answer": "When the two edge dislocations are far apart, the total energy equals the sum of their individual energies, regardless of whether they are like or unlike dislocations, given by W=W1+W2=2αGb2" }, { "idx": 3173, "question": "The density of $\\\\mathrm{CaF}_{2}$ is $\\\\rho=3.18~\\\\mathrm{g/cm}^{3}$, the lattice constant is $a=0.5463\\\\mathrm{nm}$, and the relative atomic masses of Ca and F are 40.08 and 19.00, respectively. Calculate the number of Schottky vacancies in the unit cell of CaF2.", "answer": "To maintain electrical neutrality, in the $\\\\mathrm{CaF}_{2}$ ionic crystal, for every Schottky defect formed, one $\\\\mathrm{Ca}^{2+}$ and two $\\\\mathbf{F}^{-}$ must be lost simultaneously. Let the number of Schottky defects in the unit cell be $x$, then $$ x\\\\rho=\\\\frac{\\\\left(4-x\\\\right)\\\\left(40.08+19\\\\times2\\\\right)/6.023\\\\times10^{23}}{5.463^{3}\\\\times10^{-24}}\\\\approx3.18\\\\left({\\\\mathrm{g}}/{\\\\mathrm{cm}^{3}}\\\\right) $$ Therefore, $$ x=4-\\\\frac{6.023\\\\times10^{23}\\\\times3.18\\\\times5.463^{3}\\\\times10^{-24}}{40.08+19.00\\\\times2}\\\\approx6.10\\\\times10^{-4}\\\\uparrow $$" }, { "idx": 3174, "question": "Given that the dislocation line Φt of a newly formed Frank partial dislocation lies on the (111) plane, with b = a/3 [111] perpendicular to the (111) plane, explain why this dislocation is a sessile dislocation.", "answer": "The plane determined by b and t is certainly not the close-packed plane of the face-centered cubic structure, hence this dislocation cannot glide and is a sessile dislocation." }, { "idx": 3175, "question": "Estimate the strain energy of a 1cm long edge dislocation (r0=1nm, R=1cm, μ=5×10^10Pa, b=0.25nm, ν=1/3)", "answer": "The strain energy per unit length of an edge dislocation is WE=(W/L)E=μb^2/[4π(1−ν)]ln(R/r0). The strain energy of a 1cm edge dislocation is WI={[5×10^10×(0.25×10^-9)^2]/[4×3.14×(1−1/3)]}×10^-2×ln[(1×10^-2)/(1×10^-9)]=6×10^-11J." }, { "idx": 3176, "question": "Indicate the radius of the region that accounts for half of the energy (r0=1nm, R=1cm)", "answer": "Let the radius of the region accounting for half of the dislocation energy be r, Wr/W1=ln(r/r0)/ln[(1×10^-2)/r0]=ln[r/(1×10^-9)]/ln[(1×10^-2)/(1×10^-9)]=1/2. Solving gives r=10^-5.5=10^-6×10^0.5=3.16×10^-6m=3.16μm." }, { "idx": 3177, "question": "The lattice constant of α-Fe is $a=0.28664\\mathrm{nm}.$ When two adjacent grains form a symmetric tilt boundary with a tilt angle of 1°, calculate the spacing $D$ of the edge dislocations.", "answer": "The Burgers vector of the unit dislocation in a body-centered cubic structure is $b=\\frac{a}{2}\\textcircled{111},$ so $b={\\frac{\\sqrt{3}}{2}}a.$ Using the formula $$ D={\\frac{b}{\\theta}}={\\frac{\\sqrt{3}}{2}}\\times0.28664/0.0175\\approx14.183{\\mathrm{~nm}} $$" }, { "idx": 3178, "question": "Describe the kinetic conditions of crystalline phase transformation", "answer": "The kinetic condition is that the undercooling of the liquid phase is greater than the critical undercooling required for nucleation, i.e., ΔT>ΔT*." }, { "idx": 3179, "question": "Explain the concept of critical undercooling", "answer": "In an undercooled liquid, the degree of undercooling at which an embryo can form with a radius equal to the critical nucleus radius is called the critical undercooling (ΔT*). Clearly, when the actual undercooling ΔT<ΔT*, the largest embryo size in the undercooled liquid is still smaller than the critical nucleus radius, making nucleation difficult; only when ΔT>ΔT* can homogeneous nucleation occur. Therefore, critical undercooling is required for nucleation." }, { "idx": 3180, "question": "Explain the concept of dynamic undercooling", "answer": "During crystal growth, a certain degree of undercooling in the liquid ahead of the solid/liquid interface is required to satisfy (dN/dt)F > (dN/dt)M. This undercooling is called dynamic undercooling (ΔTc = Tm − Ti), which is a necessary condition for crystal growth." }, { "idx": 3181, "question": "Given that the unit dislocation a/2 [101] can combine with the Shockley partial dislocation [121] to form a Frank partial dislocation, determine the Burgers vector of the newly formed Frank partial dislocation.", "answer": "The Burgers vector of the newly formed Frank partial dislocation is b, where b = a/2 [101] + a/6 [121] = a/3 [111]." }, { "idx": 3182, "question": "Analyze whether a screw dislocation with Burgers vector $b=\\frac{a}{2}$ [overline{{1}}10] moving on the (111) plane in a face-centered cubic crystal can cross-slip onto one of the (111), (11overline{1}), or (overline{1}11) planes when obstructed. Why?", "answer": "For the screw dislocation with $b=\\frac{a}{2}[\\overline{{1}}10]$, the dislocation line $t=\\left[\\overline{{1}}10\\right]$. Since [\\overline{{1}}10] $[\\overline{{1}}10]=0$, the [\\overline{{1}}10] direction lies on the (11overline{1}) plane, meaning $t$ is also on the (11overline{1}) plane. Therefore, the intersection line of (111) and (11overline{1}) planes is [\\overline{{1}}10], so the screw dislocation with $b=\\frac{a}{2}$ [\\overline{{1}}10] can cross-slip onto the (11overline{1}) plane." }, { "idx": 3183, "question": "Explain the concept of undercooling", "answer": "The temperature difference between the actual crystallization temperature and the theoretical crystallization temperature is called the undercooling (ΔT=Tm−Tn). It is required by the thermodynamic conditions of phase transformation. Only when ΔT>0 can the condition that the free energy of the solid phase is lower than that of the liquid phase be achieved. The free energy difference between the liquid and solid phases is the driving force for crystallization." }, { "idx": 3184, "question": "Describe the thermodynamic conditions of crystallization phase transition", "answer": "Analyzing the change in the system's free energy during crystallization phase transition reveals that the thermodynamic condition for crystallization is ΔG<0. From the change in free energy per unit volume ΔGv=-LmΔT/Tm, it can be seen that only when ΔT>0 can ΔGv<0 be achieved. That is, only undercooling can make ΔGv<0." }, { "idx": 3185, "question": "Discuss the energy conditions of crystalline phase transformation", "answer": "From the critical nucleus formation work ΔGc=1/3σA, it can be seen that when a critical nucleus is formed, 1/3 of the surface energy must still be provided by the energy fluctuations in the liquid." }, { "idx": 3186, "question": "Directly observing the aluminum sample, the dislocation density within the grains is ρ=5×10^13/m^2. If the angle between subgrains is 5°, estimate the dislocation spacing at the interface (the lattice constant of aluminum a=2.8×10^-10m).", "answer": "Using the formula D≈b/θ, where b is the magnitude of the Burgers vector of the dislocation and D is the dislocation spacing, the Burgers vector b=[101], so b_Al=a/2×√2=2.8×10^-10×√2/2≈1.98×10^-10m. θ=5/57.3=0.087, substituting into the expression for D, we obtain D=2.28×10^-9m." }, { "idx": 3187, "question": "Describe the structural conditions of crystalline phase transition", "answer": "The structural fluctuations present in the liquid are the basis for the formation of crystal nuclei during crystallization, therefore, structural fluctuations are the necessary structural conditions for the crystallization process." }, { "idx": 3188, "question": "Direct observation of an aluminum specimen shows a dislocation density within the grains of ρ=5×10^13/m^2. If the angle between subgrains is 5°, estimate the average size of the subgrains (the lattice constant of aluminum is a=2.8×10^-10m).", "answer": "Assuming all dislocations are concentrated on the subgrain boundaries and each subgrain is a regular hexagon with side length a and area S. S=1/2×a×√3/2×a×6=3√3/2×a^2. The number of subgrains per unit area is n=1/S. Substituting the values of D and S gives 1/2×6a×1/2.28×10^-9×1/(3/2×√3×a^2)=5×10^13. Finally, a=1.01×10^-5m is obtained, so the average subgrain size d=2a=2.02×10^-5m." }, { "idx": 3189, "question": "Under the same negative temperature gradient, why does Pb crystallize into dendritic crystals, while the crystallization interface of Si is flat?", "answer": "Under the same negative temperature gradient, Pb is a metallic element with a rough interface. Therefore, it grows into dendritic crystals continuously in the direction perpendicular to the liquid-solid interface. In contrast, Si is a non-metallic element with a smooth interface, and it grows into crystals with flat interfaces through discontinuous lateral growth." }, { "idx": 3190, "question": "In the symmetric tilt grain boundary of face-centered cubic metal $\\\\mathrm{Cu}$, the spacing between two positive edge dislocations is $D=1000\\\\mathrm{nm}$. Assuming the extra half-plane of the edge dislocation is the (110) plane and $d_{110}=0.1278\\\\mathrm{~nm}$, find the tilt angle $\\\\theta$ of the tilt grain boundary.", "answer": "The unit dislocation of the face-centered cubic structure is $b=\\\\frac{a}{2}\\\\langle110\\\\rangle$. Since the {110} plane has an extra atomic plane, $b=2d_{110}=2\\\\times0.1278=0.2556\\\\mathrm{nm}$. The tilt angle $\\\\theta$ of the tilt grain boundary can be calculated using the following deformation formula: $$ \\\\theta\\\\approx\\\\frac{b}{D}=\\\\frac{0.2556}{1000}=2.556\\\\times10^{-4}\\\\times\\\\frac{180}{\\\\pi}\\\\approx0.0146^{\\\\circ} $$" }, { "idx": 3191, "question": "Given: the melting point of aluminum Tm=993K, the volumetric heat of fusion Lm=1.836×10^9 J/m^3, the solid-liquid interface specific surface energy σ=93 mJ/m^2, and the atomic volume V0=1.66×10^-29 m^3. Considering the solidification of liquid aluminum at 1 atm, when the undercooling ΔT=19℃, calculate the critical nucleus size r*.", "answer": "The critical nucleus size r* = (2σ Tm) / (Lm ΔT) = (2 × 93 × 10^-3 × 993) / (1.836 × 10^9 × 19) = 9.45 × 10^-8 m = 94.5 nm" }, { "idx": 3192, "question": "What are the characteristics of the outermost crystal zone in the ingot structure?", "answer": "The outermost layer is a fine equiaxed grain zone. Its formation is due to the lower temperature of the mold wall, which results in a larger undercooling of the liquid and thus a higher nucleation rate." }, { "idx": 3193, "question": "Given: The solid-liquid interface specific surface energy of aluminum σ=93 mJ/m^2, and the free energy change per unit volume ΔGV=-3.51×10^7 J/m^3. Calculate the free energy change ΔGr* (nucleation work) at the critical nucleus size r* during the transition from liquid to solid.", "answer": "ΔGr* = (16π σ^3) / (3 ΔGV^2) = (16π × (93 × 10^-3)^3) / (3 × (-3.51 × 10^7)^2) = 3.47 × 10^-15 J" }, { "idx": 3194, "question": "Briefly describe the mechanism of crystal growth.", "answer": "The mechanism of crystal growth refers to the microscopic growth mode of crystals, which is related to the structure of the liquid/solid interface. For substances with rough interfaces, approximately $50\\\\%$ of the atomic positions on the interface are vacant. These vacancies can accept atoms, allowing liquid atoms to individually occupy the vacancies and connect with the crystal. The interface moves perpendicularly along its normal direction, exhibiting continuous growth. For crystals with smooth interfaces, growth does not occur through the attachment of individual atoms. Instead, it proceeds via homogeneous nucleation, forming a two-dimensional nucleus one atomic layer thick on the crystallographic facet interface. This creates a step between the new nucleus and the original interface. Individual atoms can then fill in the step, enabling lateral growth of the two-dimensional nucleus. Once the layer is filled, a new two-dimensional nucleus forms on the new interface, and the process repeats. If the smooth interface of a crystal has an exposed screw dislocation, the interface becomes a spiral surface, forming a step that never disappears. Atoms attach to this step, allowing the crystal to grow." }, { "idx": 3195, "question": "What are the characteristics of the central equiaxed grain zone in the ingot structure?", "answer": "The center is the equiaxed grain zone. Its formation is due to the further increase in mold wall temperature and the further decrease in liquid undercooling, resulting in the heat dissipation directionality of the remaining liquid becoming less obvious and the liquid being in a state of uniform cooling. At the same time, unmelted impurities, broken dendrites, etc., tend to concentrate in the remaining liquid, all of which promote the formation of equiaxed grains." }, { "idx": 3196, "question": "Analyze the basic conditions for the formation of single crystals.", "answer": "The basic condition for forming a single crystal is to ensure that only one nucleus is produced (or only one nucleus can grow) when the liquid metal crystallizes, and it grows into a single crystal." }, { "idx": 3197, "question": "Given: The volumetric latent heat of fusion for aluminum Lm=1.836×10^9 J/m^3, melting point Tm=993K, undercooling ΔT=19℃. Calculate the change in volumetric free energy ΔGV during the transformation from liquid to solid.", "answer": "ΔGV = (-Lm ΔT) / Tm = (-1.836 × 10^9 × 19) / 993 = -3.51 × 10^7 J/m^3" }, { "idx": 3198, "question": "Given the liquid-solid interfacial energy of pure liquid nickel σ=2.53×10^-5 J/cm^2, the critical nucleus radius r*=1 nm, the melting point of pure nickel Tm=1726 K, the heat of fusion ΔLm=18075 J/mol, the molar volume Vs=6.6 cm^3/mol, and the undercooling ΔT=319 K, calculate the critical nucleation work ΔG*.", "answer": "ΔG*=(16πσ^3·Tm^2·Vs^2)/(3·Lm^2·ΔT^2)=(16×3.14×(2.53×10^-5)^3×1726^2×6.6^2)/(3×18075^2×319^2)=1.06×10^-18 (J)" }, { "idx": 3199, "question": "How does vibration crystallization refine grain size?", "answer": "Vibration, on one hand, provides the energy required for nucleation, and on the other hand, can fracture growing crystals, thereby increasing more crystallization nuclei and refining the grain size." }, { "idx": 3200, "question": "Does the ingot structure necessarily have three crystal zones?", "answer": "It should be pointed out that not all ingot structures have three crystal zones. Due to different solidification conditions, an ingot may only have one type of crystal zone or only two types of crystal zones." }, { "idx": 3201, "question": "What are the characteristics of the intermediate crystal zone in the ingot structure?", "answer": "The intermediate zone consists of columnar crystals. Its formation is primarily due to the increase in mold wall temperature, where the growth rate of crystal nuclei exceeds the nucleation rate, and heat dissipation is more favorable in the direction perpendicular to the mold wall. In the fine crystal zone, preferentially oriented grains grow into columnar crystals." }, { "idx": 3202, "question": "It is known that liquid pure nickel undergoes homogeneous nucleation at an undercooling of 319K under 1.013×10^5Pa (1 atm). Given the critical nucleus radius is 1nm, the melting point of pure nickel is 1726K, the heat of fusion ΔLm=18075J/mol, and the molar volume Vs=6.6cm^3/mol, calculate the liquid-solid interfacial energy σ of pure nickel.", "answer": "σ=(r*·Lm·ΔT)/(2·Tm·Vs)=(1×10^-7×18075×319)/(2×1726×6.6)=2.53×10^-5(J/cm^2)" }, { "idx": 3203, "question": "According to solidification theory, what are the basic approaches to refining grains?", "answer": "From solidification theory, it is known that the number of grains per unit volume during crystallization, z, depends on two factors: the nucleation rate N and the crystal growth rate Vg, i.e., z∝N/Vg. The basic approaches include increasing undercooling, adding nucleating agents (modification treatment), and vibration crystallization." }, { "idx": 3204, "question": "How does increasing undercooling refine grains?", "answer": "Increasing the undercooling ΔT causes both N and Vg to increase, but the growth rate of N is greater than that of Vg. Therefore, the value of N/Vg increases, meaning z becomes larger." }, { "idx": 3205, "question": "How does adding a nucleating agent (modification treatment) refine the grain size?", "answer": "After adding a nucleating agent, it can promote heterogeneous nucleation in the undercooled liquid. It not only increases the substrates required for heterogeneous nucleation but also reduces the critical nucleus volume, both of which will increase the number of nuclei, thereby refining the grain size." }, { "idx": 3206, "question": "A Cu-30%Zn brass cold-rolled sheet undergoes 25% cold deformation, resulting in a thickness of 1cm. Find the original thickness Lo.", "answer": "From (Lo - 1)/Lo = 25%, solving gives Lo = 4/3 cm ≈ 1.333 cm." }, { "idx": 3207, "question": "Compress the above-mentioned cold-rolled sheet from 1cm thickness to 0.6cm, and find the true linear strain e.", "answer": "e = ln(0.6/(4/3)) ≈ -80%." }, { "idx": 3208, "question": "A specimen with an original gauge length of L0 is stretched to L. Determine the engineering linear strain and true linear strain during the stretching process.", "answer": "Engineering linear strain ε = (L - L0) / L0; true linear strain e = ln(L / L0)" }, { "idx": 3209, "question": "Compress the above-mentioned cold-rolled sheet from 1cm thickness to 0.6cm, calculate the total engineering strain ε.", "answer": "ε = (0.6 - 4/3)/(4/3) ≈ -55%." }, { "idx": 3210, "question": "Can close-packed hexagonal metal magnesium produce cross-slip? What is the slip direction?", "answer": "In addition to the (0001) $<11$ $\\overline{2}0>$ slip system, magnesium also has $|10\\overline{{1}}1\\}<11\\overline{{2}}0$ > slip systems, and the slip direction is always $<11\\overline{{2}}0:$, so cross-slip can occur." }, { "idx": 3211, "question": "A 20m long aluminum rod with a diameter of 14.0mm is drawn through a die with an aperture of 12.7mm. Calculate the true strain experienced by this aluminum rod.", "answer": "True strain e=ln(L/L0)=ln(24.3/20)=19.47%" }, { "idx": 3212, "question": "A 20m long aluminum rod with a diameter of 14.0mm is drawn through a die with an aperture of 12.7mm. Calculate the engineering linear strain experienced by this aluminum rod.", "answer": "The engineering strain ε=(L-L0)/L0=21.5%" }, { "idx": 3213, "question": "To obtain metallic glass, why is it generally necessary to choose a binary system with a steep liquidus line and thus a low eutectic temperature?", "answer": "Metallic glass is obtained by ultra-rapid cooling methods, which suppress the liquid-solid crystallization process, resulting in an unusual amorphous structure. Glass is an undercooled liquid. The viscosity of this liquid is high, and atomic mobility is low, making crystallization difficult. For example, polymer materials (silicates, plastics, etc.) can achieve a glassy state under normal cooling conditions. Metals, however, are different. Due to the low viscosity of liquid metals, they crystallize rapidly upon cooling below the liquidus line, requiring extremely high cooling rates (estimated >10^10 K/s) to achieve a glassy state. To obtain metallic glass at lower cooling rates, the stability of the liquid phase must be increased, allowing it to exist over a wider temperature range. Experiments have shown that when the liquidus line is steep, resulting in a low eutectic temperature, the stability of the liquid phase is enhanced. Therefore, such binary systems (e.g., Au-Si, Fe-C, Fe-P, Fe-Si, etc.) are chosen. To improve performance, other elements (e.g., Ni, Mo, Cr, Co, etc.) can be added. Such metallic glasses can be obtained at cooling rates of 10^5 to 10^6 K/s." }, { "idx": 3214, "question": "Compress the above stretched specimen from L to L0, and calculate the engineering linear strain and true linear strain during the compression process.", "answer": "Engineering linear strain ε = (L0 - L) / L; true linear strain e = ln(L0 / L)" }, { "idx": 3215, "question": "A 20m long aluminum rod with a diameter of 14.0mm is drawn through a die with an aperture of 12.7mm. Determine the dimensions of the aluminum rod after drawing.", "answer": "Since the volume remains unchanged before and after drawing, i.e., F0·L0=F·L, therefore (π(14.0)^2)/4·20×10^3=(π(12.7)^2)/4·L. Thus, L=24.3×10^3mm=24.3m" }, { "idx": 3216, "question": "Given: The critical nucleus size of aluminum r*=94.5 nm, atomic volume V0=1.66×10^-29 m^3. Calculate the number of atoms Nr* in a nucleus with radius r*.", "answer": "Nr* = (4/3 π r*^3) / V0 = (4/3 π × (94.5 × 10^-9)^3) / (1.66 × 10^-29) = 2.12 × 10^8" }, { "idx": 3217, "question": "The possible slip planes for a body-centered cubic crystal are {110}, {112}, {123}. If the slip direction is [111], what are the specific slip systems?", "answer": "The specific slip systems are: (110)[1\\overline{1}1]; (\\overline{1}12)[1\\overline{1}1]; and (132)[1\\overline{1}1]." }, { "idx": 3218, "question": "Given that the critical resolved shear stress of pure aluminum is τ_c = 0.79 MPa, the question is: To produce slip in the [110] direction on the (1̅11) plane, what magnitude of force should be applied in the [001] direction?", "answer": "The slip plane is (1̅11), the slip direction is [110], and the force axis is [001]. Then, cosφ = 1/√3; cosλ = 0/√2 = 0. Since λ = 90°, σ_S = τ_c / (cosλ · cosφ) = ∞. Therefore, when the force axis is in the [001] direction, the (111)[110] slip system will not be activated." }, { "idx": 3219, "question": "The surface of a copper single crystal is parallel to {001}. If a tensile force is applied along the [001] direction, and the measured critical resolved shear stress is $\\overline{{\\tau_{c}}}=0.7$ MPa, determine the stress at which the material yields.", "answer": "The force is applied along the [001] direction, which is in a soft orientation with 8 equivalent slip systems. Taking one of them, (111)[overline{1}01], we have: $$ \\cos\\lambda={\\frac{1}{\\sqrt{2}}};\\cos\\varphi={\\frac{1}{\\sqrt{3}}} $$ $$ \\sigma_{S}={\\frac{\\tau_{c}}{\\cos\\lambda\\cdot\\cos\\varphi}}={\\frac{0.7}{{\\frac{1}{\\sqrt{2}}}\\cdot{\\frac{1}{\\sqrt{3}}}}}=1.715~{\\mathrm{MPa}} $$" }, { "idx": 3220, "question": "A face-centered cubic single crystal is stretched with [131] as the force axis. When the tensile stress is 1×10^7Pa, determine the resolved shear stress on the (111)[1\\\\overline{1}0] slip system.", "answer": "For the (111)[1\\\\overline{1}0] slip system, cosφ=5√33/33; cosλ=-√22/11. Therefore, τ3=1×10^7×(5√33/33)×(-√22/11)=3.7×10^6Pa." }, { "idx": 3221, "question": "A face-centered cubic single crystal is stretched with [131] as the force axis. When the tensile stress is 1×10^7Pa, determine the resolved shear stress on the (111)[0\\overline{1}1] slip system.", "answer": "Using the formula τ=σ*cosφcosλ, for the (111)[0\\overline{1}1] slip system, cosφ=(1+1+3)/(√3*√11)=5√33/33; cosλ=-2/(√2*√11)=-√22/11. Therefore, τ1=1×10^7×(5√33/33)×(-√22/11)=3.7×10^6Pa." }, { "idx": 3222, "question": "A face-centered cubic single crystal is stretched with [131] as the force axis. When the tensile stress is 1×10^7Pa, determine the resolved shear stress on the (111)[10overline{1}] slip system.", "answer": "For the (111)[10overline{1}] slip system, since cosλ=0, τ2=0." }, { "idx": 3223, "question": "In the compression test of a magnesium single crystal at room temperature, the [0001] direction coincides with the compression axis. Assuming that the critical resolved shear stress for twinning on the (10\\\\overline{1}2) plane is 10 times that for slip on the (0001) plane, denoted as $\\\\tau_{\\\\mathrm{e}}$, determine whether the crystal will undergo twinning or slip when the compressive stress is sufficiently large, and explain why.", "answer": "If the compression axis coincides with the [0001] direction, since the slip directions of different slip systems in magnesium are the same, all being <11\\\\overline{2}0> directions, and the slip directions are all perpendicular to the [0001] direction, $\\\\cos\\\\lambda=0$. Therefore, $\\\\sigma_{S}=\\\\frac{\\\\sigma_{C}}{\\\\cos\\\\lambda\\\\cos\\\\phi}=\\\\infty$. No matter how large the pressure is, slip cannot occur. However, the twinning plane is {10\\\\overline{1}2}, and the twinning direction is {10\\\\overline{1}2}. When compressed along the [0001] direction, there is a resolved shear stress. When the external force reaches a certain value, twinning will occur along {10\\\\overline{1}2}<\\\\overline{1}011>{10\\\\overline{1}2}. In this example, the resolved shear stress for twinning has already reached 10 times $\\\\tau_{C}$, so twinning deformation can occur." }, { "idx": 3224, "question": "A copper single crystal is subjected to tension with the tensile axis along the [001] direction and σ=10^6 Pa. Calculate the force acting on a screw dislocation line with Burgers vector b=(a/2)[1̄01] on the (111) plane. Given a_Cu=0.36 nm.", "answer": "The tensile stress is applied along the [001] direction. On the (111) slip plane, the resolved shear stress along the [1̄01] direction is τ=σcosφcosλ, where cosφ=1/√3, cosλ=1/√2, and σ=10^6 Pa. Thus, τ=10^6×(1/√3)×(1/√2)≈4.08×10^5 Pa. The force per unit length on the dislocation line is F_d=τb=4.08×10^5×(√2/2)×0.36×10^(-9)≈1.04×10^(-4) N/m." }, { "idx": 3225, "question": "A copper single crystal is subjected to tensile stress with the tensile axis along the [001] direction and σ=10^6 Pa. Calculate the force on the edge dislocation line with Burgers vector b=(a/2)[1̄01] on the (111) plane. Given a_Cu=0.36 nm.", "answer": "The tensile stress is applied along the [001] direction. On the (111) slip plane, the resolved shear stress along the [1̄01] direction is τ=σcosφcosλ, where cosφ=1/√3, cosλ=1/√2, and σ=10^6 Pa. Thus, τ=10^6×(1/√3)×(1/√2)≈4.08×10^5 Pa. The force per unit length on the dislocation line is F_d=τb=4.08×10^5×(√2/2)×0.36×10^(-9)≈1.04×10^(-4) N/m." }, { "idx": 3226, "question": "What are the practical applications of fracture toughness K_IC in mechanical design?", "answer": "1. Given the working stress σ and the fracture toughness K_IC of the material, the maximum allowable crack size a_c in the workpiece can be estimated. 2. Given the crack size a and the working stress σ, the stress intensity factor K_I during operation can be determined to provide a theoretical basis for reasonable material selection. 3. Given K_IC and the existing crack size a, the maximum allowable stress σ can be determined. If the working stress is less than this stress, the crack will not propagate; otherwise, the crack will become unstable." }, { "idx": 3227, "question": "Describe the process of extended dislocation cross-slip from the (111) plane to the (111) plane in a face-centered cubic structure.", "answer": "When the extended dislocation on the (111) plane encounters obstacles during slip, it can constrict to form a screw-type perfect dislocation. The dislocation reaction is $$\\\\frac{a}{6}[\\\\bar{1}2\\\\bar{1}]+\\\\frac{a}{6}[\\\\bar{2}11]-\\\\frac{a}{2}[\\\\bar{1}10]$$The formed screw dislocation has a Burgers vector $b=\\\\frac{a}{2}$ [\\\\overline{1}10], and the dislocation line $t=[\\\\overline{{1}}10]$. It can cross-slip onto the (11\\\\overline{1}) plane and spread out, forming an extended dislocation on the (11\\\\overline{1}) plane, i.e., $\\\\frac{a}{2}[\\\\overline{{{110}}}]\\\\rightarrow\\\\frac{a}{6}[\\\\overline{{{21}}}\\\\overline{{{1}}}]+\\\\frac{a}{6}[\\\\overline{{{1}}}21]$. This extended dislocation can continue to move on the (11\\\\overline{1}) plane or constrict again to cross-slip back onto the (111) plane and spread out once more." }, { "idx": 3228, "question": "Compare the roles of twinning and slip in the plastic deformation process", "answer": "The direct contribution of twinning to plastic deformation is much smaller than that of slip, but twinning alters the crystal orientation, turning hard-oriented slip systems into soft-oriented ones, thereby activating further slip in the crystal." }, { "idx": 3229, "question": "What is called fracture toughness K_IC?", "answer": "K_IC is the critical value of the stress intensity factor for a Type I crack, representing the fracture toughness in terms of stress intensity factor under linear elastic conditions, with the unit MPa*m^0.5." }, { "idx": 3230, "question": "Describe the similarities between twinning and slip", "answer": "The similarities between slip and twinning are that both are fundamental modes of crystal plastic deformation, both occur as shear under shear stress along certain crystallographic planes and directions, and the crystal structure type remains unchanged before and after deformation." }, { "idx": 3231, "question": "In the actual crystallization process of metals, nucleation occurs in two ways: homogeneous nucleation and heterogeneous nucleation. Since homogeneous nucleation requires higher nucleation energy, heterogeneous nucleation is predominantly observed.", "answer": "√" }, { "idx": 3232, "question": "Some solid solutions with certain atomic ratios are disordered solid solutions at high temperatures, and may transform into ordered solid solutions when cooled below a certain critical temperature. Once the ordering transition occurs, it can lead to abrupt changes in certain properties.", "answer": "√" }, { "idx": 3233, "question": "Describe the differences between twinning and slip", "answer": "The differences are as follows: twinning causes a uniform shear in a portion of the crystal, while slip is concentrated only on certain slip planes. During slip, the slipped and unslipped parts of the crystal have the same phase, whereas the twinned part and the matrix have different phases, exhibiting a special mirror-symmetry relationship. In twinning deformation, the atomic displacement is less than the atomic spacing in the twinning direction, being a fractional multiple of the atomic spacing; during slip deformation, the distance atoms move is an integer multiple of the atomic spacing in the slip direction. Similar to slip, twinning elements are also related to the crystal structure, but the twinning plane and twinning direction in the same structure can differ from the slip plane and slip direction. The critical resolved shear stress for twinning is much higher than that for slip. The stress-strain curve for twinning deformation differs from that of slip, showing serrated fluctuations, mainly because the shear stress required for twinning 'nucleation' is greater than the stress needed for twinning boundary propagation. Generally, slip occurs first, and twinning deformation occurs only when slip becomes difficult." }, { "idx": 3234, "question": "Using the three-axis system to index the crystallographic plane indices and direction indices of a hexagonal close-packed structure, the resulting indices for the same family of planes or directions are different.", "answer": "√" }, { "idx": 3235, "question": "During the crystallization of liquid metal, the degree of undercooling required to form an embryo equal to the critical nucleus radius in the undercooled liquid is called the critical undercooling.", "answer": "√" }, { "idx": 3236, "question": "At room temperature, the plastic deformation modes of metal polycrystals are A. Slip, twinning, creep B. Slip, twinning, kinking C. Slip, climb, cross-slip D. Slip, twinning, grain boundary sliding", "answer": "B" }, { "idx": 3237, "question": "The relationship between the yield strength and grain size of No. 10 steel was measured as follows: when the grain diameter was $400\\\\mu m$, $\\\\partial_{\\\\mathfrak{s}}=86$ MPa, and when the grain diameter was $5\\\\mu\\\\mathrm{m}$, $\\\\sigma_{\\\\mathrm{s}}=242$ MPa. Determine the yield strength when the average grain diameter is $50~\\\\mu\\\\mathrm{m}$.", "answer": "Using the Hall-Petch formula $\\\\sigma_{S}=\\\\sigma_{0}+k d^{-\\\\frac{1}{2}}$, the system of equations is obtained as: $$186=\\\\sigma_{0}+k(400\\\\times10^{-6})^{-\\\\frac{1}{2}}$$ $$1242=\\\\sigma_{0}+k(5\\\\times10^{-6})^{-\\\\frac{1}{2}}$$ Solving gives: $$k=0.393;\\\\sigma_{0}=66.25\\\\mathrm{MPa}$$ Substituting $d=50\\\\times10^{-6}(\\\\mathrm{m})$ into the Hall-Petch formula: $$\\\\sigma_{S}=66.25+0.393(50\\\\times10^{-6})^{-\\\\frac{1}{2}}\\\\approx121.83~\\\\mathrm{MPa}$$" }, { "idx": 3238, "question": "The main mechanism of high-temperature recovery is A. Dislocation slip and cross-slip B. Dislocation climb and polygonization C. Polygonization and subgrain coalescence D. Bulge nucleation and subgrain coalescence", "answer": "B" }, { "idx": 3239, "question": "Among the following alloy phase structures, the one with high melting point, high hardness, great brittleness, and simple crystal structure is A. Topologically close-packed phase B. Interstitial compound with complex lattice structure C. Electron compound D. Interstitial phase", "answer": "D" }, { "idx": 3240, "question": "Using the Peierls-Nabarro formula, explain why slip in crystals usually occurs on the closest-packed planes and in the closest-packed directions.", "answer": "During room temperature deformation, since the grain boundary strength is higher than that within the grain, the finer the grains, the more grain boundaries are contained per unit volume, resulting in a better strengthening effect. According to the Hall-Petch formula, $\\sigma_{s}=\\sigma_{0}+k d^{-\\frac{1}{2}},$ the smaller the grain diameter $d$, the higher $\\sigma_{S}$ becomes, which is known as fine-grain strengthening. Each grain in a polycrystal is surrounded by other grains, so deformation is not isolated and requires neighboring grains to coordinate and adapt to the shape changes of plastically deformed grains. Plastic deformation must involve multiple slip systems from the outset. The finer the grains, the better the deformation coordination, and thus the better the plasticity. Additionally, finer grains reduce the severity of stress concentrations caused by dislocation pile-ups, which can delay crack initiation. The tortuous grain boundaries hinder crack propagation, contributing to improved strength and plasticity." }, { "idx": 3241, "question": "The driving force for atomic diffusion is A. Concentration gradient of components B. Temperature gradient C. Chemical potential gradient of components", "answer": "C" }, { "idx": 3242, "question": "When polycrystalline metal is heated to a relatively high temperature and held, the grains will grow. The grain growth mode is A. Subgrain coalescence and growth B. Grain boundary bowing outward and growth C. Grain boundary moving toward the curvature center D. Movement of Y junctions", "answer": "C" }, { "idx": 3243, "question": "In the cubic system, the relationship between the crystal plane (hkl) and the crystal direction [hkl] is A. [hkl]//(hkl) B. [hkl] is perpendicular to (hkl) C. No definite relationship", "answer": "B" }, { "idx": 3244, "question": "Which of the following statements about dislocation loops is correct? A. A dislocation loop cannot be entirely edge dislocation everywhere. B. A dislocation loop can be entirely edge dislocation everywhere. C. A dislocation loop must simultaneously contain both edge dislocation and screw dislocation. D. A dislocation loop must simultaneously contain edge dislocation, screw dislocation, and mixed dislocation.", "answer": "B" }, { "idx": 3245, "question": "The correct order of coordination numbers for the following alloy phases from small to large is A.SiC、NaCl、a-Fe、Cu B.a-Fe、SiC、Cu、NaClC.a-Fe、Cu、SiC、NaCl D.SiC、a=Fe、NaCl、Cu", "answer": "A" }, { "idx": 3246, "question": "Which of the following statements about zone melting is correct? A. For alloys with $K_{0}<1$, the solute is enriched at the end, and the beginning is purified; for alloys with $K_{0}>1$, the solute is enriched at the beginning, and the end is purified. B. For alloys with $K_{0}<1$, the solute is enriched at the beginning, and the end is purified; for alloys with $K_{0}>1$, the solute is enriched at the end, and the beginning is purified. C. Regardless of $K_{0}<1$ or $K_{0}>1$, the solute is enriched at the beginning. D. Regardless of $K_{0}<1$ or $K_{0}>1$, the solute is enriched at the end.", "answer": "A" }, { "idx": 3247, "question": "Carburize industrial pure iron at 920°C. If the carbon concentration on the workpiece surface remains constant, i.e., w_C=1.2%, the diffusion coefficient D=1.5×10^-11 m^2/s, and carburize for 10h. If the carburized layer depth is defined as the distance from the surface to where the carbon mass fraction is 0.2%, find the carburized layer depth?", "answer": "Substitute C_x=0.2 into the formula C_x=1.2[1-erf(6.8×10^2x)]. Referencing the error function table, Z≈6.8×10^2x=0.9784, the layer depth is x=0.00144m=1.44mm." }, { "idx": 3248, "question": "Carburize industrial pure iron at 920°C. If the carbon concentration on the workpiece surface remains constant, i.e., w_C=1.2%, with a diffusion coefficient D=1.5×10^-11 m^2/s, and carburize for 10 hours. Determine the surface carbon concentration distribution?", "answer": "Let the carburizing depth be x. Substituting C_0=0, C_s=1.2, D=1.5×10^-11 m^2/s, and t=36000s into the formula (C_s-C_x)/(C_s-C_0)=erf(x/(2√(Dt))), the surface carbon concentration distribution is obtained as C_x=1.2[1-erf(6.8×10^2x)]." }, { "idx": 3249, "question": "Please design an experimental plan to measure the recrystallization activation energy of a metal sheet cold-rolled by 75%.", "answer": "Subject the cold-rolled specimen to isothermal recrystallization at different temperatures, and for each recrystallization temperature, determine the time t required to achieve a certain recrystallized volume fraction x_V. Since the recrystallization rate of cold-deformed metal V_rec ∝ 1/t = A′ e^(-Q_R/RT), take the logarithm of both sides: ln(1/t) = ln A′ - Q_R/RT The relationship between ln(1/t) and 1/T is linear. Plot this line using experimental data, and the slope of the line can be used to determine the recrystallization activation energy Q_R." }, { "idx": 3250, "question": "Describe the recovery mechanism at medium temperatures of 0.3~0.5Tm", "answer": "At medium-temperature recovery of 0.3~0.5Tm, dislocations glide on slip planes, causing dislocations of opposite signs to meet and annihilate, reducing dislocation density, rearranging and recombining within dislocation tangles, and regularizing subgrains." }, { "idx": 3251, "question": "Describe the recovery mechanisms at high temperatures above 0.5Tm", "answer": "At high temperatures above 0.5Tm, in addition to dislocation slip, recovery can also occur through climb, with the primary mechanism being polygonization, forming low-angle grain boundaries. After polygonization, subgrain coalescence and growth still exist. Subgrain coalescence can be achieved through the movement of Y-nodes, which requires dislocation climb, slip, and cross-slip to accomplish." }, { "idx": 3252, "question": "Adding thorium oxide particles with a size of 10-50nm to nickel, after 40% rolling, the material exhibits high high-temperature strength. Explain the reason.", "answer": "According to the recrystallization theory of two-phase alloys, if the second-phase particles are very fine (less than $0.3\\\\mu m$) and the spacing is small (less than $1\\\\:\\\\mu\\\\mathrm{m}$), the second-phase particles will inhibit the formation of recrystallization nuclei. In this case, the added second-phase particles are smaller than $0.3\\\\mu m$, so adding a certain volume fraction of thorium oxide can improve high-temperature performance. This is because before the formation of recrystallization nuclei, the growth of sub-nuclei encounters the obstruction of second-phase particles, inhibiting the formation of recrystallization nuclei and preventing recrystallization from occurring, so that recrystallization does not occur even before melting. Therefore, during high-temperature operation, the dislocation density remains high, resulting in high deformation resistance. Additionally, thorium oxide particles hinder the movement of dislocations, increasing the resistance to plastic deformation, thereby enhancing high-temperature performance." }, { "idx": 3253, "question": "For an Fe-3% (Si) alloy containing MnS particles, when the particle radius is $0.05\\\\mu m$ and the volume fraction is approximately 1%, annealing below 850℃ results in the cessation of normal grain growth when the average grain diameter of the matrix is $6\\\\mu m$. Analyze the reason for this phenomenon.", "answer": "According to the formula for calculating the limiting average grain size: $$\\\\overline{{D}}_{\\\\mathrm{lim}}=\\\\frac{4r}{3\\\\varphi}=\\\\frac{4\\\\times0.05}{3\\\\times0.01}=6.7~\\\\mu\\\\mathrm{m}$$ The dispersed particles exert a pinning effect on grain boundary migration. When the driving force for grain boundary migration provided by grain boundary energy equals the resistance to grain boundary migration caused by the dispersed particles, the limiting grain size is reached. Further holding does not increase the grain size, as calculated by $\\\\overline{{D}}_{\\\\mathrm{lim}}=6.7~\\\\mu m$. During annealing, the average matrix diameter is 6 μm, which is already close to $\\\\overline{{D}}_{\\\\mathrm{lim}}$, so grain growth becomes extremely slow and nearly stops." }, { "idx": 3254, "question": "What are the conditions for the occurrence of secondary recrystallization?", "answer": "The conditions for abnormal grain growth are that normal grain growth is strongly hindered by dispersed phase particles, texture, surface energy grooves, etc., and the number of grains that can grow is small, resulting in a significant difference in grain size, which leads to abnormal growth." }, { "idx": 3255, "question": "What is called secondary recrystallization?", "answer": "The growth of grains after recrystallization is further divided into normal growth and abnormal growth. The abnormal growth of grains is also called discontinuous growth or secondary recrystallization. Secondary recrystallization is a special mode of grain growth where the growth of most grains in the matrix is suppressed, while a few grains grow rapidly, significantly increasing the size difference between grains until these rapidly growing grains come into complete contact with each other." }, { "idx": 3256, "question": "Describe the recovery mechanism at low temperatures of 0.1~0.3Tm", "answer": "Recovery at 0.1~0.3Tm low temperatures primarily involves the movement of point defects, where vacancies and interstitial atoms migrate to grain boundaries or dislocations and annihilate, the recombination of vacancies and interstitial atoms, and the aggregation of vacancies to form vacancy pairs or vacancy clusters, leading to a significant reduction in point defect density." }, { "idx": 3257, "question": "After high-temperature recovery, several edge dislocations form a subgrain boundary with a misorientation of $0.057^{\\\\circ}$. Assuming there is no interaction between these dislocations, what is the ratio of the distortion energy after forming the subgrain to the original energy? (Given the core radius of the dislocation $r_{0}\\\\approx$ $b\\\\approx10^{-8}\\\\mathrm{cm}$, and the effective radius of the dislocation stress field before forming the subgrain $R=10^{-4}\\\\mathrm{cm}$.)", "answer": "Given the core radius of the dislocation $r_{0}=b\\\\approx10^{-8}~\\\\mathrm{cm}$ and the effective radius of the dislocation stress field $R=10^{-4}$ $\\\\mathrm{{cm}}$, since the total length of dislocations remains unchanged before and after forming the subgrain, the ratio of the energy per unit dislocation line before and after forming the subgrain is equal to the ratio of the energy before and after polygonization. The energy per unit dislocation line is $$ W_{E}=\\\\frac{G b^{2}}{4\\\\pi(1-\\\\nu)}\\\\ln\\\\frac{R}{r_{0}}=\\\\frac{G b^{2}}{4\\\\pi(1-\\\\nu)}\\\\ln10^{4} $$ After polygonization, the edge dislocations are arranged perpendicular to the slip plane with a spacing of $D$, so the effective radius of the dislocation $R$ becomes $D$. From the formula $R=D=\\\\frac{b}{\\\\theta}=\\\\frac{10^{-8}}{10^{-3}}=10^{-5}$, the energy per unit dislocation line after polygonization is $$ W_{E}^{\\\\prime}={\\\\frac{G b^{2}}{4\\\\pi(1-\\\\nu)}}\\\\cdot\\\\ln{\\\\frac{10^{-5}}{10^{-8}}}={\\\\frac{G b^{2}}{4\\\\pi(1-\\\\nu)}}\\\\cdot\\\\ln10^{3} $$ $$ \\\\frac{W_{E}^{\\\\prime}}{W_{E}}=\\\\frac{\\\\ln{10^{3}}}{\\\\ln{10^{4}}}=\\\\frac{3}{4} $$ This shows that after polygonization, the dislocation energy decreases, reducing the stored energy and also the driving force for recrystallization." }, { "idx": 3258, "question": "Given that brass with a zinc mass fraction of 30% takes 1 hour to complete recrystallization at a constant temperature of 400°C, and 2 hours at 390°C, calculate how much time is needed to complete recrystallization at a constant temperature of 420°C.", "answer": "Recrystallization of cold-deformed metal is also a thermally activated process, and the recrystallization rate follows the Arrhenius equation $$ V_{\\overline{{\\{\\}}}}=A^{\\prime}\\cdot\\mathrm{e}^{-Q/R T} $$ Since $V_{\\overrightarrow{\\vert\\overrightarrow{\\vert\\cdot}\\vert}}$ is inversely proportional to the time $\\scriptstyle t$ required to achieve a certain volume fraction $x_{\\upsilon}$, i.e., $V_{\\mathrm{\\#}}\\propto1/t$, we have $$ \\frac{1}{t_{1}}=A\\mathrm{e}^{-\\frac{Q}{R T_{1}}};\\frac{1}{t_{2}}=A\\mathrm{e}^{-\\frac{Q}{R T_{2}}};\\frac{1}{t_{3}}=A\\mathrm{e}^{-\\frac{Q}{R T_{3}}} $$ From the above three equations, dividing them pairwise yields $$ \\frac{t_{2}}{t_{1}}=\\mathrm{e}^{-\\frac{Q}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)};\\quad\\frac{t_{3}}{t_{1}}=\\mathrm{e}^{-\\frac{Q}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{3}}\\right)} $$ Taking the natural logarithm and then dividing gives $$ \\frac{\\ln(t_{2}/t_{1})}{\\ln(t_{3}/t_{1})}=\\frac{1/T_{1}-1/T_{2}}{1/T_{1}-1/T_{3}} $$ Substituting $T_{1}=673~\\mathrm{K},t_{1}=1~\\mathrm{h};~T_{2}=663~\\mathrm{K},t_{2}=2~\\mathrm{h};~T_{3}=693~\\mathrm{K}$ into the above equation, we obtain $t_{3}=0.27$ h." }, { "idx": 3259, "question": "How to prolong the lifespan of tungsten filaments by adding second-phase particles?", "answer": "Finely dispersed high-melting-point second-phase particles can be added to the tungsten filaments. The second phase can effectively hinder grain growth, with the limiting grain size Dlim=4r/3f. When the particle size r is smaller and the volume fraction f of the added particles is larger, the limiting grain size becomes smaller, preventing the formation of large grains that span the filament. This mitigates grain boundary weakening at high temperatures, thereby preventing brittle fracture." }, { "idx": 3260, "question": "A factory was conducting solution treatment on the jaw plates of a crusher made of high manganese steel. After heating at 1100°C, the plates were suspended with cold-drawn steel wires and transported by a crane to the quenching water tank. During transportation, the steel wire suddenly broke. This wire was new and had no defects. Analyze the cause of the steel wire fracture.", "answer": "Due to work hardening, the cold-drawn steel wire has high strength and strong load-bearing capacity. However, when heated by the red-hot jaw plate, once the temperature rises above $T_{\\overrightarrow{\\mathbb{H}}}$, recrystallization occurs, causing a decrease in strength. As a result, the wire can no longer bear the weight of the jaw plate, leading to fracture." }, { "idx": 3261, "question": "What is superplasticity?", "answer": "When metallic materials are stretched under certain conditions, their elongation can reach over 200%, or even exceed 1000%. This property is called superplasticity. During superplastic deformation, the true stress-true strain conforms to the relation: σT(εT,T)=C·ε̇T^m, where m is called the strain rate sensitivity constant. When m=0.3-0.8, the material can exhibit superplasticity." }, { "idx": 3262, "question": "How to extend the life of tungsten wire by adding oxides of potassium, aluminum, silicon, etc.?", "answer": "Adding oxides of potassium, aluminum, silicon, etc., which vaporize during sintering to form bubbles. During processing, these bubbles elongate along the direction of processing. These small bubbles distributed along the axial direction can also act as obstacles to grain boundary migration, allowing grains to grow only parallel to the axis of the tungsten wire. This results in elongated large grains along the axis, preventing the formation of large grains that span the cross-section of the tungsten wire, thereby improving high-temperature performance." }, { "idx": 3263, "question": "How to achieve superplasticity?", "answer": "Superplasticity can be divided into three categories: structural superplasticity, transformation superplasticity, and other types of superplasticity. Structural superplasticity should meet three conditions: (1) Ultra-fine, equiaxed, and stable grains, with grain size generally not exceeding 10μm. (2) Deformation temperature is typically in the range of (0.5∼0.7)Tm. (3) A certain strain rate, with the optimal superplastic deformation rate being 10^-4∼10^-2s^-1 or 10^-3∼10^-2min^-1." }, { "idx": 3264, "question": "As an economic measure, using pure lead instead of lead-tin alloy to make solder for brazing iron, how is the wettability of pure lead on iron?", "answer": "The wettability of pure lead on iron is poor." }, { "idx": 3265, "question": "As an economic measure, using pure lead instead of lead-tin alloy to make solder for brazing iron, how does the melting point of pure lead compare to that of lead-tin alloy?", "answer": "The melting point of pure lead is higher than that of Sn-Pb alloy." }, { "idx": 3266, "question": "Why is the carburizing temperature for steel parts generally chosen to be within the γ-Fe phase region?", "answer": "Because the solubility of carbon in γ-Fe is high, a higher carbon concentration gradient can be achieved in the surface layer during carburizing, facilitating the process. Additionally, the high temperature in the γ-Fe region accelerates the diffusion process." }, { "idx": 3267, "question": "In a ternary system during diffusion, can a three-phase coexistence region appear within the diffusion layer? Why?", "answer": "A three-phase coexistence region cannot exist within the diffusion layer of a ternary system. The reasons are as follows: If three-phase equilibrium coexistence occurs in a ternary system, the compositions of the three phases are fixed, and the chemical potentials of the same component in different phases are equal, resulting in a zero chemical potential gradient, making diffusion impossible." }, { "idx": 3268, "question": "The nitriding temperature of steel is generally chosen to be close to but slightly below the eutectoid temperature (590°C) of the Fe-N system, why?", "answer": "The reason is that the diffusion coefficient of $N$ in α-Fe is higher than that in γ-Fe." }, { "idx": 3269, "question": "In a ternary system during diffusion, can a two-phase coexistence region appear within the diffusion layer? Why?", "answer": "A two-phase coexistence region can exist within the diffusion layer of a ternary system. The reasons are as follows: In a ternary system during two-phase coexistence, due to the degree of freedom being 2, when the temperature is constant, the composition of the coexisting phases can change. This disrupts the chemical potential balance of the same component atoms in the two phases, leading to diffusion." }, { "idx": 3270, "question": "What factors constitute the resistance to phase transformation?", "answer": "The resistance to phase transformation consists of strain energy and interface energy." }, { "idx": 3271, "question": "What would be the result if carburizing of steel parts is not conducted in the γ phase region?", "answer": "Since the maximum carbon solubility (mass fraction) in α-Fe is only 0.0218%, for steels with a carbon mass fraction greater than 0.0218%, the carbon concentration gradient in the parts during carburizing would be zero, making carburizing impossible. Even for pure iron, carburizing in the α phase region results in a very small concentration gradient in the iron, and a high carbon layer cannot be obtained on the surface. Additionally, due to the low temperature, the diffusion coefficient is also very small, making the carburizing process extremely slow and practically meaningless." }, { "idx": 3272, "question": "What are the main characteristics of diffusion-type phase transformations?", "answer": "The fundamental characteristics of diffusion-type phase transformations are: $\\textcircled{1}$ Atomic diffusion occurs during the phase transformation, and the transformation rate is controlled by diffusion, i.e., determined by the diffusion speed. $\\textcircled{2}$ In alloy phase transformations, the compositions of the new phase and the parent phase are often different. $\\textcircled{3}$ There is only volume change caused by the difference in specific volume between the new phase and the parent phase, with no shape alteration. Phase transformations such as allotropic transformation in pure metals, polymorphic transformation in solid solutions, precipitation transformation, eutectoid transformation, spinodal decomposition, and ordering transformation all belong to diffusion-type phase transformations." }, { "idx": 3273, "question": "Explain the influence of crystal defects on the nucleation of solid-state phase transformations in metals.", "answer": "During solid-state phase transformations, various crystal defects present in the parent phase, such as grain boundaries, dislocations, and vacancies, significantly promote the transformation. The new phase often preferentially nucleates at defect sites, and crystal defects have a substantial impact on processes such as nucleus growth and component diffusion." }, { "idx": 3274, "question": "What is the actual grain size of austenite?", "answer": "The size of austenite grains obtained under a specific heat treatment condition is generally referred to as the actual grain size." }, { "idx": 3275, "question": "What is the reason for the formation of a 'precipitate-free zone' near grain boundaries in age-hardening alloys?", "answer": "In the case of the decomposition of a supersaturated solid solution, when rapidly cooled from a high temperature, along with the solute atoms being supersaturatedly retained in the solid solution, a large number of supersaturated vacancies are also retained. These vacancies, on one hand, promote the diffusion of solute atoms and, on the other hand, act as nucleation sites for precipitates, facilitating heterogeneous nucleation and causing the precipitates to disperse throughout the matrix. When observing the distribution of precipitates in age-hardening alloys, a 'precipitate-free zone' is often seen near grain boundaries, where no precipitates are visible. This is because the supersaturated vacancies near the grain boundaries diffuse to the grain boundaries and disappear, so no heterogeneous nucleation or precipitation occurs in this region." }, { "idx": 3276, "question": "What is the reason for the appearance of transition phases during the solid-state phase transformation of metals?", "answer": "During solid-state phase transformations, metastable phases often form first to reduce surface energy, thus frequently resulting in transition phases. For example, when the undercooling is significant, the critical size for the formation of the new phase is very small, and the new phase per unit volume has a large surface area, so the interfacial energy poses a substantial barrier to nucleation. In such cases, transition phases with low interfacial energy and coherent interfaces are more likely to form to reduce the nucleation work and facilitate nucleation. An example is the ε-carbide in lower bainite in steel, which forms a coherent interface with the matrix. During the tempering of martensite in steel, to reduce the interfacial energy generated by carbide formation, ε-carbide, which is coherent with the martensite, forms as a transition phase at lower tempering temperatures. As the tempering temperature increases, the value of ΔG_B rises, and cementite, which is incoherent with the matrix, gradually forms." }, { "idx": 3277, "question": "What is the initial grain size of austenite?", "answer": "In the heating transformation, when the transformation from pearlite to austenite has just been completed, the size of the austenite grains is called the initial grain size of austenite." }, { "idx": 3278, "question": "Describe the formation process of austenite", "answer": "Austenite formation process: a) Formation of austenite nuclei; b) Growth of austenite nuclei; c) Dissolution of residual cementite; d) Homogenization of austenite composition." }, { "idx": 3279, "question": "When diffusion occurs via the vacancy mechanism, each atomic jump corresponds to a reverse jump of a vacancy without creating new vacancies, yet the diffusion activation energy includes the vacancy formation energy. Is this statement correct? Please provide the correct explanation.", "answer": "This statement is incorrect. The macroscopic diffusion flux in solids is not the result of directional jumps of individual atoms, and the diffusion activation energy is not the energy barrier that must be overcome for each jump during atomic migration. The jumps of atoms in solids are random in nature, and the diffusion flux is the macroscopic manifestation of the statistical outcome of random jumps of diffusing particles (such as atoms or ions) in the solid. When diffusion in a crystal occurs via the vacancy mechanism, any atomic jump between two equilibrium positions must simultaneously satisfy two conditions: (1) The atom must possess energy higher than a certain critical value ΔG₆, i.e., the atomic jump activation energy, to overcome the resistance hindering the jump; (2) There must be a vacancy adjacent to the atom's equilibrium position. According to statistical thermodynamics theory, at a given temperature T, the probability Pf that any atom in the crystal has energy higher than ΔGf, i.e., the percentage of atoms with energy higher than ΔGf, is Pf = exp(−ΔGf / kT). The equilibrium vacancy concentration cv in the crystal, or the probability Pv that any atomic equilibrium position is vacant, is Pv = exp(−ΔGv / kT). Clearly, the probability P that an atom in the crystal undergoes a jump at any instant is P = PfPv = exp(−(ΔGf + ΔGv) / kT) = exp(−Q / RT), where Q = ΔGf + ΔGv is the diffusion activation energy for the vacancy diffusion mechanism. P also equals the percentage of atoms undergoing jumps at that instant." }, { "idx": 3280, "question": "What is the inherent grain size of austenite?", "answer": "According to the former national metallurgical industry standard, the grain size measured after heating the steel to 930°C ± 10°C, holding for 8 hours, and then cooling is called the inherent grain size. This grain size is used to indicate the tendency of austenite grain growth during heating." }, { "idx": 3281, "question": "Compare the differences in the microstructure transformation zones between the CCT diagram and TTT diagram of eutectoid carbon steel.", "answer": "The CCT diagram of eutectoid carbon steel only has the high-temperature pearlite transformation zone and the low-temperature martensite transformation zone, without the intermediate-temperature bainite transformation zone. In the CCT diagram, the Ps curve (pearlite start transformation line) and Pf curve (pearlite finish transformation line) shift to the lower right." }, { "idx": 3282, "question": "Why is the activation energy for atoms crossing an incoherent interface much smaller than that for crossing a semicoherent interface?", "answer": "$u=\\delta V_{0}\\mathrm{exp}\\big(-\\Delta\\mathrm{G/kT}\\big)$. For an incoherent interface, the $\\Delta G$ value equals the activation energy for grain boundary diffusion; whereas for a semicoherent interface, it can be considered roughly equal to the activation energy of atoms in the $\\beta$ phase (actually slightly smaller). Therefore, the activation energy for atoms crossing an incoherent interface is much smaller than that for crossing a semicoherent interface." }, { "idx": 3283, "question": "What is the significance of the TTT diagram for eutectoid carbon steel in practical heat treatment?", "answer": "The isothermal transformation C-curve (TTT curve) can determine the process parameters during isothermal heat treatment of steel, namely the isothermal temperature, isothermal time, and the microstructure after heat treatment." }, { "idx": 3284, "question": "Methods for controlling austenite grain size", "answer": "Methods for controlling austenite grain size: a) Establish appropriate heating specifications, including controlling heating temperature and holding time, rapid short-term heating; b) Control carbon content within a certain range and add a certain amount of alloying elements that hinder austenite grain growth, such as: Al, V, Ti, Zr, Nb, etc.; c) Consider the influence of the original microstructure, such as lamellar pearlite being more prone to coarsening than spheroidized pearlite during heating." }, { "idx": 3285, "question": "Under what conditions can austenite transform into lamellar pearlite?", "answer": "When the austenitizing temperature is high, the holding time is long, the composition of austenite is uniform, and the cooling rate is fast, lamellar pearlite is prone to form." }, { "idx": 3286, "question": "What is pearlite interlamellar spacing?", "answer": "The distance between the centers of two adjacent cementite (or ferrite) plates in lamellar pearlite, or the total thickness of a pair of ferrite and cementite plates, is called pearlite interlamellar spacing, denoted as S0." }, { "idx": 3287, "question": "What are the main characteristics of diffusionless phase transformations?", "answer": "Characteristics of diffusionless phase transformations: $\\textcircled{1}$ There is a shape change caused by uniform shear, as the atoms undergo collective coordinated motion during the phase transformation, resulting in a change in the crystal's external shape. If a polished specimen surface is prepared in advance, a relief effect will appear on the polished surface after this transformation occurs. The presence of this relief can be observed under a metallographic microscope. $\\textcircled{2}$ The phase transformation does not require diffusion, and the chemical composition of the new phase is the same as that of the parent phase. $\\textcircled{3}$ There is a specific crystallographic orientation relationship between the new phase and the parent phase. $\\textcircled{4}$ The phase boundary moves extremely fast, approaching the speed of sound. The martensitic transformation in steel and some alloys (Fe-Ni, Cu-Al, Ni-Ti) is a diffusionless phase transformation. Some pure metals (such as zirconium, titanium, lithium, cobalt) also undergo diffusionless phase transformations during allotropic transformations at low temperatures." }, { "idx": 3288, "question": "What is martensite?", "answer": "The supersaturated solid solution formed by carbon in α-Fe is called martensite." }, { "idx": 3289, "question": "What is the significance of the CCT diagram of eutectoid carbon steel in practical heat treatment?", "answer": "The CCT curve can determine the heat treatment process parameters of steel during continuous cooling, such as the critical quenching cooling rate, quenching medium, and microstructure after heat treatment." }, { "idx": 3290, "question": "What is the effect of pearlite interlamellar spacing and pearlite colonies on mechanical properties?", "answer": "The lower the formation temperature, the smaller the size of pearlite colonies and interlamellar spacing, the finer the pearlite structure, and the higher the strength." }, { "idx": 3291, "question": "What are the main characteristics of various types of solid-state phase transformations in metals?", "answer": "
Classification of solid-state phase transformationsCharacteristics of phase transformations
Allotropic transformation in pure metalsChanges from one crystal structure to another with temperature or pressure variation, involving nucleation and growth processes
Polymorphic transformation in solid solutionsSimilar to allotropic transformation, such as Y in Fe-Ni alloys
Precipitation transformationDecomposition of supersaturated solid solutions, precipitating metastable or stable secondary phases
Eutectoid transformationA single phase decomposes into two phases with different structures through eutectoid transformation, such as Y→α+FeC in Fe-C alloys
Peritectoid transformationTwo phases with different structures transform into another phase through peritectoid transformation, such as α+Y→β in Ag-Al alloys, often leaving residual α phase
Martensitic transformationNo change in composition between the new and old phases during transformation; atoms undergo coordinated rearrangement (shear) without diffusion, maintaining strict orientation relationships and coherency, with surface relief effects observable on polished surfaces
Massive transformationChanges in crystal structure of metals or alloys without composition change between new and old phases, characterized by nucleation and growth with minimal diffusion, rapid growth via incoherent interface migration, producing irregular massive products, as seen in pure iron, low-carbon steel, Cu-Al alloys, and Cu-Ga alloys
Bainitic transformationOccurs in steel and many non-ferrous alloys, combining features of martensitic and diffusional transformations, resulting in composition changes; in steel, bainitic transformation is considered to proceed via coherent shear of iron atoms and diffusion of carbon atoms
Spinodal decompositionA non-nucleation decomposition process where a solid solution decomposes into two phases with the same crystal structure but different compositions (continuously varying within a certain range)
Order-disorder transformationTransition of alloy elements from random to ordered arrangement without structural change
" }, { "idx": 3292, "question": "Under what conditions can austenite transform into granular pearlite?", "answer": "When the austenitizing temperature is low, the holding time is short, the heating transformation is incomplete, the composition of austenite is non-uniform, and subsequently during cooling, the pearlite transformation isothermal temperature is high, the isothermal time is sufficiently long, or the cooling rate is extremely slow, granular pearlite is prone to form." }, { "idx": 3293, "question": "What is a pearlite colony?", "answer": "In lamellar pearlite, the region where ferrite and cementite are alternately arranged in layers with roughly the same orientation is called a pearlite colony or pearlite nodule." }, { "idx": 3294, "question": "Describe the main characteristics of martensitic transformation.", "answer": "The main characteristics (key points) of martensitic transformation are: ① The transformation is diffusionless. This is based on two pieces of evidence: a) No compositional change occurs before and after the transformation, meaning the chemical composition of austenite and martensite is identical; b) Martensite can form at very low temperatures at high speeds. ② The transformation is shear-type, exhibiting surface relief phenomena. ③ It possesses certain crystallographic orientation relationships and habit planes. ④ The transformation occurs over a temperature range. ⑤ Rapid growth. ⑥ The transformation is incomplete." }, { "idx": 3295, "question": "What factors influence the interlamellar spacing of pearlite and the pearlite colony?", "answer": "They are mainly influenced by the formation temperature of pearlite. The lower the temperature, the smaller the pearlite colony size and the smaller the interlamellar spacing of pearlite, such as S0=8.02/ΔGT×10^3(nm)." }, { "idx": 3296, "question": "Compare the diffusivity during bainitic transformation, pearlitic transformation, and martensitic transformation", "answer": "In pearlitic transformation, both iron and carbon atoms can diffuse; in bainitic transformation, carbon atoms diffuse while iron atoms do not; in martensitic transformation, neither iron nor carbon atoms diffuse." }, { "idx": 3297, "question": "Is the statement that the strength and hardness of martensite mainly depend on the mass fraction of carbon correct? Why?", "answer": "Incorrect. The hardness of martensite primarily depends on the carbon content (mass fraction), but the strength of martensite not only depends on its hardness but also on the morphology of martensite and the size of martensite laths (or plates)." }, { "idx": 3298, "question": "Compare the formation temperatures of bainitic transformation, pearlitic transformation, and martensitic transformation", "answer": "Pearlitic transformation occurs in the high-temperature region (below A), bainitic transformation occurs in the medium-temperature region (below B), and martensitic transformation occurs in the low-temperature region (below Ms)." }, { "idx": 3299, "question": "Compare the transformation processes and leading phases of bainitic transformation, pearlitic transformation, and martensitic transformation", "answer": "The leading phase of pearlitic transformation is FeC, the leading phase of bainitic transformation is the a phase, and the leading phase of martensitic transformation is not clearly specified." }, { "idx": 3300, "question": "Compare the coherency during the transformations of bainite, pearlite, and martensite", "answer": "Pearlite transformation has no coherency, while bainite and martensite transformations have shear coherency and produce surface relief." }, { "idx": 3301, "question": "Compare the lattice shear during bainitic transformation, pearlitic transformation, and martensitic transformation", "answer": "Pearlitic transformation involves no lattice shear, while bainitic transformation and martensitic transformation involve lattice shear." }, { "idx": 3302, "question": "Is the statement that the plasticity and toughness of martensite mainly depend on its substructure correct? Why?", "answer": "Yes. The plasticity and toughness of martensite mainly depend on its substructure." }, { "idx": 3303, "question": "What is the microstructural transformation of quenched steel during tempering at 20-100°C?", "answer": "Carbon segregation in martensite" }, { "idx": 3304, "question": "Compare the distribution of alloying elements in bainitic transformation, pearlitic transformation, and martensitic transformation", "answer": "In pearlitic transformation, alloying elements are redistributed through diffusion, while in bainitic transformation and martensitic transformation, alloying elements do not diffuse." }, { "idx": 3305, "question": "Compare the mechanical properties of upper bainite and lower bainite", "answer": "Generally, the hardness and strength of lower bainite are higher than those of upper bainite, and the plasticity and toughness of lower bainite are also higher than those of upper bainite. Lower bainite has good comprehensive mechanical properties and better toughness." }, { "idx": 3306, "question": "Compare the hardness of the transformation products of bainite transformation, pearlite transformation, and martensite transformation", "answer": "The pearlite transformation product has low hardness, the bainite transformation product has medium hardness, and the martensite transformation product has high hardness." }, { "idx": 3307, "question": "What is the microstructural transformation of quenched steel during tempering at 250-400°C?", "answer": "Transformation of carbides" }, { "idx": 3308, "question": "Compare the microstructures of the transformation products of bainitic transformation, pearlitic transformation, and martensitic transformation", "answer": "The product of pearlitic transformation is a+FeC (lamellar), the product of upper bainitic transformation is y→a+FeC (non-lamellar), the product of lower bainitic transformation is Ya+e-carbide (non-lamellar), and the two most typical products of martensitic transformation are lath-shaped and plate-shaped." }, { "idx": 3309, "question": "Compare the completeness of isothermal transformation for bainite transformation, pearlite transformation, and martensite transformation", "answer": "Pearlite transformation can be completely transformed, some bainite transformations can be completely transformed while others cannot, and martensite transformation cannot be completely transformed." }, { "idx": 3310, "question": "Explain the application of upper bainite and lower bainite in production", "answer": "a) Isothermal quenching of bainite (lower bainite) is often used to replace quenching + low-temperature tempering; b) Ultra-high strength steel obtains a composite structure of bainite (lower) + martensite to achieve the best combination of strength and toughness." }, { "idx": 3311, "question": "What is the microstructure transformation of quenched steel during tempering at 100-250°C?", "answer": "Decomposition of martensite" }, { "idx": 3312, "question": "What is the microstructural transformation of quenched steel during tempering at 200-300°C?", "answer": "Transformation of retained austenite" }, { "idx": 3313, "question": "What is temper brittleness?", "answer": "When some steels are tempered within a certain temperature range, their impact toughness significantly decreases compared to being tempered at lower temperatures. This embrittlement phenomenon is called temper brittleness." }, { "idx": 3314, "question": "What is the microstructure transformation of quenched steel during tempering at 400-600°C?", "answer": "Cementite aggregation and growth, and recovery and recrystallization of α phase" }, { "idx": 3315, "question": "Describe the microstructure type, obtaining conditions, characteristics, and mechanical properties of medium-temperature tempered steel", "answer": "Medium-temperature tempering Tempered troostite 350-500°C High elastic limit" }, { "idx": 3316, "question": "Describe the types of low-temperature tempered structures in steel, the conditions for obtaining them, their structural characteristics, and mechanical properties", "answer": "Low-temperature tempering Tempered martensite 150-250°C High hardness, high wear resistance" }, { "idx": 3317, "question": "Describe the microstructure type, obtaining conditions, characteristics, and mechanical properties of steel after high-temperature tempering", "answer": "High-temperature tempering Tempered sorbite 500-650℃ Good combination of strength and toughness" }, { "idx": 3318, "question": "Compare the approximate diffusion distances/mm of carbon atoms during bainitic transformation, pearlitic transformation, and martensitic transformation", "answer": "In pearlitic transformation, the diffusion distance of carbon atoms is greater than 10mm; in bainitic transformation, the diffusion distance of carbon atoms is 0~10mm; in martensitic transformation, the diffusion distance of carbon atoms is 0mm." }, { "idx": 3319, "question": "Indicate the main purpose of normalizing 20 steel gears and the microstructure after normalizing", "answer": "Normalizing of 20 steel: improve machinability." }, { "idx": 3320, "question": "Explain the factors affecting the depth of the hardened layer.", "answer": "The depth of the hardened layer is related to factors such as the hardenability of the steel, the size of the workpiece, and the cooling capacity of the quenching medium." }, { "idx": 3321, "question": "Indicate the main purpose of normalizing T12 steel file and the structure after normalizing", "answer": "T12 steel file: Normalizing eliminates the network carbide and prepares for spheroidizing annealing." }, { "idx": 3322, "question": "What is the hardenability of steel?", "answer": "The hardenability of steel refers to the ability of steel to obtain martensite during quenching." }, { "idx": 3323, "question": "How to suppress temper brittleness?", "answer": "Methods to suppress temper brittleness: For the first type of temper brittleness, due to its irreversibility, it can only be avoided by tempering outside the embrittlement temperature range; if tempering must be performed within this temperature range, isothermal quenching can be used; adding Si to shift the low-temperature temper embrittlement range to higher temperatures, etc. Methods to suppress high-temperature temper brittleness: a) Rapid cooling after high-temperature tempering; b) Reducing the content of impurity elements in the steel; c) Adding appropriate amounts of Mo or W to the steel." }, { "idx": 3324, "question": "What is the hardenability of steel?", "answer": "The hardenability of steel refers to the maximum hardness that can be achieved after quenching." }, { "idx": 3325, "question": "What are the types and characteristics of temper brittleness?", "answer": "There are mainly two types of temper brittleness, namely low-temperature temper brittleness and high-temperature temper brittleness. Low-temperature temper brittleness, also known as the first type of temper brittleness, occurs to varying degrees in almost all industrial steels and is independent of the cooling rate after tempering. Therefore, it is also called irreversible temper brittleness (i.e., it occurs whether cooled rapidly or slowly). High-temperature temper brittleness, also known as the second type of temper brittleness, is characterized by its occurrence only in steels with specific compositions. Whether temper brittleness occurs depends on the cooling rate after tempering, i.e., it appears with slow cooling after tempering but not with rapid cooling, hence it is also called reversible temper brittleness." }, { "idx": 3326, "question": "Explain the factors affecting hardenability.", "answer": "Hardenability mainly depends on the carbon content of martensite." }, { "idx": 3327, "question": "Describe the purpose of quenching", "answer": "Quenching can significantly improve the strength and hardness of steel. Combined with tempering at different temperatures, it can achieve a balance of strength, hardness, and toughness to meet various requirements." }, { "idx": 3328, "question": "What are the types of quenching methods", "answer": "The quenching methods include: 1. Single-liquid quenching; 2. Dual-liquid quenching; 3. Graded quenching; 4. Isothermal quenching." }, { "idx": 3329, "question": "There is a 20 steel workpiece with a diameter of φ12mm, which is carburized and then air-cooled, followed by normal quenching and tempering. Analyze the microstructure from the surface to the core of the workpiece after carburizing and air-cooling, and after quenching and tempering.", "answer": "After air-cooling: P + Fe3CI, P, P + F; After quenching and tempering: M⊥+ FeC, tempered M, tempered M + F." }, { "idx": 3330, "question": "After quenching T8 steel wire to 275°C and holding for 1s, indicate its phase and composition.", "answer": "Undercooled austenite (w_C=0.77%)" }, { "idx": 3331, "question": "After quenching T8 steel wire to 20°C and holding, indicate its phases and composition.", "answer": "M+A′ (w_C=0.77% in M, A′)" }, { "idx": 3332, "question": "Explain the factors affecting hardenability.", "answer": "Factors affecting hardenability: a) Hardenability depends on the critical cooling rate of the steel during quenching, i.e., the cooling rate corresponding to the 'nose tip' of the continuous cooling 'C' curve, which is determined by the distance between the 'C' curve and the vertical axis. The main factors influencing the shape and position of the 'C' curve are alloying elements. Except for Co and Al, the addition of all alloying elements shifts the 'C' curve to the right, increasing hardenability. Hardenability does not vary with the shape and size of the workpiece or the cooling capacity of the medium." }, { "idx": 3333, "question": "When complete austenitization of steel is required, is the original structure better with coarse granular pearlite or fine lamellar pearlite?", "answer": "Fine lamellar pearlite is better." }, { "idx": 3334, "question": "Explain why the initial structure with fine lamellar pearlite is better using the austenite isothermal formation mechanism.", "answer": "Increasing the dispersion of carbides can accelerate the transformation from pearlite to austenite." }, { "idx": 3335, "question": "After quenching T8 steel wire to 550°C and holding for 1s, indicate its phase and composition.", "answer": "B⊥[F(w_C=0.0218%)+Fe3C(w_C=6.69%)]" }, { "idx": 3336, "question": "Determine the heat treatment conditions (temperature, time, cooling rate, etc.) for normalizing φ25 hot-rolled eutectoid steel.", "answer": "Normalizing: 780°C holding for 1h, air cooling." }, { "idx": 3337, "question": "Indicate the main purpose of normalizing a small shaft made of 45 steel and the microstructure after normalizing", "answer": "Small shaft made of 45 steel: Using normalizing instead of quenching and tempering as the final heat treatment can achieve good comprehensive mechanical properties similar to those obtained through quenching and tempering." }, { "idx": 3338, "question": "After heating T8 steel wire to 870°C and holding for 1h, indicate its phase and composition.", "answer": "Austenite, w_C=0.77%" }, { "idx": 3339, "question": "Determine the heat treatment conditions (temperature, time, cooling rate, etc.) for φ25 hot-rolled eutectoid steel during annealing.", "answer": "Annealing: hold at 760°C for 1h, furnace cooling." }, { "idx": 3340, "question": "There is a batch of approximately 10,000 pieces of 45 steel gears, which are subjected to relatively low contact stress but require good wear resistance on the teeth and minimal heat treatment deformation. What surface treatment should be selected for these parts? Why?", "answer": "High-frequency surface quenching. After high-frequency surface quenching, the surface hardness is 1-2 HRC higher than conventional quenching, providing good wear resistance. Additionally, since high-frequency surface quenching results in minimal heat treatment deformation, this process is selected." }, { "idx": 3341, "question": "Determine the heat treatment conditions (temperature, time, cooling rate, etc.) for spheroidizing annealing of φ25 hot-rolled eutectoid steel.", "answer": "Spheroidizing annealing: hold at 780°C for 1h, air cool; hold at 680-700°C for 1h, air cool." }, { "idx": 3342, "question": "If a $\\phi5\\$ hot-rolled steel specimen is heated to 650℃, held isothermally for 15s, and then quenched in water, can the isothermal transformation curve be used to analyze the final microstructure obtained?", "answer": "No. The isothermal transformation curve describes the transformation of undercooled austenite. The hot-rolled eutectoid steel heated to 650℃ does not undergo austenitization." }, { "idx": 3343, "question": "In eutectoid steel, both pearlite and tempered martensite (assuming tempered at 400°C) are composed of ferrite and carbides. Why do the morphology of ferrite and the number of internal defects in tempered martensite differ from those in pearlite?", "answer": "The pearlite in eutectoid carbon steel is lamellar, consisting of F plates (with a low number of internal defects) + Fe3C plates. After tempering at 400°C, the martensite structure becomes tempered troostite, F45 + Fe3C45, which is acicular ferrite and fine granular cementite. The ferrite in pearlite is lamellar with a low number of crystal defects (dislocations, twins), whereas the tempered martensite contains acicular ferrite, where the number of crystal defects is much higher than in pearlite due to quenching." }, { "idx": 3344, "question": "When hypereutectoid steel is slowly cooled from the austenite region, secondary cementite precipitates along the austenite grain boundaries in a network pattern. How can the network cementite be eliminated by heating above Ac cm followed by quenching and subsequent treatment?", "answer": "Heating above Ac cm followed by quenching results in martensite and retained austenite, which is then tempered at high temperature to obtain spheroidized pearlite." }, { "idx": 3345, "question": "Among the various types of steel currently used in industry, cold-drawn high-carbon steel wire has the highest strength (up to 3000 MPa). This type of wire is generally obtained by first undergoing austenitization, followed by isothermal treatment in a lead bath at 500°C, and then cold drawing. Analyze the reasons for the high strength of this material.", "answer": "When high-carbon steel undergoes isothermal treatment at 550°C after austenitization, the interlamellar spacing of pearlite can be made very small. Subsequent cold drawing not only increases the dislocation density in the ferrite of the pearlite, causing work hardening and subgrain refinement, but also further reduces the interlamellar spacing of the pearlite, thereby achieving high strength." }, { "idx": 3346, "question": "In eutectoid steel, both pearlite and tempered martensite (assuming tempered at 400°C) consist of ferrite and carbide. Why do the morphology and dispersion of carbides in tempered martensite differ from those in pearlite?", "answer": "The carbides in pearlite exhibit a lamellar morphology with low dispersion, whereas the carbides in martensite are fine particles with high dispersion." }, { "idx": 3347, "question": "To predict the normalizing effect of φ25 eutectoid steel bars, how applicable is the iron-carbon phase diagram?", "answer": "The iron-carbon phase diagram can only be used to determine the microstructure under equilibrium cooling conditions and is not applicable for predicting the normalizing effect." }, { "idx": 3348, "question": "Compare the advantages and disadvantages of various quenching methods", "answer": "1. Single-liquid quenching: This method is widely used, simple to operate, and easy to mechanize. The drawback is that some steel parts are prone to deformation and cracking when water-quenched, while oil quenching may result in insufficient hardness. 2. Dual-liquid quenching: This method is difficult to control. 3. Graded quenching: During quenching, the internal temperature of the workpiece is uniform, and the structural transformation occurs almost simultaneously, thereby reducing internal stress and significantly lowering the tendency for deformation and cracking. However, it is only suitable for workpieces with strict deformation requirements and smaller dimensions. 4. Isothermal quenching: Workpieces treated by this method exhibit high strength and hardness, as well as good plasticity and toughness. Additionally, the quenching stress is low, and deformation is minimal. It is mostly used for workpieces with complex shapes, smaller dimensions, and higher precision requirements." }, { "idx": 3349, "question": "To predict the normalizing effect of φ25 eutectoid steel bars, how applicable is the isothermal transformation curve?", "answer": "Using the isothermal transformation curve can only provide an approximate estimation of the normalizing effect." }, { "idx": 3350, "question": "To predict the normalizing effect of φ25 eutectoid steel bars, how applicable is the continuous cooling transformation curve?", "answer": "To predict the microstructure and hardness after cooling, the continuous cooling transformation curve should be used, as normalizing is a continuous cooling process." }, { "idx": 3351, "question": "Suppose the microstructure of an alloy consists of multiple two-phase lamellar eutectic domains with different orientations and uniform distribution. The actual interlamellar spacing in each domain is equal. How should this interlamellar spacing be determined under a metallographic microscope?", "answer": "Measure the minimum interlamellar spacing." }, { "idx": 3352, "question": "In eutectoid steel, both pearlite and tempered martensite (assuming tempered at 400°C) are composed of ferrite and carbide. Why is the strength of tempered martensite higher than that of the strongest pearlite?", "answer": "The morphology of ferrite and the number of internal defects differ between the two. In pearlite, the ferrite is lamellar with fewer crystal defects (dislocations, twins), whereas in tempered martensite, the ferrite is acicular, and the number of crystal defects is much higher due to quenching. The morphology and dispersion of carbides also differ. In pearlite, the carbides are lamellar with low dispersion, while in martensite, the carbides are fine particles with high dispersion. Thus, the strength of tempered martensite is higher than that of the strongest pearlite." }, { "idx": 3353, "question": "Suppose the microstructure of a certain alloy consists of spherical second-phase particles and a matrix. If the actual diameters of the second-phase particles are all equal, how should the particle size be determined under a metallographic microscope?", "answer": "Measure the diameter of the largest particle." }, { "idx": 3354, "question": "Using the fact that pearlite, proeutectoid ferrite, and proeutectoid cementite preferentially precipitate at austenite grain boundaries, propose a method for determining the austenite grain size of steel with a carbon mass fraction of 0.8%.", "answer": "After austenitization, isothermally hold near the knee of the C-curve until partial pearlite transformation occurs (e.g., around 10%), then quench. Since pearlite preferentially nucleates and grows at austenite grain boundaries, the grain boundaries will appear as dark pearlite, while the grain interiors will be less etchable martensite. This allows convenient measurement of austenite grain size under a microscope. Alternatively, this microstructure can be obtained by cooling at an appropriate rate after austenitization (e.g., oil quenching for specimens about 10mm thick). For better control, a rod-shaped specimen can be austenitized with one end immersed in water while the rest cools in air. This creates a range of cooling rates along the rod's length, ensuring there will always be a section where pearlite precipitates along austenite grain boundaries with martensite in the grain interiors." }, { "idx": 3355, "question": "Using the fact that pearlite, proeutectoid ferrite, and proeutectoid cementite preferentially precipitate at austenite grain boundaries, propose a method for determining the austenite grain size of steel with a carbon mass fraction of 1.2%.", "answer": "Proeutectoid cementite precipitates along austenite grain boundaries. Due to the low content of proeutectoid cementite and its tendency to form a network structure during slow cooling, the steel can be fully austenitized and then furnace-cooled to approximately 650°C before air cooling." }, { "idx": 3356, "question": "A heat-treated part is made of 55 steel with an effective thickness of 10 mm, requiring a quenched and tempered hardness of HRC 32-37. Using oil quenching and tempering can also achieve the hardness specified in the drawing. Analyze whether this treatment method is reasonable.", "answer": "Generally speaking, it is not reasonable. Because oil quenching will result in a large amount of lamellar pearlite transformation products. Although the hardness after tempering falls within the required range, the overall mechanical properties are inferior to those of tempered martensite." }, { "idx": 3357, "question": "Classify according to the interaction between alloying elements and carbon, and indicate which elements are weak carbide-forming elements", "answer": "Weak carbide-forming elements: manganese" }, { "idx": 3358, "question": "Using the fact that pearlite, proeutectoid ferrite, and proeutectoid cementite preferentially precipitate at austenite grain boundaries, propose a method for determining the austenite grain size of steel with a carbon mass fraction of 0.4%.", "answer": "After austenitizing the specimen at the required temperature, isothermally hold it below A1 but above the knee of the C curve to allow proeutectoid ferrite to precipitate in a network along the austenite grain boundaries, then cool. Under microscopic observation, the network of ferrite precipitated along the austenite grain boundaries represents the original austenite grain boundaries, based on which the austenite grain size at the required temperature can be determined." }, { "idx": 3359, "question": "Classify according to the interaction between alloying elements and carbon, and indicate which elements do not form carbides", "answer": "Non-carbide forming elements: nickel, silicon, cobalt, aluminum, copper, boron, nitrogen, etc." }, { "idx": 3360, "question": "When hypereutectoid steel is slowly cooled from the austenite region, secondary cementite will precipitate along the austenite grain boundaries in a network form. How can the network cementite be eliminated by heating above Ac cm followed by air cooling and subsequent treatment?", "answer": "Heating above Ac cm followed by air cooling to obtain a pseudo-eutectoid structure, then performing high-temperature tempering or spheroidizing annealing." }, { "idx": 3361, "question": "What are the ways to improve the strength of materials", "answer": "The strengthening effects of alloying elements in steel mainly include the following four methods: solid solution strengthening, grain boundary strengthening, second-phase strengthening, and dislocation strengthening. By applying these four methods individually or in combination, the strength of steel can be effectively improved." }, { "idx": 3362, "question": "Discuss the factors affecting the plasticity of materials", "answer": "The main factors affecting the plasticity of steel include: (1) the influence of solute atoms; (2) the influence of grain size; (3) the influence of the second phase; (4) the influence of dislocation strengthening, etc." }, { "idx": 3363, "question": "Discuss the factors affecting the strength of materials", "answer": "The mechanical properties of materials are related to their chemical composition, internal microstructure, inclusions and surface microstructure, as well as stress state." }, { "idx": 3364, "question": "Classify according to the interaction between alloying elements and carbon, and indicate which elements are strong carbide-forming elements. What are the performance characteristics of strong carbides?", "answer": "Strong carbide-forming elements: titanium, zirconium, vanadium, niobium. Strong carbides are relatively stable, with high melting points, high hardness, and relatively high brittleness." }, { "idx": 3365, "question": "What is alloy cementite, and how does its performance compare with cementite", "answer": "Alloy cementite is formed when alloying elements dissolve into cementite. Alloy cementite has higher stability than ordinary cementite and exhibits a more significant strengthening effect on the steel matrix." }, { "idx": 3366, "question": "Why can shot peening and surface rolling significantly improve the fatigue limit of materials?", "answer": "Shot peening and surface rolling can create a very thin work-hardened layer on the surface of the workpiece, which not only has a higher yield limit but also contains significant compressive stress (400-500 MPa). This weakens the effect of tensile stress on the surface and delays the formation of surface cracks, thereby significantly improving the fatigue limit of the material." }, { "idx": 3367, "question": "Describe the role of alloying elements in ordinary low-alloy steels", "answer": "Adding alloying elements such as manganese is to improve the strength of the steel." }, { "idx": 3368, "question": "What elements are commonly used to improve hardenability?", "answer": "Cr, Ni, Mn, Si, B" }, { "idx": 3369, "question": "Discuss the approaches to improve the plasticity of materials", "answer": "Approaches to improve plasticity: (1) Adding small amounts of alloying elements such as titanium and vanadium to steel can fix carbon and nitrogen and form carbides or nitrides, thereby improving the plasticity of steel. (2) Refining grain size is beneficial for enhancing plasticity. (3) When strengthening with a second phase, the quantity, size, shape, and distribution of carbides can be controlled by combining alloying with tempering and spheroidization treatments; reducing the number of inclusions in steel and controlling their morphology. (4) When strengthening through cold deformation, adding trace elements such as titanium, vanadium, and zirconium to fix interstitial atoms and prevent their segregation to dislocations can improve the plasticity of steel to some extent." }, { "idx": 3370, "question": "Why are carburizing steels mostly used to manufacture gears for automobile and tractor gearboxes and rear axle gears?", "answer": "Their working conditions are demanding, requiring high toughness in the core and high hardness on the surface." }, { "idx": 3371, "question": "What is the reason why alloying elements improve hardenability?", "answer": "It shifts the C-curve to the right and reduces the critical cooling rate for quenching" }, { "idx": 3372, "question": "What are the ways to improve the strength and toughness of mild steel?", "answer": "The way to improve is solid solution strengthening." }, { "idx": 3373, "question": "What is the reason why alloying elements improve the tempering stability of steel?", "answer": "Non-carbide forming elements have the effect of delaying the decomposition of martensite. This is because they can dissolve into ε-carbide and stabilize it, slowing down the aggregation rate of carbides. Compared to carbon steel, the final decomposition temperature of martensite may be delayed to 350-500℃, which is 100-150℃ higher than that of carbon steel." }, { "idx": 3374, "question": "What elements are commonly used to improve the tempering stability of steel?", "answer": "Common elements: Si, Co." }, { "idx": 3375, "question": "Compare the heat treatment process characteristics of hot work die steel and alloy quenched and tempered steel.", "answer": "The heat treatment processes for both are quenching + high temperature tempering." }, { "idx": 3376, "question": "What are the characteristics of the heat treatment process for high-speed steel?", "answer": "The most prominent features of the process are the very high quenching heating temperature, high tempering temperature, multiple tempering cycles, and the use of preheating during quenching heating." }, { "idx": 3377, "question": "Why are 40CrNiMo and 37SiMnCrMoV steels (25mm in diameter) in the normalized state difficult to machine? Please consider the most economical method to improve machinability.", "answer": "The high carbon content and the presence of alloying elements result in excessive hardness in the normalized state, making machining difficult. The most economical method to improve machinability is annealing." }, { "idx": 3378, "question": "How to select the microstructure state of steel for machine parts based on their service conditions?", "answer": "For less critical parts where the comprehensive mechanical properties are not highly demanded, normalize to obtain ferrite + pearlite. For parts requiring good surface wear resistance and high contact fatigue resistance, while the entire part bears impact loads and the core demands higher toughness, carburizing, quenching, and low-temperature tempering can be applied, resulting in a surface of martensite with high hardness and wear resistance, and a core of ferrite + pearlite with higher toughness. For parts requiring higher comprehensive mechanical properties, apply quenching and tempering to obtain tempered sorbite. Carbon steels and alloy steels with relatively high carbon content can be used to manufacture springs, and the heat treatment process of quenching and medium-temperature tempering should be adopted to achieve tempered troostite. Bearings should be made of rolling bearing steel, treated with quenching and low-temperature tempering to obtain tempered martensite with high hardness and wear resistance." }, { "idx": 3379, "question": "What is the role of carbon and alloying elements in high-speed steel?", "answer": "The purpose of high carbon content is to form carbides with carbide-forming elements such as Cr, W, Mo, and V, and to ensure a strong martensitic matrix to enhance the hardness and wear resistance of the steel. W, Mo, and V primarily improve the red hardness of the steel, as the carbides formed by these elements have high hardness and produce a 'secondary hardening' effect, thereby significantly enhancing the steel's red hardness, hardness, and wear resistance. Cr mainly improves the hardenability of the steel." }, { "idx": 3380, "question": "Describe the role of carbon in low-alloy steel", "answer": "Low carbon content is to ensure the plasticity, toughness and weldability of the steel." }, { "idx": 3381, "question": "Why are medium carbon (alloy) steels mostly used to manufacture gearbox gears for machine tools?", "answer": "Their working conditions are much better than those of automobiles, requiring certain toughness in the core and higher hardness on the surface." }, { "idx": 3382, "question": "How to select the carbon content in steel for machine parts based on their service conditions?", "answer": "For non-critical parts with low requirements for comprehensive mechanical properties, medium carbon steel can be selected. For parts requiring good surface wear resistance and high contact fatigue resistance, while the entire part withstands impact loads and the core requires high toughness, carburizing steel, i.e., low carbon steel or low carbon alloy steel, should be selected. For parts requiring high comprehensive mechanical properties, quenched and tempered steel, i.e., medium carbon steel or medium carbon alloy steel, should be selected. Carbon steel and alloy steel with higher carbon content can be used to manufacture springs. Bearings should be made of rolling bearing steel." }, { "idx": 3383, "question": "Compare the alloying characteristics of hot-work die steel and alloy quenched and tempered steel, and analyze the similarities and differences in the roles of alloying elements.", "answer": "The main alloying elements (Si, Mn, Cr, Ni, B, etc.) in quenched and tempered steel primarily enhance the hardenability of the steel; the auxiliary elements (W, Mo, V, Ti) mainly refine the austenite grain size, thereby refining the ferrite grain size in tempered sorbite. Hot-work die steel often contains alloying elements such as Cr, Mn, Si, Mo, W, and V to improve the steel's hardenability, tempering stability, and wear resistance, while also suppressing secondary temper brittleness. Cr, Si, and W enhance fatigue resistance." }, { "idx": 3384, "question": "What is the approximate quenching process in the manufacturing route of round dies made from 9SiCr steel?", "answer": "Quenching process: Heating temperature 850-870°C (oil quenching)." }, { "idx": 3385, "question": "The large screwdriver requires the shank to be fine pearlite and the tip to be tempered martensite, with only one external heat source. How should it be processed?", "answer": "Overall normalizing, local quenching + low-temperature tempering." }, { "idx": 3386, "question": "What is the purpose of quenching and tempering in the process route of making round dies from 9SiCr steel?", "answer": "Quenching and tempering are to obtain tempered martensite, ensuring high hardness and high wear resistance after heat treatment." }, { "idx": 3387, "question": "The bolts on the gear hobbing machine should have been made of 45 steel, but T12 steel was mistakenly used, and the annealing process for 45 steel was followed. What structure will be obtained in this case? What will be the performance?", "answer": "Annealing: The structure will be lamellar pearlite and networked carbides. Steel with such a structure is prone to deformation and cracking during quenching." }, { "idx": 3388, "question": "The bolts on the gear hobbing machine should have been made of 45 steel, but T12 steel was mistakenly used. The high-temperature tempering process for 45 steel was still applied. What microstructure will be obtained in this case? What will be the performance?", "answer": "High-temperature tempering: The microstructure consists of coarse tempered sorbite and retained austenite; Performance: Poor toughness." }, { "idx": 3389, "question": "What is the general process of spheroidizing annealing in the manufacturing process of round dies made from 9SiCr steel?", "answer": "Spheroidizing annealing process: heating temperature 790-810°C, isothermal temperature 700-720°C." }, { "idx": 3390, "question": "What is the approximate tempering process in the process route of round dies made of 9SiCr steel?", "answer": "Tempering process: 160-180°C." }, { "idx": 3391, "question": "A batch of carbon tool steel workpieces were found to have insufficient hardness after quenching. It is estimated that either surface decarburization occurred or the cooling rate during quenching was too low. How to quickly determine the cause of the problem.", "answer": "Metallographic examination." }, { "idx": 3392, "question": "What is the purpose of spheroidizing annealing in the process route of making round dies with 9SiCr steel?", "answer": "Spheroidizing annealing is to eliminate forging stress, obtain spheroidized pearlite and carbides, reduce hardness to facilitate machining, prepare the microstructure for quenching, and minimize deformation and cracking during quenching." }, { "idx": 3393, "question": "Briefly describe the alloying principle of stainless steel.", "answer": "Alloying principle: Adding alloying elements to form a stable, dense, and strong protective film on the surface of the steel, obtaining a single-phase structure in the steel, and increasing the electrode potential of the solid solution." }, { "idx": 3394, "question": "The bolts on the gear hobbing machine should have been made of 45 steel, but T12 steel was mistakenly used instead, and the quenching process for 45 steel was still applied. What microstructure will be obtained in this case? What will the properties be like?", "answer": "Quenching: The microstructure will consist of martensite, a significant amount of retained austenite, and a small amount of carbide. Properties: Martensite is hard and brittle, retained austenite has low hardness, resulting in non-uniform properties." }, { "idx": 3395, "question": "Why is Cr12MoV steel not stainless steel?", "answer": "Because Cr is the main factor that makes steel corrosion-resistant. It can increase the electrode potential of steel. If its mass fraction is small (below 13 %), the electrode potential cannot be significantly increased, nor can a single-phase structure be formed, so the corrosion resistance of the steel cannot be significantly improved." }, { "idx": 3396, "question": "After the wire drawing die made of high-carbon high-chromium steel wears out, the inner hole has a slight oversize. What heat treatment can be used to reduce the inner hole diameter?", "answer": "Water quenching." }, { "idx": 3397, "question": "Why can't Cr12MoV steel be turned into stainless steel through heat treatment?", "answer": "Because Cr is the main factor that makes steel corrosion-resistant. It can increase the electrode potential of the steel. If its mass fraction is small (below 13 %), the electrode potential cannot be significantly increased, nor can a single-phase structure be formed, so the corrosion resistance of the steel cannot be significantly improved." }, { "idx": 3398, "question": "Analyze the special effects of alloying elements (such as Cr, Mo, W, etc.) on improving the thermal strength of steel", "answer": "Alloying elements (Cr, Mo, W, etc.) can increase the recrystallization temperature of steel, thereby improving its thermal strength." }, { "idx": 3399, "question": "How is the range of high, medium, and low carbon steels usually divided?", "answer": "High carbon steel: wC>0.60%, medium carbon steel: wC=0.25%~0.60%, low carbon steel: wC≤0.25%." }, { "idx": 3400, "question": "Analyze the special role of alloying elements (such as Cr, Si, Al, etc.) in improving the thermal stability of steel", "answer": "Alloying elements (Cr, Si, Al, etc.) form dense protective oxide films at high temperatures, preventing the oxidation of steel and thereby enhancing thermal stability." }, { "idx": 3401, "question": "What elements is carbon steel composed of?", "answer": "Carbon steel is mainly composed of Fe and C." }, { "idx": 3402, "question": "What are the characteristics of carbon steel and alloy steel in terms of performance?", "answer": "Carbon steel is inferior to alloy steel in mechanical, physical, and chemical properties such as strength, wear resistance, toughness, and corrosion resistance." }, { "idx": 3403, "question": "Why do the dimensions of some measuring tools change during storage and use?", "answer": "The reason for the dimensional changes of measuring tools during storage and use is: due to the excessive amount of retained austenite after quenching and tempering, stress relaxation occurs during storage and use, causing the retained austenite to transform, thereby leading to dimensional changes." }, { "idx": 3404, "question": "What are the commonly used elements in alloy steel?", "answer": "Alloy steel is mainly composed of elements such as Si, Mn, Cr, Ni, Mo, and V, in addition to Fe and C." }, { "idx": 3405, "question": "Explain the meaning of the steel grade 4Cr13", "answer": "4Cr13 is a martensitic stainless steel, wc=0.4%, w=13%" }, { "idx": 3406, "question": "How is the range of high, medium, and low alloy steels usually divided?", "answer": "High alloy steel: wMe>10%, medium alloy steel: wMe=5%~10%, low alloy steel: wMe<5%." }, { "idx": 3407, "question": "What measures can be taken to ensure the long-term dimensional stability of measuring tools?", "answer": "The method of cold treatment after quenching can be adopted to reduce the amount of retained austenite and achieve long-term dimensional stability." }, { "idx": 3408, "question": "Explain the meaning of the steel grade 16Mn", "answer": "16Mn is a low-alloy structural steel, wc=0.16%, 10Mn=1.4%" }, { "idx": 3409, "question": "Explain the meaning of the steel grade 20CrMnTi", "answer": "20CrMnTi is a carburizing steel with wc=0.20%, wMn=1%, wCr=0.3%, wC=1.1%, wTi=0.09%" }, { "idx": 3410, "question": "Compare the alloying direction of high-temperature structural steel and room-temperature structural steel", "answer": "The alloying direction of high-temperature structural steel focuses on adding elements that improve thermal strength and thermal stability (such as Cr, Mo, W, Si, Al, etc.), while the alloying direction of room-temperature structural steel pays more attention to improving strength, toughness, and other room-temperature properties." }, { "idx": 3411, "question": "Explain the meaning of the steel grade 08F", "answer": "08F is a plain carbon structural steel, wc=0.08% rimmed steel" }, { "idx": 3412, "question": "What is the meaning of the commonly referred terms 'ordinary steel', 'quality steel', and 'high-quality steel'?", "answer": "It mainly refers to the difference in the mass fraction of phosphorus and sulfur. Ordinary steel: $w_{P}\\\\leqslant0.045\\\\%$ $w_{\\\\mathrm{S}}\\\\leqslant0.05\\\\%;$ quality steel: $\\\\scriptstyle w_{\\\\mathrm{P}}\\\\leq0.035\\\\%,w_{\\\\mathrm{S}}\\\\leq0.035\\\\%;$ high-quality steel: $w_{\\\\mathrm{P}}{\\\\le}0.025\\\\%$ $w_{\\\\mathrm{S}}{\\\\leq}0.025\\\\%$" }, { "idx": 3413, "question": "Explain the meaning of the steel grade T12A", "answer": "T12A is a carbon tool steel, wc=1.2% high-grade quality steel" }, { "idx": 3414, "question": "Explain the meaning of the steel grade 9SiCr", "answer": "9SiCr is a low-alloy tool steel with wc=0.9%, ws=1%, Wc=1.1%, wMn=0.45%" }, { "idx": 3415, "question": "Explain the meaning of the steel grade 1Cr18Ni9Ti", "answer": "1Cr18Ni9Ti is an austenitic stainless steel, wc=0.1%, wc=18%, WN=9%, wv=0.7%" }, { "idx": 3416, "question": "Explain the meaning of the steel grade 3Cr2W8", "answer": "3Cr2W8V is a hot extrusion die steel with wc=0.3%, wcr=2.5%, ww=8%, and wv=0.3%" }, { "idx": 3417, "question": "Explain the meaning of the steel grade Cr12MoV", "answer": "Cr12MoV is a cold work die steel with wc=1.55%, wc=12%, WM=0.5%, wy=0.25%" }, { "idx": 3418, "question": "Explain the meaning of the steel grade 00Cr18Ni10", "answer": "00Cr18Ni10 is an austenitic stainless steel, with Wc≤0.03%, WCr=18%, WNi=10%" }, { "idx": 3419, "question": "Explain the meaning of the steel grade W6Mo5Cr4V2", "answer": "W6Mo5Cr4V2 is a high-speed steel, wc=0.85%, ww=6%, WMo=5%, wc=4%, wv=3%" }, { "idx": 3420, "question": "Which commonly used alloying elements belong to the γ-phase field narrowing elements?", "answer": "Niobium, boron, and zirconium narrow the γ-phase field." }, { "idx": 3421, "question": "Explain the meaning of the steel grade 5CrMnMo", "answer": "5CrMnMo is a hot forging die steel, wc=0.5%, 0.75Cr, 10Mm=1.4%, wMo=0.25%" }, { "idx": 3422, "question": "Explain the meaning of the steel grade GCr15SiMn", "answer": "GCr15SiMn is a rolling bearing steel, wc=1.5%, Wc=1%, wsi=0.5%, WM=1.05%" }, { "idx": 3423, "question": "Which commonly used alloying elements belong to the γ-phase expanding elements?", "answer": "Nickel, manganese, cobalt, carbon, nitrogen, and copper expand the γ-phase region." }, { "idx": 3424, "question": "Which of these elements can form infinite solid solutions with Fe? What are the conditions?", "answer": "Manganese, nickel, and cobalt can form infinite solid solutions with γ-Fe. Whether alloying elements expand or contract the γ region mainly depends on factors such as their lattice type, atomic size, electronic structure, and electrochemical properties." }, { "idx": 3425, "question": "Which commonly used alloying elements belong to the elements that close the γ region?", "answer": "Silicon, chromium, tungsten, molybdenum, phosphorus, vanadium, titanium, and aluminum close the γ region." }, { "idx": 3426, "question": "Explain the meaning of the steel grade 55Si2Mn", "answer": "55Si2Mn is a spring steel, wc=0.55%, ws=1.75%, WMn=0.75%" }, { "idx": 3427, "question": "What is the practical significance of these elements' influence on the γ region?", "answer": "Its engineering practical significance: For example, to ensure that the steel has good corrosion resistance (such as stainless steel), it is necessary to obtain a single-phase structure at room temperature. This is achieved by applying the above-mentioned principles, controlling the types and amounts of alloying elements to obtain single-phase austenite or ferrite and other single structures in the steel under room temperature conditions." }, { "idx": 3428, "question": "Explain the meaning of the steel grade ZGMn13", "answer": "ZGMn13 is a high manganese wear-resistant steel, wc=1.15%, wMn=13%" }, { "idx": 3429, "question": "What are the differences in heat treatment processes between alloy steel and carbon steel?", "answer": "Due to the influence of alloying elements on critical temperature, point E, and point S, the heat treatment temperature of alloy steel differs from that of carbon steel." }, { "idx": 3430, "question": "What are the forms of alloying elements in steel?", "answer": "The forms of alloying elements in steel are: (1) dissolved in ferrite, austenite, or martensite, existing as solute; (2) forming strengthening phases; (3) existing in a free state; (4) forming non-metallic inclusions with impurities such as oxygen, nitrogen, and sulfur in the steel." }, { "idx": 3431, "question": "Explain the practical significance of commonly added alloying elements in steel by combining their effects on the heating transformation of steel.", "answer": "Effect on heating transformation: Alloying elements (except nickel and cobalt) can slow down the austenitization process and inhibit austenite grain growth; Significance: Except for manganese steel, alloy steels are less prone to overheating during heating, which is beneficial for obtaining fine martensite after quenching, appropriately increasing heating temperature to enhance hardenability, and reducing the tendency of deformation and cracking during quenching." }, { "idx": 3432, "question": "What is the influence law of alloying elements on the critical temperature, E point, and S point of the iron-carbon phase diagram?", "answer": "After the addition of alloying elements, they affect the critical temperature, E point, and S point. Elements that expand the austenite region generally lower the A3 and A1 temperatures and shift the S point and E point to the lower left; elements that shrink the austenite region generally raise the A3 and A1 temperatures and shift the S point and E point to the upper left." }, { "idx": 3433, "question": "Explain the meaning of the steel grade 38CrMoAlA", "answer": "38CrMoAlA is a nitriding-specific steel (quenched and tempered steel), wc=0.38%, wc=1.00%, WM=0.2%, 0A=0.9%, wsi=0.3%, wM=0.45%" }, { "idx": 3434, "question": "Explain the practical significance of adding common alloying elements to steel based on their influence on the kinetic curves during the cooling transformation of steel.", "answer": "Influence on cooling transformation: The addition of alloying elements (except cobalt and aluminum) shifts the position of the isothermal transformation C-curve of undercooled austenite to the right. Strong carbide-forming elements not only shift the C-curve to the right but also alter its shape, resulting in the separation of the pearlite and bainite transformation curves. Therefore, the effects of adding alloying elements are: (1) improving hardenability; (2) carbide-forming elements can also enhance the wear resistance, tempering stability, and red hardness of steel." }, { "idx": 3435, "question": "What is the difference between the microstructure of alloy steel and carbon steel after heat treatment?", "answer": "There are also differences in the microstructure after heat treatment, for example, 4Cr13, due to Cr shifting the E point to the left, its microstructure becomes that of a hypereutectoid steel." }, { "idx": 3436, "question": "Combining ordinary low-alloy high-strength steels of different strength levels, explain the characteristics of their compositional changes?", "answer": "While using Mn for solid solution strengthening, the carbon mass fraction was adjusted and the types of alloying elements were increased." }, { "idx": 3437, "question": "Combining ordinary low-alloy high-strength steels of different strength grades, explain the role of each element in them?", "answer": "The role of alloying elements in hot-rolled steel: solution strengthening; in normalized steel: in addition to solution strengthening, it also plays a role in dispersion strengthening." }, { "idx": 3438, "question": "What are the carbon contents of engineering structural steel, carburizing steel, quenched and tempered steel, spring steel, and bearing steel?", "answer": "The mass fraction of carbon in engineering structural steel is relatively low, mostly ranging from 0.1% to 0.2%. The mass fraction of carbon in carburizing steel is generally 0.1% to 0.25%. The mass fraction of carbon in most quenched and tempered steels ranges from 0.25% to 0.5%. The mass fraction of carbon in spring steel is generally 0.6% to 0.9%. The mass fraction of carbon in bearing steel ranges from 0.95% to 1.15%." }, { "idx": 3439, "question": "What are the main reasons for the different carbon contents in engineering structural steel, carburizing steel, quenched and tempered steel, spring steel, and bearing steel?", "answer": "Engineering structural steel is mainly used for engineering structures, where the strength requirement is not high, but the weldability requirement is high, so the mass fraction of carbon is low. The carbon content of carburizing steel cannot be high to ensure high toughness in the core. Quenched and tempered steel should have good comprehensive mechanical properties after quenching and tempering, meaning not only high strength but also good toughness, so medium carbon steel is used. Spring steel requires high elastic limit and fatigue strength, so the mass fraction of carbon cannot be too low, but if it is too high, brittleness increases. Bearing steel requires high hardness and high wear resistance, so a high mass fraction of carbon is used." }, { "idx": 3440, "question": "If 20CrMnTi is used as carburizing steel, what are the differences in its mechanical properties, process performance, and process characteristics? Why?", "answer": "20CrMnTi is a medium-hardenability carburizing steel used for manufacturing larger-sized parts that bear moderate loads. The carburizing temperature is 930~950°C, the quenching temperature is 870~890°C (oil quenching), and the tempering temperature is 190°C. The differences mainly result from the types and contents of alloying elements." }, { "idx": 3441, "question": "If 20Cr is used as carburizing steel, what are the differences in its mechanical properties, process performance, and process characteristics? Why?", "answer": "20Cr is a low hardenability carburizing steel, suitable for manufacturing small wear-resistant parts with low stress. The carburizing temperature is 900~920°C, the quenching temperature is 860~890°C (water quenching or oil quenching), and the tempering temperature is 180°C. The differences are mainly due to the influence of the types and content of alloying elements." }, { "idx": 3442, "question": "If Cr12MoV steel is selected to manufacture cold work molds, what heat treatment process should be formulated? Why?", "answer": "For Cr12MoV steel used in cold work molds, the heat treatment process is the primary hardening method: quenching at 980～1030‰ and tempering at 200-270°C. Reason: The primary hardening method of Cr12MoV steel can provide the steel with high hardness and wear resistance, minimal heat treatment deformation, and most molds made of Cr12MoV steel without special requirements adopt this method." }, { "idx": 3443, "question": "What are the differences in heat treatment between engineering structural steel, carburizing steel, quenched and tempered steel, spring steel, and bearing steel? Why?", "answer": "Engineering structural steel is used in the hot-rolled normalized state. Carburizing steel undergoes carburizing, quenching + low-temperature tempering. Quenched and tempered steel uses medium-carbon steel and is subjected to quenching and tempering. Spring steel undergoes quenching + medium-temperature tempering. Bearing steel undergoes quenching + low-temperature tempering. Engineering structural steel is mainly used for engineering structures where high strength is not required, but good weldability is essential, hence it is used in the hot-rolled normalized state. Carburized parts require certain toughness in the core, so carburizing, quenching + low-temperature tempering is applied. Quenched and tempered parts should possess good comprehensive mechanical properties after treatment, meaning not only high strength but also good toughness, hence quenching and tempering is used. Spring parts require high elastic limit and fatigue strength, so quenching + medium-temperature tempering is applied. Bearings require high hardness and wear resistance, hence quenching + low-temperature tempering is used." }, { "idx": 3444, "question": "If Cr12MoV steel is selected to manufacture a stamping die with certain hot hardness requirements, what heat treatment process should be formulated? Why?", "answer": "When Cr12MoV steel is used to make a stamping die with certain hot hardness requirements, the secondary hardening method can be adopted, i.e., quenching at 1050～1100℃ followed by multiple tempering at 500-520℃, which can increase the hardness to 60～62HRC. Reason: The secondary hardening method of Cr12MoV steel is suitable for applications with higher working temperatures (400～500℃) or requiring hot hardness." }, { "idx": 3445, "question": "Explain the microstructural changes of high-speed steel during heat treatment", "answer": "Annealing not only reduces hardness, facilitating machining, but also obtains a granular structure with uniformly distributed carbides, preparing the microstructure for subsequent quenching; quenching results in martensite + granular carbides + a considerable amount of retained austenite; triple tempering yields tempered martensite + granular carbides." }, { "idx": 3446, "question": "Since high-speed steel already has good red hardness, why is high-speed steel generally not used for die-casting molds, and why was the steel grade 3Cr2W8V developed?", "answer": "Although high-speed steel has high hardness, high wear resistance, and red hardness, its toughness and thermal fatigue properties are inferior to those of 3Cr2W8V, so high-speed steel is not used for die-casting molds." }, { "idx": 3447, "question": "Describe the basis for formulating the heat treatment process of high-speed steel", "answer": "The basis is the chemical composition and the transformation curve of undercooled austenite in high-speed steel." }, { "idx": 3448, "question": "List several types of ledeburitic steel you are familiar with?", "answer": "High-speed steel, Cr12, Cr12MoV, etc. all belong to ledeburitic steel." }, { "idx": 3449, "question": "If 20Cr2Ni4W is used as carburizing steel, what are the differences in its mechanical properties, process performance, and process characteristics? Why?", "answer": "20Cr2Ni4W is a high-hardenability carburizing steel used for manufacturing heavy-duty, large-section components requiring high wear resistance and good toughness. The carburizing temperature is 900~950°C, the quenching temperature is 880°C, and a secondary quenching is performed at 780°C, with a tempering temperature of 190°C. The differences are mainly due to the influence of the types and contents of alloying elements." }, { "idx": 3450, "question": "5CrNiMo, 3Cr2W8V, and H11 (or H13) all belong to hot-work die steels. Are there any differences in their usage? Why?", "answer": "Yes, there are differences. Their total alloying element contents are low-alloy (5CrNiMo), medium-alloy (H11), and high-alloy (3Cr2W8V), respectively. 5CrNiMo is used for hot forging dies, H11 replaces 3Cr2W8V for small and medium-sized mechanical forging dies and hot extrusion dies with relatively low operating temperatures, while 3Cr2W8V is used for hot extrusion dies. The reason for their differences in usage lies in the varying types and amounts of alloying elements, which result in different levels of thermal strength, thermal fatigue resistance, and high-temperature oxidation resistance, hence their different applications." }, { "idx": 3451, "question": "What is the most notable characteristic of ledeburitic steel?", "answer": "Its most notable characteristic is the presence of eutectic ledeburite structure in the as-cast microstructure, with high hardness, high wear resistance, and red hardness." }, { "idx": 3452, "question": "What is the most prominent weakness of ledeburitic steel?", "answer": "The weakness is the coarse carbides, uneven distribution, and high brittleness." }, { "idx": 3453, "question": "If Cr12MoV steel is selected to manufacture measuring tools, what heat treatment process should be formulated? Why?", "answer": "When Cr12MoV steel is used to make measuring tools, it can be quenched at 1050～1100°C, tempered multiple times at 550～600°C, and then subjected to nitriding or carbonitriding to achieve high wear resistance, corrosion resistance, and dimensional stability. Reason: Cr12MoV steel, after quenching + high-temperature tempering (quenching and tempering) followed by nitriding or carbonitriding, is commonly used for making measuring tools due to its high wear resistance, corrosion resistance, and relatively high dimensional stability." }, { "idx": 3454, "question": "What are the types of heat-resistant steels?", "answer": "Commonly used heat-resistant steels are roughly divided into three categories according to their microstructure in the normalized state: pearlitic steel, martensitic steel, and austenitic steel." }, { "idx": 3455, "question": "Based on the alloy composition of oxidation-resistant steel, analyze how the steel meets different thermal stability requirements.", "answer": "Alloying elements such as chromium, silicon, and aluminum are added to oxidation-resistant steel. These elements have a high affinity for oxygen, so they are preferentially oxidized, forming a dense, high-melting-point oxide film that firmly covers the steel surface. This isolates the metal from external oxidizing gases, preventing further oxidation. The amount of alloying elements added varies, resulting in different levels of oxidation resistance." }, { "idx": 3456, "question": "Why is it still necessary to develop ledeburitic steel?", "answer": "Because it has high hardness, high wear resistance, and red hardness, this steel is developed." }, { "idx": 3457, "question": "Based on the role of alloying elements used in stainless steel, analyze the main characteristics of martensitic stainless steel? What are the main shortcomings? How to prevent or overcome them?", "answer": "Martensitic stainless steel belongs to chromium stainless steel. As the carbon content increases, the strength, hardness, and wear resistance of the steel improve, but the corrosion resistance decreases, because carbon forms chromium carbides with chromium, preventing its potential from rising. To enhance corrosion resistance and mechanical properties, quenching and tempering are usually performed." }, { "idx": 3458, "question": "Based on the role of alloying elements used in stainless steel, analyze the main characteristics of ferritic stainless steel? What are the main shortcomings? How to prevent or overcome them?", "answer": "Ferritic stainless steel also belongs to chromium stainless steel. Its structure remains unchanged from room temperature to high temperature (960～1100‰), and it has strong resistance to atmospheric and acid corrosion. However, if the grain coarsens during heating, it cannot be refined by heat treatment methods and can only be improved through plastic deformation and recrystallization." }, { "idx": 3459, "question": "Based on the role of alloying elements used in stainless steel, analyze the main characteristics of austenitic stainless steel? What are the main shortcomings? How to prevent or overcome them?", "answer": "Austenitic stainless steel belongs to nickel-chromium steel, which has excellent corrosion resistance and heat resistance. However, intergranular corrosion may occur at 450~850°C. It is usually prevented by reducing the mass fraction of carbon, adding elements that can form stable carbides (such as Ti), and applying appropriate heat treatment." }, { "idx": 3460, "question": "What are the different application scenarios for austenitic stainless steel, wear-resistant steel, and quenched tool steel?", "answer": "Wear-resistant steel is used to make working parts that withstand strong impact and pressure, such as bulldozer blades and railway switches. Quenched tool steel is used in situations that do not require withstanding strong impact pressure but demand high wear resistance." }, { "idx": 3461, "question": "How is tool steel further classified?", "answer": "Tool steel is divided into: cutting tool steel, die steel, and measuring tool steel. Cutting tool steel is divided into: carbon tool steel, low-alloy tool steel, and high-speed steel. Die steel is divided into: cold work die steel and hot work die steel." }, { "idx": 3462, "question": "How to overcome the weaknesses of ledeburitic steel?", "answer": "By repeatedly forging to break down the coarse carbides and distribute them uniformly, which can also reduce brittleness." }, { "idx": 3463, "question": "Discuss the microstructural characteristics and processing techniques of ZGMn13", "answer": "ZGMn13 is a wear-resistant steel with an austenitic microstructure at room temperature. Whether in the as-cast, forged, or hot-rolled condition, carbides precipitate along the austenite grain boundaries, reducing the steel's toughness and wear resistance. Therefore, water toughening treatment must be performed. After water toughening, it exhibits good toughness. During subsequent use, under strong compressive stress, martensitic transformation occurs, thereby increasing its wear resistance. Hence, this type of steel is used in applications subject to strong compressive stress." }, { "idx": 3464, "question": "Discuss the microstructural characteristics and processing techniques of Cr12-type cold-work die steel", "answer": "Cr12 is a type of cold-work die steel, with Cr as the main alloying element, which significantly enhances hardenability and wear resistance. Due to the high content of alloying elements, the as-cast microstructure contains eutectic carbides. Therefore, repeated forging is required to break down the carbides and improve their distribution and brittleness. After forging, spheroidizing annealing is performed, followed by quenching + low-temperature tempering to obtain tempered martensite, which provides high hardness and wear resistance." }, { "idx": 3465, "question": "What is the difference between the purpose of quenching for austenitic stainless steel and wear-resistant steel compared to that of general steel?", "answer": "The purpose of quenching for austenitic stainless steel and wear-resistant steel is to obtain a single-phase, uniform austenitic structure, preventing the formation of secondary phases that could cause intergranular corrosion. Therefore, the quenching of austenitic stainless steel is referred to as solution treatment, while for wear-resistant steel, the quenching process, which results in a single austenitic structure with high plasticity and toughness, is also known as water toughening treatment." }, { "idx": 3466, "question": "How is structural steel further subdivided?", "answer": "Structural steel is divided into: engineering structural steel and mechanical component steel. Engineering structural steel is further divided into: ordinary carbon structural steel and ordinary low-alloy steel. Mechanical component steel is divided into: carburizing steel, quenched and tempered steel, spring steel, and rolling bearing steel." }, { "idx": 3467, "question": "Why does alloy steel have less heat treatment deformation than carbon steel?", "answer": "The addition of alloying elements (except C) increases the hardenability of the steel, so when obtaining the same structure, alloy steel can choose a slower cooling medium, resulting in less heat treatment deformation." }, { "idx": 3468, "question": "For a machine tool spindle made of 40Cr steel, the core requires good strength and toughness (200-300HB), while the journal needs to be hard and wear-resistant (HRC54-58). What preparatory heat treatment should be selected?", "answer": "Preparatory heat treatment: Quenching + high-temperature tempering, tempered sorbite." }, { "idx": 3469, "question": "Is it reasonable to use Q235 steel after quenching and tempering treatment? Why?", "answer": "No, it is not reasonable. Because Q235 is an ordinary-quality carbon structural steel with high sulfur and phosphorus content, it is generally used in the hot-rolled state. Due to its low carbon content, high S and P content, poor quality, and low hardenability, even in the quenched and tempered state, its performance will not improve significantly." }, { "idx": 3470, "question": "Why does alloy steel have better mechanical properties than carbon steel?", "answer": "The addition of alloying elements and appropriate heat treatment can improve or enhance the mechanical properties of steel." }, { "idx": 3471, "question": "How is steel classified by use?", "answer": "Steel is classified by use into structural steel, tool steel, and special performance steel." }, { "idx": 3472, "question": "How are special property steels further classified?", "answer": "Special property steels are divided into: stainless steel, heat-resistant steel, and wear-resistant steel." }, { "idx": 3473, "question": "Why do alloy tool steels have higher wear resistance and hot hardness than carbon steels?", "answer": "Alloy tool steels contain certain alloying elements that form alloy carbides with the carbon in the steel. These alloy carbides have high hardness and high melting points, which is why alloy tool steels exhibit higher wear resistance and hot hardness than carbon steels." }, { "idx": 3474, "question": "A factory used 9Mn2V steel to manufacture plastic molds, with the original design requiring a hardness of HRC 53~58. The process involved oil quenching at 790°C and tempering at 200~220°C. However, the molds frequently experienced brittle fracture during use. Later, the process was changed to heating at 790°C, followed by isothermal treatment in a nitrite salt bath at 260~280°C for 4 hours, then air cooling. Although the hardness decreased to HRC 50, the service life significantly improved, and brittle fracture no longer occurred. Analyze the reasons for this.", "answer": "After quenching and low-temperature tempering, the 9Mn2V steel primarily forms a tempered structure of plate martensite. Since the substructure of plate martensite consists of twins and contains microcracks formed during its formation, it exhibits high brittleness. Isothermal quenching, on the other hand, produces bainite, whose matrix ferrite has a substructure of high-density dislocations and no microcracks, resulting in significantly reduced brittleness." }, { "idx": 3475, "question": "For a machine tool spindle made of 40Cr steel, the core requires good strength and toughness (200-300HB), while the journal area needs to be hard and wear-resistant (HRC54-58). What final heat treatment should be selected?", "answer": "Final heat treatment: Surface quenching of the journal to obtain surface tempered martensite and core tempered sorbite." }, { "idx": 3476, "question": "There is a W18Cr4V steel disc milling cutter, please arrange its processing route.", "answer": "Processing route: Blanking and forging → Spheroidizing annealing → Cutting → Quenching + High-temperature tempering (three times) → Finishing → Assembly." }, { "idx": 3477, "question": "For a machine tool spindle made of 40Cr steel, the core requires good strength and toughness (200-300HB), while the journal needs to be hard and wear-resistant (HRC54-58). Describe the microstructure after preparatory heat treatment.", "answer": "Microstructure after preparatory heat treatment: tempered sorbite." }, { "idx": 3478, "question": "Compare the compositional characteristics of four alloy tool steels: 9SiCr, Cr12, 5CrMnMo, and W18Cr4V.", "answer": "
Steel gradeCompositional characteristics
9SiCrwc=0.9%, wsiWcr<1.5%
Cr12wc=2.15%, wc=12%
5CrMnMowc=0.5%, wc=0.75%, wM=0.9% WMo=0.21%
W18Cr4Vwc=0.75%, ww=18%, wc=4%, wv=1.2%
" }, { "idx": 3479, "question": "What is the difference between the wear resistance principle of wear-resistant steel and that of quenched tool steel?", "answer": "The wear resistance principle of wear-resistant steel is that the single austenite structure is subjected to strong impact and pressure during work, resulting in stress-induced martensite and work hardening, which greatly increases the wear resistance of the steel. Quenched tool steel obtains high hardness and high wear resistance tempered martensite through quenching + low-temperature tempering, giving the steel high wear resistance." }, { "idx": 3480, "question": "Compare the heat treatment processes of four alloy tool steels: 9SiCr, Cr12, 5CrMnMo, and W18Cr4V", "answer": "
Steel gradeHeat treatment
9SiCrQuenching + low tempering
Cr12Quenching + low tempering
5CrMnMoQuenching + high tempering
W18Cr4VQuenching + high tempering
" }, { "idx": 3481, "question": "Compare the properties of four alloy tool steels: 9SiCr, Cr12, 5CrMnMo, and W18Cr4V", "answer": "
Steel gradeProperties
9SiCrHigh strength and high wear resistance
Cr12High strength and high wear resistance
5CrMnMoBest comprehensive mechanical properties
W18Cr4VHigh strength and high wear resistance
" }, { "idx": 3482, "question": "Explain the purposes of each hot working process.", "answer": "Spheroidizing annealing: Eliminate forging stress, spheroidize carbides to reduce hardness and improve machinability, preparing for quenching. Quenching: To obtain martensite, maintaining high hardness and wear resistance. Tempering: Adjust properties, eliminate quenching stress, and reduce retained austenite (A′)." }, { "idx": 3483, "question": "Compare the uses of four alloy tool steels: 9SiCr, Cr12, 5CrMnMo, and W18Cr4V.", "answer": "
Steel gradeUse
9SiCrTaps, dies, drill bits
Cr12Large-sized cold work dies
5CrMnMoSmall hot forging dies
W18Cr4VHigh-speed cutting dies
" }, { "idx": 3484, "question": "Is tempering high-speed steel at 560°C considered quenching and tempering treatment? Why?", "answer": "No. Quenching and tempering treatment refers to high-temperature tempering (typically 500~650°C) after quenching to obtain tempered sorbite structure, which provides good comprehensive mechanical properties. However, tempering high-speed steel at 560°C aims to achieve dispersion hardening and secondary quenching phenomena, improving hardness and wear resistance, with the microstructure being tempered martensite + a small amount of retained austenite + carbides." }, { "idx": 3485, "question": "For a machine tool spindle made of 40Cr steel, the core requires good strength and toughness (200-300HB), while the journal area needs to be hard and wear-resistant (HRC54-58). Describe the microstructure after the final heat treatment.", "answer": "Microstructure after final heat treatment: tempered martensite on the surface, tempered sorbite in the core." }, { "idx": 3486, "question": "Can one prolonged tempering at 560°C replace three tempering processes at 560°C?", "answer": "No. Because a single tempering is difficult to completely eliminate retained austenite, multiple tempering processes are required to reduce the retained austenite to the minimum amount. Moreover, each subsequent tempering can eliminate the internal stress generated by the transformation of austenite into martensite during the previous tempering." }, { "idx": 3487, "question": "What is the role of carbon in high-speed steel?", "answer": "The role of carbon in high-speed steel: high carbon content, generally with a mass fraction of 0.7% to 1.5%, is to ensure the formation of alloy carbides with alloying elements, obtaining a martensite matrix plus carbides during quenching to improve the hardness and wear resistance of the steel." }, { "idx": 3488, "question": "Why is the quenching temperature of W18Cr4V steel as high as 1280°C?", "answer": "The two elements that have the greatest impact on the hot hardness of high-speed steel are W and V. Their solubility in austenite only increases significantly above 1000°C. At 1270~1280°C, the austenite contains wW=7%~8%, wCr=4%, and wV=1%. If the temperature is higher, the austenite grains will rapidly grow and coarsen, and the retained austenite in the quenched state will also increase rapidly, thereby reducing the hardness of the high-speed steel and increasing brittleness. This is the main reason why the quenching temperature is generally set at 1270~1280°C." }, { "idx": 3489, "question": "Why does W18Cr4V steel need to undergo three tempering cycles at 560°C after quenching?", "answer": "Because W18Cr4V steel contains about 20%~25% retained austenite in the quenched state, which is difficult to fully eliminate with a single tempering. Three tempering cycles can reduce the retained austenite to a minimum (about 15% remains after the first tempering, 3%~5% after the second, and 1%~2% after the third). Each subsequent tempering also relieves the internal stress generated by the transformation of austenite into martensite during the previous tempering. The tempered microstructure consists of tempered martensite + a small amount of retained austenite + carbides." }, { "idx": 3490, "question": "What is the role of alloying elements in high-speed steel?", "answer": "The role of alloying elements in high-speed steel: W improves the red hardness of the steel, Cr enhances hardenability, Mo reduces the second type of temper brittleness, Co delays the precipitation and aggregation of carbides during tempering, that is, improves tempering stability." }, { "idx": 3491, "question": "Can austenitic stainless steel be strengthened by heat treatment?", "answer": "No. Because after quenching (solution treatment), austenitic stainless steel undergoes no phase transformation, and all second-phase particles dissolve into the austenite. As a result, its hardness and strength drop to the lowest level after quenching, so it cannot be strengthened by heat treatment." }, { "idx": 3492, "question": "How to improve the corrosion resistance of steel?", "answer": "Methods to improve the corrosion resistance of steel: (a) Increase the electrode potential of the metal. (b) Make the metal easier to passivate. (c) Obtain a single-phase structure with uniform chemical composition, microstructure, and metal purity, aiming to avoid the formation of microcells." }, { "idx": 3493, "question": "What are the characteristics of cast iron and carbon steel in terms of mechanical properties?", "answer": "Cast iron has low tensile strength, compressive strength similar to steel with the same matrix, and poor plasticity and toughness; carbon steel has higher strength and better plasticity and toughness." }, { "idx": 3494, "question": "How to distinguish between 45 steel and HT150 metals through metallographic examination", "answer": "Metallographic examination" }, { "idx": 3495, "question": "What measures should be taken to improve the strength of stainless steel?", "answer": "Measures: Work hardening. Some martensitic stainless steels (such as 4Cr13) can be strengthened through heat treatment (quenching + low-temperature tempering)." }, { "idx": 3496, "question": "What method is used to strengthen austenitic stainless steel in production?", "answer": "Work hardening is commonly used to strengthen it in production." }, { "idx": 3497, "question": "How to distinguish between 45 steel and HT150 metals through macroscopic fracture analysis", "answer": "Macroscopic fracture analysis" }, { "idx": 3498, "question": "What are the compositional characteristics of stainless steel?", "answer": "The compositional characteristics of stainless steel are: (a) Low carbon content: Most w_C = (0.1-0.2)%. C has a strong affinity with Cr, and C and Cr can form a series of complex compounds, reducing the corrosion resistance of the steel. (b) Stainless steel generally contains a high amount of chromium (the mass fraction of Cr in martensitic and ferritic stainless steels is greater than 13%, and the mass fraction of Cr in austenitic stainless steel is greater than 18%) and a high amount of nickel (the mass fraction of Ni in austenitic stainless steel is greater than 8%). (c) The carbon content is generally low, with the mass fraction of C in austenitic and ferritic stainless steels being relatively low (<0.1%)." }, { "idx": 3499, "question": "Why is it often easy to form white iron on the surface layer and thin-walled areas of the same casting?", "answer": "Because the undercooling degree is large and the cooling rate is fast at the surface layer and thin-walled areas, it is easy to obtain white iron." }, { "idx": 3500, "question": "Compare the composition of HT150 and annealed 20 steel", "answer": "Composition: a) HT150: w_C=(2.5~4)%, w_Si=(1~2.5)%, w_Mn=(0.5~1.3)%, w_P≤0.3%, w_S≤0.15%. b) 20 steel: w_C=(0.17~0.24)%, w_Mn=(0.35~0.65)%, w_Si=(0.17~0.37)%, w_P≤0.035%, w_S≤0.035%. Composition difference: The mass fractions of C, Si, P, and S are higher in cast iron." }, { "idx": 3501, "question": "What is the main difference between cast iron and carbon steel?", "answer": "The main difference between carbon steel and cast iron lies in the carbon content and the form of carbon present. The carbon content in cast iron is greater than 2.11%, while that in carbon steel is greater than 0.0218% but less than 2.11%. In carbon steel, apart from a portion dissolved in ferrite, all other carbon exists in the form of Fe3C. In cast iron, a small amount of carbon is dissolved in ferrite, while most of the carbon exists in the form of graphite (gray cast iron) or Fe3C (white cast iron). The microstructure is not entirely the same, leading to differences in properties (both processing and service performance)." }, { "idx": 3502, "question": "What are the process characteristics of high-speed steel?", "answer": "Process characteristics of high-speed steel: a) Two-stage preheating: reduces thermal stress, minimizes deformation and cracking. b) High quenching temperature: ensures full dissolution of W and V into austenite, allowing alloy carbides to precipitate during tempering, thereby ensuring red hardness. c) Graded quenching: reduces deformation and cracking. d) 560‰ tempering: achieves maximum hardness with secondary hardening and secondary quenching phenomena. e) Triple tempering: reduces retained austenite from over 20% after quenching to approximately 1%–2%." }, { "idx": 3503, "question": "Is Cr12MoV a stainless steel?", "answer": "Cr12MoV is not a stainless steel, it is a cold work die steel. The w_Cr of stainless steel is ≥13%." }, { "idx": 3504, "question": "Compare the compressive strength of HT150 and annealed 20 steel", "answer": "The tensile strength and hardness of the two are not much different." }, { "idx": 3505, "question": "Why does cast iron with a chemical composition of 'three lows' (low carbon, silicon, and manganese content) and 'one high' (high sulfur content) tend to form white iron during production?", "answer": "Because both carbon and silicon are graphitization-promoting elements, their low content facilitates the formation of white iron structure. Although manganese is a graphitization-inhibiting element, its low mass fraction cannot counteract the effect of sulfur, making the sulfur's role in inhibiting graphitization more pronounced and thus favoring the production of white cast iron. Since sulfur is a graphitization-inhibiting element, low carbon, silicon, and manganese combined with high sulfur content easily lead to the formation of white iron." }, { "idx": 3506, "question": "Compare the microstructures of HT150 and annealed 20 steel", "answer": "Microstructure: a) HT150: F+P+G (graphite); b) 20 steel: F+P." }, { "idx": 3507, "question": "Compare the welding performance of HT150 and annealed 20 steel", "answer": "The welding performance of 20 steel is better than that of HT150." }, { "idx": 3508, "question": "Compare the hardness of HT150 and annealed 20 steel", "answer": "The tensile strength and hardness are not significantly different." }, { "idx": 3509, "question": "Compare the anti-friction properties of HT150 and annealed 20 steel", "answer": "HT150 has better anti-friction properties than 20 steel." }, { "idx": 3510, "question": "In the graphitization process of cast iron, if the first stage is completely graphitized and the second stage is partially graphitized, what kind of microstructure cast iron is obtained?", "answer": "F+P+G (graphite)" }, { "idx": 3511, "question": "In the graphitization process of cast iron, if the first stage is completely graphitized and the second stage is not graphitized, what type of cast iron structure is obtained?", "answer": "P+G (graphite)" }, { "idx": 3512, "question": "Compare the machinability of HT150 and annealed 20 steel", "answer": "The machinability of HT150 is better than that of 20 steel." }, { "idx": 3513, "question": "Please indicate the type of cast iron and heat treatment method that should be used for the machine tool bed, and why?", "answer": "Machine tool bed: gray cast iron, (as-cast) HT250, stress relief annealing." }, { "idx": 3514, "question": "Compare the casting properties of HT150 and annealed 20 steel", "answer": "The casting properties of HT150 are better than those of 20 steel." }, { "idx": 3515, "question": "Compare the tensile strength of HT150 and annealed 20 steel", "answer": "The tensile strength of 20 steel is much higher than that of HT150." }, { "idx": 3516, "question": "Compare the forgeability of HT150 and annealed 20 steel", "answer": "The forgeability of 20 steel is better than that of HT150 (HT150 cannot be forged)." }, { "idx": 3517, "question": "In the graphitization process of cast iron, if the first stage is completely graphitized and the second stage is completely graphitized, what type of cast iron structure is obtained?", "answer": "F+G (graphite)" }, { "idx": 3518, "question": "Please specify the type of cast iron and heat treatment method that should be used for diesel engine crankshafts, and why?", "answer": "Crankshaft: ductile iron, QT800-2, quenching and tempering treatment." }, { "idx": 3519, "question": "Please indicate the type of cast iron and heat treatment method that should be used for plowshares, and why?", "answer": "Plowshare: white cast iron." }, { "idx": 3520, "question": "Please indicate the type of cast iron and heat treatment method that should be used for the hydraulic pump housing, and why?", "answer": "Hydraulic pump housing: malleable cast iron, KΠ2650-02, graphitizing annealing." }, { "idx": 3521, "question": "Please specify the type of cast iron and heat treatment method that should be used for ball mill liners, and why?", "answer": "Ball mill liners: alloy nodular cast iron (such as medium manganese cast iron)." }, { "idx": 3522, "question": "Can aluminum alloys be strengthened through martensitic transformation like steel? Why?", "answer": "No. Because aluminum alloys only undergo solubility changes during heating and cooling in the solid state, without allotropic transformation. Therefore, they can only be strengthened by quenching + aging." }, { "idx": 3523, "question": "Explain the alloying principles of aluminum alloys", "answer": "Alloying principles of aluminum alloys: Commonly added elements: Cu, Mg, Zn, Si, Mn" }, { "idx": 3524, "question": "Explain the role of Cu element in aluminum alloys", "answer": "Copper in aluminum not only significantly enhances the room temperature strength of aluminum alloys through solid solution strengthening and precipitation strengthening, but also improves the heat resistance of aluminum alloys. Therefore, copper is a primary alloying element in high-strength aluminum alloys and heat-resistant aluminum alloys." }, { "idx": 3525, "question": "Explain the role of Mn element in aluminum alloys", "answer": "Manganese has low solubility in aluminum, resulting in limited solid solution strengthening capability. The second phase MnAl6 in the Al-Mn system has electrochemical properties similar to aluminum, providing good corrosion resistance. Therefore, manganese is often added to rust-proof aluminum alloys, with its wMn generally not exceeding 2%." }, { "idx": 3526, "question": "Explain the role of Si element in aluminum alloys", "answer": "Similar to manganese, silicon has low solubility in aluminum, resulting in limited solid solution strengthening and insignificant precipitation strengthening effects. Therefore, it primarily relies on excess phase strengthening. The binary Al-Si system alloy has a low eutectic point, making it suitable for casting, and serves as the foundational alloy series for cast aluminum alloys, with wSi typically ranging from 10% to 13%. Silicon and magnesium can form Mg2Si precipitates in aluminum, which exhibit excellent strengthening effects. Hence, silicon can also be added as a precipitation strengthening element to magnesium-aluminum alloys, with its addition amount usually not exceeding wSi (1.0~1.2)%." }, { "idx": 3527, "question": "Explain the role of Mg element in aluminum alloys", "answer": "Magnesium has a good solid solution strengthening effect in aluminum, which can increase the strength of aluminum while also reducing its density. The precipitation strengthening effect of magnesium-aluminum alloys is not significant, but they exhibit good corrosion resistance and can be used as corrosion-resistant alloys. Magnesium cannot serve as the main additive element for high-strength aluminum alloys alone; it must be combined with other elements to fully utilize its effects." }, { "idx": 3528, "question": "What is the difference between natural aging and artificial aging of aluminum alloys?", "answer": "The process of spontaneous strengthening of the alloy at room temperature is called natural aging, while the aging process carried out at a certain heating temperature is called artificial aging." }, { "idx": 3529, "question": "Can aluminum alloys be surface strengthened by carburizing or nitriding? Why?", "answer": "No. Because the solubility of C and N in aluminum is very low, especially since aluminum has a strong affinity for oxygen, forming a very dense oxide film on the surface. This prevents the active atoms from being absorbed by the surface during carburizing or nitriding. Even if the active atoms are absorbed by the surface, the dense oxide film hinders the diffusion of these absorbed atoms toward the core. Therefore, aluminum alloys cannot be surface strengthened by carburizing or nitriding." }, { "idx": 3530, "question": "Taking A1-4Cu alloy as an example, explain the changes in microstructure and properties during the aging process.", "answer": "Aging refers to the phenomenon where the supersaturated solid solution of aluminum alloy obtained after quenching decomposes over time at a certain temperature, leading to an increase in the strength and hardness of the alloy." }, { "idx": 3531, "question": "Explain the role of Zn element in aluminum alloys", "answer": "Zinc has a high solubility in aluminum and exhibits strong solid solution strengthening capability. A small amount of zinc (wZn=(0.4~0.8)%) can enhance the strength and corrosion resistance of aluminum alloys. In multi-component aluminum alloys, zinc serves as an element that forms precipitation strengthening phases, significantly improving the precipitation strengthening effect of the alloy." }, { "idx": 3532, "question": "What is the principle for selecting natural aging or artificial aging?", "answer": "The principles for choosing between artificial aging and natural aging are: ① Determine the aging method (natural or artificial) based on the working temperature of the part; ② Consider the required aging strengthening effect for the part; ③ Take into account the type of aluminum alloy, batch size of the workpiece, production efficiency, etc." }, { "idx": 3533, "question": "How many types of titanium alloys are there?", "answer": "Titanium alloys are classified into α titanium alloys, α+β titanium alloys, and β titanium alloys" }, { "idx": 3534, "question": "Briefly describe the generation of solid solution strengthening and provide examples", "answer": "Solid solution strengthening refers to the increase in strength by adding alloying elements to form solid solutions with aluminum. Commonly used alloying elements include Cu, Mg, Zn, Si, etc. These elements can form limited solid solutions with aluminum and have significant solubility, resulting in effective solid solution strengthening, thus serving as the primary alloying elements in aluminum alloys." }, { "idx": 3535, "question": "Briefly describe the generation of dispersion strengthening and provide examples", "answer": "Excess phase (second phase) strengthening: When the content of alloying elements in an alloy exceeds the solubility limit, there will be some undissolved second phases present in the matrix (solid solution), also known as excess phases. In aluminum alloys, excess phases are mostly hard and brittle intermetallic compounds, which also hinder dislocation movement, increasing the strength and hardness of the alloy while reducing its plasticity and toughness." }, { "idx": 3536, "question": "Briefly describe the differences between solution strengthening, dispersion strengthening, and age strengthening", "answer": "Solution strengthening improves strength by forming solid solutions with alloying elements and the matrix; dispersion strengthening enhances strength by impeding dislocation movement through undissolved second phases (excess phases); age strengthening is achieved by heat treatment to form supersaturated solid solutions and precipitate transition phases or transition zones. Both solution strengthening and age strengthening involve solid solutions, but age strengthening requires subsequent heat treatment processes, while dispersion strengthening does not rely on heat treatment but rather on the formation of second phases due to alloying elements exceeding solubility limits." }, { "idx": 3537, "question": "Compare the microstructure, properties, and heat treatment characteristics of brass.", "answer": "
MaterialMicrostructurePropertiesHeat Treatment
Brassα (Cu-Zn solid solution) or α+βExcellent corrosion resistance, thermal conductivity, good cold (or hot) working propertiesStress relief annealing
" }, { "idx": 3538, "question": "Compare the microstructure, properties, and heat treatment characteristics of bronze.", "answer": "
MaterialMicrostructurePropertiesHeat treatment
Bronzea (Cu-Sn) or (a+0) eutectoidExcellent casting performance, corrosion resistance, hot and cold pressure processing performance, certain wear resistance and antifriction properties, can be used as bearing alloyQuenching + aging (beryllium bronze)
" }, { "idx": 3539, "question": "How to choose tin-based and lead-based babbitt alloys?", "answer": "Tin-based bearing alloys are suitable for high-speed bearings. Lead-based bearing alloys are mostly used for small and low-speed ordinary machinery." }, { "idx": 3540, "question": "Briefly describe the occurrence of age strengthening and provide examples", "answer": "Age strengthening (precipitation strengthening): The main heat treatment method for strengthening aluminum alloys is solution treatment (quenching) followed by aging. To achieve strong precipitation hardening effects, certain conditions must be met: the elements added to aluminum should have a high ultimate solubility, and this solubility should significantly decrease with temperature reduction; after quenching, a supersaturated solid solution should form, which can precipitate uniform and dispersed coherent or semi-coherent transition zones or transition phases during aging, creating a strong strain field in the matrix." }, { "idx": 3541, "question": "Compare the microstructure, properties, and heat treatment characteristics of cupronickel.", "answer": "
MaterialMicrostructurePropertiesHeat treatment
Cupronickelα (Cu-Ni solid solution)Good strength, excellent plasticity allowing hot and cold deformation, good corrosion resistance, high resistivityStress relief annealing
" }, { "idx": 3542, "question": "What are the characteristics of the microstructure, properties, and applications of typical α titanium alloys?", "answer": "Microstructure: α or α + trace intermetallic compounds (annealed microstructure); Property characteristics: lower room temperature strength compared to other types of titanium alloys, but highest creep strength among titanium alloys at high temperatures (500~600°C), excellent corrosion resistance, good weldability, and retains good plasticity and toughness even at ultra-low temperatures (-253°C); Applications: parts with low strength requirements working below 500°C, pressure vessel materials in aerospace." }, { "idx": 3543, "question": "What are the characteristics of the microstructure, properties, and applications of typical β titanium alloys?", "answer": "Microstructure: β (quenched structure); Property characteristics: high strength, excellent stamping performance, can be strengthened by quenching and aging; Applications: parts working below 350°C, compressor blades, aircraft components" }, { "idx": 3544, "question": "What are the characteristics of Babbitt alloy in terms of lubrication principle?", "answer": "It belongs to friction under the lubricating film. In terms of lubrication principle, it relies on the lubricating film formed between the shaft and the bearing bush to reduce the friction coefficient, thereby minimizing wear between the shaft and the bearing bush." }, { "idx": 3545, "question": "The twinning plane of a face-centered cubic crystal is ( ) A.(112} B.{110} C.111", "answer": "C" }, { "idx": 3546, "question": "What are the differences between tin-based and lead-based Babbitt alloys?", "answer": "Tin-based bearing alloy: It has a small coefficient of expansion, good embeddability and friction reduction, excellent toughness, thermal conductivity, and corrosion resistance, making it suitable for high-speed bearings. Lead-based bearing alloy: Its strength, hardness, wear resistance, and toughness are lower than those of tin-based alloys, but it is inexpensive and commonly used in small and low-speed ordinary machinery." }, { "idx": 3547, "question": "What are the characteristics of the microstructure, properties, and applications of typical α+β titanium alloys?", "answer": "Microstructure: α+β (annealed microstructure); Property characteristics: high strength, good plasticity, stable microstructure at 400°C, high creep strength, good plasticity and resistance to seawater and thermal stress corrosion at low temperatures; Applications: parts working below 400°C, manufacturing of aircraft engine blades, rocket engines" }, { "idx": 3548, "question": "Why is Babbitt alloy wear-resistant?", "answer": "Because the structural characteristic of tin-based bearing alloy is the distribution of hard particles on a soft matrix. If the structure of the bearing alloy consists of hard particles distributed on a soft matrix, during operation, the soft matrix will wear and become recessed, causing the hard particles to protrude from the matrix. This reduces the contact area between the shaft and the bearing bush, while the recesses can store lubricating oil, lowering the friction coefficient between the shaft and the bearing bush and reducing wear. Additionally, the soft matrix can withstand impact and vibration, allowing the shaft and bearing bush to fit well together and also embed small foreign hard particles, ensuring the shaft journal is not scratched, thus providing excellent wear resistance." }, { "idx": 3549, "question": "In a crystal, the defect formed by simultaneously creating a vacancy and an interstitial atom is called ( ). A. Schottky defect B. Frenkel defect C. Interstitial defect", "answer": "B" }, { "idx": 3550, "question": "The difference between cast iron and carbon steel lies in the presence or absence of ( )", "answer": "A" }, { "idx": 3551, "question": "In a body-centered cubic structure with lattice constant a, can a dislocation with Burgers vector $a$ [100] decompose into $\\frac{a}{2}$ [111] + $\\frac{a}{2}$ [1-1-1]? ( )", "answer": "A" }, { "idx": 3552, "question": "Someone said: 'A precipitation-hardening alloy can be softened by water quenching at an appropriate temperature.' Is this statement correct? Explain.", "answer": "A precipitation-hardening alloy cannot be softened by water quenching at an appropriate temperature. The reason is that when heated to a high temperature, it forms a single solid solution structure. Rapid cooling results in a single solid solution, which, although not very hard at this stage, will subsequently undergo precipitation hardening. Therefore, a precipitation-hardening alloy can only be softened through an aging process." }, { "idx": 3553, "question": "Please list several examples of strengthening alloys during the casting process", "answer": "For example, the inoculation treatment of gray cast iron (adding about $4\\%$ ferrosilicon or calcium-silicon alloy to the molten iron during casting for inoculation treatment), and the modification treatment of aluminum-silicon alloys (adding a modifier $\\frac{2}{3}\\mathrm{{NaF}+}$ $\\frac13\\mathrm{NaCl}$ to the alloy melt before pouring)" }, { "idx": 3554, "question": "In substitutional solid solutions, the general mode of atomic diffusion is (). \\n\\nA. Atomic exchange mechanism B. Interstitial mechanism C. Vacancy mechanism", "answer": "C" }, { "idx": 3555, "question": "Fick's first law describes the characteristics of steady-state diffusion, where the concentration does not vary with ( ). \\n\\nA. distance B. time C. temperature", "answer": "B" }, { "idx": 3556, "question": "When a critical nucleus forms, the decrease in volume free energy can only compensate for ( ) of the surface energy.", "answer": "B" }, { "idx": 3557, "question": "In the Kirkendall effect, the main reason for marker drift is ( ) in the diffusion couple.", "answer": "C" }, { "idx": 3558, "question": "In cubic crystals, the (110) and (211) planes belong to the same () zone.", "answer": "D" }, { "idx": 3559, "question": "In a binary phase diagram, the lever rule for calculating the relative amounts of two phases can only be applied in ().", "answer": "B" }, { "idx": 3560, "question": "In the concentration triangle of a ternary phase diagram, alloys whose compositions lie on the () have equal contents of the two components represented by the other two vertices.", "answer": "A" }, { "idx": 3561, "question": "When a pure metal undergoes allotropic transformation during cooling from high temperature to room temperature with volume expansion, the atomic coordination number of the low-temperature phase is () compared to that of the high-temperature phase.", "answer": "A" }, { "idx": 3562, "question": "The packing density of a simple cubic crystal is ( ). \\n\\nA. $100\\\\%$ B.65% C.52% D.58%", "answer": "C" }, { "idx": 3563, "question": "When the deformed material is heated up again, recovery and recrystallization phenomena occur, then the significant decrease in point defect concentration happens during ().", "answer": "A" }, { "idx": 3564, "question": "For two parallel screw dislocations, when their Burgers vectors are in the same direction, the interaction force between them is (). \\n\\nA. zero B. repulsive C. attractive", "answer": "B" }, { "idx": 3565, "question": "The dislocations that can undergo cross-slip must be (). \\n\\nA. Edge dislocation B. Screw dislocation C. Mixed dislocation", "answer": "B" }, { "idx": 3566, "question": "The dislocation that cannot undergo climb motion is (). \\n\\nA. Shockley partial dislocation B. Frank partial dislocation C. Edge full dislocation", "answer": "A" }, { "idx": 3567, "question": "The fundamental reason for diffusion to occur in materials is ( ). \\n\\nA. Temperature changes B. Concentration gradient exists C. Chemical potential gradient exists", "answer": "C" }, { "idx": 3568, "question": "What are the necessary conditions for an alloy to undergo precipitation hardening?", "answer": "The condition for an alloy to undergo precipitation hardening is that it exhibits solubility changes during heating and cooling, meaning it forms a single solid solution at high temperatures and a two-phase structure at room temperature, such as in aluminum-copper alloys." }, { "idx": 3569, "question": "The relationship between the probability of annealing twin formation and the stacking fault energy of a crystal is ( ). \\n\\nA. Unrelated, only depends on annealing temperature and time\\nB. Crystals with low stacking fault energy have a higher probability of forming annealing twins\\nC. Crystals with high stacking fault energy have a higher probability of forming annealing twins", "answer": "B" }, { "idx": 3570, "question": "Temperature indicators are sometimes produced from a coiled metal strip that uncoils a specific amount when the temperature increases. How does this work?", "answer": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change." }, { "idx": 3571, "question": "From what kind of material would the temperature indicator be made?", "answer": "The temperature indicator is made from bimetallic materials, which consist of two materials with different coefficients of thermal expansion bonded together." }, { "idx": 3572, "question": "What are the important properties that the material in the temperature indicator must possess?", "answer": "In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough modulus of elasticity so that no permanent deformation of the material occurs." }, { "idx": 3573, "question": "What properties should the head of a carpenter's hammer possess?", "answer": "The striking face and claws of the hammer should be hard-the metal should not dent or deform when driving or removing nails. Yet these portions must also possess some impact resistance, particularly so that chips do not flake off the striking face and cause injuries." }, { "idx": 3574, "question": "How would you manufacture a hammer head?", "answer": "The head for a carpenter's hammer is produced by forging, a metalworking process; a simple steel shape is heated and formed in several steps while hot into the required shape. The head is then heat treated to produce the required mechanical and physical properties." }, { "idx": 3575, "question": "Aluminum foil used for storing food weighs about 0.3g per square inch. How many atoms of aluminum are contained in one square inch of foil?", "answer": "6.69 × 10^{21} atoms" }, { "idx": 3576, "question": "Using the densities and atomic weights, calculate the number of atoms per cubic centimeter in lead.", "answer": "3.3 × 10^22 atoms/cm^3" }, { "idx": 3577, "question": "Using the densities and atomic weights, calculate the number of atoms per cubic centimeter in lithium.", "answer": "4.63 × 10^22 atoms/cm^3" }, { "idx": 3578, "question": "Using data, calculate the number of iron atoms in one ton (2000 pounds).", "answer": "9.79 × 10^{27} atoms/ton" }, { "idx": 3579, "question": "Using data, calculate the volume in cubic centimeters occupied by one mole of boron.", "answer": "4.7 cm^{3}" }, { "idx": 3580, "question": "In order to plate a steel part having a surface area of 200 in.^2 with a 0.002 in. thick layer of nickel, how many atoms of nickel are required?", "answer": "5.98 x 10^23 atoms" }, { "idx": 3581, "question": "In order to plate a steel part having a surface area of 200 in.^2 with a 0.002 in. thick layer of nickel, how many moles of nickel are required?", "answer": "0.994 mol ni required" }, { "idx": 3582, "question": "Suppose an element has a valence of 2 and an atomic number of 27 . Based only on the quantum numbers, how many electrons must be present in the 3 d energy level?", "answer": "7 electrons." }, { "idx": 3583, "question": "Calculate the fraction of bonding of MgO that is ionic.", "answer": "the fraction of bonding of mgo that is ionic is 0.734. so bonding is mostly ionic." }, { "idx": 3584, "question": "Would you expect MgO or magnesium to have the higher modulus of elasticity? Explain.", "answer": "( MgO has ionic bonds, which are strong compared to the metallic bonds in Mg. A higher force will be required to cause the same separation between the ions in MgO compared to the atoms in Mg. Therefore, MgO should have the higher modulus of elasticity. In Mg, E=6 × 10^{6} psi; in MgO, E=30 × 10^{6} psi." }, { "idx": 3585, "question": "Calculate the atomic radius in cm for a BCC metal with a0=0.3294 nm and one atom per lattice point.", "answer": "1.426 x 10^-8 cm" }, { "idx": 3586, "question": "Calculate the atomic radius in cm for an FCC metal with a0=4.0862 A and one atom per lattice point.", "answer": "1.4447 x 10^-8 cm" }, { "idx": 3587, "question": "Determine the crystal structure for a metal with a0=4.9489 Å, r=1.75 Å and one atom per lattice point.", "answer": "fcc" }, { "idx": 3588, "question": "Determine the crystal structure for a metal with a0=0.42906 nm, r=0.1858 nm and one atom per lattice point.", "answer": "bcc" }, { "idx": 3589, "question": "The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate the lattice parameter.", "answer": "5.3355 x 10^-8 cm" }, { "idx": 3590, "question": "The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate the atomic radius of potassium.", "answer": "2.3103 x 10^-8 cm" }, { "idx": 3591, "question": "The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm³. The atomic weight of thorium is 232 g/mol. Calculate the lattice parameter.", "answer": "5.0856 x 10⁻⁸ cm" }, { "idx": 3592, "question": "The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm³. The atomic weight of thorium is 232 g/mol. Calculate the atomic radius of thorium.", "answer": "1.7980 x 10⁻⁸ cm" }, { "idx": 3593, "question": "A metal having a cubic structure has a density of 2.6g / {cm}^{3}, an atomic weight of 87.62g / mol, and a lattice parameter of 6.0849 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal.", "answer": "fcc." }, { "idx": 3594, "question": "A metal having a cubic structure has a density of 1.892g / {cm}^{3}, an atomic weight \\mathrm{of} 132.91g / mol, and a lattice parameter of € 1.13 Å. One atom is associated with each lattice point. Determine the crystal structure of the \\mathrm{metal}.", "answer": "bcc." }, { "idx": 3595, "question": "Indium has a tetragonal structure with a_{0}=0.32517nm and c_{0}=0.49459nm. The density is 7.286g / {cm}^{3} and the atomic weight is 114.82g / mol. Does indium have the simple tetragonal or body-centered tetragonal structure?", "answer": "body-centered tetragonal structure." }, { "idx": 3596, "question": "Bismuth has a hexagonal structure, with a0=0.4546 nm and c0=1.186 nm. The density is 9.808 g/cm3 and the atomic weight is 208.98 g/mol. Determine the volume of the unit cell.", "answer": "0.21226 nm3 or 2.1226 x 10^-22 cm3" }, { "idx": 3597, "question": "Bismuth has a hexagonal structure, with a0=0.4546 nm and c0=1.186 nm. The density is 9.808 g/cm3 and the atomic weight is 208.98 g/mol. Determine the number of atoms in each unit cell.", "answer": "6 atoms/cell" }, { "idx": 3598, "question": "Gallium has an orthorhombic structure, with a0=0.45258 nm, b0=0.45186 nm, and c0=0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3 and the atomic weight is 69.72 g/mol. Determine the number of atoms in each unit cell.", "answer": "8 atoms/cell." }, { "idx": 3599, "question": "Gallium has an orthorhombic structure, with a0=0.45258 nm, b0=0.45186 nm, and c0=0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3 and the atomic weight is 69.72 g/mol. Determine the packing factor in the unit cell.", "answer": "0.387." }, { "idx": 3600, "question": "Beryllium has a hexagonal crystal structure, with a0=0.22858 nm and c0=0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine the number of atoms in each unit cell.", "answer": "2 atoms/cell." }, { "idx": 3601, "question": "Beryllium has a hexagonal crystal structure, with a0=0.22858 nm and c0=0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine the packing factor in the unit cell.", "answer": "0.77" }, { "idx": 3602, "question": "A typical paper clip weighs 0.59g and consists of BCC iron. Calculate the number of unit cells in the paper clip.", "answer": "3.185 × 10^21 cells" }, { "idx": 3603, "question": "A typical paper clip weighs 0.59g and consists of BCC iron. Calculate the number of iron atoms in the paper clip.", "answer": "6.37 × 10^21 atoms" }, { "idx": 3604, "question": "Determine the planar density for BCC lithium in the (100) plane.", "answer": "planar density = 0.0812 x 10^16 points/cm^2" }, { "idx": 3605, "question": "Determine the packing fraction for BCC lithium in the (100) plane.", "answer": "packing fraction = 0.589" }, { "idx": 3606, "question": "Determine the planar density for BCC lithium in the (110) plane.", "answer": "planar density = 0.1149 x 10^16 points/cm^2" }, { "idx": 3607, "question": "Determine the packing fraction for BCC lithium in the (110) plane.", "answer": "packing fraction = 0.833" }, { "idx": 3608, "question": "Determine the planar density for BCC lithium in the (111) plane.", "answer": "planar density = 0.0469 x 10^16 points/cm^2" }, { "idx": 3609, "question": "Determine the packing fraction for BCC lithium in the (111) plane.", "answer": "packing fraction = 1/2 / 0.866 a0^2 sqrt(3) a0 / 4" }, { "idx": 3610, "question": "Which, if any, of these planes is close packed in BCC lithium?", "answer": "there is no close-packed plane in bcc structures." }, { "idx": 3611, "question": "Suppose that FCC rhodium is produced as a 1-mm thick sheet, with the (111) plane parallel to the surface of the sheet. How many (111) interplanar spacings d_{111} thick is the sheet? See Appendix A for necessary data.", "answer": "the sheet is 4.563 × 10^{6} d_{111} spacings thick." }, { "idx": 3612, "question": "In a FCC unit cell, how many d_{111} are present between the 0,0,0 point and the 1,1,1 point?", "answer": "3" }, { "question": "Determine the minimum radius of an atom that will just fit into the tetrahedral interstitial site in FCC nickel.", "answer": "0.2797 Å", "idx": 3613 }, { "question": "Determine the minimum radius of an atom that will just fit into the octahedral interstitial site in BCC lithium.", "answer": "0.629 Å", "idx": 3614 }, { "idx": 3615, "question": "What is the radius of an atom that will just fit into the octahedral site in FCC copper without disturbing the crystal structure?", "answer": "0.529 Å." }, { "idx": 3616, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{Y}_{2} \\mathrm{O}_{3}.", "answer": "6" }, { "idx": 3617, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{UO}_{2}.", "answer": "6" }, { "idx": 3618, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{BaO}.", "answer": "8" }, { "idx": 3619, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{Si}_{4} N_{4}.", "answer": "4" }, { "idx": 3620, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{GeO}_{2}.", "answer": "4" }, { "idx": 3621, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{MnO}.", "answer": "6" }, { "idx": 3622, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{MgS}.", "answer": "6" }, { "idx": 3623, "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{KBr}.", "answer": "6" }, { "idx": 3624, "question": "Would you expect NiO to have the cesium chloride, sodium chloride, or zinc blende structure?", "answer": "sodium chloride structure." }, { "idx": 3625, "question": "For NiO with sodium chloride structure, determine the lattice parameter.", "answer": "4.02 Å." }, { "idx": 3626, "question": "For NiO with sodium chloride structure, determine the density.", "answer": "7.64 \\mathrm{g/cm}^3." }, { "idx": 3627, "question": "For NiO with sodium chloride structure, determine the packing factor.", "answer": "0.678." }, { "idx": 3628, "question": "Would you expect \\mathrm{UO}_{2} to have the sodium chloride, zinc blende, or fluoride structure?", "answer": "fluorite structure." }, { "idx": 3629, "question": "For \\mathrm{UO}_{2} with fluorite structure, determine the lattice parameter.", "answer": "5.2885 Å." }, { "idx": 3630, "question": "For \\mathrm{UO}_{2} with fluorite structure, determine the density.", "answer": "12.13 {g/cm}^{3}." }, { "idx": 3631, "question": "For \\mathrm{UO}_{2} with fluorite structure, determine the packing factor.", "answer": "The packing factor is pf = 0.624." }, { "idx": 3632, "question": "Would you expect BeO to have the sodium chloride, zinc blende, or fluorite structure?", "answer": "BeO has the zinc blende structure." }, { "idx": 3633, "question": "Based on the zinc blende structure of BeO, determine the lattice parameter.", "answer": "The lattice parameter is 3.8567 Å." }, { "idx": 3634, "question": "Based on the zinc blende structure of BeO, determine the density.", "answer": "The density is 2.897 g/cm³." }, { "idx": 3635, "question": "Based on the zinc blende structure of BeO, determine the packing factor.", "answer": "The packing factor is 0.684." }, { "idx": 3636, "question": "Would you expect CsBr to have the sodium chloride, zinc blende, fluorite, or cesium chloride structure?", "answer": "CsBr is expected to have the cesium chloride structure." }, { "idx": 3637, "question": "Based on the cesium chloride structure, determine the lattice parameter for CsBr.", "answer": "The lattice parameter a0 = 4.1916 Å." }, { "idx": 3638, "question": "Based on the cesium chloride structure, determine the density for CsBr.", "answer": "The density ρ = 4.8 g/cm3." }, { "idx": 3639, "question": "Based on the cesium chloride structure, determine the packing factor for CsBr.", "answer": "The packing factor pf = 0.693." }, { "idx": 3640, "question": "Calculate the planar packing fraction (ppf) on the (110) plane for ZnS (with the zinc blende structure).", "answer": "0.492" }, { "idx": 3641, "question": "Calculate the planar packing fraction (ppf) on the (110) plane for CaF2 (with the fluorite structure).", "answer": "0.699" }, { "idx": 3642, "question": "MgO, which has the sodium chloride structure, has a lattice parameter of 0.396 nm. Determine the planar density and the planar packing fraction for the (111) plane of MgO. What ions are present on this plane?", "answer": "(111) plane:\n- planar density (p.d.): 0.1473 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.202\n- ions present: mg^2+ and o^2-" }, { "idx": 3643, "question": "MgO, which has the sodium chloride structure, has a lattice parameter of 0.396 nm. Determine the planar density and the planar packing fraction for the (222) plane of MgO. What ions are present on this plane?", "answer": "(222) plane:\n- planar density (p.d.): 0.1473 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.806\n- ions present: mg^2+ and o^2-" }, { "idx": 3644, "question": "A diffracted x-ray beam is observed from the (220) planes of iron at a 2 \\theta angle of 99.1^{\\circ} when x-rays of 0.15418nm wavelength are used. Calculate the lattice parameter of the iron.", "answer": "0.2865nm" }, { "idx": 3645, "question": "Calculate the number of vacancies per {cm}^{3} expected in copper at 1080^{\\circ} C (just below the melting temperature). The activation energy for vacancy formation is 20,000 cal/mol.", "answer": "4.97 × 10^{19} vacancies/cm^3" }, { "idx": 3646, "question": "The fraction of lattice points occupied by vacancies in solid aluminum at 660^{\\circ} C is 10^{-3}. What is the activation energy required to create vacancies in aluminum?", "answer": "12800 cal/mol" }, { "idx": 3647, "question": "The density of a sample of HCP beryllium is 1.844 g/cm³ and the lattice parameters are a₀=0.22858 nm and c₀=0.35842 nm. Calculate the fraction of the lattice points that contain vacancies.", "answer": "0.0008" }, { "idx": 3648, "question": "The density of a sample of HCP beryllium is 1.844 g/cm³ and the lattice parameters are a₀=0.22858 nm and c₀=0.35842 nm. Calculate the total number of vacancies in a cubic centimeter.", "answer": "0.986 × 10²⁰ vacancies/cm³" }, { "idx": 3649, "question": "BCC lithium has a lattice parameter of 3.5089 × 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the number of vacancies per cubic centimeter.", "answer": "1.157 × 10^20 vacancies/cm³" }, { "idx": 3650, "question": "BCC lithium has a lattice parameter of 3.5089 × 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the density of Li.", "answer": "0.532 g/cm³" }, { "idx": 3651, "question": "FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate the density.", "answer": "the density is 11.335 g/cm3." }, { "idx": 3652, "question": "FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate the number of vacancies per gram of Pb.", "answer": "the number of vacancies per gram of pb is 5.82 × 10^18 vacancies/g." }, { "idx": 3653, "question": "A niobium alloy is produced by introducing tungsten substitutional atoms in the BCC structure; eventually an alloy is produced that has a lattice parameter of 0.32554nm and a density of 11.95g / {cm}^{3}. Calculate the fraction of the atoms in the alloy that are tungsten.", "answer": "0.345" }, { "idx": 3654, "question": "Tin atoms are introduced into a FCC copper crystal, producing an alloy with a lattice parameter of 3.7589 × 10^{-8} cm and a density of 8.772g / {cm}^{3}. Calculate the atomic percentage of tin present in the alloy.", "answer": "11.95%." }, { "idx": 3655, "question": "We replace 7.5 atomic percent of the chromium atoms in its BCC crystal with tantalum. X-ray diffraction shows that the lattice parameter is 0.29158nm. Calculate the density of the alloy.", "answer": "8.265g / {cm}^{3}" }, { "idx": 3656, "question": "Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find the density.", "answer": "7.89 g/cm³" }, { "idx": 3657, "question": "Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find the packing factor.", "answer": "0.681" }, { "idx": 3658, "question": "The density of BCC iron is 7.882 g/cm³ and the lattice parameter is 0.2886 nm when hydrogen atoms are introduced at interstitial positions. Calculate the atomic fraction of hydrogen atoms.", "answer": "0.004" }, { "idx": 3659, "question": "The density of BCC iron is 7.882 g/cm³ and the lattice parameter is 0.2886 nm when hydrogen atoms are introduced at interstitial positions. Calculate the number of unit cells required on average that contain hydrogen atoms.", "answer": "123.5" }, { "idx": 3660, "question": "Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate the number of anion vacancies per cm3.", "answer": "1.61 × 10^21 vacancies/cm^3." }, { "idx": 3661, "question": "Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate the density of the ceramic.", "answer": "4.205 g/cm^3." }, { "idx": 3662, "question": "ZnS has the zinc blende structure. If the density is 3.02 g/cm3 and the lattice parameter is 0.59583 nm, determine the number of Schottky defects per unit cell.", "answer": "0.0535 defects per unit cell." }, { "idx": 3663, "question": "ZnS has the zinc blende structure. If the density is 3.02 g/cm3 and the lattice parameter is 0.59583 nm, determine the number of Schottky defects per cubic centimeter.", "answer": "2.517 × 10^20 defects per cm3." }, { "idx": 3664, "question": "Calculate the length of the Burgers vector in BCC niobium", "answer": "b = 2.853 Å" }, { "idx": 3665, "question": "Calculate the length of the Burgers vector in FCC silver", "answer": "b = 2.889 Å" }, { "idx": 3666, "question": "Calculate the length of the Burgers vector in diamond cubic silicon", "answer": "b = 3.840 Å" }, { "idx": 3667, "question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [T] slip direction.", "answer": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 0." }, { "idx": 3668, "question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [T] slip direction.", "answer": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 2040 psi (active)." }, { "idx": 3669, "question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Which slip system(s) will become active first?", "answer": "the slip systems that will become active first are λ011 and λ101." }, { "idx": 3670, "question": "A single crystal of a BCC metal is oriented so that the direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (110) slip plane.", "answer": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (110) slip plane: σ = ∞" }, { "idx": 3671, "question": "A single crystal of a BCC metal is oriented so that the direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (011) slip plane.", "answer": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (011) slip plane: σ = 29,412 psi" }, { "idx": 3672, "question": "A single crystal of a BCC metal is oriented so that the direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (101) slip plane.", "answer": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (101) slip plane: σ = 29,412 psi" }, { "idx": 3673, "question": "The strength of titanium is found to be 65,000 psi when the grain size is 17 x 10^-6 m and 82,000 psi when the grain size is 0.8 x 10^-6 m. Determine the constants in the Hall-Petch equation.", "answer": "the constants in the hall-petch equation are k = 19.4 psi/sqrt(d) and σ_o = 60,290 psi." }, { "idx": 3674, "question": "The strength of titanium is found to be 65,000 psi when the grain size is 17 x 10^-6 m and 82,000 psi when the grain size is 0.8 x 10^-6 m. Determine the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m.", "answer": "the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m is 103,670 psi." }, { "idx": 3675, "question": "For an ASTM grain size number of 8, calculate the number of grains per square inch at a magnification of 100.", "answer": "128 grains/in.^2" }, { "idx": 3676, "question": "For an ASTM grain size number of 8, calculate the number of grains per square inch with no magnification.", "answer": "1.28 × 10^6 grains/in.^2" }, { "idx": 3677, "question": "Determine the ASTM grain size number if 20 grains/square inch are observed at a magnification of 400 .", "answer": "the astm grain size number is 9.3." }, { "idx": 3678, "question": "Determine the ASTM grain size number if 25 grains/square inch are observed at a magnification of 50 .", "answer": "the astm grain size number is n = 3.6" }, { "idx": 3679, "question": "Calculate the angle \\theta of a small-angle grain boundary in FCC aluminum when the dislocations are 5000 Å apart. (See Figure 4-18 and equation in Problem 4-46.)", "answer": "\\theta = 0.0328^{\\circ}" }, { "idx": 3680, "question": "Atoms are found to move from one lattice position to another at the rate of 5 × 10^{5} jumps/s at 400^{\\circ} C when the activation energy for their movement is 30,000 cal/mol. Calculate the jump rate at 750^{\\circ} C.\n\\section*{", "answer": "the jump rate at 750^{\\circ} C is 1.08 × 10^{9} jumps/s." }, { "idx": 3681, "question": "The number of vacancies in a material is related to temperature by an Arrhenius equation. If the fraction of lattice points containing vacancies is 8 × 10^{-5} \\mathrm{at} 600^{\\circ} C, determine the fraction at 1000^{\\circ} C.", "answer": "the fraction at 1000^{\\circ} C is 0.00155." }, { "idx": 3682, "question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6x10^-15 cm^2/s at 727C and is 1x10^-9 cm^2/s at 1400C. Calculate the activation energy q.", "answer": "the activation energy q is 59,230 cal/mol." }, { "idx": 3683, "question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6x10^-15 cm^2/s at 727C and is 1x10^-9 cm^2/s at 1400C. Calculate the constant d0.", "answer": "the constant d0 is 0.055 cm^2/s." }, { "idx": 3684, "question": "A 0.2-mm thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 10^8 Si atoms and the other surface contains 500 Sb atoms per 10^8 Si atoms. The lattice parameter for Si is 5.407 A. Calculate the concentration gradient in (a) atomic percent Sb per cm.", "answer": "-0.02495 at% sb/cm" }, { "idx": 3685, "question": "A 0.2-mm thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 10^8 Si atoms and the other surface contains 500 Sb atoms per 10^8 Si atoms. The lattice parameter for Si is 5.407 A. Calculate the concentration gradient in (b) Sb atoms/cm^3·cm.", "answer": "-1.246 × 10^19 sb atoms/cm^3·cm" }, { "idx": 3686, "question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025mm away contains 20 atomic percent zinc. The lattice parameter for the FCC alloy is 3.63 x 10^-8 cm. Determine the concentration gradient in (a) atomic percent Zn per cm.", "answer": "-2000 at% zn/cm" }, { "idx": 3687, "question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025mm away contains 20 atomic percent zinc. The lattice parameter for the FCC alloy is 3.63 x 10^-8 cm. Determine the concentration gradient in (b) weight percent Zn per cm.", "answer": "-2032 wt% zn/cm" }, { "idx": 3688, "question": "A 0.001 -in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650^{\\circ} C .5 × 10^{8} H atoms / {cm}^{3} are in equilibrium with the hot side of the foil, while 2 × 10^{3} H atoms / {cm}^{3} are in equilibrium with the cold side. Determine the concentration gradient of hydrogen.", "answer": "-1969 × 10^{8} h atoms/cm^3·cm" }, { "idx": 3689, "question": "A 0.001 -in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650^{\\circ} C .5 × 10^{8} H atoms / {cm}^{3} are in equilibrium with the hot side of the foil, while 2 × 10^{3} H atoms / {cm}^{3} are in equilibrium with the cold side. Determine the flux of hydrogen through the foil.", "answer": "0.33 × 10^{8} h atoms/cm^2·s" }, { "idx": 3690, "question": "A 1-mm sheet of FCC iron is used to contain nitrogen in a heat exchanger at 1200^{\\circ} C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. Determine the flux of nitrogen through the foil in N atoms / {cm}^{2}·s.", "answer": "the flux of nitrogen through the foil is 7.57 × 10^{12} n atoms/cm^2 \\cdot s." }, { "idx": 3691, "question": "A 4-cm-diameter, 0.5-mm-thick spherical container made of BCC iron holds nitrogen at 700^{\\circ} C. The concentration at the inner surface is 0.05 atomic percent and at the outer surface is 0.002 atomic percent. Calculate the number of grams of nitrogen that are lost from the container per hour.", "answer": "the nitrogen lost from the container per hour is 1.245 × 10^{-3} \\, \\text{g/h}." }, { "idx": 3692, "question": "A BCC iron structure is to be manufactured that will allow no more than 50g of hydrogen to be lost per year through each square centimeter of the iron at 400^{\\circ} C. If the concentration of hydrogen at one surface is 0.05 H atom per unit cell and is 0.001 H atom per unit cell at the second surface, determine the minimum thickness of the iron.", "answer": "the minimum thickness of the iron is 0.179 cm." }, { "idx": 3693, "question": "Determine the maximum allowable temperature that will produce a flux of less than 2000 H atoms / {cm}^{2}·s through a BCC iron foil when the concentration gradient is -5 × 10^{16} atoms / {cm}^{3}·{cm}. (Note the negative sign for the flux.)", "answer": "the maximum allowable temperature is 75k (-198^{\\circ} C)." }, { "idx": 3694, "question": "Compare the rate at which oxygen ions diffuse in \\mathrm{Al}_{2} \\mathrm{O}_{3} with the rate at which aluminum ions diffuse in \\mathrm{Al}_{2} \\mathrm{O}_{3} at 1500^{\\circ} C. Explain the difference.", "answer": "the diffusion rate of oxygen ions in \\mathrm{al}_{2}\\mathrm{o}_{3} at 1500^{\\circ}C is 3.47 × 10^{-16} {cm}^{2}/s, while the diffusion rate of aluminum ions is 2.48 × 10^{-13} {cm}^{2}/s. the difference in diffusion rates is due to the ionic radii: the oxygen ion has a larger radius (1.32 Å) compared to the aluminum ion (0.51 Å), making it easier for the smaller aluminum ion to diffuse in the ceramic." }, { "idx": 3695, "question": "Compare the diffusion coefficients of carbon in BCC iron at the allotropic transformation temperature of 912^{\\circ} C and explain the difference.", "answer": "the diffusion coefficient of carbon in bcc iron at 912^{\\circ} C is 1.51 × 10^{-6} cm^{2} / s, while in fcc iron it is (1.92 × 10^{-7} cm^{2} / s). the packing factor of the bcc lattice (0.68) is less than that of the fcc lattice; consequently, atoms are expected to be able to diffuse more rapidly in the bcc iron." }, { "idx": 3696, "question": "Iron containing 0.05 % C is heated to 912 degrees C in an atmosphere that produces 1.20 % C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if the iron is BCC.", "answer": "For BCC iron, the carbon content at 0.05 cm beneath the surface is 0.95 % C." }, { "idx": 3697, "question": "Iron containing 0.05 % C is heated to 912 degrees C in an atmosphere that produces 1.20 % C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if the iron is FCC.", "answer": "For FCC iron, the carbon content at 0.05 cm beneath the surface is 0.95 % C." }, { "idx": 3698, "question": "What temperature is required to obtain 0.50% C at a distance of 0.5mm beneath the surface of a 0.20% C steel in 2h, when 1.10% C is present at the surface? Assume that the iron is FCC.", "answer": "the required temperature is 1180k or 907^{\\circ} C." }, { "idx": 3699, "question": "A 0.15% C steel is to be carburized at 1100^{\\circ} C, giving 0.35% C at a distance of 1mm beneath the surface. If the surface composition is maintained at 0.90% C, what time is required?", "answer": "the required time is 51 \\mathrm{min}." }, { "idx": 3700, "question": "A 0.02% C steel is to be carburized at 1200^{\\circ} C in 4h, with a point 0.6mm beneath the surface reaching 0.45% C. Calculate the carbon content required at the surface of the steel.", "answer": "the carbon content required at the surface of the steel is 0.53% C." }, { "idx": 3701, "question": "A 1.2% C tool steel held at 1150^{\\circ} C is exposed to oxygen for 48h. The carbon content at the steel surface is zero. To what depth will the steel be decarburized to less than 0.20% C ?", "answer": "0.177 cm" }, { "idx": 3702, "question": "A 0.80% C steel must operate at 950^{\\circ} C in an oxidizing environment, where the carbon content at the steel surface is zero. Only the outermost 0.02 cm of the steel part can fall below 0.75% C. What is the maximum time that the steel part can operate?", "answer": "the maximum time that the steel part can operate is 2.9 min." }, { "idx": 3703, "question": "A steel with \\mathrm{BBC} crystal structure containing 0.001%N is nitrided at 550^{\\circ} C for 5h. If the nitrogen content at the steel surface is 0.08%, determine the nitrogen content at 0.25mm from the surface.", "answer": "the nitrogen content at 0.25mm from the surface is 0.049% n." }, { "idx": 3704, "question": "What time is required to nitride a 0.002N steel to obtain 0.12%N at a distance of 0.002 in. beneath the surface at 625^{\\circ} C ? The nitrogen content at the surface is 0.15%.\n\\[\n\\begin{array}{l}\n", "answer": "12.8 min" }, { "idx": 3705, "question": "We currently can successfully perform a carburizing heat treatment at 1200^{\\circ} C in 1h. In an effort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950^{\\circ} C. What time will be required to give us a similar carburizing treatment?", "answer": "9.95 h" }, { "idx": 3706, "question": "During freezing of a Cu-Zn alloy, we find that the composition is nonuniform. By heating the alloy to 600^{\\circ} C for 3 hours, diffusion of zinc helps to make the composition more uniform. What temperature would be required if we wished to perform this homogenization treatment in 30 minutes?", "answer": "667°c" }, { "idx": 3707, "question": "A ceramic part made of MgO is sintered successfully at 1700 degrees C in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees C. Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions?", "answer": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen." }, { "idx": 3708, "question": "A ceramic part made of MgO is sintered successfully at 1700 degrees C in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees C. What time will be required at the lower temperature?", "answer": "the time required at the lower temperature is 955 min or 15.9 h." }, { "idx": 3709, "question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55,000 psi. Determine whether the wire will plastically deform.", "answer": "the wire will plastically deform (48,100 psi)." }, { "idx": 3710, "question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55,000 psi. Determine whether the wire will experience necking.", "answer": "no necking will occur (48,100 psi)." }, { "idx": 3711, "question": "A force of 100000 N is applied to a 10mm × 20mm iron bar having a yield strength of 400 MPa. Determine whether the bar will plastically deform.", "answer": "the bar will plastically deform because the applied stress of 500 MPa exceeds the yield strength of 400 MPa." }, { "idx": 3712, "question": "A force of 100000 N is applied to a 10mm × 20mm iron bar having a tensile strength of 480 MPa. Determine whether the bar will experience necking.", "answer": "the bar will experience necking because the applied stress of 500 MPa exceeds the tensile strength of 480 MPa." }, { "idx": 3713, "question": "Calculate the maximum force that a 0.2 -in. diameter rod of \\mathrm{Al}_{2} \\mathrm{O}_{3}, having a yield strength of 35,000 psi, can withstand with no plastic deformation. Express your answer in pounds and Newtons.", "answer": "the maximum force the rod can withstand is 1100 lb or 4891 n." }, { "idx": 3714, "question": "A force of 20,000 N will cause a 1 cm × 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the modulus of elasticity in GPa.", "answer": "the modulus of elasticity is 44.4 GPa." }, { "idx": 3715, "question": "A force of 20,000 N will cause a 1 cm × 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the modulus of elasticity in psi.", "answer": "the modulus of elasticity is 6.44 × 10^6 psi." }, { "idx": 3716, "question": "A polymer bar's dimensions are 1 in . × 2 in. × 15 in. The polymer has a modulus of elasticity of 600,000 psi. What force is required to stretch the bar elastically to 15.25 in.?", "answer": "the force required is 200,000 lb." }, { "idx": 3717, "question": "An aluminum plate 0.5 cm thick is to withstand a force of 50,000N with no permanent deformation. If the aluminum has a yield strength of 125 MPa, what is the minimum width of the plate?", "answer": "the minimum width of the plate is 8 cm." }, { "idx": 3718, "question": "(a) A 3-in.-diameter rod of copper is to be reduced to a 2-in.-diameter rod by being pushed through an opening. To account for the elastic strain, what should be the diameter of the opening? The modulus of elasticity for the copper is 17 × 10^{6} psi and the yield strength is 40,000 psi.", "answer": "the diameter of the opening should be 1.995 in." }, { "idx": 3719, "question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16x10^6 psi, and Poisson's ratio of 0.30. Determine the length of the bar when a 500-lb load is applied.", "answer": "the final length of the bar is 12.00298 in." }, { "idx": 3720, "question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16x10^6 psi, and Poisson's ratio of 0.30. Determine the diameter of the bar when a 500-lb load is applied.", "answer": "the final diameter of the bar is 0.39997 in." }, { "idx": 3721, "question": "A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate the flexural strength.", "answer": "flexural strength = 76,800 psi" }, { "idx": 3722, "question": "A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate the flexural modulus, assuming that no plastic deformation occurs.", "answer": "flexural modulus = 22.14 x 10^6 psi" }, { "idx": 3723, "question": "A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09mm is recorded. Calculate the force that caused the fracture. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.", "answer": "the force that caused the fracture is 1327 n." }, { "idx": 3724, "question": "A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09mm is recorded. Calculate the flexural strength. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.", "answer": "the flexural strength is 276 mpa." }, { "idx": 3725, "question": "A thermosetting polymer containing glass beads is required to deflect 0.5mm when a force of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports.", "answer": "the minimum distance between the supports is 41 mm." }, { "idx": 3726, "question": "Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.", "answer": "the applied stress is 61.5 MPa, which is less than the flexural strength of 85 MPa; the polymer is not expected to fracture." }, { "idx": 3727, "question": "(b) The flexural modulus of alumina is 45 × 10^{6} psi and its flexural strength is 46,000 psi. A bar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs.", "answer": "the deflection at the moment the bar breaks is 0.0278 in." }, { "idx": 3728, "question": "A Brinell hardness measurement, using a 10-mm-diameter indenter and a 500-kg load, produces an indentation of 4.5mm on an aluminum plate. Determine the Brinell hardness number (HB) of the metal.", "answer": " hb = 29.8 " }, { "idx": 3729, "question": "When a 3000-kg load is applied to a 10-mm-diameter ball in a Brinell test of a steel, an indentation of 3.1mm is produced. Estimate the tensile strength of the steel.", "answer": "the tensile strength of the steel is 194,000 psi." }, { "idx": 3730, "question": "A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the composite is 45 MPa / m and the tensile strength is 550 MPa. Will the flaw cause the composite to fail before the tensile strength is reached? Assume that f=1.", "answer": "the applied stress required for the crack to cause failure is 11,354 MPa. the tensile strength of the composite is 550 MPa. any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws." }, { "idx": 3731, "question": "An aluminum alloy that has a plane strain fracture toughness of 25,000 psi \\sqrt{m}. fails when a stress of 42,000 psi is applied. Observation of the fracture surface indicates that fracture began at the surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f=1.1.", "answer": "the size of the flaw that initiated fracture is 0.093 \\text{ in}." }, { "idx": 3732, "question": "A polymer that contains internal flaws 1mm in length fails at a stress of 25 MPa. Determine the plane strain fracture toughness of the polymer. Assume that f=1.", "answer": "the plane strain fracture toughness of the polymer is 0.99 \\text{mpa} \\sqrt{m}." }, { "idx": 3733, "question": "A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain fracture toughness of 5,000 psi / in. To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05 in. long. Assuming that f=1.4, does our nondestructive test have the required sensitivity? Explain.", "answer": "the length of internal flaws is 2a = 0.013 in. our nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size required for failure. thus our ndt test is not satisfactory." }, { "idx": 3734, "question": "To survive for one million cycles under conditions that provide for equal compressive and tensile stresses, what is the fatigue strength, or maximum stress amplitude, required?", "answer": "The fatigue strength at one million cycles is 22 mpa." }, { "idx": 3735, "question": "What are the maximum stress, the minimum stress, and the mean stress on the part during its use under conditions that provide for equal compressive and tensile stresses?", "answer": "The maximum stress is +22 mpa, the minimum stress is -22 mpa, and the mean stress is 0 mpa." }, { "idx": 3736, "question": "What effect would the frequency of the stress application have on the fatigue strength, maximum stress, minimum stress, and mean stress?", "answer": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time." }, { "idx": 3737, "question": "The activation energy for self-diffusion in copper is 49,300 cal / mol. A copper specimen creps as t 0.002 N / fn .-h when a stress of 15,000 psi is applied at 600^{\\circ} C. If the creep rate of copper is dependent on self-diffusion, determine the creep rate if the temperature is 800^{\\circ} C.", "answer": "the creep rate is 0.4, N / in·h." }, { "idx": 3738, "question": "When a stress of 20000 psi is applied to a material heated to 900^{\\circ} C, rupture occurs in 25000h. If the activation energy for rupture is 35000 cal / mol, determine the rupture time if the temperature is reduced to 800^{\\circ} C.", "answer": "the rupture time at 800^{\\circ} C is 101660h." }, { "question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the activation energy Q_{r} for rupture in the temperature range 980 to 1090^{\\circ} C.", "answer": "the activation energy q_{r} is 117,000 \text{ cal/mol}.", "idx": 3739 }, { "question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the constant m for rupture in the temperature range 980 to 1090^{\\circ} C.", "answer": "the constant m for rupture is 3.9.", "idx": 3740 }, { "idx": 3741, "question": "A 0.25-in.-thick copper plate is to be cold worked 63%. Find the final thickness.", "answer": " t_{f} = 0.0925 in." }, { "idx": 3742, "question": "A 0.25-in.-diameter copper bar is to be cold worked 63%. Find the final diameter.", "answer": "the final diameter is 0.152 in." }, { "idx": 3743, "question": "A 2-in.-diameter copper rod is reduced to 1.5 in. diameter, then reduced again to a final diameter of 1 in. In a second case, the 2-in.-diameter rod is reduced in one step from 2 in. to a 1 in. diameter. Calculate the% CW for both cases.", "answer": "75% in both cases." }, { "idx": 3744, "question": "Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate the critical radius of the nucleus required. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.", "answer": "the critical radius of the nucleus required is 6.65 × 10^-8 cm." }, { "idx": 3745, "question": "Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.", "answer": "the number of nickel atoms in the nucleus is 109 atoms." }, { "idx": 3746, "question": "Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate the critical radius of the nucleus required. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.", "answer": "10.128 × 10⁻⁸ cm" }, { "idx": 3747, "question": "Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.", "answer": "350 atoms" }, { "idx": 3748, "question": "Suppose that solid nickel was able to nucleate homogeneously with an undercooling of only 22^{\\circ} C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid FCC nickel is 0.356nm.", "answer": "1.136 × 10^{6}." }, { "idx": 3749, "question": "Suppose that solid iron was able to nucleate homogeneously with an undercooling of only 15^{\\circ} C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid BCC iron is 2.92 Å.", "answer": "7.676 × 10^{6} atoms." }, { "idx": 3750, "question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 10 degrees Celsius undercooling. The specific heat of iron is 5.78 J/cm^3·°C.", "answer": "0.0333" }, { "idx": 3751, "question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 100 degrees Celsius undercooling. The specific heat of iron is 5.78 J/cm^3·°C.", "answer": "0.333" }, { "idx": 3752, "question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates homogeneously. The specific heat of iron is 5.78 J/cm^3·°C.", "answer": "all dendritically." }, { "idx": 3753, "question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 10 degrees Celsius undercooling. The specific heat of silver is 3.25 J/cm^3·°C.", "answer": "0.0237" }, { "idx": 3754, "question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 100 degrees Celsius undercooling. The specific heat of silver is 3.25 J/cm^3·°C.", "answer": "0.337" }, { "idx": 3755, "question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates homogeneously. The specific heat of silver is 3.25 J/cm^3·°C.", "answer": "0.842" }, { "idx": 3756, "question": "A 2-in. cube solidifies in 4.6 min. Calculate (a) the mold constant in Chvorinov's rule. Assume that n=2.", "answer": "the mold constant in chvorinov's rule is 41.48 min/in^2." }, { "idx": 3757, "question": "A 2-in. cube solidifies in 4.6 min. Calculate (b) the solidification time for a 0.5 in. x 0.5 in. x 6 in. bar cast under the same conditions. Assume that n=2.", "answer": "the solidification time for the bar is 0.60 min." }, { "idx": 3758, "question": "A 5-cm diameter sphere solidifies in 1050s. Calculate the solidification time for a 0.3 cm × 10 cm × 20 cm plate cast under the same conditions. Assume that n=2.", "answer": "the solidification time for the plate is 31.15s." }, { "idx": 3759, "question": "Calculate the diameter of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H / D of the riser is 1.5.", "answer": "the diameter of the riser d ≥ 6.67 in." }, { "idx": 3760, "question": "Calculate the height of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H / D of the riser is 1.5.", "answer": "the height of the riser h ≥ 10 in." }, { "idx": 3761, "question": "Calculate the volume of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H / D of the riser is 1.5.", "answer": "the volume of the riser v ≥ 349 in.^3" }, { "idx": 3762, "question": "Calculate the diameter of the cylindrical riser required to prevent shrinkage in a 1 in. x 6 in. x 6 in. casting if the H/D of the riser is 1.0.", "answer": "the diameter of the riser must be at least 2.25 in." }, { "idx": 3763, "question": "Calculate the height of the cylindrical riser required to prevent shrinkage in a 1 in. x 6 in. x 6 in. casting if the H/D of the riser is 1.0.", "answer": "the height of the riser must be at least 2.25 in." }, { "idx": 3764, "question": "Calculate the volume of the cylindrical riser required to prevent shrinkage in a 1 in. x 6 in. x 6 in. casting if the H/D of the riser is 1.0.", "answer": "the volume of the riser must be at least 8.95 in^3." }, { "idx": 3765, "question": "A 4-in-diameter sphere of liquid copper is allowed to solidify, producing a spherical shrinkage cavity in the center of the casting. Determine the volume and diameter of the shrinkage cavity in the copper casting.", "answer": "shrinkage volume: 1.709 in.^{3}; diameter of shrinkage cavity: 1.30 in." }, { "idx": 3766, "question": "A 4-in-diameter sphere of liquid iron is allowed to solidify, producing a spherical shrinkage cavity in the center of the casting. Determine the volume and diameter of the shrinkage cavity in the iron casting.", "answer": "shrinkage volume: 1.139 in.^{3}; diameter of shrinkage cavity: 1.30 in." }, { "idx": 3767, "question": "A 4-in. cube of a liquid metal is allowed to solidify. A spherical shrinkage cavity with a diameter of 1.49 in. is observed in the solid casting. Determine the percent volume change that occurs during solidification.", "answer": "the percent volume change that occurs during solidification is 2.7%." }, { "idx": 3768, "question": "A 2 cm x 4 cm x 6 cm magnesium casting is produced. After cooling to room temperature, what is the volume of the shrinkage cavity at the center of the casting?", "answer": "the volume of the shrinkage cavity at the center of the casting is 46.03 cm3." }, { "idx": 3769, "question": "A 2 cm x 4 cm x 6 cm magnesium casting is produced. After cooling to room temperature, what is the percent shrinkage that must have occurred during solidification?", "answer": "the percent shrinkage that must have occurred during solidification is 4.1%." }, { "idx": 3770, "question": "A 2 in. × 8 in. × 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. Determine the percent shrinkage that must have occurred during solidification.", "answer": "The percent shrinkage that must have occurred during solidification is 3.4%." }, { "idx": 3771, "question": "A 2 in. × 8 in. × 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. Determine the number of shrinkage pores in the casting if all of the shrinkage occurs as pores with a diameter of 0.05 in.", "answer": "The number of shrinkage pores in the casting is 83,354 pores." }, { "idx": 3772, "question": "Liquid magnesium is poured into a 2 cm × 2 cm × 24 cm mold and, as a result of directional solidification, all of the solidification shrinkage occurs along the length of the casting. Determine the length of the casting immediately after solidification is completed.", "answer": "the length of the casting immediately after solidification is completed is 23.04 cm." }, { "idx": 3773, "question": "A liquid cast iron has a density of 7.65 g/cm3. Immediately after solidification, the density of the solid cast iron is found to be 7.71 g/cm3. Determine the percent volume change that occurs during solidification.", "answer": "the percent volume change during solidification is 0.77%." }, { "idx": 3774, "question": "Does the cast iron expand or contract during solidification?", "answer": "the cast iron contracts during solidification." }, { "idx": 3775, "question": "The solubility of hydrogen in liquid aluminum at 715^{\\circ} C is found to be 1 cm^{3} / 100g \\mathrm{Al}. If all of this hydrogen precipitated as gas bubbles during solidification and remained trapped in the casting, calculate the volume percent gas in the solid aluminum.", "answer": "2.699%" }, { "idx": 3776, "question": "Based on Hume-Rothery's conditions, would the system Au-Ag be expected to display unlimited solid solubility? Explain.", "answer": "Yes" }, { "idx": 3777, "question": "Based on Hume-Rothery's conditions, would the system Al-Cu be expected to display unlimited solid solubility? Explain.", "answer": "No" }, { "idx": 3778, "question": "Based on Hume-Rothery's conditions, would the system Al-Au be expected to display unlimited solid solubility? Explain.", "answer": "No" }, { "idx": 3779, "question": "Based on Hume-Rothery's conditions, would the system U-W be expected to display unlimited solid solubility? Explain.", "answer": "No" }, { "idx": 3780, "question": "Based on Hume-Rothery's conditions, would the system Mo-Ta be expected to display unlimited solid solubility? Explain.", "answer": "Yes" }, { "idx": 3781, "question": "Based on Hume-Rothery's conditions, would the system Nb-W be expected to display unlimited solid solubility? Explain.", "answer": "No" }, { "idx": 3782, "question": "Based on Hume-Rothery's conditions, would the system Mg-Zn be expected to display unlimited solid solubility? Explain.", "answer": "No" }, { "idx": 3783, "question": "Based on Hume-Rothery's conditions, would the system Mg-Cd be expected to display unlimited solid solubility? Explain.", "answer": "Yes" }, { "idx": 3784, "question": "Suppose 1 at% of the following elements is added to copper (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? For copper: r_Cu=1.278 Å (a) Au: r=1.442, φr=+12.8% (b) Mn: r=1.12, φr=-12.4% (c) Sr: r=2.151, φr=+68.3% (d) Si: r=1.176, φr=-8.0% (e) Co: r=1.253, φr=-2.0%", "answer": "The Cu-Sr alloy would be expected to be strongest (largest size difference)." }, { "idx": 3785, "question": "Is any of the alloying elements expected to have unlimited solid solubility in copper? For copper: r_Cu=1.278 Å (a) Au: r=1.442, φr=+12.8% (b) Mn: r=1.12, φr=-12.4% (c) Sr: r=2.151, φr=+68.3% (d) Si: r=1.176, φr=-8.0% (e) Co: r=1.253, φr=-2.0%", "answer": "The Cu-Au alloy satisfies Hume-Rothery's conditions and might be expected to display complete solid solubility-in fact it freezes like an isomorphous series of alloys, but a number of solid-state transformations occur at lower temperatures." }, { "idx": 3786, "question": "Suppose 1 at% of the following elements is added to copper (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? (a) Au (b) Mn (c) Sr (d) Si (e) Co", "answer": "(c)" }, { "idx": 3787, "question": "Which of these alloying elements is expected to have unlimited solid solubility in copper? (a) Au (b) Mn (c) Sr (d) Si (e) Co", "answer": "(a)" }, { "idx": 3788, "question": "Which of the following oxides is expected to have the largest solid solubility in \\mathrm{Al}_{2} \\mathrm{O}_{3} ?\n(a) \\mathrm{Y}_{2} \\mathrm{O}_{3} (b) \\mathrm{Cr}_{2} \\mathrm{O}_{3} (c) \\mathrm{Fe}_{2} \\mathrm{O}_{3}", "answer": "\\mathrm{cr}_{2} \\mathrm{o}_{3} is expected to have the largest solid solubility in \\mathrm{al}_{2} \\mathrm{o}_{3}; in fact, they are completely soluble in one another." }, { "idx": 3789, "question": "How many grams of nickel must be added to 500 grams of copper to produce an\nalloy that contains 50 wt% \\alpha at 1300^{\\circ} C ?", "answer": "541.7g ni" }, { "idx": 3790, "question": "How many grams of MgO must be added to 1kg of NiO to produce a ceramic that has a solidus temperature of 2200^{\\circ} C ?", "answer": "332g of mgo" }, { "idx": 3791, "question": "We would like to produce a MgO-\\mathrm{FeO} ceramic that is 30 wt% solid at 2000^{\\circ} C. Determine the original composition of the ceramic in wt%.", "answer": "56.9 wt% feo" }, { "idx": 3792, "question": "A Nb-W alloy held at 2800°C is partly liquid and partly solid. If possible, determine the composition of each phase in the alloy.", "answer": "l: 49% w; α: 70% w" }, { "idx": 3793, "question": "A Nb-W alloy held at 2800°C is partly liquid and partly solid. If possible, determine the amount of each phase in the alloy.", "answer": "not possible unless we know the original composition of the alloy." }, { "idx": 3794, "question": "A Nb-W alloy contains 55% α at 2600°C. Determine the composition of each phase.", "answer": "l: 22% w; α: 42% w" }, { "idx": 3795, "question": "A Nb-W alloy contains 55% α at 2600°C. Determine the original composition of the alloy.", "answer": "33% w" }, { "idx": 3796, "question": "Suppose a 1200-lb bath of a Nb-40 wt % W alloy is held at 2800 degrees C. How many pounds of tungsten can be added to the bath before any solid forms?", "answer": "212 additional pounds of W must be added before any solid forms." }, { "idx": 3797, "question": "Suppose a 1200-lb bath of a Nb-40 wt % W alloy is held at 2800 degrees C. How many pounds of tungsten must be added to cause the entire bath to be solid?", "answer": "1200 additional pounds of W must be added to cause the entire bath to be solid." }, { "idx": 3798, "question": "Suppose a crucible made of pure nickel is used to contain 500g of liquid copper at 1150^{\\circ} C. Describe what happens to the system as it is held at this temperature for several hours. Explain.", "answer": "cu dissolves ni until the cu contains enough ni that it solidifies completely. when 10% ni is dissolved, freezing begins with x = 55.5 \\text{g ni}. when 18% ni is dissolved, the bath is completely solid with x = 109.8 \\text{g ni}." }, { "idx": 3799, "question": "A NiO-60 mol% MgO ceramic is allowed to solidify. Determine the composition of the first solid to form under equilibrium conditions.", "answer": "80% mgo" }, { "idx": 3800, "question": "A NiO-60 mol% MgO ceramic is allowed to solidify. Determine the composition of the last liquid to solidify under equilibrium conditions.", "answer": "35% mgo" }, { "idx": 3801, "question": "A Nb-35% W alloy is allowed to solidify. Determine the composition of the first solid to form under equilibrium conditions.", "answer": "55% w" }, { "idx": 3802, "question": "A Nb-35% W alloy is allowed to solidify. Determine the composition of the last liquid to solidify under equilibrium conditions.", "answer": "18% w" }, { "idx": 3803, "question": "An intermetallic compound is found for 10 wt% \\mathrm{Si} in the Cu-Si phase diagram. Determine the formula for the compound.", "answer": "\\mathrm{sicu}_{4}" }, { "idx": 3804, "question": "Consider a Pb-15% Sn alloy. During solidification, determine the composition of the first solid to form.", "answer": "8% Sn" }, { "idx": 3805, "question": "Consider a Pb-15% Sn alloy. During solidification, determine the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy.", "answer": "liquidus temperature =290 c, solidus temperature =240 c, solvus temperature =170 c, freezing range =50 c" }, { "idx": 3806, "question": "Consider a Pb-15% Sn alloy. During solidification, determine the amounts and compositions of each phase at 260 c.", "answer": "l: 30% sn, α: 12% sn; % l=17%, % α=83%" }, { "idx": 3807, "question": "Consider a Pb-15% Sn alloy. During solidification, determine the amounts and compositions of each phase at 183 c.", "answer": "α: 15% sn; 100% α" }, { "idx": 3808, "question": "Consider a Pb-15% Sn alloy. During solidification, determine the amounts and compositions of each phase at 25 c.", "answer": "α: 2% pb, β: 100% sn; % α=87%, % β=13%" }, { "idx": 3809, "question": "Consider a Pb-35% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.", "answer": "hypoeutectic" }, { "idx": 3810, "question": "Consider a Pb-35% Sn alloy. Determine the composition of the first solid to form during solidification.", "answer": "14% sn" }, { "idx": 3811, "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 184°C.", "answer": "α: 19% sn, l: 61.9% sn, % α = 63%, % l = 37%" }, { "idx": 3812, "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each base at 182°C.", "answer": "α: 19% sn, β: 97.5% sn, % α = 80%, % β = 20%" }, { "idx": 3813, "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each microconstituent at 182°C.", "answer": "primary α: 19% sn, % primary α = 63%, eutectic: 61.9% sn, % eutectic = 37%" }, { "idx": 3814, "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 25°C.", "answer": "α: 2% sn, β: 100% sn, % α = 66%, % β = 34%" }, { "idx": 3815, "question": "Consider a Pb-70% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.", "answer": "hypereutectic" }, { "idx": 3816, "question": "Consider a Pb-70% Sn alloy. Determine the composition of the first solid to form during solidification.", "answer": "98% sn" }, { "idx": 3817, "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 184 degrees C.", "answer": "beta: 97.5% sn, l: 61.9% sn, % beta = 22.8%, % l = 77.2%" }, { "idx": 3818, "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 182 degrees C.", "answer": "alpha: 19% sn, beta: 97.5% sn, % alpha = 35%, % beta = 65%" }, { "idx": 3819, "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each microconstituent at 182 degrees C.", "answer": "primary beta: 97.5% sn, % primary beta = 22.8%, eutectic: 61.9% sn, % eutectic = 77.2%" }, { "idx": 3820, "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 25 degrees C.", "answer": "alpha: 2% sn, beta: 100% sn, % alpha = 30%, % beta = 70%" }, { "idx": 3821, "question": "Consider an Al-4% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.", "answer": "hypo eutectic" }, { "idx": 3822, "question": "Consider an Al-4% Si alloy. Determine the composition of the first solid to form during solidification.", "answer": "1% si" }, { "idx": 3823, "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 578 degrees C.", "answer": "alpha: 1.65% si, l: 12.6% si, % alpha=78.5%, % l=21.5%" }, { "idx": 3824, "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 576 degrees C.", "answer": "alpha: 1.65% si, beta: 99.83% si, % alpha=97.6%, % beta=2.4%" }, { "idx": 3825, "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each microconstituent at 576 degrees C.", "answer": "primary alpha: 1.65% si, % primary alpha=78.5%, eutectic: 12.6% si, % eutectic=21.5%" }, { "idx": 3826, "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 25 degrees C.", "answer": "alpha: 0% si, beta: 100% si, % alpha=96%, % beta=4%" }, { "idx": 3827, "question": "Consider a Al-25% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.", "answer": "hyper eutectic" }, { "idx": 3828, "question": "Consider a Al-25% Si alloy. Determine the composition of the first solid to form during solidified.", "answer": "100% si" }, { "idx": 3829, "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 578 degrees C.", "answer": "β: 99.83% si, l: 12.6% si, %l: 85.8%, %β: 14.2%" }, { "idx": 3830, "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 576 degrees C.", "answer": "α: 1.65% si, β: 99.83% si, %α: 76.2%, %β: 23.8%" }, { "idx": 3831, "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each microconstituent at 576 degrees C.", "answer": "primary β: 99.83% si, % primary β: 14.2%; eutectic: 12.6% si, % eutectic: 85.8%" }, { "idx": 3832, "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 25 degrees C.", "answer": "α: 0% si, β: 100% si, %α: 75%, %β: 25%" }, { "idx": 3833, "question": "A Pb-Sn alloy contains 45% alpha and 55% beta at 100C. Determine the composition of the alloy.", "answer": "56.15% sn" }, { "idx": 3834, "question": "Is the alloy hypoeutectic or hypereutectic?", "answer": "hypoeutectic" }, { "idx": 3835, "question": "An Al-Si alloy contains 85 % α and 15 % β at 500°C. Determine the composition of the alloy.", "answer": "15.85% Si" }, { "idx": 3836, "question": "Is the alloy hypoeutectic or hypereutectic?", "answer": "hypereutectic" }, { "idx": 3837, "question": "A Pb-Sn alloy contains 23% primary \\alpha and 77% eutectic microconstituent. Determine the composition of the alloy.", "answer": "x = 52% sn" }, { "idx": 3838, "question": "An Al-Si alloy contains 15% primary \\beta and 85% eutectic microconstituent. Determine the composition of the alloy.", "answer": "25.71% si" }, { "idx": 3839, "question": "Observation of a microstructure shows that there is 28% eutectic and 72% primary β in an Al-Li alloy. Determine the composition of the alloy and whether it is hypoeutectic or hypereutectic.", "answer": "17.46% li, hypereutectic" }, { "idx": 3840, "question": "Observation of a microstructure shows that there is 28% eutectic and 72% primary β in an Al-Li alloy. How much α and β are in the eutectic microconstituent?", "answer": "% α_est = 64% and % β_est = 36%" }, { "idx": 3841, "question": "Calculate the total amount of alpha and beta in a Pb-50% Sn alloy at 182°C.", "answer": "total amount of alpha = 60.5%, beta = 39.5%." }, { "idx": 3842, "question": "Calculate the amount of each microconstituent in a Pb-50% Sn alloy at 182°C.", "answer": "microconstituents: primary alpha = 27.7%, eutectic = 72.3%." }, { "idx": 3843, "question": "What fraction of the total alpha in the alloy is contained in the eutectic microconstituent?", "answer": "fraction of total alpha in the eutectic microconstituent f = 0.54." }, { "idx": 3844, "question": "Recommend an artificial age-hardening heat treatment for a Cu-1.2% Be alloy (see Figure 12-34). Include appropriate temperatures.", "answer": "For the Cu-1.2% Be alloy, solution treat between 530°C and 870°C (typical at 780°C), quench, and age below 530°C (typical at 330°C)." }, { "idx": 3845, "question": "Compare the amount of the γ2 precipitate that forms by artificial aging at 400°C with the amount of the precipitate that forms by natural aging for a Cu-1.2% Be alloy.", "answer": "The amount of γ2 precipitate at 400°C is 5.4%, and at room temperature, it is 8.5%." }, { "idx": 3846, "question": "For an Fe-0.35% C alloy, determine the temperature at which austenite first begins to transform on cooling.", "answer": "795°C" }, { "idx": 3847, "question": "For an Fe-0.35% C alloy, determine the primary microconstituent that forms.", "answer": "primary α-ferrite" }, { "idx": 3848, "question": "For an Fe-0.35% C alloy, determine the composition and amount of each phase present at 728°C.", "answer": "α: 0.0218% C, 56.1% γ: 0.77% C, 43.9%" }, { "idx": 3849, "question": "For an Fe-0.35% C alloy, determine the composition and amount of each phase present at 726°C.", "answer": "α: 0.0218% C, 93.3% Fe3C: 6.67% C, 4.9%" }, { "idx": 3850, "question": "For an Fe-0.35% C alloy, determine the composition and amount of each microconstituent present at 726°C.", "answer": "primary α: 0.0218% C, 56.1% pearlite: 0.77% C, 43.9%" }, { "idx": 3851, "question": "For an Fe-1.15% C alloy, determine the temperature at which austenite first begins to transform on cooling.", "answer": "880 degrees c" }, { "idx": 3852, "question": "For an Fe-1.15% C alloy, determine the primary microconstituent that forms.", "answer": "primary Fe3C" }, { "idx": 3853, "question": "For an Fe-1.15% C alloy, determine the composition and amount of each phase present at 728 degrees C.", "answer": "Fe3C: 6.67% C, 6.4%; gamma: 0.77% C, 93.6%" }, { "idx": 3854, "question": "For an Fe-1.15% C alloy, determine the composition and amount of each phase present at 720 degrees C.", "answer": "alpha: 0.0218% C, 83%; Fe3C: 6.67% C, 17%" }, { "idx": 3855, "question": "For an Fe-1.15% C alloy, determine the composition and amount of each microconstituent present at 726 degrees C.", "answer": "primary Fe3C: 6.67% C, 6.4%; pearlite: 0.77% C, 93.6%" }, { "idx": 3856, "question": "A steel contains 8 % cementite and 92 % ferrite at room temperature. Estimate the carbon content of the steel.", "answer": "the carbon content of the steel is 0.53 % c." }, { "idx": 3857, "question": "Is the steel hypoeutectoid or hypereutectoid?", "answer": "the steel is hypoeutectoid." }, { "idx": 3858, "question": "A steel contains 18 % cementite and 82 % ferrite at room temperature. Estimate the carbon content of the steel.", "answer": "the carbon content of the steel is 1.20 % c." }, { "idx": 3859, "question": "Is the steel hypoeutectoid or hypereutectoid?", "answer": "the steel is hypereutectoid." }, { "idx": 3860, "question": "A steel contains 18 % pearlite and 82 % primary ferrite at room temperature. Estimate the carbon content of the steel.", "answer": "the carbon content of the steel is 0.156 %c." }, { "idx": 3861, "question": "Is the steel hypoeutectoid or hypereutectoid?", "answer": "the steel is hypoeutectoid." }, { "idx": 3862, "question": "A steel contains 94 % pearlite and 6 % primary cementite at room temperature. Estimate the carbon content of the steel.", "answer": "the carbon content of the steel is 1.124 % c." }, { "idx": 3863, "question": "Is the steel hypoeutectoid or hypereutectoid?", "answer": "the steel is hypereutectoid." }, { "idx": 3864, "question": "A steel contains 55% \\alpha and 45% \\gamma at 750^{\\circ} C. Estimate the carbon content of the steel.", "answer": "the carbon content of the steel is 0.281% C." }, { "idx": 3865, "question": "A steel contains 96% \\gamma and 4% \\mathrm{Fe}_{3} C at 800^{\\circ} C. Estimate the carbon content of the steel.\n\\[\n\\begin{array}{l}\n\\text {", "answer": "the carbon content of the steel is 1.15% C." }, { "idx": 3866, "question": "A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the temperature.", "answer": "the temperature is about 760 degrees c." }, { "idx": 3867, "question": "A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the overall carbon content of the steel.", "answer": "the overall carbon content of the steel is 0.212% c." }, { "idx": 3868, "question": "An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate the transformation temperature.", "answer": "transformation temperature =615 C" }, { "idx": 3869, "question": "An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate the interlamellar spacing in the pearlite.", "answer": "1 / lambda=60,000 or lambda=1.67 × 10^-5 cm" }, { "idx": 3870, "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 38", "answer": "600 degrees C" }, { "idx": 3871, "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 42", "answer": "400 degrees C" }, { "idx": 3872, "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 48", "answer": "340 degrees C" }, { "idx": 3873, "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 52", "answer": "300 degrees C" }, { "idx": 3874, "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 800^{\\circ} C for 1h, quenched to 350^{\\circ} C and held for 750s, and finally quenched to room temperature.", "answer": "HRC =47 and the microstructure is all baninite." }, { "idx": 3875, "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 700^{\\circ} C, quenched to 650^{\\circ} C and held for 500s, and finally quenched to room temperature.", "answer": "HRC =25 and the microstructure is all pearlite." }, { "idx": 3876, "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 300^{\\circ} C, quenched to 300^{\\circ} C and held for 10s, and finally quenched to room temperature.", "answer": "HRC=66 and the microstructure is all martensite." }, { "idx": 3877, "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 1000^{\\circ} C, quenched to 3010^{\\circ} C and held for 10 s, quenched to room temperature, and then reheated to 400^{\\circ} C before finally cooling to room temperature again.", "answer": "HRC=42 and the microstructure is all tempered martensite." }, { "idx": 3878, "question": "Describe the complete heat treatment required to produce a quenched and tempered eutectoid steel having a tensile strength of at least 125,000 psi. Include appropriate temperatures.", "answer": "Austenitize at approximately 750^{\\circ} C,\nQuench to below 130^{\\circ} C (the M_{\\gamma} temperature)\nTemper at 620^{\\circ} C or less." }, { "idx": 3879, "question": "What is the austenitizing temperature required to produce a quenched and tempered eutectoid steel with a HRC hardness of less than 50?", "answer": "Austenitize at approximately 75 degrees Celsius." }, { "idx": 3880, "question": "What is the quenching temperature required to produce a quenched and tempered eutectoid steel with a HRC hardness of less than 50?", "answer": "Quench to below the M_gamma (less than 130 degrees Celsius)." }, { "idx": 3881, "question": "What is the tempering temperature range required to produce a quenched and tempered eutectoid steel with a HRC hardness of less than 50?", "answer": "Temper at a temperature higher than 330 degrees Celsius, but less than 727 degrees Celsius." }, { "idx": 3882, "question": "Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of a material property? (a) Density (b) Annealing (c) Forging (d) Single-crystal (e) Crystalline", "answer": "(a) Density" }, { "idx": 3883, "question": "Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of material processing?(a) Extrusion(b) Crystalline(c) Amorphous(d) Glassy (e) Elastic Modulus", "answer": "(a)Extrusion" }, { "idx": 3884, "question": "Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of material structure? (a) Single-phase (b) Elastic Modulus (c) Sintering (d) Magnetic Permeability (e) Brittle", "answer": "(a)Single-phase" }, { "idx": 3885, "question": "Which class of material is generally associated with the highest density values at room temperature? (a) Composites (b) Ceramics (c) Metals (d) Polymers", "answer": "(c) Metals" }, { "idx": 3886, "question": "By how many orders of magnitude (powers of ten, approximately) does density vary for metals? (a) 0.13 (b) 1.3 (c) 13 (d) 130", "answer": "(b)1.3" }, { "idx": 3887, "question": "The atomic mass of an atom may be expressed as the sum of the masses of (a) Electrons (b) Neutrons (c) Protons (d) Choose all that apply.", "answer": "(b) Neutrons and (c) Protons" }, { "idx": 3888, "question": "The nucleus of an atom contains electrons?", "answer": "No, the nucleus of an atom does not contain electrons." }, { "idx": 3889, "question": "The nucleus of an atom contains neutrons?", "answer": "Yes, the nucleus of an atom contains neutrons." }, { "idx": 3890, "question": "The nucleus of an atom contains protons?", "answer": "Yes, the nucleus of an atom contains protons." }, { "idx": 3891, "question": "The atomic number of an electrically neutral atom is equal to the number of:(a) protons (b) electrons (c) neutrons (d)Choose all that apply.", "answer": "(b)electrons." }, { "idx": 3892, "question": "Hafnium has six naturally occurring isotopes: 0.16% of { }^{174} Hf, with an atomic weight of 173.940 amu; 5.26% of { }^{176} Hf, with an atomic weight of 175.941 amu ; 18.60% of { }^{177} Hf, with an atomic weight of 176.943 amu ; 27.28% of { }^{178} Hf, with an atomic weight of 177.944 amu ; 13.62% of { }^{179} Hf, with an atomic weight of 178.946 amu;. and 35.08% of { }^{180} Hf, with an atomic weight of 179.947 amu. Calculate the average atomic weight of Hf. Give your answer to three decimal places.", "answer": "the average atomic weight of hf is 178.485 amu." }, { "idx": 3893, "question": "Bromium has two naturally occurring isotopes: { }^{79} Br, with an atomic weight of 78.918 amu, and { }^{81} Br, with an atomic weight of 80.916 amu. If the average atomic weight for Br is 79.903 amu, calculate the fraction-of-occurrences of these two isotopes.Give your answer to three decimal places.", "answer": "the fraction-of-occurrence of { }^{79}\\mathrm{br} is 0.507, and the fraction-of-occurrence of { }^{81}\\mathrm{br} is 0.493." }, { "idx": 3894, "question": "An element that has the electron configuration 1 s^{2} 2 s^{2} 2 p^{6} has how many electrons? Enter numeric values only.", "answer": "9" }, { "idx": 3895, "question": "The electrons that occupy the outermost filled shell are called electrons.", "answer": "The electrons that occupy the outermost filled shell are called valence electrons." }, { "idx": 3896, "question": "When all the electrons in an atom occupy the lowest possible energy states, the atom is said to be in its: (a)ground state(b)ionized state (c) cold state (d) regular state", "answer": "(a)ground state" }, { "idx": 3897, "question": "How many p electrons at the outermost orbital do the Group VIIA elements have?", "answer": "5 p electrons." }, { "idx": 3898, "question": "To what group in the periodic table would an element with atomic number 119 belong?(a) Group 0 (or 18) (b) Group IA (or 1) (c) Group IIA (or 2) (d) Group VIIA (or 17)", "answer": "(b) Group IA (or 1)" }, { "idx": 3899, "question": "Calculate the energy of attraction between a cation with a valence of +2 and an anion with a valence of -2 , the centers of which are separated by a distance of 3.7nm.", "answer": "-2.5 × 10^{-19} J" }, { "idx": 3900, "question": "Materials whose constituent particles are bound by which type of bond are generally expected to have the lowest melting temperatures?(a) Covalent(b) Metallic(c) Ionic(d) Van der Waals(e) Hydrogen", "answer": "(d) Van der Waals" }, { "idx": 3901, "question": "Calculate % IC of the interatomic bonds for the intermetallic compound TiAl3.", "answer": "0%" }, { "idx": 3902, "question": "On the basis of this result what type of interatomic bonding would you expect to be found in TiAl3 ? (a) van der Waals (b) ionic (c) metallic (d) covalent", "answer": " (c)metallic." }, { "idx": 3903, "question": "Which of the following microstructures is expected to be most similar to a single crystal in terms of structure and properties? Assume all of the options offer the same volumes and only consider grain boundaries as a crystalline defect for this question. (a) Textured polycrystal with about 10,000 grains (b) Random polycrystal with about 1,000,000 grains (c) Random polycrystal with about 1,000,000,000 grains (d) Amorphous", "answer": "(a) Textured polycrystal with about 10,000 grains" }, { "idx": 3904, "question": "For a metal that has the simple cubic crystal structure, calculate the atomic radius if the metal has a density of 2.05g / {cm}^{3} and an atomic weight of 77.84g / mol.", "answer": "the atomic radius is 0.122nm." }, { "idx": 3905, "question": "Some metal is known to have a cubic unit cell with an edge length of 0.437 nm. In addition, it has a density of 4.37 g/cm3 and an atomic weight of 54.85 g/mol. Determine the crystal structure of the metal.", "answer": "The metal has an FCC crystal structure." }, { "idx": 3906, "question": "Some metal is known to have a cubic unit cell with an edge length of 0.437 nm. In addition, it has a density of 4.37 g/cm3 and an atomic weight of 54.85 g/mol. Determine the atomic radius of the metal.", "answer": "The metal has an atomic radius of 0.155 nm." }, { "idx": 3907, "question": "Some metal is known to have a cubic unit cell with an edge length of 0.437 nm. In addition, it has a density of 4.37 g/cm3 and an atomic weight of 54.85 g/mol. Indicate the letter of the metal listed in the following table that has these characteristics.", "answer": "The metal with these characteristics is metal C, which has an atomic radius of 0.155 nm and an FCC crystal structure." }, { "idx": 3908, "question": "If the atomic radius of a metal that has the body-centered cubic crystal structure is 0.181nm, calculate the volume of its unit cell.", "answer": "the volume of the unit cell is 0.0271 nm^3." }, { "idx": 3909, "question": "If the atomic radius of a metal that has the face-centered cubic crystal structure is 0.123nm, calculate the volume of its unit cell.", "answer": "\\[\nv_{c}=0.258nm^{3}\n\\]" }, { "idx": 3910, "question": "Rhodium has an atomic radius of 0.1345nm, a density of (12.41g / {cm}^{3}) and an atomic weight of 102.91 \\mathrm{g} / mol. What is rhodium's crystal structure?(a) Simple cubic(b) \\mathrm{BCC}(c) \\mathrm{FCC}", "answer": "the crystal structure for rhodium is \\mathrm{fcc}." }, { "idx": 3911, "question": "Niobium (Nb) has a BCC crystal structure, an atomic radius of 0.143nm and an atomic weight of 92.91g / mol. Calculate the theoretical density for nb.", "answer": "the theoretical density for nb is 8.48g / {cm}^{3}. the experimental density for nb is 8.57g / {cm}^{3}." }, { "idx": 3912, "question": "Consider the ideal barium titanate \\left(\\mathrm{BaTiO}_{3}\\right) structure. What is the coordination number of the \\mathrm{Ti}^{4+} ion in terms of surrounding \\mathrm{O}^{2-} ions?\n[a] 1\n[b] 2\n[c] 3\n[d] 4\n[e] 5\n[f] 6\n[g] 7\n[h] 8", "answer": "According to Fig. 3.10, the central \\mathrm{Ti}^{4+} ion is surrounded by six \\mathrm{O}^{2-} ions, one residing at each of the six cube faces." }, { "idx": 3913, "question": "Consider the ideal barium titanate \\left(\\mathrm{BaTiO}_{3}\\right) structure. What is the coordination number of the \\mathrm{Ba}^{2+} ion in terms of surrounding \\mathrm{Ti}^{4+} ions?\n[a] 4\n[b] 6\n[c] 8\n[d] 10\n[e] 12", "answer": "Careful consideration of Fig. 3.10, and consideration of the perovskite compound stoichiometry reveals that 12 \\mathrm{Ti}^{4+} ions surround each \\mathrm{Ba}^{2+} ion. A change in perspective is recommended to help visualize this, where the unit cell cube corners feature \\mathrm{O}^{2-} ions in the corners, \\mathrm{Ti}^{4+} ions centered along each cube edge, and \\mathrm{Ba}^{2+} ions centered in the middle of the cube." }, { "idx": 3914, "question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?\n\\[\na=b=c\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic", "answer": "For both cubic and rhombohedral crystal systems all the unit cell edge lengths are equal." }, { "idx": 3915, "question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?\n\\[\na=b \\neq c\n\\](a) Cubic(b) Hexagonal(c) Tetragonal\n\\mathrm{d}) Rhombohedral(e) Orthorhombic(f) Monoclinic (g) Triclinic", "answer": "For both hexagonal and tetragonal crystal systems two of the unit cell edge lengths are equal to one another, but unequal to the third length." }, { "idx": 3916, "question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell e edge lengths?a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic", "answer": "For orthorhombic, monoclinic and triclinic crystal systems, all three of the unit cell edge lengths are unequal to one another." }, { "idx": 3917, "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?\n\\[\n\\alpha \\neq \\beta \\neq \\gamma \\neq 90^{\\circ}\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic", "answer": "Triclinic is the only crystal system for which none of the interaxial angles are equal to one another and also not equal to 90^{\\circ}." }, { "idx": 3918, "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?(a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic\ng) Triclinic", "answer": "Cubic, tetragonal and orthorhombic crystal systems all have the three interaxial angles equal to 90^{\\circ}." }, { "idx": 3919, "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?\n\\[\n\\alpha=\\beta=90^{\\circ}, \\gamma=120^{\\circ}\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic", "answer": "Only the hexagonal crystal system has two of the interaxial angles equal to 90^{\\circ}, while the third angle is equal to 120^{\\circ}." }, { "idx": 3920, "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship? \\[\n\\alpha=\\beta=\\gamma \\neq 90^{\\circ}\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f)\nMonoclinic(g) Triclinic", "answer": "Only the rhombohedral crystal system has all three interaxial angles equal to one other, but not equal to 90^{\\circ}." }, { "idx": 3921, "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic", "answer": "Only the monoclinic crystal system has two of the interaxial angles equal to one another and to 90^{\\circ}, which is different than the third interaxial angle." }, { "idx": 3922, "question": "Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen. For this polymer, determine the concentration of F (in wt %) that must be added if this substitution occurs for 18.6 % of all of the original hydrogen atoms. Atomic weights for several elements are included in the following table: Carbon 12.01 g/mol, Chlorine 35.45 g/mol, Fluorine 19.00 g/mol, Hydrogen 1.008 g/mol, Oxygen 16.00 g/mol.", "answer": "the concentration of F that must be added is 18.0 wt %." }, { "idx": 3923, "question": "Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen. For this polymer, determine the concentration of F (in wt %) that must be added to completely fluorinate the material, i.e. to produce polytetrafluoroethylene (PTFE). Atomic weights for several elements are included in the following table: Carbon 12.01 g/mol, Chlorine 35.45 g/mol, Fluorine 19.00 g/mol, Hydrogen 1.008 g/mol, Oxygen 16.00 g/mol.", "answer": "the concentration of F required to completely fluorinate the material is 76.0 wt %." }, { "idx": 3924, "question": "Which of the following may form linear polymers?(a) Rubber(b) Epoxy(c) Polyethylene(d) Phenol-formaldehyde(e) Polystyrene(f) Nylon", "answer": "Polyethylene, polystyrene and nylon may form linear polymers." }, { "idx": 3925, "question": "Which of the following form network polymers?(a) Rubber(b) Epoxy(c) Polyethylene\nd) Phenol-formaldehyde(e) Polystyrene(f) Nylon\n\\", "answer": "Epoxy and phenol-formaldehyde form network polymers." }, { "idx": 3926, "question": "For most polymers, which configuration predominates?(a) Head-to-head(b) Head-to-tail", "answer": "For most polymers, the head-to-tail configuration predominates." }, { "idx": 3927, "question": "Is it possible to produce a polymer that is 100% crystalline? (a) True (b) False", "answer": "False. It is not possible to produce a polymer that is 100% crystalline. The maximum crystallinity that can be obtained is about 95%, with the remaining material being amorphous." }, { "idx": 3928, "question": "The number of vacancies in some hypothetical metal increases by a factor of 5 when the temperature is increased from 1040 K to 1150 K. Calculate the energy (in kj/mol ) for vacancy formation assuming that the density of the metal remains the same over this temperature range.", "answer": "the energy for vacancy formation is 115 kj/mol." }, { "idx": 3929, "question": "The number of vacancies present in some metal at 864^{\\circ} C is 1.1 × 10^{24}{m}^{-3}. Calculate the number of vacancies at 463^{\\circ} C given that the energy for vacancy formation is 1.25 \\mathrm{eV} / atom; assume that the density at both temperatures is the same", "answer": "the number of vacancies at 463^{\\circ} C is 5.45 × 10^{22}{m}^{-3}." }, { "idx": 3930, "question": "In metals, there are significantly more vacancies than self-interstitials.(a) True(b) False", "answer": "True. In metals, there are significantly more vacancies than self-interstitials; the reason for this is that the atom is significantly larger than the interstitial position in which it is situated, and, consequently significant lattice strains result." }, { "idx": 3931, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for MgO.", "answer": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility." }, { "idx": 3932, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for CaO.", "answer": "For CaO, the ionic radii of the Mn2+ and Ca2+ are 0.067 nm and 0.100 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.100 nm - 0.067 nm) / 0.100 nm × 100 = 33%. This Δr% value is much larger than the ±15% range, and, therefore, CaO is not expected to experience any appreciable solubility in MnO." }, { "idx": 3933, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for BeO.", "answer": "For BeO, the ionic radii of the Mn2+ and Be2+ are 0.067 nm and 0.035 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.067 nm - 0.035 nm) / 0.067 nm × 100 = 48%. This Δr% value is much larger than the ±15% range, and, therefore, BeO is not expected to experience any appreciable solubility in MnO." }, { "idx": 3934, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for NiO.", "answer": "For NiO, the ionic radii of the Mn2+ and Ni2+ are 0.067 nm and 0.069 nm, respectively. Therefore, the percentage difference in ionic radii Δr% is determined as follows: Δr% = (0.069 nm - 0.067 nm) / 0.069 nm × 100 = 3%, which value is, of course, within the acceptable range for a high degree of solubility." }, { "idx": 3935, "question": "Which of these elements would you expect to form a substitutional solid solution having complete solubility with copper? The elements are: Ni, O, H, Pt, Pd, Co, C, Zn, Ag, Al, Cr, Fe. The criteria for complete solubility are: 1) the difference in atomic radii between Cu and the other element must be less than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same.", "answer": "Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures." }, { "idx": 3936, "question": "Which of these elements would you expect to form a substitutional solid solution of incomplete solubility with copper? The elements are: Pd, Al, Cr, Fe, H, Pt, Ni, Zn, C, Ag, O, Co. The criteria for incomplete solubility are: these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+.", "answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+." }, { "idx": 3937, "question": "Which of these elements would you expect to form an interstitial solid solution with copper? The elements are: C, Pd, Pt, H, Cr, O, Zn, Ag, Al, Co, Fe, Ni. The criteria for interstitial solid solutions are: these elements have atomic radii that are significantly smaller than the atomic radius of Cu.", "answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu." }, { "idx": 3938, "question": "The concentration of carbon in an iron-carbon alloy is 0.57 wt%. What is the concentration in kilograms of carbon per cubic meter of alloy? The densities of iron and carbon are 7.87 and 2.25 \\mathrm{g} / {cm}^{3}, respectively.", "answer": "the concentration of carbon is 11.8 kg / m^{3}." }, { "idx": 3939, "question": "13}\nSome hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and \\mathrm{B} are 4.27 and 6.35g / {cm}^{3}, respectively, whereas their respective atomic weights are 61.4 and 125.7g / mol, determine whether the crystal structure for this alloy is simple cubic, facercentered cubic, or body-centered cubic. Assume a unit cell edge length of 0.395nm.", "answer": "the crystal structure is body-centered cubic." }, { "idx": 3940, "question": "17}\nIron and vanadium both have the BCC crystal structure andV forms a substitutional solid solution in \\mathrm{Fe} for concentrations up to approximately 20 wt%V at room temperature. Determine the concentration in weight percent ofV that must be added to iron to yield a unit cell edge length of 0.289nm.", "answer": "the concentration of \\mathrm{v} that must be added to iron to yield a unit cell edge length of 0.289nm is 12.9 wt%." }, { "idx": 3941, "question": "Which of the following is a (are) linear defect(s)?(a) An edge dislocation(b) A Frenkel defect(c) A Schottky defect", "answer": "Edge and screw dislocations are linear defects." }, { "idx": 3942, "question": "A photomicrograph was taken of a specimen at a magnification of 100 ×, and it was determined that the average number of grains per square inch was 200 . What is this specimen's ASTM grain size number?", "answer": "the specimen's astm grain size number is 4.9." }, { "idx": 3943, "question": "Diffusion by which mechanism occurs more rapidly in metal alloys?(a) Vacancy diffusion(b) Interstitial diffusion", "answer": "In metal alloys, interstitial diffusion takes place more rapidly than vacancy diffusion because the interstitial atoms are smaller and are more mobile. Also, there are more vacant adjacent interstitial sites than there are vacancies." }, { "idx": 3944, "question": "As temperature decreases, the fraction of total number of atoms that are capable of diffusive motion(a) increases.(b) decreases.", "answer": "As temperature decreases, the fraction of the total number of atoms that are capable of diffusive motion decreases." }, { "idx": 3945, "question": "If [m] atoms of helium pass through a [a] square meter plate area every [t] hours, and if this flux is constant with time, compute the flux of helium in units of atoms per square meter per second.", "answer": "\\[\nj = \\frac{[m]}{[a] × [t] × 3600} \\text{ atoms per square meter per second}\n\\]" }, { "idx": 3946, "question": "If water molecules pass through a membrane with a steady state flux of [j] mole /\\left(m^{2}\\right. day ), how long will it take, in hours, for [m] kg of water to pass through a [\\mathrm{a}] square centimeter of the membrane?", "answer": "\\[\nt = \\frac{[m] × 1000 × 24}{[a] × [j] × 18 × 10000} \\text{ hours}\n\\]" }, { "idx": 3947, "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the contact lens thickness likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.", "answer": "If the contact lens thickness is increased and all other factors are preserved, the concentration gradient decrease and the oxygen diffusion flux should decrease." }, { "idx": 3948, "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the diffusivity of oxygen gas by decreasing the contact lens porosity likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.", "answer": "If the lens features more voids, we expect the oxygen to permeate the membrane faster since oxygen should be able to transit voids faster than the bulk lens material." }, { "idx": 3949, "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the ambient temperature likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.", "answer": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate." }, { "idx": 3950, "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the ambient partial pressure of oxygen gas likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.", "answer": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen." }, { "idx": 3951, "question": "For a steel alloy it has been determined that a carburizing heat treatment of 16h duration at 757^{\\circ} C will raise the carbon concentration to 0.5 wt% at a point 2.3mm from the surface. Estimate the time necessary to achieve the same concentration at a 8mm position for an identical steel and at a carburizing temperature of 1130^{\\circ} C. Assume that D_{0} is 4.6 × 10^{-5}{m}^{2} / s and Q_{d} is 104kJ / mol.", "answer": "the time necessary to achieve the same concentration at a 8mm position for an identical steel and at a carburizing temperature of 1130^{\\circ} C is 17.9h." }, { "idx": 3952, "question": "The diffusion coefficient for aluminum in silicon is D_{\\mathrm{Al}} in_{\\mathrm{Si}}=3 × 10^{-16} cm^{2} / s at 300 K (note that 300 K is about room temperature).\nWhat is a reasonable value for D_{\\mathrm{Al} \\text { in } \\mathrm{Si}} at 600 K ?\nNote: Rather than performing a specific calculation, you should be able to justify your answer from the options below based on the mathematical temperature dependence of the diffusion coefficient.\n(a) D<3 × 10^{16} cm^{2} / s\n(b) D=3 × 10^{16} cm^{2} / s\n(c) D=6 × 10^{16} cm^{2} / s\n(d) D=1.5 × 10^{16} cm^{2} /{s}\n(e) D>6 × 10^{16} cm^{2} /\n(f) D=6 × 10^{-17} cm^{2} / s", "answer": "We expect the diffusion coefficient to increase if the temperature of this system is increased. Therefore, options (a), (b), (d), and (f) are eliminated.\nFurthermore, we expect that since the diffusion coefficient is exponentially dependent on temperature, the diffusivity should increase by more than a factor of two if the absolute temperature is doubled." }, { "idx": 3953, "question": "A specimen of some metal having a rectangular cross section 11.2mm × 12.4mm is pulled in tension with a force of 31200N, which produces only elastic deformation. Given that the elastic modulus of this metal is 63 \\mathrm{GPa}, calculate the resulting strain.", "answer": "the resulting strain is 2.92 × 10^{-3}." }, { "idx": 3954, "question": "For a bronze alloy, the stress at which plastic deformation begins is 277 MPa and the modulus of elasticity is 117 GPa. What is the maximum load that may be applied to a specimen having a cross-sectional area of 327 mm^2 without plastic deformation?", "answer": "the maximum load that may be applied without plastic deformation is 86,700 N." }, { "idx": 3955, "question": "For a bronze alloy, the stress at which plastic deformation begins is 277 MPa and the modulus of elasticity is 117 GPa. If the original specimen length is 148 mm, what is the maximum length to which it may be stretched without causing plastic deformation?", "answer": "the maximum length to which the specimen may be stretched without causing plastic deformation is 131.30 mm." }, { "idx": 3956, "question": "A steel bar 100mm (4.0 in.) long and having a square cross section 20mm (0.8 in.) on an edge is pulled in tension with a load of 89,000N\\left(20,000 lb_{0}\\right), and experiences an elongation of 0.10mm (4.0 × 10^{-3} in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.", "answer": "the elastic modulus of the steel is 223 \\text{ gpa} (31.3 × 10^6 \\text{ psi})." }, { "idx": 3957, "question": "13: Typical relationship between E and G}\nFor most metals, the relationship between elastic and shear moduli is approximately which of the following?(a) G=0.1 E(b) G=0.2 E(c) G=0.3 E(d) G=0.4 E(e) G=0.5 E", "answer": "the relationship between elastic and shear moduli for most metals is approximately g=0.4 e." }, { "idx": 3958, "question": "A cylindrical rod 380mm (15.0 in.) long and having a diameter of 10.0mm (0.40 in.), is to be subjected to a tensile load of 24,500 N (5500 lbf). Which of the four metals or alloys listed below will not experience plastic deformation under this load? The materials are: Aluminum alloy (Yield Strength: 255 MPa), Brass alloy (Yield Strength: 345 MPa), Copper (Yield Strength: 250 MPa), Steel alloy (Yield Strength: 450 MPa).", "answer": "The materials that will not experience plastic deformation under the load of 24,500 N are the Brass alloy and Steel alloy, as their yield strengths (345 MPa and 450 MPa, respectively) are greater than the applied stress." }, { "idx": 3959, "question": "A cylindrical rod 380mm (15.0 in.) long and having a diameter of 10.0mm (0.40 in.), is to be subjected to a tensile load of 24,500 N (5500 lbf). Which of the four metals or alloys listed below will experience an elongation of no more than 0.9mm (0.035 in.) under this load? The materials are: Aluminum alloy (Modulus of Elasticity: 70 GPa), Brass alloy (Modulus of Elasticity: 100 GPa), Copper (Modulus of Elasticity: 110 GPa), Steel alloy (Modulus of Elasticity: 207 GPa).", "answer": "The materials that will experience an elongation of no more than 0.9mm under the load of 24,500 N are the Brass alloy, Copper, and Steel alloy, as their calculated elongations (0.57 mm, 0.52 mm, and 0.57 mm, respectively) are less than or equal to 0.9 mm." }, { "idx": 3960, "question": "Based on the previous two sub-questions, which of the four metals or alloys listed below are possible candidates for the cylindrical rod that will neither experience plastic deformation nor an elongation of more than 0.9mm under a tensile load of 24,500 N?", "answer": "The only candidate material that satisfies both conditions is the Steel alloy, with an elongation of 0.57 mm." }, { "idx": 3961, "question": "A tensile test is performed on a specimen of some metal alloy, and it is found that a true plastic strain of 0.12 is produced when a true stress of 280 MPa is applied. For this alloy, the value of the strain hardening exponent is 0.3 . On the basis of these data what true plastic strain would be expected for a total true plastic stress of 330 MPa ?", "answer": "the true plastic strain expected for a total true plastic stress of 330 \\text{mpa} is 0.383." }, { "idx": 3962, "question": "A cylindrical specimen of a metal alloy 48.8mm long and 9.09mm in diameter is stressed in tension. A true stress of 327 MPa causes the specimen to plastically elongate to a length of 55 mm. If it is known that the strain-hardening exponent for this alloy is 0.3 , calculate the true stress (in MPa )necessary to plastically elongate a specimen of this same material from a length of 48.8mm to a length of 57.6 mm.", "answer": "390 MPa" }, { "idx": 3963, "question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5mm (0.10 in.) and the support point separation distance is 30mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied.", "answer": "Since the flexural strength for this test is 379 MPa (53,500 psi), which is less than the reported flexural strength of 390 MPa, fracture is not predicted. However, there is some chance of fracture due to variability in material strength." }, { "idx": 3964, "question": "Calculate the value of flexural strength for this test given a cylindrical specimen of aluminum oxide with a radius of 2.5mm (0.10 in.) and a support point separation distance of 30mm (1.2 in.) when a load of 620 N (140 lbf) is applied.", "answer": "The flexural strength for this test is 379 MPa (53,500 psi)." }, { "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{cc} \\hline \\sigma_{\\mathrm{fs}} (MPa) & \\mathbf{P} \\ \\hline 100 & 0.05 \\ 50 & 0.20 \\ \\hline \\end{tabular} (a) Compute the flexural strength for a completely nonporous specimen of this material.", "answer": "the flexural strength for a completely nonporous specimen of this material is 126 mpa.", "idx": 3965 }, { "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{cc} \\hline \\sigma_{\\mathrm{fs}} (MPa) & \\mathbf{P} \\ \\hline 100 & 0.05 \\ 50 & 0.20 \\ \\hline \\end{tabular} (b) Compute the flexural strength for a 0.10 volume fraction porosity.", "answer": "the flexural strength for a 0.10 volume fraction porosity is 79.4 mpa.", "idx": 3966 }, { "idx": 3967, "question": "Mechanical twinning occurs in metals having which type(s) of crystal structure(s)?(a) \\mathrm{BCC}(b) \\mathrm{FCC}(c) \\mathrm{HCP}", "answer": "Mechanical twinning occurs in metals having B C C and H C P crystal structures." }, { "idx": 3968, "question": "How does grain size influence strength of a polycrystalline material?(a) Strengthfine-grained < strength _{\\text {course-grained }}(b) Strength _{\\text {fine-grained }}= strength _{\\text {course-grained }}(c) Strengthfine-grained > strength _{\\text {course-grained }}", "answer": "A fine-grained material is stronger than a coarse-grained material." }, { "idx": 3969, "question": "Reducing the grain size of metal improves toughness.(a) True(b) False", "answer": "True. Reducing the grain size of a metal improves its toughness." }, { "idx": 3970, "question": "As dislocation density increases, the resistance to dislocation movement(a) increases.(b) decreases.", "answer": "As dislocation density increases, the resistance to dislocation movement increases. This phenomenon is responsible for cold working." }, { "idx": 3971, "question": "On the average, dislocation-dislocation strain interactions are(a) repulsive.(b) attractive.", "answer": "On the average, dislocation-dislocation strain interactions are repulsive." }, { "idx": 3972, "question": "As a metal is strain hardened, its ductility(a) increases(b) decreases", "answer": "As a metal is strain hardened, its ductility decreases." }, { "idx": 3973, "question": "Most metals strain harden at room temperature.(a) True(b) False", "answer": "True. Most metals strain harden at room temperature." }, { "idx": 3974, "question": "During the recovery of a cold-worked material, is some of the internal strain energy relieved?", "answer": "Some of the internal strain energy is relieved." }, { "idx": 3975, "question": "During the recovery of a cold-worked material, is all of the internal strain energy relieved?", "answer": "All of the internal strain energy is not relieved." }, { "idx": 3976, "question": "During the recovery of a cold-worked material, is there some reduction in the number of dislocations?", "answer": "There is some reduction in the number of dislocations." }, { "idx": 3977, "question": "During the recovery of a cold-worked material, is there a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state?", "answer": "There is not a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state." }, { "idx": 3978, "question": "During the recovery of a cold-worked material, is the electrical conductivity recovered to its precold-worked state?", "answer": "The electrical conductivity is recovered to its precold-worked state." }, { "idx": 3979, "question": "During the recovery of a cold-worked material, is the thermal conductivity recovered to its precold-worked state?", "answer": "The thermal conductivity is recovered to its precold-worked state." }, { "idx": 3980, "question": "During the recovery of a cold-worked material, does the metal become more ductile, as in its precold-worked state?", "answer": "The metal does not become more ductile, as in its precold-worked state." }, { "idx": 3981, "question": "During the recovery of a cold-worked material, are grains with high strains replaced with new, unstrained grains?", "answer": "Grains with high strains are not replaced with new, unstrained grains." }, { "idx": 3982, "question": "During the recrystallization of a cold-worked material, is some of the internal strain energy relieved?", "answer": "All of the internal strain energy is relieved." }, { "idx": 3983, "question": "During the recrystallization of a cold-worked material, is there some reduction in the number of dislocations?", "answer": "There is significant reduction in the number of dislocations." }, { "idx": 3984, "question": "During the recrystallization of a cold-worked material, does the metal become more ductile, as in its precold-worked state?", "answer": "The metal becomes more ductile, as in its precold-worked state." }, { "idx": 3985, "question": "During the recrystallization of a cold-worked material, are grains with high strains replaced with new, unstrained grains?", "answer": "Grains with high strains are replaced with new, unstrained grains." }, { "idx": 3986, "question": "Grain growth requirements\n}\nGrain growth must always be preceded by recovery and recrystallization.(a) True(b) False", "answer": "False. Grain growth does not always need to be preceded by recovery and recrystallization; it may occur in materials that have not been cold worked." }, { "idx": 3987, "question": "A hypothetical metal alloy has a grain diameter of 2.4 × 10^{-2} mm. After a heat treatment at 575^{\\circ} C for 500min, the grain diameter has increased to 7.3 × 10^{-2} mm. Compute the time required for a specimen of this same material (i.e., \\mathrm{d}_{0}=2.4 × 10^{-2}mm ) to achieve a grain diameter of 5.5 × 10^{-2}mm while being heated at 575^{\\circ} C. Assume the N grain diameter exponent has a value of 2.2.", "answer": "the time required for the specimen to achieve a grain diameter of 5.5 × 10^{-2} mm is 246min." }, { "idx": 3988, "question": "Tensile strengths and number-average molecular weights for two polymers are as follows:\nTensile strength Number average molecular weight\n(MPa)\n(\\mathrm{g} / mol)\n37.7\n36800\n131\n62400\nEstimate the tensile strength (in MPa) for a number-average molecular weight of 51500g / mol.", "answer": "the tensile strength for a number-average molecular weight of 51500g/mol is 95.2 MPa." }, { "idx": 3989, "question": "Tensile strengths and number-average molecular weights for two polymers are as follows\n\\[\n\\begin{array}{l}\n\\text { Tensile strength Number average molecular weight } \\\\\n(MPa) \\\\\n138 \\\\\n184 \\\\\n\\text { (g/mol) } \\\\\n12600 \\\\\n28100\n\\end{array}\n\\]\nEstimate number average molecular weight (in \\mathrm{g} / mol ) at a tensile strength of 141 MPa.", "answer": "the estimated number average molecular weight at a tensile strength of 141 \\, MPa is 14600 \\, \\mathrm{g/mol}." }, { "idx": 3990, "question": "The bonding forces between adhesive and adherend surfaces are thought to be(a) Electrostatic(b) Covalent(c) Chemical", "answer": "The bonding forces between adhesive and adherend surfaces are thought to be electrostatic." }, { "idx": 3991, "question": "Deformation of a semicrystalline polymer by drawing produces which of the following in the direction of drawing?", "answer": "Increase in strength in the direction of drawing." }, { "idx": 3992, "question": "Deformation of a semicrystalline polymer by drawing produces which of the following perpendicular to the direction of drawing?", "answer": "Decrease in strength perpendicular to the direction of drawing." }, { "idx": 3993, "question": "How does deformation by drawing of a semicrystalline polymer affect its tensile strength?(a) Increases(b) Decreases", "answer": "Deformation by drawing increases the tensile strength of a semicrystalline polymer. This effect is due to the highly oriented chain structure that is produced by drawing, which gives rise to higher interchain secondary bonding forces." }, { "idx": 3994, "question": "How does increasing the degree of crystallinity of a semicrystalline polymer affect its tensile strength?(a) Increases\n\\mathrm{b}) Decreases", "answer": "Increasing the degree of crystallinity of a semicrystalline polymer leads to an increase in its tensile strength. This is due to enhanced interchain bonding and forces in crystalline regions; in response to applied stresses, interchain motions become more restrained as degree of crystallinity increases." }, { "idx": 3995, "question": "How does increasing the molecular weight of a semicrysatlline polymer affect its tensile strength?(a) Increases(b) Decreases\n\\", "answer": "The tensile strength of a semicrystalline polymer increases with increasing molecular weight. This effect is explained by the increased chain entanglements at higher molecular weights." }, { "idx": 3996, "question": "Which kind of fracture (ductile or brittle) is associated with intergranular crack propagation?", "answer": "Intergranular fracture is brittle." }, { "idx": 3997, "question": "Which kind of fracture (ductile or brittle) is associated with transgranular crack propagation?", "answer": "Transgranular fracture is brittle." }, { "idx": 3998, "question": "The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 7 when 28% of the crack length is removed.", "answer": "the ratio of the etched and original crack tip radii is: \\[\n\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}} = 32\n\\]" }, { "idx": 3999, "question": "A structural component in the shape of a flat plate 27.3mm thick is to be fabricated from a metal alloy for which the yield strength and plane strain fracture toughness values are 535 MPa and 31.3 MPa·m^{-1 / 2}, respectively. For this particular geometry, the value of Y is 1.8 . Assuming a design stress of 0.5 times the yield strength, calculate the critical length of a surface flaw.", "answer": "the critical length of a surface flaw is 2.0 mm." }, { "idx": 4000, "question": "How would the plane strain fracture toughness of a metal be expected to change with rising temperature?(a) Increase(b) Decrease(c) Remain constant", "answer": "The magnitude of the plane strain fracture toughness, K_{i c}, diminishes with decreasing temperature; therefore as temperature increases, K_{i c}, should also increase." }, { "idx": 4001, "question": "Does increasing temperature favor brittle fracture in polymers?", "answer": "No, increasing temperature does not favor brittle fracture in polymers." }, { "idx": 4002, "question": "Does increasing strain rate favor brittle fracture in polymers?", "answer": "Yes, increasing strain rate favors brittle fracture in polymers." }, { "idx": 4003, "question": "Does the presence of a sharp notch favor brittle fracture in polymers?", "answer": "Yes, the presence of a sharp notch favors brittle fracture in polymers." }, { "idx": 4004, "question": "Does decreasing specimen thickness favor brittle fracture in polymers?", "answer": "No, decreasing specimen thickness does not favor brittle fracture in polymers." }, { "idx": 4005, "question": "A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa·{m}^{1 / 2}. If it is known that the largest surface crack is 0.5mm long, and that the value of Y is 1.1 , which of the following can be said about this plate when a tensile stress of 1200 MPa is applied?(a) The plate will definitely fracture.(b) The plate will definitely not fracture. c) It is not possible to determine whether or not the plate will fracture.", "answer": "the plate will definitely fracture." }, { "idx": 4006, "question": "The effect of a stress raiser is more significant for which of the following types of materials?(a) Brittle materials(b) Ductile materials", "answer": "The effect of a stress raiser is more significant for brittle materials. For a ductile material, plastic deformation ensues when the maximum stress exceeds the yield strength. This leads to a more uniform distribution of stress in the vicinity of the stress raiser and to the development of a maximum stress concentration factor less than the theoretical value. Such yielding and stress redistribution do not occur to any appreciable extent around flaws and discontinuities in brittle materials; therefore, essentially the theoretical stress concentration will result." }, { "idx": 4007, "question": "A structural component is fabricated from an alloy that has a plane strain fracture toughness of 45 MPa. It has been determined that this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm ) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane strain fracture toughness of 100.0 MPa ? The geometry factor Y is the same in both cases.", "answer": "the maximum allowable surface crack length without fracture is 1.55 mm." }, { "idx": 4008, "question": "A cylindrical specimen 7.5mm in diameter of an S-590 alloy is to be exposed to a tensile load of 9000 N. At approximately what temperature will the steady-state creep be 10^{-2}h^{-1} ?", "answer": "approximately 815^\\circ C." }, { "idx": 4009, "question": "The mineral olivine is a solid solution of the silicate compounds forsterite \\left(Mg_{2} \\mathrm{SiO}_{4}\\right) and fayalite \\left(\\mathrm{Fe}_{2} \\mathrm{SiO}_{4}\\right).\nHow many chemical components are there in a sample of olivine?\n(a) 1\n(b) 2\n(c) 3\n(d) 4", "answer": "The components of a system are defined to be the smallest set of independently variable chemical constituents that are necessary to describe the composition of each phase that is present in a system. Components are chemically distinct. In this case, we only need to specify two things. It is convenient to simply use the compounds themselves as the components, but other options exist. For instance, specifying the amount of Mg and \\mathrm{Fe} in the system automatically fixes the \\mathrm{Si} and \\mathrm{O} content, assuming the compounds are present in their stoichiometric ratios (assuming they are defect free)." }, { "idx": 4010, "question": "Seawater, which covers the majority of the earth, is composed primarily of molecules of H_{2} \\mathrm{O} and equal numbers of \\mathrm{Na}^{+}ions and \\mathrm{Cl}^{-}ions. Suppose we have a thoroughly mixed solution (containing these species only) at 25^{\\circ} C. How many components and how many phases are in such a system?\n(a) 1 component, 1 phase\n(b) 1 component, 2 phase\n(c) 1 component, 3 phase\n(d) 1 component, 4 phase\n(e) 2 component, 1 phase\n(f) 2 component, 2 phase\n(g) 2 component, 3 phase\n(h) 2 component, 4 phase\n(i) 3 component, 1 phase\n(j) 3 component, 2 phase\n(k) 3 component, 3 phase(l) 3 component, 4 phase\n(m) 4 component, 1 phase\n(n) 4 component, 2 phase\n(o) 4 component, 3 phase\n(p) 4 component, 4 phase", "answer": "(e) 2 component, 1 phase" }, { "idx": 4011, "question": "Is solid ductile cast iron (ferrite solid solution + embedded graphite spheres) a two-phase material system?", "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes." }, { "idx": 4012, "question": "Is solid sodium chloride (salt, NaCl) a two-phase material system?", "answer": "No, it is a single-phase compound, not a solution. The ratio of the two ions is in a fixed stoichiometry and they adopt an orderly arrangement in the crystal." }, { "idx": 4013, "question": "Is liquid bronze (Cu + Sn liquid solution) a two-phase material system?", "answer": "No, it is a liquid solution, and solutions are single-phase by definition." }, { "idx": 4014, "question": "Is solid gray cast iron (ferrite solid solution + embedded graphite flakes) a two-phase material system?", "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes." }, { "idx": 4015, "question": "Is solid aluminum featuring dissolved silicon a two-phase material system?", "answer": "No, it is a solid solution since silicon is implied to substitute for aluminum. Solutions are single-phase by definition." }, { "idx": 4016, "question": "Is solid lead-tin solder (a mixture of Pb-rich and Sn-rich solid solutions) a two-phase material system?", "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes." }, { "idx": 4017, "question": "Is partially melted aluminum a two-phase material system?", "answer": "No, the physical boundary merely separates volumes that are structurally distinct (crystalline and liquid)." }, { "idx": 4018, "question": "Is frozen water with trapped air bubbles a two-phase material system?", "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes." }, { "idx": 4019, "question": "Is epoxy embedded with carbon fibers a two-phase material system?", "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes." }, { "idx": 4020, "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must have a very [w] sizes. (w = similar, different)", "answer": "The solute and host species must have a very similar sizes." }, { "idx": 4021, "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must attempt to pack with [x] crystal structure. (x = a similar (or the same), a different)", "answer": "The solute and host species must attempt to pack with a similar (or the same) crystal structure." }, { "idx": 4022, "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must feature [y] valence electron configuration. (y = a similar (or the same), a different)", "answer": "The solute and host species must feature a similar (or the same) valence electron configuration." }, { "idx": 4023, "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must feature [z] ability to attract electrons (electronegativity). (z = a similar, a different)", "answer": "The solute and host species must feature a similar ability to attract electrons (electronegativity)." }, { "idx": 4024, "question": "For a solution, which of the following is present in the higher concentration?(a) Solvent(b) Solute", "answer": "Solvent. By definition, solvent is the element/compound that is present in a solution in the greatest amount." }, { "idx": 4025, "question": "A liquidus line separates which of the following combinations of phase fields?(a) Liquid and Liquid +\\alpha(b) \\alpha and Liquid +\\alpha(c) \\alpha and \\alpha+\\beta(d) Liquid +\\alpha and \\alpha+\\beta", "answer": "A liquidus line separates Liquid and Liquid +\\alpha phase fields." }, { "idx": 4026, "question": "A solvus line separates which of the following pairs of phase fields?(a) Liquid and Liquid +\\alpha(b) \\alpha \\operatorname{and} Liquid +\\alpha(c) \\alpha and \\alpha+\\beta d) Liquid +\\alpha and \\alpha+\\beta", "answer": "A solvus line separates \\alpha and \\alpha+\\beta phase fields." }, { "idx": 4027, "question": "Which of the following kinds of information may be determined with the aid of a phase diagram? The phase(s) present at a specified temperature and composition.", "answer": "With the aid of a phase diagram the following may be determined: The phase(s) present at a specified temperature and composition." }, { "idx": 4028, "question": "Which of the following kinds of information may be determined with the aid of a phase diagram? The composition(s) of phase(s) present at a specified temperature and composition.", "answer": "With the aid of a phase diagram the following may be determined: The composition(s) of phase(s) present at a specified temperature and composition." }, { "idx": 4029, "question": "Which of the following kinds of information may be determined with the aid of a phase diagram? The fraction(s) of phase(s) present at specified temperature and composition.", "answer": "With the aid of a phase diagram the following may be determined: The fraction(s) of phase(s) present at a specified temperature and composition." }, { "idx": 4030, "question": "At a eutectic point on a binary temperature-composition phase diagrams, how many phases are present when the system is at equilibrium?(a) 0(b) 1(c) 2(d) 3", "answer": "At a eutectic point on a binary phase diagram, three phases (Liquid, \\alpha, and \\beta ) are present when the system is at equilibrium." }, { "idx": 4031, "question": "An intermetallic compound is found in the magnesium-gallium system that has a composition of 41.1 wt% Mg - 58.9 wt% Ga. Specify the formula for this compound.(a) MgGa (b) Mg_{2} \\mathrm{Ga} (c) \\mathrm{MgGa}_{2} (d) Mg_{3} \\mathrm{Ga}_{2}", "answer": "(b) Mg_{2}Ga" }, { "idx": 4032, "question": "A eutectoid reaction involves which of the following phases? (a) One liquid and one solid (b) One liquid and two solid (c) Two liquids and one solid (d) Three solid", "answer": "(d) Three solid" }, { "idx": 4033, "question": "A peritectic reaction involves which of the following combinations of phase fields? (a) One liquid and one solid (b) One liquid and two solid (c) two liquids and one solid (d) Three solid", "answer": "(b) One liquid and two solid" }, { "idx": 4034, "question": "For a congruent phase transformations there are (a) no composition alterations.(b) compositional alterations.", "answer": "(a) no composition alterations" }, { "idx": 4035, "question": "Which of the following have a significant influence on a material's electrical resistivity?(a) impurity concentration(b) temperature(c) grain size(d) cold work(e) vacancy concentration", "answer": "A material's electrical resistivity will depend on the following:\n- Impurity concentration\n- Vacancy concentration\n- Temperature\n- Cold work" }, { "idx": 4036, "question": "Conductivity in a metal is almost always reduced by the introduction of defects into the lattice.\nThe factor primarily affected by defects is:\n[a] free electron concentration\n[b] electron charge\n[c] electron mobility\n[d] electron spin", "answer": "[c] electron mobility" }, { "idx": 4037, "question": "Compute the number of electrons that each aluminum atom donates, on average, to a bulk piece of aluminum metal. Room temperature data for aluminum:\nThe resistivity of aluminum is 2.63 × 10^{-8} \\Omega·m\nThe electron mobility of aluminum is 0.0012{m}^{2} /(V·s)\nThe mass density of aluminum is 2.7g / {cm}^{3}\nThe atomic weight of aluminum is 27g / mol", "answer": "3.29 electron/atom" }, { "idx": 4038, "question": "Compute the number of electrons that each atom donates, on average, to a bulk piece of hypothetical metal. Room temperature data for the metal:\nThe resistivity of the metal is [\\mathrm{r}] \\Omega·{cm}\nThe electron mobility of the metal is [m] {cm}^{2} /(V·s)\nThe mass density of the metal is [\\mathrm{d}] \\mathrm{g} / {cm}^{3}\nThe atomic weight of the metal is [\\mathrm{w}] \\mathrm{g} / mol", "answer": "the number of free electrons donated by each atom, on average, is: \\[\n\\frac{[w]}{\\left([r]\\left(1.6 × 10^{-19}\\right)[m]\\right)\\left([d]\\left(6.022 × 10^{23}\\right)\\right)}\n\\]" }, { "idx": 4039, "question": "When an electric field is applied, in which direction are the free electrons accelerated?(a) Opposite to the direction of the electric field.(b) In the same direction as the electric field.", "answer": "When an electric field is applied, the free electrons are accelerated in the direction opposite to that of the field." }, { "idx": 4040, "question": "The electrical conductivity of an intrinsic semiconductor is(a) characteristic of the high-purity metal.(b) due to the presence of impurities.", "answer": "The electrical conductivity of an intrinsic semiconductor is characteristic of the high-purity material." }, { "idx": 4041, "question": "How do the electrical conductivities of metals compare with those of semiconductors?(a) \\sigma_{\\text {metals }}>\\sigma_{\\text {semiconductors }}(b) \\sigma_{\\text {metals }}=\\sigma_{\\text {semiconductors }}(c) \\sigma_{\\text {metals }}<\\sigma_{\\text {semiconductors }}", "answer": "The electrical conductivities of metals are greater than those of semiconductors." }, { "idx": 4042, "question": "How does increasing temperature affect the concentration of both electrons and holes in an intrinsic semiconductor?(a) Increases the concentration.(b) Decreases the concentration.(c) May increase and/or decrease the concentration, depending on the temperature range.", "answer": "Increasing temperature increases the concentration of both electrons and holes in an intrinsic semiconductor." }, { "idx": 4043, "question": "Which type of charge carrier will be introduced into a semiconductor by the presence of an acceptor impurity?(a) Electron(b) Hole", "answer": "The presence of an acceptor impurity introduces holes into a semiconductor." }, { "idx": 4044, "question": "Which type of charge carrier will be introduced into a semiconductor by the presence of a donor impurity?(a) Impurity(b) Hole", "answer": "The presence of a donor impurity introduces electrons into a semiconductor." }, { "idx": 4045, "question": "The electrical conductivity of an extrinsic semiconductor is(a) Characteristic of the high-purity material.(b) due to the presence of impurities.", "answer": "The electrical conductivity of an extrinsic semiconductor is due to the presence of impurities." }, { "idx": 4046, "question": "For an n-type semiconductor, which type of charge carrier is present in the greater concentration?(a) Hole(b) Electron", "answer": "For an n-type semiconductor, electrons (i.e., negative charge carriers) are present in a greater concentration than holes." }, { "idx": 4047, "question": "For an n-type semiconductor(a) Concentration _{\\text {electrons }}< concentration _{\\text {holes }}(b) Concentration _{\\text {electrons }}= concentration _{\\text {holes }}(c) Concentration _{\\text {electrons }}< concentration _{\\text { holes }}", "answer": "For an n-type semiconductor, the concentration of electrons is much greater than the concentration of holes." }, { "idx": 4048, "question": "For a p-type semiconductor, which type of charge carrier is present in the greater concentration?(a) Holes(b) Electrons", "answer": "For a p-type semiconductor, holes, (i.e., positive charge carriers) are present in a greater concentration than electrons." }, { "idx": 4049, "question": "For a p-type semiconductor(a) Concentration _{\\text {electrons }}< concentration _{\\text {holes }}(b) Concentration _{\\text {electrons }}= concentration _{\\text {holes }}(c) Concentration _{\\text {electrons }}< concentration _{\\text { holes }}", "answer": "For a p-type semiconductor, the concentration of electrons is much lower than the concentration of holes." }, { "idx": 4050, "question": "In order for a semiconductor to exhibit extrinsic electrical characteristics, relatively high impurity concentrations are required.(a) True(b) False", "answer": "False. Even when minute impurity concentrations (e.g., 1 atom in 10^{12} ) are present in a semiconductor, its electrical characteristics will be extrinsic." }, { "idx": 4051, "question": "The following electrical characteristics have been determined for both intrinsic and n-type extrinsic indium phosphide (InP) at room temperature:\n\\[\n\\begin{array}{llll}\n\\sigma(\\Omega·\\mathbf{m})^{-1} & \\boldsymbol{n}\\left(\\mathbf{m}^{-3}\\right) & \\boldsymbol{p}\\left(\\mathbf{m}^{-3}\\right) \\\\\n\\text { Intrinsic } & 2.5 × 10^{-6} & 3.0 × 10^{13} & 3.0 × 10^{13} \\\\\n\\text { Extrinsic } & 3.6 × 10^{-5} & 4.5 × 10^{14} & 2.0 × 10^{12}\n\\end{array}\n\\]\nCalculate electron and hole mobilities.\n\\mu_{\\mathrm{e}}=\n\\mu_{h}=", "answer": "\\mu_{\\mathrm{e}} = 0.50 \\, m^{2} / \\mathrm{v}·s \\mu_{h} = 0.02 \\, m^{2} / \\mathrm{v}·s" }, { "idx": 4052, "question": "Which of the following are preferred for semiconducting devices?(a) Single crystals(b) Polycrystalline materials", "answer": "Single crystals are preferred for semiconducting devices because grain boundaries are deleterious to the performance of electronic phenomena." }, { "idx": 4053, "question": "As temperature increases, the electrical conductivities of polymers and ionic ceramics(a) Increase(b) Decrease", "answer": "As temperature increases, the electrical conductivities of polymers and ionic ceramics increase." }, { "idx": 4054, "question": "Most polymers and ionic ceramics have energy band gap structures that are most similar to those of(a) Insulators(b) Semiconductors(c) Metals", "answer": "Most polymers and ionic ceramics have wide energy band gaps; thus, they are most like those of insulators." }, { "idx": 4055, "question": "The charge carriers in ionic ceramics and polymers can be(a) Electrons(b) Holes(c) Anion(d) Cations", "answer": "The charge carriers in ionic ceramics and polymers can be electrons, holes, anions, and/or cations." }, { "idx": 4056, "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.", "answer": "F" }, { "idx": 4057, "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.", "answer": "T" }, { "idx": 4058, "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.", "answer": "F" }, { "idx": 4059, "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.", "answer": "F" }, { "idx": 4060, "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: The relatively low melting point of aluminum is often considered a significant limitation for structural applications.", "answer": "T" }, { "idx": 4061, "question": "[a] Aluminum alloys are generally viable as lightweight structural materials in humid environments because they are not very susceptible to corrosion by water vapor.", "answer": "T" }, { "idx": 4062, "question": "[b] Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.", "answer": "F" }, { "idx": 4063, "question": "[c] Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.", "answer": "T" }, { "idx": 4064, "question": "[d] Compared to other metals, like steel, pure aluminum is very resistant to failure via fatigue.", "answer": "F" }, { "idx": 4065, "question": "[e] Aluminum exhibits one of the highest melting points of all metals, which makes it difficult and expensive to cast.", "answer": "F" }, { "idx": 4066, "question": "[a] Copper has a higher elastic modulus than aluminum.", "answer": "T" }, { "idx": 4067, "question": "[b] The density of copper is closer to that of aluminum than it is to iron.", "answer": "F" }, { "idx": 4068, "question": "[c] Bronze is an alloy of copper and zinc.", "answer": "F" }, { "idx": 4069, "question": "[d] Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates.", "answer": "T" }, { "idx": 4070, "question": "[e] Copper is relatively resistant to corrosion by neutral and even mildly basic water, making it useful for freshwater plumbing applications.", "answer": "T" }, { "idx": 4071, "question": "Select T / F for the following statement regarding copper & copper alloys: Copper is much more abundant in the earth's crust compared to iron or aluminum.", "answer": "F" }, { "idx": 4072, "question": "Select T / F for the following statement regarding copper & copper alloys: Copper is one of just a few metals that can be found in metallic form in nature.", "answer": "T" }, { "idx": 4073, "question": "Select T / F for the following statement regarding copper & copper alloys: Pure and/or annealed copper is more difficult to machine compared to its work-hardened form or its alloys.", "answer": "T" }, { "idx": 4074, "question": "Select T / F for the following statement regarding copper & copper alloys: Copper is a minor component (by weight) of most brass & bronze alloys.", "answer": "F" }, { "idx": 4075, "question": "Select T / F for the following statement regarding copper & copper alloys: Amongst metals and alloys copper is one of the best conductors of heat.", "answer": "T" }, { "idx": 4076, "question": "[a] Nickel is majority component (by mass) in certain superalloys such as Waspaloy TM.", "answer": "T" }, { "idx": 4077, "question": "[b] Tungsten is the lowest density metal that has structural use.", "answer": "F" }, { "idx": 4078, "question": "[c] Tantalum offers extremely good corrosion resistance, especially at low temperatures.", "answer": "T" }, { "idx": 4079, "question": "[d] Magnesium metal is very similar to aluminum, in terms of its physical and mechanical properties.", "answer": "T" }, { "idx": 4080, "question": "[e] Beryllium metal is commonly used as an alloying agent in copper metal.", "answer": "T" }, { "idx": 4081, "question": "Increasing the alumina \\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right) content of fireclays results in(a) an increase in maximum service temperature.(b) a decrease in maximum service temperature.", "answer": "Increasing the alumina \\left(\\mathrm{Al}_{2} \\mathrm{O}\\right)_{3} content of fireclays results in an increase in maximum service temperature." }, { "idx": 4082, "question": "The high-temperature performance of silica refractories is compromised by the presence of even small concentrations of alumina \\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\end{array}(a) True(b) False", "answer": "True. The presence of even small amounts of alumina \\left(\\mathrm{Al}_{2} \\mathrm{O}_{3)}\\right) in silica refractory ceramics compromises their high-temperature performance." }, { "idx": 4083, "question": "As the porosity of refractory ceramic bricks increases, what happens to the strength?", "answer": "Strength decreases." }, { "idx": 4084, "question": "As the porosity of refractory ceramic bricks increases, what happens to the chemical resistance?", "answer": "Chemical resistance decreases." }, { "idx": 4085, "question": "As the porosity of refractory ceramic bricks increases, what happens to the thermal insulation?", "answer": "Thermal insulation increases." }, { "idx": 4086, "question": "The presence of silica \\left(\\mathrm{SiO}_{2}\\right) in basic refractory ceramics is beneficial to their high-temperature performance.(a) True(b) False", "answer": "False. The presence of silica \\left(\\mathrm{SiO}_{2}\\right) in basic refractory ceramics is deleterious on their hightemperature performance." }, { "idx": 4087, "question": "Basic refractory ceramics are often used for the containment of slags that are rich in(a) silica(b) \\mathrm{CaO}(c) MgO", "answer": "Basic refractory ceramics are often used for the containment of slags that are rich in \\mathrm{CaO} and/or MgO." }, { "idx": 4088, "question": "Silica refractory ceramics are often used for the containment of slags that are rich in(a) silica(b) \\mathrm{CaO}(c) Mg \\mathrm{O}", "answer": "Silica refractory ceramics are often used for the containment of slags that are rich in silica." }, { "idx": 4089, "question": "Such that the bond strength across the fiber-epoxy interface is [s] MPa, and the shear yield strength of the epoxy is [y] MPa, compute the minimum fiber length, in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. The tensile strength of these carbon fibers is [f] MPa.", "answer": "the minimum fiber length to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite is \\frac{[f][d]}{2[s] × 1000} millimeters." }, { "idx": 4090, "question": "For an aligned fibrous composite, when a stress is applied in a direction that is parallel to the fibers, what is the reinforcement efficiency?(a) 0(b) \\frac{1}{5}(c) \\frac{3}{8}(d) \\frac{3}{4}(e) 1", "answer": "For an aligned fibrous composite when the stress is applied parallel to the fibers, the reinforcement efficiency is 1 ." }, { "idx": 4091, "question": "For an aligned fibrous composite, when a stress is applied perpendicular to the fibers, what is the reinforcement efficiency?(a) 0(b) \\frac{1}{5}(c) \\frac{3}{8}(d)\n\\frac{3}{4}(e) 1", "answer": "begin{tabular}{l|l|l|l|l|l|l|l|l|l|l|}\n\\hline\n\\end{tabular}\nFor an aligned fibrous composite when a stress is applied perpendicular to the fibers, the reinforcement efficiency is 0 ." }, { "idx": 4092, "question": "For a fibrous composite with fibers that are randomly and uniformly oriented within a specific plane, when a stress is applied in any direction within the plane of the fibers, what is the reinforcement efficiency?(a) 0(b) \\frac{1}{5}(c) \\frac{3}{8}(d) \\frac{3}{4}(e) 1", "answer": "For a fibrous composite with fibers that are randomly and uniformly oriented within a specific plane,when a stress is applied in any direction within the plane of the fibers, the reinforcement efficiency is \\frac{3}{8}." }, { "idx": 4093, "question": "For a fibrous composite with fibers that are uniformly distributed and randomly oriented in all directions, when a stress is applied in any direction, what is the reinforcement efficiency?(a) 0(b) \\frac{1}{{ }_{5}}(c) \\frac{3}{8}(d) \\frac{3}{{ }_{4}}(e) 1", "answer": "For a fibrous composite with fibersthat are uniformly distributed and randomly oriented in all directions, when a stress is applied in any direction, the reinforcement efficiency is 1 / 5." }, { "idx": 4094, "question": "A stress-strain test is performed on an aligned fibrous composite such that the force is applied in the longitudinal direction. During the initial stage of the test, which phase bears most of the load?(a) Fibers(b) Matrix", "answer": "During a stress-strain test that is performed on an aligned fibrous composite, the fibers bear more of the applied load than the matrix in the initial stage of the test." }, { "idx": 4095, "question": "Once the fibers fail in a fibrous composite, catastrophic failure of the piece takes place.(a) True(b) False", "answer": "False. Once the fibers fail in a composite, catastrophic failure of the piece does not take place. Since the broken fibers are still embedded within the matrix, they are still capable of sustaining a diminished load." }, { "idx": 4096, "question": "How are continuous fibers typically oriented in fibrous composites?(a) Aligned(b) Partially oriented(c) Randomly oriented", "answer": "Continuous fibers are typically aligned in fibrous composites." }, { "idx": 4097, "question": "How are discontinuous fibers typically oriented in fibrous composites?(a) Aligned(b) Partially oriented", "answer": "Discontinuous fibers may be aligned, partially oriented, and randomly oriented in fibrous composites." }, { "idx": 4098, "question": "If the fiber orientation is random, which type of fibers is normally used?(a) Discontinuous(b) Continuous", "answer": "Discontinuous fibers are normally used when the fiber orientation is random." }, { "idx": 4099, "question": "[a] Composites are single-phase materials by definition.", "answer": "F" }, { "idx": 4100, "question": "[b] The term \"composite\" applies to materials that feature polymeric materials only.", "answer": "F" }, { "idx": 4101, "question": "[c] Structural composites are, in general, highly regarded for their specific strengths.", "answer": "T" }, { "idx": 4102, "question": "[d] Composites featuring continuous and aligned fibers for reinforcement generally offer properties that are highly isotropic compared to most metals (random polycrystals).", "answer": "F" }, { "idx": 4103, "question": "Which of the following materials are typically used as fibers?(a) Graphite/carbon(b) Silicon carbide(c) Silicon nitride(d) Aluminum oxide(e) Glass(f) Boron(g) Steel\nh) Tungsten\ni) Molybdenum", "answer": "Graphite, silicon carbide, glass, boron, and aluminum oxide are typically used as fibers." }, { "idx": 4104, "question": "Match the fiber type 'Whiskers' with its description.", "answer": "Whiskers are single crystals with extremely large length-to-diameter ratios." }, { "idx": 4105, "question": "Match the fiber type 'Fibers' with its description.", "answer": "Fibers are polycrystalline or amorphous materials with small diameters." }, { "idx": 4106, "question": "Match the fiber type 'Wires' with its description.", "answer": "Wires are large-diameter metals." }, { "idx": 4107, "question": "For a composite material, which phase normally has the higher elastic modulus?(a) Fiber phase(b) Matrix phase", "answer": "The fiber phase normally has a higher elastic modulus than the matrix phase." }, { "idx": 4108, "question": "For a composite material, how does the ductility of the matrix phase normally compare with the ductility of the dispersed phase?(a) more ductile(b) less ductile", "answer": "A composite's matrix phase is normally more ductile than the dispersed phase." }, { "idx": 4109, "question": "Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strengths.(a) True(b) False", "answer": "True. Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strength." }, { "idx": 4110, "question": "Which of aramid and metal fibers have higher strength-to-weight ratios?(a) Aramid fibers(b) Metal fibers", "answer": "Aramid fibers have higher strength-to-weight ratios than metal fibers." }, { "idx": 4111, "question": "Carbon fiber-reinforced composites have relatively high strengths?", "answer": "Carbon fiber-reinforced composites have relatively high strengths." }, { "idx": 4112, "question": "Carbon fiber-reinforced composites have relatively high stiffnesses?", "answer": "Carbon fiber-reinforced composites have relatively high stiffnesses." }, { "idx": 4113, "question": "Carbon fiber-reinforced composites have high service temperatures (>200 degrees C)?", "answer": "Carbon fiber-reinforced composites have high service temperatures (>200 degrees C)." }, { "idx": 4114, "question": "Which of the following materials are typically used as whiskers?(a) graphite/carbon(b) silicon carbide(c) silicon nitride(d) aluminum oxide(e) glass(f) boron(g) steel\nh) tungsten\ni) molybdenum", "answer": "Graphite, silicon carbide, silicon nitride, and aluminum oxide are typically used as whiskers." }, { "idx": 4115, "question": "Which of the following materials are typically used as wires in composites?(a) graphite/carbon(b) silicon carbide(c) silicon nitride\n\\mathrm{d}) aluminum oxide(e) glass(f) boron(g) steel\nh) tungsten\ni) molybdenum", "answer": "Steel, tungsten, and molybdenum are typically used as wires in composites." }, { "idx": 4116, "question": "Compared to other ceramic materials, do ceramic-matrix composites have better fracture toughnesses?", "answer": "Ceramic-matrix composites have higher fracture toughnesses than other ceramic materials." }, { "idx": 4117, "question": "Compared to other ceramic materials, do ceramic-matrix composites have better oxidation resistance?", "answer": "The answer is not provided in the given information." }, { "idx": 4118, "question": "Compared to other ceramic materials, do ceramic-matrix composites have better stability at elevated temperatures?", "answer": "The answer is not provided in the given information." }, { "idx": 4119, "question": "Which of the following materials are typically used as stabilizers in transformation-toughened ceramic-matrix composites?(a) \\mathrm{CaO}(b) MgO(c) \\mathrm{Y}_{2} \\mathrm{O}_{3}(d) \\mathrm{CeO}(e) \\mathrm{Al}_{2} \\mathrm{O}_{3}(f) \\mathrm{SiC}", "answer": "(\\mathrm{CaO}, MgO, \\mathrm{Y}_{2} \\mathrm{O}_{3}, and \\mathrm{CeO} are typically used as stabilizers in transformation-toughened ceramicmatrix composites." }, { "idx": 4120, "question": "Do carbon-carbon composites exhibit high tensile moduli at elevated temperatures?", "answer": "Carbon-carbon composites have high tensile moduli at elevated temperatures." }, { "idx": 4121, "question": "Do carbon-carbon composites exhibit high tensile strengths at elevated temperatures?", "answer": "Carbon-carbon composites have high tensile strengths at elevated temperatures." }, { "idx": 4122, "question": "Do carbon-carbon composites exhibit resistance to creep?", "answer": "Carbon-carbon composites are highly resistant to creep." }, { "idx": 4123, "question": "Do carbon-carbon composites exhibit large fracture toughness values?", "answer": "Carbon-carbon composites have large fracture toughness values." }, { "idx": 4124, "question": "Do carbon-carbon composites exhibit high thermal conductivities?", "answer": "Carbon-carbon composites have high thermal conductivities." }, { "idx": 4125, "question": "Do carbon-carbon composites exhibit low coefficients of thermal expansion?", "answer": "Carbon-carbon composites have low coefficients of thermal expansion." }, { "idx": 4126, "question": "Do carbon-carbon composites exhibit resistance to oxidation at elevated temperatures?", "answer": "The answer does not specify resistance to oxidation at elevated temperatures." }, { "idx": 4127, "question": "Are carbon-carbon composites low cost?", "answer": "The answer does not specify low cost." }, { "idx": 4128, "question": "Laminar composites have high strengths in all directions (in three dimensions).(a) True(b) False", "answer": "False. Laminar composites have high strengths in all directions only in their two-dimensional planes." }, { "idx": 4129, "question": "The strong outer sheets of sandwich panels are separated by a layer of material that is(a) less dense than the outer sheet material(b) more dense than the outer sheet material", "answer": "The strong outer sheets of sandwich panels are separated by a layer of material that is less dense than the outer sheet material." }, { "idx": 4130, "question": "To what temperature would 23.0kg of some material at 100^{\\circ} C be raised if 255kJ of heat is supplied? Assume a C_{p} value of 423 J/ kg-K for this material.\n(A) 26.2^{\\circ} C\n(B) 73.8^{\\circ} C\n(C) 126^{\\circ} C\n(D) 152^{\\circ} C", "answer": "the final temperature is 126^{\\circ} C, which corresponds to answer C." }, { "idx": 4131, "question": "A rod of some material 0.50{m} long elongates 0.40mm on heating from 50^{\\circ} C to 151^{\\circ} C. What is the value of the linear coefficient of thermal expansion for this material?\n(A) 5.30 × 10^{-6}\\left({ }^{\\circ} C\\right)^{-1}\n(B) 7.92 × 10^{-6}\\left({ }^{\\circ} C C\\right)^{-1}\n(C) 1.60 × 10^{-5}\\left({ }^{\\circ} C\\right)^{-1}\n(D) 1.24 × 10^{-6}\\left({ }^{\\circ} C_{-}^{1}\\right.", "answer": "the linear coefficient of thermal expansion for this material is 7.92 × 10^{-6} \\left(^{\\circ}c\\right)^{-1}, which is answer b." }, { "idx": 4132, "question": "A 10 meter long square bar of 316 stainless steel (edge length of 5 cm, with a modulus of 193 \\mathrm{GPa} and a yield point of 290 MPa ) is bolted securely in place when its installation temperature was around [1]^{\\circ} C. What is the expected thermal stress in the bar when its service temperature reaches [F]^{o} C ? Enter a negative indicate compressive stress, if necessary. The thermal expansion coefficient of 316 stainless steel is 16.0 × 10^{-6} \\mathrm{I} /{ }^{\\circ} C.", "answer": "sigma_{th} = (193,000)(16 × 10^{-6})([i] - [f]) MPa" }, { "idx": 4133, "question": "Which of the following 1kg samples is expected to change temperature the least if 100kJ of heat is perfectly transferred to each of them at a constant pressure of 1 atmosphere. The initial temperature of each specimen is 25^{\\circ} C. (a) Aluminum (b) Copper (c) Gold (d) Borosilicate Glass (e) Polystyrene", "answer": "polystyrene will experience the smallest temperature change." }, { "idx": 4134, "question": "If you were to locally heat identical geometry plates of the materials listed below with the same heat source, which would increase in temperature the fastest? (a) Polystyrene (b) Aluminum (c) Copper (d) Gold (e) Borosilicate Glass", "answer": "In this scenario, the material with the highest thermal conductivity will conduct heat the fastest. Therefore this material will absorb heat and increase in temperature the fastest. In this case, copper has the highest thermal conductivity." }, { "idx": 4135, "question": "This ceramic outlier has the highest room-temperature thermal conductivity of about 2,000 watts per meter per kelvin, a value that is five time higher than the best thermally conductive metals.\n(a) \\mathrm{Al}_{2} \\mathrm{O}_{3}\n(b) \\mathrm{CaF}_{2}\n(c) \\mathrm{TiO}_{2}\n(d) \\mathrm{SrTiO}_{3}\n(e) C (diamond)", "answer": "Diamond has the highest thermal conductivity of these options." }, { "idx": 4136, "question": "Which two of the following are ferromagnetic materials? \\\\(a) Aluminum oxide \\\\(b) Copper \\\\ c) Aluminum \\\\ d) Titanium \\\\ e) Iron ( \\alpha ferrite) \\\\ f) Nickel \\\\ g) MnO \\\\ h) \\mathrm{Fe}_{3} \\mathrm{O}_{4} \\\\ i) \\mathrm{NiFe}_{2} \\mathrm{O}_{4}\n}", "answer": "Iron ( \\alpha ferrite) and nickel are ferromagnetic materials." }, { "idx": 4137, "question": "Which of the following materials display(s) antiferromagnetic behavior?(a) Aluminum oxide(b) Copper(c) Aluminum(d) Titanium(e) Iron ( \\alpha ferrite)(f) Nickel(g) MnO\nh) \\mathrm{Fe}_{3} \\mathrm{O}_{4}\ni) \\mathrm{NiFe}_{2} \\mathrm{O}_{4}", "answer": "(\\mathrm{MnO} is an antiferromagnetic material." }, { "idx": 4138, "question": "Which of the following are ferrimagnetic materials?(a) Aluminum oxide(b) Copper(c) Aluminum(d) Titanium(e) Iron ( \\alpha ferrite)(f) Nickel(g) MnO\nh) \\mathrm{Fe}_{3} \\mathrm{O}_{4}\ni) \\mathrm{NiFe}_{2} \\mathrm{O}_{4}", "answer": "(\\mathrm{Fe}_{3} \\mathrm{O}_{4} and \\mathrm{NiFe}_{2} \\mathrm{O}_{4} are ferrimagnetic materials." }, { "idx": 4139, "question": "The formula for yttrium iron garnet \\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{3} \\mathrm{O}_{12}\\right) may be written in the form \\mathrm{Y}_{3}^{*} \\mathrm{Fe}_{2}^{*} \\mathrm{Fe}_{3}^{d} \\mathrm{O}_{12}^{2}, where the superscripts a, c, and d represent different sites on which the \\mathrm{Y}^{3+} and \\mathrm{Fe}^{3+} ions are located. The spin magnetic moments for the \\mathrm{Y}^{3+} and \\mathrm{Fe}^{\\text {out }} ions positioned in the \\mathrm{a} and C sites are oriented parallel to one another and antiparallel to the \\mathrm{Fe}^{3+} ions in \\mathrm{d} sites. Compute the number of Bohr magnetons associated with each \\mathrm{Y}^{3+} ion, given the following information: (1) each unit cell consists of eight formula \\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right) units; (2) the unit cell is cubic with an edge length of 1.2376nm; (3) the saturation magnetization for this material is 1.0 × 10^{4} \\mathrm{~A} / m; and (4) assume that there are 5 Bohr magnetons associated with each \\mathrm{Fe}^{3+} ion.", "answer": "the number of bohr magnetons associated with each \\mathrm{y}^{3+} ion is 1.75 \\mathrm{bm}." }, { "idx": 4140, "question": "At the Curie temperature, the saturation magnetization abruptly diminishes. Which of the following magnetic material types will have Curie temperatures?(a) Diamagnetics(b) Paramagnetics(c) Ferromagnetics(d) Antiferromagnetics(e) Ferrimagnetics", "answer": "Ferromagnetic and ferrimagnetic materials will have Curie temperatures." }, { "idx": 4141, "question": "In a polycrystalline material, each grain will always consist of just a single domain.(a) True(b) False", "answer": "False. In a polycrystalline material, each grain may consist of more than one domain." }, { "idx": 4142, "question": "With increasing temperature antiferromagnetic materials eventually become which of the following?(a) Diamagnetics(b) Paramagnetics(c) Ferromagnetics(d) Antiferromagnetics(e) Ferrimagnetics", "answer": "With increasing temperature antiferromagnetic materials eventually become paramagnetic." }, { "idx": 4143, "question": "Which type(s) of magnetic materials may be classified as either soft or hard?(a) Diamagnetic(b) Paramagnetic(c) Ferromagnetic(d) Antiferromagnetic(e) Ferrimagnetic", "answer": "Ferromagnetic and ferrimagnetic materials may be classified as either soft or hard." }, { "idx": 4144, "question": "Which of the following characteristics are displayed by soft magnetic materials in terms of hysteresis loop size?", "answer": "Soft materials have relatively small relative hysteresis loops." }, { "idx": 4145, "question": "Which of the following characteristics are displayed by soft magnetic materials in terms of magnetization and demagnetization fields?", "answer": "Magnetization and demagnetization may be achieved using relatively low applied fields." }, { "idx": 4146, "question": "Which of the following characteristics are displayed by hard magnetic materials in terms of hysteresis loop size?", "answer": "Hard materials have relatively large hysteresis loops." }, { "idx": 4147, "question": "Which of the following characteristics are displayed by hard magnetic materials in terms of magnetization and demagnetization field requirements?", "answer": "Magnetization and demagnetization require relatively high applied fields." }, { "idx": 4148, "question": "What is the order of magnitude wavelength for visible light?(a) 0.5 Angstroms(b) 0.5 nanometers(c) 0.5 micrometers(d) 0.5 millimeters(e) 0.5 meters(f) 0.5 kilometers", "answer": "The wavelength of visible light is on the order of 0.5 \\mathrm{micrometers}." }, { "idx": 4149, "question": "In the visible spectrum, a thick metal specimen will be(a) Transparent(b) Translucent(c) Opaque", "answer": "In the visible spectrum, a thick metal specimen is opaque." }, { "idx": 4150, "question": "In the visible spectrum, an electrical insulator that is a single crystal and without porosity is normally(a) Transparent(b) Opaque(c) Translucent", "answer": "In the visible spectrum, an electrical insulator that is a single crystal and without porosity\nnormally transparent." }, { "idx": 4151, "question": "Match the type of light transmission with its description: Transmits light with relative little absorption.", "answer": "A transparent material transmits light with relatively little absorption." }, { "idx": 4152, "question": "Match the type of light transmission with its description: Transmits light diffusely.", "answer": "A translucent material transmits light diffusely." }, { "idx": 4153, "question": "Match the type of light transmission with its description: Is impervious to light transmission.", "answer": "An opaque material is impervious to light transmission." }, { "idx": 4154, "question": "In the visible spectrum, a semiconductor that is a single crystal and nonporous may be(a) Transparent(b) Translucent(c) Opaque", "answer": "In the visible spectrum, a semiconductor that is a single crystal and nonporous may be transparent or opaque." }, { "idx": 4155, "question": "To which of the following electromagnetic radiation types are bulk metals opaque?(a) radio waves(b) microwaves(c) infrared radiation(d) ultraviolet radiation(e) X-rays", "answer": "Bulk metals are opaque to the following radiation types: radio waves, microwaves, infrared radiation, and ultraviolet radiation." }, { "idx": 4156, "question": "Every nonmetallic material becomes opaque to electromagnetic radiation having some wavelength.(a) True(b) False", "answer": "True. Every nonmetallic material becomes opaque to electromagnetic radiation having some wavelength. For all nonmetallic materials there is some maximum wavelength below which electronexcitations across the band gap will occur. This results in the adsorption of radiation, and, consequently, the material becomes opaque." }, { "idx": 4157, "question": "Match the following material types with their light transmission characteristics: Single crystal electrical insulators - Opaque, Translucent, Transparent", "answer": "Single crystal electrical insulators are transparent." }, { "idx": 4158, "question": "Match the following material types with their light transmission characteristics: Polycrystalline and nonporous electrical insulators - Translucent, Transparent, Opaque", "answer": "Polycrystalline and nonporous electrical insulators are translucent." }, { "idx": 4159, "question": "Match the following material types with their light transmission characteristics: Porous electrical insulators - Transparent, Opaque, Translucent", "answer": "Porous electrical insulators are opaque to visible light." }, { "idx": 4160, "question": "For noncubic crystals, the index of refraction is lowest in the crystallographic direction that has the(a) Highest atomic packing density(b) Lowest atomic packing density", "answer": "For noncubic crystals, the index of refraction is lowest in the crystallographic direc�on having the lowest atomic packing density." }, { "idx": 4161, "question": "A completely amorphous and nonporous polymer will be(a) Transparent(b) Translucent(c) Opaque", "answer": "A completely amorphous and nonporous polymer will be transparent." }, { "idx": 4162, "question": "A beam of light is shined on a thin (sub-millimeter thick) single crystal wafer of material. The light source is special since it can be tuned to provide any wavelength of visible light on demand. The specimen is illuminated such that the frequency of light is decreased over time while the transmitted intensity of the light is measured. If the sample becomes transparent when the frequency is less than [F] THz, what is the band gap of the material, in eV? Assume that an intrinsic excitation of electrons is responsible for the absorption.", "answer": "the band gap of the material is 4.135 × 10^{-3} [f] ev." }, { "idx": 4163, "question": "Select the word combination that best completes this statement.\nWhen a semiconductor is exposed to a light source, its intrinsic carrier concentration will increase if the [a] of the light is [b] than band gap of the semiconductor.\n[a]: intensity, energy, wavelength, frequency, voltage, current, resistance\n[b]: greater, less", "answer": "([a] energy\n[b] greater" }, { "idx": 4164, "question": "Match the luminescence characteristics with their descriptions.\nReemission of photons occurs in much less than one second after excitation.\n- Phosphorescence\n- Fluorescence\nReemission of photons occurs in more than one second after excitation.\n- Fluorescence\n- Phosphorescence", "answer": "Fluorescence involves reemission of photons in much less than one second after excitation; while phosphorescence involves reemission of photons in more than one second after excitation." }, { "idx": 4165, "question": "Only pure materials luminesce.(a) True(b) False", "answer": "False. Luminescent materials contain impurities." }, { "idx": 4166, "question": "Cite the difference between atomic mass and atomic weight.", "answer": "Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally occurring isotopes." }, { "idx": 4167, "question": "Chromium has four naturally-occurring isotopes: 4.34% of { }^{50} \\mathrm{Cr}, with an atomic weight of 49.9460 amu, 83.79% of { }^{52} \\mathrm{Cr}, with an atomic weight of 51.9405 amu, 9.50% of { }^{53} \\mathrm{Cr}, with an atomic weight of 52.9407 amu, and 2.37% of { }^{54} \\mathrm{Cr}, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic weight of \\mathrm{Cr} is 51.9963 amu.", "answer": "the average atomic weight of \\mathrm{cr} is 51.9963 \\, amu." }, { "idx": 4168, "question": "How many grams are there in one amu of a material?", "answer": "1.66 x 10^-24 g/amu" }, { "idx": 4169, "question": "Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance?", "answer": "2.73 x 10^26 atoms/lb-mol" }, { "idx": 4170, "question": "(a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.", "answer": "Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells." }, { "idx": 4171, "question": "(b) Cite two important additional refinements that resulted from the wave-mechanical atomic model.", "answer": "Two important refinements resulting from the wave-mechanical atomic model are (1) that electron position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and subshells--each electron is characterized by four quantum numbers." }, { "idx": 4172, "question": "Relative to electrons and electron states, what does the n quantum number specify?", "answer": "The n quantum number designates the electron shell." }, { "idx": 4173, "question": "Relative to electrons and electron states, what does the l quantum number specify?", "answer": "The l quantum number designates the electron subshell." }, { "idx": 4174, "question": "Relative to electrons and electron states, what does the mj quantum number specify?", "answer": "The mj quantum number designates the number of electron states in each electron subshell." }, { "idx": 4175, "question": "Relative to electrons and electron states, what does the ms quantum number specify?", "answer": "The ms quantum number designates the spin moment on each electron." }, { "idx": 4176, "question": "For the L shell (n=2), write the four quantum numbers for all of the electrons and note which correspond to the s and p subshells.", "answer": "For the L state, n=2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and ±1; and possible ms values are ±1/2. Therefore, for the s states, the quantum numbers are 200(1/2) and 200(-1/2). For the p states, the quantum numbers are 210(1/2), 210(-1/2), 211(1/2), 211(-1/2), 21(-1)(1/2), and 21(-1)(-1/2)." }, { "idx": 4177, "question": "For the M shell (n=3), write the four quantum numbers for all of the electrons and note which correspond to the s, p, and d subshells.", "answer": "For the M state, n=3, and 18 states are possible. Possible l values are 0, 1, and 2; possible ml values are 0, ±1, and ±2; and possible ms values are ±1/2. Therefore, for the s states, the quantum numbers are 300(1/2), 300(-1/2), for the p states they are 310(1/2), 310(-1/2), 311(1/2), 311(-1/2), 31(-1)(1/2), and 31(-1)(-1/2); for the d states they are 320(1/2), 320(-1/2), 321(1/2), 321(-1/2), 32(-1)(1/2), 32(-1)(-1/2), 322(1/2), 322(-1/2), 32(-2)(1/2), and 32(-2)(-1/2)." }, { "idx": 4178, "question": "Give the electron configuration for the ion Fe2+.", "answer": "The electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6." }, { "idx": 4179, "question": "Give the electron configuration for the ion Al3+.", "answer": "The electron configuration for an Al3+ ion is 1s2 2s2 2p6." }, { "idx": 4180, "question": "Give the electron configuration for the ion Cu+.", "answer": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10." }, { "idx": 4181, "question": "Give the electron configuration for the ion Ba2+.", "answer": "The electron configuration for a Ba2+ ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6." }, { "idx": 4182, "question": "Give the electron configuration for the ion Br-.", "answer": "The electron configuration for a Br- ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6." }, { "idx": 4183, "question": "Give the electron configuration for the ion O2-.", "answer": "The electron configuration for an O2- ion is 1s2 2s2 2p6." }, { "idx": 4184, "question": "The Na+ ion has an electron structure that is identical to which inert gas?", "answer": "The Na+ ion has an electron configuration the same as neon." }, { "idx": 4185, "question": "The Cl- ion has an electron structure that is identical to which inert gas?", "answer": "The Cl- ion has an electron configuration the same as argon." }, { "idx": 4186, "question": "With regard to electron configuration, what do all the elements in Group VIIA of the periodic table have in common?", "answer": "Each of the elements in Group VIIA has five p electrons." }, { "idx": 4187, "question": "To what group in the periodic table would an element with atomic number 114 belong?", "answer": "From the periodic table the element having atomic number 114 would belong to group IVA. Ds, having an atomic number of 110 lies below Pt in the periodic table and in the rightmost column of group VIII. Moving four columns to the right puts element 114 under \\mathrm{Pb} and in group IVA." }, { "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.", "idx": 4188 }, { "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}) electron configuration is that of an inert gas because of filled (3 s) and (3 p) subshells.", "idx": 4189 }, { "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{5}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{5}) electron configuration is that of a halogen because it is one electron deficient from having a filled (L) shell.", "idx": 4190 }, { "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons.", "idx": 4191 }, { "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.", "idx": 4192 }, { "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.", "idx": 4193 }, { "idx": 4194, "question": "What electron subshell is being filled for the rare earth series of elements on the periodic table?", "answer": "The 4 f subshell is being filled for the rare earth series of elements." }, { "idx": 4195, "question": "What electron subshell is being filled for the actinide series?", "answer": "The 5 f subshell is being filled for the actinide series of elements." }, { "idx": 4196, "question": "Calculate the force of attraction between a K^{+}and an \\mathrm{O}^{2-} ion the centers of which are separated by a distance of 1.5nm.", "answer": "the force of attraction between the k^{+} and \\mathrm{o}^{2-} ions is 2.05 × 10^{-10} N." }, { "idx": 4197, "question": "Briefly cite the main differences between ionic, covalent, and metallic bonding.", "answer": "The main differences between the various forms of primary bonding are: Ionic--there is electrostatic attraction between oppositely charged ions. Covalent--there is electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration. Metallic--the positively charged ion cores are shielded from one another, and also 'glued' together by the sea of valence electrons." }, { "idx": 4198, "question": "State the Pauli exclusion principle.", "answer": "The Pauli exclusion principle states that each electron state can hold no more than two electrons, which must have opposite spins." }, { "idx": 4199, "question": "What type(s) of bonding would be expected for brass (a copper-zinc alloy)?", "answer": "For brass, the bonding is metallic since it is a metal alloy." }, { "idx": 4200, "question": "What type(s) of bonding would be expected for rubber?", "answer": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)" }, { "idx": 4201, "question": "What type(s) of bonding would be expected for barium sulfide (BaS)?", "answer": "For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table." }, { "idx": 4202, "question": "What type(s) of bonding would be expected for solid xenon?", "answer": "For solid xenon, the bonding is van der Waals since xenon is an inert gas." }, { "idx": 4203, "question": "What type(s) of bonding would be expected for bronze?", "answer": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin)." }, { "idx": 4204, "question": "What type(s) of bonding would be expected for nylon?", "answer": "For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)" }, { "idx": 4205, "question": "What type(s) of bonding would be expected for aluminum phosphide (AlP)?", "answer": "For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table." }, { "idx": 4206, "question": "What is the difference between atomic structure and crystal structure?", "answer": "Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material." }, { "idx": 4207, "question": "If the atomic radius of aluminum is 0.143nm, calculate the volume of its unit cell in cubic meters.", "answer": "the volume of the unit cell is 6.62 × 10^{-29} m^{3}." }, { "idx": 4208, "question": "Show that the atomic packing factor for BCC is 0.68 .", "answer": "the atomic packing factor for bcc is 0.68." }, { "idx": 4209, "question": "Iron has a BCC crystal structure, an atomic radius of 0.124nm, and an atomic weight of 55.85g / mol. Compute and compare its theoretical density with the experimental value found inside the front cover.", "answer": "the theoretical density of iron is 7.90g / {cm}^{3}. the experimental value given inside the front cover is 7.87g / {cm}^{3}." }, { "idx": 4210, "question": "Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4g / {cm}^{3}, and an atomic weight of 192.2g / mol.", "answer": "the radius of an iridium atom is 0.136nm." }, { "idx": 4211, "question": "Calculate the radius of a vanadium atom, given thatV has a BCC crystal structure, a density of 5.96 \\mathrm{g} / {cm}^{3}, and an atomic weight of 50.9g / mol.", "answer": "the radius of a vanadium atom is 0.132 nm." }, { "idx": 4212, "question": "For alloy A with atomic weight 77.4 g/mol, density 8.22 g/cm³, and atomic radius 0.125 mm, determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination.", "answer": "for alloy a, its crystal structure is simple cubic." }, { "idx": 4213, "question": "For alloy B with atomic weight 107.6 g/mol, density 13.42 g/cm³, and atomic radius 0.133 mm, determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination.", "answer": "for alloy b, its crystal structure is fcc." }, { "idx": 4214, "question": "For alloy C with atomic weight 127.3 g/mol, density 9.23 g/cm³, and atomic radius 0.142 mm, determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination.", "answer": "for alloy c, its crystal structure is simple cubic." }, { "idx": 4215, "question": "The unit cell for tin has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318nm, respectively. If its density, atomic weight, and atomic radius are 7.30g / {cm}^{3}, 118.69g / mol, and 0.151nm, respectively, compute the atomic packing factor.", "answer": "the atomic packing factor (apf) for tin is 0.534." }, { "idx": 4216, "question": "List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure.", "answer": " the barium ions are situated at all corner positions. The point coordinates for these ions are as follows: 000,100,110,010,001,101,111, and 011 .\nThe oxygen ions are located at all face-centered positions; therefore, their coordinates are \\frac{1}{2} \\frac{1}{2} 0, \\frac{1}{2} \\frac{1}{2} 1, \\frac{1}{2} \\frac{1}{2}, \\frac{0}{2} \\frac{1}{2}, \\frac{1}{2} 0 \\frac{1}{2}, and \\frac{1}{2} 1 \\frac{1}{2}.\nAnd, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates \\frac{1}{2} \\frac{1}{2} \\frac{1}{2}." }, { "idx": 4217, "question": "List the point coordinates of all carbon atoms that occupy the corner positions of the diamond cubic unit cell.", "answer": "The coordinates of these atoms are as follows: 000, 100, 110, 010, 001, 101, 111, and 011." }, { "idx": 4218, "question": "List the point coordinates of all carbon atoms that reside on the face-centered positions of the diamond cubic unit cell.", "answer": "The coordinates of these atoms are as follows: 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, and 1/2 1/2 1/2." }, { "idx": 4219, "question": "List the point coordinates of all carbon atoms that are positioned within the interior of the diamond cubic unit cell.", "answer": "The coordinates of these atoms are as follows: 3/4 1/4 1/4, 1/4 3/4 3/4, 1/4 1/4 3/4, and 3/4 3/4 1/4." }, { "idx": 4220, "question": "Using the Molecule Definition Utility found in both \"Metallic Crystal Structures and Crystallography\" and \"Ceramic Crystal Structures\" modules of VMSE, located on the book's web site [www.wiley.com/college/Callister (Student Companion Site)], generate a three-dimensional unit cell for the intermetallic compound \\mathrm{AuCu}_{3} given the following: (1) the unit cell is cubic with an edge length of 0.374nm, (2) gold atoms are situated at all cube corners, and (3) copper atoms are positioned at the centers of all unit cell faces.", "answer": "First of all, open the \"Molecular Definition Utility\"; it may be found in either of \"Metallic Crystal Structures and Crystallography\" or \"Ceramic Crystal Structures\" modules.\nIn the \"Step 1\" window, it is necessary to define the atom types, colors for the spheres (atoms), and specify atom sizes. Let us enter \"Au\" as the name for the gold atoms (since \"Au\" the symbol for gold), and \"Cu\" as the name for the copper atoms. Next it is necessary to choose a color for each atom type from the selections that appear in the pull-down menu-for example, \"Yellow\" for Au and \"Red\" for Cu. In the \"Atom Size\" window, it is necessary to enter an atom/ion size. In the instructions for this step, it is suggested that the atom/ion diameter in nanometers be used. From the table found inside the front cover of the textbook, the atomic radii for gold and copper are 0.144nm and 0.128nm, respectively, and, therefore, their ionic diameters are twice these values (i.e., 0.288nm and 0.256nm ); therefore, we enter the values \" 0.288 \" and \" 0.256 \" for the two atom types. Now click on the \"Register\" button, followed by clicking on the \"Go to Step 2\" button.\nIn the \"Step 2\" window we specify positions for all of the atoms within the unit cell; their point coordinates are specified in the problem statement. Let's begin with gold. Click on the yellow sphere that is located to the right of the \"Molecule Definition Utility\" box. Again, Au atoms are situated at all eight corners of the cubic unit cell. One Au will be positioned at the origin of the coordinate system-i.e., its point coordinates are 000, and, therefore, we enter a \"0\" (zero) in each of the \" \\mathrm{x} \", \" \\mathrm{y} \", and \" \\mathrm{z} \" atom position boxes. Next we click on the \"Register Atom Position\" button. Now we enter the coordinates of another gold atom; let us arbitrarily select the one that resides at the corner of the unit cell that is one unit-cell length along the x-axis (i.e., at the 100 point coordinate). Inasmuch as it is located a distance of a units along the x-axis the value of \" 0.374 \" is entered in the \" \\mathrm{x} \" atom position box (since this is the value of a given in the problem statement); zeros are entered in each of the \" \\mathrm{y} \" and \" z \" position boxes. We repeat this procedure for the remaining six Au atoms.\nAfter this step has been completed, it is necessary to specify positions for the copper atoms, which are located at all six face-centered sites. To begin, we click on the red sphere that is located next to the \"Molecule Definition Utility\" box. The point coordinates for some of the Cu atoms are fractional ones; in these instances, the a unit cell length (i.e., 0.374) is multiplied by the fraction. For example, one Cu atom is located 1 \\frac{1}{2} \\frac{1}{2} coordinate. Therefore, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atoms positions are (1)(0.374) =0.374, \\frac{1}{2}(0.374)=0.187, and \\frac{1}{2}(0.374)=0.187, respectively.\nFor the gold atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z}^{\\prime} atom position entries for all 8 sets of point coordinates are as follows: \\[\n\\begin{array}{l}\n0,0, \\text { and } 0 \\\\\n0.374,0, \\text { and } 0 \\\\\n0,0.374, \\text { and } 0 \\\\\n0,0, \\text { and } 0.374 \\\\\n0,0.374,0.374 \\\\\n0.374,0.374 \\\\\n0.374,0 .374,0 \\\\\n0.374,0.374,0.374\n\\end{array}\n\\]\nNow, for the copper atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atom position entries for all 6 sets of point coordinates are as follows:\n\\[\n\\begin{array}{l}\n0.187,0.187,0 \\\\\n0.187,0,0.187 \\\\\n0,0.187,0.187 \\\\\n0.374,0.187,0.187 \\\\\n0.187,0.374,0.187 \\\\\n0.187,0.187,0.374\n\\end{array}\n\\]\nIn Step 3, we may specify which atoms are to be represented as being bonded to one another, and which type of bond(s) to use (single solid, single dashed, double, and triple are possibilities), or we may elect to not represent any bonds at all (in which case we are finished). If it is decided to show bonds, probably the best thing to do is to represent unit cell edges as bonds. This image may be rotated by using mouse click-and-drag\nYour image should appear as the following screen shot. Here the gold atoms appear lighter than the copper atoms. [Note: Unfortunately, with this version of the Molecular Definition Utility, it is not possible to save either the data or the image that you have generated. You may use screen capture (or screen shot) software to record and store your image.]" }, { "idx": 4221, "question": "Convert the (010) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.", "answer": "the (010) plane in the four-index miller-bravais scheme is (01 -1 0)." }, { "idx": 4222, "question": "Convert the (101) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.", "answer": "the (101) plane in the four-index miller-bravais scheme is (10 -1 1)." }, { "idx": 4223, "question": "(a) Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius R.", "answer": "the planar density expression for the hcp (0001) plane in terms of the atomic radius r is: pd_{0001} = 1 / (2 r^{2} sqrt(3))" }, { "idx": 4224, "question": "(b) Compute the planar density value for this same plane for magnesium.", "answer": "the planar density value for the (0001) plane for magnesium is: pd_{0001}(mg) = 1.128 × 10^{19} m^{-2}" }, { "idx": 4225, "question": "Explain why the properties of polycrystalline materials are most often isotropic.", "answer": "Although each individual grain in a polycrystalline material may be anisotropic, if the grains have random orientations, then the solid aggregate of the many anisotropic grains will behave isotropically." }, { "idx": 4226, "question": "The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at 69.22 degrees (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute the interplanar spacing for this set of planes.", "answer": "the interplanar spacing for the (220) set of planes for iridium is 0.1357 nm." }, { "idx": 4227, "question": "The metal iridium has an FCC crystal structure. Given the interplanar spacing for the (220) set of planes is 0.1357 nm, compute the atomic radius for an iridium atom.", "answer": "the atomic radius for an iridium atom is 0.1357 nm." }, { "idx": 4228, "question": "Which of these elements would you expect to form with copper a substitutional solid solution having complete solubility? Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.", "answer": "Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures." }, { "idx": 4229, "question": "Which of these elements would you expect to form with copper a substitutional solid solution of incomplete solubility? Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.", "answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+." }, { "idx": 4230, "question": "Which of these elements would you expect to form with copper an interstitial solid solution? Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.", "answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu." }, { "idx": 4231, "question": "What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu ?", "answer": "the composition of the alloy is 29.4 \\, \\text{at}% \\, \\mathrm{zn} and 70.6 \\, \\text{at}% \\, \\mathrm{cu}." }, { "idx": 4232, "question": "What is the composition, in weight percent, of an alloy that consists of 6 at% \\mathrm{Pb} and 94 \\mathrm{at}% \\mathrm{Sn} ?", "answer": "the composition of the alloy is: 1. 10.0 \\, \\text{wt% pb} 2. 90.0 \\, \\text{wt% sn}" }, { "idx": 4233, "question": "What is the composition, in atom percent, of an alloy that contains 99.7 lb_{m} copper, 102 lb_{m} zinc, and 2.1 lb_{m} lead?\n", "answer": "the composition of the alloy in atom percent is: \\[\nc_{\\mathrm{cu}} = 50.0 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{zn}} = 49.7 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{pb}} = 0.3 \\, \\text{at}%\n\\]" }, { "idx": 4234, "question": "What is the composition, in atom percent, of an alloy that consists of 97 wt% Fe and 3 wt% Si?", "answer": "the composition of the alloy is 94.2 at% fe and 5.8 at% si." }, { "idx": 4235, "question": "Calculate the number of atoms per cubic meter in aluminum.", "answer": "6.05 × 10^{28} atoms/m^3" }, { "idx": 4236, "question": "The concentration of carbon in an iron-carbon alloy is 0.15 wt%. What is the concentration in kilograms of carbon per cubic meter of alloy?", "answer": "the concentration of carbon in the alloy is 11.8 \\mathrm{kg/m^{3}}." }, { "idx": 4237, "question": "Cite the relative Burgers vector-dislocation line orientations for edge, screw, and mixed dislocations.", "answer": "The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations." }, { "idx": 4238, "question": "For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (111) plane? Why? ", "answer": "The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] - that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.54, planar densities for FCC (100) and (111) planes are \\frac{1}{4 R^{2}} and \\frac{1}{2 R^{2} \\sqrt{3}}, respectively - that is \\frac{0.25}{R^{2}} and \\frac{0.29}{R^{2}} (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the lower surface energy." }, { "idx": 4239, "question": "For a BCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (110) plane? Why? ", "answer": "The surface energy for a crystallographic plane will depend on its packing density-that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.55, the planar densities for BCC (100) and (110) are \\frac{3}{16 R^{2}} and \\frac{3}{8 R^{2} \\sqrt{2}}, respectively-that is \\frac{0.19}{R^{2}} and \\frac{0.27}{R^{2}}. Thus, since the planar density for (110) is greater, it will have the lower surface energy." }, { "idx": 4240, "question": "For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why?", "answer": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary." }, { "idx": 4241, "question": "The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?", "answer": "The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds." }, { "idx": 4242, "question": "(a) Briefly describe a twin and a twin boundary.", "answer": "A twin boundary is an interface such that atoms on one side are located at mirror image positions of those atoms situated on the other boundary side. The region on one side of this boundary is called a twin." }, { "idx": 4243, "question": "(b) Cite the difference between mechanical and annealing twins.", "answer": "Mechanical twins are produced as a result of mechanical deformation and generally occur in BCC and HCP metals. Annealing twins form during annealing heat treatments, most often in FCC metals." }, { "idx": 4244, "question": "Briefly explain the concept of steady state as it applies to diffusion.", "answer": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time." }, { "idx": 4245, "question": "(a) Briefly explain the concept of a driving force.", "answer": "The driving force is that which compels a reaction to occur." }, { "idx": 4246, "question": "(b) What is the driving force for steady-state diffusion?", "answer": "The driving force for steady-state diffusion is the concentration gradient." }, { "idx": 4247, "question": "For a steel alloy it has been determined that a carburizing heat treatment of 10-h duration will raise the carbon concentration to 0.45 wt% at a point 2.5mm from the surface. Estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.", "answer": "the time necessary to achieve a carbon concentration of 0.45 \\text{wt}% at a 5.0 \\text{mm} position is 40 \\text{h}." }, { "idx": 4248, "question": "The preexponential and activation energy for the diffusion of iron in cobalt are 1.1 × 10^{-5}{m}^{2} / s and 253,300 J/ mol, respectively. At what temperature will the diffusion coefficient have a value of 2.1 × 10^{-14}{m}^{2} / s ?", "answer": "the temperature at which the diffusion coefficient has a value of 2.1 × 10^{-14} m^{2}/s is 1518 k (1245^{\\circ} C). alternatively, using the vmse software, the temperature is found to be 1519 k." }, { "idx": 4249, "question": "The activation energy for the diffusion of carbon in chromium is 111,000 J/ mol. Calculate the diffusion coefficient at 1100 K\\left(827^{\\circ} C\\right), given that D at 1400 K\\left(1127^{\\circ} C\\right) is 6.25 × 10^{-11}{m}^{2} / s.", "answer": "the diffusion coefficient at 1100k is 4.6 × 10^{-12}{m}^{2} / s." }, { "idx": 4250, "question": "The diffusion coefficients for iron in nickel are given at two temperatures: 1273 K with D = 9.4 x 10^-16 m^2/s and 1473 K with D = 2.4 x 10^-14 m^2/s. Determine the values of D0 and the activation energy Qd.", "answer": "the values are d0 = 2.2 x 10^-5 m^2/s and the activation energy qd = 252,400 j/mol." }, { "idx": 4251, "question": "The diffusion coefficients for iron in nickel are given at two temperatures: 1273 K with D = 9.4 x 10^-16 m^2/s and 1473 K with D = 2.4 x 10^-14 m^2/s. What is the magnitude of D at 1100°C (1373 K)?", "answer": "the magnitude of d at 1100°c (1373 k) is 5.4 x 10^-15 m^2/s." }, { "idx": 4252, "question": "The diffusion coefficients for silver in copper are given at two temperatures: T(°C) = 650, D(m2/s) = 5.5 × 10^-16 and T(°C) = 900, D(m2/s) = 1.3 × 10^-13. Determine the values of D0 and Qd.", "answer": "the values are d0 = 7.5 × 10^-5 m2/s and qd = 196,700 j/mol." }, { "idx": 4253, "question": "The diffusion coefficients for silver in copper are given at two temperatures: T(°C) = 650, D(m2/s) = 5.5 × 10^-16 and T(°C) = 900, D(m2/s) = 1.3 × 10^-13. What is the magnitude of D at 875°C?", "answer": "the magnitude of d at 875°c is 8.3 × 10^-14 m2/s." }, { "idx": 4254, "question": "Carbon is allowed to diffuse through a steel plate 15mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30kg C / m^{2} Fe, which are maintained constant. If the preexponential and activation energy are 6.2 × 10^{-7}{m}^{2} / s and 80,000 J/ mol, respectively, compute the temperature at which the diffusion flux is 1.43 × 10^{-9}kg / m^{2}-s.", "answer": "the temperature at which the diffusion flux is 1.43 × 10^{-9}kg / m^{2}-s is 1044k or 771^{\\circ} C." }, { "idx": 4255, "question": "The steady-state diffusion flux through a metal plate is 5.4 × 10^{-10}kg / m^{2}-s at a temperature of 727^{\\circ} C (1000 K) and when the concentration gradient is -350kg / m^{4}. Calculate the diffusion flux at 1027^{\\circ} C(1300 K) for the same concentration gradient and assuming an activation energy for diffusion of 125,000 J/ mol.", "answer": "the diffusion flux at 1027^{\\circ} C (1300 k) is 1.74 × 10^{-8} \\mathrm{kg/m^{2}-s}." }, { "idx": 4256, "question": "For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 x 10^6 psi). What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm^2 (0.5 in.^2) without plastic deformation?", "answer": "the maximum load that may be applied without plastic deformation is 89,375 n (20,000 lb_f)." }, { "idx": 4257, "question": "For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 x 10^6 psi). If the original specimen length is 115mm (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation?", "answer": "the maximum length to which the specimen may be stretched without causing plastic deformation is 115.28mm (4.51 in)." }, { "idx": 4258, "question": "A cylindrical rod of copper (E=110g P a, 16 × 10^{6} psi) having a yield strength of 240 MPa(35,000 psi) is to be subjected to a load of 6660N\\left(1500 lb p\\right). If the length of the rod is 380 mm(15.0 in.), what must be the diameter to allow an elongation of 0.50 mm(0.020 in.) ?", "answer": "the diameter must be 7.65mm (0.30 in)." }, { "idx": 4259, "question": "Compute the elastic modulus for titanium, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?", "answer": "(a) the elastic modulus for titanium is 106.4 gpa, which is in very good agreement with the value of 107 gpa from table 6.1." }, { "idx": 4260, "question": "Compute the elastic modulus for tempered steel, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?", "answer": "(b) the elastic modulus for tempered steel is 204.7 gpa, which is in reasonably good agreement with the value of 207 gpa from table 6.1." }, { "idx": 4261, "question": "Compute the elastic modulus for aluminum, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?", "answer": "(c) the elastic modulus for aluminum is 69.1 gpa, which is in excellent agreement with the value of 69 gpa from table 6.1." }, { "idx": 4262, "question": "Compute the elastic modulus for carbon steel, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?", "answer": "(d) the elastic modulus for carbon steel is 215 gpa, which is in reasonable agreement with the value of 207 gpa from table 6.1." }, { "idx": 4263, "question": "Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 8.0 mm(0.31 in.). A tensile force of 1000N\\left(225 lb_{\\mathrm{f}}\\right) produces an elastic reduction in diameter of 2.8 × 10^{-4} mm\\left(1.10 × 10^{-5} in\\right..). Compute the modulus of elasticity for this alloy, given that Poisson's ratio is 0.30 .", "answer": "the modulus of elasticity for this alloy is 1.705 × 10^{11} pa or 170.5 gpa." }, { "idx": 4264, "question": "A cylindrical rod 100mm long and having a diameter of 10.0mm is to be deformed using a tensile load of 27,500 N. It must not experience plastic deformation. Of the materials listed as follows, which are possible candidates? Justify your choice(s).", "answer": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion of not experiencing plastic deformation." }, { "idx": 4265, "question": "A cylindrical rod 100mm long and having a diameter of 10.0mm is to be deformed using a tensile load of 27,500 N. It must not experience a diameter reduction of more than 7.5 x 10^-3 mm. Of the materials listed as follows, which are possible candidates? Justify your choice(s).", "answer": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion regarding diameter reduction. The titanium alloy is not a candidate as it fails this criterion." }, { "idx": 4266, "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60mm (0.260 in.), and the fractured gauge length is 72.14mm (2.840 in.). Calculate the ductility in terms of percent reduction in area.", "answer": "the ductility in terms of percent reduction in area is 73.4%." }, { "idx": 4267, "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60mm (0.260 in.), and the fractured gauge length is 72.14mm (2.840 in.). Calculate the ductility in terms of percent elongation.", "answer": "the ductility in terms of percent elongation is 42%." }, { "idx": 4268, "question": "For some metal alloy, a true stress of 415 MPa(60,175 psi) produces a plastic true strain of 0.475 . How much will a specimen of this material elongate when a true stress of 325 MPa(46,125 psi) is applied if the original length is 300mm (11.8 in.)? Assume a value of 0.25 for the strain-hardening exponent n.", "answer": "the specimen will elongate by 58.8 mm (2.31 \\text{ in})." }, { "idx": 4269, "question": "Cite five factors that lead to scatter in measured material properties.", "answer": "The five factors that lead to scatter in measured material properties are the following: (1) test method; (2) variation in specimen fabrication procedure; (3) operator bias; (4) apparatus calibration; and (5) material inhomogeneities and/or compositional differences." }, { "idx": 4270, "question": "To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 10^4 mm^-2. Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end. How far (in miles) would this chain extend?", "answer": "for a dislocation density of 10^4 mm^-2, the chain would extend 6.2 miles." }, { "idx": 4271, "question": "Now suppose that the density is increased to 10^10 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?", "answer": "for a dislocation density of 10^10 mm^-2, the chain would extend 6.2 x 10^6 miles." }, { "idx": 4272, "question": "For edge dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.", "answer": "edge dislocation--parallel" }, { "idx": 4273, "question": "For screw dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.", "answer": "screw dislocation--perpendicular" }, { "idx": 4274, "question": "For mixed dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.", "answer": "mixed dislocation--neither parallel nor perpendicular" }, { "idx": 4275, "question": "Define a slip system.", "answer": "A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs." }, { "idx": 4276, "question": "Do all metals have the same slip system? Why or why not?", "answer": "All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure." }, { "idx": 4277, "question": "Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1^{\\circ} and 47.9^{\\circ}, respectively, with the tensile axis. If the critical resolved shear stress is 20.7 MPa (3000 psi), will an applied stress of 45 MPa(6500 psi) cause the single crystal to yield? If not, what stress will be necessary?", "answer": "the resolved shear stress is 22.0 MPa (3181 psi). since this is greater than the critical resolved shear stress of 20.7 MPa (3000 psi), the single crystal will yield." }, { "idx": 4278, "question": "A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of 28.1 degrees with the tensile axis. Three possible slip directions make angles of 62.4 degrees, 72.0 degrees, and 81.1 degrees with the same tensile axis. Which of these three slip directions is most favored?", "answer": "the most favored slip direction is at an angle of 62.4 degrees with the tensile axis." }, { "idx": 4279, "question": "A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of 28.1 degrees with the tensile axis. The most favored slip direction makes an angle of 62.4 degrees with the tensile axis. If plastic deformation begins at a tensile stress of 1.95 MPa (280 psi), determine the critical resolved shear stress for aluminum.", "answer": "the critical resolved shear stress for aluminum is 0.80 mpa (114 psi)." }, { "idx": 4280, "question": "The critical resolved shear stress for iron is 27 MPa(4000 psi). Determine the maximum possible yield strength for a single crystal of \\mathrm{Fe} pulled in tension.", "answer": "the maximum possible yield strength for a single crystal of \\mathrm{fe} pulled in tension is 54 MPa (8000 psi)." }, { "idx": 4281, "question": "What is the difference between deformation by twinning and deformation by slip relative to mechanism?", "answer": "With slip deformation there is no crystallographic reorientation, whereas with twinning there is a reorientation." }, { "idx": 4282, "question": "What is the difference between deformation by twinning and deformation by slip relative to conditions of occurrence?", "answer": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems." }, { "idx": 4283, "question": "What is the difference between deformation by twinning and deformation by slip relative to final result?", "answer": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning." }, { "idx": 4284, "question": "Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.", "answer": "Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction." }, { "idx": 4285, "question": "Briefly explain why HCP metals are typically more brittle than FCC and BCC metals.", "answer": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP." }, { "idx": 4286, "question": "Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16mm and 11 mm, respectively. The second specimen, with an initial radius of 12 mm, must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.", "answer": "the second specimen's radius after deformation is 8.25 mm." }, { "idx": 4287, "question": "Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows:\n\\begin{tabular}{ccc}\n\\hline & Circular (diameter, \\mathbf{m m} ) & Rectangular (\\mathbf{m m}) \\\\\n\\hline Original dimensions & 15.2 & 125 × 175 \\\\\nDeformed dimensions & 11.4 & 75 × 200 \\\\\n\\hline\n\\end{tabular}\nWhich of these specimens will be the hardest after plastic deformation, and why?", "answer": "the hardest specimen will be the deformed circular one with a cold work percentage of 43.8%." }, { "idx": 4288, "question": "A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. If its cold-worked radius is 10mm (0.40 in.), what was its radius before deformation?", "answer": "the radius before deformation was 10.6mm (0.424 in.)." }, { "idx": 4289, "question": "Explain the differences in grain structure for a metal that has been cold worked and one that has been cold worked and then recrystallized.", "answer": "During cold-working, the grain structure of the metal has been distorted to accommodate the deformation. Recrystallization produces grains that are equiaxed and smaller than the parent grains." }, { "idx": 4290, "question": "What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 × 10^{-4} mm\\left(10^{-5} in.\\right) and a crack length of 2.5 × 10^{-2} mm\\left(10^{-3} in.\\right) when a tensile stress of 170 MPa(25,000 psi) is applied?", "answer": "the magnitude of the maximum stress at the tip of the internal crack is 2404 \\text{ mpa} (354,000 \\text{ psi})." }, { "idx": 4291, "question": "Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.25 mm(0.01 in.) and having a tip radius of curvature of 1.2 × 10^{-3} mm\\left(4.7 × 10^{-5} in\\right..) when a stress of 1200 MPa(174,000 psi) is applied.", "answer": "the theoretical fracture strength is 3.5 × 10^{4} MPa (5.1 × 10^{6} psi)." }, { "idx": 4292, "question": "If the specific surface energy for soda-lime glass is 0.30 J/ m^{2}, using data contained in Table 12.5, compute the critical stress required for the propagation of a surface crack of length 0.05 mm.", "answer": "the critical stress required for the propagation of a surface crack of length 0.05 mm is 16.2 MPa." }, { "idx": 4293, "question": "A polystyrene component must not fail when a tensile stress of 1.25 MPa(180 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/ m^{2}\\left(2.86 × 10^{-3}\\right. in.-\\left.lb_{\\mathrm{f}} / in^{2}\\right). Assume a modulus of elasticity of 3.0 \\mathrm{GPa}\\left(0.435 × 10^{6} psi\\right).", "answer": "the maximum allowable surface crack length is 6.1 × 10^{-4} \\, m (0.61mm or 0.024 in.)." }, { "idx": 4294, "question": "A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPa \\sqrt{m}(41 ksi \\sqrt{in}.) is exposed to a stress of 1000 MPa(145,000 psi). Will this specimen experience fracture if it is known that the largest surface crack is 0.75 mm(0.03 in.) long? Why or why not? Assume that the parameter Y has a value of 1.0 .", "answer": "the specimen will experience fracture because it can only tolerate a stress of 927 \\text{mpa} (133,500 \\text{psi}) before fracture, which is less than the applied stress of 1000 \\text{mpa} (145,000 \\text{psi})." }, { "idx": 4295, "question": "Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa \\sqrt{m} (31.9 ksi \\sqrt{m}. It has been determined that fracture results at a stress of 250 MPa (36,250 psi) when the maximum (or critical) internal crack length is 2.0 mm(0.08 in.). For this same component and alloy, will fracture occur at a stress level of 325 MPa(47,125 psi) when the maximum internal crack length is 1.0 mm(0.04 in).? Why or why not?", "answer": "fracture will not occur since the value 32.2 MPa \\sqrt{m} is less than the fracture toughness k_{ic} of the material, 35 MPa \\sqrt{m}." }, { "idx": 4296, "question": "Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa \\sqrt{m} (36.4 k \\left.\\mathrm{si} \\sqrt{m}\\right). It has been determined that fracture results at a stress of 365 MPa(53,000 psi) when the maximum internal crack length is 2.5 mm(0.10 in.). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm(0.16 in.).", "answer": "the stress level at which fracture will occur for a critical internal crack length of 4.0mm is 288 MPa (41,500 psi)." }, { "idx": 4297, "question": "A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa \\sqrt{m}(50 ksi \\sqrt{in}.). If, during service use, the plate is exposed to a tensile stress of 200 MPa(29,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.", "answer": "the minimum length of a surface crack that will lead to fracture is 0.024 \\, m (24 mm)." }, { "idx": 4298, "question": "Calculate the maximum internal crack length allowable for a 7075-T651 aluminum alloy (Table 8.1) component that is loaded to a stress one half of its yield strength. Assume that the value of Y is 1.35 .", "answer": "the maximum internal crack length allowable is 0.0033 \\, m (3.3 mm)." }, { "idx": 4299, "question": "A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 77.0 MPa \\sqrt{m}(70.1 ksi \\sqrt{m}.) and a yield strength of 1400 MPa(205,000 psi). The flaw size resolution limit of the flaw detection apparatus is 4.0 mm(0.16 in.) If the design stress is one half of the yield strength and the value of Y is 1.0 , determine whether or not a critical flaw for this plate is subject to detection.", "answer": "the critical flaw size a_c is 3.9 mm. therefore, the critical flaw is not subject to detection since it is less than the 4.0 mm resolution limit." }, { "idx": 4300, "question": "Cite five factors that may lead to scatter in fatigue life data.", "answer": "Five factors that lead to scatter in fatigue life data are (1) specimen fabrication and surface preparation, (2) metallurgical variables, (3) specimen alignment in the test apparatus, (4) variation in mean stress, and (5) variation in test cycle frequency." }, { "idx": 4301, "question": "Briefly explain the difference between fatigue striations and beachmarks in terms of size.", "answer": "With regard to size, beachmarks are normally of macroscopic dimensions and may be observed with the naked eye; fatigue striations are of microscopic size and it is necessary to observe them using electron microscopy." }, { "idx": 4302, "question": "Briefly explain the difference between fatigue striations and beachmarks in terms of origin.", "answer": "With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue striation corresponds to the advance of a fatigue crack during a single load cycle." }, { "idx": 4303, "question": "List four measures that may be taken to increase the resistance to fatigue of a metal alloy.", "answer": "Four measures that may be taken to increase the fatigue resistance of a metal alloy are:\n(1) Polish the surface to remove stress amplification sites.\n(2) Reduce the number of internal defects (pores, etc.) by means of altering processing and fabrication techniques.\n(3) Modify the design to eliminate notches and sudden contour changes.\n(4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot peening." }, { "idx": 4304, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for nickel.", "answer": "for ni: 418 degrees c (785 degrees f)" }, { "idx": 4305, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for copper.", "answer": "for cu: 270 degrees c (518 degrees f)" }, { "idx": 4306, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for iron.", "answer": "for fe: 450 degrees c (845 degrees f)" }, { "idx": 4307, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for tungsten.", "answer": "for w: 1200 degrees c (2190 degrees f)" }, { "idx": 4308, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for lead.", "answer": "for pb: -33 degrees c (-27 degrees f)" }, { "idx": 4309, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for aluminum.", "answer": "for al: 100 degrees c (212 degrees f)" }, { "idx": 4310, "question": "Cite three variables that determine the microstructure of an alloy.", "answer": "Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy." }, { "idx": 4311, "question": "What thermodynamic condition must be met for a state of equilibrium to exist?", "answer": "In order for a system to exist in a state of equilibrium the free energy must be a minimum for some specified combination of temperature, pressure, and composition." }, { "idx": 4312, "question": "For alloys of two hypothetical metals A and B, there exist an α, A-rich phase and a β, B-rich phase. From the mass fractions of both phases for two different alloys provided in the table below, (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for the α phase at this temperature.", "answer": "the composition of the phase boundary for the α phase is 90 wt% A-10 wt% B." }, { "idx": 4313, "question": "For alloys of two hypothetical metals A and B, there exist an α, A-rich phase and a β, B-rich phase. From the mass fractions of both phases for two different alloys provided in the table below, (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for the β phase at this temperature.", "answer": "the composition of the phase boundary for the β phase is 20.2 wt% A-79.8 wt% B." }, { "idx": 4314, "question": "A hypothetical A-B alloy of composition 55 wt% B-45 wt% A at some temperature is found to consist of mass fractions of 0.5 for both \\alpha and \\beta phases. If the composition of the \\beta phase is 90 wt% \\mathrm{~B}-10 wt% \\mathrm{~A}, what is the composition of the \\alpha phase?", "answer": "the composition of the \\alpha phase is 20 \\text{ wt}% \\text{ b- } 80 \\text{ wt}% \\text{ a}." }, { "idx": 4315, "question": "For 11.20 kg of a magnesium-lead alloy of composition 30 wt% Pb-70 wt% Mg, is it possible, at equilibrium, to have α and Mg2Pb phases having respective masses of 7.39 kg and 3.81 kg?", "answer": "yes, it is possible to have a 30 wt% pb-70 wt% mg alloy with masses of 7.39 kg and 3.81 kg for the α and mg2pb phases, respectively." }, { "idx": 4316, "question": "What will be the approximate temperature of the alloy?", "answer": "the approximate temperature of the alloy is 190 degrees c." }, { "idx": 4317, "question": "Briefly describe the phenomenon of coring and why it occurs.", "answer": "Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys, with higher concentrations of the component having the lower melting temperature at the grain boundaries. It occurs, during solidification, as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase." }, { "idx": 4318, "question": "Cite one undesirable consequence of coring.", "answer": "One undesirable consequence of a cored structure is that, upon heating, the grain boundary regions will melt first and at a temperature below the equilibrium phase boundary from the phase diagram; this melting results in a loss in mechanical integrity of the alloy." }, { "idx": 4319, "question": "Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases.", "answer": "Upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers\nof the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum." }, { "idx": 4320, "question": "What is the difference between a phase and a microconstituent?", "answer": "A \"phase\" is a homogeneous portion of the system having uniform physical and chemical characteristics, whereas a \"microconstituent\" is an identifiable element of the microstructure (that may consist of more than one phase)." }, { "idx": 4321, "question": "Two intermetallic compounds, A B and A B_{2}, exist for elements A and B. If the compositions for A B and A B_{2} are 34.3 wt% \\mathrm{~A}-65.7 wt% \\mathrm{~B} and 20.7 wt% \\mathrm{~A}-79.3 wt% \\mathrm{~B}, respectively, and element A is potassium, identify element B.", "answer": "element b is arsenic (as), with an atomic weight of 74.92 \\, \\text{g/mol} . the two intermetallic compounds are \\text{kas} and \\text{kas}_2 ." }, { "idx": 4322, "question": "What is the principal difference between congruent and incongruent phase transformations?", "answer": "The principal difference between congruent and incongruent phase transformations is that for congruent no compositional changes occur with any of the phases that are involved in the transformation. For incongruent there will be compositional alterations of the phases." }, { "idx": 4323, "question": "Compute the mass fraction of α ferrite in pearlite.", "answer": "the mass fraction of α ferrite in pearlite is 0.89." }, { "idx": 4324, "question": "Compute the mass fraction of cementite in pearlite.", "answer": "the mass fraction of cementite in pearlite is 0.11." }, { "idx": 4325, "question": "What is the distinction between hypoeutectoid and hypereutectoid steels?", "answer": "A \"hypoeutectoid\" steel has a carbon concentration less than the eutectoid; on the other hand, a \"hypereutectoid\" steel has a carbon content greater than the eutectoid." }, { "idx": 4326, "question": "In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each?", "answer": "For a hypoeutectoid steel, the proeutectoid ferrite is a microconstituent that formed above the eutectoid temperature. The eutectoid ferrite is one of the constituents of pearlite that formed at a temperature below the eutectoid. The carbon concentration for both ferrites is 0.022 wt% C." }, { "idx": 4327, "question": "What is the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.94 ?", "answer": "the carbon concentration of the iron-carbon alloy is 0.42 \\text{ wt% c}." }, { "idx": 4328, "question": "What is the proeutectoid phase for an iron-carbon alloy in which the mass fractions of total ferrite and total cementite are 0.92 and 0.08 , respectively? Why?", "answer": "the proeutectoid phase is \\alpha ferrite because the alloy composition c_{0} is less than 0.76 wt% C." }, { "idx": 4329, "question": "Consider 1.0 kg of austenite containing 1.15 wt % C, cooled to below 727 C (1341 F). What is the proeutectoid phase?", "answer": "the proeutectoid phase is fe3c." }, { "idx": 4330, "question": "Consider 1.0 kg of austenite containing 1.15 wt % C, cooled to below 727 C (1341 F). How many kilograms each of total ferrite and cementite form?", "answer": "the total ferrite formed is 0.83 kg, and the total cementite formed is 0.17 kg." }, { "idx": 4331, "question": "Consider 1.0 kg of austenite containing 1.15 wt % C, cooled to below 727 C (1341 F). How many kilograms each of pearlite and the proeutectoid phase form?", "answer": "the pearlite formed is 0.93 kg, and the proeutectoid phase (cementite) formed is 0.07 kg." }, { "idx": 4332, "question": "Consider 2.5 kg of austenite containing 0.65 wt % C, cooled to below 727 C (1341 F). What is the proeutectoid phase?", "answer": "ferrite is the proeutectoid phase." }, { "idx": 4333, "question": "Consider 2.5 kg of austenite containing 0.65 wt % C, cooled to below 727 C (1341 F). How many kilograms each of total ferrite and cementite form?", "answer": "the total ferrite formed is 2.27 kg, and the total cementite formed is 0.23 kg." }, { "idx": 4334, "question": "Consider 2.5 kg of austenite containing 0.65 wt % C, cooled to below 727 C (1341 F). How many kilograms each of pearlite and the proeutectoid phase form?", "answer": "the pearlite formed is 2.12 kg, and the proeutectoid ferrite formed is 0.38 kg." }, { "idx": 4335, "question": "Compute the mass fraction of proeutectoid ferrite that forms in an iron-carbon alloy containing 0.25 wt% C.", "answer": "the mass fraction of proeutectoid ferrite is 0.69." }, { "idx": 4336, "question": "Compute the mass fraction of pearlite that forms in an iron-carbon alloy containing 0.25 wt% C.", "answer": "the mass fraction of pearlite is 0.31." }, { "idx": 4337, "question": "The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these two microconstituents are 0.286 and 0.714 , respectively. Determine the concentration of carbon in this alloy.\n\\[\n\\text {", "answer": "0.55% \\text{c}" }, { "idx": 4338, "question": "The mass fractions of total ferrite and total cementite in an iron-carbon alloy are 0.88 and 0.12 , respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why?", "answer": "the alloy is hypereutectoid since c_0 is greater than 0.76 \\, \\text{wt% c}." }, { "idx": 4339, "question": "The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these microconstituents are 0.20 and 0.80 , respectively. Determine the concentration of carbon in this alloy.", "answer": "the concentration of carbon in this alloy is 0.61% \\text{c}." }, { "idx": 4340, "question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of proeutectoid ferrite form?", "answer": "0.98 kg of proeutectoid ferrite forms." }, { "idx": 4341, "question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of eutectoid ferrite form?", "answer": "0.90 kg of eutectoid ferrite forms." }, { "idx": 4342, "question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of cementite form?", "answer": "0.114 kg of cementite forms." }, { "idx": 4343, "question": "Compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid ironcarbon alloy.", "answer": "the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy is 0.232." }, { "idx": 4344, "question": "Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 and 0.049 , respectively? Why or why not?", "answer": "yes, it is possible to have an iron-carbon alloy with mass fractions of total ferrite w_{\\alpha}=0.846 and proeutectoid cementite w_{\\mathrm{fe}_{3} C^{\\prime}}=0.049. the alloy composition for both conditions is 1.05 wt% C. since the composition values are equal, such an alloy exists." }, { "idx": 4345, "question": "Is it possible to have an iron-carbon alloy for which the mass fractions of total cementite and pearlite are 0.039 and 0.417 , respectively? Why or why not?", "answer": "no, such an iron-carbon alloy is not possible because the required carbon compositions for the given mass fractions of total cementite (0.28 wt% C) and pearlite (0.33 wt% C) are different." }, { "idx": 4346, "question": "Compute the mass fraction of eutectoid ferrite in an iron-carbon alloy that contains 0.43 wt% C.", "answer": "the mass fraction of eutectoid ferrite in the iron-carbon alloy is 0.493." }, { "idx": 4347, "question": "Is it possible to determine the composition of an iron-carbon alloy if the mass fraction of eutectoid cementite is 0.104?", "answer": "yes, it is possible to determine the alloy composition, and there are two possible answers." }, { "idx": 4348, "question": "What is the composition of the alloy if eutectoid cementite exists in addition to proeutectoid cementite?", "answer": "the alloy composition is 1.11 wt% c." }, { "idx": 4349, "question": "What is the composition of a hypoeutectoid alloy where all cementite is eutectoid cementite and the mass fraction of total cementite is 0.104?", "answer": "the composition can be determined using the lever rule." }, { "idx": 4350, "question": "Is it possible to determine the composition of an iron-carbon alloy if the mass fraction of eutectoid ferrite is 0.82?", "answer": "yes, it is possible to determine the alloy composition; and, in fact, there are two possible answers." }, { "idx": 4351, "question": "What is the first possible composition of the alloy if the mass fraction of eutectoid ferrite is 0.82?", "answer": "for the first case, the composition is c0 = 0.70 wt % c." }, { "idx": 4352, "question": "What is the second possible composition of the alloy if the mass fraction of eutectoid ferrite is 0.82?", "answer": "for the second case, the composition is c0 = 1.22 wt % c." }, { "idx": 4353, "question": "What is the first stage involved in the formation of particles of a new phase? Briefly describe it.", "answer": "The first stage is nucleation. The nucleation process involves the formation of normally very small particles of the new phase(s) which are stable and capable of continued growth." }, { "idx": 4354, "question": "What is the second stage involved in the formation of particles of a new phase? Briefly describe it.", "answer": "The second stage is growth. The growth stage is simply the increase in size of the new phase particles." }, { "idx": 4355, "question": "If copper (which has a melting point of 1085^{\\circ} C ) homogeneously nucleates at 849^{\\circ} C, calculate the critical radius given values of -1.77 × 10^{9} J/ m^{3} and 0.200 J/ m^{2}, respectively, for the latent heat of fusion and the surface free energy.", "answer": "1.30 nm" }, { "idx": 4356, "question": "The kinetics of the austenite-to-pearlite transformation obey the Avrami relationship. Using the fraction transformed-time data given here, determine the total time required for 95% of the austenite to transform to pearlite:\n\\begin{tabular}{cc}\n\\hline Fraction Transformed & Time (s) \\\\\n\\hline 0.2 & 12.6 \\\\\n0.8 & 28.2 \\\\\n\\hline\n\\end{tabular}", "answer": "35.7(s)" }, { "idx": 4357, "question": "The fraction recrystallized-time data for the recrystallization at 600^{\\circ} C of a previously deformed steel are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of 22.8min.\n\\begin{tabular}{cc}\n\\hline \\begin{tabular}{c} \nFraction \\\\\nRecrystallized\n\\end{tabular} & Time (min) \\\\\n\\hline 0.20 & 13.1 \\\\\n0.70 & 29.1 \\\\\n\\hline\n\\end{tabular}", "answer": "0.51" }, { "idx": 4358, "question": "In terms of heat treatment and the development of microstructure, what is one major limitation of the iron-iron carbide phase diagram related to nonequilibrium phases?", "answer": "The nonequilibrium martensite does not appear on the diagram." }, { "idx": 4359, "question": "In terms of heat treatment and the development of microstructure, what is one major limitation of the iron-iron carbide phase diagram related to time-temperature relationships?", "answer": "The diagram provides no indication as to the time-temperature relationships for the formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium ferrite and cementite phases." }, { "idx": 4360, "question": "Briefly describe the phenomena of superheating and supercooling.", "answer": "Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation." }, { "idx": 4361, "question": "Why do these phenomena occur?", "answer": "These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling." }, { "idx": 4362, "question": "Briefly cite the differences between pearlite, bainite, and spheroidite relative to microstructure.", "answer": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles." }, { "idx": 4363, "question": "Briefly cite the differences between pearlite, bainite, and spheroidite relative to mechanical properties.", "answer": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite." }, { "idx": 4364, "question": "What is the driving force for the formation of spheroidite?", "answer": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area." }, { "idx": 4365, "question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 200°C/s.", "answer": "At a rate of 200°C/s, only martensite forms." }, { "idx": 4366, "question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 100°C/s.", "answer": "At a rate of 100°C/s, both martensite and pearlite form." }, { "idx": 4367, "question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 20°C/s.", "answer": "At a rate of 20°C/s, only fine pearlite forms." }, { "idx": 4368, "question": "What is one important difference between continuous cooling transformation diagrams for plain carbon and alloy steels regarding the presence of a bainite nose?", "answer": "For an alloy steel, a bainite nose will be present, which nose will be absent for plain carbon alloys." }, { "idx": 4369, "question": "What is one important difference between continuous cooling transformation diagrams for plain carbon and alloy steels regarding the position of the pearlite-proeutectoid noses?", "answer": "The pearlite-proeutectoid noses for plain carbon steel alloys are positioned at shorter times than for the alloy steels." }, { "idx": 4370, "question": "Briefly explain why there is no bainite transformation region on the continuous cooling transformation diagram for an iron-carbon alloy of eutectoid composition.", "answer": "There is no bainite transformation region on the continuous cooling transformation diagram for an ironcarbon alloy of eutectoid composition because by the time a cooling curve has passed into the bainite region, the entirety of the alloy specimen will have transformed to pearlite." }, { "idx": 4371, "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 10°C/s.", "answer": "only martensite forms." }, { "idx": 4372, "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 1°C/s.", "answer": "both martensite and bainite form." }, { "idx": 4373, "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 0.1°C/s.", "answer": "martensite, proeutectoid ferrite, and bainite form." }, { "idx": 4374, "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 0.01°C/s.", "answer": "martensite, proeutectoid ferrite, pearlite, and bainite form." }, { "idx": 4375, "question": "On the basis of diffusion considerations, explain why fine pearlite forms for the moderate cooling of austenite through the eutectoid temperature, whereas coarse pearlite is the product for relatively slow cooling rates.", "answer": "For moderately rapid cooling, the time allowed for carbon diffusion is not as great as for slower cooling rates. Therefore, the diffusion distance is shorter, and thinner layers of ferrite and cementite form (i.e., fine pearlite forms)." }, { "idx": 4376, "question": "Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite.", "answer": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite." }, { "idx": 4377, "question": "Cite two reasons why martensite is so hard and brittle.", "answer": "Two reasons why martensite is so hard and brittle are: (1) there are relatively few operable slip systems for the body-centered tetragonal crystal structure, and (2) virtually all of the carbon is in solid solution, which produces a solid-solution hardening effect." }, { "idx": 4378, "question": "Rank the following iron-carbon alloys and associated microstructures from the highest to the lowest tensile strength:\n(a) 0.25 wt% C with spheroidite,\n(b) 0.25 wt% C with coarse pearlite,\n(c) 0.60 wt% C with fine pearlite, and\n(d) 0.60 wt% C with coarse pearlit.", "answer": "(c)(d)(b)(a)" }, { "idx": 4379, "question": "Briefly explain why the hardness of tempered martensite diminishes with tempering time at constant temperature.", "answer": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness." }, { "idx": 4380, "question": "Briefly explain why the hardness of tempered martensite diminishes with increasing temperature at constant tempering time.", "answer": "The hardness of tempered martensite diminishes with increasing temperature at constant tempering time because higher temperatures accelerate the rate of cementite particle growth. This growth reduces the ferrite-cementite phase boundary area, which acts as barriers to dislocation motion. Consequently, the reduction in phase boundary area results in a decrease in hardness." }, { "idx": 4381, "question": "Briefly describe the microstructural difference between spheroidite and tempered martensite.", "answer": "Both tempered martensite and spheroidite have sphere-like cementite particles within a ferrite matrix; however, these particles are much larger for spheroidite." }, { "idx": 4382, "question": "Explain why tempered martensite is much harder and stronger.", "answer": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion." }, { "idx": 4383, "question": "List the four classifications of steels.", "answer": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)" }, { "idx": 4384, "question": "For Low Carbon Steels, briefly describe the properties and typical applications.", "answer": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans." }, { "idx": 4385, "question": "For Medium Carbon Steels, briefly describe the properties and typical applications.", "answer": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts." }, { "idx": 4386, "question": "For High Carbon Steels, briefly describe the properties and typical applications.", "answer": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades." }, { "idx": 4387, "question": "For High Alloy Steels (Stainless and Tool), briefly describe the properties and typical applications.", "answer": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools." }, { "idx": 4388, "question": "Cite three reasons why ferrous alloys are used so extensively.", "answer": "Ferrous alloys are used extensively because: (1) Iron ores exist in abundant quantities. (2) Economical extraction, refining, and fabrication techniques are available. (3) The alloys may be tailored to have a wide range of properties." }, { "idx": 4389, "question": "Cite three characteristics of ferrous alloys that limit their utilization.", "answer": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities." }, { "idx": 4390, "question": "What is the function of alloying elements in tool steels?", "answer": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds." }, { "idx": 4391, "question": "On the basis of microstructure, briefly explain why gray iron is brittle and weak in tension.", "answer": "Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration." }, { "idx": 4392, "question": "Compare white and nodular cast irons with respect to composition and heat treatment.", "answer": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix." }, { "idx": 4393, "question": "Compare white and nodular cast irons with respect to microstructure.", "answer": "White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix." }, { "idx": 4394, "question": "Compare white and nodular cast irons with respect to mechanical characteristics.", "answer": "White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility." }, { "idx": 4395, "question": "Is it possible to produce malleable cast iron in pieces having large cross-sectional dimensions? Why or why not?", "answer": "It is not possible to produce malleable iron in pieces having large cross-sectional dimensions. White cast iron is the precursor of malleable iron, and a rapid cooling rate is necessary for the formation of white iron, which may not be accomplished at interior regions of thick cross-sections." }, { "idx": 4396, "question": "What is the principal difference between wrought and cast alloys?", "answer": "The principal difference between wrought and cast alloys is as follows: wrought alloys are ductile enough so as to be hot or cold worked during fabrication, whereas cast alloys are brittle to the degree that shaping by deformation is not possible and they must be fabricated by casting." }, { "idx": 4397, "question": "Why must rivets of a 2017 aluminum alloy be refrigerated before they are used?", "answer": "Rivets of a 2017 aluminum alloy must be refrigerated before they are used because, after being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven." }, { "idx": 4398, "question": "What is the chief difference between heat-treatable and non-heat-treatable alloys?", "answer": "The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments." }, { "idx": 4399, "question": "Cite advantages of cold working.", "answer": "The advantages of cold working are: (1) A high quality surface finish. (2) The mechanical properties may be varied. (3) Close dimensional tolerances." }, { "idx": 4400, "question": "Cite disadvantages of cold working.", "answer": "The disadvantages of cold working are: (1) High deformation energy requirements. (2) Large deformations must be accomplished in steps, which may be expensive. (3) A loss of ductility." }, { "idx": 4401, "question": "Cite advantages of hot working.", "answer": "The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low." }, { "idx": 4402, "question": "Cite disadvantages of hot working.", "answer": "The disadvantages of hot working are: (1) A poor surface finish. (2) A variety of mechanical properties is not possible." }, { "idx": 4403, "question": "Cite advantages of forming metals by extrusion as opposed to rolling.", "answer": "The advantages of extrusion as opposed to rolling are as follows: (1) Pieces having more complicated cross-sectional geometries may be formed. (2) Seamless tubing may be produced." }, { "idx": 4404, "question": "Cite some disadvantages of forming metals by extrusion as opposed to rolling.", "answer": "The disadvantages of extrusion over rolling are as follows: (1) Nonuniform deformation over the cross-section. (2) A variation in properties may result over a cross-section of an extruded piece." }, { "idx": 4405, "question": "List four situations in which casting is the preferred fabrication technique.", "answer": "Four situations in which casting is the preferred fabrication technique are:\n(1) For large pieces and/or complicated shapes.\n(2) When mechanical strength is not an important consideration.\n(3) For alloys having low ductilities.\n(4) When it is the most economical fabrication technique." }, { "idx": 4406, "question": "Compare sand casting technique", "answer": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast." }, { "idx": 4407, "question": "Compare die casting technique", "answer": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast." }, { "idx": 4408, "question": "Compare investment casting technique", "answer": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low." }, { "idx": 4409, "question": "Compare lost foam casting technique", "answer": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes." }, { "idx": 4410, "question": "Compare continuous casting technique", "answer": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section." }, { "idx": 4411, "question": "Describe one problem that might exist with a steel weld that was cooled very rapidly.", "answer": "If a steel weld is cooled very rapidly, martensite may form, which is very brittle. In some situations, cracks may form in the weld region as it cools." }, { "idx": 4412, "question": "Cite three sources of internal residual stresses in metal components.", "answer": "Three sources of residual stresses in metal components are plastic deformation processes, nonuniform cooling of a piece that was cooled from an elevated temperature, and a phase transformation in which parent and product phases have different densities." }, { "idx": 4413, "question": "What are two possible adverse consequences of internal residual stresses in metal components?", "answer": "Two adverse consequences of these stresses are distortion (or warpage) and fracture." }, { "idx": 4414, "question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.20 wt% C iron-carbon alloy during a normalizing heat treatment.", "answer": "At least 905°C (1660°F)" }, { "idx": 4415, "question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.76 wt% C iron-carbon alloy during a normalizing heat treatment.", "answer": "At least 782°C (1440°F)" }, { "idx": 4416, "question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.95 wt% C iron-carbon alloy during a normalizing heat treatment.", "answer": "At least 840°C (1545°F)" }, { "idx": 4417, "question": "Give the approximate temperature at which it is desirable to heat a 0.25 wt% C iron-carbon alloy during a full anneal heat treatment.", "answer": "About 880°C (1510°F)" }, { "idx": 4418, "question": "Give the approximate temperature at which it is desirable to heat a 0.45 wt% C iron-carbon alloy during a full anneal heat treatment.", "answer": "About 830°C (1525°F)" }, { "idx": 4419, "question": "Give the approximate temperature at which it is desirable to heat a 0.85 wt% C iron-carbon alloy during a full anneal heat treatment.", "answer": "About 777°C (1430°F)" }, { "idx": 4420, "question": "Give the approximate temperature at which it is desirable to heat a 1.10 wt% C iron-carbon alloy during a full anneal heat treatment.", "answer": "About 777°C (1430°F)" }, { "idx": 4421, "question": "What is the purpose of a spheroidizing heat treatment?", "answer": "The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure." }, { "idx": 4422, "question": "On what classes of alloys is a spheroidizing heat treatment normally used?", "answer": "It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong." }, { "idx": 4423, "question": "Briefly explain the difference between hardness and hardenability.", "answer": "Hardness is a measure of a material's resistance to localized surface deformation, whereas hardenability is a measure of the depth to which a ferrous alloy may be hardened by the formation of martensite. Hardenability is determined from hardness tests." }, { "idx": 4424, "question": "What influence does the presence of alloying elements (other than carbon) have on the shape of a herdenability curve? Briefly explain this effect.", "answer": "The presence of alloying elements (other than carbon) causes a much more gradual decrease in hardness with position from the quenched end for a hardenability curve. The reason for this effect is that alloying elements retard the formation of pearlitic and bainitic structures which are not as hard as martensite." }, { "idx": 4425, "question": "Name two thermal properties of a liquid medium that will influence its quenching effectiveness.", "answer": "The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity." }, { "idx": 4426, "question": "For a ceramic compound, what are the two characteristics of the component ions that determine the crystal structure?", "answer": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions." }, { "idx": 4427, "question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. Will the stacking sequence for this structure be FCC or HCP? Why?", "answer": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC." }, { "idx": 4428, "question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. Will cations fill tetrahedral or octahedral positions? Why?", "answer": "The cations will fill tetrahedral positions since the coordination number for cations is four." }, { "idx": 4429, "question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. What fraction of the positions will be occupied?", "answer": "Only one-half of the tetrahedral positions will be occupied because there are two tetrahedral sites per anion, and yet only one cation per anion." }, { "idx": 4430, "question": "Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.582nm. If the measured density is 4.82g / {cm}^{3}, how many \\mathrm{Cd}^{2+} and \\mathrm{S}^{2-} ions are there per unit cell?", "answer": "there are four \\mathrm{cd}^{2+} and four s^{2-} ions per unit cell." }, { "idx": 4431, "question": "A hypothetical A X type of ceramic material is known to have a density of 2.65g / {cm}^{3} and a unit cell of cubic symmetry with a cell edge length of 0.43nm. The atomic weights of the A and X elements are 86.6 and 40.3g / mol, respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).", "answer": "the only possible crystal structure for this material is cesium chloride, which has one formula unit per unit cell." }, { "idx": 4432, "question": "In terms of bonding, explain why silicate materials have relatively low densities.", "answer": "The silicate materials have relatively low densities because the atomic bonds are primarily covalent in nature, and, therefore, directional. This limits the packing efficiency of the atoms, and therefore, the magnitude of the density." }, { "idx": 4433, "question": "Would you expect Frenkel defects for anions to exist in ionic ceramics in relatively large concentrations? Why or why not?", "answer": "Frenkel defects for anions would not exist in appreciable concentrations because the anion is quite large and is highly unlikely to exist as an interstitial." }, { "idx": 4434, "question": "Calculate the fraction of lattice sites that are Schottky defects for sodium chloride at its melting temperature \\left(801^{\\circ} C\\right). Assume an energy for defect formation of 2.3 \\mathrm{eV}.", "answer": "the fraction of lattice sites that are schottky defects for sodium chloride at its melting temperature is 4.03 × 10^{-6}." }, { "idx": 4435, "question": "(a) Suppose that Li2O is added as an impurity to CaO. If the Li+ substitutes for Ca2+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Li+ added?", "answer": "oxygen vacancies. Two Li+ ions added, a single oxygen vacancy is formed." }, { "idx": 4436, "question": "(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl substitutes for O2-, what kind of vacancies would you expect to form? How many of the vacancies are created for every Cl- added?", "answer": "calcium vacancies. Two Cl- ions will lead to the formation of one calcium vacancy." }, { "idx": 4437, "question": "What point defects are possible for Al2O3 as an impurity in MgO?", "answer": "For every Al3+ ion that substitutes for Mg2+ in MgO, a single positive charge is added. Thus, in order to maintain charge neutrality, either a positive charge must be removed or a negative charge must be added. Negative charges are added by forming O2- interstitials, which are not likely to form. Positive charges may be removed by forming Mg2+ vacancies." }, { "idx": 4438, "question": "How many Al2+ ions must be added to form each of these defects?", "answer": "One magnesium vacancy would be formed for every two Al3+ ions added." }, { "idx": 4439, "question": "When kaolinite clay [Al2(Si2O5)(OH)4] is heated to a sufficiently high temperature, chemical water is driven off. Under these circumstances, what is the composition of the remaining product (in weight percent Al2O3)?", "answer": "the composition of the remaining product is 45.9% Al2O3 by weight." }, { "idx": 4440, "question": "When kaolinite clay [Al2(Si2O5)(OH)4] is heated to a sufficiently high temperature, chemical water is driven off. What are the liquidus and solidus temperatures of this material?", "answer": "the liquidus and solidus temperatures of this material are 1825 degrees C and 1587 degrees C, respectively." }, { "idx": 4441, "question": "Why may there be significant scatter in the fracture strength for some given ceramic material?", "answer": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material." }, { "idx": 4442, "question": "Why does fracture strength increase with decreasing specimen size?", "answer": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes." }, { "idx": 4443, "question": "The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter surface crack geometry (i.e., reduce crack length and increase the tip radius). Compute the ratio of the original and etched crack tip radii for an eightfold increase in fracture strength if two-thirds of the crack length is removed.\n\\[\n\\text {", "answer": "the ratio of the original and etched crack tip radii is 21.3." }, { "idx": 4444, "question": "A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 425N\\left(95.5 lb \\mathrm{b}_{1}\\right), the flexural strength is 105 MPa(15,000 psi), and the separation between load points is 50 mm(2.0 in).", "answer": "the minimum possible radius of the specimen without fracture is 4.0mm (0.16 in)." }, { "idx": 4445, "question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5mm (0.10 in.) and the support point separation distance is 30mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied. Justify your prediction.", "answer": "the calculated stress is 379 mpa (53,500 psi), which is less than the flexural strength of 390 mpa. therefore, fracture is not predicted." }, { "idx": 4446, "question": "Would you be 100% certain of the prediction in part (a)? Why or why not?", "answer": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials. since the calculated stress 379 mpa is relatively close to the flexural strength 390 mpa, there is some chance that fracture will occur." }, { "idx": 4447, "question": "Cite one reason why ceramic materials are, in general, harder yet more brittle than metals.", "answer": "Crystalline ceramics are harder yet more brittle than metals because they (ceramics) have fewer slip systems, and, therefore, dislocation motion is highly restricted." }, { "idx": 4448, "question": "The modulus of elasticity for beryllium oxide (BeO) having 5 vol% porosity is 310 GPa (45 x 10^6 psi). Compute the modulus of elasticity for the nonporous material.", "answer": "the modulus of elasticity for the nonporous material is 342 gpa (49.6 x 10^6 psi)." }, { "idx": 4449, "question": "The modulus of elasticity for beryllium oxide (BeO) having 5 vol% porosity is 310 GPa (45 x 10^6 psi). Compute the modulus of elasticity for 10 vol% porosity.", "answer": "the modulus of elasticity for 10 vol% porosity is 280 gpa (40.6 x 10^6 psi)." }, { "idx": 4450, "question": "The modulus of elasticity for boron carbide (B4C) having 5 vol% porosity is 290 GPa (42 x 10^6 psi). Compute the modulus of elasticity for the nonporous material.", "answer": "320 GPa (46.3 x 10^6 psi)" }, { "idx": 4451, "question": "The modulus of elasticity for boron carbide (B4C) having 5 vol% porosity is 290 GPa (42 x 10^6 psi). At what volume percent porosity will the modulus of elasticity be 235 GPa (34 x 10^6 psi)?", "answer": "the modulus of elasticity will be 235 GPa (34 x 10^6 psi) at a porosity of 15.1 vol%" }, { "idx": 4452, "question": "Cite the two desirable characteristics of glasses.", "answer": "Two desirable characteristics of glasses are optical transparency and ease of fabrication." }, { "idx": 4453, "question": "What is crystallization?", "answer": "Crystallization is the process whereby a glass material is caused to transform to a crystalline solid, usually as a result of a heat treatment." }, { "idx": 4454, "question": "Cite two properties that may be improved by crystallization.", "answer": "Two properties that may be improved by crystallization are (1) a lower coefficient of thermal expansion, and (2) higher strength." }, { "idx": 4455, "question": "For refractory ceramic materials, cite three characteristics that improve with increasing porosity.", "answer": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock." }, { "idx": 4456, "question": "For refractory ceramic materials, cite two characteristics that are adversely affected by increasing porosity.", "answer": "For refractory ceramic materials, two characteristics that are adversely affected by increasing porosity are (1) load-bearing capacity and (2) resistance to attack by corrosive materials." }, { "idx": 4457, "question": "How do the aggregate particles become bonded together in clay-based mixtures during firing?", "answer": "For clay-based aggregates, a liquid phase forms during firing, which infiltrates the pores between the unmelted particles; upon cooling, this liquid becomes a glass, that serves as the bonding phase." }, { "idx": 4458, "question": "How do the aggregate particles become bonded together in cements during setting?", "answer": "With cements, the bonding process is a chemical, hydration reaction between the water that has been added and the various cement constituents. The cement particles are bonded together by reactions that occur at the particle surfaces." }, { "idx": 4459, "question": "What is the glass transition temperature for a noncrystalline ceramic?", "answer": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve." }, { "idx": 4460, "question": "What is the melting temperature for a crystalline material?", "answer": "The melting temperature is, for a crystalline material and upon cooling, that temperature at which there is a sudden and discontinuous decrease in the specific-volume-versus-temperature curve." }, { "idx": 4461, "question": "Compare the softening points for 96% silica, borosilicate, and soda-lime glasses.", "answer": "The softening point of a glass is that temperature at which the viscosity is 4 × 10^{6} \\mathrm{~Pa}·s; from Figure 13.7, these temperatures for the 96% silica, borosilicate, and soda-lime glasses are 1540^{\\circ} C\\left(2800^{\\circ} F\\right), 830^{\\circ} C\\left(1525^{\\circ} F\\right), and 700^{\\circ} C\\left(1290^{\\circ} F\\right), respectively." }, { "idx": 4462, "question": "Explain why residual thermal stresses are introduced into a glass piece when it is cooled.", "answer": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established." }, { "idx": 4463, "question": "Are thermal stresses introduced upon heating? Why or why not?", "answer": "Yes, thermal stresses will be introduced because of thermal expansion upon heating for the same reason as for thermal contraction upon cooling." }, { "idx": 4464, "question": "Borosilicate glasses and fused silica are resistant to thermal shock. Why is this so?", "answer": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low." }, { "idx": 4465, "question": "Glass pieces may also be strengthened by chemical tempering. With this procedure, the glass surface is put in a state of compression by exchanging some of the cations near the surface with other cations having a larger diameter. Suggest one type of cation that, by replacing \\mathrm{Na}^{+}, will induce chemical tempering in a soda-lime glass.", "answer": "Chemical tempering will be accomplished by substitution, for \\mathrm{Na}^{+}, another monovalent cation with a slightly larger diameter. Both K^{+}and \\mathrm{Cs}^{+}fill these criteria, having ionic radii of 0.138 and 0.170 nm, respectively, which are larger than the ionic radius of \\mathrm{Na}^{+}(0.102nm). In fact, soda-lime glasses are tempered by a K^{+}-\\mathrm{Na}^{+}ion exchange." }, { "idx": 4466, "question": "Cite the two desirable characteristics of clay minerals relative to fabrication processes.", "answer": "Two desirable characteristics of clay minerals relative to fabrication processes are (1) they become hydroplastic (and therefore formable) when mixed with water; and (2) during firing, clays melt over a range of temperatures, which allows some fusion and bonding of the ware without complete melting and a loss of mechanical integrity and shape." }, { "idx": 4467, "question": "From a molecular perspective, briefly explain the mechanism by which clay minerals become hydroplastic when water is added.", "answer": "Clays become hydroplastic when water is added because the water molecules occupy regions between the layered molecular sheets; these water molecules essentially eliminate the secondary molecular bonds between adjacent sheets, and also form a thin film around the clay particles. The net result is that the clay particles are relatively free to move past one another, which is manifested as the hydroplasticity phenomenon." }, { "idx": 4468, "question": "What are the three main components of a whiteware ceramic such as porcelain?", "answer": "clay, quartz, and a flux." }, { "idx": 4469, "question": "What role does each component play in the forming and firing procedures?", "answer": "Quartz acts as a filler material. Clay facilitates the forming operation since, when mixed with water, the mass may be made to become either hydroplastic or form a slip. Also, since clays melt over a range of temperatures, the shape of the piece being fired will be maintained. The flux facilitates the formation of a glass having a relatively low melting temperature." }, { "idx": 4470, "question": "Why is it so important to control the rate of drying of a ceramic body that has been hydroplastically formed or slip cast?", "answer": "It is important to control the rate of drying inasmuch as if the rate of drying is too rapid, there will be nonuniform shrinkage between surface and interior regions, such that warping and/or cracking of the ceramic ware may result." }, { "idx": 4471, "question": "Cite three factors that influence the rate of drying, and explain how each affects the rate.", "answer": "Three factors that affect the rate of drying are temperature, humidity, and rate of air flow. The rate of drying is enhanced by increasing both the temperature and rate of air flow, and by decreasing the humidity of the air." }, { "idx": 4472, "question": "Cite one reason why drying shrinkage is greater for slip cast or hydroplastic products that have smaller clay particles.", "answer": "The reason that drying shrinkage is greater for products having smaller clay particles is because there is more particle surface area, and, consequently, more water will surround a given volume of particles. The drying shrinkage will thus be greater as this water is removed, and as the interparticle separation decreases." }, { "idx": 4473, "question": "Name three factors that influence the degree to which vitrification occurs in clay-based ceramic wares.", "answer": "(1) composition (2) the temperature of firing(3) the time at the firing temperature." }, { "idx": 4474, "question": "Explain how density, firing distortion, strength, corrosion resistance, and thermal conductivity are affected by the extent of vitrification.", "answer": "Density will increase with degree of vitrification since the total remaining pore volume decreases. Strength will also increase with degree of vitrification inasmuch as more of the liquid phase forms, which fills in a greater fraction of pore volume. Upon cooling, the liquid forms a glass matrix of relatively high strength. Corrosion resistance normally increases also, especially at service temperatures below that at which the glass phase begins to soften. The rate of corrosion is dependent on the amount of surface area exposed to the corrosive medium; hence, decreasing the total surface area by filling in some of the surface pores, diminishes the corrosion rate. Thermal conductivity will increase with degree of vitrification. The glass phase has a higher conductivity than the pores that it has filled." }, { "idx": 4475, "question": "Some ceramic materials are fabricated by hot isostatic pressing. Cite some of the limitations and difficulties associated with this technique.", "answer": "The principal disadvantage of hot-isostatic pressing is that it is expensive. The pressure is applied on a pre-formed green piece by a gas. Thus, the process is slow, and the equipment required to supply the gas and withstand the elevated temperature and pressure is costly." }, { "idx": 4476, "question": "Compute repeat unit molecular weight for poly(vinyl chloride)", "answer": "62.49g/mol" }, { "idx": 4477, "question": "Compute repeat unit molecular weight for poly(ethylene terephthalate)", "answer": "192.16g/mol" }, { "idx": 4478, "question": "Compute repeat unit molecular weight for polycarbonate", "answer": "254.27g/mol" }, { "idx": 4479, "question": "Compute repeat unit molecular weight for polydimethylsiloxane", "answer": "74.16g/mol" }, { "idx": 4480, "question": "The number-average molecular weight of a polypropylene is 1,000,000g / mol. Compute the degree of polymerization.", "answer": "23760" }, { "idx": 4481, "question": "Compute the repeat unit molecular weight of polystyrene.", "answer": "the repeat unit molecular weight of polystyrene is 104.14 g/mol." }, { "idx": 4482, "question": "Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000.", "answer": "the number-average molecular weight for a polystyrene with a degree of polymerization of 25,000 is 2.60 x 10^6 g/mol." }, { "idx": 4483, "question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 5% of all the original hydrogen atoms.", "answer": "the concentration of Cl that must be added is 20.3 wt%." }, { "idx": 4484, "question": "In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)?", "answer": "chlorinated polyethylene differs from poly(vinyl chloride) in that, for pvc, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random." }, { "idx": 4485, "question": "Make comparisons of thermoplastic and thermosetting polymers on the basis of mechanical characteristics upon heating.", "answer": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening." }, { "idx": 4486, "question": "Make comparisons of thermoplastic and thermosetting polymers according to possible molecular structures.", "answer": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked." }, { "idx": 4487, "question": "Is it possible to grind up and reuse phenol-formaldehyde? Why or why not?", "answer": "It is not possible to grind up and reuse phenol-formaldehyde because it is a network thermoset polymer and, therefore, is not amenable to remolding." }, { "idx": 4488, "question": "Is it possible to grind up and reuse polypropylene? Why or why not?", "answer": "Yes, it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer, will soften when reheated, and, thus, may be remolded." }, { "idx": 4489, "question": "The number-average molecular weight of a poly(styrene-butadiene) alternating copolymer is 1,350,000g / mol; determine the average number of styrene and butadiene repeat units per molecule.", "answer": "the average number of styrene and butadiene repeat units per molecule is 8530." }, { "idx": 4490, "question": "Calculate the number-average molecular weight of a random nitrile rubber [poly(acrylonitrilebutadiene) copolymer] in which the fraction of butadiene repeat units is 0.30 ; assume that this concentration corresponds to a degree of polymerization of 2000 .", "answer": "the number-average molecular weight of the nitrile rubber copolymer is \\bar{m}_{n} = 106740, \\text{g/mol}." }, { "idx": 4491, "question": "An alternating copolymer is known to have a number-average molecular weight of 250,000g / mol and a degree of polymerization of 3420 . If one of the repeat units is styrene, which of ethylene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit? ", "answer": "propylene" }, { "idx": 4492, "question": "(a) Determine the ratio of butadiene to styrene repeat units in a copolymer having a number-average molecular weight of 350,000 g/mol and degree of polymerization of 4425.", "answer": "1.0." }, { "idx": 4493, "question": "(b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block?", "answer": "random, alternating, graft, and block" }, { "idx": 4494, "question": "Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylene may have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both repeat unit types.", "answer": "the fraction of the ethylene repeat unit is f(\\text{ethylene}) = 0.69 . the fraction of the propylene repeat unit is f(\\text{propylene}) = 0.31 ." }, { "idx": 4495, "question": "Explain briefly why the tendency of a polymer to crystallize decreases with increasing molecular weight.", "answer": "The tendency of a polymer to crystallize decreases with increasing molecular weight because as the chains become longer it is more difficult for all regions along adjacent chains to align so as to produce the ordered atomic array." }, { "idx": 4496, "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Linear and syndiotactic poly(vinyl chloride); linear and isotactic polystyrene.", "answer": "Yes, for these two polymers it is possible to decide. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl side-group for polystyrene is bulkier than the Cl side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize." }, { "idx": 4497, "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (b) Network phenol-formaldehyde; linear and heavily crosslinked cis-isoprene.", "answer": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize." }, { "idx": 4498, "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (c) Linear polyethylene; lightly branched isotactic polypropylene.", "answer": "Yes, it is possible to decide for these two polymers. The linear polyethylene is more likely to crystallize. The repeat unit structure for polypropylene is chemically more complicated than is the repeat unit structure for polyethylene. Furthermore, branched structures are less likely to crystallize than are linear structures." }, { "idx": 4499, "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (d) Alternating poly(styrene-ethylene) copolymer; random poly(vinyl chloride-tetrafluoroethylene) copolymer.", "answer": "Yes, it is possible to decide for these two copolymers. The alternating poly(styrene-ethylene) copolymer is more likely to crystallize. Alternating copolymers crystallize more easily than do random copolymers." }, { "idx": 4500, "question": "Consider the diffusion of water vapor through a polypropylene (PP) sheet 2mm thick. The pressures of H_{2} \\mathrm{O} at the two faces are 1 kPa and 10 kPa, which are maintained constant. Assuming conditions of steady state, what is the diffusion flux [in \\left[{cm}^{3} STP\\right] / {cm}^{2}-s ] at 298 K ?", "answer": "the diffusion flux is 1.71 × 10^{-7} \\ \\text{cm}^{3}\\text{stp}/\\text{cm}^{2}-\\text{s}." }, { "idx": 4501, "question": "Argon diffuses through a high density polyethylene (HDPE) sheet 40mm thick at a rate of 4.0 × 10^{-7} \\left({cm}^{3} STP / {cm}^{2}-s\\right. at 325 K. The pressures of argon at the two faces are 5000 kPa and 1500 kPa, which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at 325 K ?", "answer": "the permeability coefficient of ar through hdpe at 325k is 4.57 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}-s-\\mathrm{pa}}." }, { "idx": 4502, "question": "The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the following equation:\n\\[\nP_{M}=P_{M_{0}} \\exp \\left(-\\frac{Q_{p}}{R T}\\right)\n\\]\nwhere P_{M_{0}} and Q_{p} are constants for a given gas-polymer pair. Consider the diffusion of hydrogen through a poly(dimethyl siloxane) (PDMSO) sheet 20mm thick. The hydrogen pressures at the two faces are 10 kPa and 1 kPa, which are maintained constant. Compute the diffusion flux [in \\left({cm}^{3} STP\\right) / {cm}^{2}-s ] at 350 K. For this diffusion system\n\\[\n\\begin{array}{l}\nP_{M_{0}}=1.45 × 10^{-8}\\left(cm^{3} STP\\right)\\left({cm}^{3} / {cm}^{2}-s-\\mathrm{Pa}\\right. \\\\\nQ_{p}=13.7kJ / mol\n\\end{array}\n\\]\nAlso, assume a condition of steady state diffusion", "answer": "the diffusion flux is 5.90 × 10^{-7} \\frac{\\text{cm}^{3} \\text{stp}}{\\text{cm}^{2}·\\text{s}}." }, { "idx": 4503, "question": "For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to\n\\[\n\\sigma(t)=\\sigma(0) \\exp \\left(-\\frac{t}{\\tau}\\right)\n\\]\nwhere \\sigma(t) and \\sigma(t) represent the time-dependent and initial (i.e., time =0 ) stresses, respectively, and t and \\tau denote elapsed time and the relaxation time; \\tau is a time-independent constant characteristic of the material. A specimen of some viscoelastic polymer the stress relaxation of which obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.6 ; the stress necessary to maintain this constant strain was measured as a function of time. Determine E_{\\tau}(10) for this material if the initial stress level was 2.76 MPa(400 psi), which dropped to 1.72 MPa (250 psi) after 60s.", "answer": "e_{\\tau}(10) = 4.25 \\text{ mpa (616 psi)}" }, { "idx": 4504, "question": "(a) Contrast the manner in which stress relaxation and viscoelastic creep tests are conducted.", "answer": "Stress relaxation tests are conducted by rapidly straining the material elastically in tension, holding the strain level constant, and then measuring the stress as a function of time. For viscoelastic creep tests, a stress (usually tensile) is applied instantaneously and maintained constant while strain is measured as a function of time." }, { "idx": 4505, "question": "(b) For each of these tests, cite the experimental parameter of interest and how it is determined.", "answer": "The experimental parameters of interest from the stress relaxation and viscoelastic creep tests are the relaxation modulus and creep modulus (or creep compliance), respectively. The relaxation modulus is the ratio of stress measured after 10 s and strain; creep modulus is the ratio of stress and strain taken at a specific time." }, { "idx": 4506, "question": "For thermoplastic polymers, cite five factors that favor brittle fracture.", "answer": "For thermoplastic polymers, five factors that favor brittle fracture are as follows: (1) a reduction in temperature, (2) an increase in strain rate, (3) the presence of a sharp notch, (4) increased specimen thickness, and (5) modifications of the polymer structure." }, { "idx": 4507, "question": "For the pair of polymers: random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.", "answer": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking." }, { "idx": 4508, "question": "For the pair of polymers: branched and syndiotactic polypropylene with a degree of polymerization of 5000; linear and isotactic polypropylene with a degree of polymerization of 3000, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.", "answer": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. " }, { "idx": 4509, "question": "For the pair of polymers: branched polyethylene with a number-average molecular weight of 250000 g/mol; linear and isotactic poly(vinyl chloride) with a number-average molecular weight of 200000 g/mol, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.", "answer": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity." }, { "idx": 4510, "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. Syndiotactic polystyrene having a number-average molecular weight of 600,000 g/mol; atactic polystyrene having a number-average molecular weight of 500,000 g/mol", "answer": "Yes it is possible. The syndiotactic polystyrene has the higher tensile strength. Syndiotactic polymers are more likely to crystallize than atactic ones; the greater the crystallinity, the higher the tensile strength. Furthermore, the syndiotactic also has a higher molecular weight; increasing molecular weight also enhances the strength." }, { "idx": 4511, "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. Random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; block acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked", "answer": "No it is not possible. The random acrylonitrile-butadiene copolymer has more crosslinking; increased crosslinking leads to an increase in strength. However, the block copolymeric material will most likely have a higher degree of crystallinity; and increasing crystallinity improves the strength." }, { "idx": 4512, "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. Network polyester; lightly branched polypropylene", "answer": "Yes it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polypropylene since there are many more of the strong covalent bonds for the network structure." }, { "idx": 4513, "question": "List the two molecular characteristics that are essential for elastomers.", "answer": "Two molecular characteristics essential for elastomers are: (1) they must be amorphous, having chains that are extensively coiled and kinked in the unstressed state; and (2) there must be some crosslinking." }, { "idx": 4514, "question": "Name the following polymer(s) that would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?", "answer": "This question asks us to name which, of several polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100^{\\circ} C\\left(212^{\\circ} F\\right). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of 100^{\\circ} C or above, and would be suitable for this application." }, { "idx": 4515, "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Isotactic polystyrene that has a density of 1.12 g/cm3 and a weight-average molecular weight of 150,000 g/mol; syndiotactic polystyrene that has a density of 1.10 g/cm3 and a weight-average molecular weight of 125,000 g/mol", "answer": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight." }, { "idx": 4516, "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Linear polyethylene that has a degree of polymerization of 5,000; linear and isotactic polypropylene that has a degree of polymerization of 6,500", "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization." }, { "idx": 4517, "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Branched and isotactic polystyrene that has a degree of polymerization of 4,000; linear and isotactic polypropylene that has a degree of polymerization of 7,500", "answer": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The polystyrene has a bulkier side group than the polypropylene; on the basis of this effect alone, the polystyrene should have the greater Tm. However, the polystyrene has more branching and a lower degree of polymerization; both of these factors lead to a lowering of the melting temperature." }, { "idx": 4518, "question": "Briefly explain the difference in molecular chemistry between silicone polymers and other polymeric materials.", "answer": "The backbone chain of most polymers consists of carbon atoms that are linked together. For the silicone polymers, this backbone chain is composed of silicon and oxygen atoms that alternate positions." }, { "idx": 4519, "question": "List two important characteristics for polymers that are to be used in fiber applications.", "answer": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity." }, { "idx": 4520, "question": "Cite five important characteristics for polymers that are to be used in thin-film applications.", "answer": "Five important characteristics for polymers that are to be used in thin-film applications are: (1) low density; (2) high flexibility; (3) high tensile and tear strengths; (4) resistance to moisture/chemical attack; and (5) low gas permeability.\n}" }, { "idx": 4521, "question": "Cite the primary differences between addition and condensation polymerization techniques.", "answer": "For addition polymerization, the reactant species have the same chemical composition as the monomer species in the molecular chain. This is not the case for condensation polymerization, wherein there is a chemical reaction between two or more monomer species, producing the repeating unit. There is often a low molecular weight by-product for condensation polymerization; such is not found for addition polymerization." }, { "idx": 4522, "question": "Estimate the maximum thermal conductivity value for a cermet that contains 85 vol % titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co, respectively.", "answer": "the maximum thermal conductivity k_max is 33.3 W/m-K." }, { "idx": 4523, "question": "Estimate the minimum thermal conductivity value for a cermet that contains 85 vol % titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co, respectively.", "answer": "the minimum thermal conductivity k_min is 29.7 W/m-K." }, { "idx": 4524, "question": "A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.60 and 0.40 , respectively, estimate the upper limit for the specific stiffness of this composite given the data that follow.\n\\begin{tabular}{lcc}\n\\hline & Specific Gravity & \\begin{tabular}{c} \nModulus of Elasticity \\\\\n(GPa)\n\\end{tabular} \\\\\n\\hline Copper & 8.9 & 110 \\\\\nTungsten & 19.3 & 407 \\\\\n\\hline\n\\end{tabular}", "answer": "the upper limit for the specific stiffness of this composite is 19.0 \\text{ gpa}. an alternate approach yields a specific stiffness of 17.6 \\text{ gpa}." }, { "idx": 4525, "question": "(a) What is the distinction between cement and concrete?", "answer": "Concrete consists of an aggregate of particles that are bonded together by a cement." }, { "idx": 4526, "question": "(b) Cite three important limitations that restrict the use of concrete as a structural material.", "answer": "Three limitations of concrete are: (1) it is a relatively weak and brittle material; (2) it experiences relatively large thermal expansions (contractions) with changes in temperature; and (3) it may crack when exposed to freeze-thaw cycles." }, { "idx": 4527, "question": "(c) Briefly explain three techniques that are utilized to strengthen concrete by reinforcement.", "answer": "Three reinforcement strengthening techniques are: (1) reinforcement with steel wires, rods, etc.; (2) reinforcement with fine fibers of a high modulus material; and (3) introduction of residual compressive stresses by prestressing or postensioning." }, { "idx": 4528, "question": "Cite one similarity between precipitation hardening and dispersion strengthening.", "answer": "The similarity between precipitation hardening and dispersion strengthening is the strengthening mechanism--i.e., the precipitates/particles effectively hinder dislocation motion." }, { "idx": 4529, "question": "Cite two differences between precipitation hardening and dispersion strengthening.", "answer": "The two differences are: (1) the hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening; and (2) the strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening." }, { "idx": 4530, "question": "A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows: Modulus of Elasticity for Aramid fiber is 131 GPa (19x10^6 psi) and Tensile Strength is 3600 MPa (520,000 psi); Modulus of Elasticity for Polycarbonate is 2.4 GPa (3.5x10^5 psi) and Tensile Strength is 65 MPa (9425 psi). Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa (6500 psi). For this composite, compute the longitudinal tensile strength.", "answer": "the longitudinal tensile strength is 1100 mpa (160,000 psi)." }, { "idx": 4531, "question": "A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows: Modulus of Elasticity for Aramid fiber is 131 GPa (19x10^6 psi) and Tensile Strength is 3600 MPa (520,000 psi); Modulus of Elasticity for Polycarbonate is 2.4 GPa (3.5x10^5 psi) and Tensile Strength is 65 MPa (9425 psi). Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa (6500 psi). For this composite, compute the longitudinal modulus of elasticity.", "answer": "the longitudinal modulus of elasticity is 41 gpa (5.95x10^6 psi)." }, { "idx": 4532, "question": "For a continuous and oriented fiber-reinforced composite, the modulus of elasticity in the longitudinal direction is 19.7 GPa (2.8 x 10^5 psi). If the volume fraction of fibers is 0.25, determine the modulus of elasticity of the fiber phase.", "answer": "the modulus of elasticity of the fiber phase e_f is 70.4 GPa (10.2 x 10^6 psi)." }, { "idx": 4533, "question": "For a continuous and oriented fiber-reinforced composite, the modulus of elasticity in the transverse direction is 3.66 GPa (5.3 x 10^5 psi). If the volume fraction of fibers is 0.25, determine the modulus of elasticity of the matrix phase.", "answer": "the modulus of elasticity of the matrix phase e_m is 2.79 GPa (4.04 x 10^5 psi)." }, { "idx": 4534, "question": "In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94% of a load applied in the longitudinal direction. Using the data provided, determine the volume fraction of fibers that will be required.", "answer": "the required volume fraction of fibers is 0.393." }, { "idx": 4535, "question": "In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94% of a load applied in the longitudinal direction. What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 MPa (4350 psi).", "answer": "the tensile strength of this composite is 1354 mpa (196,400 psi)." }, { "idx": 4536, "question": "For a polymer-matrix fiber-reinforced composite, list three functions of the matrix phase.", "answer": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation." }, { "idx": 4537, "question": "For a polymer-matrix fiber-reinforced composite, compare the desired mechanical characteristics of matrix and fiber phases.", "answer": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong." }, { "idx": 4538, "question": "For a polymer-matrix fiber-reinforced composite, cite two reasons why there must be a strong bond between fiber and matrix at their interface.", "answer": "There must be a strong interfacial bond between fiber and matrix in order to: (1) maximize the stress transmittance between matrix and fiber phases; and (2) minimize fiber pull-out, and the probability of failure." }, { "idx": 4539, "question": "What is the distinction between matrix and dispersed phases in a composite material?", "answer": "The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase." }, { "idx": 4540, "question": "Contrast the mechanical characteristics of matrix and dispersed phases for fiber-reinforced composites.", "answer": "In general, the matrix phase is relatively weak, has a low elastic modulus, but is quite ductile. On the other hand, the fiber phase is normally quite strong, stiff, and brittle." }, { "idx": 4541, "question": "Briefly explain the difference between oxidation and reduction electrochemical reactions.", "answer": "Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or electrons) and becomes an anion." }, { "idx": 4542, "question": "Which reaction occurs at the anode and which at the cathode?", "answer": "Oxidation occurs at the anode; reduction at the cathode." }, { "idx": 4543, "question": "(a) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in each of the following solutions: (i) HCl", "answer": "In HCl, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 2 H+ + 2 e- -> H2 (reduction)." }, { "idx": 4544, "question": "(a) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in each of the following solutions: (ii) an HCl solution containing dissolved oxygen", "answer": "In an HCl solution containing dissolved oxygen, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 4 H+ + O2 + 4 e- -> 2 H2O (reduction)." }, { "idx": 4545, "question": "(a) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in each of the following solutions: (iii) an HCl solution containing dissolved oxygen and, in addition, Fe2+ ions", "answer": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are Mg -> Mg2+ + 2 e- (oxidation), 4 H+ + O2 + 4 e- -> 2 H2O (reduction), and Fe2+ + 2 e- -> Fe (reduction)." }, { "idx": 4546, "question": "(b) In which of these solutions would you expect the magnesium to oxidize most rapidly? Why?", "answer": "The magnesium would probably oxidize most rapidly in the HCl solution containing dissolved oxygen and Fe2+ ions because there are two reduction reactions that will consume electrons from the oxidation of magnesium." }, { "idx": 4547, "question": "Demonstrate that the value of F in Equation 17.19 is 96,500 C/mol.", "answer": "The faraday constant F is 96,500 C/mol." }, { "idx": 4548, "question": "Demonstrate that at 25 C (298 K), (RT)/(nF) ln x = (0.0592)/(n) log x.", "answer": "At 25 C (298 K), (RT)/(nF) ln x = (0.0592)/(n) log x. This gives units in volts since a volt is a J/C." }, { "idx": 4549, "question": "Is a voltage generated between the two cell halves of a Zn/Zn2+ concentration cell where both electrodes are pure zinc, with Zn2+ concentrations of 1.0 M and 10^-2 M?", "answer": "Yes, a voltage is generated." }, { "idx": 4550, "question": "What is the magnitude of the voltage generated in the Zn/Zn2+ concentration cell with Zn2+ concentrations of 1.0 M and 10^-2 M?", "answer": "The magnitude of the voltage is 0.0592 v." }, { "idx": 4551, "question": "Which electrode will be oxidized in the Zn/Zn2+ concentration cell with Zn2+ concentrations of 1.0 M and 10^-2 M?", "answer": "The electrode in the cell half with the Zn2+ concentration of 10^-2 M will be oxidized." }, { "idx": 4552, "question": "A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 10 in^{2} and that approximately 2.6kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 \\mathrm{mpy} for this alloy in seawater, estimate the time of submersion in years. The density of steel is 7.9g / {cm}^{3}.", "answer": "the time of submersion is 10 \\mathrm{yr}." }, { "idx": 4553, "question": "A thick steel sheet of area 400 cm^{2} is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of 375g due to corrosion. To what rate of corrosion, in both mpy and mm / \\mathrm{yr}, does this correspond?", "answer": "the corrosion rate is 1.2mm / \\mathrm{yr}. the corrosion rate is 46.7 \\mathrm{mpy}." }, { "idx": 4554, "question": "Cite the major differences between activation and concentration polarizations.", "answer": "Activation polarization is the condition wherein a reaction rate is controlled by one step in a series of steps that takes place at the slowest rate. For corrosion, activation polarization is possible for both oxidation and reduction reactions. Concentration polarization occurs when a reaction rate is limited by diffusion in a solution. For corrosion, concentration polarization is possible only for reduction reactions." }, { "idx": 4555, "question": "Under what conditions is activation polarization rate controlling?", "answer": "Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high." }, { "idx": 4556, "question": "Under what conditions is concentration polarization rate controlling?", "answer": "Concentration polarization is rate controlling when the reaction rate is high and/or the concentration of active species in the liquid solution is low." }, { "idx": 4557, "question": "Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal M (in mol/cm2-s) given the following corrosion data: For Metal M: V(M/M2+)=-0.47 V, i0=5x10-10 A/cm2, β=+0.15; For Hydrogen: V(H+/H2)=0 V, i0=2x10-9 A/cm2, β=-0.12.", "answer": "the rate of corrosion of metal m is 2.64x10-13 mol/cm2-s." }, { "idx": 4558, "question": "Compute the corrosion potential for this reaction given the following corrosion data: For Metal M: V(M/M2+)=-0.47 V, i0=5x10-10 A/cm2, β=+0.15; For Hydrogen: V(H+/H2)=0 V, i0=2x10-9 A/cm2, β=-0.12.", "answer": "the corrosion potential for this reaction is -0.169 v." }, { "idx": 4559, "question": "Briefly describe the phenomenon of passivity.", "answer": "Passivity is the loss of chemical reactivity, under particular environmental conditions, of normally active metals and alloys." }, { "idx": 4560, "question": "Name two common types of alloy that passivate.", "answer": "Stainless steels and aluminum alloys often passivate." }, { "idx": 4561, "question": "Why does chromium in stainless steels make them more corrosion resistant in many environments than plain carbon steels?", "answer": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms." }, { "idx": 4562, "question": "Briefly explain why cold-worked metals are more susceptible to corrosion than noncold-worked metals.", "answer": "Cold-worked metals are more susceptible to corrosion than noncold-worked metals because of the increased dislocation density for the latter. The region in the vicinity of a dislocation that intersects the surface is at a higher energy state, and, therefore, is more readily attacked by a corrosive solution." }, { "idx": 4563, "question": "Briefly explain why, for a small anode-to-cathode area ratio, the corrosion rate will be higher than for a large ratio.", "answer": "For a small anode-to-cathode area ratio, the corrosion rate will be higher than for A large ratio. The reason for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i) according to Equation." }, { "idx": 4564, "question": "What are inhibitors?", "answer": "Inhibitors are substances that, when added to a corrosive environment in relatively low concentrations, decrease the environment's corrosiveness." }, { "idx": 4565, "question": "What possible mechanisms account for their effectiveness?", "answer": "Possible mechanisms that account for the effectiveness of inhibitors are: (1) elimination of a chemically active species in the solution; (2) attachment of inhibitor molecules to the corroding surface so as to interfere with either the oxidation or reduction reaction; and (3) the formation of a very thin and protective coating on the corroding surface." }, { "idx": 4566, "question": "Briefly describe the sacrificial anode technique used for galvanic protection.", "answer": "A sacrificial anode is electrically coupled to the metal piece to be protected, which anode is also situated in the corrosion environment. The sacrificial anode is a metal or alloy that is chemically more reactive in the particular environment. It (the anode) preferentially oxidizes, and, upon giving up electrons to the other metal, protects it from electrochemical corrosion." }, { "idx": 4567, "question": "Briefly describe the impressed current technique used for galvanic protection.", "answer": "An impressed current from an external dc power source provides excess electrons to the metallic structure to be protected." }, { "idx": 4568, "question": "Compute the electrical conductivity of a 5.1-mm (0.2-in.) diameter cylindrical silicon specimen 51mm (2 in.) long in which a current of 0.1 A passes in an axial direction. A voltage of 12.5V is measured across two probes that are separated by 38mm (1.5 in.).", "answer": "the electrical conductivity is σ = 14.9 (ω·m)^-1." }, { "idx": 4569, "question": "Compute the resistance over the entire 51mm (2 in.) of the specimen.", "answer": "the resistance is r = 168 ω." }, { "idx": 4570, "question": "Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the electric field is 1000 V/m.", "answer": "the drift velocity of electrons in germanium is 380 m/s." }, { "idx": 4571, "question": "Under these circumstances, how long does it take an electron to traverse a 25-mm (1-in.) length of crystal?", "answer": "the time required for an electron to traverse a 25-mm length of crystal is 6.6 × 10^-5 s." }, { "idx": 4572, "question": "At room temperature the electrical conductivity and the electron mobility for copper are 6.0 x 10^7 (Ω·m)^-1 and 0.0030 m^2/V·s, respectively. Compute the number of free electrons per cubic meter for copper at room temperature.", "answer": "the number of free electrons per cubic meter for copper at room temperature is 1.25 x 10^29 m^-3." }, { "idx": 4573, "question": "At room temperature the electrical conductivity and the electron mobility for copper are 6.0 x 10^7 (Ω·m)^-1 and 0.0030 m^2/V·s, respectively. What is the number of free electrons per copper atom? Assume a density of 8.9 g/cm^3.", "answer": "the number of free electrons per copper atom is 1.48." }, { "idx": 4574, "question": "At room temperature the electrical conductivity of \\mathrm{PbTe} is 500(\\Omega·m)^{-1}, whereas the electron and hole mobilities are 0.16 and 0.075{m}^{2} / V·s, respectively. Compute the intrinsic carrier concentration for \\mathrm{PbTe} at room temperature.", "answer": "the intrinsic carrier concentration for \\mathrm{pbte} at room temperature is 1.33 × 10^{22}{m}^{-3}." }, { "idx": 4575, "question": "Define the term 'intrinsic' as it pertains to semiconducting materials and provide an example.", "answer": "intrinsic--high purity (undoped) Si, GaAs, CdS, etc." }, { "idx": 4576, "question": "Define the term 'extrinsic' as it pertains to semiconducting materials and provide an example.", "answer": "extrinsic--P-doped Ge, B-doped Si, S-doped GaP, etc." }, { "idx": 4577, "question": "Define the term 'compound' as it pertains to semiconducting materials and provide an example.", "answer": "compound--GaAs, InP, CdS, etc." }, { "idx": 4578, "question": "Define the term 'elemental' as it pertains to semiconducting materials and provide an example.", "answer": "elemental--Ge and Si." }, { "idx": 4579, "question": "Explain why no hole is generated by the electron excitation involving a donor impurity atom.", "answer": "No hole is generated by an electron excitation involving a donor impurity atom because the excitation comes from a level within the band gap, and thus, no missing electron is created within the normally filled valence band." }, { "idx": 4580, "question": "Explain why no free electron is generated by the electron excitation involving an acceptor impurity atom.", "answer": "No free electron is generated by an electron excitation involving an acceptor impurity atom because the electron is excited from the valence band into the impurity level within the band gap; no free electron is introduced into the conduction band." }, { "idx": 4581, "question": "Germanium to which 5 × 10^22 m^-3 Sb atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier exists for each Sb atom). Is this material n-type or p-type?", "answer": "this germanium material to which has been added 5 × 10^22 m^-3 sb atoms is n-type since sb is a donor in ge." }, { "idx": 4582, "question": "Germanium to which 5 × 10^22 m^-3 Sb atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier exists for each Sb atom). Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and 0.05 m^2/V-s, respectively.", "answer": "the electrical conductivity of this material is 800 (ω·m)^-1." }, { "idx": 4583, "question": "The following electrical characteristics have been determined for both intrinsic and p-type extrinsic indium phosphide ( \\mathrm{InP} ) at room temperature:\n\\begin{tabular}{lccc}\n\\hline & \\boldsymbol{\\sigma}\\left(\\boldsymbol{\\Omega}·\\mathbf{m}\\right)^{-1} & \\mathbf{n}\\left(\\mathbf{m}^{-3}\\right) & \\mathbf{p}\\left(\\mathbf{m}^{-3}\\right) \\\\\n\\hline Intrinsic & 2.5 × 10^{-6} & 3.0 × 10^{13} & 3.0 × 10^{12} \\\\\nExtrinsic (n-type) & 3.6 × 10^{-5} & 4.5 × 10^{14} & 2.0 × 10^{12} \\\\\n\\hline\n\\end{tabular}\nCalculate electron and hole mobilities.", "answer": "the electron and hole mobilities for \\mathrm{inp} are \\mu_{e} = 0.50 \\, m^{2}/\\mathrm{v}\\cdots and \\mu_{h} = 0.02 \\, m^{2}/\\mathrm{v}\\cdots." }, { "idx": 4584, "question": "Some metal alloy is known to have electrical conductivity and electron mobility values of 1.5 × 10^{7} (\\Omega·m)^{-1} and 0.0020{m}^{2} / V·s, respectively. Through a specimen of this alloy that is 35mm thick is passed a current of 45 A. What magnetic field would need to be imposed to yield a Hall voltage of -1.0 × 10^{-7}V ?", "answer": "the required magnetic field is 0.58 \\mathrm{tesla}." }, { "idx": 4585, "question": "What is one function that a transistor may perform in an electronic circuit?", "answer": "In an electronic circuit, a transistor may be used to amplify an electrical signal." }, { "idx": 4586, "question": "What is another function that a transistor may perform in an electronic circuit?", "answer": "In an electronic circuit, a transistor may act as a switching device in computers." }, { "idx": 4587, "question": "At temperatures between 775^{\\circ} C\\left(1048 K\\right) and 1100^{\\circ} C\\left(1373 K\\right), the activation energy and preexponential for the diffusion coefficient of \\mathrm{Fe}^{2+} in \\mathrm{FeO} are 102,000 J/ mol and 7.3 × 10^{-8}{m}^{2} / s, respectively. Compute the mobility for an \\mathrm{Fe}^{2+} ion at 1000^{\\circ} C(1273 K).", "answer": "the mobility for an \\mathrm{fe}^{2+} ion at 1000^{\\circ} C (1273 k) is 8.64 × 10^{-11}{m}^{2} / \\mathrm{v}·s." }, { "idx": 4588, "question": "A parallel-plate capacitor using a dielectric material having an \\varepsilon_{r} of 2.5 has a plate spacing of 1 mm (0.04 in.). If another material having a dielectric constant of 4.0 is used and the capacitance is to be unchanged, what must be the new spacing between the plates?", "answer": "the new spacing between the plates must be 1.6 mm." }, { "idx": 4589, "question": "(a) For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field.", "answer": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field." }, { "idx": 4590, "question": "(b) For solid lead titanate (PbTiO3), gaseous neon, diamond, solid KCl, and liquid NH3 what kind(s) of polarization is (are) possible? Why?", "answer": "(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state." }, { "idx": 4591, "question": "For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/mol-K, and the Debye temperature is 375 K. Estimate the specific heat at 50 K.", "answer": "the specific heat at 50 K is 139 J/kg-K." }, { "idx": 4592, "question": "For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/mol-K, and the Debye temperature is 375 K. Estimate the specific heat at 425 K.", "answer": "the specific heat at 425 K is 923 J/kg-K." }, { "idx": 4593, "question": "(a) Briefly explain why Cv rises with increasing temperature at temperatures near 0 K.", "answer": "The reason that Cv rises with increasing temperature at temperatures near 0 K is because, in this temperature range, the allowed vibrational energy levels of the lattice waves are far apart relative to the available thermal energy, and only a portion of the lattice waves may be excited. As temperature increases, more of the lattice waves may be excited by the available thermal energy, and, hence, the ability of the solid to absorb energy (i.e., the magnitude of the heat capacity) increases." }, { "idx": 4594, "question": "(b) Briefly explain why Cv becomes virtually independent of temperature at temperatures far removed from 0 K.", "answer": "At temperatures far removed from 0 K, Cv becomes independent of temperature because all of the lattice waves have been excited and the energy required to produce an incremental temperature change is nearly constant." }, { "idx": 4595, "question": "A 0.1{m}(3.9 in.) rod of a metal elongates 0.2 mm(0.0079 in.) on heating from 20 to 100^{\\circ} C(68 \\mathrm{to} 212^{\\circ} F ). Determine the value of the linear coefficient of thermal expansion for this material.", "answer": "the linear coefficient of thermal expansion for this material is 25.0 × 10^{-6} \\, (\\degree c)^{-1}." }, { "idx": 4596, "question": "Briefly explain why the thermal conductivities are higher for crystalline than noncrystalline ceramics.", "answer": "Thermal conductivities are higher for crystalline than for noncrystalline ceramics because, for noncrystalline, phonon scattering, and thus the resistance to heat transport, is much more effective due to the highly disordered and irregular atomic structure." }, { "idx": 4597, "question": "Briefly explain why metals are typically better thermal conductors than ceramic materials.", "answer": "Metals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons." }, { "idx": 4598, "question": "Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric materials, rendering them more thermally insulative.", "answer": "Porosity decreases the thermal conductivity of ceramic and polymeric materials because the thermal conductivity of a gas phase that occupies pore space is extremely small relative to that of the solid material. Furthermore, contributions from gaseous convection are generally insignificant." }, { "idx": 4599, "question": "Briefly explain how the degree of crystallinity affects the thermal conductivity of polymeric materials and why.", "answer": "Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal transport when a crystalline structure prevails." }, { "idx": 4600, "question": "For some ceramic materials, why does the thermal conductivity first decrease and then increase with rising temperature?", "answer": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature." }, { "idx": 4601, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Pure copper; aluminum bronze (95 wt% Cu-5 wt% Al).", "answer": "Pure copper will have a larger conductivity than aluminum bronze because the impurity atoms in the latter will lead to a greater degree of free electron scattering." }, { "idx": 4602, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Fused silica; quartz.", "answer": "Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline (whereas quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline materials." }, { "idx": 4603, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Linear polyethylene; branched polyethylene.", "answer": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity." }, { "idx": 4604, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Random poly(styrene-butadiene) copolymer; alternating poly(styrene-butadiene) copolymer.", "answer": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c)." }, { "idx": 4605, "question": "Explain the two sources of magnetic moments for electrons.", "answer": "The two sources of magnetic moments for electrons are the electron's orbital motion around the nucleus, and also, its spin." }, { "idx": 4606, "question": "Do all electrons have a net magnetic moment? Why or why not?", "answer": "Each electron will have a net magnetic moment from spin, and possibly, orbital contributions, which do not cancel for an isolated atom." }, { "idx": 4607, "question": "Do all atoms have a net magnetic moment? Why or why not?", "answer": "All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments." }, { "idx": 4608, "question": "Confirm that there are 2.2 Bohr magnetons associated with each iron atom, given that the saturation magnetization is 1.70 × 10^{6} \\mathrm{~A} / m, that iron has a BCC crystal structure, and that the unit cell edge length is 0.2866 nm.", "answer": "2.16 bohr magnetons/atom" }, { "idx": 4609, "question": "There is associated with each atom in paramagnetic and ferromagnetic materials a net magnetic moment. Explain why ferromagnetic materials can be permanently magnetized whereas paramagnetic ones cannot.", "answer": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains." }, { "idx": 4610, "question": "The formula for yttrium iron garnet \\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right) may be written in the form \\mathrm{Y}_{3}^{+} \\mathrm{Fe}_{4}^{a} \\mathrm{Fe}_{d}^{a} \\mathrm{O}_{12}, where the superscripts a, c, and d represent different sites on which the Y^{3+} and \\mathrm{Fe}^{3+} ions are located. The spin magnetic moments for the Y^{3+} and \\mathrm{Fe}^{3+}\n moments for the Y^{3+} and \\mathrm{Fe}^{3-} ions positioned in the a and c sites are oriented parallel to one another and antiparallel to the \\mathrm{Fe}^{3+} ions in d sites. Compute the number of Bohr magnetons associated with each Y^{3+} ion, given the following information: (1) each unit cell consists of eight formula \\left(Y_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right), units; (2) the unit cell is cubic with an edge length of 1.2376nm; (3) the saturation magnetization for this material is 1.0 × 10^{4} \\mathrm{~A} / m; and (4) assume that there are 5 Bohr magnetons associated with each \\mathrm{Fe}^{3+} ion.", "answer": "the number of bohr magnetons associated with each y^{3+} ion is 1.75 \\, \\text{bm}." }, { "idx": 4611, "question": "Why does the magnitude of the saturation magnetization decrease with increasing temperature for ferromagnetic materials?", "answer": "The saturation magnetization decreases with increasing temperature because the atomic thermal vibrational motions counteract the coupling forces between the adjacent atomic dipole moments, causing some magnetic dipole misalignment." }, { "idx": 4612, "question": "Why does ferromagnetic behavior cease above the Curie temperature for ferromagnetic materials?", "answer": "Ferromagnetic behavior ceases above the Curie temperature because the atomic thermal vibrations are sufficiently violent so as to completely destroy the mutual spin coupling forces." }, { "idx": 4613, "question": "An iron bar magnet having a coercivity of 4000 \\mathrm{~A} / m is to be demagnetized. If the bar is inserted within a cylindrical wire coil 0.15{m} long and having 100 turns, what electric current is required to generate the necessary magnetic field?", "answer": "the required electric current is 6 \\mathrm{a}." }, { "idx": 4614, "question": "Visible light having a wavelength of 6 x 10^-7 m appears orange. Compute the frequency of a photon of this light.", "answer": "the frequency of the photon is 5 x 10^14 s^-1." }, { "idx": 4615, "question": "Visible light having a wavelength of 6 x 10^-7 m appears orange. Compute the energy of a photon of this light.", "answer": "the energy of the photon is 3.31 x 10^-19 j (2.07 ev)." }, { "idx": 4616, "question": "What are the characteristics of opaque materials in terms of their appearance and light transmittance?", "answer": "Opaque materials are impervious to light transmission; it is not possible to see through them." }, { "idx": 4617, "question": "What are the characteristics of translucent materials in terms of their appearance and light transmittance?", "answer": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material." }, { "idx": 4618, "question": "What are the characteristics of transparent materials in terms of their appearance and light transmittance?", "answer": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them." }, { "idx": 4619, "question": "In ionic materials, how does the size of the component ions affect the extent of electronic polarization?", "answer": "In ionic materials, the larger the size of the component ions the greater the degree of electronic polarization." }, { "idx": 4620, "question": "Can a material have an index of refraction less than unity? Why or why not?", "answer": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum. This is not possible." }, { "idx": 4621, "question": "Briefly describe the phenomenon of dispersion in a transparent medium.", "answer": "Dispersion in a transparent medium is the phenomenon wherein the index of refraction varies slightly with the wavelength of the electromagnetic radiation." }, { "idx": 4622, "question": "Briefly explain how reflection losses of transparent materials are minimized by thin surface coatings.", "answer": "The thickness and dielectric constant of a thin surface coating are selected such that there is destructive interference between the light beam that is reflected from the lens-coating interface and the light beam that is reflected from the coating-air interface; thus, the net intensity of the total reflected beam is very low." }, { "idx": 4623, "question": "The index of refraction of corundum \\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right) is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.757 and 1.779 in the direction of light propagation.", "answer": "the reflectivity at the boundary is 3.87 × 10^{-5}." }, { "idx": 4624, "question": "Briefly explain what determines the characteristic color of a metal.", "answer": "The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected." }, { "idx": 4625, "question": "Briefly explain what determines the characteristic color of a transparent nonmetal.", "answer": "The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material." }, { "idx": 4626, "question": "Briefly explain why some transparent materials appear colored while others are colorless.", "answer": "For a transparent material that appears colorless, any absorption within its interior is the same for all visible wavelengths. On the other hand, if there is any selective absorption of visible light (usually by electron excitations), the material will appear colored, its color being dependent on the frequency distribution of the transmitted light beam." }, { "idx": 4627, "question": "Briefly explain why amorphous polymers are transparent, while predominantly crystalline polymers appear opaque or, at best, translucent.", "answer": "Amorphous polymers are normally transparent because there is no scattering of a light beam within the material. However, for semicrystalline polymers, visible light will be scattered at boundaries between amorphous and crystalline regions since they have different indices of refraction. This leads to translucency or, for extensive scattering, opacity, except for semicrystalline polymers having very small crystallites." }, { "idx": 4628, "question": "Zinc has five naturally occurring isotopes: 48.63% of { }^{64} Zn with an atomic weight of 63.929 amu; 27.90% of { }^{66} Zn with an atomic weight of 65.926 amu^{\\text {; }} 4.10% of { }^{67} Zn with an atomic weight of 66.927 amu ; 18.75% of { }^{68} Zn with an atomic weight of 67.925 amu ; and 0.62% of { }^{70} Zn with an atomic weight of 69.925 amu. Calculate the average atomic weight of Zn.", "answer": "the average atomic weight of \\mathrm{zn} is 65.400 amu." }, { "idx": 4629, "question": "Indium has two naturally occurring isotopes: { }^{113} In with an atomic weight of 112.904 amu, and { }^{115} \\mathrm{In} with an atomic weight of 114.904 amu. If the average atomic weight for In is 114.818 amu, calculate the fraction-of-occurrences of these two isotopes.", "answer": "the fraction-of-occurrence of { }^{113}in is 0.043. the fraction-of-occurrence of { }^{115}in is 0.957." }, { "idx": 4630, "question": "Potassium iodide (KI) exhibits predominantly ionic bonding. The K+ ion has an electron structure that is identical to which inert gas?", "answer": "The K+ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon." }, { "idx": 4631, "question": "Potassium iodide (KI) exhibits predominantly ionic bonding. The I- ion has an electron structure that is identical to which inert gas?", "answer": "The I- ion is an iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon." }, { "idx": 4632, "question": "With regard to electron configuration, what do all the elements in Group IIA of the periodic table have in common?", "answer": "Each of the elements in Group IIA has two s electrons." }, { "idx": 4633, "question": "To what group in the periodic table would an element with atomic number 112 belong?", "answer": "From the periodic table the element having atomic number 112 would belong to group IIB. According to the periodic table, Ds, having an atomic number of 110 lies below Pt in the periodic table and in the right-most column of group VIII. Moving two columns to the right puts element 112 under \\mathrm{Hg} and in group IIB.\nThis element has been artificially created and given the name Copernicium with the symbol Cn. It was named after Nicolaus Copernicus, the Polish scientist who proposed that the earth moves around the sun (and not vice versa)." }, { "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}) electron configuration is that of a halogen because it is one electron deficient from having a filled (p) subshell.", "idx": 4634 }, { "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.", "idx": 4635 }, { "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}) electron configuration is that of an inert gas because of filled (4 s) and (4 p) subshells.", "idx": 4636 }, { "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.", "idx": 4637 }, { "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.", "idx": 4638 }, { "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.", "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons.", "idx": 4639 }, { "question": "Differentiate (E_{N}) with respect to (r), and then set the resulting expression equal to zero, because the curve of (E_{N}) versus (r) is a minimum at (E_{0}).", "answer": "The derivative of (E_{N}) with respect to (r) is set to zero to find the minimum.", "idx": 4640 }, { "question": "Solve for (r) in terms of (A, B), and (n), which yields (r_{0}), the equilibrium interionic spacing.", "answer": "The equilibrium interionic spacing (r_{0}) is solved in terms of (A, B), and (n).", "idx": 4641 }, { "question": "Determine the expression for (E_{0}) by substitution of (r_{0}) into Equation.", "answer": "The bonding energy (e_{0}) is given by: (e_{0}=-\frac{a}{\\left(\frac{a}{nb}\right)^{1 /(1-n)}}+\frac{b}{\\left(\frac{a}{nb}\right)^{n /(1-n)}})", "idx": 4642 }, { "idx": 4643, "question": "Explain why hydrogen fluoride (HF) has a higher boiling temperature than hydrogen \\\\ chloride (\\mathrm{HCl})\\left(19.4 \\mathrm{vs} .-85^{\\circ} C\\right), even though \\mathrm{HF} has a lower molecular weight.", "answer": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature." }, { "idx": 4644, "question": "What type(s) of bonding would be expected for solid xenon?", "answer": "For solid xenon, the bonding is van der Waals since xenon is an inert gas." }, { "idx": 4645, "question": "What type(s) of bonding would be expected for calcium fluoride (CaF2)?", "answer": "For CaF2, the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of Ca and F in the periodic table." }, { "idx": 4646, "question": "What type(s) of bonding would be expected for bronze?", "answer": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin)." }, { "idx": 4647, "question": "What type(s) of bonding would be expected for cadmium telluride (CdTe)?", "answer": "For CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of Cd and Te in the periodic table." }, { "idx": 4648, "question": "What type(s) of bonding would be expected for rubber?", "answer": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)" }, { "idx": 4649, "question": "What type(s) of bonding would be expected for tungsten?", "answer": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table." }, { "idx": 4650, "question": "Which of the following electron configurations is for an inert gas? \\\\ (A) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} \\\\ (B) 1 s^{2} 2 s^{2} 2 p^{6} s 3 s^{2} \\\\ (C) 1 s^{2} 2 s^{2} 2 p^{6} .3 s^{2} 3 p^{6} 4 s^{1} \\\\ (D) 1 s^{2} 2 s^{2} 2 p^{6}·3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}", "answer": "The correct answer is A. The 1 s^{2} 2 s^{2} 2 p^{6} 13 s^{2} 3 p^{6} electron configuration is that of an inert gas because of filled 3 s and 3 p subshells." }, { "idx": 4651, "question": "What type(s) of bonding would be expected for bronze (a copper-tin alloy)?\n(A) Ionic bonding\n(B) Metallic bonding\n(C) Covalent bonding with some van der Waals bonding\n(D) van der Waals bonding", "answer": "The correct answer is B. For bronze, the bonding is metallic because it is a metal alloy." }, { "idx": 4652, "question": "What type(s) of bonding would be expected for rubber?\n(A) Ionic bonding\n(B) Metallic bonding\n(C) Covalent bonding with some van der Waals bonding\n(D) van der Waals bonding", "answer": "FE}\nThe correct answer is C. For rubber, the bonding is covalent with some van der Waals bonding. (Rubber is composed primarily of carbon and hydrogen atoms.)" }, { "idx": 4653, "question": "If the atomic radius of lead is 0.175nm, calculate the volume of its unit cell in cubic meters.", "answer": "the volume of the unit cell is 1.213 × 10^{-28} m^{3}." }, { "idx": 4654, "question": "Calculate the radius of a palladium (P d) atom, given that P d has an FCC crystal structure, a density of 12.0g / {cm}^{3}, and an atomic weight of 106.4g / mol.", "answer": "the radius of a palladium (pd) atom is 0.138nm." }, { "idx": 4655, "question": "Calculate the radius of a tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6g / {cm}^{3}, and an atomic weight of 180.9g / mol.", "answer": "the radius of a tantalum (ta) atom is 0.143 \\, \\text{nm}." }, { "idx": 4656, "question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm^3. What is the volume of its unit cell in cubic meters?", "answer": "the volume of the ti unit cell is 1.058 x 10^-28 m^3." }, { "idx": 4657, "question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm^3. If the c/a ratio is 1.58, compute the values of c and a.", "answer": "the values of a and c are 0.296 nm and 0.468 nm, respectively." }, { "idx": 4658, "question": "Magnesium (Mg) has an HCP crystal structure and a density of 1.74 g/cm3. What is the volume of its unit cell in cubic centimeters?", "answer": "the volume of the unit cell is 1.39 × 10^-22 cm3." }, { "idx": 4659, "question": "Magnesium (Mg) has an HCP crystal structure and a density of 1.74 g/cm3. If the c/a ratio is 1.624, compute the values of c and a.", "answer": "the values of a and c are 0.321 nm and 0.521 nm, respectively." }, { "idx": 4660, "question": "The unit cell for uranium (U) has orthorhombic symmetry, with a, b, and c lattice parameters of 0.286,0.587, and 0.495nm, respectively. If its density, atomic weight, and atomic radius are 19.05g / {cm}^{3}, 238.03g / mol, and 0.1385nm, respectively, compute the atomic packing factor.\n\\title{", "answer": "the atomic packing factor (apf) for uranium is 0.536." }, { "idx": 4661, "question": "Indium (In) has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm, respectively. If the atomic packing factor and atomic radius are 0.693 and 0.1625 nm, respectively, determine the number of atoms in each unit cell.", "answer": "4.0 atoms/unit cell" }, { "idx": 4662, "question": "Indium (In) has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm, respectively. The atomic weight of indium is 114.82 g/mol; compute its theoretical density.", "answer": "7.31 g/cm3" }, { "question": "Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is 0.1143 nm, determine the unit cell volume.", "answer": "4.87 × 10^{-23} cm3", "idx": 4663 }, { "question": "Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is 0.1143nm, calculate the theoretical density of Be and compare it with the literature value.", "answer": "the theoretical density of be is 1.84 g/cm3, and the literature value is 1.85 g/cm3", "idx": 4664 }, { "idx": 4665, "question": "Magnesium ( Mg ) has an HCP crystal structure, a c/a ratio of 1.624 , and a density of 1.74 \\mathrm{g} / {cm}^{3}. Compute the atomic radius for Mg.", "answer": "the atomic radius for \\mathrm{mg} is 0.160\\;nm." }, { "idx": 4666, "question": "Cobalt (Co) has an HCP crystal structure, an atomic radius of 0.1253nm, and a c/a ratio of 1.623. Compute the volume of the unit cell for Co.", "answer": "the volume of the unit cell for co is 6.64 × 10^{-23} {cm}^3 or 6.64 × 10^{-2} nm^3." }, { "idx": 4667, "question": "One crystalline form of silica \\left(\\mathrm{SiO}_{2}\\right) has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.700nm. If the measured density is 2.32g / {cm}^{3}, how many \\mathrm{Si}^{4+} and \\mathrm{O}^{2}-ions are there per unit cell?", "answer": "there are 8 \\mathrm{si}^{4+} ions and 16 \\mathrm{o}^{2-} ions per \\mathrm{sio}_{2} unit cell." }, { "idx": 4668, "question": "A hypothetical AX type of ceramic material is known to have a density of 2.10g / {cm}^{3} and a unit cell of cubic symmetry with a cell edge length of 0.57nm. The atomic weights of the A and X elements are 28.5 and 30.0g / mol, respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: sodium chloride, cesium chloride, or zinc blende? Justify your choice(s).", "answer": "of the three possible crystal structures, only sodium chloride and zinc blende have four formula units per unit cell, and therefore, are possibilities." }, { "idx": 4669, "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. Which type of interstitial site will the Fe2+ ions occupy? Why?", "answer": "From Table, the cation-anion radius ratio is 0.077 nm / 0.140 nm = 0.550. Since this ratio is between 0.414 and 0.732, the Fe2+ ions will occupy octahedral sites." }, { "idx": 4670, "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. Which type of interstitial site will the Ti4+ ions occupy? Why?", "answer": "The titanium-oxygen ionic radius ratio is 0.0601 nm / 0.140 nm = 0.436. Since this ratio is between 0.414 and 0.723, the Ti4+ ions will also occupy octahedral sites." }, { "idx": 4671, "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. What fraction of the total tetrahedral sites will be occupied?", "answer": "Since both Fe2+ and Ti4+ ions occupy octahedral sites, no tetrahedral sites will be occupied." }, { "idx": 4672, "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. What fraction of the total octahedral sites will be occupied?", "answer": "For every FeTiO3 formula unit, there are three O2- ions, and, therefore, three octahedral sites; since there is one ion each of Fe2+ and Ti4+ per formula unit, two-thirds of these octahedral sites will be occupied." }, { "question": "To what equation does the expression for interplanar spacing (d_{\text {hkl }}) reduce for crystals having cubic symmetry, given that (a,(b), and c are the lattice parameters?", "answer": "For the crystals having cubic symmetry, a=b=c. Making this substitution into the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{a^{2}} = \frac{h^{2}+k^{2}+l^{2}}{a^{2}}).", "idx": 4673 }, { "question": "To what equation does the expression for interplanar spacing (d_{\text {hkl }}) reduce for crystals having tetragonal symmetry, given that a, b and c are the lattice parameters?", "answer": "For crystals having tetragonal symmetry, a=b \neq c. Replacing b with a in the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}} = \frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}).", "idx": 4674 }, { "idx": 4675, "question": "The number-average molecular weight of a polystyrene is 500,000g / mol. Compute the degree of polymerization.\n\\title{", "answer": "the degree of polymerization is 4800." }, { "idx": 4676, "question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 8% of all the original hydrogen atoms.", "answer": "the concentration of Cl that must be added is 29.0 wt%." }, { "idx": 4677, "question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)?", "answer": "chlorinated polyethylene differs from poly(vinyl chloride) in that, for pvc, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random." }, { "idx": 4678, "question": "Compare thermoplastic and thermosetting polymers on the basis of mechanical characteristics upon heating.", "answer": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening." }, { "idx": 4679, "question": "Compare thermoplastic and thermosetting polymers according to possible molecular structures.", "answer": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked." }, { "idx": 4680, "question": "The number-average molecular weight of a poly(acrylonitrile-butadiene) alternating copolymer is 1,000,000g / mol; determine the average number of acrylonitrile and butadiene repeat units per molecule.", "answer": "the average number of acrylonitrile and butadiene repeat units per molecule is 9333." }, { "idx": 4681, "question": "Calculate the number-average molecular weight of a random poly(isobutylene-isoprene) copolymer in which the fraction of isobutylene repeat units is 0.25 ; assume that this concentration corresponds to a degree of polymerization of 1500 .", "answer": "the number-average molecular weight of the poly(isobutylene-isoprene) copolymer is \\bar{m}_{n} = 97,700g/mol." }, { "idx": 4682, "question": "An alternating copolymer is known to have a number-average molecular weight of 100,000g / mol and a degree of polymerization of 2210 . If one of the repeat units is ethylene, which of styrene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit? Why?", "answer": "the other repeat unit is vinyl chloride, with a molecular weight of 62.49 \\mathrm{g/mol}. this matches the calculated value of 62.45 \\mathrm{g/mol} for the unknown repeat unit." }, { "idx": 4683, "question": "(a) Determine the ratio of butadiene to acrylonitrile repeat units in a copolymer having a number-average molecular weight of 250,000 g/mol and a degree of polymerization of 4640.", "answer": "the ratio of butadiene to acrylonitrile repeat units in the copolymer is fb/fa = 4.0." }, { "idx": 4684, "question": "(b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?", "answer": "of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit types is alternating; the ratio must be 1:1. therefore, on the basis of the result in part (a), the possibilities for this copolymer are random, graft, and block." }, { "idx": 4685, "question": "Crosslinked copolymers consisting of 35 wt% ethylene and 65 wt% propylene may have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both repeat unit types.\n\\title{", "answer": "the fraction of the ethylene repeat unit is 0.45. the fraction of the propylene repeat unit is 0.55." }, { "idx": 4686, "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Linear and atactic poly(vinyl chloride); linear and isotactic polypropylene", "answer": "No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more likely to crystallize." }, { "idx": 4687, "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (b) Linear and syndiotactic polypropylene; crosslinked cis-polyisoprene", "answer": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones." }, { "idx": 4688, "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (c) Network phenol-formaldehyde; linear and isotactic polystyrene", "answer": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily." }, { "idx": 4689, "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (d) Block poly(acrylonitrile-isoprene) copolymer; graft poly(chloroprene-isobutylene) copolymer", "answer": "Yes, it is possible to decide for these two copolymers. The block poly(acrylonitrileisoprene) copolymer is more likely to crystallize than the graft poly(chloroprene-isobutylene) copolymer. Block copolymers crystallize more easily than graft ones." }, { "idx": 4690, "question": "For some hypothetical metal, the equilibrium number of vacancies at 900^{\\circ} C is 2.3 × 10^{25} m^{-3}. If the density and atomic weight of this metal are 7.40g / {cm}^{3} and 85.5g / mol, respectively, calculate the fraction of vacancies for this metal at 900^{\\circ} C.", "answer": "the fraction of vacancies for this metal at 900^{\\circ} C is 4.41 × 10^{-4}." }, { "idx": 4691, "question": "(a) Calculate the fraction of atom sites that are vacant for copper (Cu) at its melting temperature of 1084°C (1357 K). Assume an energy for vacancy formation of 0.90 eV/atom.", "answer": "the fraction of atom sites that are vacant for copper at 1357 k is 4.56 × 10^-4." }, { "idx": 4692, "question": "(b) Repeat this calculation at room temperature (298 K). Assume an energy for vacancy formation of 0.90 eV/atom.", "answer": "the fraction of atom sites that are vacant for copper at 298 k is 6.08 × 10^-16." }, { "idx": 4693, "question": "(c) What is ratio of N_V/N(1357 K) and N_V/N(298 K) ?", "answer": "the ratio of n_v/n(1357 k) to n_v/n(298 k) is 7.5 × 10^11." }, { "idx": 4694, "question": "Calculate the number of vacancies per cubic meter in gold (\\mathrm{Au}) at 900^{\\circ} C. The energy for vacancy formation is 0.98 \\mathrm{eV} / atom. Furthermore, the density and atomic weight for Au are 18.63 \\mathrm{g} / {cm}^{3} (at 900^{\\circ} C ) and 196.9g / mol, respectively.", "answer": "the number of vacancies per cubic meter in gold at 900^{\\circ} C is 3.52 × 10^{24}{m}^{-3}." }, { "idx": 4695, "question": "Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at its melting temperature \\left(645^{\\circ} C\\right). Assume an energy for defect formation of 1.86 \\mathrm{eV}.", "answer": "the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature is 7.87 × 10^{-6}." }, { "idx": 4696, "question": "In your own words, briefly define the term stoichiometric.", "answer": "Stoichiometric means having exactly the ratio of anions to cations as specified by the chemical formula for the compound." }, { "idx": 4697, "question": "Under conditions where cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. Name one crystalline defect that you would expect to form in order to maintain charge neutrality.", "answer": "For a Cu2+ O2- compound in which a small fraction of the copper ions exist as Cu+, for each Cu+ formed there is one less positive charge introduced (or one more negative charge). In order to maintain charge neutrality, we must either add an additional positive charge or subtract a negative charge. This may be accomplished be either creating Cu2+ interstitials or O2- vacancies." }, { "idx": 4698, "question": "Under conditions where cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. How many Cu+ ions are required for the creation of each defect?", "answer": "There will be two Cu+ ions required for each of these defects." }, { "idx": 4699, "question": "Under conditions where cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. How would you express the chemical formula for this nonstoichiometric material?", "answer": "The chemical formula for this nonstoichiometric material is Cu1+ O or CuO1-x, where x is some small fraction." }, { "idx": 4700, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for FeO.", "answer": "For FeO, the ionic radii of the Mg2+ and Fe2+ are 0.072 nm and 0.077 nm, respectively. The percentage difference in ionic radii is 6.5%, which is within the acceptable range for a high degree of solubility. Additionally, both MgO and FeO have rock salt crystal structures. Therefore, FeO and MgO are expected to form high degrees of solid solubility, and experimentally, they exhibit 100% solubility." }, { "idx": 4701, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for BaO.", "answer": "For BaO, the ionic radii of the Mg2+ and Ba2+ are 0.072 nm and 0.136 nm, respectively. The percentage difference in ionic radii is 47%, which is much larger than the acceptable range. Therefore, BaO is not expected to experience any appreciable solubility in MgO. Experimentally, the solubility of BaO in MgO is very small." }, { "idx": 4702, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for PbO.", "answer": "For PbO, the ionic radii of the Mg2+ and Pb2+ are 0.072 nm and 0.120 nm, respectively. The percentage difference in ionic radii is 40%, which is much larger than the acceptable range. Therefore, PbO is not expected to experience any appreciable solubility in MgO." }, { "idx": 4703, "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for CoO.", "answer": "For CoO, the ionic radii of the Mg2+ and Co2+ are 0.072 nm and 0.072 nm, respectively. The percentage difference in ionic radii is 0%, which is within the acceptable range for a high degree of solubility. Therefore, CoO and MgO are expected to form high degrees of solid solubility, likely 100%." }, { "idx": 4704, "question": "Suppose that CaO is added as an impurity to Li2O. If the Ca2+ substitutes for Li+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca2+ added?", "answer": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed." }, { "idx": 4705, "question": "Suppose that CaO is added as an impurity to CaCl2. If the O2- substitutes for Cl-, what kind of vacancies would you expect to form? How many of these vacancies are created for every O2- added?", "answer": "For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy." }, { "idx": 4706, "question": "Which of these elements would you expect to form a substitutional solid solution having complete solubility with nickel? The four Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same.", "answer": "Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures." }, { "idx": 4707, "question": "Which of these elements would you expect to form a substitutional solid solution of incomplete solubility with nickel? The four Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same.", "answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Ni are greater than ±15%, and/or have a valence different than 2+." }, { "idx": 4708, "question": "Which of these elements would you expect to form an interstitial solid solution with nickel? The four Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same.", "answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Ni." }, { "idx": 4709, "question": "What is the composition, in atom percent, of an alloy that consists of 92.5 wt% \\mathrm{Ag} and 7.5 wt% Cu ?", "answer": "the composition of the alloy is 87.9 \\text{ at}% \\text{ ag} and 12.1 \\text{ at}% \\text{ cu}." }, { "idx": 4710, "question": "What is the composition, in atom percent, of an alloy that consists of 5.5 wt% \\mathrm{~Pb} and 94.5 wt% \\mathrm{Sn} ?", "answer": "the composition of the alloy is 3.23 \\text{ at% pb} and 96.77 \\text{ at% sn}." }, { "idx": 4711, "question": "What is the composition, in weight percent, of an alloy that consists of 5 at% Cu and 95 at% Pt?", "answer": "the composition in weight percent is: 1.68 wt% cu \n98.32 wt% pt" }, { "idx": 4712, "question": "Calculate the composition, in weight percent, of an alloy that contains 105kg of iron, 0.2 kg of carbon, and 1.0kg of chromium.\n\\title{", "answer": "the composition, in weight percent, of the alloy is: c_{\\mathrm{fe}} = 98.87 \\, \\text{wt}% c_{C} = 0.19 \\, \\text{wt}% c_{\\mathrm{cr}} = 0.94 \\, \\text{wt}%" }, { "idx": 4713, "question": "What is the composition, in atom percent, of an alloy that contains 44.5 lb m of \\mathrm{Ag}, 83.7 lb_{m} of \\mathrm{Au}, and 5.3 lb_{m} of Cu ?", "answer": "the composition of the alloy in atom percent is: c_{\\mathrm{ag}} = 44.8 \\, \\text{at}% c_{\\mathrm{au}} = 46.2 \\, \\text{at}% c_{\\mathrm{cu}} = 9.0 \\, \\text{at}% " }, { "idx": 4714, "question": "Calculate the number of atoms per cubic meter in \\mathrm{Pb}.", "answer": "3.30 × 10^{28} atoms/m^3" }, { "idx": 4715, "question": "Calculate the number of atoms per cubic meter in chromium.", "answer": "the number of atoms per cubic meter in chromium is 8.33 × 10^{28} \\ \\text{atoms/m}^3." }, { "idx": 4716, "question": "The concentration of \\mathrm{Si} in an \\mathrm{Fe}-\\mathrm{Si} alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?", "answer": "the concentration of \\mathrm{si} is 19.6kg / m^{3}." }, { "idx": 4717, "question": "Determine the approximate density of a Ti-6Al-4V titanium alloy that has a composition of 90 wt% \\mathrm{Ti}, 6 wt% \\mathrm{Al}, and 4 wt% V.", "answer": "the approximate density of the ti-6al-4v titanium alloy is 4.38 \\mathrm{g/cm}^3." }, { "idx": 4718, "question": "Calculate the unit cell edge length for an 80 wt% \\mathrm{Ag}-20 wt% \\mathrm{Pd} alloy. All of the palladium is in solid solution, the crystal structure for this alloy is FCC, and the roomtemperature density of P d is 12.02g / {cm}^{3}.", "answer": "the unit cell edge length for an 80 wt% \\mathrm{ag}-20 wt% \\mathrm{pd} alloy is 0.4050nm." }, { "idx": 4719, "question": "Some hypothetical alloy is composed of 25 wt% of metal A and 75 wt% of metal B. If the densities of metals A and B are 6.17 and 8.00g / {cm}^{3}, respectively, and their respective atomic weights are 171.3 and 162.0g / mol, determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic. Assume a unit cell edge length of 0.332nm.", "answer": "the crystal structure is simple cubic with 1.00 atom per unit cell." }, { "idx": 4720, "question": "Molybdenum (Mo) forms a substitutional solid solution with tungsten (W). Compute the number of molybdenum atoms per cubic centimeter for a molybdenum-tungsten alloy that contains 16.4 wt% Mo and 83.6 wt% W. The densities of pure molybdenum and tungsten are 10.22 and 19.30g / {cm}^{3}, respectively.", "answer": "1.73 × 10^{22} \\text{ atoms/cm}^3" }, { "idx": 4721, "question": "Niobium forms a substitutional solid solution with vanadium. Compute the number of niobium atoms per cubic centimeter for a niobium-vanadium alloy that contains 24 wt% nb and 76 wt% V. The densities of pure niobium and vanadium are 8.57 and 6.10g / {cm}^{3}, respectively.", "answer": "the number of niobium atoms per cubic centimeter is 1.02 × 10^{22} \\text{atoms/cm}^3." }, { "idx": 4722, "question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the fraction of unit cells that contain carbon atoms.\n\\title{", "answer": "the fraction of unit cells that contain carbon atoms is 9.31 × 10^{-3} atoms/unit cell. there is one carbon atom per 107.5 unit cells." }, { "idx": 4723, "question": "Gold (Au) forms a substitutional solid solution with silver (Ag). Compute the weight percent of gold that must be added to silver to yield an alloy that contains 5.5 × 10^{21} \\mathrm{Au} atoms per cubic centimeter. The densities of pure \\mathrm{Au} and \\mathrm{Ag} are 19.32 and 10.49g / {cm}^{3}, respectively.", "answer": "15.9 wt%" }, { "idx": 4724, "question": "Germanium (Ge) forms a substitutional solid solution with silicon (Si). Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 2.43 × 10^{21} Ge atoms per cubic centimeter. The densities of pure \\mathrm{Ge} and \\mathrm{Si} are 5.32 and 2.33g / {cm}^{3}, respectively.", "answer": "11.7 wt%" }, { "idx": 4725, "question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch at a magnification of 100x?", "answer": "32 grains/in.^2" }, { "idx": 4726, "question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch without any magnification?", "answer": "320000 grains/in.^2" }, { "idx": 4727, "question": "Determine the ASTM grain size number if 30 grains per square inch are measured at a magnification of 250 ×.", "answer": "the astm grain size number is 8.6." }, { "idx": 4728, "question": "Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 75 ×.", "answer": "the astm grain size number is 4.8." }, { "idx": 4729, "question": "Briefly explain the difference between self-diffusion and interdiffusion.", "answer": "Self-diffusion is atomic migration in pure metals--i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal." }, { "idx": 4730, "question": "Self-diffusion involves the motion of atoms that are all of the same type; therefore, it is not subject to observation by compositional changes, as with interdiffusion. Suggest one way in which self-diffusion may be monitored.", "answer": "Self-diffusion may be monitored by using radioactive isotopes of the metal being studied. The motion of these isotopic atoms may be detected by measurement of radioactivity level." }, { "idx": 4731, "question": "Compare interstitial and vacancy atomic mechanisms for diffusion.", "answer": "With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism." }, { "idx": 4732, "question": "Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.", "answer": "Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom." }, { "idx": 4733, "question": "A sheet of steel 5.0-mm thick has nitrogen atmospheres on both sides at 900^{\\circ} C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 1.85 × 10^{-10}{m}^{2} / s, and the diffusion flux is found to be 1.0 × 10^{-7} kg / m^{2}·s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 2kg / m^{3}. How far into the sheet from this high-pressure side will the concentration be 0.5kg / m^{3} ? Assume a linear concentration profile.", "answer": "the concentration will be 0.5 \\mathrm{kg/m}^{3} at a distance of 2.78 mm from the high-pressure side." }, { "idx": 4734, "question": "For a steel alloy it has been determined that a carburizing heat treatment of 15h duration will raise the carbon concentration to 0.35 wt% at a point 2.0mm from the surface. Estimate the time necessary to achieve the same concentration at a 6.0-mm position for an identical steel and at the same carburizing temperature.", "answer": "the time necessary to achieve the same concentration at a 6.0-mm position is 135 h." }, { "idx": 4735, "question": "The preexponential and activation energy for the diffusion of chromium in nickel are 1.1 × 10^{-4}{m}^{2} / s and 272,000 J/ mol, respectively. At what temperature will the diffusion coefficient have a value of 1.2 × 10^{-14}{m}^{2} / s ?", "answer": "the temperature at which the diffusion coefficient has a value of 1.2 × 10^{-14} m^{2} / s is 1427 k or 1154^{\\circ} C. alternatively, using the vmse software, the temperature is found to be 1430 k." }, { "idx": 4736, "question": "The activation energy for the diffusion of copper in silver is 193,000 J/ mol. Calculate the diffusion coefficient at 1200 K\\left(927^{\\circ} C\\right), given that D at 1000 K\\left(727^{\\circ} C\\right) is 1.0 × 10^{-14}{m}^{2} / s.", "answer": "the diffusion coefficient at 1200k is 4.8 × 10^{-13}{m}^{2} / s." }, { "idx": 4737, "question": "Carbon is allowed to diffuse through a steel plate 10-mm thick. The concentrations of carbon at the two faces are 0.85 and 0.40kg C / {cm}^{3} Fe, which are maintained constant. If the preexponential and activation energy are 5.0 × 10^{-7}{m}^{2} / s and 77,000 J/ mol, respectively, compute the temperature at which the diffusion flux is 6.3 × 10^{-10}kg / m^{2}·s.", "answer": "the temperature at which the diffusion flux is 6.3 × 10^{-10} kg/m^2·s is 884 k or 611^{\\circ} C." }, { "idx": 4738, "question": "The steady-state diffusion flux through a metal plate is 7.8 × 10^{-8}kg / m^{2}·s at a temperature of 1200^{\\circ} C(1473 K) and when the concentration gradient is -500kg / m^{4}. Calculate the diffusion flux at 1000^{\\circ} C(1273 K) for the same concentration gradient and assuming an activation energy for diffusion of 145,000 J/ mol.", "answer": "the diffusion flux at 1000^{\\circ} C(1273k) is 1.21 × 10^{-8}kg / m^{2}·s." }, { "idx": 4739, "question": "Consider the diffusion of some hypothetical metal Y into another hypothetical metal Z at 950^{\\circ} C; after 10h the concentration at the 0.5mm position (in metal Z) is 2.0 wt% Y. At what position will the concentration also be 2.0 wt% Y after a 17.5h heat treatment again at 950^{\\circ} C ? Assume preexponential and activation energy values of 4.3 × 10^{-4}{m}^{2} / s and 180,000 J/ mol, respectively, for this diffusion system.", "answer": "the position will be 0.66 mm." }, { "idx": 4740, "question": "Carbon dioxide diffuses through a high-density polyethylene (HDPE) sheet 50mm thick at a rate of 2.2 × 10^{-8}\\left(cm^{3} STP / {cm}^{2}\\right.-s at 325 K. The pressures of carbon dioxide at the two faces are 4000 kPa and 2500 kPa, which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at 325 K ?", "answer": "the permeability coefficient at 325k is 0.73 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}·s·\\mathrm{pa}}." }, { "idx": 4741, "question": "A cylindrical specimen of a nickel alloy having an elastic modulus of 207 \\mathrm{GPa}\\left(30 × 10^{6} psi\\right) and an original diameter of 10.2 mm(0.40 in.) experiences only elastic deformation when a tensile load of 8900N\\left(2000 lb f_{\\mathrm{f}}\\right) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm(0.010 in.).", "answer": "the maximum length of the specimen before deformation is 0.475{m} (475mm or 18.7 in.)." }, { "idx": 4742, "question": "An aluminum bar 125mm (5.0 in.) long and having a square cross section 16.5mm (0.65 in.) on an edge is pulled in tension with a load of 66,700N(15,000 lb) and experiences an elongation of 0.43 mm(1.7 × 10^{-2} in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.", "answer": "the modulus of elasticity of the aluminum is 71.2 gpa (10.4 × 10^{6} psi)." }, { "idx": 4743, "question": "Consider a cylindrical nickel wire 2.0 mm(0.08 in.) in diameter and 3 × 10^{4}mm (1200 in.) long. Calculate its elongation when a load of 300N\\left(67 lb^{2}\\right) is applied. Assume that the deformation is totally elastic.", "answer": "the elongation of the nickel wire is 13.8mm (0.53 in.)." }, { "idx": 4744, "question": "For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 x 10^6 psi). What is the maximum load that can be applied to a specimen with a cross-sectional area of 130 mm^2 (0.2 in.^2) without plastic deformation?", "answer": "the maximum load that can be applied without plastic deformation is 44,850 n (10,000 lb_f)." }, { "idx": 4745, "question": "For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 x 10^6 psi). If the original specimen length is 76mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?", "answer": "the maximum length to which the specimen can be stretched without causing plastic deformation is 76.25mm (3.01 in.)." }, { "idx": 4746, "question": "A cylindrical rod of steel (E=207 \\mathrm{GPa}, 30 × 10^{6} psi) having a yield strength of 310 MPa(45,000 psi) is to be subjected to a load of 11,100N(2500 lb). If the length of the rod is 500 mm(20.0 in.), what must be the diameter to allow an elongation of 0.38 mm(0.015 in.)?", "answer": "the diameter must be 9.5 mm (0.376 in.)." }, { "idx": 4747, "question": "A cylindrical specimen of a metal alloy 10 mm(0.4 in.) in diameter is stressed elastically in tension. A force of 15,000N(3,370 lb r) produces a reduction in specimen diameter of 7 × 10^{-3} mm(2.8 × 10^{-4} in.). Compute Poisson's ratio for this material if its elastic modulus is 100 \\mathrm{GPa} \\left(14.5 × 10^{6} psi\\right).", "answer": "poisson's ratio for the material is v = 0.367 ." }, { "idx": 4748, "question": "Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10.0 mm(0.39 in.). A tensile force of 1500N(340 lb) produces an elastic reduction in diameter of 6.7 × 10^{-4} mm(2.64 ×\n10^{-5} in.). Compute the elastic modulus of this alloy, given that Poisson's ratio is 0.35 .", "answer": "100 gpa." }, { "idx": 4749, "question": "A cylindrical rod 500mm (20.0 in.) long and having a diameter of 12.7mm (0.50 in.) is to be subjected to a tensile load of 29,000 N (6500 lb). Which of the four metals or alloys listed in the table can experience neither plastic deformation nor an elongation of more than 1.3mm (0.05 in.)?", "answer": "Of the four metal alloys listed, only brass and steel satisfy the stipulated criteria. Brass has an elongation of 1.15 mm, and steel has an elongation of 0.56 mm, both of which are less than the maximum allowed elongation of 1.3 mm." }, { "idx": 4750, "question": "Why is aluminum alloy not a candidate for the cylindrical rod subjected to a tensile load of 29,000 N (6500 lb) with a maximum allowed elongation of 1.3mm (0.05 in.)?", "answer": "Aluminum alloy is not a candidate because its elongation of 1.64mm exceeds the limit." }, { "idx": 4751, "question": "Why is copper not considered for the cylindrical rod subjected to a tensile load of 29,000 N (6500 lb) with a maximum allowed elongation of 1.3mm (0.05 in.)?", "answer": "Copper is not considered due to its lower yield strength." }, { "idx": 4752, "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13mm (0.320 in.). Calculate the ductility in terms of percent reduction in area.", "answer": "the ductility in terms of percent reduction in area is 60%." }, { "idx": 4753, "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The fractured gauge length is 74.17mm (2.920 in.). Calculate the ductility in terms of percent elongation.", "answer": "the ductility in terms of percent elongation is 46%." }, { "idx": 4754, "question": "The following true stresses produce the corresponding true plastic strains for a brass alloy:\n\\begin{tabular}{cc}\n\\hline True Stress (psi) & True Strain \\\\\n\\hline 60,000 & 0.15 \\\\\n70,000 & 0.25 \\\\\n\\hline\n\\end{tabular}\nWhat true stress is necessary to produce a true plastic strain of 0.21 ?", "answer": "the true stress necessary to produce a true plastic strain of 0.21 is 66,400 \\, \\text{psi} (460 mpa)." }, { "idx": 4755, "question": "A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 5560N (1250 lb), the flexural strength is 105 MPa(15,000 psi), and the separation between load points is 45mm (1.75 in.).", "answer": "the minimum allowable specimen radius is 9.1mm (0.36 in)." }, { "idx": 4756, "question": "A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm(0.20 in.); the specimen fractured at a load of 3000N (675 lb f) when the distance between the support points was 40mm (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of 15mm (0.6 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at 40mm (1.6 in.)?", "answer": "the specimen with a square cross section would be expected to fracture at a load of 17,200 n (3870 lb f)." }, { "idx": 4757, "question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 300 MPa (43,500 psi). If the specimen radius is 5.0mm (0.20 in.) and the support point separation distance is 15.0mm (0.61 in.), would you expect the specimen to fracture when a load of 7500 N (1690 lbf) is applied? Justify your answer.", "answer": "the flexural stress is 286.5 mpa. since this value is less than the flexural strength of 300 mpa, fracture is not predicted." }, { "idx": 4758, "question": "Would you be 100% certain of the answer in part (a)? Why or why not?", "answer": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and the calculated stress is relatively close to the flexural strength, so there is some chance that fracture will occur." }, { "idx": 4759, "question": "Compute the modulus of elasticity for the nonporous material given that the modulus of elasticity for spinel (MgAl2O4) having 5 vol% porosity is 240 GPa (35 x 10^6 psi).", "answer": "the modulus of elasticity for the nonporous material is 265 gpa (38.6 x 10^6 psi)." }, { "idx": 4760, "question": "Compute the modulus of elasticity for 15 vol% porosity given that the modulus of elasticity for spinel (MgAl2O4) having 5 vol% porosity is 240 GPa (35 x 10^6 psi).", "answer": "the modulus of elasticity for 15 vol% porosity is 195 gpa (28.4 x 10^6 psi)." }, { "idx": 4761, "question": "Compute the modulus of elasticity for the nonporous titanium carbide (TiC) given that the modulus of elasticity for titanium carbide (TiC) having 5 vol% porosity is 310 GPa (45 x 10^6 psi).", "answer": "the modulus of elasticity for the nonporous material is 342 gpa (49.6 x 10^6 psi)." }, { "idx": 4762, "question": "At what volume percent porosity will the modulus of elasticity for titanium carbide (TiC) be 240 GPa (35 x 10^6 psi)?", "answer": "the volume percent porosity at which the modulus of elasticity is 240 gpa (35 x 10^6 psi) is 17.1 vol%." }, { "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{ll} \\hline (\\sigma_{f_{S}}(\\mathbf{M P a})) & (\boldsymbol{P}) \\ \\hline 70 & 0.10 \\ 60 & 0.15 \\ \\hline \\end{tabular} (a) Compute the flexural strength for a completely nonporous specimen of this material.", "answer": "the flexural strength for a completely nonporous specimen of this material is 95.3 mpa.", "idx": 4763 }, { "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{ll} \\hline (\\sigma_{f_{S}}(\\mathbf{M P a})) & (\boldsymbol{P}) \\ \\hline 70 & 0.10 \\ 60 & 0.15 \\ \\hline \\end{tabular} (b) Compute the flexural strength for a 0.20 volume fraction porosity.", "answer": "the flexural strength for a 0.20 volume fraction porosity is 51.5 mpa.", "idx": 4764 }, { "idx": 4765, "question": "For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to\n\\[\n\\sigma(t)=\\sigma(0) \\exp \\left(-\\frac{t}{\\tau}\\right)\n\\]\nwhere \\sigma(t) and \\sigma(0) represent the time-dependent and initial (i.e., time =0 ) stresses, respectively, and t and \\tau denote elapsed time and the relaxation time, respectively; \\tau is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation was suddenly pulled in tension to a measured strain of 0.5 ; the stress necessary to maintain this constant strain was measured as a function of time. Determine E_{f}(10) for this material if the initial stress level was 3.5 MPa(500 psi), which dropped to 0.5 MPa(70 psi) after 30s.", "answer": "e_{f}(10) = 3.66 \\text{mpa} (522 \\text{psi})" }, { "idx": 4766, "question": "The following table gives a number of Rockwellg hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values.\n\\begin{tabular}{rrr}\n47.3 & 48.7 & 47.1 \\\\\n52.1 & 50.0 & 50.4 \\\\\n45.6 & 46.2 & 45.9 \\\\\n49.9 & 48.3 & 46.4 \\\\\n47.6 & 51.1 & 48.5 \\\\\n50.4 & 46.7 & 49.7\n\\end{tabular}", "answer": "the average hardness value is 48.4 hrg. the standard deviation of the hardness values is 1.95 hrg." }, { "idx": 4767, "question": "The following table gives a number of yield strength values (in MPa) that were measured on the same aluminum alloy. Compute average and standard deviation yield strength values.\n\\begin{tabular}{rrr}\n274.3 & 277.1 & 263.8 \\\\\n267.5 & 258.6 & 271.2 \\\\\n255.4 & 266.9 & 257.6 \\\\\n270.8 & 260.1 & 264.3 \\\\\n261.7 & 279.4 & 260.5\n\\end{tabular}", "answer": "the average yield strength is 265.9 \\text{ mpa}. the standard deviation of the yield strength is 7.34 \\text{ mpa}." }, { "idx": 4768, "question": "What is the first criterion upon which factors of safety are based?", "answer": "The first criterion is consequences of failure." }, { "idx": 4769, "question": "What is the second criterion upon which factors of safety are based?", "answer": "The second criterion is previous experience." }, { "idx": 4770, "question": "What is the third criterion upon which factors of safety are based?", "answer": "The third criterion is accuracy of measurement of mechanical forces and/or material properties." }, { "idx": 4771, "question": "What is the fourth criterion upon which factors of safety are based?", "answer": "The fourth criterion is economics." }, { "idx": 4772, "question": "Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60^{\\circ} and 35^{\\circ}, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa(900 psi), will an applied stress of 12 MPa (1750 psi) cause the single crystal to yield? If not, what stress will be necessary?", "answer": "the applied stress of 12 MPa will not cause the single crystal to yield. the necessary stress for yielding to occur is 15.1 MPa (2200 psi)." }, { "idx": 4773, "question": "A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degrees with the tensile axis. Three possible slip directions make angles of 30 degrees, 48 degrees, and 78 degrees with the same tensile axis. Which of these three slip directions is most favored?", "answer": "the most favored slip direction is at an angle of 30 degrees with the tensile axis." }, { "idx": 4774, "question": "A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degrees with the tensile axis. The most favored slip direction makes an angle of 30 degrees with the same tensile axis. If plastic deformation begins at a tensile stress of 2.5 MPa (355 psi), determine the critical resolved shear stress for zinc.", "answer": "the critical resolved shear stress for zinc is 0.91 mpa (130 psi)." }, { "idx": 4775, "question": "The critical resolved shear stress for copper (Cu) is 0.48 MPa(70 psi). Determine the maximum possible yield strength for a single crystal of Cu pulled in tension.", "answer": "the maximum possible yield strength for a single crystal of Cu pulled in tension is \\sigma_{y} = 0.96 MPa (140 psi)." }, { "idx": 4776, "question": "The lower yield point for an iron that has an average grain diameter of 1 × 10^{-2}mm is 230 MPa(33,000 psi). At a grain diameter of 6 × 10^{-3} mm, the yield point increases to 275 MPa (40,000 psi). At what grain diameter will the lower yield point be 310 MPa(45,000 psi) ?\n\\underline{\\text {", "answer": "the grain diameter at which the lower yield point is 310 MPa is 4.34 × 10^{-3} mm." }, { "idx": 4777, "question": "Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as rectangular. Their original and deformed dimensions are as follows:\n\\begin{tabular}{lcc}\n\\hline & Circular (diameter, mm ) & Rectangular (mm) \\\\\n\\hline Original dimensions & 18.0 & 20 × 50 \\\\\nDeformed dimensions & 15.9 & 13.7 × 55.1 \\\\\n\\hline\n\\end{tabular}\nWhich of these specimens will be the hardest after plastic deformation, and why?", "answer": "the deformed rectangular specimen will be harder since it has the greater% \\text{cw} (24.5% compared to 22.0% for the circular specimen)." }, { "idx": 4778, "question": "Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \\tau_{\\text {CrSS }} is a function of the dislocation density \\rho_{D} as\n\\[\n\\tau_{\\text {crss }}=\\tau_{0}+A \\sqrt{\\rho_{D}}\n\\]\nwhere \\tau_{0} and A are constants. For copper, the critical resolved shear stress is 0.69 MPa(100 psi) at a dislocation density of 10^{4} mm^{-2}. If it is known that the value of \\tau_{0} for copper is 0.069 MPa (10 psi), compute \\tau_{\\text {crss }} at a dislocation density of 10^{6} mm^{-2}.", "answer": "the critical resolved shear stress \\tau_{\\text{crss}} at a dislocation density of 10^{6} mm^{-2} is 6.28 MPa (910 psi)." }, { "idx": 4779, "question": "Briefly cite the differences between the recovery and recrystallization processes.", "answer": "For recovery, there is some relief of internal strain energy by dislocation motion; however, there are virtually no changes in either the grain structure or mechanical characteristics. During recrystallization, on the other hand, a new set of strain-free grains forms, and the material becomes softer and more ductile." }, { "idx": 4780, "question": "What is the driving force for recrystallization?", "answer": "The driving force for recrystallization is the difference in internal energy between the strained and unstrained material." }, { "idx": 4781, "question": "What is the driving force for grain growth?", "answer": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases." }, { "idx": 4782, "question": "The average grain diameter for a brass material was measured as a function of time at 650 degrees C, which is shown in the following table at two different times: Time (min) 40, Grain Diameter (mm) 5.6 x 10^-2; Time (min) 100, Grain Diameter (mm) 8.0 x 10^-2. What was the original grain diameter?", "answer": "the original grain diameter was 0.031 mm." }, { "idx": 4783, "question": "The average grain diameter for a brass material was measured as a function of time at 650 degrees C, which is shown in the following table at two different times: Time (min) 40, Grain Diameter (mm) 5.6 x 10^-2; Time (min) 100, Grain Diameter (mm) 8.0 x 10^-2. What grain diameter would you predict after 200 min at 650 degrees C?", "answer": "the predicted grain diameter after 200 min at 650 degrees C is 0.109 mm." }, { "idx": 4784, "question": "An undeformed specimen of some alloy has an average grain diameter of 0.050 mm. You are asked to reduce its average grain diameter to 0.020 mm. Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.", "answer": "Yes, it is possible to reduce the average grain diameter of an undeformed alloy specimen from 0.050mm to 0.020 mm. In order to do this, plastically deform the material at room temperature (i.e., cold work it), and then anneal at an elevated temperature in order to allow recrystallization and some grain growth to occur until the average grain diameter is 0.020 mm." }, { "idx": 4785, "question": "A non-cold-worked brass specimen of average grain size 0.01mm has a yield strength of 150 MPa(21,750 psi). Estimate the yield strength of this alloy after it has been heated to 500^{\\circ} C for 1000s, if it is known that the value of \\sigma_{0} is 25 MPa(3625 psi).", "answer": "the yield strength of the brass alloy after heating to 500^{\\circ} C for 1000s is 124 MPa (18,000 psi)." }, { "idx": 4786, "question": "Normal butane and isobutane have boiling temperatures of -0.5^{\\circ} C and -12.3^{\\circ} C\\left(31.1^{\\circ} F\\right. and 9.9^{\\circ} F ), respectively. Briefly explain this behavior on the basis of their molecular structures.", "answer": "Normal butane has a higher melting temperature as a result of its molecular structure . There is more of an opportunity for van der Waals bonds to form between two molecules in close proximity to one another than for isobutane because of the linear nature of each normal butane molecule." }, { "idx": 4787, "question": "The tensile strength and number-average molecular weight for two poly(methyl methacrylate) materials are as follows:\n\\begin{tabular}{cc}\n\\hline Tensile Strength (MPa) & Number-Average Molecular Weight (g/mol) \\\\\n\\hline 50 & 30,000 \\\\\n150 & 50,000 \\\\\n\\hline\n\\end{tabular}\nEstimate the tensile strength at a number-average molecular weight of 40,000 g/mol.", "answer": "the estimated tensile strength at a number-average molecular weight of 40,000 g/mol is 112.5 mpa." }, { "idx": 4788, "question": "The tensile strength and number-average molecular weight for two polyethylene materials are as follows:\n\\begin{tabular}{cc}\n\\hline Tensile Strength (MPa) & Number-Average Molecular Weight (g/mol) \\\\\n\\hline 90 & 20,000 \\\\\n180 & 40,000 \\\\\n\\hline\n\\end{tabular}\nEstimate the number-average molecular weight that is required to give a tensile strength of 140 MPa.", "answer": "the required number-average molecular weight is 27700 g/mol" }, { "idx": 4789, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Branched and atactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol; linear and isotactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol", "answer": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight." }, { "idx": 4790, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene copolymer with 10% of possible sites crosslinked", "answer": "Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material. A higher degree of crystallinity favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus." }, { "idx": 4791, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Branched polyethylene with a number-average molecular weight of 100,000 g/mol; atactic polypropylene with a number-average molecular weight of 150,000 g/mol", "answer": "No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene.Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight." }, { "idx": 4792, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Linear and isotactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol; branched and atactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol", "answer": "Yes, it is possible. The linear and isotactic material will have the higher tensile strength. Both linearity and isotacticity favor a higher degree of crystallinity than do branching and atacticity; and tensile strength increases with increasing degree of crystallinity. Furthermore, the molecular weight of the linear/isotactic material is higher (100,000 g/mol versus 75,000 g/mol), and tensile strength increases with increasing molecular weight." }, { "idx": 4793, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Graft acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked", "answer": "No, it is not possible. Alternating copolymers tend to be more crystalline than graft copolymers, and tensile strength increases with degree of crystallinity. However, the graft material has a higher degree of crosslinking, and tensile strength increases with the percentage of crosslinks." }, { "idx": 4794, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Network polyester; lightly branched polytetrafluoroethylene", "answer": "Yes, it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polytetrafluoroethylene since there are many more of the strong covalent bonds for the network structure." }, { "idx": 4795, "question": "What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10^{-4} mm\\left(7.5 × 10^{-6} in.\\right) and a crack length of 3.8 × 10^{-2} mm\\left(1.5 × 10^{-3}\\right. in.) when a tensile stress of 140 MPa(20,000 psi) is applied?", "answer": "the magnitude of the maximum stress is 2800 \\text{ mpa} (400,000 \\text{ psi})." }, { "idx": 4796, "question": "Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.5 mm(0.02 in.) and a tip radius of curvature of 5 × 10^{-3} mm\\left(2 × 10^{-4} in\\right), when a stress of 1035 MPa(150,000 psi) is applied.", "answer": "the theoretical fracture strength of the material is 20.7 gpa (3 × 10^{6} psi)." }, { "idx": 4797, "question": "A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa \\sqrt{m} (50 ksi \\sqrt{m}.) is exposed to a stress of 1030 MPa(150,000 psi). Will this specimen experience fracture if the largest surface crack is 0.5mm ( 0.02 in.) long? Why or why not? Assume that the parameter Y has a value of 1.0 .", "answer": "the specimen will not experience fracture because it can tolerate a critical stress of 1380 \\text{ mpa} before fracture, which is greater than the applied stress of 1030 \\text{ mpa}." }, { "idx": 4798, "question": "An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa sqrt(m) (36.4 ksi sqrt(in)). It has been determined that fracture results at a stress of 300 MPa (43,500 psi) when the maximum (or critical) internal crack length is 4.0mm (0.16 in.). For this same component and alloy, will fracture occur at a stress level of 260 MPa (38,000 psi) when the maximum internal crack length is 6.0mm (0.24 in.)? Why or why not?", "answer": "fracture will occur since the value 42.4 MPa sqrt(m) is greater than the fracture toughness K_IC of the material, 40 MPa sqrt(m)." }, { "idx": 4799, "question": "Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa \\sqrt{m}(23.7 ksi \\sqrt{m}.). It has been determined that fracture results at a stress of 112 MPa(16,240 psi) when the maximum internal crack length is 8.6 mm(0.34 in.). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm(0.24 in.).", "answer": "the stress level at which fracture will occur for a critical internal crack length of 6.0mm is 134 MPa (19,300 psi)." }, { "idx": 4800, "question": "A structural component is fabricated from an alloy that has a plane-strain fracture toughness of 62 MPa \\sqrt{m}. It has been determined that this component fails at a stress of 250 MPa when the maximum length of a surface crack is 1.6 mm. What is the maximum allowable surface crack length (in mm ) without fracture for this same component exposed to a stress of 250 MPa and made from another alloy that has a plane strain fracture toughness of 51 MPa \\sqrt{m} ?", "answer": "the maximum allowable surface crack length without fracture is 1.08 mm." }, { "idx": 4801, "question": "A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa \\sqrt{m}(75.0 ksi \\sqrt{m}.). If the plate is exposed to a tensile stress of 345 MPa(50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.", "answer": "the minimum length of a surface crack that will lead to fracture is 0.0182 \\text{ m} (18.2 mm)." }, { "idx": 4802, "question": "A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture toughness of 98.9 MPa \\sqrt{m} ( 90 ksi \\sqrt{in}.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm(0.12 in.). If the design stress is one-half the yield strength and the value of Y is 1.0 , determine whether a critical flaw for this plate is subject to detection.", "answer": "the critical flaw size a_c is 16.8 mm. therefore, the critical flaw is subject to detection since it exceeds the 3.0 mm resolution limit." }, { "idx": 4803, "question": "Why may there be significant scatter in the fracture strength for some given ceramic material?", "answer": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material." }, { "idx": 4804, "question": "Why does the fracture strength increase with decreasing specimen size?", "answer": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes." }, { "idx": 4805, "question": "The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter surface crack geometry (i.e., reduce the crack length and increase the tip radius). Compute the ratio of the etched and original crack-tip radii for a fourfold increase in fracture strength if half of the crack length is removed.", "answer": "the ratio of the etched and original crack-tip radii is 8." }, { "idx": 4806, "question": "What is the maximum carbon content possible for a plain carbon steel that must have an impact energy of at least 200 Jat -50^{\\circ} C ?", "answer": "the maximum carbon concentration is 0.11% C." }, { "idx": 4807, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for tin.", "answer": "for sn, -71°c or -96°f." }, { "idx": 4808, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for molybdenum.", "answer": "for mo, 776°c or 1429°f." }, { "idx": 4809, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for iron.", "answer": "for fe, 451°c or 845°f." }, { "idx": 4810, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for gold.", "answer": "for au, 262°c or 504°f." }, { "idx": 4811, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for zinc.", "answer": "for zn, 4°c or 39°f." }, { "idx": 4812, "question": "Give the approximate temperature at which creep deformation becomes an important consideration for chromium.", "answer": "for cr, 586°c or 1087°f." }, { "idx": 4813, "question": "A cylindrical specimen 13.2mm in diameter of an S-590 alloy is to be exposed to a tensile load of 27,000N. At approximately what temperature will the steady-state creep be 10^{-3} h^{-1} ?", "answer": "the temperature would be approximately 775^\\circ c." }, { "idx": 4814, "question": "Steady-state creep rate data are given in the following table for a nickel alloy at 538^{\\circ} C \\\\ (811 K):\n\\begin{tabular}{lc}\n\\hline \\dot{\\varepsilon}_{s}\\left(h^{-1}\\right) & \\sigma(M P(a) \\\\\n\\hline 10^{-7} & 22.0 \\\\\n10^{-6} & 36.1 \\\\\n\\hline\n\\end{tabular}\nCompute the stress at which the steady-state creep is 10^{-5}h^{-1} (also at 538^{\\circ} C ).", "answer": "the stress at which the steady-state creep is 10^{-5}h^{-1} is 59.4 MPa." }, { "idx": 4815, "question": "Steady-state creep rate data are given in the following table for some alloy taken at 200^{\\circ} C(473 K) :\n\\begin{tabular}{ll}\n\\hlines\\left(h^{-1}\\right) & \\sigma[MPa(psi)] \\\\\n\\hline 2.5 × 10^{-3} & 55(8000) \\\\\n2.4 × 10^{-2} & 69(10,000) \\\\\n\\hline\n\\end{tabular}\nIf it is known that the activation energy for creep is 140,000 J/ mol, compute the steady-state creep rate at a temperature of 250^{\\circ} C(523 K) and a stress level of 48 MPa(7000 psi).", "answer": "the steady-state creep rate at 250^{\\circ} C and 48 MPa is 1.94 × 10^{-2} h^{-1}." }, { "idx": 4816, "question": "Steady-state creep data taken for an iron at a stress level of 140 MPa(20,000 psi) are given here:\n\\begin{tabular}{cc}\n\\hline\\dot{\\epsilon}_{s}\\left(h^{-1}\\right) & T(K) \\\\\n\\hline 6.6 × 10^{-4} & 1090 \\\\\n8.8 × 10^{-2} & 1200 \\\\\n\\hline\n\\end{tabular}\nIf it is known that the value of the stress exponent n for this alloy is 8.5 , compute the steady-state creep rate at 1300 K and a stress level of 83 MPa(12,000 psi).", "answer": "the steady-state creep rate at 1300k and a stress level of 83 MPa is 4.31 × 10^{-2}h^{-1}." }, { "idx": 4817, "question": "If ice homogeneously nucleates at -40^{\\circ} C, calculate the critical radius given values of - 3.1 × 10^{8} J/ m^{3} and 25 × 10^{-3} J/ m^{2}, respectively, for the latent heat of fusion and the surface free energy.", "answer": "the critical radius is 1.10 nm." }, { "idx": 4818, "question": "It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, and that the value of n in the exponential is 5.0. If, at some temperature, the fraction recrystallized is 0.30 after 100 min, determine the rate of recrystallization at this temperature.", "answer": "the rate of recrystallization at this temperature is 8.76 × 10^{-3}, \\text{min}^{-1}." }, { "idx": 4819, "question": "The kinetics of the austenite-to-pearlite transformation obeys the Avrami relationship. Using the fraction transformed-time data given here, determine the total time required for 95% of the austenite to transform to pearlite:\n\\begin{tabular}{cc}\n\\hline Fraction Transformed & Time (s) \\\\\n\\hline 0.2 & 280 \\\\\n0.6 & 425 \\\\\n\\hline\n\\end{tabular}", "answer": "the total time required for 95% of the austenite to transform to pearlite is 603s." }, { "idx": 4820, "question": "The fraction recrystallized-time data for the recrystallization at 350^{\\circ} C of a previously deformed aluminum are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of 116.8min.\n\\begin{tabular}{cc}\n\\hline Fraction Recrystallized & Time (min) \\\\\n\\hline 0.30 & 95.2 \\\\\n0.80 & 126.6 \\\\\n\\hline\n\\end{tabular}", "answer": "the fraction recrystallized after a total time of 116.8min is 0.65." }, { "idx": 4821, "question": "Briefly cite the differences among pearlite, bainite, and spheroidite relative to microstructure.", "answer": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles." }, { "idx": 4822, "question": "Briefly cite the differences among pearlite, bainite, and spheroidite relative to mechanical properties.", "answer": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite." }, { "idx": 4823, "question": "What is the principal difference between natural and artificial aging processes?", "answer": "For precipitation hardening, natural aging is allowing the precipitation process to occur at the ambient temperature; artificial aging is carried out at an elevated temperature." }, { "idx": 4824, "question": "Which of the following polymers would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?", "answer": "This question asks us to name, which, of five polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100^{\\circ} C\\left(212^{\\circ} F\\right). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of 100^{\\circ} C or above, and would be suitable for this application." }, { "idx": 4825, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Branched polyethylene having a number-average molecular weight of 850,000 g/mol; linear polyethylene having a number-average molecular weight of 850,000 g/mol", "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration." }, { "idx": 4826, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Polytetrafluoroethylene having a density of 2.14 g/cm3 and a weight-average molecular weight of 600,000 g/mol; PTFE having a density of 2.20 g/cm3 and a weight-average molecular weight of 600,000 g/mol", "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. Of these two polytetrafluoroethylene polymers, the PTFE with the higher density (2.20 g/cm3) will have the higher percent crystallinity, and, therefore, a higher melting temperature than the lower density PTFE. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration." }, { "idx": 4827, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Linear and syndiotactic poly(vinyl chloride) having a number-average molecular weight of 500,000 g/mol; linear polyethylene having a number-average molecular weight of 225,000 g/mol", "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration." }, { "idx": 4828, "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Linear and syndiotactic polypropylene having a weight-average molecular weight of 500,000 g/mol; linear and atactic polypropylene having a weight-average molecular weight of 750,000 g/mol", "answer": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature." }, { "question": "Compute the electrical conductivity of a cylindrical silicon specimen 7.0mm (0.28 in.) diameter and 57mm (2.25 in.) in length in which a current of 0.25 A passes in an axial direction. A voltage of 24V is measured across two probes that are separated by 45mm (1.75 in.).", "answer": "12.2 (omega·m)^{-1}.", "idx": 4829 }, { "question": "Compute the resistance over the entire 57mm (2.25 in.) of the specimen.", "answer": "the resistance over the entire specimen length is 121.4 omega.", "idx": 4830 }, { "idx": 4831, "question": "What is the distinction between electronic and ionic conduction? \\\\", "answer": "When a current arises from a flow of electrons, the conduction is termed electronic; for \\\\ ionic conduction, the current results from the net motion of charged ions." }, { "idx": 4832, "question": "In terms of electron energy band structure, discuss reasons for the difference in electrical \\\\ conductivity among metals, semiconductors, and insulators.", "answer": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities." }, { "idx": 4833, "question": "How does the electron structure of an isolated atom differ from that of a solid material?", "answer": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid." }, { "idx": 4834, "question": "Briefly state what is meant by the drift velocity of a free electron.", "answer": "The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric field." }, { "idx": 4835, "question": "Briefly state what is meant by the mobility of a free electron.", "answer": "The mobility is the proportionality constant between the drift velocity and the electric field. It is also a measure of the frequency of scattering events (and is inversely proportional to the frequency of scattering)." }, { "idx": 4836, "question": "(a) Calculate the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 500 V/m.", "answer": "the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 500 V/m is 72.5 m/s." }, { "idx": 4837, "question": "(b) Under these circumstances, how long does it take an electron to traverse a 25-mm length of crystal?", "answer": "under these circumstances, it takes an electron 3.45 x 10^-4 s to traverse a 25-mm length of crystal." }, { "idx": 4838, "question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 x 10^7 (Ω·m)^-1 and 0.0012 m^2/V·s, respectively. Compute the number of free electrons per cubic meter for aluminum at room temperature.", "answer": "the number of free electrons per cubic meter for aluminum at room temperature is 1.98 x 10^29 m^-3." }, { "idx": 4839, "question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 x 10^7 (Ω·m)^-1 and 0.0012 m^2/V·s, respectively. What is the number of free electrons per aluminum atom? Assume a density of 2.7 g/cm^3.", "answer": "the number of free electrons per aluminum atom is 3.28 electrons/al atom." }, { "idx": 4840, "question": "At room temperature the electrical conductivity of P b S is 25\\left(\\Omega·m\\right)^{-1}, whereas the electron and hole mobilities are 0.06 and 0.02{m}^{2} / V·s, respectively. Compute the intrinsic carrier concentration for PbS at room temperature.", "answer": "the intrinsic carrier concentration for pbs at room temperature is 1.95 × 10^{21}{m}^{-3}." }, { "idx": 4841, "question": "An n-type semiconductor is known to have an electron concentration of 5 × 10^{17}{m}^{-3}. If the electron drift velocity is 350{m} / s in an electric field of 1000V / m, calculate the conductivity of this material.", "answer": "the conductivity of the material is 0.028, (\\omega·m)^{-1}." }, { "idx": 4842, "question": "Predict whether Nitrogen (N) will act as a donor or an acceptor when added to Silicon (Si). Assume that the impurity elements are substitutional.", "answer": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom." }, { "idx": 4843, "question": "Predict whether Boron (B) will act as a donor or an acceptor when added to Germanium (Ge). Assume that the impurity elements are substitutional.", "answer": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom." }, { "idx": 4844, "question": "Predict whether Sulfur (S) will act as a donor or an acceptor when added to Indium Antimonide (InSb). Assume that the impurity elements are substitutional.", "answer": "Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb; an S atom has one more valence electron than an Sb atom." }, { "idx": 4845, "question": "Predict whether Indium (In) will act as a donor or an acceptor when added to Cadmium Sulfide (CdS). Assume that the impurity elements are substitutional.", "answer": "Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for Cd; an In atom has one more valence electron than a Cd atom." }, { "idx": 4846, "question": "Predict whether Arsenic (As) will act as a donor or an acceptor when added to Zinc Telluride (ZnTe). Assume that the impurity elements are substitutional.", "answer": "Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom." }, { "idx": 4847, "question": "Germanium to which 10^24 m^-3 As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). Is this material n-type or p-type?", "answer": "this germanium material to which has been added 10^24 m^-3 as atoms is n-type since as is a donor in ge." }, { "idx": 4848, "question": "Germanium to which 10^24 m^-3 As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and 0.05 m^2/V-s, respectively.", "answer": "the conductivity is equal to 1.60 × 10^4 (ω·m)^-1." }, { "idx": 4849, "question": "A metal alloy is known to have electrical conductivity and electron mobility values of 1.2 × 10^{7}(\\Omega·m)^{-1} and 0.0050{m}^{2} / V·s, respectively. A current of 40 \\mathrm{~A} is passed through a specimen of this alloy that is 35mm thick. What magnetic field would need to be imposed to yield a Hall voltage of -3.5 × 10^{-7}V ?", "answer": "the required magnetic field is 0.735, \\text{tesla}." }, { "idx": 4850, "question": "At temperatures between 540^{\\circ} C(813 K) and 727^{\\circ} C(1000 K), the activation energy and preexponential for the diffusion coefficient of \\mathrm{Na}^{+}in \\mathrm{NaCl} are 173,000 J/ mol and 4.0 × 10^{-4} m^{2} / s, respectively. Compute the mobility for an \\mathrm{Na}^{+}ion at 600^{\\circ} C(873 K).", "answer": "the mobility for a \\mathrm{na}^{+} ion at 600^{\\circ} C (873 k) is 2.34 × 10^{-13} m^{2}/\\mathrm{v}-s." }, { "idx": 4851, "question": "A parallel-plate capacitor using a dielectric material having an \\varepsilon_{\\mathrm{r}} of 2.2 has a plate spacing of 2mm ( 0.08 in.). If another material having a dielectric constant of 3.7 is used and the capacitance is to be unchanged, what must be the new spacing between the plates?", "answer": "the new spacing between the plates must be 3.36 mm." }, { "idx": 4852, "question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates. What is the capacitance of this capacitor?", "answer": "the capacitance of this capacitor is 100 pf." }, { "idx": 4853, "question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates. Compute the electric field that must be applied for 2 × 10−8 C to be stored on each plate.", "answer": "the electric field that must be applied is 2.0 × 105 v/m." }, { "idx": 4854, "question": "The polarization P of a dielectric material positioned within a parallel-plate capacitor is to be 4.0 × 10^-6 C/m^2. What must be the dielectric constant if an electric field of 10^5 V/m is applied?", "answer": "the dielectric constant εr must be 5.52." }, { "idx": 4855, "question": "The polarization P of a dielectric material positioned within a parallel-plate capacitor is to be 4.0 × 10^-6 C/m^2. What will be the dielectric displacement D?", "answer": "the dielectric displacement D will be 4.89 × 10^-6 c/m^2." }, { "idx": 4856, "question": "For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field.", "answer": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field." }, { "idx": 4857, "question": "For gaseous argon, solid LiF, liquid H2O, and solid Si, what kind(s) of polarization is (are) possible? Why?", "answer": "Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Both electronic and ionic polarizations are found in solid LiF, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid H2O. The H2O molecules have permanent dipole moments that are easily oriented in the liquid state. Only electronic polarization is to be found in solid Si; this material does not have molecules with permanent dipole moments, nor is it an ionic material." }, { "idx": 4858, "question": "List the four classifications of steels.", "answer": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool)." }, { "idx": 4859, "question": "For Low Carbon Steels, briefly describe the properties and typical applications.", "answer": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans." }, { "idx": 4860, "question": "For Medium Carbon Steels, briefly describe the properties and typical applications.", "answer": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts." }, { "idx": 4861, "question": "For High Carbon Steels, briefly describe the properties and typical applications.", "answer": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades." }, { "idx": 4862, "question": "For High Alloy Steels (Stainless and Tool), briefly describe the properties and typical applications.", "answer": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools." }, { "idx": 4863, "question": "Compare gray and malleable cast irons with respect to composition and heat treatment", "answer": "Gray iron--2.5 to 4.0 wt% C and 1.0 to 3.0 wt% Si. For most gray irons there is no heat treatment after solidification. Malleable iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. White iron is heated in a nonoxidizing atmosphere and at a temperature between 800 and 900°C for an extended time period." }, { "idx": 4864, "question": "Compare gray and malleable cast irons with respect to microstructure", "answer": "Gray iron--Graphite flakes are embedded in a ferrite or pearlite matrix. Malleable iron--Graphite clusters are embedded in a ferrite or pearlite matrix." }, { "idx": 4865, "question": "Compare gray and malleable cast irons with respect to mechanical characteristics", "answer": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility." }, { "idx": 4866, "question": "What are the distinctive features, limitations, and applications of titanium alloys?", "answer": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries." }, { "idx": 4867, "question": "What are the distinctive features, limitations, and applications of refractory metals?", "answer": "Distinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths. Limitation: some experience rapid oxidation at elevated temperatures. Applications: extrusion dies, structural parts in space vehicles, incandescent light filaments, x-ray tubes, and welding electrodes." }, { "idx": 4868, "question": "What are the distinctive features, limitations, and applications of superalloys?", "answer": "Distinctive features: able to withstand high temperatures and oxidizing atmospheres for long time periods. Applications: aircraft turbines, nuclear reactors, and petrochemical equipment." }, { "idx": 4869, "question": "What are the distinctive features, limitations, and applications of noble metals?", "answer": "Distinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile. Limitation: expensive. Applications: jewelry, dental restoration materials, coins, catalysts, and thermocouples." }, { "idx": 4870, "question": "For the MgO-Al2O3 system, what is the maximum temperature that is possible without the formation of a liquid phase?", "answer": "This problem asks that we specify, for the MgO-Al2O3 system, the maximum temperature without the formation of a liquid phase; it is approximately 2800 degrees Celsius." }, { "idx": 4871, "question": "For the MgO-Al2O3 system, at what composition or over what range of compositions will this maximum temperature be achieved?", "answer": "This maximum temperature is possible for pure MgO." }, { "idx": 4872, "question": "What are the characteristics of sand casting technique?", "answer": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast." }, { "idx": 4873, "question": "What are the characteristics of die casting technique?", "answer": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast." }, { "idx": 4874, "question": "What are the characteristics of investment casting technique?", "answer": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low." }, { "idx": 4875, "question": "What are the characteristics of lost-foam casting technique?", "answer": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes." }, { "idx": 4876, "question": "What are the characteristics of continuous casting technique?", "answer": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section." }, { "idx": 4877, "question": "What is the distinction between dye and pigment colorants?", "answer": "The distinction between dye and pigment colorants is that a dye dissolves within and becomes a part of the polymer structure, whereas a pigment does not dissolve, but remains as a separate phase." }, { "idx": 4878, "question": "Cite four factors that determine what fabrication technique is used to form polymeric materials.", "answer": "Four factors that determine what fabrication technique is used to form polymeric materials are: (1) whether the polymer is thermoplastic or thermosetting; (2) if thermoplastic, the softening temperature; (3) atmospheric stability; and (4) the geometry and size of the finished product." }, { "idx": 4879, "question": "What is the process of compression molding used to form plastic materials?", "answer": "For compression molding, both heat and pressure are applied after the polymer and necessary additives are situated between the mold members." }, { "idx": 4880, "question": "What is the process of transfer molding used to form plastic materials?", "answer": "For transfer molding, the solid materials (normally thermosetting in nature) are first melted in the transfer chamber prior to being forced into the die." }, { "idx": 4881, "question": "What is the process of injection molding used to form plastic materials?", "answer": "For injection molding (normally used for thermoplastic materials), the raw materials are impelled by a ram through a heating chamber, and finally into the die cavity." }, { "idx": 4882, "question": "Why must fiber materials that are melt-spun be thermoplastic?", "answer": "Fiber materials that are melt spun must be thermoplastic because they must be capable of forming a viscous liquid when heated, which is not possible for thermosets." }, { "idx": 4883, "question": "Why must fiber materials that are drawn be thermoplastic?", "answer": "Fiber materials that are drawn must be thermoplastic because during drawing, mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively brittle, they are not easily elongated." }, { "idx": 4884, "question": "Estimate the maximum thermal conductivity value for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix. Assume thermal conductivities of 27 and 67 W/m·K for TiC and Ni, respectively.", "answer": "the maximum thermal conductivity k_max is 31.0 W/m·K." }, { "idx": 4885, "question": "Estimate the minimum thermal conductivity value for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix. Assume thermal conductivities of 27 and 67 W/m·K for TiC and Ni, respectively.", "answer": "the minimum thermal conductivity k_min is 28.7 W/m·K." }, { "idx": 4886, "question": "A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.70 and 0.30 , respectively, estimate the upper limit for the specific stiffness of this composite, given the data that follow.\n\\begin{tabular}{lcc}\n\\hline & Specific Gravity & Modulus of Elasticity (GPa) \\\\\n\\hline Copper & 8.9 & 110 \\\\\nTungsten & 19.3 & 407 \\\\\n\\hline\n\\end{tabular}", "answer": "the upper limit for the specific stiffness of this composite is 19.65 \\text{ gpa}. alternatively, the specific stiffness can be estimated as 18.47 \\text{ gpa}." }, { "idx": 4887, "question": "For a continuous and oriented fiber-reinforced composite, the moduli of elasticity in the longitudinal and transverse directions are 33.1 and 3.66 \\mathrm{GPa}\\left(4.8 × 10^{6}\\right. and \\left.5.3 × 10^{5} psi\\right), respectively. If the volume fraction of fibers is 0.30 , determine the moduli of elasticity of fiber and matrix phases.", "answer": "the moduli of elasticity of the matrix and fiber phases are: \\[\ne_{m} = 2.6 \\, \\text{gpa} (3.77 × 10^{5} \\, \\text{psi})\n\\] \\[\ne_{f} = 104 \\, \\text{gpa} (15 × 10^{6} \\, \\text{psi})\n\\]" }, { "idx": 4888, "question": "In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. Using the data provided, determine the volume fraction of fibers required.", "answer": "the volume fraction of fibers required is 0.258." }, { "idx": 4889, "question": "In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 50 MPa (7250 psi).", "answer": "the tensile strength of this composite is 1070 mpa (155,000 psi)." }, { "idx": 4890, "question": "Cite several reasons why fiberglass-reinforced composites are used extensively.", "answer": "Reasons why fiberglass-reinforced composites are utilized extensively are: (1) glass fibers are very inexpensive to produce; (2) these composites have relatively high specific strengths; and (3) they are chemically inert in a wide variety of environments." }, { "idx": 4891, "question": "Cite several limitations of fiberglass-reinforced composites.", "answer": "Several limitations of these composites are: (1) care must be exercised in handling the fibers inasmuch as they are susceptible to surface damage; (2) they are lacking in stiffness in comparison to other fibrous composites; and (3) they are limited as to maximum temperature use." }, { "idx": 4892, "question": "What is a hybrid composite?", "answer": "A hybrid composite is a composite that is reinforced with two or more different fiber materials in a single matrix." }, { "idx": 4893, "question": "List two important advantages of hybrid composites over normal fiber composites.", "answer": "Two advantages of hybrid composites are: (1) better overall property combinations, and (2) failure is not as catastrophic as with single-fiber composites." }, { "idx": 4894, "question": "(a) Write an expression for the modulus of elasticity for a hybrid composite in which all fibers of both types are oriented in the same direction.", "answer": "the expression for the modulus of elasticity for a hybrid composite with all fibers aligned in the same direction is: e_cl = e_m v_m + e_f1 v_f1 + e_f2 v_f2" }, { "idx": 4895, "question": "(b) Using this expression, compute the longitudinal modulus of elasticity of a hybrid composite consisting of aramid and glass fibers in volume fractions of 0.25 and 0.35 , respectively, within a polyester resin matrix [E_m=4.0 GPa (6x10^5 psi)].", "answer": "the longitudinal modulus of elasticity of the hybrid composite is: e_cl = 59.7 gpa (8.67x10^6 psi)" }, { "idx": 4896, "question": "Briefly describe laminar composites.", "answer": "Laminar composites are a series of sheets or panels, each of which has a preferred high-strength direction. These sheets are stacked and then cemented together such that the orientation of the high-strength direction varies from layer to layer." }, { "idx": 4897, "question": "What is the prime reason for fabricating laminar composites?", "answer": "These composites are constructed in order to have a relatively high strength in virtually all directions within the plane of the laminate." }, { "idx": 4898, "question": "Is a voltage generated between the two cell halves of an Fe/Fe2+ concentration cell where both electrodes are pure iron, with Fe2+ concentrations of 0.5 M and 2 × 10^-2 M?", "answer": "Yes, a voltage is generated." }, { "idx": 4899, "question": "What is the magnitude of the voltage generated in an Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 × 10^-2 M?", "answer": "The magnitude of the voltage is 0.0414 V." }, { "idx": 4900, "question": "Which electrode will be oxidized in an Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 × 10^-2 M?", "answer": "Oxidation occurs in the cell half with the lower Fe2+ concentration (2 × 10^-2 M)." }, { "idx": 4901, "question": "For the aluminum and cast iron pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.", "answer": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode." }, { "idx": 4902, "question": "For the Inconel and nickel pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.", "answer": "For the Inconel-nickel couple, corrosion is unlikely inasmuch as both alloys appear within the same set of brackets (in both active and passive states)." }, { "idx": 4903, "question": "For the cadmium and zinc pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.", "answer": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode." }, { "idx": 4904, "question": "For the brass and titanium pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.", "answer": "For the brass-titanium pair, corrosion is possible, and brass will corrode." }, { "idx": 4905, "question": "For the low-carbon steel and copper pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.", "answer": "For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode." }, { "idx": 4906, "question": "A piece of corroded metal alloy plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 800 cm^2 and that approximately 7.6kg had corroded away during the submersion. Assuming a corrosion penetration rate of 4mm / yr for this alloy in seawater, estimate the time of submersion in years. The density of the alloy is 4.5g/cm^3.", "answer": "the time of submersion is 5.27 yr." }, { "idx": 4907, "question": "Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal M (in mol/cm^2-s). The following corrosion data are known about the metal and solution: For Metal M: V_(MM^2+) = -0.90 V, i_0 = 10^-12 A/cm^2, beta = +0.10. For Hydrogen: V_(H^-/H_2) = 0 V, i_0 = 10^-10 A/cm^2.", "answer": "the rate of corrosion of metal M is 3.27 x 10^-13 mol/cm^2-s." }, { "idx": 4908, "question": "Compute the corrosion potential for this reaction. The following corrosion data are known about the metal and solution: For Metal M: V_(MM^2+) = -0.90 V, i_0 = 10^-12 A/cm^2, beta = +0.10. For Hydrogen: V_(H^-/H_2) = 0 V, i_0 = 10^-10 A/cm^2.", "answer": "the corrosion potential for this reaction is -0.420 V." }, { "idx": 4909, "question": "Why does chromium in stainless steels make them more corrosion resistant than plain carbon steels in many environments?", "answer": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms." }, { "idx": 4910, "question": "For a concentration cell, briefly explain why corrosion occurs at the region having the lower concentration.", "answer": "Inasmuch as \\left[\\mathrm{M}_{\\mathrm{L}}^{2+}\\right] \\prec\\left[\\mathrm{M}_{H}^{2+}\\right] then the natural logarithm of the \\left[\\mathrm{M}^{2+}\\right] ratio is negative, which yields a positive value for \\Delta V. This means that the electrochemical reaction is spontaneous as written, or that oxidation occurs at the electrode having the lower \\mathrm{M}^{2+} concentration." }, { "idx": 4911, "question": "For copper, the heat capacity at constant volume C_V at 20 K is 0.38 J/mol-K and the Debye temperature is 340 K. Estimate the specific heat at 40 K.", "answer": "the specific heat for copper at 40 K is 47.8 J/kg-K." }, { "idx": 4912, "question": "For copper, the heat capacity at constant volume C_V at 20 K is 0.38 J/mol-K and the Debye temperature is 340 K. Estimate the specific heat at 400 K.", "answer": "the specific heat for copper at 400 K is 392 J/kg-K." }, { "idx": 4913, "question": "A 0.4-m (15.7-in.) rod of a metal elongates 0.48mm (0.019 in.) on heating from 20^{\\circ} C to 100^{\\circ} C\\left(68^{\\circ} F\\right. to \\left.212^{\\circ} F\\right). Determine the value of the linear coefficient of thermal expansion for this material.", "answer": "the linear coefficient of thermal expansion for this material is 15.0 × 10^{-6} \\, ({}^{\\circ} C)^{-1} [8.40 × 10^{-6} \\, ({}^{\\circ} \\mathrm{f})^{-1}]." }, { "idx": 4914, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Pure silver; sterling silver (92.5 wt% Ag-7.5 wt% Cu)", "answer": "Pure silver will have a larger conductivity than sterling silver because the impurity atoms in the latter will lead to a greater degree of free electron scattering." }, { "idx": 4915, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Fused silica; polycrystalline silica", "answer": "Polycrystalline silica will have a larger conductivity than fused silica because fused silica is noncrystalline and lattice vibrations are more effectively scattered in noncrystalline materials." }, { "idx": 4916, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Linear and syndiotactic poly(vinyl chloride) (DP =1000); linear and syndiotactic polystyrene (DP=1000)", "answer": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity." }, { "idx": 4917, "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Atactic polypropylene (Mw=10^6 g/mol); isotactic polypropylene (Mw=10^5 g/mol)", "answer": "The isotactic polypropylene will have a larger thermal conductivity than the atactic polypropylene because isotactic polymers have higher degrees of crystallinity. The influence of crystallinity on conductivity is explained in part (c)." }, { "idx": 4918, "question": "Briefly explain why thermal stresses may be introduced into a structure by rapid heating or cooling.", "answer": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced." }, { "idx": 4919, "question": "For cooling, what is the nature of the surface stresses?", "answer": "For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser degree than the cooler surface." }, { "idx": 4920, "question": "For heating, what is the nature of the surface stresses?", "answer": "For heating, the surface stresses will be compressive in nature since the interior expands to a lesser degree than the hotter surface." }, { "idx": 4921, "question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m. Compute the magnetic permeability for this material.", "answer": "1.260 x 10^-6 h/m." }, { "idx": 4922, "question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m. Compute the magnetic susceptibility for this material.", "answer": "2.387 x 10^-3." }, { "idx": 4923, "question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m. What type(s) of magnetism would you suggest is (are) being displayed by this material? Why?", "answer": "This material would display both diamagnetic and paramagnetic behavior. All materials are diamagnetic, and since χm is positive and on the order of 10^-3, there would also be a paramagnetic contribution." }, { "idx": 4924, "question": "A net magnetic moment is associated with each atom in paramagnetic and ferromagnetic materials. Explain why ferromagnetic materials can be permanently magnetized whereas paramagnetic ones cannot.", "answer": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions - domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains." }, { "idx": 4925, "question": "The chemical formula for copper ferrite may be written as \\left(\\mathrm{CuFe}_{2} \\mathrm{O}_{4}\\right)_{8} because there are eight formula units per unit cell. If this material has a saturation magnetization of 1.35 × 10^{5} A/m and a density of 5.40g /cm^3, estimate the number of Bohr magnetons associated with each Cu^{2+} ion.", "answer": "1.07 bohr magnetons/cu^{2+} ion" }, { "idx": 4926, "question": "The formula for samarium iron garnet (Sm3Fe5O12) may be written in the form Sm3^aFe2^cFe3^dO12, where the superscripts a, c, and d represent different sites on which the Sm^3+ and Fe^3+ ions are located. The spin magnetic moments for the Sm^3+ and Fe^3+ ions positioned in a and c sites are oriented parallel to one another and antiparallel to the Fe^3+ ions in d sites. Compute the number of Bohr magnetons associated with each Sm^3+ ion, given the following information: (1) each unit cell consists of eight formula (Sm3Fe5O12) units; (2) the unit cell is cubic with an edge length of 1.2529 nm; (3) the saturation magnetization for this material is 1.35 × 10^5 A/m; and (4) there are 5 Bohr magnetons associated with each Fe^3+ ion.", "answer": "2.86 bm." }, { "idx": 4927, "question": "An iron bar magnet having a coercivity of 7000 A/m is to be demagnetized. If the bar is inserted within a cylindrical wire coil 0.25m long and having 150 turns, what electric current is required to generate the necessary magnetic field?", "answer": "11.7A" }, { "idx": 4928, "question": "Cite the differences between type I and type II superconductors.", "answer": "For type I superconductors, with increasing magnetic field the material is completely diamagnetic and superconductive below H_{C}, while at H_{C} conduction becomes normal and complete magnetic flux penetration takes place. On the other hand, for type II superconductors upon increasing the magnitude of the magnetic field, the transition from the superconducting to normal conducting states is gradual between lower-critical and upper-critical fields; so also is magnetic flux penetration gradual. Furthermore, type II superconductors generally have higher critical temperatures and critical magnetic fields." }, { "idx": 4929, "question": "Briefly describe the Meissner effect.", "answer": "The Meissner effect is a phenomenon found in superconductors wherein, in the superconducting state, the material is diamagnetic and completely excludes any external magnetic field from its interior. In the normal conducting state complete magnetic flux penetration of the material occurs." }, { "idx": 4930, "question": "Visible light having a wavelength of 5 x 10^-7 m appears green. Compute the frequency of a photon of this light.", "answer": "the frequency of the photon is 6 x 10^14 s^-1." }, { "idx": 4931, "question": "Visible light having a wavelength of 5 x 10^-7 m appears green. Compute the energy of a photon of this light.", "answer": "the energy of the photon is 3.98 x 10^-19 j (2.48 ev)." }, { "idx": 4932, "question": "What are the characteristics of opaque materials in terms of their appearance and light transmittance?", "answer": "Opaque materials are impervious to light transmission; it is not possible to see through them." }, { "idx": 4933, "question": "What are the characteristics of translucent materials in terms of their appearance and light transmittance?", "answer": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material." }, { "idx": 4934, "question": "What are the characteristics of transparent materials in terms of their appearance and light transmittance?", "answer": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them." }, { "idx": 4935, "question": "Can a material have a positive index of refraction less than unity? Why or why not?", "answer": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum . This is not possible." }, { "idx": 4936, "question": "The index of refraction of quartz is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.544 and 1.553 in the direction of light propagation.", "answer": "the reflectivity at the boundary is 8.45 × 10^{-6}." }, { "idx": 4937, "question": "Zinc selenide has a band gap of 2.58 eV. Over what range of wavelengths of visible light is it transparent?", "answer": "pure znse is transparent to visible light having wavelengths between 0.48 and 0.7 \\mu m." } ]