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MatBench/layer2/rubbish/821_single_select_includes_process.json
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[
{
"idx": 1,
"question": "How does a coiled metal strip temperature indicator work when the temperature increases?",
"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": 1,
"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 bonded materials with different coefficients of thermal expansion."
},
{
"idx": 1,
"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": 2,
"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": 2,
"question": "How would you manufacture a hammer head for a carpenter's hammer?",
"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": 3,
"question": "(a) Aluminum foil used for storing food weighs about \\(0.3 \\mathrm{~g}\\) per square inch. How many atoms of aluminum are contained in one square inch of foil?",
"answer": "In a one square inch sample:\n\\[\n\\text { number }=\\frac{\\left(0.3 \\mathrm{~g}\\right)\\left(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol}\\right)}{26.981 \\mathrm{~g} / \\mathrm{mol}}=6.69 \\times 10^{21} \\text { atoms }\n\\]"
},
{
"idx": 4,
"question": "Using the density (11.36 g/cm3) and atomic weight (207.19 g/mol), calculate the number of atoms per cubic centimeter in lead.",
"answer": "(11.36 g/cm3)(1 cm3)(6.02 x 10^23 atoms/mol) / 207.19 g/mol = 3.3 x 10^22 atoms/cm3"
},
{
"idx": 4,
"question": "Using the density (0.534 g/cm3) and atomic weight (6.94 g/mol), calculate the number of atoms per cubic centimeter in lithium.",
"answer": "(0.534 g/cm3)(1 cm3)(6.02 x 10^23 atoms/mol) / 6.94 g/mol = 4.63 x 10^22 atoms/cm3"
},
{
"idx": 5,
"question": "(a)Using data, calculate the number of iron atoms in one ton (2000 pounds).",
"answer": "(\\frac{(2000 \\mathrm{lb})(454 \\mathrm{~g} / \\mathrm{lb})(6.02 \\times 10^{23} \\text { atoms } / 0 \\mathrm{~mol})}{55.847 \\mathrm{~g} / \\mathrm{mol}}=9.79 \\times 10^{27} \\mathrm{atoms} / \\mathrm{ton}\\)"
},
{
"idx": 6,
"question": "(b) Using data, calculate the volume in cubic centimeters occupied by one mole of boron.",
"answer": "(\\frac{(1 \\mathrm{~mol})(10.81 \\mathrm{~g} / \\mathrm{mol})}{2.3 \\mathrm{~g} / \\mathrm{cm}^{3}}=4.7 \\mathrm{~cm}^{3}\\)"
},
{
"idx": 7,
"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": "Volume = (200 in.²)(0.002 in.)(2.54 cm/in.)³ = 6.555 cm³\n(6.555 cm³)(8.902 g/cm³)(6.02 × 10²³ atoms/mol) / (58.71 g/mol) = 5.98 × 10²³ atoms"
},
{
"idx": 7,
"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": "Volume = (200 in.²)(0.002 in.)(2.54 cm/in.)³ = 6.555 cm³\n(6.555 cm³)(8.902 g/cm³) / (58.71 g/mol) = 0.994 mol Ni required"
},
{
"idx": 8,
"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": "We can let \\(x\\) be the number of electrons in the \\(3 d\\) energy level. Then:\n\\[\n1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{4} 3 s^{2} \\quad \\text { (must be } 2 \\text { electrons in } 4 s \\text { for valence }=2)\n\\]\nSince \\(27-(2+2+6+2+6+2)=7=x\\) there must be 7 electrons in the \\(3 d\\) level."
},
{
"idx": 9,
"question": "Calculate the fraction of bonding of \\(\\mathrm{MgO}\\) that is ionic.",
"answer": "(\\quad E_{\\mathrm{Mg}}=1.2 \\quad E_{0}=3.5\\)\n\\(\\angle_{\\text {covalent }}=\\exp \\left((-0.25)(3.5-1.2)^{2}\\right]=\\exp (-1.3225)=0.266\\)\n\\(\\int_{\\text {fanc }}=1-0.266=0.734 \\therefore\\) So bonding is mostly ionic"
},
{
"idx": 10,
"question": "Would you expect \\(\\mathrm{MgO}\\) or magnesium to have the higher modulus of elasticity?\nExplain.",
"answer": "(\\quad \\mathrm{MgO}\\) has ionic bonds, which are strong compared to the metallic bonds in \\(\\mathrm{Mg}\\). A higher force will be required to cause the same separation between the ions in \\(\\mathrm{MgO}\\) compared to the atoms in \\(\\mathrm{Mg}\\). Therefore, \\(\\mathrm{MgO}\\) should have the higher modulus of elasticity. In \\(\\mathrm{Mg}, E=6 \\times 10^{6} \\mathrm{psi}\\); in \\(\\mathrm{MgO}, E=30 \\times 10^{6} \\mathrm{psi}\\)."
},
{
"idx": 11,
"question": "Calculate the atomic radius in cm for a BCC metal with a0=0.3294 nm and one atom per lattice point.",
"answer": "For BCC metals, r=((sqrt(3))a0)/4=((sqrt(3))(0.3294 nm))/4=0.1426 nm=1.426×10^(-8) cm."
},
{
"idx": 11,
"question": "Calculate the atomic radius in cm for an FCC metal with a0=4.0862 Å and one atom per lattice point.",
"answer": "For FCC metals, r=((sqrt(2))a0)/4=((sqrt(2))(4.0862 Å))/4=1.4447 Å=1.4447×10^(-8) cm."
},
{
"idx": 12,
"question": "Determine the crystal structure for a metal with a0=4.9489 AA, r=1.75 AA and one atom per lattice point.",
"answer": "We want to determine if x in the calculations equals sqrt(2) (for FCC) or sqrt(3) (for BCC): (x)(4.9489 AA)=(4)(1.75 AA). x=sqrt(2), therefore FCC."
},
{
"idx": 12,
"question": "Determine the crystal structure for a metal with a0=0.42906 nm, r=0.1858 nm and one atom per lattice point.",
"answer": "We want to determine if x in the calculations equals sqrt(2) (for FCC) or sqrt(3) (for BCC): (x)(0.42906 nm)=(4)(0.1858 nm). x=sqrt(3), therefore BCC."
},
{
"idx": 13,
"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": "Using Equation 3-5:\n0.855 g/cm3 = (2 atoms/cell)(39.09 g/mol) / (a0)^3(6.02 x 10^23 atoms/mol)\na0^3 = 1.5189 x 10^-22 cm3 or a0 = 5.3355 x 10^-8 cm"
},
{
"idx": 13,
"question": "The lattice parameter of potassium with BCC structure is 5.3355 x 10^-8 cm. Calculate the atomic radius of potassium.",
"answer": "From the relationship between atomic radius and lattice parameter:\nr = (sqrt(3))(5.3355 x 10^-8 cm) / 4 = 2.3103 x 10^-8 cm"
},
{
"idx": 14,
"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": "From Equation 3-5: 11.72 g/cm³ = (4 atoms/cell)(232 g/mol) / (a₀³)(6.02 × 10²³ atoms/mol). a₀³ = 1.315297 × 10⁻²² cm³ or a₀ = 5.0856 × 10⁻⁸ cm."
},
{
"idx": 14,
"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. The lattice parameter is 5.0856 × 10⁻⁸ cm. Calculate the atomic radius of thorium.",
"answer": "From the relationship between atomic radius and lattice parameter: r = (√2)(5.0856 × 10⁻⁸ cm) / 4 = 1.7980 × 10⁻⁸ cm."
},
{
"idx": 15,
"question": "A metal having a cubic structure has a density of \\(2.6 \\mathrm{~g} / \\mathrm{cm}^{3}\\), an atomic weight of \\(87.62 \\mathrm{~g} / \\mathrm{mol}\\), and a lattice parameter of \\(6.0849 \\AA\\). One atom is associated with each lattice point. Determine the crystal structure of the metal.",
"answer": "(\\quad 2.6 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{(x \\text { atoms } / \\mathrm{cell})(87.52 \\mathrm{~g} / \\mathrm{mol})}{(6.08 49 \\times 10^{-8} \\mathrm{~cm})^{3}(5.02 \\times 10^{23} \\text { atoms } / \\mathbf{m o l})}\\)\n\\(x=4\\), therefore FCC"
},
{
"idx": 16,
"question": "A metal having a cubic structure has a density of \\(1.892 \\mathrm{~g} / \\mathrm{cm}^{3}\\), an atomic weight \\(\\mathrm{of}\\) \\(132.91 \\mathrm{~g} / \\mathrm{mol}\\), and a lattice parameter of \\(€ 1.13 \\AA\\). One atom is associated with each lattice point. Determine the crystal structure of the \\(\\mathrm{metal}\\).",
"answer": "(\\quad 1.892 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{(x\\text { atoms } / \\mathrm{cell})(132.91 \\mathrm{~g} / \\mathrm{mol})}{(6.13 \\times 10^{-8} \\mathrm{~cm})^{3}(6.102 \\times 10^{23} \\text { atoms } / \\mathrm{\\mathrm{mol}})}\\)\n\\(x=2\\), therefore BCC"
},
{
"idx": 17,
"question": "Indium has a tetragonal structure with \\(a_{0}=0.32517 \\mathrm{~nm}\\) and \\(c_{0}=0.49459 \\mathrm{~nm}\\). The density is \\(7.286 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and the atomic weight is \\(114.82 \\mathrm{~g} / \\mathrm{mol}\\). Does indium have the simple tetragonal or body-centered tetragonal structure?",
"answer": "[\n\\begin{array}{l}\n7.286 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{(\\text { atoms } / \\mathrm{cell})(114.82 \\mathrm{~g} / \\mathrm{mol})}{(3.2517 \\times 10^{-8} \\mathrm{~cm})^{2}(4.9459 \\times 10^{-8} \\mathrm{~cm})(6.02 \\times 10^{23} \\text { atoms } / 0 \\mathrm{~mol})} \\\\\nx=2, \\text { therefore BCT (body-centered tetragonal) }\n\\end{array}\n\\]"
},
{
"idx": 18,
"question": "Bismuth has a hexagonal structure, with a0=0.4546 nm and c0=1.186 nm. Determine the volume of the unit cell.",
"answer": "The volume of the unit cell is V=a1^2 c cos30. V=(0.4546 nm)^2(1.186 nm)(cos 30)=0.21226 nm^3=2.1226×10^-22 cm^3."
},
{
"idx": 18,
"question": "Bismuth has a density of 9.808 g/cm^3 and an atomic weight of 208.98 g/mol. The volume of the unit cell is 2.1226×10^-22 cm^3. Determine the number of atoms in each unit cell.",
"answer": "If x is the number of atoms per unit cell, then: 9.808 g/cm^3=(x atoms/cell)(208.98 g/mol)/((2.1226×10^-22 cm^3)(6.02×10^23 atoms/mol)). x=6 atoms/cell."
},
{
"idx": 19,
"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": "The volume of the unit cell is V=a0 b0 c0 or V=(0.45258 nm)(0.45186 nm)(0.76570 nm)=0.1566 nm3 =1.566×10^-22 cm3. From the density equation: 5.904 g/cm3 = (x atoms/cell)(69.72 g/mol)/(1.566×10^-22 cm3)(6.02×10^23 atoms/mol). Solving for x gives x=8 atoms/cell."
},
{
"idx": 19,
"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": "The volume of the unit cell is V=a0 b0 c0 or V=(0.45258 nm)(0.45186 nm)(0.76570 nm)=0.1566 nm3. The packing factor (PF) is calculated as: PF=(8 atoms/cell)(4π/3)(0.1218 nm)^3/0.1566 nm3=0.387."
},
{
"idx": 20,
"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": "V=(0.22858 nm)2(0.35842 nm) cos 30=0.01622 nm3=16.22×10−24 cm3\nFrom the density equation:\n1.848 g/cm3 = (x atoms/cell)(9.01 g/mol) / (16.22×10−24 cm3)(6.02×1023 atoms/mol)\nx = 2 atoms/cell"
},
{
"idx": 20,
"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": "The packing factor (PF) is:\nPF = (2 atoms/cell)(4π/3)(0.1143 nm)3 / 0.01622 nm3 = 0.77"
},
{
"idx": 21,
"question": "A typical paper clip weighs 0.59 g and consists of BCC iron. The lattice parameter for BCC iron is 2.866 × 10^-8 cm. Calculate the number of unit cells in the paper clip, given the density of iron is 7.87 g/cm^3.",
"answer": "The volume of a unit cell is (2.866 × 10^-8 cm)^3 = 2.354 × 10^-23 cm^3. The number of unit cells is: number = (0.59 g) / [(7.87 g/cm^3)(2.354 × 10^-23 cm^3/cell)] = 3.185 × 10^21 cells."
},
{
"idx": 21,
"question": "A typical paper clip weighs 0.59 g and consists of BCC iron. Given there are 2 atoms per unit cell in BCC iron and the number of unit cells is 3.185 × 10^21, calculate the number of iron atoms in the paper clip.",
"answer": "The number of atoms is: number = (3.185 × 10^21 cells)(2 atoms/cell) = 6.37 × 10^21 atoms."
},
{
"idx": 22,
"question": "Determine the planar density for BCC lithium in the (100) plane (a_o = 3.5089 Å).",
"answer": "planar density = 1 / (3.5089 × 10^-8 cm)^2 = 0.0812 × 10^16 points/cm^2"
},
{
"idx": 22,
"question": "Determine the packing fraction for BCC lithium in the (100) plane (a_o = 3.5089 Å).",
"answer": "packing fraction = π[√3 a_o / 4]^2 / a_o^2 = 0.589"
},
{
"idx": 22,
"question": "Determine the planar density for BCC lithium in the (110) plane (a_o = 3.5089 Å).",
"answer": "planar density = 2 / √2 (3.5089 × 10^-8 cm)^2 = 0.1149 × 10^16 points/cm^2"
},
{
"idx": 22,
"question": "Determine the packing fraction for BCC lithium in the (110) plane (a_o = 3.5089 Å).",
"answer": "packing fraction = 2 √3 a_o / 4 / √2 a_o^2 = 0.833"
},
{
"idx": 22,
"question": "Determine the planar density for BCC lithium in the (111) plane (a_o = 3.5089 Å).",
"answer": "planar density = 1/2 / 0.866 (3.5089 × 10^-8 cm)^2 = 0.0469 × 10^16 points/cm^2"
},
{
"idx": 22,
"question": "Determine the packing fraction for BCC lithium in the (111) plane (a_o = 3.5089 Å).",
"answer": "packing fraction = 1/2 / 0.866 a_o^2 √3 a_o / 4"
},
{
"idx": 22,
"question": "Which, if any, of the (100), (110), and (111) planes in BCC lithium is close packed?",
"answer": "There is no close-packed plane in BCC structures."
},
{
"idx": 23,
"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": "(\\quad d_{111}=\\frac{a_{0}}{\\sqrt{1^{2}+1^{2}}+1^{2}}=\\frac{3.796 \\AA}{\\sqrt{3}}=2.1916 \\AA\\)\nthickness \\(=(1 \\mathrm{~mm} / 10 \\mathrm{~mm} / \\mathrm{cm})=4.563 \\times 10^{6} d_{111}\\) spacings"
},
{
"idx": 24,
"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": "The distance between the \\(0,0,0\\) and \\(1,1,1\\) points is \\(\\sqrt{3} a_{0}\\). The interplanar spacing is\n\\[\nd_{111}=a_{0} \\sqrt{1^{2}+1^{2}+1^{2}}=a_{0} \\sqrt{3}\n\\]\nTherefore the number of interplanar spacings is\nnumber of \\(d_{111}\\) spacings \\(=\\sqrt{3} a_{0} / a_{0} / \\sqrt{3})=3"
},
{
"idx": 25,
"question": "Determine the minimum radius of an atom that will just fit into the tetrahedral interstitial site in FCC nickel (σ_c = 3.5167 Å).",
"answer": "For the tetrahedral site in FCC nickel (σ_c = 3.5167 Å): r_Ni = (√2 × 3.5167 Å) / 4 = 1.243 Å. r / r_Ni = 0.225 for a tetrahedral site. Therefore: r = (1.243 Å)(0.225) = 0.2797 Å."
},
{
"idx": 25,
"question": "Determine the minimum radius of an atom that will just fit into the octahedral interstitial site in BCC lithium (a_o = 3.5089 Å).",
"answer": "For the octahedral site in BCC lithium (a_o = 3.5089 Å): r_Li = (√3 × 3.5089 Å) / 4 = 1.519 Å. r / r_Li = 0.414 for an octahedral site. Therefore: r = (1.519 Å)(0.414) = 0.629 Å."
},
{
"idx": 26,
"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": "(\\quad r_{\\mathrm{Cu}}=1.278 \\AA\\)\n\\[\n\\begin{array}{l}\nr / r_{\\mathrm{Cu}}=0.414 \\text { for an octahedral site. Therefore: }\n\\end{array}\n\\]\n\\[\nr=(1.278 \\AA)(0.414)=0.529 \\AA\n\\]"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for Y2O3.",
"answer": "rY3+/rO2- = 0.89/1.32 = 0.67, CN = 6"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for UO2.",
"answer": "rU4+/rO2- = 0.97/1.32 = 0.73, CN = 6"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for BaO.",
"answer": "rO2-/rBa2+ = 1.32/1.34 = 0.99, CN = 8"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for Si3N4.",
"answer": "rN3-/rSi4+ = 0.15/0.42 = 0.36, CN = 4"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for GeO2.",
"answer": "rGe4+/rO2- = 0.53/1.32 = 0.40, CN = 4"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for MnO.",
"answer": "rMn2+/rO2- = 0.80/1.32 = 0.61, CN = 6"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for MgS.",
"answer": "rMg2+/rS2- = 0.66/1.84 = 0.36, CN = 6"
},
{
"idx": 27,
"question": "Using the ionic radii, determine the coordination number expected for KBr.",
"answer": "rK+/rBr- = 1.33/1.96 = 0.68, CN = 6"
},
{
"idx": 28,
"question": "Would you expect NiO to have the cesium chloride, sodium chloride, or zinc blende structure? (Given: r_Ni+2 = 0.69 Å, r_O-2 = 1.32 Å, r_Ni+2 = 0.52, CN = 6)",
"answer": "A coordination number of 8 is expected for the CsCl structure, and a coordination number of 4 is expected for ZnS. But a coordination number of 6 is consistent with the NaCl structure."
},
{
"idx": 28,
"question": "Based on the NaCl structure for NiO, determine the lattice parameter. (Given: r_Ni+2 = 0.69 Å, r_O-2 = 1.32 Å)",
"answer": "a0 = 2(0.69) + 2(1.32) = 4.02 Å"
},
{
"idx": 28,
"question": "Based on the NaCl structure for NiO, determine the density. (Given: a0 = 4.02 Å, atomic weights: Ni = 58.71 g/mol, O = 16 g/mol)",
"answer": "ρ = (4 of each ion(cell)(58.71 + 16 g/mol)) / ((4.02 × 10^-8 cm)^3 (6.02 × 10^23 atoms/mol)) = 7.64 g/cm^3"
},
{
"idx": 28,
"question": "Based on the NaCl structure for NiO, determine the packing factor. (Given: a0 = 4.02 Å, r_Ni+2 = 0.69 Å, r_O-2 = 1.32 Å)",
"answer": "PF = (4π/3)(4 ions/cell)[(0.69)^3 + (1.32)^3] / (4.02)^3 = 0.678"
},
{
"idx": 29,
"question": "Would you expect UO2 to have the sodium chloride, zinc blende, or fluoride structure? Given the ionic radii r_U+4 = 0.97 Å and r_O-2 = 1.32 Å, and the valence of U = +4 and O = -2.",
"answer": "The radius ratio r_U+4/r_O-2 = 0.97/1.32 = 0.735 predicts a coordination number of 8. To balance the charge, there must be twice as many oxygen ions as uranium ions. The fluorite structure satisfies these requirements, with U in FCC positions (4) and O in tetrahedral positions (8)."
},
{
"idx": 29,
"question": "Determine the lattice parameter (a0) for UO2 with the fluorite structure, given r_U+4 = 0.97 Å and r_O-2 = 1.32 Å.",
"answer": "For the fluorite structure, sqrt(3) * a0 = 4(r_U+4 + r_O-2) = 4(0.97 + 1.32) = 9.16 Å. Therefore, a0 = 5.2885 Å."
},
{
"idx": 29,
"question": "Calculate the density (ρ) of UO2 with the fluorite structure, given a0 = 5.2885 Å, atomic weight of U = 238.03 g/mol, and atomic weight of O = 16 g/mol.",
"answer": "ρ = [4(238.03 g/mol) + 8(16 g/mol)] / [(5.2885 × 10^-8 cm)^3 * (6.02 × 10^23 atoms/mol)] = 12.13 g/cm^3."
},
{
"idx": 29,
"question": "Determine the packing factor (PF) for UO2 with the fluorite structure, given a0 = 5.2885 Å, r_U+4 = 0.97 Å, and r_O-2 = 1.32 Å.",
"answer": "PF = [(4π/3)(4(0.97)^3 + 8(1.32)^3)] / (5.2885)^3 = 0.624."
},
{
"idx": 30,
"question": "Would you expect BeO to have the sodium chloride, zinc blende, or fluorite structure? (Given: r_Be+2 = 0.35 Å, r_O-2 = 1.32 Å)",
"answer": "r_Be/r_O = 0.265, CN = 4, therefore Zinc Blende structure."
},
{
"idx": 30,
"question": "Determine the lattice parameter of BeO with zinc blende structure. (Given: r_Be+2 = 0.35 Å, r_O-2 = 1.32 Å)",
"answer": "sqrt(3) a_0 = 4 r_Be+2 + 4 r_O-2 = 4(0.35 + 1.32) = 6.68 or a_0 = 3.8567 Å."
},
{
"idx": 30,
"question": "Calculate the density of BeO with zinc blende structure. (Given: a_0 = 3.8567 Å, atomic weights: Be = 9.01 g/mol, O = 16 g/mol)",
"answer": "ρ = 4(9.01 + 16 g/mol) / (3.8567 × 10^-8 cm)^3 (6.02 × 10^23 atoms/mol) = 2.897 g/cm^3."
},
{
"idx": 30,
"question": "Calculate the packing factor of BeO with zinc blende structure. (Given: a_0 = 3.8567 Å, r_Be+2 = 0.35 Å, r_O-2 = 1.32 Å)",
"answer": "PF = (4π/3)(4)(0.35)^3 + 8(1.32)^3 / (3.8567)^3 = 0.684."
},
{
"idx": 31,
"question": "Would you expect CsBr to have the sodium chloride, zinc blende, fluorite, or cesium chloride structure? (Given r_Cs+1=1.67 Å, r_Br-1=1.96 Å)",
"answer": "r_Cs+1=1.67 Å, r_Br-1=1.96 Å. The radius ratio r_Cs+1/r_Br-1=0.852, coordination number=8, therefore CsCl structure."
},
{
"idx": 31,
"question": "Determine the lattice parameter for CsBr with the cesium chloride structure. (Given r_Cs+1=1.67 Å, r_Br-1=1.96 Å)",
"answer": "For CsCl structure, √3 a0=2 r_Cs+1+2 r_Br-1=2(1.96+1.67)=7.26 Å or a0=4.1916 Å"
},
{
"idx": 31,
"question": "Calculate the density of CsBr with the cesium chloride structure. (Given r_Cs+1=1.67 Å, r_Br-1=1.96 Å, a0=4.1916 Å)",
"answer": "ρ=((4.1916×10^-8 cm)^3 (6.02×10^23 atoms/mol))/((4π/3)[(1.96)^3+(1.67)^3])=4.8 g/cm^3"
},
{
"idx": 31,
"question": "Calculate the packing factor for CsBr with the cesium chloride structure. (Given a0=4.1916 Å)",
"answer": "PF=(4.1916)^3/(4.1916)^3=0.693"
},
{
"idx": 32,
"question": "Sketch the ion arrangement on the (110) plane of ZnS (with the zinc blende structure).",
"answer": "The ion arrangement on the (110) plane of ZnS (zinc blende structure) consists of alternating Zn²⁺ and S²⁻ ions arranged in a rectangular pattern. The Zn²⁺ ions have a radius of 0.074 nm and the S²⁻ ions have a radius of 0.184 nm. The lattice parameter a₀ for ZnS is calculated as 0.596 nm."
},
{
"idx": 32,
"question": "Sketch the ion arrangement on the (110) plane of CaF₂ (with the fluorite structure).",
"answer": "The ion arrangement on the (110) plane of CaF₂ (fluorite structure) consists of Ca²⁺ ions surrounded by F⁻ ions in a more complex rectangular pattern. The Ca²⁺ ions have a radius of 0.099 nm and the F⁻ ions have a radius of 0.133 nm. The lattice parameter a₀ for CaF₂ is calculated as 0.536 nm."
},
{
"idx": 32,
"question": "Compare the ion arrangement on the (110) plane of ZnS (zinc blende structure) to that of CaF₂ (fluorite structure).",
"answer": "The (110) plane of ZnS shows alternating Zn²⁺ and S²⁻ ions in a simpler rectangular arrangement, while the (110) plane of CaF₂ shows Ca²⁺ ions surrounded by F⁻ ions in a more complex pattern. The ZnS arrangement has two types of ions (Zn²⁺ and S²⁻), whereas CaF₂ has one type of cation (Ca²⁺) and two types of anion positions (F⁻)."
},
{
"idx": 32,
"question": "Calculate the planar packing fraction (PPF) for the (110) plane of ZnS (zinc blende structure).",
"answer": "The planar packing fraction (PPF) for the (110) plane of ZnS is calculated as follows: PPF = [2π(0.074 nm)² + 2π(0.184 nm)²] / [√2 × (0.596 nm)²] = 0.492."
},
{
"idx": 32,
"question": "Calculate the planar packing fraction (PPF) for the (110) plane of CaF₂ (fluorite structure).",
"answer": "The planar packing fraction (PPF) for the (110) plane of CaF₂ is calculated as follows: PPF = [2π(0.099 nm)² + 4π(0.133 nm)²] / [√2 × (0.536 nm)²] = 0.699."
},
{
"idx": 32,
"question": "Compare the planar packing fraction (PPF) on the (110) planes for ZnS (zinc blende structure) and CaF₂ (fluorite structure).",
"answer": "The planar packing fraction (PPF) for the (110) plane of ZnS is 0.492, while for CaF₂ it is 0.699. This indicates that the (110) plane of CaF₂ has a higher packing density compared to ZnS."
},
{
"idx": 33,
"question": "Determine the planar density for the (111) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
"answer": "The area of the (111) plane is 0.866 a_o^2. Given a_o = 3.96 Å (0.396 nm), the planar density (P.D.) is calculated as: P.D. = 2 Mg / (0.866)(3.96 × 10^-8 cm)^2 = 0.1473 × 10^16 points/cm^2."
},
{
"idx": 33,
"question": "Determine the planar packing fraction for the (111) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
"answer": "The planar packing fraction (PPF) for the (111) plane is calculated as: PPF = 2 π(0.66)^2 / (0.866)(3.96)^2 = 0.202."
},
{
"idx": 33,
"question": "Determine the planar density for the (222) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
"answer": "The planar density (P.D.) for the (222) plane is the same as for the (111) plane: P.D. = 0.1473 × 10^16 points/cm^2."
},
{
"idx": 33,
"question": "Determine the planar packing fraction for the (222) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
"answer": "The planar packing fraction (PPF) for the (222) plane is calculated as: PPF = 2 π(1.32)^2 / (0.866)(3.96)^2 = 0.806."
},
{
"idx": 33,
"question": "What ions are present on the (111) plane of MgO, which has the sodium chloride structure?",
"answer": "The (111) plane of MgO contains Mg^2+ ions."
},
{
"idx": 33,
"question": "What ions are present on the (222) plane of MgO, which has the sodium chloride structure?",
"answer": "The (222) plane of MgO contains O^2- ions."
},
{
"idx": 34,
"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 \\(0.15418 \\mathrm{~nm}\\) wavelength are used. Calculate the lattice parameter of the iron.",
"answer": "(\\quad \\sin \\theta=\\lambda / 2 d_{220}\\)\n\\[\n\\begin{aligned}\n\\sin (99.1 / 2) & =\\frac{0.15418 \\sqrt{2^{2}+2^{2}+0^{2}}}{2 a_{\\mathrm{o}}} \\\\\na_{\\mathrm{o}} & =\\frac{0.15418 \\sqrt{8}}{2 \\sin [49.55)}=0.2865 \\mathrm{~nm}\n\\end{aligned}\n\\]"
},
{
"idx": 35,
"question": "Calculate the number of vacancies per \\(\\mathrm{cm}^{3}\\) expected in copper at \\(1080^{\\circ} \\mathrm{C}\\) (just below the melting temperature). The activation energy for vacancy formation is 20,000 cal/mol.",
"answer": "(\\quad n=\\frac{(4 \\text { atoms } / \\mathrm{u} . c .)}{(3.6151 \\times 10^{-8} \\mathrm{~cm})^{3}}=8.47 \\times 10^{22} \\mathrm{atoms} / \\mathrm{cm}^{3}\\)\n\\[\n\\begin{aligned}\nn_{v} & =8.47 \\times 10^{22} \\exp \\left[-20,000 /(1.987)(1353)\\right] \\\\\n& =8.47 \\times 10^{22} \\exp \\left(-7.4393\\right)=4.97 \\times 10^{19} \\text { vacancies } / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 36,
"question": "The fraction of lattice points occupied by vacancies in solid aluminum at \\(660^{\\circ} \\mathrm{C}\\) is \\(10^{-3}\\). What is the activation energy required to create vacancies in aluminum?",
"answer": "(\\quad n_{v} / n=10^{-3}=\\exp \\left[-Q /(1.987)(933)\\right]\\)\n\\[\n\\begin{array}{l}\n\\ln \\left(10^{-3}\\right)=-6.9078--Q /(1.987)(933) \\\\\nQ=12800 \\mathrm{cal} / \\mathrm{mol}\n\\end{array}\n\\]"
},
{
"idx": 37,
"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 (a) the fraction of the lattice points that contain vacancies.",
"answer": "V_u,c=(0.22858 nm)²(0.35842 nm) cos 30=0.01622 nm³ =1.622×10⁻²³ cm³ 1.844 g/cm³=(x)(9.01 g/mol)/(1.622×10⁻²³ cm³)(6.02×10²³ atoms/mol) x=1.9984 fraction=(2-1.9984)/2=0.0008"
},
{
"idx": 37,
"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 (b) the total number of vacancies in a cubic centimeter.",
"answer": "number=0.0016 vacancies/uc / 1.622×10⁻²³ cm³=0.986×10²⁰ vacancies/cm³"
},
{
"idx": 38,
"question": "BCC lithium has a lattice parameter of 3.5089 x 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the number of vacancies per cubic centimeter.",
"answer": "(1 vacancy) / (200 x (3.5089 x 10^-8 cm)^3) = 1.157 x 10^20 vacancies/cm^3"
},
{
"idx": 38,
"question": "BCC lithium has a lattice parameter of 3.5089 x 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the density of Li given that in 200 unit cells there are 399 Li atoms.",
"answer": "Atoms per cell = 399 / 200. Density = ((399 / 200) x 6.94 g/mol) / ((3.5089 x 10^-8 cm)^3 x 6.02 x 10^23 atoms/mol) = 0.532 g/cm^3"
},
{
"idx": 39,
"question": "FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate the density.",
"answer": "The number of atoms/cell = (499 / 500)(4 sites/cell)\n\nρ = (499 / 500)(4)(207.19 g/mol) / (4.949 × 10^-8 cm)^3(6.02 × 10^23 atoms/mol) = 11.335 g/cm^3"
},
{
"idx": 39,
"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 500 Pb atoms occupy 500 / 4 = 125 unit cells:\n\n(1 vacancy / 125 cells) × (1 / 11.335 g/cm^3) = 5.82 × 10^18 vacancies/g"
},
{
"idx": 40,
"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 \\(0.32554 \\mathrm{~nm}\\) and a density of \\(11.95 \\mathrm{~g} / \\mathrm{cm}^{3}\\). Calculate the fraction of the atoms in the alloy that are tungsten.",
"answer": "(\\quad 11.95 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{\\left(x_{\\mathrm{W}}\\right)(183.85 \\mathrm{~g} / \\mathrm{mol})+\\left(2-x_{\\mathrm{W}}\\right)(92.91 \\mathrm{~g} / \\mathrm{mol})}{\\left(3.2554 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}(16.02 \\times 10^{23} \\text { atoms } / / \\mathrm{mol})}\\)\n\\[\n\\begin{array}{l}\n248.186=183.85 x_{\\mathrm{W}}+185.82-92.91 x_{\\mathrm{W}} \\\\\n90.94 x_{\\mathrm{W}}=62.366 \\quad \\text { or } \\quad x_{\\mathrm{W}}=0.69 \\mathrm{~W} \\text { atoms } / \\mathrm{cell}\n\\end{array}\n\\]There are 2 atoms per cell in BCC metals. Thus:\n\\[\nf_{\\mathrm{w}}=0.69 / 2=0.345\n\\]"
},
{
"idx": 41,
"question": "Tin atoms are introduced into a FCC copper crystal, producing an alloy with a lattice parameter of \\(3.7589 \\times 10^{-8} \\mathrm{~cm}\\) and a density of \\(8.772 \\mathrm{~g} / \\mathrm{cm}^{3}\\). Calculate the atomic percentage of tin present in the alloy.",
"answer": "(\\quad 8.772 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{\\left(\\mathrm{S}_{\\mathrm{ox}}\\right)(118.69 \\mathrm{~g} / \\mathrm{mol})+\\left(4-\\mathrm{S}_{\\mathrm{ox}}\\right)(63.54 \\mathrm{~g} / \\mathrm{mol})}{\\left(3.7589 \\times 10^{-8} \\mathrm{~cm}^{3}\\right)^{2}(6.02 \\times 10^{23}\\) atoms \\(/ \\mathrm{mol})}\\)\n\\[\n280.5=55.15 x_{\\mathrm{Sn}}+254.16 \\quad \\text { or } \\quad x_{\\mathrm{Sn}}=0.478 \\mathrm{Sn} \\text { atoms } / \\mathrm{cell}\n\\]\nThere are 4 atoms per cell in FCC metals; therefore the at\\% \\(\\mathrm{Sn}\\) is:\n\\[\n(0.478 / 4)=11.95 \\%\n\\]"
},
{
"idx": 42,
"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 \\(0.29158 \\mathrm{~nm}\\). Calculate the density of the alloy.",
"answer": "(\\quad \\rho=\\frac{(2)(0.925)(51.996 \\mathrm{~g} / \\mathrm{mol})+2(0.075)(180.95 \\mathrm{~g} / \\mathrm{mol})}{(2.9158 \\times 10^{-8} \\mathrm{~cm})^{3}(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol})}=8.265 \\mathrm{~g} / \\mathrm{cm}^{3}\\)"
},
{
"idx": 43,
"question": "For the Fe-C alloy with one carbon atom introduced for every 100 iron atoms in an interstitial position in BCC iron and a lattice parameter of 0.2867 nm, find the density.",
"answer": "There is one carbon atom per 100 iron atoms, or 1 C / 50 unit cells, or 1/50 C per unit cell: ρ = ((2)(55.847 g/mol) + (1/50)(12 g/mol)) / ((2.867 × 10^-8 cm)^3 (6.02 × 10^23 atoms/mol)) = 7.89 g/cm^3"
},
{
"idx": 43,
"question": "For the Fe-C alloy with one carbon atom introduced for every 100 iron atoms in an interstitial position in BCC iron and a lattice parameter of 0.2867 nm, find the packing factor.",
"answer": "Packing Factor = (2(4π/3)(1.241)^3 + (1/50)(4π/3)(0.77)^3) / (2.867)^3 = 0.681"
},
{
"idx": 44,
"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": "7.882 g/cm³ = (2(55.847 g/mol) + x(1.00797 g/mol)) / ((2.866 × 10⁻⁸ cm)³(6.022 × 10²³ atoms/mol)). Solving gives x = 0.0081 H atoms/cell. The total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus: f_H = 0.0081 / 2.0081 = 0.004."
},
{
"idx": 44,
"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": "Since there is 0.0081 H/cell, then the number of cells containing H atoms is: cells = 1 / 0.0081 = 123.5 or 1 H in 123.5 cells."
},
{
"idx": 45,
"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": "In 10 unit cells, we expect 40 Mg + 40 O ions, but due to the defect: 40 O - 1 = 39. 1 vacancy / (10 cells)(3.96 x 10^-8 cm)^3 = 1.61 x 10^21 vacancies/cm3."
},
{
"idx": 45,
"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": "(39/40)(4)(24.312 g/mol) + (39/40)(4)(16 g/mol) / (3.96 x 10^-8 cm)^3 (6.02 x 10^23 atoms/mol) = 4.205 g/cm3."
},
{
"idx": 46,
"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": "Let x be the number of each type of ion in the unit cell. There normally are 4 of each type. 3.02 g/cm³ = (x(65.38 g/mol) + x(32.064 g/mol)) / ((5.9583 × 10⁻⁸ cm)³ × (6.02 × 10²³ ions/mol)) → x = 3.9465. 4 - 3.9465 = 0.0535 defects/u.c."
},
{
"idx": 46,
"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": "Number of unit cells/cm³ = 1 / (5.9583 × 10⁻⁸ cm)³ = 4.704 × 10²¹. Schottky defects per cm³ = (4.704 × 10²¹)(0.0535) = 2.517 × 10²⁰."
},
{
"idx": 47,
"question": "Calculate the length of the Burgers vector in BCC niobium with a lattice parameter of 3.294 Å",
"answer": "The repeat distance, or Burgers vector, is half the body diagonal, or: b = (sqrt(2)) * (sqrt(3)) * (3.294 Å) = 2.853 Å"
},
{
"idx": 47,
"question": "Calculate the length of the Burgers vector in FCC silver with a lattice parameter of 4.0862 Å",
"answer": "The repeat distance, or Burgers vector, is half of the face diagonal, or: b = (sqrt(2)) * (sqrt(2)) * (4.0862 Å) = 2.889 Å"
},
{
"idx": 47,
"question": "Calculate the length of the Burgers vector in diamond cubic silicon with a lattice parameter of 5.4307 Å",
"answer": "The slip direction is , where the repeat distance is half of the face diagonal: b = (sqrt(2)) * (sqrt(Σ)) * (5.4307 Å) = 3.840 Å"
},
{
"idx": 48,
"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 [110] slip direction.",
"answer": "phi=54.76 degrees, lambda=90 degrees, tau=5000 cos 54.76 cos 90, tau=0 psi"
},
{
"idx": 48,
"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 [011] slip direction.",
"answer": "phi=54.76 degrees, lambda=45 degrees, tau=5000 cos 54.76 cos 45, tau=2040 psi"
},
{
"idx": 48,
"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 [101] slip direction.",
"answer": "phi=54.76 degrees, lambda=45 degrees, tau=5000 cos 54.76 cos 45, tau=2040 psi"
},
{
"idx": 48,
"question": "Which slip system(s) will become active first for the given single crystal of an FCC metal under 5000 psi applied stress?",
"answer": "The slip systems with [011] and [101] slip directions will become active first, both having a resolved shear stress of 2040 psi."
},
{
"idx": 49,
"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": "CRSS = 12,000 psi = σ cos φ cos λ; λ = 54.76°; φ = 90°; σ = 12,000 psi / (cos 90° cos 54.76°) = ∞"
},
{
"idx": 49,
"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": "CRSS = 12,000 psi = σ cos φ cos λ; λ = 54.76°; φ = 45°; σ = 12,000 psi / (cos 45° cos 54.76°) = 29,412 psi"
},
{
"idx": 49,
"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": "CRSS = 12,000 psi = σ cos φ cos λ; λ = 54.76°; φ = 45°; σ = 12,000 psi / (cos 45° cos 54.76°) = 29,412 psi"
},
{
"idx": 50,
"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": "65000 = σ_o + K / (sqrt(17 x 10^-6)) = σ_o + 242.5 K\n82000 = σ_o + K / (sqrt(0.8 x 10^-6)) = σ_o + 1118.0 K\nBy solving the two simultaneous equations:\nK = 19.4 psi / sqrt(d)\nσ_o = 60,290 psi"
},
{
"idx": 50,
"question": "Using the Hall-Petch equation constants determined previously (σ_o = 60,290 psi and K = 19.4 psi / sqrt(d)), find the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m.",
"answer": "σ = 60,290 + 19.4 / sqrt(0.2 x 10^-6) = 103,670 psi"
},
{
"idx": 51,
"question": "For an ASTM grain size number of 8, calculate the number of grains per square inch at a magnification of 100.",
"answer": "N=2^(n-1) N=2^(n-1)=2^7=128 grains/in.^2"
},
{
"idx": 51,
"question": "For an ASTM grain size number of 8, calculate the number of grains per square inch with no magnification.",
"answer": "No magnification means that the magnification is 1: (27)(100/1)^2=1.28x10^6 grains/in.^2"
},
{
"idx": 52,
"question": "Determine the ASTM grain size number if 20 grains/square inch are observed at a magnification of 400 .",
"answer": "(\\quad(20)(400 / 100)^{2}=2^{n-1} \\quad \\log (320)=(n-1) \\log (2)\\)\n\\[\n2.505=(n-1)(0.301) \\text { or } n=9.3\n\\]"
},
{
"idx": 53,
"question": "Determine the ASTM grain size number if 25 grains/square inch are observed at a magnification of 50 .",
"answer": "(\\quad 25(50 / 100)^{2}=2^{n-1} \\quad \\operatorname{log}(6.25)=(n-1) \\log (2)\\)\n\\[\n0.796=(n-1)(0.301) \\text { or } n=3.6\n\\]"
},
{
"idx": 54,
"question": "Calculate the angle \\(\\theta\\) of a small-angle grain boundary in FCC aluminum when the dislocations are \\(5000 \\AA\\) apart. (See Figure 4-18 and equation in Problem 4-46.)",
"answer": "(\\quad b=(1 / 2)(\\sqrt{2})(4.04958)=2.8635 \\AA\\) and \\(D=5000 \\AA\\)\n\\[\n\\begin{array}{l}\n\\sin (\\theta / 2)=\\frac{2.8635}{(2)(5000)}=0.000286 \\\\\n\\theta / 2=0.0164 \\\\\n\\theta=0.0328^{\\circ}\n\\end{array}\n\\]"
},
{
"idx": 55,
"question": "Atoms are found to move from one lattice position to another at the rate of \\(5 \\times 10^{5}\\) jumps/s at \\(400^{\\circ} \\mathrm{C}\\) when the activation energy for their movement is 30,000 cal/mol. Calculate the jump rate at \\(750^{\\circ} \\mathrm{C}\\).\n\\section*{",
"answer": "\n\\[\n\\begin{array}{l}\n\\text { Rate }=\\frac{5 \\times 10^{5}}{x}=\\frac{c_{0} \\exp \\left[-30,000 /(1.987)(673)\\right]}{c_{0} \\exp \\left[-30,000 /(1.897)(1023)\\right]}=\\exp (-22.434+14.759) \\\\\n\\frac{5 \\times 10^{5}}{x}=\\exp (-7.675)=4.64 \\times 10^{-4} \\\\\nx=\\frac{5 \\times 10^{5}}{4.64 \\times 10^{-4}}=1.08 \\times 10^{9} \\text { jumps/s }\n\\end{array}\n\\]"
},
{
"idx": 56,
"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 \\times 10^{-5} \\mathrm{at} 600^{\\circ} \\mathrm{C}\\), determine the fraction at \\(1000^{\\circ} \\mathrm{C}\\).",
"answer": "(\\quad 8 \\times 10^{-5}=\\exp \\left[-Q /(1.987)(873)\\right] \\quad Q=16,364 \\mathrm{cal} / \\mathrm{mol}\\)\n\\(f=n_{y} / n=\\exp \\left[-16,364 /(1.987)(1273)\\right]=0.00155\\)"
},
{
"idx": 57,
"question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6 × 10^-15 cm²/s at 727°C and is 1 × 10^-9 cm²/s at 1400°C. Calculate the activation energy (Q).",
"answer": "6 × 10^-15 / 1 × 10^-9 = D0 exp[-Q/(1.987)(1000)] / D0 exp[-Q/(1.987)(1673)]\n6 × 10^-6 = exp[-Q(0.000503 - 0.00030)] = exp[-0.000203 Q]\n-12.024 = -0.000203 Q or Q = 59,230 cal/mol"
},
{
"idx": 57,
"question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6 × 10^-15 cm²/s at 727°C and is 1 × 10^-9 cm²/s at 1400°C. Calculate the constant D0.",
"answer": "1 × 10^-9 = D0 exp[-59,230/(1.987)(1673)] = D0 exp(-17.818)\n1 × 10^-9 = 1.828 × 10^-8 D0 or D0 = 0.055 cm²/s"
},
{
"idx": 58,
"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 Å. Calculate the concentration gradient in (a) atomic percent Sb per cm.",
"answer": "Δc/Δx = ((1/10^8 - 500/10^8)/0.02 cm) × 100% = -0.02495 at%/cm"
},
{
"idx": 58,
"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 Å. Calculate the concentration gradient in (b) Sb atoms/cm^3·cm.",
"answer": "a0 = 5.4307 Å, Vunit cell = 160.16 × 10^-24 cm^3; c1 = (8 Si atoms/unit cell)(1 Sb/10^8 Si)/160.16 × 10^-24 cm^3/unit cell = 0.04995 × 10^16 Sb atoms/cm^3; c2 = (8 Si atoms/unit cell)(500 Sb/10^8 Si)/160.16 × 10^-24 cm^3/unit cell = 24.975 × 10^16 Sb atoms/cm^3; Δc/Δx = (0.04995 × 10^16 - 24.975 × 10^16)/0.02 cm = -1.246 × 10^19 Sb atoms/cm^3·cm"
},
{
"idx": 59,
"question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025 mm 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": "Delta c / Delta x = (20% - 25%) / (0.025 mm x 0.1 cm/mm) = -2000 at% Zn/cm"
},
{
"idx": 59,
"question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025 mm 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": "wt% Zn = (20 x 65.38 g/mol) / (20 x 65.38 + 80 x 63.54) x 100 = 20.46; wt% Zn = (25 x 65.38 g/mol) / (25 x 65.38 + 75 x 63.54) x 100 = 25.54; Delta c / Delta x = (20.46% - 25.54%) / 0.0025 cm = -2032 wt% Zn/cm"
},
{
"idx": 60,
"question": "A 0.001-in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650 degrees C. 5 x 10^8 H atoms/cm^3 are in equilibrium with the hot side of the foil, while 2 x 10^3 H atoms/cm^3 are in equilibrium with the cold side. Determine the concentration gradient of hydrogen.",
"answer": "Delta c / Delta x = (2 x 10^3 - 5 x 10^8) / ((0.001 in)(2.54 cm/in)) = -1969 x 10^8 H atoms/cm^3.cm"
},
{
"idx": 60,
"question": "A 0.001-in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650 degrees C. 5 x 10^8 H atoms/cm^3 are in equilibrium with the hot side of the foil, while 2 x 10^3 H atoms/cm^3 are in equilibrium with the cold side. Determine the flux of hydrogen through the foil.",
"answer": "J = -D(Delta c / Delta x) = -0.0012 exp[-3600 / (1.987)(923)] (-1969 x 10^8) = 0.33 x 10^8 H atoms/cm^2.s"
},
{
"idx": 61,
"question": "Given a 1-mm sheet of FCC iron used to contain nitrogen in a heat exchanger at 1200°C, with nitrogen concentrations of 0.04 atomic percent at one surface and 0.005 atomic percent at the second surface, calculate the concentration gradient (Δc/Δx) in N atoms/cm³·cm.",
"answer": "Δc/Δx = (0.00005 - 0.0004)(4 atoms per cell)/(3.589 × 10^-8 cm)^3 / (1 mm)(0.1 cm/mm) = -3.03 × 10^20 N atoms/cm³·cm"
},
{
"idx": 61,
"question": "Using the concentration gradient calculated previously (-3.03 × 10^20 N atoms/cm³·cm) for nitrogen diffusion through a 1-mm FCC iron sheet at 1200°C, determine the flux of nitrogen (J) in N atoms/cm²·s. The diffusion coefficient (D) is given by D = 0.0034 exp[-34,600/(1.987)(1473)] cm²/s.",
"answer": "J = -D(Δc/Δx) = -0.0034 exp[-34,600/(1.987)(1473)] × (3.03 × 10^20) = 7.57 × 10^12 N atoms/cm²·s"
},
{
"idx": 62,
"question": "A 4-cm-diameter, \\(0.5-\\mathrm{mm}\\)-thick spherical container made of BCC iron holds nitrogen at \\(700^{\\circ} \\mathrm{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": "(\\quad \\Delta c / \\Delta x=\\frac{[0.00002-0.0005)[2 \\text { atoms } / \\mathrm{cell}] /\\left(2.866 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}}{(0.5 \\mathrm{~mm})(0.1 \\mathrm{~cm} / \\mathrm{nm})}\\)\n\\[\n\\begin{aligned}\n& =-8.16 \\times 10^{20} \\mathrm{~N} / \\mathrm{cm}^{3} \\cdot \\mathrm{cm} \\\\\nJ & =-0.0047 \\exp [-18.300 /(1.987)(973)][-8.16 \\times 10^{20}]=2.97 \\times 10^{14} \\mathrm{~N} / \\mathrm{cm}^{2} \\cdot \\mathrm{s} \\\\\nA_{\\text {sphere }} & =4 \\pi t^{2}=4 \\pi(2 \\mathrm{~cm})^{2}=50.27 \\mathrm{~cm}^{2} \\quad \\mathrm{t}=3600 \\mathrm{~s} / \\mathrm{h} \\\\\n\\mathrm{N} & \\mathrm{atoms} / \\mathrm{t}=(2.97 \\times 10^{14})(50.27)(3600)=5.37 \\times 10^{19} \\mathrm{~N} \\text { atoms } / \\mathrm{h} \\\\\n& \\mathrm{N} \\text { loss }=\\frac{(5.37 \\times 10^{19} \\text { atoms })\\left(14.007 \\mathrm{~g} / \\mathrm{mol}\\right)}{(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol})}=1.245 \\times 10^{-3} \\mathrm{~g} / \\mathrm{h}\n\\end{aligned}\n\\]"
},
{
"idx": 63,
"question": "Calculate the concentration of hydrogen atoms (c1) at the first surface of the BCC iron structure, given that there are 0.05 H atoms per unit cell and the lattice parameter is 2.866×10⁻⁸ cm.",
"answer": "c1 = 0.05 H / (2.866×10⁻⁸ cm)³ = 212.4×10¹⁹ H atoms/cm³"
},
{
"idx": 63,
"question": "Calculate the concentration of hydrogen atoms (c2) at the second surface of the BCC iron structure, given that there are 0.001 H atoms per unit cell and the lattice parameter is 2.866×10⁻⁸ cm.",
"answer": "c2 = 0.001 H / (2.866×10⁻⁸ cm)³ = 4.25×10¹⁹ H atoms/cm³"
},
{
"idx": 63,
"question": "Determine the concentration gradient (Δc/Δx) of hydrogen atoms across the iron structure, given the concentrations c1 and c2, and the unknown thickness Δx.",
"answer": "Δc/Δx = (4.25×10¹⁹ - 212.4×10¹⁹) / Δx = -2.08×10²¹ / Δx"
},
{
"idx": 63,
"question": "Convert the allowable hydrogen loss rate (50 g/cm² per year) into a flux (J) in H atoms/cm²·s, given the atomic weight of hydrogen (1.00797 g/mol) and Avogadro's number (6.02×10²³ atoms/mol).",
"answer": "J = (50 g/cm²·y)(6.02×10²³ atoms/mol) / (1.00797 g/mol)(31.536×10⁶ s/y) = 9.47×10¹⁷ H atoms/cm²·s"
},
{
"idx": 63,
"question": "Calculate the minimum thickness (Δx) of the iron structure required to limit the hydrogen flux to 9.47×10¹⁷ H atoms/cm²·s, given the diffusion coefficient for hydrogen in BCC iron at 400°C (D = 0.0012 exp[-3600/(1.987)(673)]).",
"answer": "J = -D(Δc/Δx) → 9.47×10¹⁷ = (2.08×10²¹/Δx)(0.0012 exp[-3600/(1.987)(673)]) → Δx = 0.179 cm"
},
{
"idx": 64,
"question": "Determine the maximum allowable temperature that will produce a flux of less than \\(2000 \\mathrm{H}\\) atoms \\(/ \\mathrm{cm}^{2} \\cdot \\mathrm{s}\\) through a BCC iron foil when the concentration gradient is \\(-5 \\times 10^{16}\\) atoms \\(/ \\mathrm{cm}^{3} \\cdot \\mathrm{cm}\\). (Note the negative sign for the flux.)",
"answer": "(2000 \\mathrm{H}\\) atoms \\(/ \\mathrm{cm}^{3} \\cdot \\mathrm{s}=-0.0012 \\exp [-3600 / 1.9877 /[-5 \\times 10^{16} \\mathrm{atoms} / \\mathrm{cm}^{3} \\cdot \\mathrm{cm}]\\)\n\\[\n\\begin{array}{l}\n\\ln (3.33 \\times 10^{-11})=-3600 / 1.987 T \\\\\nT=-3600 / \\text { (-(-24.12)(1.987)) }=75 \\mathrm{~K}=-198^{\\circ} \\mathrm{C}\n\\end{array}\n\\]"
},
{
"idx": 65,
"question": "Calculate the diffusion rate of oxygen ions (D_O^-2) in Al2O3 at 1500°C.",
"answer": "D_O^-2 = 1900 exp[-152,000/(1.987)(1773)] = 3.47 × 10^-16 cm^2/s"
},
{
"idx": 65,
"question": "Calculate the diffusion rate of aluminum ions (D_Al^+3) in Al2O3 at 1500°C.",
"answer": "D_Al^+3 = 28 exp[-114,000/(1.987)(1773)] = 2.48 × 10^-13 cm^2/s"
},
{
"idx": 65,
"question": "Compare the diffusion rates of oxygen ions and aluminum ions in Al2O3 at 1500°C and explain the difference based on their ionic radii.",
"answer": "The diffusion rate of aluminum ions (2.48 × 10^-13 cm^2/s) is higher than that of oxygen ions (3.47 × 10^-16 cm^2/s). The ionic radius of the oxygen ion is 1.32 Å, compared with the aluminum ionic radius of 0.51 Å; consequently it is much easier for the smaller aluminum ion to diffuse in the ceramic."
},
{
"idx": 66,
"question": "Calculate the diffusion coefficient of carbon in BCC iron at the allotropic transformation temperature of 912°C (1185 K).",
"answer": "D_BCC = 0.011 exp [-20,900 / (1.987)(1185)] = 1.51 × 10^-6 cm^2/s"
},
{
"idx": 66,
"question": "Calculate the diffusion coefficient of carbon in FCC iron at the allotropic transformation temperature of 912°C (1185 K).",
"answer": "D_FCC = 0.23 exp [-32,900 / (1.987)(1185)] = 1.92 × 10^-7 cm^2/s"
},
{
"idx": 66,
"question": "Explain why the diffusion coefficient of carbon in BCC iron is higher than in FCC iron at the allotropic transformation temperature.",
"answer": "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": 67,
"question": "Iron containing 0.05 % C is heated to 912°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": "t=(24 h)(3600 s/h)=86,400 s\nD_BCC=0.011 exp [-20,900 /(1.987)(1185)]=1.54 × 10^-6 cm^2/s\nBCC: (1.2-c_x)/(1.2-0.05)=erf[0.05 /(2 sqrt((1.54 × 10^-6)(86,400)))]=erf[0.0685]=0.077\nc_x=0.95 % C"
},
{
"idx": 67,
"question": "Iron containing 0.05 % C is heated to 912°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": "t=(24 h)(3600 s/h)=86,400 s\nD_FCC=0.23 exp [-32,900 /(1.987)(1185)]=1.97 × 10^-7 cm^2/s\nFCC: (1.2-c_x)/(1.2-0.05)=erf[0.05 /(2 sqrt((1.97 × 10^-7)(86,400)))]=erf[0.192]=0.2139\nc_x=0.95 % C"
},
{
"idx": 67,
"question": "Explain the difference in carbon content between BCC and FCC iron structures.",
"answer": "Faster diffusion occurs in the looser packed BCC structure, leading to the higher carbon content at point 'X'."
},
{
"idx": 68,
"question": "What temperature is required to obtain \\(0.50 \\% \\mathrm{C}\\) at a distance of \\(0.5 \\mathrm{~mm}\\) beneath the surface of a \\(0.20 \\% \\mathrm{C}\\) steel in \\(2 \\mathrm{~h}\\), when \\(1.10 \\% \\mathrm{C}\\) is present at the surface? Assume that the iron is FCC.",
"answer": "(\\frac{1.1-0.5}{1.1-0.2}=0.667=\\mathrm{erf}[0.05 / 2 \\sqrt{D t}]\\)\n\\[\n\\begin{array}{l}\n0.05 / 2 \\sqrt{D t}=0.685 \\quad \\text { or } \\quad \\sqrt{D t}=0.0365 \\quad \\text { or } \\quad D t=0.00133 \\\\\nt=(2 \\mathrm{~h})(3600 \\mathrm{~s} / \\mathrm{s})=7200 \\mathrm{~s} \\\\\nD=0.00133 / 7200=1.85 \\times 10^{-7}=0.23 \\exp [-32,900 /(1.987)] \\\\\n\\exp (-16,558 / T)=8.043 \\times 10^{-7} \\\\\nT=1180 \\mathrm{~K}=907^{\\circ} \\mathrm{C}\n\\end{array}\n\\]"
},
{
"idx": 69,
"question": "What is the value of the error function argument (z) when a 0.15% C steel is carburized at 1100°C to achieve 0.35% C at 1 mm beneath the surface with a surface composition of 0.90% C?",
"answer": "The error function argument is calculated as (0.9-0.35)/(0.9-0.15) = 0.733 = erf[0.1 / 2√(Dt)]."
},
{
"idx": 69,
"question": "What is the value of 0.1 / 2√(Dt) corresponding to the error function value of 0.733?",
"answer": "0.1 / 2√(Dt) = 0.786."
},
{
"idx": 69,
"question": "What is the value of √(Dt) derived from the equation 0.1 / 2√(Dt) = 0.786?",
"answer": "√(Dt) = 0.0636."
},
{
"idx": 69,
"question": "What is the value of Dt calculated from √(Dt) = 0.0636?",
"answer": "Dt = 0.00405 cm²."
},
{
"idx": 69,
"question": "What is the diffusion coefficient (D) for carbon in steel at 1100°C?",
"answer": "D = 0.23 exp[-32,900 / (1.987)(1373)] = 1.332 × 10⁻⁶ cm²/s."
},
{
"idx": 69,
"question": "What time (t) is required to achieve the desired carbon concentration profile, given Dt = 0.00405 cm² and D = 1.332 × 10⁻⁶ cm²/s?",
"answer": "t = 0.00405 / 1.332 × 10⁻⁶ = 3040 s = 51 min."
},
{
"idx": 70,
"question": "A \\(0.02 \\% \\mathrm{C}\\) steel is to be carburized at \\(1200^{\\circ} \\mathrm{C}\\) in \\(4 \\mathrm{~h}\\), with a point \\(0.6 \\mathrm{~mm}\\) beneath the surface reaching \\(0.45 \\% \\mathrm{C}\\). Calculate the carbon content required at the surface of the steel.",
"answer": "(\\frac{c_{s}-0.45}{c_{s}-0.02}=\\operatorname{erf}[0.06 / 2 \\sqrt{D t}]\\)\n\\[\n\\begin{array}{l}\nD=0.23 \\exp \\left[-32.900 /(1 .987)(1473)\\right]=3.019 \\times 10^{-6} \\mathrm{~cm}^{2} / \\text { s} \\\\\nt=(4 \\mathrm{~h})(3600)=14,400 \\mathrm{~s} \\\\\n\\sqrt{D t}=\\sqrt{(3.019 \\times 10^{-6})(14,400)}=0.2085 \\\\\n\\operatorname{erf}[0.06 /(2)(0.2085)]=\\operatorname{erf}(0.144)=0.161 \\\\\n\\frac{c_{s}-0.45}{c_{s}-0.02}=0.161 \\text { or } c_{s}=0.53 \\% \\mathrm{C}\n\\end{array}\n\\]"
},
{
"idx": 71,
"question": "What is the diffusion coefficient (D) for a 1.2% C tool steel held at 1150°C?",
"answer": "D = 0.23 exp(-32,900 / (1.987)(1423)) ≈ 0.234 × 10^-6 cm²/s"
},
{
"idx": 71,
"question": "What is the total time (t) in seconds for the steel exposed to oxygen for 48 hours?",
"answer": "t = (48 h)(3600 s/h) = 17.28 × 10^4 s"
},
{
"idx": 71,
"question": "What is the value of √(Dt) for the given diffusion coefficient and time?",
"answer": "√(Dt) = 0.5929"
},
{
"idx": 71,
"question": "To what depth (x) will the steel be decarburized to less than 0.20% C, given the surface carbon content is zero and the initial carbon content is 1.2% C?",
"answer": "(0 - 0.2)/(0 - 1.2) = 0.1667 ∴ x / 2√(Dt) = 0.149\nx = (0.149)(2)(0.5929) = 0.177 cm"
},
{
"idx": 72,
"question": "A \\(0.80 \\% \\mathrm{C}\\) steel must operate at \\(950^{\\circ} \\mathrm{C}\\) in an oxidizing environment, where the carbon content at the steel surface is zero. Only the outermost \\(0.02 \\mathrm{~cm}\\) of the steel part can fall below \\(0.75 \\% \\mathrm{C}\\). What is the maximum time that the steel part can operate?",
"answer": "(\\frac{0-0.75}{0-0.8}=0.9375=\\operatorname{erf}[x / 2 \\sqrt{D t}] \\therefore x / 2 \\sqrt{D t}=1.384\\)\n\\[\n\\begin{array}{l}\n0.02 / 2 \\sqrt{D t}=1.384 \\text { or } \\sqrt{D t}=0.007226 \\text { or } \\quad D t=5.22 \\times 10^{-5} \\\\\nD=0.23 \\exp \\left[-32.900 /(l .987)(1223)\\right]=3.03 \\times 10^{-7} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\nt=5.22 \\times 10^{-5} / 3.03 \\times 10^{-7}=172 \\mathrm{~s}=2.9 \\mathrm{~min}\n\\end{array}\n\\]"
},
{
"idx": 73,
"question": "A steel with \\(\\mathrm{BBC}\\) crystal structure containing \\(0.001 \\% \\mathrm{~N}\\) is nitrided at \\(550^{\\circ} \\mathrm{C}\\) for \\(5 \\mathrm{~h}\\). If the nitrogen content at the steel surface is \\(0.08 \\%\\), determine the nitrogen content at \\(0.25 \\mathrm{~mm}\\) from the surface.",
"answer": "(\\frac{0.08-c_{s}}{0.08-0.001}=\\operatorname{erf}[0.025 / 2 \\sqrt{D t}] \\quad \\mathrm{t}=(5 \\mathrm{~h}) 600 \\mathrm{~s} / \\mathrm{h})=1.8 \\times 10^{4} \\mathrm{~s}\\)\n\\[\n\\begin{aligned}\n\\mathrm{D} & =0.0047 \\exp [-18,300 /(1.987)(823)] \\\\\n& =6.488 \\times 10^{-8} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\n\\sqrt{D t} & =0.0342 \\\\\n\\operatorname{erf}[0.025 /(2)(0.0342)] & =\\operatorname{erf}(0.3655)=0.394 \\\\\n\\frac{0.08-c_{s}}{0.079} & =0.394 \\text { or } c_{s}=0.049 \\% \\mathrm{~N}\n\\end{aligned}\n\\]"
},
{
"idx": 74,
"question": "What time is required to nitride a \\(0.002 \\mathrm{~N}\\) steel to obtain \\(0.12 \\% \\mathrm{~N}\\) at a distance of 0.002 in. beneath the surface at \\(625^{\\circ} \\mathrm{C}\\) ? The nitrogen content at the surface is \\(0.15 \\%\\).\n\\[\n\\begin{array}{l}\n\\text {",
"answer": "\\frac{0.15-0.12}{0.15-0.002}=0.2027=\\operatorname{erf}[x / 2 \\sqrt{D t}] \\therefore x / 2 \\sqrt{D t}=0.2256 \\\\\nD=0.0047 \\exp [-18,300 /(1.9 87)(898)]=1.65 \\times 10^{-7} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\nx=0.002 \\text { in. }=0.00508 \\mathrm{~cm} \\\\\n\\frac{0.00508}{2 \\sqrt{(1.65 \\times 10^{-7})} t}=\\frac{0.2256}{2 \\sqrt{(1.65 \\times 10^{-7})} t}\n\\end{array}\n\\]\n\\[\nD t=1.267 \\times 10^{-4} \\text { or } t=1.267 \\times 10^{-4} / 1.65 \\times 10^{-7}=768 \\mathrm{~s}=12.8 \\mathrm{~min}\n\\]"
},
{
"idx": 75,
"question": "We currently can successfully perform a carburizing heat treatment at \\(1200^{\\circ} \\mathrm{C}\\) in \\(1 \\mathrm{~h}\\). In an effort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to \\(950^{\\circ} \\mathrm{C}\\). What time will be required to give us a similar carburizing treatment?",
"answer": "(\\quad D_{1200}=0.23 \\exp [-32,900 /(1.987)(1473)]=3.019 \\times 10^{-6} \\mathrm{~cm}^{2} / \\mathrm{s}\\)\n\\[\n\\begin{array}{l}\nD_{550}=0.23 \\exp [-32,900 /(1.9 87)(1223)]=3.034 \\times 10^{-7} \\mathrm{~cm}^{2} / \\text { s} \\\\\nt_{1200}=1 \\mathrm{~h} \\\\\nt_{590}=D_{1200} t_{1200} / D_{550}=\\frac{(3.019 \\times 10^{-6})(1)}{3.034 \\times 10^{-7}}=9.95 \\mathrm{~h}\n\\end{array}\n\\]"
},
{
"idx": 76,
"question": "During freezing of a \\(\\mathrm{Cu}-\\mathrm{Zn}\\) alloy, we find that the composition is nonuniform. By heating the alloy to \\(600^{\\circ} \\mathrm{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": "(\\quad D_{600}=0.78 \\exp [-43,900 /(1.987)(873)]=7.9636 \\times 10^{-12} \\quad t_{600}=3 \\mathrm{~h}\\)\n\\[\n\\begin{array}{c}\nD_{x}=D_{600} t_{600} / t_{x}=(7.9636 \\times 10^{-12})(3) / 0.5 \\\\\nD_{x}=4.778 \\times 10^{-11}=0.78 \\exp [-43,900 / 1.987 T \\\\\n\\ln (6.1258 \\times 10^{-11})=-23.516=-43.900 / 1.987 T \\\\\nT=940 \\mathrm{~K}=667^{\\circ} \\mathrm{C}\n\\end{array}\n\\]"
},
{
"idx": 77,
"question": "A ceramic part made of MgO is sintered successfully at 1700 degrees Celsius in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees Celsius. 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": 77,
"question": "A ceramic part made of MgO is sintered successfully at 1700 degrees Celsius in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees Celsius. What time will be required at the lower temperature?",
"answer": "D_1760 = 0.000043 exp[-82,100 /(1.987)(1973)] = 3.455 x 10^-14 cm^2/s. D_1560 = 0.000043 exp[-82,100 /(1.987)(1773)] = 3.255 x 10^-15 cm^2/s. t_1560 = D_1760 t_1760 / D_1560 = (3.455 x 10^-14)(90) / 3.255 x 10^-15 = 955 min = 15.9 h."
},
{
"idx": 78,
"question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi. Determine whether the wire will plastically deform.",
"answer": "First determine the stress acting on the wire: sigma = F / A = 850 lb / (pi / 4)(0.15 in)^2 = 48,100 psi. Because sigma is greater than the yield strength of 45,000 psi, the wire will plastically deform."
},
{
"idx": 78,
"question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a tensile strength of 55,000 psi. Determine whether the wire will experience necking.",
"answer": "First determine the stress acting on the wire: sigma = F / A = 850 lb / (pi / 4)(0.15 in)^2 = 48,100 psi. Because sigma is less than the tensile strength of 55,000 psi, no necking will occur."
},
{
"idx": 79,
"question": "A force of 100,000 N is applied to a 10 mm × 20 mm iron bar having a yield strength of 400 MPa. Determine whether the bar will plastically deform.",
"answer": "First determine the stress acting on the bar: σ = F / A = 100,000 N / (10 mm)(20 mm) = 500 N/mm² = 500 MPa. Because σ is greater than the yield strength of 400 MPa, the bar will plastically deform."
},
{
"idx": 79,
"question": "A force of 100,000 N is applied to a 10 mm × 20 mm iron bar having a tensile strength of 480 MPa. Determine whether the bar will experience necking.",
"answer": "First determine the stress acting on the bar: σ = F / A = 100,000 N / (10 mm)(20 mm) = 500 N/mm² = 500 MPa. Because σ is greater than the tensile strength of 480 MPa, the bar will experience necking."
},
{
"idx": 80,
"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": "(F=\\mathrm{O} A=(35,000 \\mathrm{psi})(\\pi / 4)(0.2 \\mathrm{in})^{2}=1100 \\mathrm{lb}\\)\n\\[\nF=(1100 \\mathrm{lb})(4.448 \\mathrm{~N} / \\mathrm{lb})=4891 \\mathrm{~N}\n\\]"
},
{
"idx": 81,
"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 strain.",
"answer": "The strain e is ε=(10.045 cm−10 cm)/10 cm=0.0045 cm/cm"
},
{
"idx": 81,
"question": "A force of 20,000 N is applied to a 1 cm × 1 cm bar of magnesium. Calculate the stress in MPa.",
"answer": "The stress σ is σ=20,000 N/(10 mm×10 mm)=200 N/mm²=200 MPa"
},
{
"idx": 81,
"question": "Using the calculated stress of 200 MPa and strain of 0.0045 cm/cm, determine the modulus of elasticity in GPa.",
"answer": "E=σ/ε=200 MPa/0.0045 cm/cm=44,444 MPa=44.4 GPa"
},
{
"idx": 81,
"question": "Convert the modulus of elasticity from 44,444 MPa to psi.",
"answer": "E=(44,444 MPa)(145 psi/MPa)=6.44×10⁶ psi"
},
{
"idx": 82,
"question": "A polymer bar's dimensions are \\(1 \\mathrm{in} . \\times 2\\) in. \\(\\times 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 strain \\(\\varepsilon\\) is \\(\\varepsilon=(15.25\\) in. -15 in. \\() /(15\\) in. \\()=0.01667\\) in./in.\nThe stress \\(\\sigma\\) is \\(\\sigma=E \\varepsilon=(600,000 \\mathrm{psi})(0.01667 \\mathrm{in} / \\mathrm{in})=10,000 \\mathrm{psi}\\)\nThe force is then \\(F=\\sigma A=(10,000 \\mathrm{psi})(1\\) in. \\() / 2\\) in.) \\(=200,000 \\mathrm{lb}\\)"
},
{
"idx": 83,
"question": "An aluminum plate \\(0.5 \\mathrm{~cm}\\) thick is to withstand a force of \\(50,000 \\mathrm{~N}\\) with no permanent deformation. If the aluminum has a yield strength of \\(125 \\mathrm{MPa}\\), what is the minimum width of the plate?",
"answer": "The area is \\(A=F / \\sigma=50,000 \\mathrm{~N} / 125 \\mathrm{~N} / \\mathrm{mm}^{2}=400 \\mathrm{~mm}^{2}\\)\nThe minimum width is \\(w=A / t=(400 \\mathrm{~mm}^{2})(0.1 \\mathrm{~cm} / \\mathrm{mm})^{2} / 0.5 \\mathrm{~cm}\\) \\(=8 \\mathrm{~cm}\\)"
},
{
"idx": 84,
"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 \\times 10^{6} \\mathrm{psi}\\) and the yield strength is 40,000 psi.",
"answer": "(a) The strain is \\(\\varepsilon=\\sigma / E=40,000 \\mathrm{psi} / 17 \\times 10^{6} \\mathrm{psi}=0.00235 \\mathrm{in} . / \\mathrm{in}\\).\nThe strain is also \\(\\varepsilon=(2 \\mathrm{in} .-d_{0}) / d_{0}=0.00235 \\mathrm{in} . / \\mathrm{in}^{\\text {. }}\\)\n\\(2-d_{0}=0.00235 d_{0}\\)\n\\(d_{0}=2 / 1.00235=1.995\\) in.\nThe opening in the die must be smaller than the final diameter."
},
{
"idx": 85,
"question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16 x 10^6 psi, and Poisson's ratio of 0.30. Determine the length of the bar when a 500-lb load is applied.",
"answer": "The stress is σ=F/A=500 lb/(π/4)(0.4 in.)^2=3,979 psi. The applied stress is much less than the yield strength; therefore Hooke's law can be used. The strain is ε=σ/E=3,979 psi/(16 x 10^6 psi)=0.00024868 in./in. (ℓ_f-ℓ_0)/ℓ_0=(ℓ_f-12 in.)/12 in.=0.00024868 in./in. ℓ_f=12.00298 in."
},
{
"idx": 85,
"question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16 x 10^6 psi, and Poisson's ratio of 0.30. Determine the diameter of the bar when a 500-lb load is applied.",
"answer": "From Poisson's ratio, μ=-ε_lat/ε_long=0.3. ε_lat=-(0.3)(0.00024868)=-0.0000746 in./in. (d_f-d_0)/d_0=(d_f-0.4 in.)/0.4 in.=-0.0000746 in./in. d_f=0.39997 in."
},
{
"idx": 86,
"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 = 3 F L / 2 w h^2 = (3)(400 lb)(4 in) / (2)(0.5 in)(0.25 in)^2 = 76,800 psi"
},
{
"idx": 86,
"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 = F L^3 / 4 w h^3 delta = (400 lb)(4 in)^3 / (4)(0.5 in)(0.25 in)^3(0.037 in) = 22.14 x 10^6 psi"
},
{
"idx": 87,
"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 0.09 mm 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 F required to produce a deflection of 0.09 mm is F = (flexural modulus)(4 w h^3 δ) / L^3 F = (480,000 MPa)(4)(15 mm)(6 mm)^3(0.09 mm) / (75 mm)^3 F = 1327 N"
},
{
"idx": 87,
"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 0.09 mm is recorded. Calculate the flexural strength. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.",
"answer": "Flexural strength = 3 F L / 2 w h^2 = (3)(1327 N)(75 mm) / (2)(15 mm)(6 mm)^2 = 276 MPa"
},
{
"idx": 88,
"question": "A thermosetting polymer containing glass beads is required to deflect 0.5 mm 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 L between the supports can be calculated from the flexural modulus. L^3 = 4 w h^2 δ (flexural modulus) / F. L^3 = (4)(20 mm)(5 mm)^3(0.5 mm)(6.9 GPa)(1000 MPa/GPa) / 500 N. L^3 = 69,000 mm^3 or L = 41 mm."
},
{
"idx": 88,
"question": "Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.",
"answer": "The stress acting on the bar when a deflection of 0.5 mm is obtained is σ = 3 F L / 2 w h^2 = (3)(500 N)(41 mm) / (2)(20 mm)(5 mm)^2 = 61.5 MPa. The applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture."
},
{
"idx": 89,
"question": "The flexural strength of alumina 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 force required to break the bar.",
"answer": "F = (2)(1 in.)(0.3 in.)^2(46,000 psi)/(3)(7 in.) = 394 lb"
},
{
"idx": 89,
"question": "The flexural modulus of alumina is 45 x 10^6 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. The force required to break the bar is 394 lb.",
"answer": "δ = (394 lb)(7 in.)^3/(4)(1 in.)(0.3 in.)^3(45 x 10^6 psi) = 0.0278 in."
},
{
"idx": 90,
"question": "A Brinell hardness measurement, using a 10-mm-diameter indenter and a 500-kg load, produces an indentation of \\(4.5 \\mathrm{~mm}\\) on an aluminum plate. Determine the Brinell hardness number \\((\\mathrm{HB})\\) of the metal.",
"answer": "(\\quad H B=\\frac{500 \\mathrm{~kg}}{(\\pi / 2)(10 \\mathrm{~mm})[10-\\sqrt{10^{2}-4.5^{2}}]}=29.8\\)"
},
{
"idx": 91,
"question": "When a \\(3000-\\mathrm{kg}\\) load is applied to a \\(10-\\mathrm{mm}\\)-diameter ball in a Brinell test of a steel, an indentation of \\(3.1 \\mathrm{~mm}\\) is produced. Estimate the tensile strength of the steel.",
"answer": "(\\quad H B=\\frac{3000 \\mathrm{~kg}}{(\\pi / 2)(10 \\mathrm{\\mathrm{mm}})[10-\\sqrt{10^{2}-3.1^{2}}]}=388\\)\nTensile strength \\(=500 \\mathrm{HB}=(500)(388)=194,000 \\mathrm{psi}\\)"
},
{
"idx": 92,
"question": "A ceramic matrix composite contains internal flaws as large as \\(0.001 \\mathrm{~cm}\\) in length. The plane strain fracture toughness of the composite is \\(45 \\mathrm{MPa} / \\mathrm{m}\\) and the tensile strength is \\(550 \\mathrm{MPa}\\). Will the flaw cause the composite to fail before the tensile strength is reached? Assume that \\(f=1\\).",
"answer": "Since the crack is internal, \\(2 a=0.001 \\mathrm{~cm}=0.00001 \\mathrm{~m}\\). Therefore\n\\[\n\\begin{array}{l}\na=0.000005 \\mathrm{~m} \\\\\nK_{\\mathrm{fc}}=f \\sigma \\sqrt{\\pi a} \\quad \\text { or } \\quad \\sigma=K_{\\mathrm{fc}} / f \\sqrt{\\pi a} \\\\\n\\sigma=(45 \\mathrm{MPa} \\sqrt{\\mathrm{m}}) /(1) \\sqrt{\\pi(0.000005 \\mathrm{~m})}=11,354 \\mathrm{MPa}\n\\end{array}\n\\]The applied stress required for the crack to cause failure is much larger than the tensile strength of \\(550 \\mathrm{MPa}\\). Any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws."
},
{
"idx": 93,
"question": "An aluminum alloy that has a plane strain fracture toughness of \\(25,000 \\mathrm{psi} \\sqrt{\\mathrm{m}}\\). fails when a stress of \\(42,000 \\mathrm{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": "(\\quad K_{k_{c}}=f \\sigma \\sqrt{\\pi a} \\quad\\) or \\(\\quad a=(1 / \\pi)\\left[K_{k_{c}} / \\right.\\) for \\(\\left.{ }^{2}\\right]\\)\n\\[\na=(1 / \\pi)[25,000 \\mathrm{psi} \\sqrt{\\mathrm{m}},(1.1)(42,000 \\mathrm{psi})]^{2}=0.093 \\mathrm{in} .\n\\]"
},
{
"idx": 94,
"question": "A polymer that contains internal flaws \\(1 \\mathrm{~mm}\\) in length fails at a stress of \\(25 \\mathrm{MPa}\\). Determine the plane strain fracture toughness of the polymer. Assume that \\(f=1\\).",
"answer": "Since the flaws are internal, \\(2 a=1 \\mathrm{~mm}=0.001 \\mathrm{~m}\\); thus \\(a=0.0005 \\mathrm{~m}\\)\n\\[\nK_{k_{c}}=f \\sigma \\sqrt{\\pi a}=(1)(25 \\mathrm{MPa}) \\sqrt{\\pi(0.0005 \\mathrm{~m})}=0.99 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\n\\]"
},
{
"idx": 95,
"question": "What is the maximum applied stress for the ceramic part given a yield strength of 75,000 psi and a safety factor of one third?",
"answer": "The applied stress is sigma = (1/3) * 75,000 psi = 25,000 psi."
},
{
"idx": 95,
"question": "What is the critical flaw size (a) for the ceramic part given a plane strain fracture toughness of 5,000 psi sqrt(in), a geometric factor f of 1.4, and the applied stress of 25,000 psi?",
"answer": "a = (1/pi) * [K_IC / (f * sigma)]^2 = (1/pi) * [5,000 psi sqrt(in) / (1.4 * 25,000 psi)]^2 = 0.0065 in. The length of internal flaws is 2a = 0.013 in."
},
{
"idx": 95,
"question": "Does the nondestructive test that detects flaws greater than 0.05 in. long have the required sensitivity given the critical flaw size of 0.013 in.?",
"answer": "The nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size of 0.013 in. required for failure. Thus, the NDT test is not satisfactory."
},
{
"idx": 96,
"question": "What is the fatigue strength, or maximum stress amplitude, required to survive for one million cycles under conditions that provide for equal compressive and tensile stresses?",
"answer": "From the figure, the fatigue strength at one million cycles is 22 MPa."
},
{
"idx": 96,
"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": 96,
"question": "What effect would the frequency of the stress application have on your answers regarding 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": 96,
"question": "The high-strength steel in Figure 7-21 is subjected to a stress alternating at 200 revolutions per minute between 600 MPa and 200 MPa (both tension). Calculate the growth rate of a surface crack when it reaches a length of 0.2 mm in both m/cycle and m/s. Assume that f=1.0.",
"answer": "For the steel, C=1.62×10^-12 and n=3.2. The change in the stress intensity factor ΔK is ΔK−√A S √α u=(1.2)(600 MPa−200 MPa)√α(0.0002 m)=12.03 MPa√m. The crack growth rate is da/dN=1.62×10^-12(ΔK)^3.2=1.62×10^-12(12.03)^3.2=4.638×10^-9 m/cycle. da/dt=(4.638×10^-9 m/cycle)(200 cycles/min)/60 s/min=1.55×10^-8 m/s."
},
{
"idx": 96,
"question": "The high-strength steel in Figure 7-21, which has a critical fracture toughness of 80 MPa√m, is subjected to an alternating stress varying from -900 MPa (compression) to +900 MPa (tension). It is to survive for 10^5 cycles before failure occurs. Calculate (a) the size of a surface crack required for failure to occur and (b) the largest initial surface crack size that will permit this to happen. Assume that f=1.",
"answer": "(a) Only the tensile portion of the applied stress is considered in Δσ. Based on the applied stress of 900 MPa and the fracture toughness of 80 MPa√m, the size of a surface crack required for failure to occur is K=fσ√πa_c or a_c=(1/π)[K/fσ]^2. a_c=(1/π)[80 MPa√m/(1)(900 MPa)]^2=0.0025 m=2.5 mm. (b) The largest initial surface crack tolerable to prevent failure within 10^5 cycles is N=10^5 cycles=2[(0.0025 m)^(2-3.2)/2−a_i^(2-3.2)/2]/(2-3.2)(1.62×10^-12)(1)^3.2(900)^3.2(π)^3.2/2. 10^5=2[36.41−(a_i)^-0.60]/(−1.2)(1.62×10^-12)(1)(2.84×10^9)(6.244). (a_i)^-0.6=1760. a_i=3.9×10^-6 m=0.0039 mm."
},
{
"idx": 97,
"question": "The activation energy for self-diffusion in copper is \\(49,300 \\mathrm{cal} / \\mathrm{mol}\\). A copper specimen creps as \\(t\\) 0.002 \\(\\mathrm{n} / \\mathrm{fn} .-\\mathrm{h}\\) when a stress of 15,000 psi is applied at \\(600^{\\circ} \\mathrm{C}\\). If the creep rate of copper is dependent on self-diffusion, determine the creep rate if the temperature is \\(800^{\\circ} \\mathrm{C}\\).",
"answer": "The creep rate is governed by an Arrhenius relationship of the form rate \\(=A \\exp (-0 / R T)\\). From the information given,\n\\[\n\\begin{array}{l}\nx=\\frac{A \\exp \\left[-49,300 /(1.987)(800+273)\\right]}{A \\exp \\left[-49,300 /(1.987) 600+273)\\right]}=\\frac{9.07 \\times 10^{-11}}{4.54 \\times 10^{-13}} \\\\\nx=(0.002)(9.07 \\times 10^{-11} / 4.54 \\times 10^{-13})=0.4 \\mathrm{n} / \\mathrm{in} \\cdot \\mathrm{h}\n\\end{array}\n\\]"
},
{
"idx": 98,
"question": "When a stress of 20000 psi is applied to a material heated to \\(900^{\\circ} \\mathrm{C}\\), rupture occurs in \\(25000 \\mathrm{~h}\\). If the activation energy for rupture is \\(35000 \\mathrm{cal} / \\mathrm{mol}\\), determine the rupture time if the temperature is reduced to \\(800^{\\circ} \\mathrm{C}\\).",
"answer": "The rupture time is related to temperature by an Arrhenius relationship of the form \\(t_{c}=A \\exp (+O R T)\\); the argument of the exponential is positive because the rupture time is inversely related to the rate. From the information given\n\\[\n\\begin{array}{l}\n\\frac{t_{c}}{25000 \\mathrm{~h}}=\\frac{A \\exp \\left[35000 /(1.987)(800+273)\\right]}{\\frac{A \\exp \\left[35000 /(1.98 7)(900+273)\\right]}{A \\exp \\left[35000 /(1.987) /(900+273)\\right]}}=\\frac{1.35 \\times 10^{7}}{3.32 \\times 10^{6}} \\\\\nt_{c}=(25000)(1.35 \\times 10^{7} / 3.32 \\times 10^{6})=101660 \\mathrm{~h}\n\\end{array}\n\\]"
},
{
"idx": 99,
"question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the activation energy Qr for rupture in the temperature range 980 to 1090 degrees Celsius.",
"answer": "The appropriate equation is tr=K sigma'' exp(Qr/RT). From Figure 7-27(a), we can determine the rupture time versus temperature for a fixed stress, sigma=1000 psi: tr=2,400 h at 1090 degrees Celsius, tr=14,000 h at 1040 degrees Celsius, tr=100,000 h at 980 degrees Celsius. From this data, the equation becomes tr=K' exp(Qr/RT) and we can find Qr by simultaneous equations or graphically. Qr=117,000 cal/mol."
},
{
"idx": 99,
"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 degrees Celsius.",
"answer": "The appropriate equation is tr=K sigma'' exp(Qr/RT). We can determine the rupture time versus applied stress for a constant temperature, 1090 degrees Celsius: tr=10^5 h for sigma=450 psi, tr=10^4 h for sigma=800 psi, tr=10^3 h for sigma=1200 psi, tr=10^2 h for sigma=2100 psi. With this approach, the equation becomes tr=K'' sigma^m, where m is obtained graphically or by simultaneous equations: m=3.9."
},
{
"idx": 100,
"question": "A 0.25-in.-thick copper plate is to be cold worked \\(63 \\%\\). Find the final thickness.",
"answer": "(See Figure 8-7.) \\(\\quad 63=\\frac{0.25-f}{0.25} \\times 100 \\% \\quad\\) or \\(\\quad t_{f}=0.0925\\) in."
},
{
"idx": 101,
"question": "A 0.25-in.-diameter copper bar is to be cold worked \\(63 \\%\\). Find the final diameter.",
"answer": "(\\quad 63=\\frac{(0.25)^{2}-d_{\\mathrm{r}}^{2}}{(0.25)^{2}} \\times 100 \\% \\quad\\) or \\(\\quad d_{\\mathrm{r}}^{2}=0.023 \\quad\\) or \\(\\quad d_{f}=0.152\\) in."
},
{
"idx": 102,
"question": "A 2-in.-diameter copper rod is reduced to 1.5 in. diameter, then reduced again to a final diameter of 1 in. Calculate the % CW for this case.",
"answer": "% CW = ((2)^2 - (1)^2) / (2)^2 * 100 = 75%"
},
{
"idx": 102,
"question": "A 2-in.-diameter copper rod is reduced in one step from 2 in. to a 1 in. diameter. Calculate the % CW for this case.",
"answer": "% CW = ((2)^2 - (1)^2) / (2)^2 * 100 = 75%"
},
{
"idx": 103,
"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": "From Table 9-1, ΔT_max = 480°C\nr^s = (2)(255×10^-7 J/cm^2)(1453+273)/(2756 J/cm^3)(480) = 6.65×10^-8 cm"
},
{
"idx": 103,
"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": "a_0 = 3.56 Å, V = 45.118×10^-24 cm^3\nV_nucleus = (4π/3)(6.65×10^-8 cm)^3 = 1232×10^-24 cm^3\nnumber of unit cells = 1232/45.118 = 27.3\natoms per nucleus = (4 atoms/cell)(27.3 cells) = 109 atoms"
},
{
"idx": 104,
"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": "r^s = (2)(204×10^-7 J/cm^2)(1538+273)/(1737 J/cm^3)(420) = 10.128×10^-8 cm"
},
{
"idx": 104,
"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": "V = (4π/3)(10.128)^3 = 4352 Å^3 = 4352×10^-24 cm^3. V_lc = (2.92 Å)^3 = 24.897 Å^3 = 24.897×10^-24 cm^3. Number of unit cells = 4352/24.897 = 175. Atoms per nucleus = (175 cells)(2 atoms/cell) = 350 atoms"
},
{
"idx": 105,
"question": "Suppose that solid nickel was able to nucleate homogeneously with an undercooling of only \\(22^{\\circ} \\mathrm{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 \\(0.356 \\mathrm {~nm}\\).",
"answer": "(\\quad r^{*}=\\frac{(2)(255 \\times 10^{-7} \\mathrm{~J} / \\mathrm{cm}^{2})(1453+273)}{(2756 \\mathrm{~J} / \\mathrm{cm}^{3})(22)}=145.18 \\times 10^{-8} \\mathrm{~cm}\\)\n\\[\nV_{\\mathrm{uc}}=45.118 \\times 10^{-24} \\mathrm{~cm}^{3} \\quad \\text { (see Problem 9-10) }\n\\]\n\\(V_{\\text {nuc }}=(4 \\pi / 3)(145.18 \\times 10^{-8} \\mathrm{~cm})^{3}=1.282 \\times 10^{-17} \\mathrm{~cm}^{3}\\)\nnumber of unit cells \\(=1.282 \\times 10^{-17} / 45.118 \\times 10^{-24}=2.84 \\times 10^{5}\\)\natoms per nucleus \\(=(4\\) atoms/cells \\()(2.84 \\times 10^{5}\\) cell \\()=1.136 \\times 10^{6}\\)"
},
{
"idx": 106,
"question": "Suppose that solid iron was able to nucleate homogeneously with an undercooling of only \\(15^{\\circ} \\mathrm{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 \\AA\\).",
"answer": "(\\quad r^{*}=\\frac{(2)(204 \\times 10^{-7} \\mathrm{~J} / \\mathrm{cm})^{2})(1538+273)}{(1737 \\mathrm{~J} / \\mathrm{cm})^{3}}=283.6 \\times 10^{-8} \\mathrm{~cm}\\)\n\\[\n\\begin{array}{l}\nV_{\\mathrm{uc}}=24.897 \\times 10^{-24} \\mathrm{~cm}^{3} \\text { (see Problem 9-10) } \\\\\nV_{\\mathrm{nuc}}=(4 \\pi / 3)(283.6 \\times 10^{-8} \\mathrm{~cm})^{3}=95,544,850 \\times 10^{-24} \\mathrm{~cm}^{3} \\\\\n\\text { number of unit cells }=95,544,850 / 24.897=3.838 \\times 10^{6} \\\\\n\\text { atoms per nucleus }=(2 \\text { atoms/cells })(3.838 \\times 10^{6} \\text { cell })=7.676 \\times 10^{6}\n\\end{array}\n\\]"
},
{
"idx": 107,
"question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 10°C undercooling. The specific heat of iron is 5.78 J/cm³·°C and the heat of fusion is 1737 J/cm³.",
"answer": "f = (c ΔT) / ΔH_f = (5.78 J/cm³·°C * 10°C) / 1737 J/cm³ = 0.0333"
},
{
"idx": 107,
"question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 100°C undercooling. The specific heat of iron is 5.78 J/cm³·°C and the heat of fusion is 1737 J/cm³.",
"answer": "f = (c ΔT) / ΔH_f = (5.78 J/cm³·°C * 100°C) / 1737 J/cm³ = 0.333"
},
{
"idx": 107,
"question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates homogeneously. The specific heat of iron is 5.78 J/cm³·°C and the heat of fusion is 1737 J/cm³.",
"answer": "f = (c ΔT) / ΔH_f = (5.78 J/cm³·°C * 420°C) / 1737 J/cm³ = 1.4, therefore, all solidification occurs dendritically."
},
{
"idx": 108,
"question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 10°C undercooling. The specific heat of silver is 3.25 J/cm³·°C and the heat of fusion is 965 J/cm³.",
"answer": "f = (c ΔT) / ΔH_f = (3.25 J/cm³·°C)(10°C) / 965 J/cm³ = 0.0337"
},
{
"idx": 108,
"question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 100°C undercooling. The specific heat of silver is 3.25 J/cm³·°C and the heat of fusion is 965 J/cm³.",
"answer": "f = (c ΔT) / ΔH_f = (3.25 J/cm³·°C)(100°C) / 965 J/cm³ = 0.337"
},
{
"idx": 108,
"question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates homogeneously. The specific heat of silver is 3.25 J/cm³·°C and the heat of fusion is 965 J/cm³.",
"answer": "f = (c ΔT) / ΔH_f = (3.25 J/cm³·°C)(250°C) / 965 J/cm³ = 0.842"
},
{
"idx": 109,
"question": "Calculate the mold constant in Chvorinov's rule for a 2-in. cube that solidifies in 4.6 min. Assume that n=2.",
"answer": "We can find the volume and surface area of the cube: V=(2)^3=8 in^3, A=6(2)^2=24 in^2, t=4.6=B(8/24)^2. B=4.6/(0.333)^2=41.48 min/in^2."
},
{
"idx": 109,
"question": "Calculate the solidification time for a 0.5 in. x 0.5 in. x 6 in. bar cast under the same conditions, using the mold constant B=41.48 min/in^2. Assume that n=2.",
"answer": "For the bar: V=(0.5)(0.5)(6)=1.5 in^3, A=2(0.5)(0.5)+4(0.5)(6)=12.5 in^2. t=(41.48)(1.5/12.5)^2=0.60 min."
},
{
"idx": 110,
"question": "A 5-cm diameter sphere solidifies in \\(1050 \\mathrm{~s}\\). Calculate the solidification time for a \\(0.3 \\mathrm{~cm} \\times 10 \\mathrm{~cm} \\times 20 \\mathrm{~cm}\\) plate cast under the same conditions. Assume that \\(n=2\\).",
"answer": "(\\quad t=1050 \\mathrm{~s}=B\\left[\\left(\\frac{(4 \\pi / 3)(2.5)^{3}}{4 \\pi(2.5)^{2}}\\right)^{2}\\right]=B[2.5 / 3]^{2}\\) or \\(B=1512 \\mathrm{~s} / \\mathrm{cm}^{2}\\)\n\\[\nt=\\frac{(1512)(0.3 \\times 10 \\times 20)^{2}}{[2(0.3)(10)+2(0.3)(20)+2(10)(20)]^{2}}=1512[60 / 8]^{2}=31.15 \\mathrm{~s}\n\\]"
},
{
"idx": 111,
"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": "(V A)_C = (4)(10)(20) / [2(4)(10) + 2(4)(20) + 2(10)(20)] = 800 / 640 = 1.25\n(V A)_F = [(π/4)D²H] / [2(π/4)D² + πDH] = [(π/4)(3/2)D³] / [(π/2)D² + (3π/2)D²] = (3D/8) / 2 = 3D/16 ≈ 1.25\nD ≥ 6.67 in., H ≥ 10 in., V ≥ 349 in.³"
},
{
"idx": 111,
"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": "(V A)_C = (4)(10)(20) / [2(4)(10) + 2(4)(20) + 2(10)(20)] = 800 / 640 = 1.25\n(V A)_F = [(π/4)D²H] / [2(π/4)D² + πDH] = [(π/4)(3/2)D³] / [(π/2)D² + (3π/2)D²] = (3D/8) / 2 = 3D/16 ≈ 1.25\nD ≥ 6.67 in., H ≥ 10 in., V ≥ 349 in.³"
},
{
"idx": 111,
"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": "(V A)_C = (4)(10)(20) / [2(4)(10) + 2(4)(20) + 2(10)(20)] = 800 / 640 = 1.25\n(V A)_F = [(π/4)D²H] / [2(π/4)D² + πDH] = [(π/4)(3/2)D³] / [(π/2)D² + (3π/2)D²] = (3D/8) / 2 = 3D/16 ≈ 1.25\nD ≥ 6.67 in., H ≥ 10 in., V ≥ 349 in.³"
},
{
"idx": 112,
"question": "Calculate the volume of the casting with dimensions 1 in. × 6 in. × 6 in.",
"answer": "V = (1)(6)(6) = 36 in^3"
},
{
"idx": 112,
"question": "Calculate the surface area of the casting with dimensions 1 in. × 6 in. × 6 in.",
"answer": "A = 2(1)(6) + 2(1)(6) + 2(6)(6) = 96 in^2"
},
{
"idx": 112,
"question": "Determine the volume-to-surface area ratio (V/A) of the casting.",
"answer": "(V/A)_C = 36 / 96 = 0.375"
},
{
"idx": 112,
"question": "Calculate the minimum diameter (D) of the cylindrical riser with H/D = 1.0 to prevent shrinkage, given (V/A)_F ≥ (V/A)_C.",
"answer": "(V/A)_F = D/6 ≥ 0.375 → D ≥ 2.25 in."
},
{
"idx": 112,
"question": "Calculate the minimum height (H) of the cylindrical riser with H/D = 1.0.",
"answer": "H = D ≥ 2.25 in."
},
{
"idx": 112,
"question": "Calculate the minimum volume (V) of the cylindrical riser with D = 2.25 in. and H = 2.25 in.",
"answer": "V = (π/4)D^2H = (π/4)(2.25)^2(2.25) ≥ 8.95 in^3"
},
{
"idx": 113,
"question": "For a 4-in-diameter sphere of liquid copper (Cr) allowed to solidify, producing a spherical shrinkage cavity in the center of the casting, calculate the volume and diameter of the shrinkage cavity. Given: Shrinkage percentage for Cr = 5.1%, r_sphere = 2 in.",
"answer": "V_shrinkage = (4π/3)(2)^3(0.051) = 1.709 in^3. Diameter of shrinkage cavity = 1.30 in."
},
{
"idx": 113,
"question": "For a 4-in-diameter sphere of liquid iron (Fe) allowed to solidify, producing a spherical shrinkage cavity in the center of the casting, calculate the volume and diameter of the shrinkage cavity. Given: Shrinkage percentage for Fe = 3.4%, r_sphere = 2 in.",
"answer": "V_shrinkage = (4π/3)(2)^3(0.034) = 1.139 in^3. Diameter of shrinkage cavity = 1.06 in."
},
{
"idx": 114,
"question": "A 4-in. cube of a liquid metal is allowed to solidify. A spherical shrinkage cavity with a diameter of \\(1.49 \\mathrm{in}\\). is observed in the solid casting. Determine the percent volume change that occurs during solidification.",
"answer": "(\\quad V_{\\text {liquid }}=(4 \\mathrm{in} \\cdot \\mathrm{s})^{3}=64 \\mathrm{in} . \\mathrm{s}\\)\n\\(V_{\\text {triangle }}=(4 \\pi / 3)(1.49 / 2)^{3}=1.732\\) in. \\(^{3}\\)\n\\(V_{\\text {solid }}=64-1.732=62.268\\) in. \\(^{3}\\)\n\\(\\%\\) volume change \\(=\\frac{64-62.268}{64} \\times 100=2.7 \\%\\)"
},
{
"idx": 115,
"question": "A 2 cm × 4 cm × 6 cm magnesium casting is produced. After cooling to room temperature, calculate the volume cavity at the center of the casting. The density of the magnesium is 1.738 g/cm3.",
"answer": "V_initial = (2)(4)(6) = 48 cm3. V_initial = 80 g / 1.738 g/cm3 = 46.03 cm3."
},
{
"idx": 115,
"question": "A 2 cm × 4 cm × 6 cm magnesium casting is produced. After cooling to room temperature, calculate the percent shrinkage that must have occurred during solidification. The density of the magnesium is 1.738 g/cm3.",
"answer": "% shrinkage = (48 - 46.03) / 48 × 100% = 4.1%."
},
{
"idx": 116,
"question": "A 2 in. x 8 in. x 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. The density of the iron is 7.87 g/cm3. Determine the percent shrinkage that must have occurred during solidification.",
"answer": "v_actual = (43.9 lb)(454 g) / 7.87 g/cm3 = 2532.5 cm3\nV_intended = (2)(8)(10) = 160 in3 x (2.54 cm/in)3 = 2621.9 cm3\nshrinkage = (2621.9 - 2532.5) / 2621.9 x 100% = 3.4%"
},
{
"idx": 116,
"question": "A 2 in. x 8 in. x 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. The density of the iron is 7.87 g/cm3. 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": "V_shrinkage = 2621.9 cm3 - 2532.5 cm3 = 89.4 cm3\nV_pore = (0.05 in / 2)(2.54 cm/in) = 0.0635 cm\n# pores = 89.4 cm3 / [(4π/3)(0.0635 cm)3] = 83,354 pores"
},
{
"idx": 117,
"question": "Liquid magnesium is poured into a \\(2 \\mathrm{~cm} \\times 2 \\mathrm{~cm} \\times 24 \\mathrm{~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": "(V_{\\text {initial }}=(2)(2)(24)=96 \\mathrm{~cm}^{3}\\)\n\\(\\%\\) contraction \\(=4\\) or \\(\\quad 0.04 \\times 96=3.84 \\mathrm{~cm}^{3}\\)\n\\(V_{\\text {initial }}=96-3.84=92.16 \\mathrm{~cm}^{3}=(2)(2)(\\mathrm{L})\\)\nLength (L) \\(=23.04 \\mathrm{~cm}\\)"
},
{
"idx": 118,
"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": "(1 / 7.65 - 1 / 7.71) / (1 / 7.65) * 100% = (0.1307 cm3 - 0.1297 cm3) / 0.1307 cm3 * 100% = 0.77%"
},
{
"idx": 118,
"question": "Does the cast iron expand or contract during solidification?",
"answer": "The casting contracts."
},
{
"idx": 119,
"question": "The solubility of hydrogen in liquid aluminum at \\(715^{\\circ} \\mathrm{C}\\) is found to be \\(1 \\mathrm{~cm}^{3} / 100 \\mathrm{~g}\\) \\(\\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": "(\\quad\\left(1 \\mathrm{~cm}^{3} \\mathrm{H}_{2} / 1100 \\mathrm{~g} \\mathrm{Al}\\right)\\left(2.699 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)=0.02699 \\mathrm{~cm}^{3} \\mathrm{H}_{2} / \\mathrm{cm}^{3} \\mathrm{Al}=2.699 \\%\\)"
},
{
"idx": 120,
"question": "In the unary phase diagram for SiO2 shown in Figure 10-19, locate the triple point where solid, liquid, and vapor coexist and give the temperature and the type of solid present.",
"answer": "The solid-liquid-vapor triple point occurs at 1713 degrees Celsius; the solid phase present at this point is beta-cristobalite."
},
{
"idx": 120,
"question": "In the unary phase diagram for SiO2 shown in Figure 10-19, what do the other triple points indicate?",
"answer": "The other triple points describe the equilibrium between two solids and a vapor phase."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the Au-Ag system be expected to display unlimited solid solubility? Explain.",
"answer": "r_Au=1.442 v=+1 FCC, r_Ag=1.445 v=+1 FCC, radius difference=0.2%. The Au-Ag system has the required radius ratio, the same crystal structures, and the same valences. It might be expected to display complete solid solubility. The Au-Ag system does have isomorphous phase diagrams."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the Al-Cu system be expected to display unlimited solid solubility? Explain.",
"answer": "r_Al=1.432 v=+3 FCC, r_Cu=1.278 v=+1 FCC, radius difference=10.7%. The Al-Cu system does not meet the Hume-Rothery conditions for unlimited solid solubility."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the Al-Au system be expected to display unlimited solid solubility? Explain.",
"answer": "r_Al=1.432 v=+3 FCC, r_Au=1.442 v=+1 FCC, radius difference=0%. The Al-Au system does not meet the Hume-Rothery conditions for unlimited solid solubility due to different valences."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the U-W system be expected to display unlimited solid solubility? Explain.",
"answer": "r_U=1.38 v=+4 Ortho, r_W=1.371 v=+4 FCC, radius difference=0.7%. The U-W system does not meet the Hume-Rothery conditions for unlimited solid solubility due to different crystal structures."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the Mo-Ta system be expected to display unlimited solid solubility? Explain.",
"answer": "r_Mo=1.363 v=+4, r_Ta=1.43 v=+5, radius difference=4.7%. The Mo-Ta system has the required radius ratio, the same crystal structures, and similar valences. It might be expected to display complete solid solubility. The Mo-Ta system does have isomorphous phase diagrams."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the Nb-W system be expected to display unlimited solid solubility? Explain.",
"answer": "r_Nb=1.426 v=+4, r_W=1.371 v=+4, radius difference=3.9%. The Nb-W system does not meet the Hume-Rothery conditions for unlimited solid solubility due to different crystal structures."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the Mg-Zn system be expected to display unlimited solid solubility? Explain.",
"answer": "r_Mg=1.604 v=+2, r_Zn=1.332 v=+2, radius difference=17%. The Mg-Zn system does not meet the Hume-Rothery conditions for unlimited solid solubility due to large radius difference."
},
{
"idx": 121,
"question": "Based on Hume-Rothery's conditions, would the Mg-Cd system be expected to display unlimited solid solubility? Explain.",
"answer": "r_Mg=1.604 v=+2, r_Cd=1.490 v=+2, radius difference=7.1%. The Mg-Cd system has the required radius ratio, the same crystal structures, and the same valences. It might be expected to display complete solid solubility. The Mg-Cd alloys all solidify like isomorphous alloys, though solid-state phase transformations complicate the diagram."
},
{
"idx": 122,
"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": "The Cu-Sr alloy would be expected to be strongest (largest size difference)."
},
{
"idx": 122,
"question": "Is any of the alloying elements expected to have unlimited solid solubility in copper? (a) Au (b) Mn (c) Sr (d) Si (e) Co",
"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": 123,
"question": "Suppose 1 at% of the following elements is added to aluminum (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the least reduction in electrical conductivity? For aluminum: r=1.432 AA (FCC structure with valence of 3). Consider the following elements: (a) Li: r=1.519 AA, phi r=6.1%, BCC valence=1; (b) Ba: r=2.176 AA, phi r=-52.0%, BCC valence=2; (c) Be: r=1.143 AA, phi r=-20.2%, HCP valence=2; (d) Cd: r=1.49 AA, phi r=4.1%, HCP valence=2; (e) Ga: r=1.218 AA, phi r=14.9%, Orthorhombic valence=3",
"answer": "The cadmium would be expected to give the smallest reduction in electrical conductivity, since the Cd atoms are most similar in size to the aluminum atoms."
},
{
"idx": 123,
"question": "Is any of the alloy elements (Li, Ba, Be, Cd, Ga) expected to have unlimited solid solubility in aluminum? For aluminum: r=1.432 AA (FCC structure with valence of 3). Consider the following elements: (a) Li: r=1.519 AA, phi r=6.1%, BCC valence=1; (b) Ba: r=2.176 AA, phi r=-52.0%, BCC valence=2; (c) Be: r=1.143 AA, phi r=-20.2%, HCP valence=2; (d) Cd: r=1.49 AA, phi r=4.1%, HCP valence=2; (e) Ga: r=1.218 AA, phi r=14.9%, Orthorhombic valence=3",
"answer": "None are expected to have unlimited solid solubility, due either to difference in valence, atomic radius, or crystal structure."
},
{
"idx": 124,
"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} \\quad\\) (b) \\(\\mathrm{Cr}_{2} \\mathrm{O}_{3} \\quad\\) (c) \\(\\mathrm{Fe}_{2} \\mathrm{O}_{3}\\)",
"answer": "The ionic radius of \\(\\mathrm{Al}^{3+}=0.51 \\AA\\)\n(a) \\(r_{\\mathrm{Y} 21}=0.89 \\quad \\phi r=\\frac{0.63-0.51}{0.51} \\times 100=74.5 \\%\\)\n(b) \\(r_{\\mathrm{Cr} 21}=0.63 \\quad \\phi r=23.5 \\%\\)\n(c) \\(r_{\\mathrm{Fe} 21}=0.64 \\quad \\phi r=25.5 \\%\\)\nWe would expect \\(\\mathrm{Cr}_{2} \\mathrm{O}_{3}\\) to have a high solubility in \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\); in fact, they are completely soluble in one another."
},
{
"idx": 125,
"question": "What is the required weight percentage of nickel (Ni) in the alloy to achieve 50 wt% α phase at 1300°C, given the phase compositions L: 46 wt% Ni and α: 58 wt% Ni?",
"answer": "The required weight percentage of nickel (Ni) in the alloy is calculated using the lever rule: (x - 46)/(58 - 46) * 100 = 50% α, solving for x gives x = 52 wt% Ni."
},
{
"idx": 125,
"question": "How many grams of nickel (Ni) must be added to 500 grams of copper (Cu) to produce an alloy with 52 wt% Ni?",
"answer": "The number of grams of nickel (Ni) required is calculated by: (x)/(x + 500) * 100% = 52, solving for x gives x = 541.7 g Ni."
},
{
"idx": 126,
"question": "What is the molecular weight of MgO and NiO?",
"answer": "MW_MgO = 40.312 g/mol, MW_NiO = 74.71 g/mol"
},
{
"idx": 126,
"question": "What mole percentage of MgO is needed to achieve a solidus temperature of 2200°C in the MgO-NiO ceramic system?",
"answer": "38 mol% MgO is needed to obtain the correct solidus temperature."
},
{
"idx": 126,
"question": "How to calculate the weight percentage of MgO required in the mixture, given the molecular weights and mole percentage?",
"answer": "wt% MgO = (38 * 40.312) / (38 * 40.312 + 62 * 74.71) * 100% = 24.9%"
},
{
"idx": 126,
"question": "How many grams of MgO must be added to 1 kg of NiO to achieve 24.9 wt% MgO in the mixture?",
"answer": "x / (x + 1000) * 100% = 24.9% or x = 332 g of MgO"
},
{
"idx": 127,
"question": "We would like to produce a \\(\\mathrm{MgO}-\\mathrm{FeO}\\) ceramic that is \\(30 \\mathrm{wt} \\%\\) solid at \\(2000^{\\circ} \\mathrm{C}\\). Determine the original composition of the ceramic in wt\\%.",
"answer": "(\\quad L: 65 \\mathrm{wt} \\% \\mathrm{FeO} \\quad S: 38 \\mathrm{wt} \\% \\mathrm{FeO}\\)\n\\[\n30 \\mathrm{wt} \\%=\\frac{65-x}{65-38} \\times 100 \\% \\text { or } x=56.9 \\mathrm{wt} \\% \\mathrm{FeO}\n\\]"
},
{
"idx": 128,
"question": "A Nb-W alloy held at 2800 degrees Celsius is partly liquid and partly solid. If possible, determine the composition of each phase in the alloy.",
"answer": "L: 49 wt% W; alpha: 70 wt% W"
},
{
"idx": 128,
"question": "A Nb-W alloy held at 2800 degrees Celsius 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": 129,
"question": "A Nb-W alloy contains 55% alpha at 2600 degrees C. Determine the composition of each phase.",
"answer": "L: 22 wt% W, alpha: 42 wt% W"
},
{
"idx": 129,
"question": "A Nb-W alloy contains 55% alpha at 2600 degrees C. Determine the original composition of the alloy.",
"answer": "0.55 = (x - 22)/(42 - 22) or x = 33 wt% W"
},
{
"idx": 130,
"question": "Suppose a 1200-lb bath of a Nb-40 wt% W alloy is held at 2800°C. How many pounds of tungsten can be added to the bath before any solid forms?",
"answer": "Solid starts to form at 2800°C when 49 wt% W is in the alloy. In 1200 lb of the original Nb-40% W alloy, there are (0.4)(1200)=480 lb W and 720 lb Nb. The total amount of tungsten that must be in the final alloy is: 0.49 = x/(x+720) or x=692 lb W total or 692-480=212 additional pounds of W must be added."
},
{
"idx": 130,
"question": "Suppose a 1200-lb bath of a Nb-40 wt% W alloy is held at 2800°C. How many pounds of tungsten must be added to cause the entire bath to be solid?",
"answer": "To be completely solid at 2800°C, the alloy must contain 70 wt% W. The total amount of tungsten required in the final alloy is: 0.70 = x/(x+720) or x=1680 lb W total or 1680-480=1200 additional pounds of W must be added."
},
{
"idx": 131,
"question": "Suppose a crucible made of pure nickel is used to contain \\(500 \\mathrm{~g}\\) of liquid copper at \\(1150^{\\circ} \\mathrm{C}\\). Describe what happens to the system as it is held at this temperature for several hours. Explain.",
"answer": "Cu dissolves \\(\\mathrm{Ni}\\) until the \\(\\mathrm{Cu}\\) contains enough \\(\\mathrm{Ni}\\) that it solidifies completely. When \\(10 \\% \\mathrm{Ni}\\) is dissolved, freezing begins:\n\\[\n0.10=\\frac{x}{x+500} \\quad \\text { or } \\quad x=55.5 \\mathrm{~g} \\mathrm{Ni}\n\\]\nWhen \\(18 \\% \\mathrm{Ni}\\) dissolved, the bath is completely solid:\n\\[\n0.18=\\frac{x}{x+500} \\quad \\text { or } \\quad \\begin{aligned}\nx & =109.8 \\mathrm{~g} \\mathrm{Ni}\n\\end{aligned}\n\\]"
},
{
"idx": 132,
"question": "A NiO-60 mol% MgO ceramic is allowed to solidify under equilibrium conditions. Determine the composition of the first solid to form.",
"answer": "1st α: 80% MgO"
},
{
"idx": 132,
"question": "A NiO-60 mol% MgO ceramic is allowed to solidify under equilibrium conditions. Determine the composition of the last liquid to solidify.",
"answer": "Last L: 35% MgO"
},
{
"idx": 133,
"question": "A Nb-35% W alloy is allowed to solidify under equilibrium conditions. Determine the composition of the first solid to form.",
"answer": "1st alpha: 55% W"
},
{
"idx": 133,
"question": "A Nb-35% W alloy is allowed to solidify under equilibrium conditions. Determine the composition of the last liquid to solidify.",
"answer": "Last L: 18% W"
},
{
"idx": 134,
"question": "An intermetallic compound is found for \\(10 \\mathrm{wt} \\% \\mathrm{Si}\\) in the Cu-Si phase diagram. Determine the formula for the compound.",
"answer": "at\\% \\(\\mathrm{Si}=\\frac{10 \\mathrm{~g} / 28.08 \\mathrm{~g} / \\mathrm{mol}}{10 / 28.08+90 / 63.54}=0.20 \\quad\\) or \\(\\quad \\mathrm{SiCu}_{4}\\)"
},
{
"idx": 135,
"question": "Consider a Pb-15% Sn alloy. During solidification, determine the composition of the first solid to form.",
"answer": "8% Sn"
},
{
"idx": 135,
"question": "Consider a Pb-15% Sn alloy. Determine the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy.",
"answer": "liquidus = 290 degrees C, solidus = 240 degrees C, solvus = 170 degrees C, freezing range = 50 degrees C"
},
{
"idx": 135,
"question": "Consider a Pb-15% Sn alloy. Determine the amounts and compositions of each phase at 260 degrees C.",
"answer": "L: 30% Sn; alpha: 12% Sn; % L = ((15 - 12)/(30 - 12)) x 100% = 17%; % alpha = 83%"
},
{
"idx": 135,
"question": "Consider a Pb-15% Sn alloy. Determine the amounts and compositions of each phase at 183 degrees C.",
"answer": "alpha: 15% Sn; 100% alpha"
},
{
"idx": 135,
"question": "Consider a Pb-15% Sn alloy. Determine the amounts and compositions of each phase at 25 degrees C.",
"answer": "alpha: 2% Pb; beta: 100% Sn; % alpha = ((100 - 15)/(100 - 2)) x 100 = 87%; % beta = 13%"
},
{
"idx": 136,
"question": "Consider a Pb-35% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.",
"answer": "hypoeutectic"
},
{
"idx": 136,
"question": "Consider a Pb-35% Sn alloy. Determine the composition of the first solid to form during solidification.",
"answer": "14% Sn"
},
{
"idx": 136,
"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 = (61.9 - 35)/(61.9 - 19) × 100% = 63%, % L = 37%"
},
{
"idx": 136,
"question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 182°C.",
"answer": "α: 19% Sn, β: 97.5% Sn; % α = (97.5 - 35)/(97.5 - 19) × 100% = 80%, % β = 20%"
},
{
"idx": 136,
"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": 136,
"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 = (100 - 35)/(100 - 2) × 100% = 66%, % β = 34%"
},
{
"idx": 137,
"question": "Consider a Pb-70% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.",
"answer": "hypereutectic"
},
{
"idx": 137,
"question": "Consider a Pb-70% Sn alloy. Determine the composition of the first solid to form during solidification.",
"answer": "98% Sn"
},
{
"idx": 137,
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 184°C.",
"answer": "β: 97.5% Sn L: 61.9% Sn % β=(70-61.9)/(97.5-61.9)×100%=22.8% % L=77.2%"
},
{
"idx": 137,
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 182°C.",
"answer": "α: 19% Sn β: 97.5% Sn % α=(97.5-70)/(97.5-19)×100%=35% % β=65%"
},
{
"idx": 137,
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each microconstituent at 182°C.",
"answer": "primary β: 97.5% Sn % primary β=22.8% eutectic: 61.9% Sn % eutectic=77.2%"
},
{
"idx": 137,
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 25°C.",
"answer": "α: 2% Sn β: 100% Sn % α=(100-70)/(100-2)×100%=30% % β=70%"
},
{
"idx": 138,
"question": "Consider an Al-4% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.",
"answer": "hypo eutectic"
},
{
"idx": 138,
"question": "Consider an Al-4% Si alloy. Determine the composition of the first solid to form during solidification.",
"answer": "1% Si"
},
{
"idx": 138,
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 578 degrees Celsius.",
"answer": "α: 1.65% Si, L: 12.6% Si. % α = (12.6 - 4)/(12.6 - 1.65) = 78.5%, % L = 21.5%"
},
{
"idx": 138,
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 576 degrees Celsius.",
"answer": "α: 1.65% Si, β: 99.83% Si. % α = (99.83 - 4)/(99.83 - 1.65) = 97.6%, % β = 2.4%"
},
{
"idx": 138,
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each microconstituent at 576 degrees Celsius.",
"answer": "primary α: 1.65% Si, % primary α = 78.5%. eutectic: 12.6% Si, % eutectic = 21.5%"
},
{
"idx": 138,
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 25 degrees Celsius.",
"answer": "α: 0% Si, β: 100% Si. % α = 100 - 4 = 96%, % β = 4%"
},
{
"idx": 139,
"question": "Consider a Al-25% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.",
"answer": "hyper eutectic"
},
{
"idx": 139,
"question": "Consider a Al-25% Si alloy. Determine the composition of the first solid to form during solidification.",
"answer": "100% Si"
},
{
"idx": 139,
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 578°C.",
"answer": "β: 99.83% Si, L: 12.6% Si\n% L = (99.83 - 25)/(99.83 - 12.6) = 85.8%, % β = 14.2%"
},
{
"idx": 139,
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 576°C.",
"answer": "α: 1.65% Si, β: 99.83% Si\n% α = (99.83 - 25)/(99.83 - 1.65) = 76.2%, % β = 23.8%"
},
{
"idx": 139,
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each microconstituent at 576°C.",
"answer": "primary β: 99.83% Si, % primary β = 14.2%\neutectic: 12.6% Si, % eutectic = 85.8%"
},
{
"idx": 139,
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 25°C.",
"answer": "α: 0% Si, β: 100% Si\n% α = (100 - 25)/(100 - 0) = 75%, % β = 25%"
},
{
"idx": 140,
"question": "A Pb-Sn alloy contains 45% α and 55% β at 100°C. Determine the composition of the alloy.",
"answer": "45% α = (98.0 - x)/(98.0 - 5) × 100 or x = 56.15% Sn"
},
{
"idx": 140,
"question": "Is the Pb-Sn alloy with 56.15% Sn hypoeutectic or hypereutectic?",
"answer": "Hypoeutectic"
},
{
"idx": 141,
"question": "An Al-Si alloy contains 85% α and 15% β at 500°C. Determine the composition of the alloy.",
"answer": "85 = (100 - x) / (100 - 1) × 100 or x = 15.85% Si"
},
{
"idx": 141,
"question": "Is the alloy hypoeutectic or hypereutectic?",
"answer": "Hypereutectic"
},
{
"idx": 142,
"question": "A Pb-Sn alloy contains \\(23 \\%\\) primary \\(\\alpha\\) and \\(77 \\%\\) eutectic microconstituent. Determine the composition of the alloy.",
"answer": "(\\quad \\%\\) primary \\(\\alpha=23=\\frac{61.9-x}{61.9-19} \\times 100 \\quad\\) or \\(\\quad x=52 \\% \\mathrm{Sn}\\)"
},
{
"idx": 143,
"question": "An Al-Si alloy contains \\(15 \\%\\) primary \\(\\beta\\) and \\(85 \\%\\) eutectic microconstituent. Determine the composition of the alloy.",
"answer": "[\n\\% \\text { eutectic }=85=\\frac{100-x}{100-12.6} \\times 100 \\quad \\text { or } \\quad x=25.71 \\% \\mathrm{Si}\n\\]"
},
{
"idx": 144,
"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": "28=(20.4-x)/(20.4-9.9)×100 or x=17.46% Li Hypereutectic"
},
{
"idx": 144,
"question": "How much α and β are in the eutectic microconstituent?",
"answer": "% αEst=(20.4-9.9)/(20.4-4)×100%=64% and % βEst=36%"
},
{
"idx": 145,
"question": "Calculate the total amount of alpha and beta in a Pb-50% Sn alloy at 182°C.",
"answer": "alpha_total = (97.5 - 50)/(97.5 - 19) × 100% = 60.5%, beta_Total = 39.5%"
},
{
"idx": 145,
"question": "Calculate the amount of each microconstituent (primary alpha and eutectic) in a Pb-50% Sn alloy at 182°C.",
"answer": "alpha_Primary = (61.9 - 50)/(61.9 - 19) × 100% = 27.7%, Eutectic = 72.3%"
},
{
"idx": 145,
"question": "What fraction of the total alpha in the alloy is contained in the eutectic microconstituent for a Pb-50% Sn alloy at 182°C?",
"answer": "alpha_in eutectic = 60.5 - 27.7 = 32.8%, f = 32.8/60.5 = 0.54"
},
{
"idx": 146,
"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, the peritectic temperature is 870 degrees C; above this temperature, liquid may form. The solvus temperature is about 530 degrees C. Therefore: 1) Solution treat between 530 degrees C and 870 degrees C (780 degrees C is typical for beryllium copper alloys) 2) Quench 3) Age below 530 degrees C (330 degrees C is typical for these alloys)"
},
{
"idx": 146,
"question": "Compare the amount of the gamma2 precipitate that forms by artificial aging at 400 degrees C with the amount of the precipitate that forms by natural aging for a Cu-1.2% Be alloy.",
"answer": "We can perform lever law calculations at 400 degrees C and at room temperature. The solubility of Be in Cu at 400 degrees C is about 0.6% Be and that at room temperature is about 0.2% Be: gamma2 at 400 degrees C = (1.2-0.6)/(11.7-0.6)*100 = 5.4% gamma2 at room T = (1.2-0.2)/(12-0.2)*100 = 8.5%"
},
{
"idx": 147,
"question": "For an Fe-0.35% C alloy, determine the temperature at which austenite first begins to transform on cooling.",
"answer": "795°C"
},
{
"idx": 147,
"question": "For an Fe-0.35% C alloy, determine the primary microconstituent that forms.",
"answer": "Primary α-ferrite"
},
{
"idx": 147,
"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, %α = (0.77 - 0.35)/(0.77 - 0.0218) × 100 = 56.1%; γ: 0.77% C, %γ = 43.9%"
},
{
"idx": 147,
"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, %α = (6.67 - 0.35)/(6.67 - 0.0218) × 100 = 95.1%; Fe3C: 6.67% C, %Fe3C = 4.9%"
},
{
"idx": 147,
"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, % primary α = 56.1%; Pearlite: 0.77% C, % Pearlite = 43.9%"
},
{
"idx": 148,
"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": 148,
"question": "For an Fe-1.15% C alloy, determine the primary microconstituent that forms.",
"answer": "primary Fe3C"
},
{
"idx": 148,
"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, % Fe3C = (1.15-0.77)/(6.67-0.77) x 100 = 6.4%; gamma: 0.77% C, % gamma = 93.6%"
},
{
"idx": 148,
"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, % alpha = (6.67-1.15)/(6.67-0.0218) x 100 = 83%; Fe3C: 6.67% C, % Fe3C = 17%"
},
{
"idx": 148,
"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, % primary Fe3C = 6.4%; pearlite: 0.77% C, % Pearlite = 93.6%"
},
{
"idx": 149,
"question": "A steel contains 8% cementite and 92% ferrite at room temperature. Estimate the carbon content of the steel.",
"answer": "alpha=0.92=(6.67-x)/(6.67-0); x=0.53% C"
},
{
"idx": 149,
"question": "Is the steel hypoeutectoid or hypereutectoid?",
"answer": "Hypoeutectoid"
},
{
"idx": 150,
"question": "A steel contains 18% cementite and 82% ferrite at room temperature. Estimate the carbon content of the steel.",
"answer": "alpha=0.82=(6.67-x)/(6.67-0), x=1.20%C"
},
{
"idx": 150,
"question": "Is the steel hypoeutectoid or hypereutectoid?",
"answer": "Hypereutectoid"
},
{
"idx": 151,
"question": "A steel contains 18% pearlite and 82% primary ferrite at room temperature. Estimate the carbon content of the steel.",
"answer": "primary α=0.82=(0.77-x)/(0.77-0.0218)\nx=0.156% C"
},
{
"idx": 151,
"question": "Is the steel hypoeutectoid or hypereutectoid?",
"answer": "Hypoeutectoid"
},
{
"idx": 152,
"question": "A steel contains 94% pearlite and 6% primary cementite at room temperature. Estimate the carbon content of the steel.",
"answer": "Pearlite = 0.94 = (6.67 - x) / (6.67 - 0.77). Solving for x gives x = 1.124% C."
},
{
"idx": 152,
"question": "Is the steel hypoeutectoid or hypereutectoid?",
"answer": "Hypereutectoid"
},
{
"idx": 153,
"question": "A steel contains \\(55 \\% \\alpha\\) and \\(45 \\% \\gamma\\) at \\(750^{\\circ} \\mathrm{C}\\). Estimate the carbon content of the steel.",
"answer": "(\\quad \\alpha=0.02 \\% \\mathrm{C} \\quad\\) and \\(\\quad \\gamma=0.6 \\% \\mathrm{C} \\quad\\) (from the tie line at \\(750^{\\circ} \\mathrm{C}\\) )\n\\[\n\\% \\alpha=55=\\frac{0.6-x}{0.6-0.02} \\times 100 \\quad x=0.281 \\% \\mathrm{C}\n\\]"
},
{
"idx": 154,
"question": "A steel contains \\(96 \\% \\gamma\\) and \\(4 \\% \\mathrm{Fe}_{3} \\mathrm{C}\\) at \\(800^{\\circ} \\mathrm{C}\\). Estimate the carbon content of the steel.\n\\[\n\\begin{array}{l}\n\\text {",
"answer": "\\quad \\gamma=0.92 \\% \\mathrm{C} \\text { and } \\quad \\mathrm{Fe}_{3} \\mathrm{C}=6.67 \\% \\mathrm{C} \\text { (from the tie line at } 800^{\\circ} \\mathrm{C}) \\\\\n\\gamma=0.96=\\frac{6.67-x}{6.67-0.92} \\quad x=1.15 \\% \\mathrm{C}\n\\end{array}\n\\]"
},
{
"idx": 155,
"question": "A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the temperature.",
"answer": "In order for γ to contain 0.5% C, the austenitizing temperature must be about 760°C (from the tie line)."
},
{
"idx": 155,
"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": "At 760°C: 0.4 = (x - 0.02) / (0.5 - 0.02) x = 0.212% C"
},
{
"idx": 156,
"question": "An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate the transformation temperature.",
"answer": "transformation temperature = 615 degrees C"
},
{
"idx": 156,
"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 x 10^-5 cm"
},
{
"idx": 157,
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 38",
"answer": "600 degrees Celsius, pearlite"
},
{
"idx": 157,
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 42",
"answer": "400 degrees Celsius, bainite"
},
{
"idx": 157,
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 48",
"answer": "340 degrees Celsius, bainite"
},
{
"idx": 157,
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 52",
"answer": "300 degrees Celsius, bainite"
},
{
"idx": 158,
"question": "What is the hardness in Rockwell C scale (HRC) of an eutectoid steel that has been heated to 800 degrees Celsius for 1 hour, quenched to 350 degrees Celsius and held for 750 seconds, and finally quenched to room temperature?",
"answer": "HRC = 47"
},
{
"idx": 158,
"question": "What is the microstructure of an eutectoid steel that has been heated to 800 degrees Celsius for 1 hour, quenched to 350 degrees Celsius and held for 750 seconds, and finally quenched to room temperature?",
"answer": "The microstructure is all bainite."
},
{
"idx": 159,
"question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to \\(700^{\\circ} \\mathrm{C}\\), quenched to \\(650^{\\circ} \\mathrm{C}\\) and held for \\(500 \\mathrm{~s}\\), and finally quenched to room temperature.",
"answer": "HRC \\(=25\\) and the microstructure is all pearlite."
},
{
"idx": 160,
"question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to \\(300^{\\circ} \\mathrm{C}\\), quenched to \\(300^{\\circ} \\mathrm{C}\\) and held for \\(10 \\mathrm{~s}\\), and finally quenched to room temperature.",
"answer": "(H R C=66\\) and the microstructure is all martensite."
},
{
"idx": 161,
"question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to \\(1000^{\\circ} \\mathrm{C}\\), quenched to \\(3010^{\\circ} \\mathrm{C}\\) and held for \\(10 \\mathrm{s}\\), quenched to room temperature, and then reheated to \\(400^{\\circ} \\mathrm{C}\\) before finally cooling to room temperature again.",
"answer": "(\\quad \\mathrm{HRC}=42\\) and the microstructure is all tempered martensite."
},
{
"idx": 162,
"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} \\mathrm{C}\\),\nQuench to below \\(130^{\\circ} \\mathrm{C}\\) (the \\(M_{\\gamma}\\) temperature)\nTemper at \\(620^{\\circ} \\mathrm{C}\\) or less."
},
{
"idx": 163,
"question": "Describe the complete heat treatment required to produce a quenched and tempered eutectoid steel haying a HRC hardness of less than 50. Include appropriate temperatures.",
"answer": "Austenitize at approximately \\(75^{\\circ} \\mathrm{C}\\),\nQuench to below the \\(M_{\\gamma}\\) (less than \\(130^{\\circ} \\mathrm{C}\\) )\nTemper at a temperature higher than \\(330^{\\circ} \\mathrm{C}\\), but less than \\(727^{\\circ} \\mathrm{C}\\)."
},
{
"idx": 164,
"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?\n(a) Density\n(b) Annealing\n(c) Forging\n(d) Single-crystal\n(e) Crystalline",
"answer": "Material structure refers to the arrangement of a material's constituent atoms, ions, molecules, etc. Material properties are typically either conceptual traits or specific numeric values that describe a material's response to applied stimuli. A few examples of numeric property examples are density, elastic modulus, hardness, tensile strength, and electrical conductivity. A few conceptual properties include brittle, ductile, and toughness.\nMaterial processing refers to conditions, stimuli, and situations that a material is exposed to, such as manufacturing steps such as melting \\& casting, annealing, forging, rolling, cutting, extrusion. It is often convenient to consider service conditions as processing parameters as well, since factors like operating temperature, ambient atmosphere, humidity, etc. may further influence a material's structure, properties, and performance.\nMaterial performance \\& behavior typically pertains to how a material responds while in service.\nExample: A polycrystalline (structure) steel specimen that has been carburized (processing) should exhibit a higher surface hardness (property) compared to un-carburized steel. The surface of the carburized steel specimen will therefore be more difficult to cut or abrade (performance / behavior).A thin steel specimen that has been quenched from a glowing red-hot temperature (processing) is expected to form martensite (structure), the hardest and most brittle form of steel (property). If the quenched specimen is subsequently loaded, it will not exhibit much strain prior to failure (performance / behavior).\nIn this case (b) and (c) are process options, (d) is a conceptual property, and (e) is a structural description."
},
{
"idx": 165,
"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?\n(a) Extrusion\n(b) Crystalline\n(c) Amorphous\n(d) Glassy\n(e) Elastic Modulus",
"answer": "Material structure refers to the arrangement of a material's constituent atoms, ions, molecules, etc. Material properties are typically either conceptual traits or specific numeric values that describe a material's response to applied stimuli. A few examples of numeric property examples are density, elastic modulus, hardness, tensile strength, and electrical conductivity. A few conceptual properties include brittle, ductile, and toughness.\nMaterial processing refers to conditions, stimuli, and situations that a material is exposed to, such as manufacturing steps such as melting \\& casting, annealing, forging, rolling, cutting, extrusion. It is often convenient to consider service conditions as processing parameters as well, since factors like operating temperature, ambient atmosphere, humidity, etc. may further influence a material's structure, properties, and performance.\nMaterial performance \\& behavior typically pertains to how a material responds while in service.Example:\nA polycrystalline (structure) steel specimen that has been carburized (processing) should exhibit a higher surface hardness (property) compared to un-carburized steel. The surface of the carburized steel specimen will therefore be more difficult to cut or abrade (performance / behavior).\nA thin steel specimen that has been quenched from a glowing red-hot temperature (processing) is expected to form martensite (structure), the hardest and most brittle form of steel (property). If the quenched specimen is subsequently loaded, it will not exhibit much strain prior to failure (performance / behavior).\nIn this case (a) is a process, (b), (c), and (d) are structural descriptions and (e) is a property."
},
{
"idx": 166,
"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?\n(a) Single-phase\n(b) Elastic Modulus\n(c) Sintering\n(d) Magnetic Permeability\n(e) Brittle",
"answer": "Material structure refers to the arrangement of a material's constituent atoms, ions, molecules, etc.\nMaterial properties are typically either conceptual traits or specific numeric values that describe a material's response to applied stimuli. A few examples of numeric property examples are density, elastic modulus, hardness, tensile strength, and electrical conductivity. A few conceptual properties include brittle, ductile, and toughness.\nMaterial processing refers to conditions, stimuli, and situations that a material is exposed to, such as manufacturing steps such as melting \\& casting, annealing, forging, rolling, cutting, extrusion. It is often convenient to consider service conditions as processing parameters as well, since factors like operating temperature, ambient atmosphere, humidity, etc. may further influence a material's structure, properties, and performance.\nMaterial performance \\& behavior typically pertains to how a material responds while in service.Example:\nA polycrystalline (structure) steel specimen that has been carburized (processing) should exhibit a higher surface hardness (property) compared to un-carburized steel. The surface of the carburized steel specimen will therefore be more difficult to cut or abrade (performance / behavior).\nA thin steel specimen that has been quenched from a glowing red-hot temperature (processing) is expected to form martensite (structure), the hardest and most brittle form of steel (property). If the quenched specimen is subsequently loaded, it will not exhibit much strain prior to failure (performance / behavior).\nIn this case (a) is a structural description, (b), (d), and (e) are properties, and (c) is a process."
},
{
"idx": 167,
"question": "Which class of material is generally associated with the highest density values at room temperature?\n(a) Composites\n(b) Ceramics\n(c) Metals\n(d) Polymers",
"answer": "Metals tend to be the most dense material class because (i) the atoms are generally heavy and (ii) the heavy atoms tend to pack fairly efficiently"
},
{
"idx": 168,
"question": "By how many orders of magnitude (powers of ten, approximately) does density vary for metals?\n(a) 0.13\n(b) 1.3\n(c) 13\n(d) 130",
"answer": "The density of metals ranges from about \\(1 \\mathrm{~g} / \\mathrm{cm}^{3}\\) to about \\(30 \\mathrm{~g} / \\mathrm{cm}^{3}\\).\nThe change from 1 to \\(10 \\mathrm{~g} / \\mathrm{cm}^{3}\\) is 1 order of magnitude of variation.\nThe change from 10 to \\(30 \\mathrm{~g} / \\mathrm{cm}^{3}\\) about 0.3 order of magnitude.\nThe variation in density for metals is therefore about 1.3 orders of magnitude."
},
{
"idx": 169,
"question": "The atomic mass of an atom may be expressed as the sum of the masses of\n- Electrons\n- Neutrons\n- Protons\nChoose all that apply.",
"answer": "The atomic mass of an atom may be expressed as the sum of the masses of protons and neutrons within the nucleus."
},
{
"idx": 170,
"question": "The nucleus of an atom contains\n- Electrons\n- Neutrons\n- Protons\nChoose all that apply.",
"answer": "An atom consists of a very small nucleus that contains protons and neutrons (surrounded by moving electrons)."
},
{
"idx": 171,
"question": "The atomic number of an electrically neutral atom is equal to the number of:\n- protons\n- electrons\n- neutrons\nChoose all that apply.",
"answer": "The atomic number of an electrically neutral atom is the number of protons, which is equal to the number of electrons."
},
{
"idx": 172,
"question": "Hafnium has six naturally occurring isotopes: \\(0.16 \\%\\) of \\({ }^{174} \\mathrm{Hf}\\), with an atomic weight of \\(173.940 \\mathrm{amu}\\); \\(5.26 \\%\\) of \\({ }^{176} \\mathrm{Hf}\\), with an atomic weight of \\(175.941 \\mathrm{amu} ; 18.60 \\%\\) of \\({ }^{177} \\mathrm{Hf}\\), with an atomic weight of \\(176.943 \\mathrm{amu} ; 27.28 \\%\\) of \\({ }^{178} \\mathrm{Hf}\\), with an atomic weight of \\(177.944 \\mathrm{amu} ; 13.62 \\%\\) of \\({ }^{179} \\mathrm{Hf}\\), with an atomic weight of 178.946 amu;. and \\(35.08 \\%\\) of \\({ }^{180} \\mathrm{Hf}\\), with an atomic weight of 179.947 amu. Calculate the average atomic weight of \\(\\mathrm{Hf}\\). Give your answer to three decimal places.",
"answer": "The average atomic weight of halfnium \\(\\bar{A}_{H f}\\) is computed by adding fraction-of-occurrence-atomic weight products for the six isotopes-i.e., using Equation. (Remember: fraction of occurrence is equal to the percent of occurrence divided by 100 .) Thus\n\\[\n\\bar{A}_{\\mathrm{Hf}}=f_{174 \\mathrm{H}} A_{174 \\mathrm{H}}+f_{176 \\mathrm{H}} A_{176 \\mathrm{H}}+f_{177 \\mathrm{H}} A_{177 \\mathrm{H}}+f_{178 \\mathrm{H}} A_{178 \\mathrm{H}}+f_{179 \\mathrm{H}} A_{179 \\mathrm{H}}+f_{180 \\mathrm{H}} A_{180 \\mathrm{H}}\n\\]\nIncluding data provided in the Chapter 02, Problem statement we solve for \\(\\bar{A}_{H f}\\) as\n\\[\n\\begin{aligned}\n\\bar{A}_{\\mathrm{Hf}}= & (0.0016)(173.940 \\mathrm{amu})+(0.0526)(175.941 \\mathrm{amu})+(0.1860)(176.943 \\mathrm{amu}) \\\\\n& +(0.2728)(177.944 \\mathrm{amu})+(0.1362)(178.946 \\mathrm{amu})+(0.3508)(179.947) \\\\\n& =178.485 \\mathrm{amu}\n\\end{aligned}\n\\]"
},
{
"idx": 173,
"question": "Bromium has two naturally occurring isotopes: \\({ }^{79} \\mathrm{Br}\\), with an atomic weight of \\(78.918 \\mathrm{amu}\\), and \\({ }^{81} \\mathrm{Br}\\), with an atomic weight of 80.916 amu. If the average atomic weight for \\(\\mathrm{Br}\\) is 79.903 amu, calculate the fraction-of-occurrences of these two isotopes.Give your answer to three decimal places.",
"answer": "The average atomic weight of indium \\(\\left(\\bar{A}_{\\mathrm{Br}}\\right)\\) is computed by adding fraction-of-occurrence-atomic weight products for the two isotopes-i.e., using Equation, or\n\\[\n\\bar{A}_{\\mathrm{Br}}=f_{79_{\\mathrm{Br}}} A_{79_{\\mathrm{B}}}+f_{81_{\\mathrm{m}}} A_{81_{\\mathrm{m}}}\n\\]Because there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000 ; or\n\\[\nf_{79_{\\mathrm{br}}}+f_{81_{\\mathrm{br}}}=1.000\n\\]\nwhich means that\n\\[\nf_{79_{\\mathrm{br}}}^{\\mathrm{r}}=1.000-f_{81_{\\mathrm{br}}}\n\\]\nSubstituting into this expression the one noted above for \\(f_{79_{\\mathrm{br}}}\\), and incorporating the atomic weight values provided in the problem statement yields\n\\[\n\\begin{aligned}\n79.903 \\mathrm{amu}= & f_{79_{\\mathrm{br}}} A_{79_{\\mathrm{br}}}+f_{81_{\\mathrm{br}}} A_{81_{\\mathrm{br}}} \\\\\n79.903 \\mathrm{amu} & =\\left(1.000-f_{81_{\\mathrm{br}}}\\right) A_{81_{\\mathrm{br}}} A_{79_{\\mathrm{br}}}+f_{51_{\\mathrm{br}}} A_{81_{\\mathrm{br}}} \\\\\n7.903 \\mathrm{amu} & =\\left(1.000-\\left.f_{81_{\\mathrm{br}}}\\right)\\left(78.918 \\mathrm{amu}\\right)+f_{51_{\\mathrm{br}}}\\left(80.916 \\mathrm{amu}\\right)\\right. \\\\\n80.916 \\mathrm{amu} & =78.918 \\mathrm{amu}-\\left(78.918 \\mathrm{amu}\\right)\\left(f_{81_{\\mathrm{br}}}\\right)+\\left(80.916 \\mathrm{amu}\\right)\\left(f_{81_{\\mathrm{br}}}\\end{aligned}\n\\]\nSolving this expression for \\(f_{81_{\\mathrm{br}}}\\) yields \\(f_{81_{\\mathrm{br}}}^{\\mathrm{f}}=0.493\\). Furthermore, because\n\\[\nf_{79_{\\mathrm{br}}}^{\\mathrm{f}}=1.000-f_{81_{\\mathrm{br}}}\n\\]\n\nthen\n\n\\[\nf_{79_{\\mathrm{br}}}=1.000-0.493=0.507\n\\]"
},
{
"idx": 174,
"question": "An element that has the electron configuration \\(1 s^{2} 2 s^{2} 2 p^{6}\\) has how many electrons?\n\nEnter numeric values only.",
"answer": "Both the 1s and 2s shells are filled; therefore, they contain a total of 4 electrons (2 electrons for each \\(s\\) orbital). Thus the total number of electrons in the atom is 4 plus the number of electrons in the \\(2 p\\) shell-in this case 5 -for a total of 9 ."
},
{
"idx": 175,
"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": 176,
"question": "When all the electrons in an atom occupy the lowest possible energy states, the atom is said to be in its:\n-ground state\n-ionized state\n-cold state\n-regular state",
"answer": "When all the electrons in an atom occupy the lowest possible energy states, the atom is said to be in its ground state."
},
{
"idx": 177,
"question": "How many \\(p\\) electrons at the outermost orbital do the Group VIIA elements have?",
"answer": "Each of the elements in Group VIIA has five \\(p\\) electrons."
},
{
"idx": 178,
"question": "To what group in the periodic table would an element with atomic number 119 belong?\n- Group 0 (or 18)\n- Group IA (or 1)\n- Group IIA (or 2)\n- Group VIIA (or 17)",
"answer": "From the periodic table the element having atomic number 119 would belong to group IA."
},
{
"idx": 179,
"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 \\(3.7 \\mathrm{~nm}\\).",
"answer": "The attractive energy between positive and negative ions may be calculated using Equation, that is\n\\[\nE_{A}=-\\frac{A}{r}\n\\]\nFurthermore, the value of \\(A\\) is given in Equation as\n\\[\nA=\\frac{1}{4 \\pi \\varepsilon_{0}}\\left(\\left|Z_{1} e\\right|\\right)\\left(\\left|Z_{2} e\\right|\\right)\n\\]\nFor this problem we will assume that \\(Z_{1}=+2\\) and \\(Z_{2}=-2\\); also form the problem statement that \\(r=\\) \\(3.7 \\mathrm{~nm}=3.7 \\times 10^{-9} \\mathrm{~m}\\). Finally, the value of the attractive energy is computed as follows:\n\\[\n\\begin{array}{l}\nE_{A}=-\\frac{1}{4 \\pi \\varepsilon_{0}}\\left(\\left|z_{1}\\right| e\\right)\\left(\\left|Z_{2}\\right| e\\right) \\\\\n=-\\frac{1}{(4)(\\pi)\\left(8.85 \\times 10^{-12} \\mathrm{~F} / \\mathrm{m}\\right)\\left(3.7 \\times 10^{-9} \\mathrm{~m}\\right)}\\left(\\left|+2\\right|\\left[1.602 \\times 10^{-19} \\mathrm{C}\\right]\\right)\\left(\\left|-2\\right|\\left[1.602 \\times 10^{-19}\\mathrm{C}\\right]\\right) \\\\\n=-2.5 \\times 10^{-19} \\mathrm{~J}\n\\end{array}\n\\]"
},
{
"idx": 180,
"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": "Van der Waals interactions are generally the weakest bond type. Therefore, materials whose particles are bound together by this bonding will tend to exhibit inferior mechanical and thermal properties (e.g. lower elastic modulus, lower melting temperature)."
},
{
"idx": 181,
"question": "Calculate % IC of the interatomic bonds for the intermetallic compound TiAl3.",
"answer": "The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation. The electronegativities for Al and Ti are both 1.5. Therefore, the percent ionic character is determined using Equation as follows: % IC = [1 - exp (-0.25)(1.5 - 1.5)^2] x 100 = 0%."
},
{
"idx": 181,
"question": "On the basis of this result what type of interatomic bonding would you expect to be found in TiAl3? (Options: van der Waals, ionic, metallic, covalent)",
"answer": "Because the percent ionic character is zero and this intermetallic compound is composed of two metals, the bonding is completely metallic."
},
{
"idx": 182,
"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.\n(a) Textured polycrystal with about 10,000 grains\n(b) Random polycrystal with about 1,000,000 grains\n(c) Random polycrystal with about 1,000,000,000 grains\n(d) Amorphous",
"answer": "A single crystal can be described as a single grain with one crystallographic orientation in space.\nTherefore, the option that has the fewest and most oriented grains will be most similar to the single crystal. In this case the textured polycrystal also features the lowest grain count so both characteristics are a better match compared to the random polycrystal options, which both feature larger grain counts. Amorphous configurations are, structurally, the least similar to a single crystal."
},
{
"idx": 183,
"question": "For a metal that has the simple cubic crystal structure, calculate the atomic radius if the metal has a density of \\(2.05 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and an atomic weight of \\(77.84 \\mathrm{~g} / \\mathrm{mol}\\).",
"answer": "The density of a metal may be calculated using Equation-that is\n\n\\[\n\\rho=\\frac{n A}{V_{C} N_{\\mathrm{A}}}\n\\]\n\nFor the simple cubic crystal structure, one atom is associated with each unit cell\n\n(i.e., \\(n=1\\) ). Furthermore, the atomic radius and unit cell length are related as \\(a=2 R\\); and since the unit cell has cubic symmetry\n\n\\[\nV_{C}=a^{3}=(2 R)^{3}=8 R^{3}\n\\]\n\nWhich means that, for this problem, Equation takes the following form:\n\n\\[\n\\rho=\\frac{n A}{8 R^{3} N_{\\mathrm{A}}}\n\\]And solving for \\(R\\) in this expression yields\n\\[\nR=\\left(\\frac{n A}{8 \\rho N_{\\mathrm{A}}}\\right)^{1 / 3}\n\\]\nSubstitution of values for parameters given in the problem statement leads to the following value of \\(R\\) :\n\\[\nR=\\left[\\frac{(\\text { latom/unit cell })(61.12 \\mathrm{~g} / \\mathrm{mol})}{(8)\\left(7.02 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol}\\right)}\\right]^{1 / 3}\n\\]\n\\[\n1.22 \\times 10^{-8} \\mathrm{~cm}=0.122 \\mathrm{~nm}\n\\]"
},
{
"idx": 184,
"question": "Some metal is known to have a cubic unit cell with an edge length of \\(0.437 \\mathrm{~nm}\\). In addition, it has a density of \\(4.37 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and an atomic weight of \\(54.85 \\mathrm{~g} / \\mathrm{mol}\\). Indicate the letter of the metal listed in the following table that has these characteristics.\n\\begin{tabular}{ccc} \nMetal & Crystal Structure & Atomic Radius (nm) \\\\\nA & BCC & 0.219 \\\\\nB & FCC & 0.309 \\\\\nC & FCC & 0.155 \\\\\nD & HCP & 0.125\n\\end{tabular}",
"answer": "First of all, since the unit cell is cubic, the HCP crystal structure is eliminated because its unit cell has hexagonal symmetry. Now all we need do is compute the atomic radius for both FCC and BCC crystal structures from the unit cell edge length provided in the problem statement, and compare these values with \\(R\\) values provided. For FCC this relationship is according to a rearranged form of Equation-that is\n\\[\nR=\\frac{a}{2 \\sqrt{2}}\n\\]\nand for BCC a rearranged form of Equationviz.\n\\[\nR=\\frac{a \\sqrt{3}}{4}\n\\]Let's begin by computing \\(R\\) for the BCC crystal structure, as follows:\n\\[\nR=\\frac{(0.428 \\mathrm{~nm}) \\sqrt{3}}{4}=0.185 \\mathrm{~nm}\n\\]\nSince this value does not correspond to any of the values provided in the table, the crystal structure is not BCC. We will now carry out the computation for FCC using the appropriate equation cited above. Hence\n\\[\nR=\\frac{0.428 \\mathrm{~nm}}{2 \\sqrt{2}}=0.151 \\mathrm{~nm}\n\\]\nThis is the value of \\(R\\) for Metal C, which has the FCC crystal structure."
},
{
"idx": 185,
"question": "If the atomic radius of a metal that has the body-centered cubic crystal structure is \\(0.181 \\mathrm{~nm}\\), calculate the volume of its unit cell.",
"answer": "For the BCC crystal structure, the atomic radius and unit cell edge length are related according to Equation-that is\n\n\\[\na=\\frac{4 R}{\\sqrt{3}}\n\\]\n\nBecause the unit cell has cubic symmetry, its volume \\(V C\\) is equal to \\(a^{3}\\), or\n\n\\[\nV_{C}=a^{3}=\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}=\\frac{64 R^{3}}{3 \\sqrt{3}}\n\\]\n\nAnd, for this problem, \\(R=0.130 \\mathrm{~nm}\\), then\n\n\\[\nV_{C}=\\frac{(64)(0.130 \\mathrm{~nm})^{3}}{3 \\sqrt{3}}=0.0271 \\mathrm{~nm}^{3}\n\\]"
},
{
"idx": 186,
"question": "If the atomic radius of a metal that has the face-centered cubic crystal structure is \\(0.123 \\mathrm{~nm}\\), calculate the volume of its unit cell.",
"answer": "Since this metal has an FCC crystal structure, the volume of its unit cell may be determined using Equation 3.6 as follows:\n\\[\nV_{C}=16 R^{3} \\sqrt{2}=(16)(0.225 \\mathrm{~nm})^{3}(\\sqrt{2})=0.258 \\mathrm{~nm}^{3}\n\\]"
},
{
"idx": 187,
"question": "Given rhodium's atomic radius of 0.1345 nm, density of 12.41 g/cm³, and atomic weight of 102.91 g/mol, calculate the number of atoms per unit cell using the formula n = (ρ * V_C * N_A) / A.",
"answer": "The number of atoms per unit cell is calculated as follows: n = (12.41 g/cm³ * V_C * 6.022 × 10²³ atoms/mol) / 102.91 g/mol = 7.26 × 10²² * V_C atoms/cm³."
},
{
"idx": 187,
"question": "Determine the unit cell volume V_C for a simple cubic (SC) crystal structure given the atomic radius R = 0.1345 nm.",
"answer": "For SC, V_C = a³ = (2R)³ = 8R³ = 8 * (1.345 × 10⁻⁸ cm)³ = 8 * 2.434 × 10⁻²⁴ cm³ = 1.947 × 10⁻²³ cm³."
},
{
"idx": 187,
"question": "Calculate the number of atoms per unit cell n for a simple cubic (SC) crystal structure using the derived V_C.",
"answer": "n = 7.26 × 10²² atoms/cm³ * 1.947 × 10⁻²³ cm³ = 1.41 atoms."
},
{
"idx": 187,
"question": "Determine the unit cell volume V_C for a body-centered cubic (BCC) crystal structure given the atomic radius R = 0.1345 nm.",
"answer": "For BCC, V_C = a³ = (4R/√3)³ = 64R³ / (3√3) = 64 * (1.345 × 10⁻⁸ cm)³ / (3√3) = 64 * 2.434 × 10⁻²⁴ cm³ / 5.196 = 2.999 × 10⁻²³ cm³."
},
{
"idx": 187,
"question": "Calculate the number of atoms per unit cell n for a body-centered cubic (BCC) crystal structure using the derived V_C.",
"answer": "n = 7.26 × 10²² atoms/cm³ * 2.999 × 10⁻²³ cm³ = 2.18 atoms."
},
{
"idx": 187,
"question": "Determine the unit cell volume V_C for a face-centered cubic (FCC) crystal structure given the atomic radius R = 0.1345 nm.",
"answer": "For FCC, V_C = a³ = (2R√2)³ = 16R³√2 = 16 * (1.345 × 10⁻⁸ cm)³ * √2 = 16 * 2.434 × 10⁻²⁴ cm³ * 1.414 = 5.508 × 10⁻²³ cm³."
},
{
"idx": 187,
"question": "Calculate the number of atoms per unit cell n for a face-centered cubic (FCC) crystal structure using the derived V_C.",
"answer": "n = 7.26 × 10²² atoms/cm³ * 5.508 × 10⁻²³ cm³ = 4.0 atoms."
},
{
"idx": 187,
"question": "Based on the calculated number of atoms per unit cell n for SC, BCC, and FCC crystal structures, determine rhodium's crystal structure.",
"answer": "The calculated n values are: SC = 1.41 atoms, BCC = 2.18 atoms, FCC = 4.0 atoms. Since only FCC matches the theoretical value of 4.0 atoms per unit cell, rhodium's crystal structure is FCC."
},
{
"idx": 188,
"question": "Niobium (Nb) has a BCC crystal structure, an atomic radius of \\(0.143 \\mathrm{~nm}\\) and an atomic weight of \\(92.91 \\mathrm{~g} / \\mathrm{mol}\\). Calculate the theoretical density for \\(\\mathrm{Nb}\\).",
"answer": "According to Equation:\n\\[\n\\rho=\\frac{n A_{\\mathrm{Nb}}}{V_{C} N_{\\mathrm{A}}}\n\\]\nFor \\(\\mathrm{BCC}, n=2\\) atoms/unit cell. Furthermore, because \\(V_{\\mathrm{C}}=a^{3}\\), and \\(a=\\frac{4 R}{\\sqrt{3}}\\) , then\n\\[\nV_{\\mathrm{C}}=\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}\n%\\varepsilon_{\\mathrm{b}}=\\frac{4 R}{\\sqrt{3}}\n\\]Thus, realizing that \\(A_{\\mathrm{Nb}}=92.91 \\mathrm{~g} / \\mathrm{mol}\\), using the above version of Equation, we compute the theoretical density of \\(\\mathrm{Nb}\\) as follows:\n\\[\n\\begin{array}{l}\n\\rho=\\frac{n A_{\\mathrm{Nb}}}{\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}} N_{\\mathrm{A}} \\\\\n=\\frac{(2 \\text { atoms/unit cell })(91.92 \\mathrm{~g} / \\mathrm{mol})}{\\left[\\frac{(4)(0.143 \\times 10^{-7} \\mathrm{~cm})}{\\sqrt{3}}\\right]^{3}} /(\\text { unit cell })(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol})\n\\end{array}\n\\]\n\\[\n=8.48 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\]\nThe experimental density for \\(\\mathrm{Nb}\\) is \\(8.57 \\mathrm{~g} / \\mathrm{cm}^{3}\\)."
},
{
"idx": 189,
"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": 190,
"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": 191,
"question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?\n\\[\na=b=c\n\\]\na) Cubic\nb) Hexagonal\nc) Tetragonal\nd) Rhombohedral\ne) Orthorhombic\nf) Monoclinic\ng) Triclinic",
"answer": "For both cubic and rhombohedral crystal systems all the unit cell edge lengths are equal."
},
{
"idx": 192,
"question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?\n\\[\na=b \\neq c\n\\]\na) Cubic\nb) Hexagonal\nc) Tetragonal\n\\(\\mathrm{d})\\) Rhombohedral\ne) Orthorhombic\nf) Monoclinic\n\ng) 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": 193,
"question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell e edge lengths?a) Cubic\nb) Hexagonal\nc) Tetragonal\nd) Rhombohedral\ne) Orthorhombic\nf) Monoclinic\ng) Triclinic",
"answer": "For orthorhombic, monoclinic and triclinic crystal systems, all three of the unit cell edge lengths are unequal to one another."
},
{
"idx": 194,
"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\\]\na) Cubic\nb) Hexagonal\nc) Tetragonal\nd) Rhombohedral\ne) Orthorhombic\nf) Monoclinic\ng) 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": 195,
"question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?\n\na) Cubic\nb) Hexagonal\nc) Tetragonal\nd) Rhombohedral\ne) Orthorhombic\nf) Monoclinic\ng） Triclinic",
"answer": "Cubic, tetragonal and orthorhombic crystal systems all have the three interaxial angles equal to \\(90^{\\circ}\\)."
},
{
"idx": 196,
"question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?\n\\[\n\\alpha=\\beta=90^{\\circ}, \\gamma=120^{\\circ}\n\\]\na) Cubic\nb) Hexagonal\nc) Tetragonal\nd) Rhombohedral\ne) Orthorhombic\nf) Monoclinic\ng) 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": 197,
"question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?\n\n\\[\n\\alpha=\\beta=\\gamma \\neq 90^{\\circ}\n\\]\na) Cubic\nb) Hexagonal\nc) Tetragonal\nd) Rhombohedral\ne) Orthorhombic\nf)\nMonoclinic\ng) Triclinic",
"answer": "Only the rhombohedral crystal system has all three interaxial angles equal to one other, but not equal to \\(90^{\\circ}\\)."
},
{
"idx": 198,
"question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?a) Cubic\nb) Hexagonal\nc) Tetragonal\nd) Rhombohedral\ne) Orthorhombic\nf) Monoclinic\ng) 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": 199,
"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: Carbon 12.01 g/mol, Hydrogen 1.008 g/mol, Fluorine 19.00 g/mol.",
"answer": "Consider 50 carbon atoms; for polyethylene, there are 100 possible side-bonding sites. For 18.6% substitution, 18.6 sites are occupied by F and 81.4 by H. The mass of 50 carbon atoms, m_C, is (50)(12.01 g/mol) = 600.5 g. The mass of hydrogen, m_H, is (81.4)(1.008 g/mol) = 82.05 g. The mass of fluorine, m_F, is (18.6)(19.00 g/mol) = 353.4 g. The concentration of fluorine, C_F, is (353.4 g) / (600.5 g + 82.05 g + 353.4 g) × 100 = 34.1 wt %."
},
{
"idx": 199,
"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: Carbon 12.01 g/mol, Fluorine 19.00 g/mol.",
"answer": "For complete fluorination, consider 50 carbon atoms with 100 side-bonding sites, all occupied by F. The mass of 50 carbon atoms, m_C, is (50)(12.01 g/mol) = 600.5 g. The mass of fluorine, m_F, is (100)(19.00 g/mol) = 1900.00 g. The concentration of fluorine, C_F, is (1900.00 g) / (600.5 g + 1900.00 g) × 100 = 76.0 wt %."
},
{
"idx": 200,
"question": "Which of the following may form linear polymers?\na) Rubber\nb) Epoxy\nc) Polyethylene\nd) Phenol-formaldehyde\ne) Polystyrene\nf) Nylon",
"answer": "Polyethylene, polystyrene and nylon may form linear polymers."
},
{
"idx": 201,
"question": "Which of the following form network polymers?\na) Rubber\nb) Epoxy\nc) Polyethylene\nd） Phenol-formaldehyde\ne) Polystyrene\nf) Nylon\n\\",
"answer": "Epoxy and phenol-formaldehyde form network polymers."
},
{
"idx": 202,
"question": "For most polymers, which configuration predominates?\na) Head-to-head\nb) Head-to-tail",
"answer": "For most polymers, the head-to-tail configuration predominates."
},
{
"idx": 203,
"question": "Is it possible to produce a polymer that is \\(100 \\%\\) crystalline?\n\na) True\n\nb) 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": 204,
"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 \\mathrm{~K}\\). Calculate the energy (in \\(\\mathrm{kJ} / \\mathrm{mol}\\) ) for vacancy formation assuming that the density of the metal remains the same over this temperature range.",
"answer": "If the factor by which the number of vacancies increases as the temperature is increased from \\(T_{2}\\) to \\(T_{1}\\) (i.e., \\(T_{1}>T_{2}\\) ) is denoted by \\(f\\), then \\(f\\) may be determined by the following expression, which incorporates Equation:\n\\[\nf=\\frac{N_{v 1}}{N_{v 2}}=\\frac{N \\exp \\left(-\\frac{Q_{v}}{k T_{1}}\\right)}{N \\exp \\left(-\\frac{Q_{v}}{k T_{2}}\\right)}=\\exp \\left[-\\frac{Q_{v}}{k}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\\right]\n\\]\nTaking natural logarithms of both sides of this equation gives the following:\n\\[\n\\ln f=-\\frac{Q_{v}}{k}\\left(\\frac{1}{T_{1}},-\\frac{1}{T_{2}}\\right)\n\\]\nAnd solving this equation for the energy of vacancy formation, \\(Q_{v}\\), leads to\n\\[\nQ_{v}=-\\frac{k \\ln f}{\\frac{1}{T_{1}}-\\frac{1}{T_{2}}}\n\\]From the problem statement we take \\(T_{1}=1170 \\mathrm{~K}\\) and \\(T_{2}=1030 \\mathrm{~K}\\) [and use the gas constant \\(R\\) (8.31 \\(\\mathrm{J} / \\mathrm{mol}-\\mathrm{K}\\) ) instead of Boltzmann's constant], then we compute the value of \\(Q_{\\mathrm{v}}\\) as follows:\n\\[\nQ_{\\mathrm{v}}=-\\frac{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})[\\ln (5)]}{1170 \\mathrm{~K}-\\frac{1}{1030 \\mathrm{~K}}}=115,000 \\mathrm{~J} / \\mathrm{mol}=115 \\mathrm{~kJ} / \\mathrm{mol}\n\\]"
},
{
"idx": 205,
"question": "What is the equation to calculate the number of vacancies per cubic meter in a metal at a given temperature?",
"answer": "The number of vacancies per cubic meter may be calculated using the equation: N_v = N exp (-Q_v / kT), where N_v is the number of vacancies, N is the total number of atomic sites per cubic meter, Q_v is the energy for vacancy formation, k is the Boltzmann constant, and T is the temperature in Kelvin."
},
{
"idx": 205,
"question": "How can the total number of atomic sites per cubic meter (N) be determined if the number of vacancies (N_v) and other parameters are known?",
"answer": "The total number of atomic sites per cubic meter (N) can be determined by solving the equation N = N_v / exp (-Q_v / kT) = N_v exp (Q_v / kT)."
},
{
"idx": 205,
"question": "Given that the density of the metal is the same at two different temperatures, how are the total number of atomic sites (N) related at these temperatures?",
"answer": "If the densities are the same at both temperatures, the total number of atomic sites per cubic meter (N) will also be the same at both temperatures. This means N_v1 exp (Q_v / kT1) = N_v2 exp (Q_v / kT2), where subscripts 1 and 2 denote the lower and higher temperatures, respectively."
},
{
"idx": 205,
"question": "How can the number of vacancies at a lower temperature (N_v1) be calculated if the number of vacancies at a higher temperature (N_v2) and other parameters are known?",
"answer": "The number of vacancies at the lower temperature (N_v1) can be calculated using the equation: N_v1 = N_v2 exp (Q_v / k) (1/T2 - 1/T1), where T1 and T2 are the lower and higher temperatures in Kelvin, respectively."
},
{
"idx": 205,
"question": "Given N_v2 = 1.1 x 10^24 m^-3 at T2 = 864°C, Q_v = 1.25 eV/atom, and k = 8.62 x 10^-5 eV/atom-K, calculate the number of vacancies (N_v1) at T1 = 463°C.",
"answer": "First, convert temperatures to Kelvin: T2 = 864 + 273 = 1137 K, T1 = 463 + 273 = 736 K. Then, calculate N_v1 using the equation: N_v1 = (1.1 x 10^24 m^-3) exp (1.25 eV/atom / 8.62 x 10^-5 eV/atom-K) (1/1137 K - 1/736 K). The result is N_v1 = 5.45 x 10^22 m^-3."
},
{
"idx": 206,
"question": "In metals, there are significantly more vacancies than self-interstitials.\na) True\nb) 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": 207,
"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 answer for MgO.",
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. 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": 207,
"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 answer for CaO.",
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. 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": 207,
"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 answer for BeO.",
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. 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": 207,
"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 answer for NiO.",
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. 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": 208,
"question": "Which of the given elements would form a substitutional solid solution with copper having complete solubility, considering atomic radius difference less than ±15%, same crystal structure, similar electronegativity, and same or nearly same valence?",
"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": 208,
"question": "Which of the given elements would form a substitutional solid solution with copper having incomplete solubility, considering differences in atomic radii, crystal structures, electronegativities, and valences?",
"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": 208,
"question": "Which of the given elements would form an interstitial solid solution with copper, considering significantly smaller atomic radii compared to copper?",
"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": 209,
"question": "The concentration of carbon in an iron-carbon alloy is \\(0.57 \\mathrm{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} / \\mathrm{cm}^{3}\\), respectively.",
"answer": "In order to compute the concentration in \\(\\mathrm{kg} / \\mathrm{m}^{3}\\) of \\(\\mathrm{C}\\) in a \\(0.15 \\mathrm{wt} \\% \\mathrm{C}-99.85 \\mathrm{wt} \\% \\mathrm{Fe}\\) alloy we must use Equation, that is\n\\[\nC_{\\mathrm{T}}^{\\star}=\\left(\\frac{C_{1}}{\\frac{C_{1}}{\\rho_{1}}+\\frac{C_{2}}{\\rho_{2}}}\\right) \\times 10^{3}\n\\]\nFor this problem, the above equation takes the form\n\\[\nC_{\\mathrm{C}}^{\\star}=\\left(\\frac{C_{\\mathrm{C}}}{\\frac{C_{\\mathrm{C}}}{\\rho_{\\mathrm{C}}}+\\frac{C_{\\mathrm{Fe}}}{\\rho_{\\mathrm{Fe}}}}\\right) \\times 10^{3}\n\\]\nAs given in the problem statement, the densities for iron and carbon are 7.87 and \\(2.25 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively, and, therefore\n\\[\n\\begin{aligned}\nC_{\\mathrm{C}}^{\\star} & =\\left(\\frac{0.15 \\mathrm{wt} \\%}{0.15 \\mathrm{wt} \\%}+\\frac{99.85 \\mathrm{wt} \\%}{7.87 \\mathrm{~g} / \\mathrm{cm}^{3}}\\right) \\times 10^{3} \\\\\n& =11.8 \\mathrm{~kg} / \\mathrm{m}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 210,
"question": "What is the average density of the alloy composed of 12.5 wt% of metal A and 87.5 wt% of metal B, given the densities of metals A and B are 4.27 g/cm³ and 6.35 g/cm³, respectively?",
"answer": "The average density of the alloy is calculated using the formula: ρ_ave = 100 / ( (C_A/ρ_A) + (C_B/ρ_B) ), where C_A = 12.5 wt%, C_B = 87.5 wt%, ρ_A = 4.27 g/cm³, and ρ_B = 6.35 g/cm³. Substituting the values: ρ_ave = 100 / ( (12.5/4.27) + (87.5/6.35) ) = 5.96 g/cm³."
},
{
"idx": 210,
"question": "What is the average atomic weight of the alloy composed of 12.5 wt% of metal A and 87.5 wt% of metal B, given the atomic weights of metals A and B are 61.4 g/mol and 125.7 g/mol, respectively?",
"answer": "The average atomic weight of the alloy is calculated using the formula: A_ave = 100 / ( (C_A/A_A) + (C_B/A_B) ), where C_A = 12.5 wt%, C_B = 87.5 wt%, A_A = 61.4 g/mol, and A_B = 125.7 g/mol. Substituting the values: A_ave = 100 / ( (12.5/61.4) + (87.5/125.7) ) = 113.5 g/mol."
},
{
"idx": 210,
"question": "Given the average density of the alloy (ρ_ave = 5.96 g/cm³), the average atomic weight (A_ave = 113.5 g/mol), the unit cell edge length (a = 0.395 nm), and Avogadro's number (N_A = 6.022 × 10²³ atoms/mol), determine the number of atoms per unit cell (n) for the alloy.",
"answer": "The number of atoms per unit cell is calculated using the formula: n = (ρ_ave * a³ * N_A) / A_ave. First, convert the unit cell edge length to cm: a = 0.395 nm = 3.95 × 10⁻⁸ cm. Then, substitute the values: n = (5.96 g/cm³ * (3.95 × 10⁻⁸ cm)³ * 6.022 × 10²³ atoms/mol) / 113.5 g/mol = 2.00 atoms/unit cell."
},
{
"idx": 210,
"question": "Based on the number of atoms per unit cell (n = 2.00), determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic.",
"answer": "For n = 1, the crystal structure is simple cubic; for n = 2, it is body-centered cubic (BCC); and for n = 4, it is face-centered cubic (FCC). Since n = 2.00, the crystal structure is body-centered cubic."
},
{
"idx": 211,
"question": "17}\nIron and vanadium both have the BCC crystal structure and \\(\\mathrm{V}\\) forms a substitutional solid solution in \\(\\mathrm{Fe}\\) for concentrations up to approximately \\(20 \\mathrm{wt} \\% \\mathrm{~V}\\) at room temperature. Determine the concentration in weight percent of \\(\\mathrm{V}\\) that must be added to iron to yield a unit cell edge length of \\(0.289 \\mathrm{~nm}\\).",
"answer": "To begin, it is necessary to employ Equation, and solve for the unit cell volume, \\(\\mathrm{V}_{\\mathrm{C}}\\), as\n\\[\nV_{C}=\\frac{n A_{\\mathrm{ave}}}{\\rho_{\\mathrm{ave}} N_{\\mathrm{A}}}\n\\]\nwhere \\(\\mathrm{A}_{\\mathrm{ave}}\\) and \\(\\rho_{\\mathrm{ave}}\\) are the atomic weight and density, respectively, of the \\(\\mathrm{Fe}-\\mathrm{V}\\) alloy. Inasmuch as both of these materials have the BCC crystal structure, which has cubic symmetry, \\(\\mathrm{V}_{\\mathrm{C}}\\) is just the cube of the unit cell length, a. That is\n\\[\n\\begin{aligned}\nV_{C} & =a^{3}=(0.289 \\mathrm{~nm})^{3} \\\\\n& =(2.89 \\times 10^{-8} \\mathrm{~cm})^{3}=2.414 \\times 10^{-23} \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]\nIt is now necessary to construct expressions for \\(\\mathrm{A}_{\\mathrm{ave}}\\) and \\(\\rho_{\\mathrm{\\text {ave }}}\\) in terms of the concentration of vanadium, \\(\\mathrm{C}_{\\mathrm{V}}\\), using Equations. For \\(\\mathrm{A}_{\\mathrm{ave}}\\) we have\n\\[\n\\begin{aligned}\nA_{\\mathrm{ave}} & =\\frac{100}{\\frac{C_{\\mathrm{V}}}{A_{\\mathrm{V}}}+\\frac{(100-C_{\\mathrm{V}})}{A_{\\mathrm{Fe}}}} \\\\\n& =\\frac{\\frac{100}{C_{\\mathrm{V}}}+\\frac{(100-C_{\\mathrm{V}})\\)}{55.85 \\mathrm{~g} / \\mathrm{mol}} \\\\\n& 50.94 \\mathrm{~g} / \\mathrm{mol}+55.85 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]whereas for \\(\\rho_{\\mathrm{ave}}\\)\n\\[\n\\begin{aligned}\n\\rho_{\\mathrm{ave}} & =\\frac{100}{\\rho_{\\mathrm{V}}}+\\frac{\\left(100-C_{\\mathrm{V}}\\right)}{\\rho_{\\mathrm{Fe}}}, \\\\\n& =\\frac{100}{\\frac{C_{\\mathrm{V}}}{6.10 \\mathrm{~g} / \\mathrm{cm}^{3}}+\\frac{\\left(100-C_{\\mathrm{V}}\\right)}{7.87 \\mathrm{~g} / \\mathrm{cm}^{3}}}\n\\end{aligned}\n\\]\nWithin the BCC unit cell there are 2 equivalent atoms, and thus, the value of \\(\\mathrm{n}\\) in Equation is 2; hence, this expression may be written in terms of the concentration of \\(\\mathrm{V}\\) in weight percent as follows:\n\\[\n\\begin{aligned}\n\\mathrm{V}_{\\mathrm{C}} & =2.414 \\times 10^{-23} \\mathrm{~cm}^{3} \\\\\n& =\\frac{n A_{\\mathrm{ave}}}{\\rho_{\\mathrm{ave}} N_{\\mathrm{A}}} \\\\\n& =\\left(\\begin{array}{c}\n100 \\\\\n\\frac{C_{\\mathrm{V}}}{50.94 \\mathrm{~g} / \\mathrm{mol}}+\\frac{\\left(100-C_{\\mathrm{V}}\\right)}{\\mathbf{5 5 . 8 5} \\mathrm{g} / \\mathrm{mol}} \\\\\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\n100 \\\\\n\\frac{c_{\\mathrm{V}}}{6.10 \\mathrm{~g} / \\textrm{cm}^{3}}+\\frac{\\left(100-C_{\\mathrm{v}}\\right)}{7.87 \\mathrm{~g} / \\mathrm{c} \\mathrm{m}^{3}}\\right)\n\\end{array}\\right)\\left(\\begin{array}{l}\n6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\n\\end{array}\\right)\n\\end{aligned}\n\\]\nAnd solving this expression for \\(\\mathrm{C}_{\\mathrm{V}}\\) leads to \\(\\mathrm{C}_{\\mathrm{V}}=12.9 \\mathrm{wt} \\%\\)."
},
{
"idx": 212,
"question": "Which of the following is a (are) linear defect(s)?\na) An edge dislocation\nb) A Frenkel defect\nc) A Schottky defect",
"answer": "Edge and screw dislocations are linear defects."
},
{
"idx": 213,
"question": "A photomicrograph was taken of a specimen at a magnification of \\(100 \\times\\), 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": "For a micrograph taken at a magnification of \\(100 \\times\\), the number of grains per square inch, \\(n\\), is related to the ASTM grain size number, \\(G\\), according to Equation , that is\n\\[\nn=2^{G-1}\n\\]\nTaking logarithms of both sides of this equation and rearranging the resulting expression and solving for \\(G\\) gives the following:\n\\[\nG=\\frac{\\log n}{\\log 2}+1\n\\]\nThus, for the situation when \\(n=15\\)\n\\[\nG=\\frac{\\log 15}{\\log 2}+1=4.9\n\\]"
},
{
"idx": 214,
"question": "Diffusion by which mechanism occurs more rapidly in metal alloys?\na) Vacancy diffusion\nb) 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": 215,
"question": "As temperature decreases, the fraction of total number of atoms that are capable of diffusive motion\na) increases.\nb) decreases.",
"answer": "As temperature decreases, the fraction of the total number of atoms that are capable of diffusive motion decreases."
},
{
"idx": 216,
"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": "The definition of flux is the quantity of something (which could be in terms of particle count, mass, etc.) that passes through a specified unit of area in a specified unit of time.\nIn this case, we are provided an atom count that passes through the known area each hour.\nTherefore, accounting for the seconds in each hour:\n\\[\nJ=\\frac{M}{A t}=\\frac{[\\mathrm{m}]}{\\left(\\left[a\\right] \\mathrm{m}^{2}\\right)\\left(\\left[t\\right] \\operatorname{hr}(3600 \\mathrm{~s} / \\mathrm{hr})\\right)}\n\\]"
},
{
"idx": 217,
"question": "If water molecules pass through a membrane with a steady state flux of \\([j]\\) mole \\(/\\left(\\mathrm{m}^{2}\\right.\\) day \\()\\), how long will it take, in hours, for \\([\\mathrm{m}] \\mathrm{kg}\\) of water to pass through a \\([\\mathrm{a}]\\) square centimeter of the membrane?",
"answer": "The definition of flux is the quantity of something (which could BE in terms of particle count, mass, etc.) that passes through a specified unit of area in A specified unit of time.\nIn this case, we are provided the flux of water molecules and are asked to compute the time required for a certain mass of water molecules to transit a specific area of membrane.\nTherefore, accounting for the number of hours in each day:\n\\[\nt=\\frac{M}{A J}=\\frac{[m] \\mathrm{kg}\\left(\\frac{1000 \\mathrm{~g}}{1 \\mathrm{~kg}}\\right)\\left(\\frac{1 \\mathrm{mole}}{18 \\mathrm{~g}}\\right)}{\\left(\\left[a] \\mathrm{cm}^{2}\\right)\\left(\\frac{1 \\mathrm{~m}^{2}}{100^{2} \\mathrm{~cm}^{2}}\\right)\\left[\\left(j\\right] \\frac{\\mathrm{mol}}{\\mathrm{m}^{2} \\mathrm{day}}\\right)^{24 \\mathrm{hr}}\\right)}\n\\]"
},
{
"idx": 218,
"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": 218,
"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": 218,
"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": 218,
"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": 219,
"question": "For a steel alloy, given a carburizing heat treatment of 16 h duration at 757°C raises the carbon concentration to 0.5 wt% at a point 2.3 mm from the surface, estimate the time necessary to achieve the same concentration at an 8 mm position for an identical steel and at a carburizing temperature of 1130°C. Assume that D0 is 4.6×10−5 m2/s and Qd is 104 kJ/mol.",
"answer": "The solution involves using Fick's second law and the temperature dependence of the diffusion coefficient. The key equation is:\n\nt2 = (t1 * x2^2 / x1^2) * exp[-(Qd/R) * (1/T1 - 1/T2)]\n\nGiven:\nt1 = 16 h\nT1 = 757°C = 1030 K\nx1 = 2.3 mm = 2.3×10−3 m\nx2 = 8 mm = 8×10−3 m\nT2 = 1130°C = 1403 K\nQd = 104 kJ/mol = 104,000 J/mol\nR = 8.31 J/mol-K\n\nSubstituting these values:\n\nt2 = [(16 h) * (8×10−3 m)^2 / (2.3×10−3 m)^2] * exp[-(104,000 J/mol / 8.31 J/mol-K) * (1/1030 K - 1/1403 K)]\n\nCalculating the exponent:\n(1/1030 - 1/1403) = (9.7087×10−4 - 7.1276×10−4) = 2.5811×10−4\n\nExponent term:\nexp[-(104,000 / 8.31) * 2.5811×10−4] = exp[-3.233] ≈ 0.0394\n\nTime ratio:\n(8/2.3)^2 = (3.478)^2 ≈ 12.10\n\nFinal calculation:\nt2 = 16 h * 12.10 * 0.0394 ≈ 7.63 h"
},
{
"idx": 220,
"question": "The diffusion coefficient for aluminum in silicon is \\(D_{\\mathrm{Al}} \\mathrm{in}_{\\mathrm{Si}}=3 \\times 10^{-16} \\mathrm{~cm}^{2} / \\mathrm{s}\\) at \\(300 \\mathrm{~K}\\) (note that \\(300 \\mathrm{~K}\\) is about room temperature).\nWhat is a reasonable value for \\(D_{\\mathrm{Al} \\text { in } \\mathrm{Si}}\\) at \\(600 \\mathrm{~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 \\times 10^{16} \\mathrm{~cm}^{2} / \\mathrm{s}\\)\n(b) \\(D=3 \\times 10^{16} \\mathrm{~cm}^{2} /\\) s\n(c) \\(D=6 \\times 10^{16} \\mathrm{~cm}^{2} / s\\)\n(d) \\(D=1.5 \\times 10^{16} \\mathrm{~cm}^{2} /{\\mathrm{s}}\\)\n(e) \\(D>6 \\times 10^{16} \\mathrm{~cm}^{2} /\\)\n(f) \\(D=6 \\times 10^{-17} \\mathrm{~cm}^{2} / \\mathrm{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": 221,
"question": "A specimen of some metal having a rectangular cross section \\(11.2 \\mathrm{~mm} \\times 12.4 \\mathrm{~mm}\\) is pulled in tension with a force of \\(31200 \\mathrm{~N}\\), which produces only elastic deformation. Given that the elastic modulus of this metal is \\(63 \\mathrm{GPa}\\), calculate the resulting strain.",
"answer": "This problem calls for us to calculate the elastic strain that results for a specimen of some metal alloy stressed in tension. Combining Equations and solving for strain yields\n\\[\n\\begin{array}{c}\n\\varepsilon=\\frac{\\sigma}{E}=\\frac{F}{A_{0}}=\\frac{F}{A_{0} E} \\\\\n=\\frac{32,600 \\mathrm{~N}}{\\left(10.9 \\times 10^{-3} \\mathrm{~mm}\\right)\\left(13.3 \\times 10^{-3} \\mathrm{~mm}\\right)\\left(77 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)}=2.92 \\times 10^{-3}\n\\end{array}\n\\]"
},
{
"idx": 222,
"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 mm2 without plastic deformation?",
"answer": "The maximum load that may be applied without yielding is calculated using the formula Fy = σy * A0. Given σy = 277 MPa = 277 × 10^6 N/m2 and A0 = 327 mm2 = 327 × 10^-6 m2, then Fy = (277 × 10^6 N/m2)(327 × 10^-6 m2) = 90,579 N."
},
{
"idx": 222,
"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 plastic deformation is calculated using the formula li = l0(σy/E + 1). Given l0 = 148 mm = 148 × 10^-3 m, σy = 277 MPa = 277 × 10^6 N/m2, and E = 117 GPa = 117 × 10^9 N/m2, then li = (148 × 10^-3 m)((277 × 10^6 N/m2)/(117 × 10^9 N/m2) + 1) = 0.14835 m = 148.35 mm."
},
{
"idx": 223,
"question": "A steel bar \\(100 \\mathrm{~mm}\\) (4.0 in.) long and having a square cross section \\(20 \\mathrm{~mm}\\) (0.8 in.) on an edge is pulled in tension with a load of \\(89,000 \\mathrm{~N}\\left(20,000 \\mathrm{lb}_{0}\\right)\\), and experiences an elongation of \\(0.10 \\mathrm{~mm}\\) (4.0 \\(\\times 10^{-3}\\) in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.",
"answer": "This problem asks us to compute the elastic modulus of steel. For a square cross-section, \\(\\mathrm{A}_{0}=b_{0}^{2}\\), where \\(b_{0}\\) is the edge length. Combining Equations and solving for E, leads to\n\\[\n\\begin{aligned}\nE & =\\frac{\\sigma}{\\varepsilon}=\\frac{\\frac{F}{A_{0}}}{\\frac{\\Delta l}{l_{0}}}=\\frac{F l_{0}}{b_{0}^{2} \\Delta l} \\\\\n& =\\frac{(89,000 \\mathrm{~N})\\left(100 \\times 10^{-3} \\mathrm{~m}\\right)}{\\left(20 \\times 10^{-3} \\mathrm{~m}\\right)^{2}\\left(0.10 \\times 10^{-3} \\mathrm{~m}\\right)} \\\\\n& =223 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}{ }^{2}=223 \\mathrm{GPa}\\left(31.3 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]"
},
{
"idx": 224,
"question": "13: Typical relationship between E and G}\nFor most metals, the relationship between elastic and shear moduli is approximately which of the following?\na) \\(G=0.1 E\\)\nb) \\(G=0.2 E\\)\nc) \\(G=0.3 E\\)\nd) \\(G=0.4 E\\)\ne) \\(G=0.5 E\\)",
"answer": "Elastic and shear moduli are related according to Equation as follows:\n\\[\nE=2 G(1+\\nu)\n\\]\nAnd solving for \\(G\\) gives\n\\[\nG=\\frac{E}{2(1+\\nu)}\n\\]\nBecause the value of Poisson's ratio for most metals is approximately 0.30 , then\n\\[\nG=\\frac{E}{2(1+0.30)}=\\frac{E}{2.6}=0.38 E\n\\]\nOr, approximately \\(0.4 E\\)."
},
{
"idx": 225,
"question": "Which of the four metals or alloys (Aluminum alloy, Brass alloy, Copper, Steel alloy) will not experience plastic deformation when a tensile load of 24,500 N is applied to a cylindrical rod with a diameter of 10.0 mm?",
"answer": "To determine which materials will not experience plastic deformation, compute the stress using the formula σ = F/A0. The cross-sectional area A0 is π*(10.0×10^-3 m/2)^2 = 7.854×10^-5 m^2. The stress σ = 24,500 N / 7.854×10^-5 m^2 = 312 MPa. The yield strengths of the materials are: Aluminum alloy (255 MPa), Brass alloy (345 MPa), Copper (250 MPa), Steel alloy (450 MPa). Only Brass alloy and Steel alloy have yield strengths greater than 312 MPa."
},
{
"idx": 225,
"question": "For the remaining candidates (Brass alloy and Steel alloy) that do not experience plastic deformation, which will not elongate more than 0.9 mm under the given load and rod dimensions (length = 380 mm, diameter = 10.0 mm)?",
"answer": "Compute the elongation for Brass alloy: Δl = (σ * l0)/E = (312 MPa * 380 mm)/(100×10^3 MPa) = 1.19 mm. Compute the elongation for Steel alloy: Δl = (312 MPa * 380 mm)/(207×10^3 MPa) = 0.57 mm. Only Steel alloy has an elongation less than 0.9 mm."
},
{
"idx": 226,
"question": "Given a metal alloy with a true plastic strain of 0.12 at a true stress of 280 MPa and a strain hardening exponent of 0.3, what is the value of the strength coefficient K?",
"answer": "To find the strength coefficient K, we use the relationship between true stress and true strain: σ_T = K * (ε_T)^n. Rearranging to solve for K gives K = σ_T / (ε_T)^n. Substituting the given values: K = 280 MPa / (0.12)^0.3 = 280 MPa / 0.6813 = 410.9 MPa."
},
{
"idx": 226,
"question": "Using the strength coefficient K = 410.9 MPa and strain hardening exponent n = 0.3, what true plastic strain would be expected for a total true plastic stress of 330 MPa?",
"answer": "To find the true plastic strain at 330 MPa, we use the relationship ε_T = (σ_T / K)^(1/n). Substituting the given values: ε_T = (330 MPa / 410.9 MPa)^(1/0.3) = (0.803)^3.333 = 0.171."
},
{
"idx": 227,
"question": "Given a cylindrical specimen of a metal alloy with initial length 48.8 mm and diameter 9.09 mm, calculate the true strain when the specimen is plastically elongated to a length of 55 mm.",
"answer": "The true strain is calculated using the formula: ε_T = ln(l_i / l_0). Substituting the given values: ε_T = ln(55 mm / 48.8 mm) = ln(1.127) ≈ 0.119."
},
{
"idx": 227,
"question": "Using the true stress of 327 MPa and the true strain of 0.119, along with the strain-hardening exponent of 0.3, calculate the strength coefficient K for the material.",
"answer": "The relationship between true stress and true strain is given by σ_T = K * (ε_T)^n. Solving for K: K = σ_T / (ε_T)^n = 327 MPa / (0.119)^0.3 ≈ 327 MPa / 0.668 ≈ 489.5 MPa."
},
{
"idx": 227,
"question": "Calculate the true strain when the specimen is elongated from the initial length of 48.8 mm to a final length of 57.6 mm.",
"answer": "The true strain is calculated as: ε_T = ln(l_i / l_0) = ln(57.6 mm / 48.8 mm) = ln(1.180) ≈ 0.166."
},
{
"idx": 227,
"question": "Using the strength coefficient K of 489.5 MPa and the strain-hardening exponent of 0.3, calculate the true stress required to plastically elongate the specimen from 48.8 mm to 57.6 mm.",
"answer": "The true stress is calculated using the formula: σ_T = K * (ε_T)^n = 489.5 MPa * (0.166)^0.3 ≈ 489.5 MPa * 0.715 ≈ 350.0 MPa."
},
{
"idx": 228,
"question": "Predict whether or not a cylindrical specimen of aluminum oxide with a reported flexural strength of 390 MPa (56,600 psi), radius of 2.5 mm (0.10 in.), and support point separation distance of 30 mm (1.2 in.) would fracture when a load of 620 N (140 lbf) is applied in a three-point bending test.",
"answer": "Using the equation σ = FL / (πR^3), where F = 620 N, L = 30 × 10^-3 m, and R = 2.5 × 10^-3 m, the flexural strength is calculated as σ = (620 N)(30 × 10^-3 m) / (π)(2.5 × 10^-3 m)^3 = 379 × 10^6 N/m^2 = 379 MPa (53,500 psi). Since this value is less than the given flexural strength of 390 MPa, fracture is not predicted. However, there is some chance of fracture due to variability in ceramic materials."
},
{
"idx": 228,
"question": "Calculate the value of flexural strength for the three-point bending test on the cylindrical specimen of aluminum oxide with radius 2.5 mm (0.10 in.) and support point separation distance 30 mm (1.2 in.) when a load of 620 N (140 lbf) is applied.",
"answer": "The flexural strength is calculated using the equation σ = FL / (πR^3), where F = 620 N, L = 30 × 10^-3 m, and R = 2.5 × 10^-3 m. The calculation yields σ = (620 N)(30 × 10^-3 m) / (π)(2.5 × 10^-3 m)^3 = 379 × 10^6 N/m^2 = 379 MPa (53,500 psi)."
},
{
"idx": 229,
"question": "Compute the flexural strength for a completely nonporous specimen of this material given the flexural strength and associated volume fraction porosity for two specimens: (100 MPa, 0.05) and (50 MPa, 0.20).",
"answer": "Given the flexural strengths at two different volume fraction porosities, we determine the flexural strength for a nonporous material using the equation ln σ_fs = ln σ_0 - nP. Using the provided data, two simultaneous equations are written as ln(100 MPa) = ln σ_0 - (0.05)n and ln(50 MPa) = ln σ_0 - (0.20)n. Solving for n and σ_0 leads to n = 4.62 and σ_0 = 126 MPa. For the nonporous material, P = 0, and σ_0 = σ_fs. Thus, σ_fs for P = 0 is 126 MPa."
},
{
"idx": 229,
"question": "Compute the flexural strength for a 0.10 volume fraction porosity given the flexural strength and associated volume fraction porosity for two specimens: (100 MPa, 0.05) and (50 MPa, 0.20).",
"answer": "We are asked for σ_fs at P = 0.10 for this material. Utilizing the equation σ_fs = σ_0 exp(-nP), we substitute σ_0 = 126 MPa and n = 4.62 to get σ_fs = (126 MPa) exp[-(4.62)(0.10)] = 79.4 MPa."
},
{
"idx": 230,
"question": "Mechanical twinning occurs in metals having which type(s) of crystal structure(s)?\na) \\(\\mathrm{BCC}\\)\nb) \\(\\mathrm{FCC}\\)\nc) \\(\\mathrm{HCP}\\)",
"answer": "Mechanical twinning occurs in metals having \\(B C C\\) and \\(H C P\\) crystal structures."
},
{
"idx": 231,
"question": "How does grain size influence strength of a polycrystalline material?\na) Strengthfine-grained \\(<\\) strength \\(_{\\text {course-grained }}\\)\nb) Strength \\(_{\\text {fine-grained }}=\\) strength \\(_{\\text {course-grained }}\\)\nc) Strengthfine-grained \\(>\\) strength \\(_{\\text {course-grained }}\\)",
"answer": "A fine-grained material is stronger than a coarse-grained material."
},
{
"idx": 232,
"question": "Reducing the grain size of metal improves toughness.\na) True\nb) False",
"answer": "True. Reducing the grain size of a metal improves its toughness."
},
{
"idx": 233,
"question": "As dislocation density increases, the resistance to dislocation movement\na) increases.\nb) decreases.",
"answer": "As dislocation density increases, the resistance to dislocation movement increases. This phenomenon is responsible for cold working."
},
{
"idx": 234,
"question": "On the average, dislocation-dislocation strain interactions are\na) repulsive.\nb) attractive.",
"answer": "On the average, dislocation-dislocation strain interactions are repulsive."
},
{
"idx": 235,
"question": "As a metal is strain hardened, its ductility\na) increases\nb) decreases",
"answer": "As a metal is strain hardened, its ductility decreases."
},
{
"idx": 236,
"question": "Most metals strain harden at room temperature.\na) True\nb) False",
"answer": "True. Most metals strain harden at room temperature."
},
{
"idx": 237,
"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": 237,
"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": 237,
"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": 237,
"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": 237,
"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": 237,
"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": 237,
"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": 237,
"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": 238,
"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": 238,
"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": 238,
"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": 238,
"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": 239,
"question": "Grain growth requirements\n}\nGrain growth must always be preceded by recovery and recrystallization.\na) True\nb) 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": 240,
"question": "Given a metal alloy with initial grain diameter d0 = 2.4 × 10^-2 mm, final grain diameter d = 7.3 × 10^-2 mm after heat treatment at 575°C for t = 500 min, and grain diameter exponent n = 2.2, compute the parameter K in the grain growth equation.",
"answer": "K = (d^2.2 - d0^2.2) / t = ((7.3 × 10^-2 mm)^2.2 - (2.4 × 10^-2 mm)^2.2) / 500 min = 5.77 × 10^-6 mm^2.2 / min"
},
{
"idx": 240,
"question": "Using the computed K = 5.77 × 10^-6 mm^2.2 / min, initial grain diameter d0 = 2.4 × 10^-2 mm, grain diameter exponent n = 2.2, and target grain diameter d = 5.5 × 10^-2 mm, calculate the required heat treatment time at 575°C.",
"answer": "t = (d^2.2 - d0^2.2) / K = ((5.5 × 10^-2 mm)^2.2 - (2.4 × 10^-2 mm)^2.2) / (5.77 × 10^-6 mm^2.2 / min) = 246 min"
},
{
"idx": 241,
"question": "Tensile strengths and number-average molecular weights for two polymers are as follows:\nTensile strength Number average molecular weight\n\\((\\mathrm{MPa})\\)\n\\((\\mathrm{g} / \\mathrm{mol})\\)\n37.7\n36800\n131\n62400\nEstimate the tensile strength (in MPa) for a number-average molecular weight of \\(51500 \\mathrm{~g} / \\mathrm{mol}\\).",
"answer": "Tensile strength is related to the number-average molecular weight according to\n\\[\n\\mathrm{TS}=\\mathrm{TS}_{\\infty}-\\frac{A}{M_{n}}\n\\]\nUsing the two data sets given, two equations having the above form may be written as follows:\n\\[\n\\begin{array}{l}\n\\mathrm{TS}_{1}=\\mathrm{TS}_{\\infty}-\\frac{A}{M_{n 1}} \\\\\n\\mathrm{TS}_{2}=\\mathrm{TS}_{\\infty}-\\frac{A}{M_{n 2}}\n\\end{array}\n\\]\nwhere each of the \" 1 \" and \" 2 \" subscripts refers to one set of data given in the problem statement. Using these data, simultaneous solution of these equations yields values for \\(\\mathrm{TS}_{\\infty}\\) and \\(A\\). The tensile strength called for may be calculated when values for these constants and the specified molecular weight are incorporated into the same equation."
},
{
"idx": 242,
"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(\\mathrm{MPa}) \\\\\n138 \\\\\n184 \\\\\n\\text { (g/mol) } \\\\\n12600 \\\\\n28100\n\\end{array}\n\\]\nEstimate number average molecular weight (in \\(\\mathrm{g} / \\mathrm{mol}\\) ) at a tensile strength of \\(141 \\mathrm{Mpa}\\).",
"answer": "Tensile strength is related to the number-average molecular weight according to\n\\[\n\\mathrm{TS}=\\mathrm{TS}_{\\infty}-\\frac{A}{M_{n}}\n\\]\nUsing the two data sets given, two equations having the above form may be written as follows:\n\\[\n\\begin{array}{l}\n\\mathrm{TS}_{1}=\\mathrm{TS}_{\\infty}-\\frac{A}{M_{n 1}} \\\\\n\\mathrm{TS}_{2}=\\mathrm{TS}_{\\infty}-\\frac{A}{M_{n 2}}\n\\end{array}\n\\]\nwhere each of the \" 1 \" and \" 2 \" subscripts refers to one set of data given in the problem statement. Using these data, simultaneous solution of these equations yields values for \\(\\mathrm{TS}_{\\infty}\\) and \\(A\\). The numberaverage molecular weight required for the tensile strength specified in the problem statement may be determined using the following rearranged form of the above equation and incorporating the values of \\(\\mathrm{TS}_{\\infty}\\) and \\(A\\) just determined.\n\\[\nM_{n}=\\frac{A}{T S_{\\infty}-T S}\n\\]"
},
{
"idx": 243,
"question": "The bonding forces between adhesive and adherend surfaces are thought to be\na) Electrostatic\nb) Covalent\nc) Chemical",
"answer": "The bonding forces between adhesive and adherend surfaces are thought to be electrostatic."
},
{
"idx": 244,
"question": "Deformation of a semicrystalline polymer by drawing produces which of the following?\na) Increase in strength in the direction of drawing.\nb) Decrease in strength in the direction of drawing.\nc) Increase in strength perpendicular to the direction of drawing.\nd) Decrease in strength perpendicular to the direction of drawing.",
"answer": "Deformation of a semicrystalline polymer by drawing produces an increase in strength in the direction of drawing, and a decrease in strength perpendicular to the direction of drawing."
},
{
"idx": 245,
"question": "How does deformation by drawing of a semicrystalline polymer affect its tensile strength?\na) Increases\nb) 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": 246,
"question": "How does increasing the degree of crystallinity of a semicrystalline polymer affect its tensile strength?\na) 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": 247,
"question": "How does increasing the molecular weight of a semicrysatlline polymer affect its tensile strength?\na) Increases\nb) 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": 248,
"question": "Which kind of fracture (ductile or brittle) is associated with intergranular crack propagation?",
"answer": "Intergranular fracture is brittle. Crack propagation occurs along grain boundaries."
},
{
"idx": 248,
"question": "Which kind of fracture (ductile or brittle) is associated with transgranular crack propagation?",
"answer": "Transgranular fracture is brittle. Crack propagation occurs through the grains."
},
{
"idx": 249,
"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": "This problem asks that we compute the crack tip radius ratio before and after etching. Let\n\\[\n\\begin{aligned}\n\\rho_{t} & =\\text { unetched crack tip radius, and } \\\\\n\\rho_{t}^{\\prime} & =\\text { etched crack tip radius }\n\\end{aligned}\n\\]\nThe maximum stress, \\(\\sigma_{m}\\) in Equation will be the same for both unetched and etched specimens, that is\n\\[\n\\sigma_{m}^{\\prime}=\\sigma_{m}\n\\]\nAlso, as noted in the problem statement, the etched crack length \\(\\left(a^{\\prime}\\right)\\) and unetched crack length \\((a)\\) are related as\n\\[\na^{\\prime}=(1-0.35) a=0.65 a\n\\]\\[\n\\sigma_{0}^{\\prime} \\text { and } \\sigma_{0} \\text {, respectively) is }\n\\]\n\\[\n\\sigma_{0}^{\\prime}=7 \\sigma_{0}\n\\]\nUsing the above notations, we write the following expression for Equation :\n\\[\n\\sigma_{m}=2 \\sigma_{0}\\left(\\frac{a}{\\rho_{t}}\\right)^{1 / 2}=\\mathscr{C}_{m}^{\\prime}=2 \\sigma_{0}^{\\prime}\\left(\\frac{a^{\\prime}}{\\rho_{t}^{\\prime}}\\right)^{1 / 2}\n\\]\nWe now solve for the \\(\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}}\\) ratio, which yields the following:\n\\[\n\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}}=\\left(\\frac{\\sigma_{0}^{\\prime}}{\\sigma_{0}}\\right)^{2}\\left(\\frac{a^{\\prime}}{a}\\right)\n\\]\nAnd when we incorporate the above \\(a-a^{\\prime}\\) and \\(\\sigma_{0}-\\sigma_{0}^{\\prime}\\) relationships into this expression, the following results:\n\\[\n\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}}= \\left(\\frac{7 \\sigma_{0}}{\\sigma_{0}}\\right)^{2}\\left(\\frac{0.65 a}{a}\\right)=32\n\\]"
},
{
"idx": 250,
"question": "Given a metal alloy with a yield strength of 535 MPa and a plane strain fracture toughness of 31.3 MPa·m^(-1/2), and a design stress of 0.5 times the yield strength, what is the design stress?",
"answer": "The design stress is calculated as 0.5 times the yield strength: 0.5 * 535 MPa = 267.5 MPa."
},
{
"idx": 250,
"question": "Using the design stress of 267.5 MPa, a plane strain fracture toughness of 31.3 MPa·m^(-1/2), and a geometry factor Y of 1.8, what is the critical length of a surface flaw (a_c)?",
"answer": "The critical length of a surface flaw is calculated using the equation: a_c = (1/π) * (K_Ic / (σ * Y))^2. Substituting the given values: a_c = (1/π) * (31.3 MPa·m^(-1/2) / (267.5 MPa * 1.8))^2 = 2.0 × 10^(-3) m = 2.0 mm."
},
{
"idx": 251,
"question": "How would the plane strain fracture toughness of a metal be expected to change with rising temperature?\na) Increase\nb) Decrease\nc) 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": 252,
"question": "Does increasing temperature favor brittle fracture in polymers?",
"answer": "No, increasing temperature does not favor brittle fracture in polymers."
},
{
"idx": 252,
"question": "Does increasing strain rate favor brittle fracture in polymers?",
"answer": "Yes, increasing strain rate favors brittle fracture in polymers."
},
{
"idx": 252,
"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": 252,
"question": "Does decreasing specimen thickness favor brittle fracture in polymers?",
"answer": "The answer does not specify the effect of decreasing specimen thickness on brittle fracture in polymers."
},
{
"idx": 253,
"question": "A plate of an alloy steel has a plane-strain fracture toughness of \\(50 \\mathrm{MPa} \\cdot \\mathrm{m}^{1 / 2}\\). If it is known that the largest surface crack is \\(0.5 \\mathrm{~mm}\\) 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 \\mathrm{MPa}\\) is applied?\na) The plate will definitely fracture.\nb) The plate will definitely not fracture.\nc) It is not possible to determine whether or not the plate will fracture.",
"answer": "From Equation,\n\\[\nK_{L_{C}}=Y \\sigma \\sqrt{\\pi a}\n\\]\nand for this problem, we determine the value of the right-hand side of the equation. If the value of \\(Y \\sigma \\sqrt{\\pi a}\\) is less than \\(K_{L_{C}}\\) for this material, then fracture will not occur; if greater than \\(K_{L_{C}}\\), then fracture will occur. For this problem:\n\\(K_{L_{C}}=50 \\mathrm{MPa} \\cdot \\mathrm{m}^{1 / 12}\\)\n\\(\\sigma=1200 \\mathrm{MPa}\\)\n\\(Y=1.1\\)\n\\(a=0.5 \\mathrm{~mm}=0.5 \\times 10^{-3} \\mathrm{~m}\\)\nTherefore,\n\\[\n\\begin{aligned}\nY \\sigma \\sqrt{\\pi a} & =(1.1)(1200 \\mathrm{MPa}) \\sqrt{(\\pi)(0.5 \\times 10^{-3} \\mathrm{~m})} \\\\\n& =52.3 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\n\\end{aligned}\n\\]\nSince this value is greater than \\(K_{L_{C}}\\left(50 \\mathrm{MPa} \\cdot \\mathrm{m}^{-1 / 2}\\right)\\), fracture will occur."
},
{
"idx": 254,
"question": "The effect of a stress raiser is more significant for which of the following types of materials?\na) Brittle materials\nb) 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": 255,
"question": "Given a structural component made from an alloy with a plane strain fracture toughness of 45 MPa√m, a failure stress of 300 MPa, and a maximum surface crack length of 0.95 mm, calculate the geometry factor Y.",
"answer": "The geometry factor Y can be calculated using the rearranged equation: Y = K_Ic / (σ √(π a_c)). Substituting the given values: K_Ic = 45 MPa√m, σ = 300 MPa, and a_c = 0.95 mm (9.5 × 10^-4 m), we get Y = 45 / (300 √(π × 9.5 × 10^-4)) = 2.75."
},
{
"idx": 255,
"question": "Using the calculated geometry factor Y = 2.75, determine the maximum allowable surface crack length without fracture for the same component made from another alloy with a plane strain fracture toughness of 100.0 MPa√m and exposed to a stress of 300 MPa.",
"answer": "The maximum allowable surface crack length a_c can be calculated using the equation: a_c = 1/π (K_Ic / (σ Y))^2. Substituting the given values: K_Ic = 100.0 MPa√m, σ = 300 MPa, and Y = 2.75, we get a_c = 1/π (100.0 / (300 × 2.75))^2 = 0.00468 m = 4.68 mm."
},
{
"idx": 256,
"question": "A cylindrical specimen \\(7.5 \\mathrm{~mm}\\) in diameter of an S-590 alloy is to be exposed to a tensile load of 9000 \\(\\mathrm{N}\\). At approximately what temperature will the steady-state creep be \\(10^{-2} \\mathrm{~h}^{-1}\\) ?",
"answer": "Let us first determine the stress imposed on this specimen using values of cross-section diameter and applied load; this is possible using Equation as\n\\[\n\\sigma=\\frac{F}{A_{0}}=\\frac{F}{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}}\n\\]\nInasmuch as \\(\\mathrm{d}_{0}=7.5 \\mathrm{~mm}\\left(7.5 \\times 10^{-3} \\mathrm{~m}\\right)\\) and \\(\\mathrm{F}=9000 \\mathrm{~N}\\), the stress is equal to\n\\[\n\\begin{aligned}\n\\sigma & =\\frac{9000 \\mathrm{~N}}{\\pi\\left(\\frac{7.5 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}} \\\\\n& =204 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=204 \\mathrm{MPa}\n\\end{aligned}\n\\]\nOr approximately \\(200 \\mathrm{MPa}\\). From Figure 9.40 the point corresponding \\(10^{-2} \\mathrm{~h}^{-1}\\) and \\(200 \\mathrm{MPa}\\) is approximately \\(815^{\\circ} \\mathrm{C}\\)."
},
{
"idx": 257,
"question": "The mineral olivine is a solid solution of the silicate compounds forsterite \\(\\left(\\mathrm{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 \\(\\mathrm{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": 258,
"question": "Seawater, which covers the majority of the earth, is composed primarily of molecules of \\(\\mathrm{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} \\mathrm{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": "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 \\(\\mathrm{H}_{2} \\mathrm{O}\\) and \\(\\mathrm{NaCl}\\) themselves as the components, but other options exist. For instance, specifying the amount of \\(\\mathrm{Na}^{+}\\)ions and \\(\\mathrm{H}\\) atoms in the system automatically fixes the \\(\\mathrm{Cl}\\) ion and \\(\\mathrm{O}\\) atom content, assuming the compounds are present in their stoichiometric ratios (assuming they are defect free).\nIt is worth noting that even though \\(\\mathrm{NaCl}\\) dissolves into its constituent ions in the water, the ions are present in the same ratio as their parent compound. In other words, the \\(\\mathrm{Na}\\) and \\(\\mathrm{Cl}\\) ion concentrations are coupled and not independent. If they were independent, the system would be described as three components, since knowing one ion concentration would not inherently imply the value of the other ion's concentration."
},
{
"idx": 259,
"question": "Is solid ductile cast iron (ferrite solid solution + embedded graphite spheres) a two-phase material system?",
"answer": "Yes, solid ductile cast iron is a two-phase system because it consists of a ferrite solid solution with embedded graphite spheres, which are chemically and structurally distinct phases separated by physical boundaries."
},
{
"idx": 259,
"question": "Is solid sodium chloride (salt, NaCl) a two-phase material system?",
"answer": "No, solid sodium chloride is a single-phase compound with a fixed stoichiometry and orderly arrangement of ions in the crystal, not a two-phase system."
},
{
"idx": 259,
"question": "Is liquid bronze (Cu + Sn liquid solution) a two-phase material system?",
"answer": "No, liquid bronze is a single-phase liquid solution, as solutions by definition are single-phase."
},
{
"idx": 259,
"question": "Is solid gray cast iron (ferrite solid solution + embedded graphite flakes) a two-phase material system?",
"answer": "Yes, solid gray cast iron is a two-phase system because it consists of a ferrite solid solution with embedded graphite flakes, which are chemically and structurally distinct phases separated by physical boundaries."
},
{
"idx": 259,
"question": "Is solid aluminum featuring dissolved silicon a two-phase material system?",
"answer": "No, solid aluminum with dissolved silicon is a single-phase solid solution, as solutions by definition are single-phase."
},
{
"idx": 259,
"question": "Is solid lead-tin solder (a mixture of Pb-rich and Sn-rich solid solutions) a two-phase material system?",
"answer": "Yes, solid lead-tin solder is a two-phase system because it consists of Pb-rich and Sn-rich solid solutions, which are chemically and structurally distinct phases separated by physical boundaries."
},
{
"idx": 259,
"question": "Is partially melted aluminum a two-phase material system?",
"answer": "No, partially melted aluminum is not a two-phase system in the context of chemically distinct phases; it merely has structurally distinct volumes (crystalline and liquid)."
},
{
"idx": 259,
"question": "Is frozen water with trapped air bubbles a two-phase material system?",
"answer": "Yes, frozen water with trapped air bubbles is a two-phase system because it consists of ice and air bubbles, which are chemically and structurally distinct phases separated by physical boundaries."
},
{
"idx": 259,
"question": "Is epoxy embedded with carbon fibers a two-phase material system?",
"answer": "Yes, epoxy embedded with carbon fibers is a two-phase system because it consists of epoxy and carbon fibers, which are chemically and structurally distinct phases separated by physical boundaries."
},
{
"idx": 260,
"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": 260,
"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": 260,
"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": 260,
"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": 261,
"question": "For a solution, which of the following is present in the higher concentration?\na) Solvent\nb) Solute",
"answer": "Solvent. By definition, solvent is the element/compound that is present in a solution in the greatest amount."
},
{
"idx": 262,
"question": "A liquidus line separates which of the following combinations of phase fields?\na) Liquid and Liquid \\(+\\alpha\\)\nb) \\(\\alpha\\) and Liquid \\(+\\alpha\\)\nc) \\(\\alpha\\) and \\(\\alpha+\\beta\\)\nd) Liquid \\(+\\alpha\\) and \\(\\alpha+\\beta\\)",
"answer": "A liquidus line separates Liquid and Liquid \\(+\\alpha\\) phase fields."
},
{
"idx": 263,
"question": "A solvus line separates which of the following pairs of phase fields?\na) Liquid and Liquid \\(+\\alpha\\)\nb) \\(\\alpha \\operatorname{and}\\) Liquid \\(+\\alpha\\)\nc) \\(\\alpha\\) and \\(\\alpha+\\beta\\)\n\nd) Liquid \\(+\\alpha\\) and \\(\\alpha+\\beta\\)",
"answer": "A solvus line separates \\(\\alpha\\) and \\(\\alpha+\\beta\\) phase fields."
},
{
"idx": 264,
"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": 264,
"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": 264,
"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": 265,
"question": "At a eutectic point on a binary temperature-composition phase diagrams, how many phases are present when the system is at equilibrium?\na) 0\nb) 1\nc) 2\nd) 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": 266,
"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.\n- \\(\\mathrm{MgGa}\\)\n- \\(\\mathrm{Mg}_{2} \\mathrm{Ga}\\)\n- \\(\\mathrm{MgGa}_{2}\\)\n- \\(\\mathrm{Mg}_{3} \\mathrm{Ga}_{2}\\)",
"answer": "Probably the easiest way to solve this problem is to begin by computing concentrations of \\(\\mathrm{Mg}\\) and \\(\\mathrm{Ga}\\) in terms of atom percents; from these values, it is possible to determine the ratio of \\(\\mathrm{Mg}\\) to \\(\\mathrm{Ga}\\) in this compound. Atom percent \\(\\left(C_{\\mathrm{Mg}}^{\\prime}\\right.\\) and \\(\\left.C_{\\mathrm{Ga}}^{\\prime}\\right)\\) are computed using Equation as follows (incorporating atomic weight values of \\(24.31 \\mathrm{~g} / \\mathrm{mol}\\) and \\(69.72 \\mathrm{~g} / \\mathrm{mol}\\), respectively for \\(\\mathrm{Mg}\\) and \\(\\mathrm{Ga}\\) ):\n\\[\n\\begin{aligned}\nC_{\\mathrm{Mg}}^{\\prime} & =\\frac{C_{\\mathrm{Mg}} A_{\\mathrm{Ga}}}{C_{\\mathrm{Mg}} A_{\\mathrm{Ga}}+C_{\\mathrm{Ga}} A_{\\mathrm{Mg}}} \\times 100 \\\\\n& =\\frac{(41.1 \\mathrm{wt} \\%)(69.72 \\mathrm{~g} / \\mathrm{mol})}{(41.1 \\mathrm{wt} \\%)(69.72 \\mathrm{g} / \\mathrm{mol})+(58.9 \\mathrm{wt} \\%)(24.31 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n& =67.7 \\mathrm{at} \\%\n\\end{aligned}\n\\]\nBecause this intermetallic compound is composed of only \\(\\mathrm{Mg}\\) and \\(\\mathrm{Ga}, C_{\\mathrm{Ga}}^{\\prime}\\) will equal 33.3 at\\%. Finally, the formula for this intermetallic compound is \\(\\mathrm{Mg}_{2} \\mathrm{Ga}\\), since the magnesium-to-gallium atom percent ratio is 2:1."
},
{
"idx": 267,
"question": "A eutectoid reaction involves which of the following phases?\na) One liquid and one solid\nb) One liquid and two solid\nc) Two liquids and one solid\nd) Three solid",
"answer": "A eutectoid reaction involves three solid phases."
},
{
"idx": 268,
"question": "A peritectic reaction involves which of the following combinations of phase fields?\na) One liquid and one solid\nb) One liquid and two solid\nc) two liquids and one solid\nd) Three solid",
"answer": "A peritectic reaction involves one liquid and two solid phases."
},
{
"idx": 269,
"question": "For a congruent phase transformations there are\na) no composition alterations.\nb) compositional alterations.",
"answer": "There are no compositional alterations for a congruent phase transformation."
},
{
"idx": 270,
"question": "Which of the following have a significant influence on a material's electrical resistivity?\na) impurity concentration\nb) temperature\nc) grain size\nd) cold work\ne) 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": 271,
"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": "The conductivity of a metal is given as the product of free electron concentration, charge of one electron, and electron mobility. The free electron concentration is essentially constant at around \\(10^{29}\\) electrons per cubic meter for a typical metal. Changes of this value are typically insignificant. The charge of an electron is a fundamental constant. Electron spin is a quantum number that is not related to electrical conduction within the scope of this text. As more defects are introduced into a crystalline lattice there is an increased probability of electron scattering. As the scattering frequency increases, the electron drift velocity decreases and the electron mobility decreases."
},
{
"idx": 272,
"question": "Compute the atom concentration of aluminum at room temperature given the mass density of aluminum is 2.7 g/cm3 and the atomic weight of aluminum is 27 g/mol.",
"answer": "The atom concentration of aluminum at room temperature is: ((2.7 g/cm3)(6.022 x 10^23 atom/mol))/(27 g/mol) = 6.022 x 10^22 atom/cm3"
},
{
"idx": 272,
"question": "Compute the free electron concentration of aluminum at room temperature given the resistivity of aluminum is 2.63 x 10^-8 Ω·m, the electron mobility of aluminum is 0.0012 m2/(V·s), and the elementary charge is 1.6 x 10^-19 C.",
"answer": "The free electron concentration at room temperature is: 1/((2.63 x 10^-8 Ω·m)(1.6 x 10^-19 C)(0.0012 m2/(V·s))) = 1.98 x 10^29 electron/m3"
},
{
"idx": 272,
"question": "Compute the number of free electrons donated by each aluminum atom, on average, given the atom concentration is 6.022 x 10^22 atom/cm3 and the free electron concentration is 1.98 x 10^29 electron/m3.",
"answer": "The number of free electrons donated by each atom, on average, is: (1.98 x 10^29 electron/m3)/(6.022 x 10^28 atom/m3) = 3.29 electron/atom"
},
{
"idx": 273,
"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 \\cdot \\mathrm{cm}\\)\nThe electron mobility of the metal is \\([\\mathrm{m}] \\mathrm{cm}^{2} /(\\mathrm{V} \\cdot \\mathrm{s})\\)\nThe mass density of the metal is \\([\\mathrm{d}] \\mathrm{g} / \\mathrm{cm}^{3}\\)\nThe atomic weight of the metal is \\([\\mathrm{w}] \\mathrm{g} / \\mathrm{mol}\\)",
"answer": "The atom concentration of aluminum at room temperature is:\n\\[\n\\rho_{\\max } N_{A}\n\\]\nThe free electron concentration at room temperature is:\n\\[\n\\frac{1}{\\rho e \\mu_{e}}\n\\]\nTherefore, the number of free electrons donated by each atom, on average, is:\n\\[\n\\frac{\\frac{1}{\\rho} e \\mu_{e}}{\\frac{\\rho_{\\max } N_{A}}{A}}=\\frac{A}{\\left(\\rho e \\mu_{e}\\right)\\left(\\rho_{\\max } N_{A}\\right)}=\\frac{[w]}{\\left([r]\\left[1.6 \\times 10^{-19}\\right][\\mathrm{m}]\\right)\\left([d]\\left[6.022 \\times 2^{23}\\right]\\right)}\n\\]"
},
{
"idx": 274,
"question": "When an electric field is applied, in which direction are the free electrons accelerated?\na) Opposite to the direction of the electric field.\nb) 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": 275,
"question": "The electrical conductivity of an intrinsic semiconductor is\na) characteristic of the high-purity metal.\nb) due to the presence of impurities.",
"answer": "The electrical conductivity of an intrinsic semiconductor is characteristic of the high-purity material."
},
{
"idx": 276,
"question": "How do the electrical conductivities of metals compare with those of semiconductors?\na) \\(\\sigma_{\\text {metals }}>\\sigma_{\\text {semiconductors }}\\)\nb) \\(\\sigma_{\\text {metals }}=\\sigma_{\\text {semiconductors }}\\)\nc) \\(\\sigma_{\\text {metals }}<\\sigma_{\\text {semiconductors }}\\)",
"answer": "The electrical conductivities of metals are greater than those of semiconductors."
},
{
"idx": 277,
"question": "How does increasing temperature affect the concentration of both electrons and holes in an intrinsic semiconductor?\na) Increases the concentration.\nb) Decreases the concentration.\nc) 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": 278,
"question": "Which type of charge carrier will be introduced into a semiconductor by the presence of an acceptor impurity?\na) Electron\nb) Hole",
"answer": "The presence of an acceptor impurity introduces holes into a semiconductor."
},
{
"idx": 279,
"question": "Which type of charge carrier will be introduced into a semiconductor by the presence of a donor impurity?\na) Impurity\nb) Hole",
"answer": "The presence of a donor impurity introduces electrons into a semiconductor."
},
{
"idx": 280,
"question": "The electrical conductivity of an extrinsic semiconductor is\na) Characteristic of the high-purity material.\nb) due to the presence of impurities.",
"answer": "The electrical conductivity of an extrinsic semiconductor is due to the presence of impurities."
},
{
"idx": 281,
"question": "For an \\(n\\)-type semiconductor, which type of charge carrier is present in the greater concentration?\na) Hole\nb) Electron",
"answer": "For an \\(n\\)-type semiconductor, electrons (i.e., negative charge carriers) are present in a greater concentration than holes."
},
{
"idx": 282,
"question": "For an \\(n\\)-type semiconductor\na) Concentration \\(_{\\text {electrons }}<\\) concentration \\(_{\\text {holes }}\\)\nb) Concentration \\(_{\\text {electrons }}=\\) concentration \\(_{\\text {holes }}\\)\nc) 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": 283,
"question": "For a \\(p\\)-type semiconductor, which type of charge carrier is present in the greater concentration?\na) Holes\nb) Electrons",
"answer": "For a \\(p\\)-type semiconductor, holes, (i.e., positive charge carriers) are present in a greater concentration than electrons."
},
{
"idx": 284,
"question": "For a \\(p\\)-type semiconductor\na) Concentration \\(_{\\text {electrons }}<\\) concentration \\(_{\\text {holes }}\\)\nb) Concentration \\(_{\\text {electrons }}=\\) concentration \\(_{\\text {holes }}\\)\nc) 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": 285,
"question": "In order for a semiconductor to exhibit extrinsic electrical characteristics, relatively high impurity concentrations are required.\na) True\nb) 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": 286,
"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 \\cdot \\mathbf{m})^{-1} & \\boldsymbol{n}\\left(\\mathbf{m}^{-3}\\right) & \\boldsymbol{p}\\left(\\mathbf{m}^{-3}\\right) \\\\\n\\text { Intrinsic } & 2.5 \\times 10^{-6} & 3.0 \\times 10^{13} & 3.0 \\times 10^{13} \\\\\n\\text { Extrinsic } & 3.6 \\times 10^{-5} & 4.5 \\times 10^{14} & 2.0 \\times 10^{12}\n\\end{array}\n\\]\nCalculate electron and hole mobilities.\n\\(\\mu_{\\mathrm{e}}=\\)\n\\(\\mu_{\\mathrm{h}}=\\)",
"answer": "In order to solve for the electron and hole mobilities for InP, we must write conductivity expressions for the two materials, of the form of Equation-i.e.,\n\\[\n\\sigma=n|e| \\mu_{e}+p|e| \\mu_{h}\n\\]\nFor the intrinsic material\n\\[\n\\begin{array}{l}\n2.5 \\times 10^{-6}(\\Omega \\cdot \\mathrm{m})^{-1}=\\left(3.0 \\times 10^{13} \\mathrm{~m}^{-3}\\right)\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right) \\mu_{e} \\\\\n\\quad+\\left(3.0 \\times 10^{13} \\mathrm{~m}{ }^{-3}\\right)\\left(1.602 \\times 10^{-10} \\mathrm{C}\\right) \\mu_{h}\n\\end{array}\n\\]\nwhich reduces to\n\\[\n0.52=\\mu_{e}+\\mu_{h}\n\\]\nWhereas, for the extrinsic InP\n\\[\n\\begin{array}{l}\n3.6 \\times 10^{-5}(\\Omega \\cdot \\mathrm{m})^{-1}=\\left(4.5 \\times 10^{14} \\mathrm{~m}^{-3}\\right)\\left(1.602 × 10^{-19} \\mathrm{C}\\right) \\mu_{e} \\\\ \n\\quad+\\left(2.0 \\times 10^{12} \\mathrm{~m}^{-3}\\right)\\left(1.602\\times 10^{-19} \\mathrm{C}\\right) \\mu_{h}\n\\end{array}\n\\]\n\nwhich may be simplified to\n\n\\[\n112.4=225 \\mu_{e}+\\mu_{h}\n\\]\n\nThus, we have two independent expressions with two unknown mobilities. Upon solving these equations simultaneously, we get \\(\\mu_{e}=0.50 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\) and \\(\\mu_{h}=0.02 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\)."
},
{
"idx": 287,
"question": "Which of the following are preferred for semiconducting devices?\n\na) Single crystals\n\nb) Polycrystalline materials",
"answer": "Single crystals are preferred for semiconducting devices because grain boundaries are deleterious to the performance of electronic phenomena."
},
{
"idx": 288,
"question": "As temperature increases, the electrical conductivities of polymers and ionic ceramics\na) Increase\nb) Decrease",
"answer": "As temperature increases, the electrical conductivities of polymers and ionic ceramics increase."
},
{
"idx": 289,
"question": "Most polymers and ionic ceramics have energy band gap structures that are most similar to those of\na) Insulators\nb) Semiconductors\nc) Metals",
"answer": "Most polymers and ionic ceramics have wide energy band gaps; thus, they are most like those of insulators."
},
{
"idx": 290,
"question": "The charge carriers in ionic ceramics and polymers can be\na) Electrons\nb) Holes\nc) Anion\nd) Cations",
"answer": "The charge carriers in ionic ceramics and polymers can be electrons, holes, anions, and/or cations."
},
{
"idx": 291,
"question": "[a] 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": 291,
"question": "[b] 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": 291,
"question": "[c] 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": 291,
"question": "[d] Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.",
"answer": "F"
},
{
"idx": 291,
"question": "[e] The relatively low melting point of aluminum is often considered a significant limitation for structural applications.",
"answer": "T"
},
{
"idx": 292,
"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": 292,
"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": 292,
"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": 292,
"question": "[d] Compared to other metals, like steel, pure aluminum is very resistant to failure via fatigue.",
"answer": "F"
},
{
"idx": 292,
"question": "[e] Aluminum exhibits one of the highest melting points of all metals, which makes it difficult and expensive to cast.",
"answer": "F"
},
{
"idx": 293,
"question": "[a] Copper has a higher elastic modulus than aluminum.",
"answer": "T"
},
{
"idx": 293,
"question": "[b] The density of copper is closer to that of aluminum than it is to iron.",
"answer": "F"
},
{
"idx": 293,
"question": "[c] Bronze is an alloy of copper and zinc.",
"answer": "F"
},
{
"idx": 293,
"question": "[d] Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates.",
"answer": "T"
},
{
"idx": 293,
"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": 294,
"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": 294,
"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": 294,
"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": 294,
"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": 294,
"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": 295,
"question": "[a] Nickel is majority component (by mass) in certain superalloys such as Waspaloy TM.",
"answer": "T"
},
{
"idx": 295,
"question": "[b] Tungsten is the lowest density metal that has structural use.",
"answer": "F"
},
{
"idx": 295,
"question": "[c] Tantalum offers extremely good corrosion resistance, especially at low temperatures.",
"answer": "T"
},
{
"idx": 295,
"question": "[d] Magnesium metal is very similar to aluminum, in terms of its physical and mechanical properties.",
"answer": "T"
},
{
"idx": 295,
"question": "[e] Beryllium metal is commonly used as an alloying agent in copper metal.",
"answer": "T"
},
{
"idx": 296,
"question": "Increasing the alumina \\(\\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right)\\) content of fireclays results in\na) an increase in maximum service temperature.\nb) 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": 297,
"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}\\)\na) True\nb) 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": 298,
"question": "As the porosity of refractory ceramic bricks increases, what happens to the strength?",
"answer": "Strength decreases."
},
{
"idx": 298,
"question": "As the porosity of refractory ceramic bricks increases, what happens to the chemical resistance?",
"answer": "Chemical resistance decreases."
},
{
"idx": 298,
"question": "As the porosity of refractory ceramic bricks increases, what happens to the thermal insulation?",
"answer": "Thermal insulation increases."
},
{
"idx": 299,
"question": "The presence of silica \\(\\left(\\mathrm{SiO}_{2}\\right)\\) in basic refractory ceramics is beneficial to their high-temperature performance.\na) True\nb) False",
"answer": "False. The presence of silica \\(\\left(\\mathrm{SiO}_{2}\\right)\\) in basic refractory ceramics is deleterious on their hightemperature performance."
},
{
"idx": 300,
"question": "Basic refractory ceramics are often used for the containment of slags that are rich in\na) silica\nb) \\(\\mathrm{CaO}\\)\nc) \\(\\mathrm{MgO}\\)",
"answer": "Basic refractory ceramics are often used for the containment of slags that are rich in \\(\\mathrm{CaO}\\) and/or \\(\\mathrm{MgO}\\)."
},
{
"idx": 301,
"question": "Silica refractory ceramics are often used for the containment of slags that are rich in\na) silica\nb) \\(\\mathrm{CaO}\\)\nc) \\(\\mathrm{Mg} \\mathrm{O}\\)",
"answer": "Silica refractory ceramics are often used for the containment of slags that are rich in silica."
},
{
"idx": 302,
"question": "Given that the bond strength across the fiber-epoxy interface is [s] MPa, and the shear yield strength of the epoxy is [y] MPa, determine which strength value (bond strength or shear yield strength) should be used as the critical shear strength (τ_c) for calculating the minimum fiber length.",
"answer": "Since τ_fiber-matrix < τ_y, matrix, the critical shear strength τ_c should be equal to the fiber-matrix bond strength, i.e., τ_c = τ_fiber-matrix = [s] MPa."
},
{
"idx": 302,
"question": "Compute the minimum fiber length (l), in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite, given the tensile strength of the carbon fibers is [f] MPa and the fiber diameter is [d] mm.",
"answer": "The minimum fiber length (l) is calculated using the formula: l = (σ_f * d) / (2 * τ_c) = ([f] MPa * [d] mm) / (2 * [s] MPa) = [f][d] / (2[s] * 1000) mm."
},
{
"idx": 303,
"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?\na) 0\nb) \\(\\frac{1}{5}\\)\nc) \\(\\frac{3}{8}\\)\nd) \\(\\frac{3}{4}\\)\ne) 1",
"answer": "For an aligned fibrous composite when the stress is applied parallel to the fibers, the reinforcement efficiency is 1 ."
},
{
"idx": 304,
"question": "For an aligned fibrous composite, when a stress is applied perpendicular to the fibers, what is the reinforcement efficiency?\na) 0\nb) \\(\\quad \\frac{1}{5}\\)\nc) \\(\\frac{3}{8}\\)\nd)\n\\(\\frac{3}{4}\\)\ne) 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": 305,
"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?\na) 0\nb) \\(\\frac{1}{5}\\)\nc) \\(\\frac{3}{8}\\)\nd) \\(\\frac{3}{4}\\)\ne) 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": 306,
"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?\na) 0\nb) \\(\\frac{1}{{ }_{5}}\\)\nc) \\(\\frac{3}{8}\\)\nd) \\(\\frac{3}{{ }_{4}}\\)\ne) 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": 307,
"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?\na) Fibers\nb) 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": 308,
"question": "Once the fibers fail in a fibrous composite, catastrophic failure of the piece takes place.\na) True\nb) 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": 309,
"question": "How are continuous fibers typically oriented in fibrous composites?\na) Aligned\nb) Partially oriented\nc) Randomly oriented",
"answer": "Continuous fibers are typically aligned in fibrous composites."
},
{
"idx": 310,
"question": "How are discontinuous fibers typically oriented in fibrous composites?\na) Aligned\nb) Partially oriented",
"answer": "Discontinuous fibers may be aligned, partially oriented, and randomly oriented in fibrous composites."
},
{
"idx": 311,
"question": "If the fiber orientation is random, which type of fibers is normally used?\na) Discontinuous\nb) Continuous",
"answer": "Discontinuous fibers are normally used when the fiber orientation is random."
},
{
"idx": 312,
"question": "Select \\(\\mathrm{T} / \\mathrm{F}\\) for each of the following statements:\n[a] Composites are single-phase materials by definition.\n[b] The term \"composite\" applies to materials that feature polymeric materials only.\n[c] Structural composites are, in general, highly regarded for their specific strengths.\n[d] Composites featuring continuous and aligned fibers for reinforcement generally offer properties that are highly isotropic compared to most metals (random polycrystals).",
"answer": "([a] \\mathrm{F}\\)}\n\\([b] \\mathrm{F}\\)\n[c] \\(\\mathrm{T}\\)\n[d] \\(\\mathrm{F}\\)"
},
{
"idx": 313,
"question": "Which of the following materials are typically used as fibers?\na) Graphite/carbon\nb) Silicon carbide\nc) Silicon nitride\nd) Aluminum oxide\ne) Glass\nf) Boron\ng) Steel\nh) Tungsten\ni) Molybdenum",
"answer": "Graphite, silicon carbide, glass, boron, and aluminum oxide are typically used as fibers."
},
{
"idx": 314,
"question": "Match the fiber type 'Whiskers' with its description.",
"answer": "Whiskers are single crystals with extremely large length-to-diameter ratios."
},
{
"idx": 314,
"question": "Match the fiber type 'Fibers' with its description.",
"answer": "Fibers are polycrystalline or amorphous materials with small diameters."
},
{
"idx": 314,
"question": "Match the fiber type 'Wires' with its description.",
"answer": "Wires are metals having relatively large diameters."
},
{
"idx": 315,
"question": "For a composite material, which phase normally has the higher elastic modulus?\na) Fiber phase\nb) Matrix phase",
"answer": "The fiber phase normally has a higher elastic modulus than the matrix phase."
},
{
"idx": 316,
"question": "For a composite material, how does the ductility of the matrix phase normally compare with the ductility of the dispersed phase?\na) more ductile\nb) less ductile",
"answer": "A composite's matrix phase is normally more ductile than the dispersed phase."
},
{
"idx": 317,
"question": "Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strengths.\na) True\nb) False",
"answer": "True. Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strength."
},
{
"idx": 318,
"question": "Which of aramid and metal fibers have higher strength-to-weight ratios?\na) Aramid fibers\nb) Metal fibers",
"answer": "Aramid fibers have higher strength-to-weight ratios than metal fibers."
},
{
"idx": 319,
"question": "Do carbon fiber-reinforced composites have relatively high strengths?",
"answer": "Carbon fiber-reinforced composites have relatively high strengths."
},
{
"idx": 319,
"question": "Do carbon fiber-reinforced composites have relatively high stiffnesses?",
"answer": "Carbon fiber-reinforced composites have relatively high stiffnesses."
},
{
"idx": 319,
"question": "Do carbon fiber-reinforced composites have high service temperatures (>200°C)?",
"answer": "Carbon fiber-reinforced composites have high service temperatures (>200°C)."
},
{
"idx": 320,
"question": "Which of the following materials are typically used as whiskers?\na) graphite/carbon\nb) silicon carbide\nc) silicon nitride\nd) aluminum oxide\ne) glass\nf) boron\ng) steel\nh) tungsten\ni) molybdenum",
"answer": "Graphite, silicon carbide, silicon nitride, and aluminum oxide are typically used as whiskers."
},
{
"idx": 321,
"question": "Which of the following materials are typically used as wires in composites?\na) graphite/carbon\nb) silicon carbide\nc) silicon nitride\n\\(\\mathrm{d})\\) aluminum oxide\ne) glass\nf) boron\ng) steel\nh) tungsten\ni) molybdenum",
"answer": "Steel, tungsten, and molybdenum are typically used as wires in composites."
},
{
"idx": 322,
"question": "Compared to other ceramic materials, ceramic-matrix composites have better/higher\na) fracture toughnesses\nb) oxidation resistance\nc) stability at elevated temperatures",
"answer": "Ceramic-matrix composites have higher fracture toughnesses than other ceramic materials."
},
{
"idx": 323,
"question": "Which of the following materials are typically used as stabilizers in transformation-toughened ceramic-matrix composites?\na) \\(\\mathrm{CaO}\\)\nb) \\(\\mathrm{MgO}\\)\nc) \\(\\mathrm{Y}_{2} \\mathrm{O}_{3}\\)\nd) \\(\\mathrm{CeO}\\)\ne) \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\)\nf) \\(\\mathrm{SiC}\\)",
"answer": "(\\mathrm{CaO}, \\mathrm{MgO}, \\mathrm{Y}_{2} \\mathrm{O}_{3}\\), and \\(\\mathrm{CeO}\\) are typically used as stabilizers in transformation-toughened ceramicmatrix composites."
},
{
"idx": 324,
"question": "Do carbon-carbon composites exhibit high tensile moduli at elevated temperatures?",
"answer": "Carbon-carbon composites have high tensile moduli at elevated temperatures."
},
{
"idx": 324,
"question": "Do carbon-carbon composites exhibit high tensile strengths at elevated temperatures?",
"answer": "Carbon-carbon composites have high tensile strengths at elevated temperatures."
},
{
"idx": 324,
"question": "Are carbon-carbon composites resistant to creep?",
"answer": "Carbon-carbon composites are highly resistant to creep."
},
{
"idx": 324,
"question": "Do carbon-carbon composites have large fracture toughness values?",
"answer": "Carbon-carbon composites have large fracture toughness values."
},
{
"idx": 324,
"question": "Do carbon-carbon composites have high thermal conductivities?",
"answer": "Carbon-carbon composites have high thermal conductivities."
},
{
"idx": 324,
"question": "Do carbon-carbon composites have low coefficients of thermal expansion?",
"answer": "Carbon-carbon composites have low coefficients of thermal expansion."
},
{
"idx": 324,
"question": "Are carbon-carbon composites resistant to oxidation at elevated temperatures?",
"answer": "The answer does not specify resistance to oxidation at elevated temperatures."
},
{
"idx": 324,
"question": "Are carbon-carbon composites low cost?",
"answer": "The answer does not specify low cost."
},
{
"idx": 325,
"question": "Laminar composites have high strengths in all directions (in three dimensions).\na) True\nb) False",
"answer": "False. Laminar composites have high strengths in all directions only in their two-dimensional planes."
},
{
"idx": 326,
"question": "The strong outer sheets of sandwich panels are separated by a layer of material that is\na) less dense than the outer sheet material\nb) 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": 327,
"question": "To what temperature would \\(23.0 \\mathrm{~kg}\\) of some material at \\(100^{\\circ} \\mathrm{C}\\) be raised if \\(255 \\mathrm{~kJ}\\) of heat is supplied? Assume a \\(\\mathrm{c}_{\\mathrm{p}}\\) value of \\(423 \\mathrm{~J} / \\mathrm{kg}-\\mathrm{K}\\) for this material.\n(A) \\(26.2^{\\circ} \\mathrm{C}\\)\n(B) \\(73.8^{\\circ} \\mathrm{C}\\)\n(C) \\(126^{\\circ} \\mathrm{C}\\)\n(D) \\(152^{\\circ} \\mathrm{C}\\)",
"answer": "The change in temperature \\((\\Delta T)\\) when \\(\\Delta Q\\) of heat is supplied to a mass of material \\(m\\) is equal to a revised form of Equation as follows:\n\\[\n\\frac{\\Delta Q}{c_{p} m}=\\Delta T=T_{f}-T_{0}\n\\]\nwhere \\(T_{0}\\) and \\(T_{f}\\) are the initial and final temperatures, respectively. Solving this equation for the final temperature yields the following:\n\\[\nT_{f}=T_{0}+\\frac{\\Delta Q}{c_{p} m}\n\\]\nUsing values provided in the problem statement-that is\n\\[\n\\begin{array}{l}\nT_{0}=100^{\\circ} \\mathrm{C} \\\\\n\\Delta Q=255 \\mathrm{~kJ}=255,000 \\mathrm{~J} \\\\\nc_{p}=423 \\mathrm{~J} / \\mathrm{kg}-\\mathrm{K} \\\\\nm=23.0 \\mathrm{~kg}\n\\end{array}\n\\]\nWe compute the final temperature as follows:\n\\[\nT_{f}=100^{\\circ} \\mathrm{C}+\\frac{255,000 \\mathrm{~J}}{(423 \\mathrm{~J} / \\mathrm{kg}-\\mathrm{K})(23.0 \\mathrm{~kg})}=126^{\\circ} \\mathrm{C}\n\\]\nwhich is answer \\(\\mathrm{C}\\)."
},
{
"idx": 328,
"question": "A rod of some material \\(0.50 \\mathrm{~m}\\) long elongates \\(0.40 \\mathrm{~mm}\\) on heating from \\(50^{\\circ} \\mathrm{C}\\) to \\(151^{\\circ} \\mathrm{C}\\). What is the value of the linear coefficient of thermal expansion for this material?\n(A) \\(5.30 \\times 10^{-6}\\left({ }^{\\circ} \\mathrm{C}\\right)^{-1}\\)\n(B) \\(7.92 \\times 10^{-6}\\left({ }^{\\circ} \\mathrm{C} \\mathrm{C}\\right)^{-1}\\)\n(C) \\(1.60 \\times 10^{-5}\\left({ }^{\\circ} \\mathrm{C}\\right)^{-1}\\)\n(D) \\(1.24 \\times 10^{-6}\\left({ }^{\\circ} \\mathrm{C}_{-}^{1}\\right.\\)",
"answer": "The relationship among change in length \\((\\Delta l)\\), coefficient of thermal expansion \\(\\left(\\alpha_{i}\\right)\\), and change in temperature \\((\\Delta T)\\) is given by Equation-that is\n\\[\n\\frac{\\Delta l}{l_{0}}=\\alpha_{i} \\Delta T=\\alpha_{l}\\left(T_{f}-T_{0}\\right)\n\\]\nwhere \\(T_{f}\\) and \\(T_{0}\\) are the final and initial temperatures, respectively. Thus, the linear coefficient of thermal expansion may be determined using a rearranged form of the above equation as follows:\n\\[\n\\alpha_{l}=\\frac{\\Delta l}{l_{0}\\left(T_{f}-T_{0}\\right)}\n\\]\nIncorporation of values of \\(\\Delta l, l_{0}, T_{f}\\) and \\(T_{0}\\) provided in the problem statements leads to the following value of \\(\\alpha_{l}\\)\n\\[\n\\alpha_{l}=\\frac{\\left(0.40 \\times 10^{-3} \\mathrm{~m}\\right)}{\\left(0.50 \\mathrm{~m}\\right)\\left(151^{\\circ} \\mathrm{C}-50^{\\circ} \\mathrm{C}\\right)}=7.92 \\times 10^{-6}\\left({ }^{\\circ} C\\right)^{-1}\n\\]\nwhich is answer B."
},
{
"idx": 329,
"question": "A 10 meter long square bar of 316 stainless steel (edge length of \\(5 \\mathrm{~cm}\\), with a modulus of \\(193 \\mathrm{GPa}\\) and a yield point of \\(290 \\mathrm{MPa}\\) ) is bolted securely in place when its installation temperature was around \\([1]^{\\circ} \\mathrm{C}\\). What is the expected thermal stress in the bar when its service temperature reaches \\([F]^{o} \\mathrm{C}\\) ? Enter a negative indicate compressive stress, if necessary. The thermal expansion coefficient of 316 stainless steel is \\(16.0 \\times 10^{-6} \\mathrm{I} /{ }^{\\circ} \\mathrm{C}\\).",
"answer": "In this scenario, the bar is installed and secured at a warmer temperature than its ultimate service temperature. Therefore, as the temperature drops the steel would like to contract and occupy less length, but it cannot do so. The bar instead endures compressive thermal stress via Hooke's Law (provided the thermal strain is elastic).\n\\[\n\\sigma_{t h}=E \\varepsilon_{t h}=E \\alpha_{L} \\Delta T=E \\alpha_{L}\\left(T_{o}-T_{f}\\right)=(193,000 \\mathrm{MPa})\\left(16 \\times 10^{-6} 1 /{ }^{\\circ} \\mathrm{C}\\right)\\left(\\left[I\\right]^{o} \\mathrm{C}-[F]^{o} \\mathrm{C}\\right)\n\\]"
},
{
"idx": 330,
"question": "Which of the following \\(1 \\mathrm{~kg}\\) samples is expected to change temperature the least if \\(100 \\mathrm{~kJ}\\) 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} \\mathrm{C}\\).\n(a) Aluminum\n(b) Copper\n(c) Gold\n(d) Borosilicate Glass\n(e) Polystyrene",
"answer": "In this scenario, the material with the highest heat capacity will change temperature the least for a given increase in enthalpy. Polystyrene features the highest heat capacity, so it will experience the smallest temperature change, according to:\n\\[\n\\Delta T=\\frac{Q}{m c_{p}}\n\\]"
},
{
"idx": 331,
"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?\n(a) Polystyrene\n(b) Aluminum\n(c) Copper\n(d) Gold\n(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": 332,
"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) \\(\\mathrm{C}\\) (diamond)",
"answer": "Diamond has the highest thermal conductivity of these options."
},
{
"idx": 333,
"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": 334,
"question": "Which of the following materials display(s) antiferromagnetic behavior?\na) Aluminum oxide\nb) Copper\nc) Aluminum\nd) Titanium\ne) Iron ( \\(\\alpha\\) ferrite)\nf) Nickel\ng) MnO\nh) \\(\\mathrm{Fe}_{3} \\mathrm{O}_{4}\\)\ni) \\(\\mathrm{NiFe}_{2} \\mathrm{O}_{4}\\)",
"answer": "(\\mathrm{MnO}\\) is an antiferromagnetic material."
},
{
"idx": 335,
"question": "Which of the following are ferrimagnetic materials?\na) Aluminum oxide\nb) Copper\nc) Aluminum\nd) Titanium\ne) Iron ( \\(\\alpha\\) ferrite)\nf) Nickel\ng) 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": 336,
"question": "Calculate the number of Bohr magnetons per unit cell (n_B) given the saturation magnetization (M_s = 1.0 x 10^4 A/m), the unit cell edge length (a = 1.2376 x 10^-9 m), and the Bohr magneton (μ_B = 9.27 x 10^-24 A·m^2/BM).",
"answer": "n_B = (M_s * a^3) / μ_B = (1.0 x 10^4 A/m) * (1.2376 x 10^-9 m)^3 / (9.27 x 10^-24 A·m^2/BM) = 2.04 Bohr magnetons/unit cell."
},
{
"idx": 336,
"question": "Determine the number of Bohr magnetons per formula unit given that there are 8 formula units per unit cell and the calculated n_B = 2.04 Bohr magnetons/unit cell.",
"answer": "Bohr magnetons per formula unit = n_B / 8 = 2.04 / 8 = 0.255 Bohr magnetons/formula unit."
},
{
"idx": 336,
"question": "Calculate the net magnetic moment contribution per formula unit from the Fe^3+ ions, given that there are 2 Fe^3+ ions on a sites and 3 Fe^3+ ions on d sites per formula unit, each with 5 Bohr magnetons, and their magnetic moments are aligned antiparallel.",
"answer": "Net magnetic moment from Fe^3+ ions = (2 * 5 BM) - (3 * 5 BM) = 10 BM - 15 BM = -5 BM per formula unit."
},
{
"idx": 336,
"question": "Compute the number of Bohr magnetons associated with each Y^3+ ion, given the net magnetic moment per formula unit from Fe^3+ ions is -5 BM, the Bohr magnetons per formula unit is 0.255 BM, and there are 3 Y^3+ ions per formula unit.",
"answer": "Number of Bohr magnetons per Y^3+ ion = (0.255 BM + 5 BM) / 3 = 5.255 BM / 3 = 1.75 BM."
},
{
"idx": 337,
"question": "At the Curie temperature, the saturation magnetization abruptly diminishes. Which of the following magnetic material types will have Curie temperatures?\na) Diamagnetics\nb) Paramagnetics\nc) Ferromagnetics\nd) Antiferromagnetics\ne) Ferrimagnetics",
"answer": "Ferromagnetic and ferrimagnetic materials will have Curie temperatures."
},
{
"idx": 338,
"question": "In a polycrystalline material, each grain will always consist of just a single domain.\na) True\nb) False",
"answer": "False. In a polycrystalline material, each grain may consist of more than one domain."
},
{
"idx": 339,
"question": "With increasing temperature antiferromagnetic materials eventually become which of the following?\na) Diamagnetics\nb) Paramagnetics\nc) Ferromagnetics\nd) Antiferromagnetics\ne) Ferrimagnetics",
"answer": "With increasing temperature antiferromagnetic materials eventually become paramagnetic."
},
{
"idx": 340,
"question": "Which type(s) of magnetic materials may be classified as either soft or hard?\na) Diamagnetic\nb) Paramagnetic\nc) Ferromagnetic\nd) Antiferromagnetic\ne) Ferrimagnetic",
"answer": "Ferromagnetic and ferrimagnetic materials may be classified as either soft or hard."
},
{
"idx": 341,
"question": "Which of the following characteristics are displayed by soft magnetic materials?\na) A relatively small hysteresis loop.\nb) A relatively large hysteresis loop.\nc) Magnetization and demagnetization may be achieved using relatively low applied fields.\nd) Magnetization and demagnetization require relatively high applied fields.",
"answer": "Soft materials have relatively small relative hysteresis loops; also, magnetization and demagnetization may be achieved using relatively low applied fields."
},
{
"idx": 342,
"question": "Which of the following characteristics are displayed by hard magnetic materials?\na) A relatively small hysteresis loop.\nb) A relatively large hysteresis loop.\nc) Magnetization and demagnetization may be achieved using relatively low applied fields.\n\\(\\mathrm{d})\\) Magnetization and demagnetization require relatively high applied fields.",
"answer": "Hard materials have relatively large hysteresis loops; also, magnetization and demagnetization require relatively high applied fields."
},
{
"idx": 343,
"question": "What is the order of magnitude wavelength for visible light?\na) 0.5 Angstroms\nb) 0.5 nanometers\nc) 0.5 micrometers\nd) 0.5 millimeters\ne) 0.5 meters\nf) 0.5 kilometers",
"answer": "The wavelength of visible light is on the order of \\(0.5 \\mathrm{micrometers}\\)."
},
{
"idx": 344,
"question": "In the visible spectrum, a thick metal specimen will be\na) Transparent\nb) Translucent\nc) Opaque",
"answer": "In the visible spectrum, a thick metal specimen is opaque."
},
{
"idx": 345,
"question": "In the visible spectrum, an electrical insulator that is a single crystal and without porosity is normally\na) Transparent\nb) Opaque\nc) Translucent",
"answer": "In the visible spectrum, an electrical insulator that is a single crystal and without porosity\nnormally transparent."
},
{
"idx": 346,
"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": 346,
"question": "Match the type of light transmission with its description: Transmits light diffusely.",
"answer": "A translucent material transmits light diffusely."
},
{
"idx": 346,
"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": 347,
"question": "In the visible spectrum, a semiconductor that is a single crystal and nonporous may be\na) Transparent\nb) Translucent\nc) Opaque",
"answer": "In the visible spectrum, a semiconductor that is a single crystal and nonporous may be transparent or opaque."
},
{
"idx": 348,
"question": "To which of the following electromagnetic radiation types are bulk metals opaque?\na) radio waves\nb) microwaves\nc) infrared radiation\nd) ultraviolet radiation\ne) X-rays",
"answer": "Bulk metals are opaque to the following radiation types: radio waves, microwaves, infrared radiation, and ultraviolet radiation."
},
{
"idx": 349,
"question": "Every nonmetallic material becomes opaque to electromagnetic radiation having some wavelength.\na) True\nb) 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": 350,
"question": "What are the light transmission characteristics of single crystal electrical insulators?",
"answer": "Single crystal electrical insulators are transparent."
},
{
"idx": 350,
"question": "What are the light transmission characteristics of polycrystalline and nonporous electrical insulators?",
"answer": "Polycrystalline and nonporous electrical insulators are translucent."
},
{
"idx": 350,
"question": "What are the light transmission characteristics of porous electrical insulators?",
"answer": "Porous electrical insulators are opaque to visible light."
},
{
"idx": 351,
"question": "For noncubic crystals, the index of refraction is lowest in the crystallographic direction that has the\na) Highest atomic packing density\nb) Lowest atomic packing density",
"answer": "For noncubic crystals, the index of refraction is lowest in the crystallographic direc<65>on having the lowest atomic packing density."
},
{
"idx": 352,
"question": "A completely amorphous and nonporous polymer will be\na) Transparent\nb) Translucent\nc) Opaque",
"answer": "A completely amorphous and nonporous polymer will be transparent."
},
{
"idx": 353,
"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 energy is equal to the energy of the photon that is exactly energetic enough to excite electrons across the band gap. Thus, in this experiment, when the frequency is decreased the photon energy decreases. At some point, the frequency will be so low that the photon will no longer have sufficient energy to ionize the semiconductor. For this frequency and below, the semiconductor will be transparent. Thus, at the absorption edge:\n\\[\nE_{g}=h f=\\left(4.135 \\times 10^{-15} \\mathrm{eV} \\mathrm{s}\\right)\\left([F] \\times 10^{12} 1 / \\mathrm{s}\\right)\n\\]"
},
{
"idx": 354,
"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": 355,
"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": 356,
"question": "Only pure materials luminesce.\na) True\nb) False",
"answer": "False. Luminescent materials contain impurities."
},
{
"idx": 357,
"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": 358,
"question": "Chromium has four naturally-occurring isotopes: \\(4.34 \\%\\) of \\({ }^{50} \\mathrm{Cr}\\), with an atomic weight of 49.9460 \\(\\mathrm{amu}, 83.79 \\%\\) of \\({ }^{52} \\mathrm{Cr}\\), with an atomic weight of \\(51.9405 \\mathrm{amu}, 9.50 \\%\\) of \\({ }^{53} \\mathrm{Cr}\\), with an atomic weight of \\(52.9407 \\mathrm{amu}\\), and \\(2.37 \\%\\) of \\({ }^{54} \\mathrm{Cr}\\), with an atomic weight of \\(53.9389 \\mathrm{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 silicon \\(\\left(\\overline{\\mathcal{A}}_{\\mathrm{Cr}}\\right)\\) is computed by adding fraction-of-occurrence/atomic weight products for the three isotopes. Thus\n\\[\n\\begin{aligned}\n\\overline{\\mathcal{A}}_{\\mathrm{Cr}} & =f_{50_{\\mathrm{Cr}}} A_{50_{\\mathrm{Cr}}}+f_{52_{\\mathrm{Cr}}} A_{52_{\\mathrm{Cr}}}+f_{53_{\\mathrm{Cr}}} A_{33_{\\mathrm{Cr}}}+f_{54_{\\mathrm{Cr}}} A_{54_{\\mathrm{Cr}}} \\\\\n& =(0.0434)(49.9460 \\mathrm{amu})+(0.8379)(51.9405 \\mathrm{amu})+(0.0950)(52.9407 \\mathrm{amu})+(0.0237)(53.9389 \\mathrm{amu})=51.9963 \\mathrm{amu}\n\\end{aligned}\n\\]"
},
{
"idx": 359,
"question": "How many grams are there in one amu of a material?",
"answer": "In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as # g/amu = (1 mol/6.022 x 10^23 atoms)(1 g/mol/1 amu/atom) = 1.66 x 10^-24 g/amu"
},
{
"idx": 359,
"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": "Since there are 453.6 g/lb_m, 1 lb-mol = (453.6 g/lb_m)(6.022 x 10^23 atoms/g-mol) = 2.73 x 10^26 atoms/lb-mol"
},
{
"idx": 360,
"question": "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": 360,
"question": "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": 361,
"question": "Relative to electrons and electron states, what does the n quantum number specify?",
"answer": "The n quantum number designates the electron shell."
},
{
"idx": 361,
"question": "Relative to electrons and electron states, what does the l quantum number specify?",
"answer": "The l quantum number designates the electron subshell."
},
{
"idx": 361,
"question": "Relative to electrons and electron states, what does the m_j quantum number specify?",
"answer": "The m_j quantum number designates the number of electron states in each electron subshell."
},
{
"idx": 361,
"question": "Relative to electrons and electron states, what does the m_s quantum number specify?",
"answer": "The m_s quantum number designates the spin moment on each electron."
},
{
"idx": 362,
"question": "For the L shell (n=2), what are the four quantum numbers for all of the electrons, and 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 (l=0), the quantum numbers are 200(1/2) and 200(-1/2). For the p states (l=1), 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": 362,
"question": "For the M shell (n=3), what are the four quantum numbers for all of the electrons, and 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 (l=0), the quantum numbers are 300(1/2) and 300(-1/2). For the p states (l=1), the quantum numbers 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 (l=2), the quantum numbers 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": 363,
"question": "Give the electron configuration for the ion Fe2+.",
"answer": "From Table 2.2, the electron configuration for an atom of iron is 1s2 2s2 2p6 3s2 3p6 3d6 4s2. In order to become an ion with a plus two charge, it must lose two electrons-in this case the two 4s. Thus, the electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6."
},
{
"idx": 363,
"question": "Give the electron configuration for the ion Al3+.",
"answer": "From Table 2.2, the electron configuration for an atom of aluminum is 1s2 2s2 2p6 3s2 3p1. In order to become an ion with a plus three charge, it must lose three electrons-in this case two 3s and the one 3p. Thus, the electron configuration for an Al3+ ion is 1s2 2s2 2p6."
},
{
"idx": 363,
"question": "Give the electron configuration for the ion Cu+.",
"answer": "From Table 2.2, the electron configuration for an atom of copper is 1s2 2s2 2p6 3s2 3p6 3d10 4s1. In order to become an ion with a plus one charge, it must lose one electron-in this case the 4s. Thus, the electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10."
},
{
"idx": 363,
"question": "Give the electron configuration for the ion Ba2+.",
"answer": "The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a transition element the electron configuration for one of its atoms is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2. In order to become an ion with a plus two charge, it must lose two electrons-in this case the two 6s. Thus, the electron configuration for a Ba2+ ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6."
},
{
"idx": 363,
"question": "Give the electron configuration for the ion Br-.",
"answer": "From Table 2.2, the electron configuration for an atom of bromine is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5. In order to become an ion with a minus one charge, it must acquire one electron-in this case another 4p. Thus, the electron configuration for a Br- ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6."
},
{
"idx": 363,
"question": "Give the electron configuration for the ion O2-.",
"answer": "From Table 2.2, the electron configuration for an atom of oxygen is 1s2 2s2 2p4. In order to become an ion with a minus two charge, it must acquire two electrons-in this case another two 2p. Thus, the electron configuration for an O2- ion is 1s2 2s2 2p6."
},
{
"idx": 364,
"question": "Sodium chloride \\((\\mathrm{NaCl})\\) exhibits predominantly ionic bonding. The \\(\\mathrm{Na}^{+}\\)and \\(\\mathrm{Cl}^{-}\\)ions have electron structures that are identical to which two inert gases?",
"answer": "The \\(\\mathrm{Na}^{+}\\)ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the same as neon .\nThe \\(\\mathrm{Cl}^{-}\\)ion is a chlorine atom that has acquired one extra electron; therefore, it has an electron configuration the same as argon."
},
{
"idx": 365,
"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": 366,
"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."
},
{
"idx": 367,
"question": "Determine whether each of the electron configurations given below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.\n(a) \\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}\\)\n(b) \\(1 s^{2} 2 s^{2} 2 p^{6} s 3 s^{2} 3 p^{6}\\)\n(c) \\(1 s^{2} 2 s^{2} 2 p^{5}\\)\n(d) \\(1 s^{2} 2 s^{2} 2 p^{6} .3 s^{2}\\)\n(e) \\(1 s^{2} 2 s^{2} 2 p^{6} \\cdot 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}\\)\n(f) \\(1 s^{2} 2 s^{2} 2 p^{6} t 3 s^{2} 3 p^{6} 4 s^{1}\\)",
"answer": "(a) The \\(1 s^{2} 2 s^{2} 2 p^{6} 23 s^{2} 3 p^{6} 3 d^{7} 4 \\mathrm{~s}^{2}\\) electron configuration is that of a transition metal because of an incomplete \\(d\\) subshell.\n(b) 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.\n(c) 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.\n(d) The \\(1 s^{2} 2 s^{2} 2 p^{63} 2 s^{2}\\) electron configuration is that of an alkaline earth metal because of two \\(s\\) electrons.\n(e) The \\(1 s^{2} 2 s^{2} 2 p^{6}{ }_{3} 2 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.\n(f) The \\(1 s^{2} 2 s^{2} 2 p^{6 3} 2 s^{2} 3 p^{6} 4 s^{1}\\) electron configuration is that of an alkali metal because of a single s electron."
},
{
"idx": 368,
"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": 368,
"question": "What electron subshell is being filled for the actinide series of elements on the periodic table?",
"answer": "The 5 f subshell is being filled for the actinide series of elements."
},
{
"idx": 369,
"question": "Calculate the force of attraction between \\(a \\mathrm{~K}^{+}\\)and an \\(\\mathrm{O}^{2-}\\) ion the centers of which are separated by \\(a\\) distance of \\(1.5 \\mathrm{~nm}\\).",
"answer": "The attractive force between two ions \\(F_{A}\\) is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation, which is just\n\\[\nF_{A}=\\frac{d E_{A}}{d r}=\\frac{d\\left(-\\frac{A}{r}\\right)}{d r}=\\frac{A}{r^{2}}\n\\]\nSince the valences of the \\(\\mathrm{K}^{+}\\)and \\(\\mathrm{O}^{2-}\\) ions \\(\\left(Z_{1}\\right.\\) and \\(\\left.Z_{2}\\right)\\) are +1 and -2 , respectively, \\(Z_{1}=1\\) and \\(Z_{2}=2\\), then\n\\[\n\\begin{aligned}\nF_{A} & =\\frac{\\left(Z_{1} \\mathrm{e}\\right)\\left(Z_{2} \\mathrm{e}\\right)}{4 \\pi v_{0} r^{2}} \\\\\n& =\\frac{(1)(2)(1.602 \\times 10^{-19} \\mathrm{C})^{2}}{(4)(\\pi)(8.85 \\times 10^{-12} \\mathrm{~F} / \\mathrm{m})(1.5 \\times 10^{-9} \\mathrm{~m})^{2}} \\\\\n& =2.05 \\times 10^{-10} \\mathrm{~N}\n\\end{aligned}\n\\]"
},
{
"idx": 370,
"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": 370,
"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": 371,
"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": 371,
"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": 371,
"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": 371,
"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": 371,
"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": 371,
"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": 371,
"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": 372,
"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": 373,
"question": "If the atomic radius of aluminum is \\(0.143 \\mathrm{~nm}\\), calculate the volume of its unit cell in cubic meters.",
"answer": "For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation 3.4 as\n\\[\nV_{C}=16 R^{3} \\sqrt{2}=(16)\\left(0.143 \\times 10^{-9} \\mathrm{~m}\\right)^{3}(\\sqrt{2})=6.62 \\times 10^{-29} \\mathrm{~m}^{3}\n\\]"
},
{
"idx": 374,
"question": "Show that the atomic packing factor for BCC is 0.68 .",
"answer": "The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or\n\\[\n\\mathrm{APF}=\\frac{V_{\\mathrm{S}}}{V_{\\mathrm{C}}}\n\\]\nSince there are two spheres associated with each unit cell for BCC\n\\[\nV_{\\mathrm{S}}=2 \\text { (sphere volume) }=2\\left(\\frac{4 \\pi R^{3}}{3}\\right)=\\frac{8 \\pi R^{3}}{3}\n\\]\nAlso, the unit cell has cubic symmetry, that is \\(V_{C}=a^{3}\\). But \\(a\\) depends on \\(R\\) according to Equation 3.3, and\n\\[\nV_{C}=\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}=\\frac{64 R^{3}}{3 \\sqrt{3}}\n\\]\nThus,\n\\[\n\\mathrm{APF}=\\frac{V_{\\mathrm{S}}}{\\mathrm{V}_{\\mathrm{C}}}=\\frac{8 \\pi R^{3} / 3}{64 R^{3} / 3 \\sqrt{3}}=0.68\n\\]"
},
{
"idx": 375,
"question": "Iron has a BCC crystal structure, an atomic radius of \\(0.124 \\mathrm{~nm}\\), and an atomic weight of \\(55.85 \\mathrm{~g} / \\mathrm{mol}\\). Compute and compare its theoretical density with the experimental value found inside the front cover.",
"answer": "This problem calls for a computation of the density of iron. According to Equation 3.5\n\\[\n\\rho=\\frac{n A_{\\mathrm{Fe}}}{V_{C} N_{\\mathrm{A}}}\n\\]\nFor \\(\\mathrm{BCC}, n=2\\) atoms/unit cell, and\n\\[\nV_{C}=\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}\n\\]\nThus,\n\\[\n\\begin{aligned}\n\\rho & =\\frac{n A_{\\mathrm{Fe}}}{\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}} N_{\\mathrm{A}} \\\\\n& =\\frac{(2 \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell})(55.85 \\mathrm{~g} / \\mathrm{mol})}{[(4)\\left(0.124 \\times 10^{-7} \\mathrm{~cm}\\right) / \\sqrt{3}\\right]^{3} /(\\text { unit cell })(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol})} \\\\\n& =7.90 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]\nThe value given inside the front cover is \\(7.87 \\mathrm{~g} / \\mathrm{cm}^{3}\\)."
},
{
"idx": 376,
"question": "Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 \\(g / \\mathrm{cm}^{3}\\), and an atomic weight of \\(192.2 \\mathrm{~g} / \\mathrm{mol}\\).",
"answer": "We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure. For FCC, \\(n=4\\) atoms/unit cell, and \\(V_{\\mathrm{C}}=16 R^{3} \\sqrt{2}\\) . Now,\n\\[\n\\begin{aligned}\n\\rho & =\\frac{n A_{\\mathrm{f}}}{V_{\\mathrm{C}} N_{\\mathrm{A}}} \\\\\n& =\\frac{n A_{\\mathrm{f}}}{\\left(16 R^{3} \\sqrt{2}\\right) N_{\\mathrm{A}}}\n\\end{aligned}\n\\]\nAnd solving for \\(R\\) from the above expression yields\n\\[\n\\begin{aligned}\nR & =\\left(\\frac{n A_{\\mathrm{f}}}{16 \\rho N_{\\mathrm{A}} \\sqrt{2}}\\right)^{1 / 3} \\\\\n& =\\left[\\frac{(4 \\text { atoms/unit cell) }(192.2 \\mathrm{~g} / \\mathrm{mol})}{(16)\\left(22.4 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol})(\\sqrt{2})}\\right]^{1 / 3} \\\\\n& =1.36 \\times 10^{-8} \\mathrm{~cm}=0.136 \\mathrm{~nm}\n\\end{aligned}\n\\]"
},
{
"idx": 377,
"question": "Calculate the radius of a vanadium atom, given that \\(V\\) has a BCC crystal structure, a density of 5.96 \\(\\mathrm{g} / \\mathrm{cm}^{3}\\), and an atomic weight of \\(50.9 \\mathrm{~g} / \\mathrm{mol}\\).",
"answer": "This problem asks for us to calculate the radius of a vanadium atom. For BCC, \\(n=2\\) atoms/unit cell, and\n\\[\nV_{C}=\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}=\\frac{64 R^{3}}{3 \\sqrt{3}}\n\\]\nSince, from Equation\n\\[\n\\begin{aligned}\n\\rho & =\\frac{n A_{V}}{V_{C} N_{\\mathrm{A}}} \\\\\n& =\\frac{n A_{V}}{\\left(\\frac{64 R^{3}}{3 \\sqrt{3}}\\right) N_{\\mathrm{A}}}\n\\end{aligned}\n\\]\nand solving for \\(R\\) the previous equation\n\\[\nR=\\left(\\frac{3 \\sqrt{3} n A_{V}}{64 \\rho N_{\\mathrm{A}}}\\right)^{1 / 3}\n\\]\nand incorporating values of parameters given in the problem statement\n\\[\n\\begin{aligned}\nR= & {\\left[\\frac{(3 \\sqrt{3})(2 \\text { atoms } / \\text { unit cell })(50.9 \\mathrm{~g} / \\mathrm{mol})}{(64)\\left(5.96 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol}\\right)}\\right]^{1 / 3}} \\\\\n& =1.32 \\times 10^{-8} \\mathrm{~cm}=0.132 \\mathrm{~nm}\n\\end{aligned}\n\\]"
},
{
"idx": 378,
"question": "Determine whether the crystal structure of alloy A (atomic weight = 77.4 g/mol, density = 8.22 g/cm³, atomic radius = 0.125 mm) is FCC, BCC, or simple cubic and justify your determination.",
"answer": "For alloy A, assuming a simple cubic crystal structure, the density is calculated as follows: ρ = (n * A_A) / ((2R)^3 * N_A) = (1 atom/unit cell * 77.4 g/mol) / ([(2)(1.25 × 10^-8 cm)]^3 / unit cell * 6.022 × 10^23 atoms/mol) = 8.22 g/cm³. Therefore, its crystal structure is simple cubic."
},
{
"idx": 378,
"question": "Determine whether the crystal structure of alloy B (atomic weight = 107.6 g/mol, density = 13.42 g/cm³, atomic radius = 0.133 mm) is FCC, BCC, or simple cubic and justify your determination.",
"answer": "For alloy B, assuming an FCC crystal structure, the density is calculated as follows: ρ = (n * A_B) / ((2R√2)^3 * N_A) = (4 atoms/unit cell * 107.6 g/mol) / ([(2√2)(1.33 × 10^-8 cm)]^3 / unit cell * 6.022 × 10^23 atoms/mol) = 13.42 g/cm³. Therefore, its crystal structure is FCC."
},
{
"idx": 378,
"question": "Determine whether the crystal structure of alloy C (atomic weight = 127.3 g/mol, density = 9.23 g/cm³, atomic radius = 0.142 mm) is FCC, BCC, or simple cubic and justify your determination.",
"answer": "For alloy C, assuming a simple cubic crystal structure, the density is calculated as follows: ρ = (n * A_C) / ((2R)^3 * N_A) = (1 atom/unit cell * 127.3 g/mol) / ([(2)(1.42 × 10^-8 cm)]^3 / unit cell * 6.022 × 10^23 atoms/mol) = 9.23 g/cm³. Therefore, its crystal structure is simple cubic."
},
{
"idx": 379,
"question": "Given the tetragonal unit cell for tin with lattice parameters a = 0.583 nm and c = 0.318 nm, density = 7.30 g/cm³, atomic weight = 118.69 g/mol, and Avogadro's number = 6.022 × 10²³ atoms/mol, calculate the number of atoms per unit cell (n).",
"answer": "n = (ρ * V_C * N_A) / A_Sn = (7.30 g/cm³ * (5.83 × 10⁻⁸ cm)² * (3.18 × 10⁻⁸ cm) * 6.022 × 10²³ atoms/mol) / 118.69 g/mol = 4.00 atoms/unit cell"
},
{
"idx": 379,
"question": "Given the atomic radius R = 0.151 nm and the number of atoms per unit cell n = 4, calculate the total sphere volume (V_S) for the tin unit cell.",
"answer": "V_S = n * (4/3 * π * R³) = 4 * (4/3 * π * (1.51 × 10⁻⁸ cm)³) = 4 * (4/3 * π * 3.45 × 10⁻²⁴ cm³) = 1.44 × 10⁻²³ cm³"
},
{
"idx": 379,
"question": "Given the tetragonal unit cell parameters a = 0.583 nm and c = 0.318 nm, calculate the unit cell volume (V_C).",
"answer": "V_C = a² * c = (5.83 × 10⁻⁸ cm)² * (3.18 × 10⁻⁸ cm) = 3.40 × 10⁻²³ cm³ * 3.18 × 10⁻⁸ cm = 1.08 × 10⁻²² cm³"
},
{
"idx": 379,
"question": "Using the total sphere volume V_S = 1.44 × 10⁻²³ cm³ and unit cell volume V_C = 1.08 × 10⁻²² cm³, calculate the atomic packing factor (APF) for tin.",
"answer": "APF = V_S / V_C = (1.44 × 10⁻²³ cm³) / (1.08 × 10⁻²² cm³) = 0.534"
},
{
"idx": 380,
"question": "List the point coordinates of the barium 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."
},
{
"idx": 380,
"question": "List the point coordinates of the oxygen ions for a unit cell of the perovskite crystal structure.",
"answer": "The oxygen ions are located at all face-centered positions; therefore, their coordinates are 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, 1/2 1 1/2, and 1/2 0 1/2."
},
{
"idx": 380,
"question": "List the point coordinates of the titanium ion for a unit cell of the perovskite crystal structure.",
"answer": "The titanium ion resides at the center of the cubic unit cell, with coordinates 1/2 1/2 1/2."
},
{
"idx": 381,
"question": "List the point coordinates of all corner position atoms in the diamond cubic unit cell.",
"answer": "The coordinates of the corner position atoms are: 000, 100, 110, 010, 001, 101, 111, and 011."
},
{
"idx": 381,
"question": "List the point coordinates of all face-centered position atoms in the diamond cubic unit cell.",
"answer": "The coordinates of the face-centered position atoms are: 1/2 1/2 0, 1/2 0 1/2, 0 1/2 1/2, 1/2 1/2 1, 1/2 1 1/2, and 1 1/2 1/2."
},
{
"idx": 381,
"question": "List the point coordinates of all interior position atoms in the diamond cubic unit cell.",
"answer": "The coordinates of the interior position atoms are: 1/4 1/4 1/4, 3/4 3/4 1/4, 3/4 1/4 3/4, and 1/4 3/4 3/4."
},
{
"idx": 382,
"question": "How to define atom types, colors, and sizes for Au and Cu in the Molecule Definition Utility for generating a unit cell of AuCu3?",
"answer": "In the 'Step 1' window of the Molecule Definition Utility, define 'Au' as the name for gold atoms and 'Cu' for copper atoms. Choose colors from the pull-down menu, e.g., 'Yellow' for Au and 'Red' for Cu. Enter atom sizes based on atomic radii: 0.288 nm for Au (2 × 0.144 nm) and 0.256 nm for Cu (2 × 0.128 nm). Click 'Register' and then 'Go to Step 2'."
},
{
"idx": 382,
"question": "How to specify the positions of gold atoms in the cubic unit cell of AuCu3 with an edge length of 0.374 nm?",
"answer": "In 'Step 2', select the yellow sphere for Au. Enter the following coordinates for the 8 corner positions: 0, 0, 0; 0.374, 0, 0; 0, 0.374, 0; 0, 0, 0.374; 0, 0.374, 0.374; 0.374, 0, 0.374; 0.374, 0.374, 0; 0.374, 0.374, 0.374. Click 'Register Atom Position' after each entry."
},
{
"idx": 382,
"question": "How to specify the positions of copper atoms in the cubic unit cell of AuCu3 with an edge length of 0.374 nm?",
"answer": "In 'Step 2', select the red sphere for Cu. Enter the following coordinates for the 6 face-centered positions: 0.187, 0.187, 0; 0.187, 0, 0.187; 0, 0.187, 0.187; 0.374, 0.187, 0.187; 0.187, 0.374, 0.187; 0.187, 0.187, 0.374. Click 'Register Atom Position' after each entry."
},
{
"idx": 382,
"question": "How to proceed after specifying all atom positions in the Molecule Definition Utility for AuCu3?",
"answer": "In 'Step 3', you may choose to represent unit cell edges as bonds or elect to not represent any bonds. The image can be rotated using mouse click-and-drag. Note that data or images cannot be saved in this version of the utility; use screen capture software to record the image."
},
{
"idx": 383,
"question": "Convert the (010) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.",
"answer": "For (010), h=0, k=1, and l=0, and, from Equation, the value of i is equal to i=-(h+k)=-(0+1)=-1. Therefore, the (010) plane becomes (01-10)."
},
{
"idx": 383,
"question": "Convert the (101) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.",
"answer": "For the (101) plane, h=1, k=0, and l=1, and computation of i using Equation leads to i=-(h+k)=-[1+0]=-1 such that (101) becomes (10-11)."
},
{
"idx": 384,
"question": "Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius R.",
"answer": "A (0001) plane for an HCP unit cell is shown below. Each of the 6 perimeter atoms in this plane is shared with three other unit cells, whereas the center atom is shared with no other unit cells; this gives rise to three equivalent atoms belonging to this plane. In terms of the atomic radius R, the area of each of the 6 equilateral triangles that have been drawn is R^(2) sqrt(3), or the total area of the plane shown is 6 R^(2) sqrt(3). And the planar density for this (0001) plane is equal to PD_(0001) = (number of atoms centered on (0001) plane) / (area of (0001) plane) = (3 atoms) / (6 R^(2) sqrt(3)) = 1 / (2 R^(2) sqrt(3))"
},
{
"idx": 384,
"question": "Compute the planar density value for the HCP (0001) plane for magnesium, given the atomic radius for magnesium is 0.160 nm.",
"answer": "From the table inside the front cover, the atomic radius for magnesium is 0.160 nm. Therefore, the planar density for the (0001) plane is PD_(0001)(Mg) = 1 / (2 R^(2) sqrt(3)) = 1 / (2 (0.160 nm)^(2) sqrt(3)) = 11.28 nm^(-2) = 1.128 x 10^(19) m^(-2)"
},
{
"idx": 385,
"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": 386,
"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": "From the data given in the problem, and realizing that 69.22 degrees = 2 theta, the interplanar spacing for the (220) set of planes for iridium may be computed using the equation d220 = (n * lambda) / (2 * sin(theta)) = (1 * 0.1542 nm) / (2 * sin(34.61 degrees)) = 0.1357 nm."
},
{
"idx": 386,
"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": "In order to compute the atomic radius we must first determine the lattice parameter, a, using the equation a = d220 * sqrt(h^2 + k^2 + l^2) = (0.1357 nm) * sqrt(8) = 0.3838 nm. And, from the equation for FCC crystal structure, R = a / (2 * sqrt(2)) = (0.3838 nm) / (2 * sqrt(2)) = 0.1357 nm."
},
{
"idx": 387,
"question": "Which of the elements listed would form a substitutional solid solution having complete solubility with copper (Cu)? Consider the criteria: atomic radius difference within ±15%, same crystal structure, similar electronegativity, and same or nearly same valence.",
"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": 387,
"question": "Which of the elements listed would form a substitutional solid solution of incomplete solubility with copper (Cu)? Consider the criteria: atomic radius difference greater than ±15%, different crystal structure, dissimilar electronegativity, or different valence.",
"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": 387,
"question": "Which of the elements listed would form an interstitial solid solution with copper (Cu)? Consider the criteria: significantly smaller atomic radius compared to 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": 388,
"question": "What is the composition, in atom percent, of an alloy that consists of \\(30 \\mathrm{wt} \\% \\mathrm{Zn}\\) and \\(70 \\mathrm{wt} \\% \\mathrm{Cu}\\) ?",
"answer": "In order to compute composition, in atom percent, of a \\(30 \\mathrm{wt} \\% \\mathrm{Zn}-70 \\mathrm{wt} \\% \\mathrm{Cu}\\) alloy, we employ Equation 4.6 as\n\\[\n\\begin{aligned}\nC_{\\mathrm{Zn}}^{\\prime} & =\\frac{C_{\\mathrm{Zn}} \\hat{A}_{\\mathrm{Cu}}}{C_{\\mathrm{Zn}} \\hat{A}_{\\mathrm{Cu}}+C_{\\mathrm{Cu}} A_{\\mathrm{Zn}}} \\times 100 \\\\\n& =\\frac{(30)(63.55 \\mathrm{~g} / \\mathrm{mol})}{(30)(63.55 \\mathrm{~g} / \\mathrm{mol})\n+(70)(65.41 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n& =29.4 \\mathrm{at} \\% \\\\\nC_{\\mathrm{Cu}}^{\\prime} & =\\frac{C_{\\mathrm{Cu}} A_{\\mathrm{Zn}}}{C_{\\mathrm{Zn}} \\hat{A}_{\\mathrm{Cu}^{\\prime}}+C_{\\mathrm{Cu}} A_{\\mathrm{Zn}}} \\times \\quad 100 \\\\\n& =\\frac{(70)(65.41 \\mathrm{~g} / \\mathrm{mol})}\n{(30)(63.55 \\mathrm{~g} / \\mathrm{mol})+(70)(65.41 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n& =70.6 \\mathrm{at} \\%\n\\end{aligned}\n\\]"
},
{
"idx": 389,
"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": "In order to compute composition, in weight percent, of a 6 at\\% \\(\\mathrm{Pb}-94\\) at\\% \\(\\mathrm{Sn}\\) alloy, we employ Equation as\n\\[\n\\begin{aligned}\nC_{\\mathrm{Pb}} & =\\frac{C_{\\mathrm{Pb}}^{\\cdot} A_{\\mathrm{Pb}}}{C_{\\mathrm{Pb}} A_{\\mathrm{Pb}}+C_{\\mathrm{Sn}}^{\\cdot} A_{\\mathrm{Sn}}} \\times 100 \\\\\n& =\\frac{(6)(207.2 \\mathrm{~g} / \\mathrm{mol})}{(6)(207.2 \\mathrm{~g} / \\mathrm{mol})+(94)(118.71 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n& =10.0 \\mathrm{wt} \\%\n\\end{aligned}\n\\]\n\\[\n\\begin{aligned}\nC_{\\mathrm{Sn}} & =\\frac{C_{\\mathrm{Sn}}^{\\cdot} A_{\\mathrm{Sn}}}{C_{\\mathrm{Pb}}^{\\cdot} A_{\\mathrm{Pb}+} C_{\\mathrm{Sn}}^{\\cdot} A_{\\mathrm{Sn}}} \\times \\frac{100}{(94)(118.71 \\mathrm{~g} / \\mathrm{mol})} \\\\\n& =\\frac{(94)(118.71 \\mathrm{~g} / \\mathrm{mo1})}{(6)(207.2 \\mathrm{~g} / \\mathrm{\\mathrm{mol})}+(94)(118.71 \\mathrm{~g} / \\mathrm{\\mathrm{mol})}} \\times 100 \\\\\n& =90.0 \\mathrm{wt} \\%\n\\end{aligned}\n\\]"
},
{
"idx": 390,
"question": "What is the mass of copper in grams, given 99.7 lbm of copper?",
"answer": "The mass of copper is (99.7 lbm)(453.6 g/lbm) = 45,224 g."
},
{
"idx": 390,
"question": "What is the mass of zinc in grams, given 102 lbm of zinc?",
"answer": "The mass of zinc is (102 lbm)(453.1 g/lbm) = 46,267 g."
},
{
"idx": 390,
"question": "What is the mass of lead in grams, given 2.1 lbm of lead?",
"answer": "The mass of lead is (2.1 lbm)(453.3 g/lbm) = 953 g."
},
{
"idx": 390,
"question": "How many moles of copper are there in 45,224 g of copper, given the atomic weight of copper is 63.55 g/mol?",
"answer": "The number of moles of copper is 45,224 g / 63.55 g/mol = 711.6 mol."
},
{
"idx": 390,
"question": "How many moles of zinc are there in 46,267 g of zinc, given the atomic weight of zinc is 65.41 g/mol?",
"answer": "The number of moles of zinc is 46,267 g / 65.41 g/mol = 707.3 mol."
},
{
"idx": 390,
"question": "How many moles of lead are there in 953 g of lead, given the atomic weight of lead is 207.2 g/mol?",
"answer": "The number of moles of lead is 953 g / 207.2 g/mol = 4.6 mol."
},
{
"idx": 390,
"question": "What is the atom percent of copper in the alloy, given 711.6 mol of copper, 707.3 mol of zinc, and 4.6 mol of lead?",
"answer": "The atom percent of copper is (711.6 mol / (711.6 mol + 707.3 mol + 4.6 mol)) × 100 = 50.0 at%."
},
{
"idx": 390,
"question": "What is the atom percent of zinc in the alloy, given 711.6 mol of copper, 707.3 mol of zinc, and 4.6 mol of lead?",
"answer": "The atom percent of zinc is (707.3 mol / (711.6 mol + 707.3 mol + 4.6 mol)) × 100 = 49.7 at%."
},
{
"idx": 390,
"question": "What is the atom percent of lead in the alloy, given 711.6 mol of copper, 707.3 mol of zinc, and 4.6 mol of lead?",
"answer": "The atom percent of lead is (4.6 mol / (711.6 mol + 707.3 mol + 4.6 mol)) × 100 = 0.3 at%."
},
{
"idx": 391,
"question": "What is the composition, in atom percent, of an alloy that consists of 97 wt\\% Fe and 3 wt\\% Si?",
"answer": "We are asked to compute the composition of an Fe-Si alloy in atom percent. Employment of Equation leads to\n\\[\n\\begin{array}{l}\nC_{\\mathrm{Fe}}^{+}=\\frac{C_{\\mathrm{Fe}} A_{\\mathrm{Si}}}{C_{\\mathrm{Fe}} A_{\\mathrm{Si}}+C_{\\mathrm{Si}} A_{\\mathrm{Fe}}} \\times 100 \\\\\n=\\frac{97(28.09 \\mathrm{~g} / \\mathrm{mol})}{97(28.09 \\mathrm{~g} / \\mathrm{mol)}+3(55.85 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n=94.2 \\mathrm{at} \\%\n\\end{array}\n\\]\n\\[\n\\begin{aligned}\nC_{\\mathrm{Si}}^{*} & =\\frac{C_{\\mathrm{Si}} A_{\\mathrm{Fe}}}{C_{\\mathrm{Si}} A_{\\mathrm{Fe}}+C_{\\mathrm{Fe}} A_{\\mathrm{Si}}} \\times 100 \\\\\n& =\\frac{3(55.85 \\mathrm{~g} / \\mathrm{mol})}{3(55.85 \\mathrm{~g} / \\mathrm{mol})+97(28.09 \\mathrm{~g} / \\mathrm{mol}))} \\times 100 \\\\\n& =5.8 \\mathrm{at} \\%\n\\end{aligned}\n\\]"
},
{
"idx": 392,
"question": "Calculate the number of atoms per cubic meter in aluminum.",
"answer": "In order to solve this problem, one must employ Equation,\n\\[\nN=\\frac{N_{\\mathrm{A}} \\rho_{\\mathrm{Al}}}{A_{\\mathrm{Al}}}\n\\]\nThe density of \\(\\mathrm{Al}\\) (from the table inside of the front cover) is \\(2.71 \\mathrm{~g} / \\mathrm{cm}^{3}\\), while its atomic weight is \\(26.98 \\mathrm{~g} / \\mathrm{mol}\\). Thus,\n\\[\n\\begin{aligned}\nN & =\\frac{\\left(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol}\\right)\\left(2.71 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)}{26.98 \\mathrm{~g} / \\mathrm{mol}} \\\\\n& =6.05 \\times 10^{22} \\text { atoms } / \\mathrm{cm}^{3}=6.05 \\times 10^{28} \\text { atoms } / \\mathrm{m}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 393,
"question": "The concentration of carbon in an iron-carbon alloy is \\(0.15 \\mathrm{wt} \\%\\). What is the concentration in kilograms of carbon per cubic meter of alloy?",
"answer": "In order to compute the concentration in \\(\\mathrm{kg} / \\mathrm{m}^{3}\\) of \\(\\mathrm{C}\\) in a \\(0.15 \\mathrm{wt} \\% \\mathrm{C}-99.85 \\mathrm{wt} \\% \\mathrm{Fe}\\) alloy we must employ Equation as\n\\[\nC_{\\mathrm{C}}^{\\prime \\prime}=\\frac{C_{\\mathrm{C}}}{\\frac{C_{\\mathrm{C}}}{\\rho_{\\mathrm{C}}}+\\frac{C_{\\mathrm{Fe}}}{\\rho_{\\mathrm{Fe}}}} \\times 10^{3}\n\\]\nFrom inside the front cover, densities for carbon and iron are 2.25 and \\(7.87 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively; and, therefore\n\\[\n\\begin{aligned}\nC_{\\mathrm{C}}^{\\prime \\prime}= & \\frac{0.15}{0.15} \\xrightarrow{99.85} \\times 10^{3} \\\\\n& 2.25 \\mathrm{~g} / \\mathrm{cm}^{3}+\\frac{7.87 \\mathrm{~g} / \\mathrm{cm}^{3}}{3}\n\\end{aligned}\n\\]\n\\[\n\\begin{aligned}\n= & 11.8 \\mathrm{~kg} / \\mathrm{m}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 394,
"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": 395,
"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": 396,
"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": 397,
"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": 397,
"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": 398,
"question": "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": 398,
"question": "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": 399,
"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": 400,
"question": "Briefly explain the concept of a driving force.",
"answer": "The driving force is that which compels a reaction to occur."
},
{
"idx": 400,
"question": "What is the driving force for steady-state diffusion?",
"answer": "The driving force for steady-state diffusion is the concentration gradient."
},
{
"idx": 401,
"question": "For a steel alloy, given that a carburizing heat treatment of 10-h duration raises the carbon concentration to 0.45 wt% at a point 2.5 mm 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": "This problem is solved using the relation x²/Dt = constant. Since D is constant at the same temperature, x²/t = constant. Thus, (2.5 mm)²/10 h = (5.0 mm)²/t₂. Solving for t₂ gives t₂ = 40 h."
},
{
"idx": 402,
"question": "The preexponential and activation energy for the diffusion of iron in cobalt are \\(1.1 \\times 10^{-5} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(253,300 \\mathrm{~J} / \\mathrm{mol}\\), respectively. At what temperature will the diffusion coefficient have a value of \\(2.1 \\times 10^{-14} \\mathrm{~m}^{2} / \\mathrm{s}\\) ?",
"answer": "For this problem we are given \\(D_{0}\\left(1.1 \\times 10^{-5}\\right)\\) and \\(Q_{d}(253,300 \\mathrm{~J} / \\mathrm{mol})\\) for the diffusion of \\(\\mathrm{Fe}\\) in \\(\\mathrm{Co}\\), and asked to compute the temperature at which \\(D=2.1 \\times 10^{-14} \\mathrm{~m}^{ 2 / s}\\). Solving for \\(T\\) from Equation yields\n\\[\n\\begin{array}{l}\nT=\\frac{Q_{d}}{R\\left(\\ln D_{0}-\\ln D\\right)} \\\\\n=\\frac{253,300 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})\\left[\\ln \\left(1.1 \\times 10^{-5} \\mathrm{~m}^{2 / s}\\right)-\\ln \\left(2.1 \\times 10^{-14} \\mathrm{~m}^{3} / \\mathrm{s}\\right)\\right]} \\\\\n=1518 \\mathrm{~K}=1245^{\\circ} \\mathrm{C}\n\\end{array}\n\\]\nNote: this problem may also be solved using the \"Diffusion\" module in the VMSE software. Open the \"Diffusion\" module, click on the \"D vs \\(1 / T\\) Plot\" submodule, and then do the following:\n1. In the left-hand window that appears, click on the \"Custom1\" box.\n2. In the column on the right-hand side of this window enter the data for this problem. In the window under \"D0\" enter preexponential value-viz. \"1.1e-5\". Next just below the \"Qd\" window enter the activation energy value-viz. \"253.3\". It is next necessary to specify a temperature range over which the data is to be plotted. The temperature at which \\(D\\) has the stipulated value is probably between \\(1000^{\\circ} \\mathrm{C}\\) and \\(1500^{\\circ} \\mathrm{C}\\), so enter \" 1000 \" in the \"T Min\" box that is beside \"C\"; and similarly for the maximum temperature—enter \" 1500 \" in the box below \"T Max\".\n3. Next, at the bottom of this window, click the \"Add Curve\" button.\n4. A \\(\\log \\mathrm{D}\\) versus \\(1 / \\mathrm{T}\\) plot then appears, with a line for the temperature dependence of the diffusion coefficient for \\(\\mathrm{Fe}\\) in \\(\\mathrm{Co}\\). At the top of this curve is a diamond-shaped cursor. Click-and-drag this cursor down the line to the point at which the entry under the \"Diff Coeff (D):\" label reads \\(2.1 \\times 10^{-14} \\mathrm{~m} 2 / \\mathrm{s}\\). The temperature at which the diffusion coefficient has this value is given under the label \"Temperature (T):\". For this problem, the value is \\(1519 \\mathrm{~K}\\)."
},
{
"idx": 403,
"question": "Given the activation energy for the diffusion of carbon in chromium is 111,000 J/mol and the diffusion coefficient D at 1400 K is 6.25 × 10^-11 m²/s, calculate the pre-exponential factor D₀.",
"answer": "D₀ = D exp (Q_d / (R T)) = (6.25 × 10^-11 m²/s) exp [111,000 J/mol / (8.31 J/mol-K × 1400 K)] = 8.7 × 10^-7 m²/s"
},
{
"idx": 403,
"question": "Using the pre-exponential factor D₀ = 8.7 × 10^-7 m²/s and activation energy Q_d = 111,000 J/mol, calculate the diffusion coefficient D at 1100 K.",
"answer": "D = D₀ exp (-Q_d / (R T)) = (8.7 × 10^-7 m²/s) exp [-111,000 J/mol / (8.31 J/mol-K × 1100 K)] = 4.6 × 10^-12 m²/s"
},
{
"idx": 404,
"question": "The diffusion coefficients for iron in nickel are given at two temperatures: T1 = 1273 K, D1 = 9.4 × 10^-16 m²/s and T2 = 1473 K, D2 = 2.4 × 10^-14 m²/s. Determine the values of D0 and the activation energy Qd.",
"answer": "Using the Arrhenius equation, we set up two simultaneous equations with Qd and D0 as unknowns: ln D1 = ln D0 - (Qd / R)(1 / T1) and ln D2 = ln D0 - (Qd / R)(1 / T2). Solving for Qd: Qd = -R (ln D1 - ln D2) / (1 / T1 - 1 / T2) = -(8.31 J/mol-K) [ln(9.4 × 10^-16) - ln(2.4 × 10^-14)] / (1 / 1273 K - 1 / 1473 K) = 252,400 J/mol. Solving for D0: D0 = D1 exp(Qd / R T1) = (9.4 × 10^-16 m²/s) exp[252,400 J/mol / (8.31 J/mol-K)(1273 K)] = 2.2 × 10^-5 m²/s."
},
{
"idx": 404,
"question": "Using the values of D0 = 2.2 × 10^-5 m²/s and Qd = 252,400 J/mol determined from the diffusion coefficients for iron in nickel at T1 = 1273 K and T2 = 1473 K, what is the magnitude of D at 1100°C (1373 K)?",
"answer": "Using the Arrhenius equation with the determined values of D0 and Qd, D at 1373 K is calculated as: D = D0 exp(-Qd / R T) = (2.2 × 10^-5 m²/s) exp[-252,400 J/mol / (8.31 J/mol-K)(1373 K)] = 5.4 × 10^-15 m²/s."
},
{
"idx": 405,
"question": "Determine the values of D0 and Qd given the diffusion coefficients for silver in copper at two temperatures: T1 = 650°C (923 K) with D1 = 5.5 × 10^-16 m²/s and T2 = 900°C (1173 K) with D2 = 1.3 × 10^-13 m²/s.",
"answer": "Using the Arrhenius equation, set up two simultaneous equations: ln D1 = ln D0 - (Qd / R)(1 / T1) and ln D2 = ln D0 - (Qd / R)(1 / T2). Solving for Qd: Qd = -R (ln D1 - ln D2) / (1 / T1 - 1 / T2) = -(8.31 J/mol-K) [ln(5.5 × 10^-16) - ln(1.3 × 10^-13)] / (1/923 K - 1/1173 K) = 196,700 J/mol. Solving for D0: D0 = D1 exp(Qd / R T1) = (5.5 × 10^-16 m²/s) exp[196,700 J/mol / (8.31 J/mol-K)(923 K)] = 7.5 × 10^-5 m²/s."
},
{
"idx": 405,
"question": "What is the magnitude of D at 875°C (1148 K) using the values of D0 = 7.5 × 10^-5 m²/s and Qd = 196,700 J/mol?",
"answer": "Using the Arrhenius equation: D = D0 exp(-Qd / R T) = (7.5 × 10^-5 m²/s) exp[-196,700 J/mol / (8.31 J/mol-K)(1148 K)] = 8.3 × 10^-14 m²/s."
},
{
"idx": 406,
"question": "Carbon is allowed to diffuse through a steel plate \\(15 \\mathrm{~mm}\\) thick. The concentrations of carbon at the two faces are 0.65 and \\(0.30 \\mathrm{~kg} \\mathrm{C} / \\mathrm{m}^{2}\\) Fe, which are maintained constant. If the preexponential and activation energy are \\(6.2 \\times 10^{-7} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(80,000 \\mathrm{~J} / \\mathrm{mol}\\), respectively, compute the temperature at which the diffusion flux is \\(1.43 \\times 10^{-9} \\mathrm{~kg} / \\mathrm{m}^{2}\\)-s.",
"answer": "Combining Equations yields\n\\[\n\\begin{aligned}\nJ & =-D \\frac{\\Delta C}{\\Delta x} \\\\\n& =-D_{0} \\frac{\\Delta C}{\\Delta x} \\exp \\left(-\\frac{Q_{d}}{R T}\\right)\n\\end{aligned}\n\\]\nSolving for \\(T\\) from this expression leads to\n\\[\nT=\\left(\\frac{Q_{d}}{R}\\right) \\frac{1}{\\ln \\left(-\\frac{D_{0} \\Delta C}{J \\Delta x}\\right)}\n\\]\nAnd incorporation of values provided in the problem statement yields\n\\[\n\\begin{aligned}\n& =\\left(\\frac{80,000 \\mathrm{~J} / \\mathrm{mol}}{8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}}\\right)\\left[\\frac{1}{\\left[\\frac{(6.2 \\times 10^{-7} \\mathrm{~m}^{2 /} \\mathrm{s}\\right)\\left(0.65 \\mathrm{~kg} / \\mathrm{m}^{3}-0.30 \\mathrm{~kg} / \\mathrm{m}^{3}\\right)}{\\left(1.43 \\times 10^{-9} \\mathrm{~kg} / m^{2}-\\mathrm{s}\\right)\\left(15 \\times 10^{-3} \\mathrm{~m}\\right)}\\right] \\\\\n& =1044 \\mathrm{~K}=771^{\\circ} \\mathrm{C}\n\\end{aligned}\n\\]"
},
{
"idx": 407,
"question": "Calculate the value of D0 from the given data at 727°C (1000 K) with a diffusion flux of 5.4×10−10 kg/m2-s and a concentration gradient of −350 kg/m4, assuming an activation energy for diffusion of 125,000 J/mol.",
"answer": "To calculate D0, use the equation J = -D0 (ΔC/Δx) exp(-Qd/RT). Solving for D0 gives D0 = - (J / (ΔC/Δx)) exp(Qd/RT). Plugging in the values: D0 = - (5.4×10−10 kg/m2-s / -350 kg/m4) exp(125,000 J/mol / (8.31 J/mol-K)(1000 K)) = 5.26×10−6 m2/s."
},
{
"idx": 407,
"question": "Calculate the diffusion flux at 1027°C (1300 K) for the same concentration gradient of −350 kg/m4 and activation energy for diffusion of 125,000 J/mol, using the previously calculated D0 value of 5.26×10−6 m2/s.",
"answer": "Using the equation J = -D0 (ΔC/Δx) exp(-Qd/RT), the diffusion flux at 1300 K is calculated as: J = - (5.26×10−6 m2/s)(-350 kg/m4) exp(-125,000 J/mol / (8.31 J/mol-K)(1300 K)) = 1.74×10−8 kg/m2-s."
},
{
"idx": 408,
"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 × 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": "This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation (F_y). Taking the yield strength to be 275 MPa, and employment of Equation leads to F_y = σ_y A_0 = (275 × 10^6 N/m^2)(325 × 10^-6 m^2) = 89,375 N (20,000 lb_f)."
},
{
"idx": 408,
"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 × 10^6 psi). If the original specimen length is 115 mm (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 sample may be deformed without plastic deformation is determined from Equations as l_i = l_0 (1 + σ/E) = (115 mm)[1 + (275 MPa)/(115 × 10^3 MPa)] = 115.28 mm (4.51 in.)."
},
{
"idx": 409,
"question": "A cylindrical rod of copper \\((E=110 G P a, 16 \\times 10^{6} \\mathrm{psi})\\) having a yield strength of \\(240 \\mathrm{MPa}(35,000\\) psi) is to be subjected to a load of \\(6660 \\mathrm{~N}\\left(1500 \\mathrm{lb} p\\right)\\). If the length of the rod is \\(380 \\mathrm{~mm}(15.0 \\mathrm{in}\\).), what must be the diameter to allow an elongation of \\(0.50 \\mathrm{~mm}(0.020 \\mathrm{in.})\\) ?",
"answer": "This problem asks us to compute the diameter of a cylindrical specimen of copper in order to allow an elongation of \\(0.50 \\mathrm{~mm}\\). Employing Equations, assuming that deformation is entirely elastic\n\\[\n\\sigma=\\frac{F}{A_{0}}=\\frac{F}{\\pi\\left(\\frac{d_{0}^{2}}{4}\\right)}=E \\frac{\\Delta l}{l_{0}}\n\\]\nOr, solving for \\(d_{0}\\)\n\\[\n\\begin{aligned}\nd_{0} & =\\sqrt{\\frac{4 l_{0} F}{\\pi E \\Delta l}} \\\\\n& =\\sqrt{\\frac{(4)\\left(380 \\times 10^{-3} \\mathrm{~m}\\right)(6660 \\mathrm{~N})}{(\\pi)\\left(110 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)\\left(0.50 \\times 10^{-3} \\mathrm{~m}\\right)}} \\\\\n& =7.65 \\times 10^{-3} \\mathrm{~m}=7.65 \\mathrm{~mm}(0.30 \\mathrm{in}) .\n\\end{aligned}\n\\]\nExcerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections \\(10^{7}\\) or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful."
},
{
"idx": 410,
"question": "Compute the elastic modulus for titanium alloy using the stress-strain behavior 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 titanium?",
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For titanium, σ2 = 404.2 MPa and ε2 = 0.0038. Therefore, E = (404.2 MPa - 0 MPa)/(0.0038 - 0) = 106,400 MPa = 106.4 GPa. The elastic modulus for titanium given in Table 6.1 is 107 GPa, which is in very good agreement with this value."
},
{
"idx": 410,
"question": "Compute the elastic modulus for tempered steel using the stress-strain behavior 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 tempered steel?",
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For tempered steel, σ2 = 962.2 MPa and ε2 = 0.0047. Therefore, E = (962.2 MPa - 0 MPa)/(0.0047 - 0) = 204,700 MPa = 204.7 GPa. The elastic modulus for steel given in Table 6.1 is 207 GPa, which is in reasonably good agreement with this value."
},
{
"idx": 410,
"question": "Compute the elastic modulus for aluminum using the stress-strain behavior 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 aluminum?",
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For aluminum, σ2 = 145.1 MPa and ε2 = 0.0021. Therefore, E = (145.1 MPa - 0 MPa)/(0.0021 - 0) = 69,100 MPa = 69.1 GPa. The elastic modulus for aluminum given in Table 6.1 is 69 GPa, which is in excellent agreement with this value."
},
{
"idx": 410,
"question": "Compute the elastic modulus for carbon steel using the stress-strain behavior 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 carbon steel?",
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For carbon steel, σ2 = 129 MPa and ε2 = 0.0006. Therefore, E = (129 MPa - 0 MPa)/(0.0006 - 0) = 215,000 MPa = 215 GPa. The elastic modulus for steel given in Table 6.1 is 207 GPa, which is in reasonable agreement with this value."
},
{
"idx": 411,
"question": "What is the modulus of elasticity (E) for a cylindrical specimen of a hypothetical metal alloy with a diameter of 8.0 mm (0.31 in.), subjected to a tensile force of 1000 N (225 lbf), given that the elastic reduction in diameter is 2.8 x 10^-4 mm (1.10 x 10^-5 in.) and Poisson's ratio is 0.30?",
"answer": "The modulus of elasticity (E) is calculated using the formula: E = (4 F v) / (π d0 Δd), where F is the tensile force, v is Poisson's ratio, d0 is the original diameter, and Δd is the reduction in diameter. Substituting the given values: F = 1000 N, v = 0.30, d0 = 8 x 10^-3 m, Δd = 2.8 x 10^-7 m, we get E = (4 x 1000 N x 0.30) / (π x 8 x 10^-3 m x 2.8 x 10^-7 m) = 1.705 x 10^11 Pa = 170.5 GPa (24.7 x 10^6 psi)."
},
{
"idx": 412,
"question": "For a cylindrical rod 100 mm long with a diameter of 10.0 mm subjected to a tensile load of 27,500 N, which of the listed materials (Aluminum alloy, Brass alloy, Steel alloy, Titanium alloy) will not experience plastic deformation? The yield strengths are: Aluminum alloy - 200 MPa, Brass alloy - 300 MPa, Steel alloy - 400 MPa, Titanium alloy - 650 MPa.",
"answer": "First, compute the stress: sigma = F / A0 = 27,500 N / (pi * (10.0 mm / 2)^2) = 350 MPa. The materials with yield strengths greater than 350 MPa are Steel alloy (400 MPa) and Titanium alloy (650 MPa). Thus, Steel and Titanium alloys will not experience plastic deformation."
},
{
"idx": 412,
"question": "For the Steel alloy (Modulus of Elasticity = 207 GPa, Poisson's Ratio = 0.30) under the same loading conditions, will the diameter reduction be less than 7.5 x 10^-3 mm?",
"answer": "Calculate the diameter reduction: Delta d = - (v * sigma * d0) / E = - (0.30 * 350 MPa * 10 mm) / 207,000 MPa = -5.1 x 10^-3 mm. Since |Delta d| = 5.1 x 10^-3 mm < 7.5 x 10^-3 mm, the Steel alloy meets the diameter reduction criterion."
},
{
"idx": 412,
"question": "For the Titanium alloy (Modulus of Elasticity = 107 GPa, Poisson's Ratio = 0.34) under the same loading conditions, will the diameter reduction be less than 7.5 x 10^-3 mm?",
"answer": "Calculate the diameter reduction: Delta d = - (v * sigma * d0) / E = - (0.34 * 350 MPa * 10 mm) / 107,000 MPa = -11.1 x 10^-3 mm. Since |Delta d| = 11.1 x 10^-3 mm > 7.5 x 10^-3 mm, the Titanium alloy does not meet the diameter reduction criterion."
},
{
"idx": 413,
"question": "A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60 mm (0.260 in.). Calculate the ductility in terms of percent reduction in area.",
"answer": "Percent reduction in area is computed using the equation: %RA = (π(d0/2)^2 - π(df/2)^2) / π(d0/2)^2 × 100, where d0 and df are the original and fracture diameters. Thus, %RA = (π(12.8 mm/2)^2 - π(6.60 mm/2)^2) / π(12.8 mm/2)^2 × 100 = 73.4%."
},
{
"idx": 413,
"question": "A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The fractured gauge length is 72.14 mm (2.840 in.). Calculate the ductility in terms of percent elongation.",
"answer": "Percent elongation is computed using the equation: %EL = (lf - l0) / l0 × 100, where l0 and lf are the original and fractured gauge lengths. Thus, %EL = (72.14 mm - 50.80 mm) / 50.80 mm × 100 = 42%."
},
{
"idx": 414,
"question": "For a metal alloy with a strain-hardening exponent n of 0.25, a true stress of 415 MPa produces a plastic true strain of 0.475. Calculate the strength coefficient K.",
"answer": "The strength coefficient K can be calculated using the equation K = σ_T / (ε_T)^n. Substituting the given values: K = 415 MPa / (0.475)^0.25 = 500 MPa (72,500 psi)."
},
{
"idx": 414,
"question": "Using the calculated strength coefficient K of 500 MPa and strain-hardening exponent n of 0.25, determine the true strain ε_T when a true stress of 325 MPa is applied.",
"answer": "The true strain ε_T can be calculated using the equation ε_T = (σ_T / K)^(1/n). Substituting the given values: ε_T = (325 MPa / 500 MPa)^(1/0.25) = 0.179."
},
{
"idx": 414,
"question": "Given the original length l0 of 300 mm and the calculated true strain ε_T of 0.179, find the final length li after elongation.",
"answer": "The final length li can be calculated using the equation li = l0 * e^(ε_T). Substituting the given values: li = 300 mm * e^(0.179) = 358.8 mm (14.11 in)."
},
{
"idx": 414,
"question": "Calculate the elongation Δl of the specimen using the original length l0 of 300 mm and the final length li of 358.8 mm.",
"answer": "The elongation Δl is calculated as Δl = li - l0 = 358.8 mm - 300 mm = 58.8 mm (2.31 in)."
},
{
"idx": 415,
"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": 416,
"question": "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": "The total length in 1000 mm^3 of material having a density of 10^4 mm^-2 is (10^4 mm^-2)(1000 mm^3) = 10^7 mm = 10^4 m = 6.2 mi."
},
{
"idx": 416,
"question": "Suppose that the dislocation 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 total length is (10^10 mm^-2)(1000 mm^3) = 10^13 mm = 10^10 m = 6.2 x 10^6 mi."
},
{
"idx": 417,
"question": "For an 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": 417,
"question": "For a 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": 417,
"question": "For a 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": 418,
"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": 418,
"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": 419,
"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 \\mathrm{MPa}\\) (3000 psi), will an applied stress of \\(45 \\mathrm{MPa}(6500 \\mathrm{psi})\\) cause the single crystal to yield? If not, what stress will be necessary?",
"answer": "This problem calls for us to determine whether or not a metal single crystal having a specific orientation and of given critical resolved shear stress will yield. We are given that \\(\\phi=43.1^{\\circ}, \\lambda=47.9^{\\circ}\\), and that the values of the critical resolved shear stress and applied tensile stress are \\(20.7 \\mathrm{MPa}(3000 \\mathrm{psi})\\) and \\(45 \\mathrm{MPa}(6500 \\mathrm{psi})\\), respectively. From Equation \n\\[\n\\tau_{R}=\\sigma \\cos \\phi \\cos \\lambda=(45 \\mathrm{MPa})\\left(\\cos 43.1^{\\circ}\\right)\\left(\\cos 47.9^{\\circ}\\right)=22.0 \\mathrm{MPa} \\quad \\text { (3181 psi) }\n\\]\nSince the resolved shear stress \\((22 \\mathrm{MPa})\\) is greater than the critical resolved shear stress \\((20.7 \\mathrm{MPa})\\), the single crystal will yield."
},
{
"idx": 420,
"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": "Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the largest cos λ. Cosines for the possible λ values are given below. cos(62.4 degrees) = 0.46, cos(72.0 degrees) = 0.31, cos(81.1 degrees) = 0.15. Thus, the slip direction is at an angle of 62.4 degrees with the tensile axis."
},
{
"idx": 420,
"question": "If plastic deformation begins at a tensile stress of 1.95 MPa (280 psi) for the aluminum crystal oriented as described, determine the critical resolved shear stress for aluminum.",
"answer": "The critical resolved shear stress is given by τ_crss = σ_y (cos φ cos λ)_max. Using φ = 28.1 degrees and λ = 62.4 degrees, τ_crss = (1.95 MPa)[cos(28.1 degrees) cos(62.4 degrees)] = 0.80 MPa (114 psi)."
},
{
"idx": 421,
"question": "The critical resolved shear stress for iron is \\(27 \\mathrm{MPa}(4000 \\mathrm{psi})\\). Determine the maximum possible yield strength for a single crystal of \\(\\mathrm{Fe}\\) pulled in tension.",
"answer": "In order to determine the maximum possible yield strength for a single crystal of \\(\\mathrm{Fe}\\) pulled in tension, we simply employ Equation as\n\\[\n\\sigma_{y}=2 \\tau_{\\text {cross }}=(2)(27 \\mathrm{MPa})=54 \\mathrm{MPa} \\quad(8000 \\mathrm{psi})\n\\]\n\n\nDeformation by Twinning\n}"
},
{
"idx": 422,
"question": "What is the difference between deformation by twinning and deformation by slip relative to crystallographic reorientation?",
"answer": "With slip deformation there is no crystallographic reorientation, whereas with twinning there is a reorientation."
},
{
"idx": 422,
"question": "What is the difference between deformation by twinning and deformation by slip relative to atomic displacements?",
"answer": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples."
},
{
"idx": 422,
"question": "What is the difference between deformation by twinning and deformation by slip relative to the conditions of occurrence in metals?",
"answer": "Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems."
},
{
"idx": 422,
"question": "What is the difference between deformation by twinning and deformation by slip relative to the final deformation result?",
"answer": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning."
},
{
"idx": 423,
"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": 424,
"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": 425,
"question": "For the first cylindrical specimen with initial radius of 16 mm and deformed radius of 11 mm, calculate the percent cold work (%CW).",
"answer": "The percent cold work (%CW) for the first specimen is calculated as follows: %CW = (A0 - Ad)/A0 × 100 = (πr0² - πrd²)/πr0² × 100 = (π(16 mm)² - π(11 mm)²)/π(16 mm)² × 100 = 52.7% CW."
},
{
"idx": 425,
"question": "For the second cylindrical specimen with initial radius of 12 mm to achieve the same deformed hardness as the first specimen (52.7% CW), compute its deformed radius (rd).",
"answer": "The deformed radius (rd) for the second specimen is computed using the equation: rd = r0 √(1 - %CW/100) = (12 mm) √(1 - 52.7% CW/100) = 8.25 mm."
},
{
"idx": 426,
"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 \\times 175\\) \\\\\nDeformed dimensions & 11.4 & \\(75 \\times 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 one that has experienced the greatest degree of cold work. Therefore, all we need do is to compute the \\%CW for each specimen using Equation. For the circular one\n\\[\n\\begin{aligned}\n\\% \\mathrm{CW} & =\\left[\\frac{A_{0}-A_{u}}{A_{0}}\\right] \\times 100 \\\\\n& =\\left[\\frac{\\pi r_{0}^{2}-\\pi r_{d}^{2}}{\\pi r_{0}^{2}}\\right] \\times 100 \\\\\n& =\\left[\\frac{\\pi\\left(\\frac{15.2 \\mathrm{~mm}}{2}\\right)^{2}}{\\pi\\left(\\frac{15.2 \\mathrm{~mm}}{2} \\right)^{2}}-\\frac{\\pi\\left(\\frac{11.4 \\mathrm{~mm}}{2}\\right)^{2}}{2}\\right] \\times 100=43.8 \\% \\mathrm{CW}\n\\end{aligned}\n\\]\nFor the rectangular one\n\\[\n\\% \\mathrm{CW}=\\left[\\frac{(125 \\mathrm{~mm})(175 \\mathrm{~mm})-(75 \\mathrm{~mm})(200 \\mathrm{~mm})}{(125 \\mathrm{~mm})(175 \\mathrm{~mm})}\\right] \\times 100=31.4 \\% \\mathrm{CW}\n\\]\nTherefore, the deformed circular specimen will be harder."
},
{
"idx": 427,
"question": "A cylindrical specimen of cold-worked copper has a ductility (\\%EL) of \\(25 \\%\\). If its cold-worked radius is \\(10 \\mathrm{~mm}\\) (0.40 in.), what was its radius before deformation?",
"answer": "This problem calls for us to calculate the precold-worked radius of a cylindrical specimen of copper that has a cold-worked ductility of \\(25 \\% \\mathrm{EL}\\). From Figure 7.19c, copper that has a ductility of \\(25 \\% \\mathrm{EL}\\) will have experienced a deformation of about \\(11 \\% \\mathrm{CW}\\). For a cylindrical specimen, Equation becomes\n\\[\n\\% \\mathrm{CW}=\\left[\\frac{\\pi r_{0}^{2}-\\pi r_{d}^{2}}{\\pi r_{0}^{2}}\\right] \\times 100\n\\]\nSince \\(r_{d}=10 \\mathrm{~mm}(0.40 \\mathrm{in}\\).), solving for \\(r_{0}\\) yields\n\\[\nr_{0}=\\frac{r_{d}}{\\sqrt{1-\\frac{96 \\mathrm{CW}}{100}}}=\\frac{10 \\mathrm{~mm}}{\\sqrt{1-\\frac{11.0}{100}}}=10.6 \\mathrm{~mm} \\quad(0.424 \\mathrm{in}) .\n\\]"
},
{
"idx": 428,
"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": 429,
"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 \\times 10^{-4} \\mathrm{~mm}\\left(10^{-5} \\mathrm{in.}\\right)\\) and a crack length of \\(2.5 \\times 10^{-2} \\mathrm{~mm}\\left(10^{-3} \\mathrm{in.}\\right)\\) when a tensile stress of 170 \\(\\mathrm{MPa}(25,000 \\mathrm{psi})\\) is applied?",
"answer": "This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an internal crack. Equation is employed to solve this problem, as\n\\[\n\\begin{aligned}\n\\sigma_{m} & =2 \\sigma_{0}\\left(\\frac{a}{\\rho_{t}}\\right)^{1 / 2} \\\\\n& =\\left(2)\\left(170 \\mathrm{MPa}\\right)\\left[\\frac{2.5 \\times 10^{-2} \\mathrm{~mm}}{\\frac{2}{2.5 \\times 10^{-4} \\mathrm{~mm}}}\\right]^{1 / 2}=2404 \\mathrm{MPa}(354,000 \\mathrm{psi})\n\\end{aligned}\n\\]"
},
{
"idx": 430,
"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 \\mathrm{~mm}(0.01 \\mathrm{in}\\).) and having a tip radius of curvature of \\(1.2 \\times 10^{-3} \\mathrm{~mm}\\left(4.7 \\times 10^{-5} \\mathrm{in}\\right.\\).) when a stress of \\(1200 \\mathrm{MPa}(174,000 \\mathrm{psi})\\) is applied.",
"answer": "In order to estimate the theoretical fracture strength of this material it is necessary to calculate \\(\\sigma_{m}\\) using Equation given that \\(\\sigma_{0}=1200 \\mathrm{MPa}, a=0.25 \\mathrm{~mm}\\), and \\(\\rho_{t}=1.2 \\times 10^{-3} \\mathrm{~mm}\\). Thus,\n\\[\n\\begin{aligned}\n\\sigma_{m} & =2 \\sigma_{0}\\left(\\frac{a}{\\rho_{t}}\\right)^{1 / 2} \\\\\n& =\\left(2)\\left(1200 \\mathrm{MPa}\\left[\\frac{0.25 \\mathrm{~mm}}{1.2 \\times 10^{-3} \\mathrm{~mm}}\\right]^{1 / 2}=3.5 \\times 10^{4} \\mathrm{MPa} \\quad\\left(5.1 \\times 10^{6} \\mathrm{psi}\\right)\\right.\n\\end{aligned}\n\\]"
},
{
"idx": 431,
"question": "If the specific surface energy for soda-lime glass is \\(0.30 \\mathrm{~J} / \\mathrm{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 \\mathrm{~mm}\\).",
"answer": "We may determine the critical stress required for the propagation of an surface crack in soda-lime glass using Equation; taking the value of \\(69 \\mathrm{GPa}\\) (Table 12.5) as the modulus of elasticity, we get\n\\[\n\\begin{aligned}\n\\sigma_{c} & =\\left[\\frac{2 E_{\\gamma_{\\mathrm{S}}}}{\\pi a}\\right]^{1 / 2} \\\\\n& =\\left[\\frac{(2)\\left(69 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)(0.30 \\mathrm{~N} / \\mathrm{m})}{(\\pi)\\left(0.05 \\times 10^{-3} \\mathrm{~m}\\right)}\\right]^{1 / 2}=16.2 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=16.2 \\mathrm{MPa}"
},
{
"idx": 432,
"question": "A polystyrene component must not fail when a tensile stress of \\(1.25 \\mathrm{MPa}(180 \\mathrm{psi})\\) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is \\(0.50 \\mathrm{~J} / \\mathrm{m}^{2}\\left(2.86 \\times 10^{-3}\\right.\\) in.-\\(\\left.\\mathrm{lb}_{\\mathrm{f}} / \\mathrm{in}^{2}\\right)\\). Assume a modulus of elasticity of \\(3.0 \\mathrm{GPa}\\left(0.435 \\times 10^{6} \\mathrm{psi}\\right)\\).",
"answer": "The maximum allowable surface crack length for polystyrene may be determined using Equation; taking \\(3.0 \\mathrm{GPa}\\) as the modulus of elasticity, and solving for \\(a\\), leads to\n\\[\n\\begin{aligned}\na= & \\frac{2 E \\gamma_{S}}{\\pi \\sigma_{C}^{2}}=\\frac{(2)\\left(3 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)(0.50 \\mathrm{~N} / \\mathrm{m})}{(\\pi)\\left(1.25 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}\\right)^{2}} \\\\\n= & 6.1 \\times 10^{-4} \\mathrm{~m}=0.61 \\mathrm{~mm}(0.024 \\mathrm{in} .)\n\\end{aligned}\n\\]"
},
{
"idx": 433,
"question": "Given a 4340 steel alloy specimen with a plane strain fracture toughness of 45 MPa sqrt(m) (41 ksi sqrt(in)), a stress of 1000 MPa (145,000 psi), and the largest surface crack of 0.75 mm (0.03 in) long with parameter Y = 1.0, calculate the critical stress (σ_c) required for fracture.",
"answer": "The critical stress (σ_c) is calculated using the formula σ_c = K_IC / (Y * sqrt(π * a)). Substituting the given values: σ_c = 45 MPa sqrt(m) / (1.0 * sqrt(π * 0.75 × 10^-3 m)) = 927 MPa (133,500 psi)."
},
{
"idx": 433,
"question": "Will the 4340 steel alloy specimen fracture under the given stress of 1000 MPa (145,000 psi) when the critical stress (σ_c) is 927 MPa (133,500 psi)? Why or why not?",
"answer": "Fracture will most likely occur because the applied stress of 1000 MPa (145,000 psi) exceeds the critical stress of 927 MPa (133,500 psi) that the specimen can tolerate before fracture."
},
{
"idx": 434,
"question": "For an aluminum alloy with a plane strain fracture toughness of 35 MPa sqrt(m), given that fracture occurs at a stress of 250 MPa when the maximum internal crack length is 2.0 mm, calculate the parameter Y.",
"answer": "Y is calculated using the formula Y = K_Ic / (σ sqrt(πa)), where K_Ic = 35 MPa sqrt(m), σ = 250 MPa, and a = 1.0 mm (half of 2.0 mm). Thus, Y = 35 MPa sqrt(m) / (250 MPa sqrt(π × 1.0 × 10^-3 m)) = 2.50."
},
{
"idx": 434,
"question": "Using the calculated Y value of 2.50, determine if fracture will occur for the same alloy under a stress level of 325 MPa and a maximum internal crack length of 1.0 mm.",
"answer": "Calculate Yσ sqrt(πa) with Y = 2.50, σ = 325 MPa, and a = 0.5 mm (half of 1.0 mm). Thus, Yσ sqrt(πa) = 2.50 × 325 MPa × sqrt(π × 0.5 × 10^-3 m) = 32.2 MPa sqrt(m). Since 32.2 MPa sqrt(m) is less than the K_Ic of 35 MPa sqrt(m), fracture will not occur."
},
{
"idx": 435,
"question": "For an aluminum alloy with a plane strain fracture toughness of 40 MPa sqrt(m), given that fracture occurs at a stress of 365 MPa when the maximum internal crack length is 2.5 mm, determine the parameter Y for these conditions.",
"answer": "Y = (K_IC) / (σ sqrt(πa)) = (40 MPa sqrt(m)) / (365 MPa sqrt(π (2.5 × 10^-3 m / 2))) = 1.75"
},
{
"idx": 435,
"question": "Using the determined parameter Y (1.75) and the same fracture toughness (40 MPa sqrt(m)), compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm.",
"answer": "σ_c = (K_IC) / (Y sqrt(πa)) = (40 MPa sqrt(m)) / (1.75 sqrt(π (4 × 10^-3 m / 2))) = 288 MPa (41,500 psi)"
},
{
"idx": 436,
"question": "A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of \\(55 \\mathrm{MPa} \\sqrt{\\mathrm{m}}(50 \\mathrm{ksi} \\sqrt{\\mathrm{in}}\\).). If, during service use, the plate is exposed to a tensile stress of \\(200 \\mathrm{MPa}(29,000 \\mathrm{psi})\\), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for \\(Y\\).",
"answer": "For this problem, we are given values of \\(K_{I c}(55 \\mathrm{MPa} \\sqrt{\\mathrm{m}}), \\sigma(200 \\mathrm{MPa})\\), and \\(Y(1.0)\\) for a large plate and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is to solve for \\(a_{c}\\) using Equation; therefore\n\\[\na_{c}=\\frac{1}{\\pi}\\left(\\frac{K_{I c}}{Y \\sigma}\\right)^{2}=\\frac{1}{\\pi}\\left[\\frac{55 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(1.0)(200 \\mathrm{MPa})}\\right]^{2}=0.024 \\mathrm{~m}=24 \\mathrm{~mm} \\quad(0.95 \\mathrm{in} .)\n\\]"
},
{
"idx": 437,
"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": "This problem asks us to calculate the maximum internal crack length allowable for the 7075-T651 aluminum alloy in Table 8.1 given that it is loaded to a stress level equal to one-half of its yield strength. For this alloy, \\(K_{I c}=24 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) (22 \\(\\mathrm{ksi} \\sqrt{\\mathrm{in}}\\) ); also, \\(\\sigma=\\sigma_{y} / 2=(495 \\mathrm{MPa}) / 2=248 \\mathrm{MPa}(36,000 \\mathrm{psi})\\). Now solving for \\(2 a_{c}\\) using Equation yields\n\\[\n2 a_{c}=\\frac{2}{\\pi}\\left(\\frac{K_{I c}}{Y \\sigma}\\right)^{2}=\\frac{2}{\\pi}\\left[\\frac{24 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(1.35)(248 \\mathrm{MPa})}\\right]^{2}=0.0033 \\mathrm{~m}=3.3 \\mathrm{~mm} \\quad(0.13 \\mathrm{in} .)\n\\]"
},
{
"idx": 438,
"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 \\mathrm{MPa} \\sqrt{\\mathrm{m}}(70.1 \\mathrm{ksi} \\sqrt{\\mathrm{m}}\\).) and a yield strength of \\(1400 \\mathrm{MPa}(205,000 \\mathrm{psi})\\). The flaw size resolution limit of the flaw detection apparatus is \\(4.0 \\mathrm{~mm}(0.16 \\mathrm{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": "This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus \\((4.0 \\mathrm{~mm})\\), the value of \\(K_{I C}\\left(77 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right)\\), the design stress \\(\\left(\\sigma_{y} / 2\\right.\\) in which \\(\\sigma_{y}=1400 \\mathrm{MPa}\\) ), and \\(Y=1.0\\). We first need to compute the value of \\(a_{c}\\) using Equation; thus\n\\[\na_{c}=\\frac{1}{\\pi}\\left(\\frac{K_{I C}}{Y \\sigma}\\right)^{2}=\\frac{1}{\\pi}\\left[\\frac{77}{(1.0)\\left(\\frac{\\mathrm{MPa} \\sqrt{\\mathrm{m}}}{1400 \\mathrm{MPa}}\\right)^{2}}\\right]^{2}=0.0039 \\mathrm{~m}=3.9 \\mathrm{~mm} \\quad(0.15 \\mathrm{in} .)\n\\]\nTherefore, the critical flaw is not subject to detection since this value of \\(a_{c}(3.9 \\mathrm{~mm})\\) is less than the \\(4.0 \\mathrm{~mm}\\) resolution limit."
},
{
"idx": 439,
"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": 440,
"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": 440,
"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": 441,
"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": 442,
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for nickel, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
"answer": "For Ni, 0.4 Tm = (0.4)(1455 + 273) = 691 K or 418 degrees Celsius (785 degrees Fahrenheit)"
},
{
"idx": 442,
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for copper, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
"answer": "For Cu, 0.4 Tm = (0.4)(1085 + 273) = 543 K or 270 degrees Celsius (518 degrees Fahrenheit)"
},
{
"idx": 442,
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for iron, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
"answer": "For Fe, 0.4 Tm = (0.4)(1538 + 273) = 725 K or 450 degrees Celsius (845 degrees Fahrenheit)"
},
{
"idx": 442,
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for tungsten, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
"answer": "For W, 0.4 Tm = (0.4)(3410 + 273) = 1473 K or 1200 degrees Celsius (2190 degrees Fahrenheit)"
},
{
"idx": 442,
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for lead, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
"answer": "For Pb, 0.4 Tm = (0.4)(327 + 273) = 240 K or -33 degrees Celsius (-27 degrees Fahrenheit)"
},
{
"idx": 442,
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for aluminum, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
"answer": "For Al, 0.4 Tm = (0.4)(660 + 273) = 373 K or 100 degrees Celsius (212 degrees Fahrenheit)"
},
{
"idx": 443,
"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": 444,
"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": 445,
"question": "For the alloy composition 60 wt% A-40 wt% B with fraction α phase 0.57 and fraction β phase 0.43, determine the composition of the phase boundary (or solubility limit) for both α and β phases at this temperature.",
"answer": "Using the lever rule expression for this alloy composition: W_α1 = 0.57 = (C_β - C_01) / (C_β - C_α) = (C_β - 60) / (C_β - C_α). This equation relates the phase boundary compositions C_α and C_β for the given alloy."
},
{
"idx": 445,
"question": "For the alloy composition 30 wt% A-70 wt% B with fraction α phase 0.14 and fraction β phase 0.86, determine the composition of the phase boundary (or solubility limit) for both α and β phases at this temperature.",
"answer": "Using the lever rule expression for this alloy composition: W_α2 = 0.14 = (C_β - C_02) / (C_β - C_α) = (C_β - 30) / (C_β - C_α). This equation relates the phase boundary compositions C_α and C_β for the given alloy."
},
{
"idx": 445,
"question": "Solve the system of equations derived from the lever rule expressions for both alloy compositions to determine the phase boundary compositions C_α and C_β.",
"answer": "Solving the system of equations: 0.57 = (C_β - 60) / (C_β - C_α) and 0.14 = (C_β - 30) / (C_β - C_α), yields the phase boundary compositions: C_α = 90 wt% A-10 wt% B and C_β = 20.2 wt% A-79.8 wt% B."
},
{
"idx": 446,
"question": "Given an A-B alloy with composition 55 wt% B-45 wt% A at a certain temperature, mass fractions of 0.5 for both α and β phases, and the composition of the β phase as 90 wt% B-10 wt% A, what is the composition of the α phase?",
"answer": "Using the lever rule for Wα: Wα = 0.5 = (Cβ - C0) / (Cβ - Cα) = (90 - 55) / (90 - Cα). Solving for Cα: Cα = 20 (or 20 wt% B-80 wt% A)."
},
{
"idx": 447,
"question": "For a 11.20 kg magnesium-lead alloy of composition 30 wt% Pb-70 wt% Mg, is it possible, at equilibrium, to have alpha and Mg2Pb phases with masses of 7.39 kg and 3.81 kg respectively?",
"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 alpha and Mg2Pb phases, respectively. The mass fractions are calculated as follows: W_alpha = m_alpha / (m_alpha + m_Mg2Pb) = 7.39 kg / (7.39 kg + 3.81 kg) = 0.66, W_Mg2Pb = 1.00 - 0.66 = 0.34."
},
{
"idx": 447,
"question": "What is the composition of the alpha phase (C_alpha) in the magnesium-lead alloy, given the mass fractions and overall composition?",
"answer": "The composition of the alpha phase (C_alpha) is determined using the lever rule: W_alpha = (C_Mg2Pb - C_0) / (C_Mg2Pb - C_alpha). Given C_Mg2Pb = 81 wt% Pb and C_0 = 30 wt% Pb, solving for C_alpha yields C_alpha = 3.7 wt% Pb."
},
{
"idx": 447,
"question": "What is the approximate temperature of the alloy given the composition of the alpha phase (C_alpha = 3.7 wt% Pb)?",
"answer": "The approximate temperature of the alloy, corresponding to the alpha phase composition of 3.7 wt% Pb along the alpha-(alpha + Mg2Pb) phase boundary, is approximately 190 degrees Celsius."
},
{
"idx": 448,
"question": "(a) 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": 448,
"question": "(b) 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": 449,
"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": 450,
"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": 451,
"question": "Given the intermetallic compound AB with composition 34.3 wt% A-65.7 wt% B, and knowing that element A is potassium (atomic weight 39.10 g/mol), determine the atomic weight of element B.",
"answer": "First, consider the AB intermetallic compound, which has equal numbers of A and B atoms, resulting in 50 at% A-50 at% B. Using the equation for atomic percent composition: CB = (CB * AA) / (CA * AB + CB * AA) * 100, where AA and AB are the atomic weights of A and B, and CA and CB are their weight percentages. Substituting the given values: 50 at% B = (65.7 wt% B * 39.10 g/mol) / (34.3 wt% A * AB + 65.7 wt% B * 39.10 g/mol) * 100. Solving this expression yields AB = 1.916 * 39.10 g/mol = 74.92 g/mol."
},
{
"idx": 451,
"question": "Identify element B from its atomic weight of 74.92 g/mol.",
"answer": "Upon consulting the periodic table, the element with an atomic weight closest to 74.92 g/mol is arsenic (As). Therefore, element B is arsenic, and the intermetallic compounds are KAs and KAs2."
},
{
"idx": 452,
"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": 453,
"question": "Compute the mass fraction of α ferrite in pearlite, given C_Fe3C = 6.70 wt% C, C_0 = 0.76 wt% C, and C_α = 0.022 wt% C.",
"answer": "The lever-rule expression for ferrite is W_α = (C_Fe3C - C_0) / (C_Fe3C - C_α). Substituting the given values: W_α = (6.70 - 0.76) / (6.70 - 0.022) = 0.89."
},
{
"idx": 453,
"question": "Compute the mass fraction of cementite in pearlite, given C_Fe3C = 6.70 wt% C, C_0 = 0.76 wt% C, and C_α = 0.022 wt% C.",
"answer": "The lever-rule expression for cementite is W_Fe3C = (C_0 - C_α) / (C_Fe3C - C_α). Substituting the given values: W_Fe3C = (0.76 - 0.022) / (6.70 - 0.022) = 0.11."
},
{
"idx": 454,
"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": 454,
"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": 455,
"question": "What is the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.94 ?",
"answer": "This problem asks that we compute the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 9.94. Application of the lever rule yields\n\\[\nW_{\\alpha}=0.94=\\frac{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{0}^{\\prime}}{C_{\\mathrm{Fe}_{2} \\mathrm{C}}-C_{\\alpha}}=\\frac{6.70-C_{0}^{\\prime}}{6.70-0.022}\n\\]\nand solving for \\(C_{0}^{\\prime}\\)\n\\[\nC_{0}^{\\prime}=0.42 \\mathrm{wt} \\% \\mathrm{C}\n\\]"
},
{
"idx": 456,
"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?",
"answer": "In this problem we are given values of W_α and W_Fe3C (0.92 and 0.08, respectively) for an iron-carbon alloy and then are asked to specify the proeutectoid phase. Employment of the lever rule for total α leads to W_α=0.92=(C_Fe3C-C_0)/(C_Fe3C-C_α)=(6.70-C_0)/(6.70-0.022). Now, solving for C_0, the alloy composition, leads to C_0=0.56 wt% C. Therefore, the proeutectoid phase is α ferrite since C_0 is less than 0.76 wt% C."
},
{
"idx": 456,
"question": "Why is the proeutectoid phase α ferrite for an iron-carbon alloy with mass fractions of total ferrite and total cementite of 0.92 and 0.08, respectively?",
"answer": "The proeutectoid phase is α ferrite because the calculated alloy composition C_0=0.56 wt% C is less than the eutectoid composition of 0.76 wt% C, which is the threshold below which the proeutectoid phase is α ferrite."
},
{
"idx": 457,
"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 will be Fe3C since 1.15 wt% C is greater than the eutectoid composition (0.76 wt% C)."
},
{
"idx": 457,
"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": "For total ferrite: Wα = (CFe3C - C0) / (CFe3C - Cα) = (6.70 - 1.15) / (6.70 - 0.022) = 0.83, which gives 0.83 kg of total ferrite. For total cementite: WFe3C = (C0 - Cα) / (CFe3C - Cα) = (1.15 - 0.022) / (6.70 - 0.022) = 0.17, which gives 0.17 kg of total cementite."
},
{
"idx": 457,
"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": "For pearlite: Wp = (6.70 - C1') / (6.70 - 0.76) = (6.70 - 1.15) / (6.70 - 0.76) = 0.93, which gives 0.93 kg of pearlite. For the proeutectoid phase (cementite): WFe3C = (C1' - 0.76) / 5.94 = (1.15 - 0.76) / 5.94 = 0.07, which gives 0.07 kg of the proeutectoid phase."
},
{
"idx": 458,
"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 since 0.65 wt% C is less than 0.76 wt% C."
},
{
"idx": 458,
"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": "For ferrite, application of the appropriate lever rule expression yields W_α = (6.70 - 0.65)/(6.70 - 0.022) = 0.91, which corresponds to (0.91)(2.5 kg) = 2.27 kg of total ferrite. Similarly, for total cementite, W_Fe3C = (0.65 - 0.022)/(6.70 - 0.022) = 0.09, or (0.09)(2.5 kg) = 0.23 kg of total cementite form."
},
{
"idx": 458,
"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": "Using the equation W_p = (0.65 - 0.022)/0.74 = 0.85, this corresponds to (0.85)(2.5 kg) = 2.12 kg of pearlite. Also, from the equation W_α' = (0.76 - 0.65)/0.74 = 0.15, or there are (0.15)(2.5 kg) = 0.38 kg of proeutectoid ferrite."
},
{
"idx": 459,
"question": "Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron-carbon alloy containing \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\).",
"answer": "The mass fractions of proeutectoid ferrite and pearlite that form in a \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\) iron-carbon alloy are considered in this problem. From Equation\n\\[\nW_{p}=\\frac{C_{0}^{\\prime}-0.022}{0.74}=\\frac{0.25-0.022}{0.74}=0.31\n\\]\nAnd, from Equation (for proeutectoid ferrite)\n\\[\nW_{Q^{\\prime}}=\\frac{0.76-C_{0}^{\\prime}}{0.74}=\\frac{0.76-0.25}{0.74}=0.69\n\\]"
},
{
"idx": 460,
"question": "Given the mass fraction of pearlite (Wp) is 0.714 in an iron-carbon alloy, determine the carbon concentration (C0) in the alloy using the equation Wp = (C0 - 0.022) / 0.74.",
"answer": "Using the equation Wp = (C0 - 0.022) / 0.74, where Wp = 0.714, we solve for C0: 0.714 = (C0 - 0.022) / 0.74. Rearranging gives C0 - 0.022 = 0.714 * 0.74. Calculating the right side: 0.714 * 0.74 = 0.52836. Thus, C0 = 0.52836 + 0.022 = 0.55036 wt% C. The carbon concentration in the alloy is approximately 0.55 wt% C."
},
{
"idx": 461,
"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": "In this problem we are given values of \\(W_{\\alpha}\\) and \\(W_{\\mathrm{Fe}_{3} \\mathrm{C}}\\) for an iron-carbon alloy ( 0.88 and 0.12 , respectively), and then are asked to specify whether the alloy is hypoeutectoid or hypereutectoid. Employment of the lever rule for total \\(\\alpha\\) leads to\n\\[\nW_{\\alpha}=0.88=\\frac{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{0}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\alpha}}=\\frac{6.70-C_{0}}{6.70-0.022}\n\\]\nNow, solving for \\(C_{0}\\), the alloy composition, leads to \\(C_{0}=0.82 \\mathrm{wt} \\% \\mathrm{C}\\). Therefore, the alloy is hypereutectoid since \\(C_{0}\\) is greater than \\(0.76 \\mathrm{wt} \\% \\mathrm{C}\\)."
},
{
"idx": 462,
"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": "We are asked in this problem to determine the concentration of carbon in an alloy for which \\(W_{\\alpha^{\\prime}}=0.20\\) and \\(W_{\\rho}=0.80\\). If we let \\(C_{0}^{\\prime}\\) equal the carbon concentration in the alloy, employment of the appropriate lever rule expression, Equation, leads to\n\\[\nW_{p}=\\frac{C_{0}^{\\prime}-0.022}{0.74}=0.80\n\\]\nSolving for \\(C_{0}^{\\prime}\\) yields \\(C_{0}^{\\prime}=0.61 \\mathrm{wt} \\% \\mathrm{C}\\)."
},
{
"idx": 463,
"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": "To compute the amount of proeutectoid ferrite that forms, the following equation is used: W_alpha' = (0.76 - C0') / 0.74 = (0.76 - 0.40) / 0.74 = 0.49. Or, (0.49)(2.0 kg) = 0.98 kg of proeutectoid ferrite forms."
},
{
"idx": 463,
"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": "To determine the amount of eutectoid ferrite, first compute the amount of total ferrite using the lever rule applied across the alpha + Fe3C phase field: W_alpha = (C_Fe3C - C0') / (C_Fe3C - C_alpha) = (6.70 - 0.40) / (6.70 - 0.022) = 0.94, which corresponds to (0.94)(2.0 kg) = 1.88 kg. The amount of eutectoid ferrite is the difference between total and proeutectoid ferrites: 1.88 kg - 0.98 kg = 0.90 kg."
},
{
"idx": 463,
"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": "To determine the amount of cementite that forms, apply the lever rule across the alpha + Fe3C phase field: W_Fe3C = (C0' - C_alpha) / (C_Fe3C - C_alpha) = (0.40 - 0.022) / (6.70 - 0.022) = 0.057, which amounts to (0.057)(2.0 kg) = 0.114 kg cementite in the alloy."
},
{
"idx": 464,
"question": "Compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid ironcarbon alloy.",
"answer": "This problem asks that we compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoidiron-carbon alloy. This requires that we utilize Equation 9.23 with \\(C_{1}^{\\prime}=2.14 \\mathrm{wt} \\% \\mathrm{C}\\), the maximum solubility of carbon in austenite. Thus,\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}^{\\prime}}=\\frac{C_{1}^{\\prime}-0.76}{5.94}=\\frac{2.14-0.76}{5.94}=0.232\n\\]"
},
{
"idx": 465,
"question": "Is it possible to have an iron-carbon alloy for which the mass fraction of total ferrite (W_α) is 0.846? Why or why not?",
"answer": "To determine if an iron-carbon alloy with W_α=0.846 is possible, we use the lever rule expression for the mass fraction of total ferrite: W_α=(C_Fe3C - C_0)/(C_Fe3C - C_α)=(6.70 - C_0)/(6.70 - 0.022)=0.846. Solving for C_0 yields C_0=1.05 wt% C. This composition is within the possible range for iron-carbon alloys, so such an alloy is possible."
},
{
"idx": 465,
"question": "Is it possible to have an iron-carbon alloy for which the mass fraction of proeutectoid cementite (W_Fe3C') is 0.049? Why or why not?",
"answer": "To determine if an iron-carbon alloy with W_Fe3C'=0.049 is possible, we use the lever rule expression for the mass fraction of proeutectoid cementite: W_Fe3C'=(C_1' - 0.76)/5.94=0.049. Solving for C_1' yields C_1'=1.05 wt% C. This composition is within the possible range for iron-carbon alloys, so such an alloy is possible."
},
{
"idx": 465,
"question": "Given that both W_α=0.846 and W_Fe3C'=0.049 yield an alloy composition of 1.05 wt% C, is such an iron-carbon alloy possible? Why or why not?",
"answer": "Since both the mass fraction of total ferrite (W_α=0.846) and the mass fraction of proeutectoid cementite (W_Fe3C'=0.049) yield the same alloy composition (C_0=C_1'=1.05 wt% C), and this composition is within the possible range for iron-carbon alloys, such an alloy is possible."
},
{
"idx": 466,
"question": "Is it possible to have an iron-carbon alloy with a mass fraction of total cementite (W_Fe3C) of 0.039? Why or why not?",
"answer": "To determine if an iron-carbon alloy with W_Fe3C = 0.039 is possible, we use the lever rule expression: W_Fe3C = (C_0 - C_a)/(C_Fe3C - C_a) = (C_0 - 0.022)/(6.70 - 0.022) = 0.039. Solving for C_0 yields C_0 = 0.28 wt% C. This value is valid as it falls within the possible composition range for iron-carbon alloys."
},
{
"idx": 466,
"question": "Is it possible to have an iron-carbon alloy with a mass fraction of pearlite (W_p) of 0.417? Why or why not?",
"answer": "To determine if an iron-carbon alloy with W_p = 0.417 is possible, we use the lever rule expression for hypoeutectoid alloys: W_p = (C_0' - 0.022)/0.74 = 0.417. Solving for C_0' yields C_0' = 0.33 wt% C. This value is valid as it falls within the possible composition range for iron-carbon alloys."
},
{
"idx": 466,
"question": "Can an iron-carbon alloy simultaneously have mass fractions of total cementite (W_Fe3C) = 0.039 and pearlite (W_p) = 0.417? Why or why not?",
"answer": "For an alloy to simultaneously have W_Fe3C = 0.039 and W_p = 0.417, the calculated compositions C_0 and C_0' must be equal. However, C_0 = 0.28 wt% C and C_0' = 0.33 wt% C are different. Therefore, such an alloy is not possible."
},
{
"idx": 467,
"question": "Compute the mass fraction of total ferrite (W_α) in an iron-carbon alloy that contains 0.43 wt% C using the lever rule and a tie line that extends across the entire α + Fe3C phase field.",
"answer": "The mass fraction of total ferrite is calculated using the lever rule as W_α = (C_Fe3C - C_0) / (C_Fe3C - C_α) = (6.70 - 0.43) / (6.70 - 0.022) = 0.939."
},
{
"idx": 467,
"question": "Compute the mass fraction of proeutectoid ferrite (W_α') in an iron-carbon alloy that contains 0.43 wt% C.",
"answer": "The mass fraction of proeutectoid ferrite is calculated as W_α' = (0.76 - C_0) / 0.74 = (0.76 - 0.43) / 0.74 = 0.446."
},
{
"idx": 467,
"question": "Compute the mass fraction of eutectoid ferrite (W_α'') in an iron-carbon alloy that contains 0.43 wt% C by subtracting the mass fraction of proeutectoid ferrite from the mass fraction of total ferrite.",
"answer": "The mass fraction of eutectoid ferrite is calculated as W_α'' = W_α - W_α' = 0.939 - 0.446 = 0.493."
},
{
"idx": 468,
"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, in fact, there are two possible answers."
},
{
"idx": 468,
"question": "What is the first possible composition of the alloy where the eutectoid cementite exists in addition to proeutectoid cementite?",
"answer": "For the first case, the mass fraction of eutectoid cementite is the difference between total cementite and proeutectoid cementite mass fractions. The expression is W_Fe3C* = (C0 - C_α)/(C_Fe3C - C_α) - (C0 - 0.76)/5.94 = 0.104. Solving for C0 yields C0 = 1.11 wt% C."
},
{
"idx": 468,
"question": "What is the second possible composition of the alloy where all of the cementite is eutectoid cementite in a hypoeutectoid alloy?",
"answer": "For the second possibility, the mass fraction of total cementite is 0.104. The lever rule expression is W_Fe3C = (C0 - C_α)/(C_Fe3C - C_α) = (C0 - 0.022)/(6.70 - 0.022). Solving for C0 in this case would complete the calculation, but the exact value is not provided in the given answer."
},
{
"idx": 469,
"question": "Is it possible to determine the composition of an iron-carbon alloy for which 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": 469,
"question": "What is the composition of the alloy if the eutectoid ferrite exists in addition to proeutectoid ferrite?",
"answer": "For this case, the mass fraction of eutectoid ferrite (W_α') is the difference between total ferrite and proeutectoid ferrite mass fractions. The expression is W_α' = W_α - W_α'. Solving for C_0 yields C_0 = 0.70 wt% C."
},
{
"idx": 469,
"question": "What is the composition of the alloy if it is a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite?",
"answer": "For this case, the mass fraction of total ferrite is 0.82. The lever rule expression is W_α = (C_Fe3C - C_0) / (C_Fe3C - C_α) = (6.70 - C_0) / (6.70 - 0.022) = 0.82. Solving for C_0 yields C_0 = 1.22 wt% C."
},
{
"idx": 470,
"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": 470,
"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": 471,
"question": "If copper (which has a melting point of \\(1085^{\\circ} \\mathrm{C}\\) ) homogeneously nucleates at \\(849^{\\circ} \\mathrm{C}\\), calculate the critical radius given values of \\(-1.77 \\times 10^{9} \\mathrm{~J} / \\mathrm{m}^{3}\\) and \\(0.200 \\mathrm{~J} / \\mathrm{m}^{2}\\), respectively, for the latent heat of fusion and the surface free energy.",
"answer": "This problem states that copper homogeneously nucleates at \\(849^{\\circ} \\mathrm{C}\\) and that we are to calculate the critical radius given the latent heat of fusion \\(\\left(-1.77 \\times 10^{9} \\mathrm{~J} \\mathrm{~m}^{3}\\right)\\) and the surface free energy \\(\\left(0.200 \\mathrm{~J} / \\mathrm{m}^{2} \\mathrm{~s}\\right)\\). Solution to this problem requires the utilization of Equation as\n\\[\n\\begin{aligned}\nr * & =\\left(-\\frac{2 \\gamma T_{m}}{\\Delta H_{f}}\\right)\\left(\\frac{1}{T_{m}-T}\\right) \\\\\n& =\\left[-\\frac{(2)\\left(0.200 \\mathrm{~J} / \\mathrm{m} ^{2}\\right)(1085+273 \\mathrm{~K})}{-1.77 \\times 10^{9} \\mathrm{~J} / m^{3}}\\right]\\left(\\begin{array}{c}\n1 \\\\\n1085^{\\circ} \\mathrm{C}-849^{\\circ} \\mathrm{C}\n\\end{array}\\right) \\\\\n& =1.30 \\times 10^{-9} \\mathrm{~m}=1.30 \\mathrm{~nm}\n\\end{aligned}\n\\]"
},
{
"idx": 472,
"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": "The first thing necessary is to set up two expressions of the form of Equation, and then to solve simultaneously for the values of \\(n\\) and \\(k\\). In order to expedite this process, we will rearrange and do some algebraic manipulation of Equation. First of all, we rearrange as follows:\n\\[\n1-y=\\exp \\left(-k t^{n}\\right)\n\\]\nNow taking natural logarithms\n\\[\n\\ln (1-y)=-k t^{n}\n\\]\nOr\n\\[\n-\\ln (1-y)=k t^{n}\n\\]\nwhich may also be expressed as\n\\[\n\\ln \\left(\\frac{1}{1-y}\\right)=k t^{n}\n\\]\nNow taking natural logarithms again, leads to\n\\[\n\\ln \\left[\\ln \\left(\\frac{1}{1-y}\\right)\\right]=\\ln k+n \\ln t\n\\]\n\n\nwhich is the form of the equation that we will now use. Using values cited in the problem statement, the two equations are thus\n\\[\n\\begin{array}{l}\n\\ln \\left\\{\\ln \\left[\\frac{1}{1-0.2}\\right]\\right\\}=\\ln k+n \\ln (12.6 \\mathrm{~s}) \\\\\n\\ln \\left\\{\\ln \\left[\\frac{1}{1-0.8}\\right]\\right\\}=\\ln k+n \\ln (28.2 \\mathrm{~s})\n\\end{array}\n\\]\nSolving these two expressions simultaneously for \\(n\\) and \\(k\\) yields \\(n=2.453\\) and \\(k=4.46 \\times 10^{-4}\\).\nNow it becomes necessary to solve for the value of \\(t\\) at which \\(y=0.95\\). One of the above equations-viz\n\\[\n-\\ln (1-y)=k t^{n}\n\\]\nmay be rewritten as\n\\[\nt^{n}=-\\frac{\\ln (1-y)}{k}\n\\]\nAnd solving for \\(t\\) leads to\n\\[\nt=\\left[-\\frac{\\ln (1-y)}{k}\\right]^{1 / n}\n\\]\nNow incorporating into this expression values for \\(n\\) and \\(k\\) determined above, the time required for \\(95 \\%\\) austenite transformation is equal to\n\\[\nt=\\left[-\\frac{\\ln (1-\\frac{0.95}{4.64 \\times 10^{-4}}}\\right]^{1 / 2.453}=35.7 \\mathrm{~s}\n\\]"
},
{
"idx": 473,
"question": "Given the fraction recrystallized-time data for the recrystallization at 600°C of a previously deformed steel, determine the values of n and k in the Avrami relationship using the data points (0.20, 13.1 min) and (0.70, 29.1 min).",
"answer": "To determine the values of n and k, we use the Avrami relationship in the form ln[ln(1/(1-y))] = ln k + n ln t. For the first data point (0.20, 13.1 min): ln[ln(1/(1-0.20))] = ln k + n ln(13.1). For the second data point (0.70, 29.1 min): ln[ln(1/(1-0.70))] = ln k + n ln(29.1). Solving these two equations simultaneously yields n = 2.112 and k = 9.75 × 10^-4."
},
{
"idx": 473,
"question": "Using the Avrami relationship with n = 2.112 and k = 9.75 × 10^-4, determine the fraction recrystallized after a total time of 22.8 min at 600°C.",
"answer": "Using the Avrami relationship y = 1 - exp(-k t^n), we substitute n = 2.112, k = 9.75 × 10^-4, and t = 22.8 min: y = 1 - exp[-(9.75 × 10^-4)(22.8)^2.112] = 0.51."
},
{
"idx": 474,
"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 iron-iron carbide phase diagram."
},
{
"idx": 474,
"question": "In terms of heat treatment and the development of microstructure, what is another major limitation of the iron-iron carbide phase diagram related to time-temperature relationships?",
"answer": "The iron-iron carbide phase 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": 475,
"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": 475,
"question": "Why do superheating and supercooling 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": 476,
"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 by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
},
{
"idx": 476,
"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": 477,
"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": 478,
"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": 478,
"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": 478,
"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": 479,
"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": 479,
"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": 480,
"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": 481,
"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 degrees Celsius per second.",
"answer": "At a cooling rate of 10 degrees Celsius per second, only martensite forms."
},
{
"idx": 481,
"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 degree Celsius per second.",
"answer": "At a cooling rate of 1 degree Celsius per second, both martensite and bainite form."
},
{
"idx": 481,
"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 degrees Celsius per second.",
"answer": "At a cooling rate of 0.1 degrees Celsius per second, martensite, proeutectoid ferrite, and bainite form."
},
{
"idx": 481,
"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 degrees Celsius per second.",
"answer": "At a cooling rate of 0.01 degrees Celsius per second, martensite, proeutectoid ferrite, pearlite, and bainite form."
},
{
"idx": 482,
"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": 483,
"question": "Why is fine pearlite harder and stronger than coarse pearlite in iron-carbon alloys with α-ferrite and cementite phases?",
"answer": "Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner in fine pearlite, resulting in more phase boundary area. The greater phase boundary area increases hardness and strength by impeding dislocation motion and restricting deformation of the ferrite phase near the boundaries."
},
{
"idx": 483,
"question": "Why is coarse pearlite harder and stronger than spheroidite in iron-carbon alloys with α-ferrite and cementite phases?",
"answer": "Coarse pearlite is harder and stronger than spheroidite because the phase boundary area between the sphere-like cementite particles and the ferrite matrix in spheroidite is less than the alternating layered microstructure in coarse pearlite. The reduced boundary area in spheroidite decreases its ability to impede dislocations and restrict deformation."
},
{
"idx": 484,
"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": 485,
"question": "Rank the following iron-carbon alloys and associated microstructures from the highest to the lowest tensile strength: (a) 0.25 wt% C with spheroidite, (b) 0.25 wt% C with coarse pearlite, (c) 0.60 wt% C with fine pearlite, and (d) 0.60 wt% C with coarse pearlite. Justify this ranking.",
"answer": "The ranking from highest to lowest tensile strength is as follows: 0.60 wt% C, fine pearlite; 0.60 wt% C, coarse pearlite; 0.25 wt% C, coarse pearlite; 0.25 wt% C, spheroidite. The 0.25 wt% C, coarse pearlite is stronger than the 0.25 wt% C, spheroidite since coarse pearlite is stronger than spheroidite; the composition of the alloys is the same. The 0.60 wt% C, coarse pearlite is stronger than the 0.25 wt% C, coarse pearlite, since increasing the carbon content increases the strength. Finally, the 0.60 wt% C, fine pearlite is stronger than the 0.60 wt% C, coarse pearlite inasmuch as the strength of fine pearlite is greater than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine pearlite."
},
{
"idx": 486,
"question": "Why does the hardness of tempered martensite diminish 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": 486,
"question": "Why does the hardness of tempered martensite diminish 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 decrease in phase boundary area results in a reduction of hardness."
},
{
"idx": 487,
"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": 487,
"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": 488,
"question": "List the four classifications of steels.",
"answer": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)."
},
{
"idx": 488,
"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": 488,
"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": 488,
"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": 488,
"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": 489,
"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": 489,
"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": 490,
"question": "What is the function of alloying elements in tool steels?",
"answer": "The alloying elements in tool steels (e.g., \\(\\mathrm{Cr}, \\mathrm{V}, \\mathrm{W}\\), and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds."
},
{
"idx": 491,
"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": 492,
"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": 492,
"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": 492,
"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": 493,
"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": 494,
"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": 495,
"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": 496,
"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": 497,
"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": 497,
"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": 497,
"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": 497,
"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": 498,
"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": 498,
"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": 499,
"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": 500,
"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": 500,
"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": 500,
"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": 500,
"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": 500,
"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": 501,
"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": 502,
"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": 502,
"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": 503,
"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": "For 0.20 wt% C, heat to at least 905°C (1660°F) since the A3 temperature is 850°C (1560°F)."
},
{
"idx": 503,
"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": "For 0.76 wt% C, heat to at least 782°C (1440°F) since the A3' temperature is 727°C (1340°F)."
},
{
"idx": 503,
"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": "For 0.95 wt% C, heat to at least 840°C (1545°F) since the Acm temperature is 785°C (1445°F)."
},
{
"idx": 504,
"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": "For 0.25 wt% C, heat to about 880°C (1510°F) since the A3 temperature is 830°C (1420°F)."
},
{
"idx": 504,
"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": "For 0.45 wt% C, heat to about 830°C (1525°F) since the A3 temperature is 780°C (1435°F)."
},
{
"idx": 504,
"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": "For 0.85 wt% C, heat to about 777°C (1430°F) since the A1 temperature is 727°C (1340°F)."
},
{
"idx": 504,
"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": "For 1.10 wt% C, heat to about 777°C (1430°F) since the A1 temperature is 727°C (1340°F)."
},
{
"idx": 505,
"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": 505,
"question": "On what classes of alloys is 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": 506,
"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": 507,
"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": 508,
"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": 509,
"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": 510,
"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": 510,
"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": 510,
"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": 511,
"question": "Cadmium sulfide (CdS) has a cubic unit cell, and from \\(x\\)-ray diffraction data it is known that the cell edge length is \\(0.582 \\mathrm{~nm}\\). If the measured density is \\(4.82 \\mathrm{~g} / \\mathrm{cm}^{3}\\), how many \\(\\mathrm{Cd}^{2+}\\) and \\(\\mathrm{S}^{2-}\\) ions are there per unit cell?",
"answer": "We are asked to determine the number of \\(\\mathrm{Cd}^{2+}\\) and \\(\\mathrm{S}^{3-}\\) ions per unit cell for cadmium sulfide (CdS). For \\(\\mathrm{CdS}, a=0.582 \\mathrm{~nm}\\) and \\(\\rho=4.82 \\mathrm{~g} / \\mathrm{cm}^{3}\\). Solving for \\(n^{\\prime}\\) from Equation, we get\n\\[\n\\begin{aligned}\nn^{\\prime} & =\\frac{\\rho V_{\\mathrm{C}} N_{\\mathrm{A}}}{A_{\\mathrm{Cd}}+A_{\\mathrm{S}}}=\\frac{\\rho a^{3} N_{\\mathrm{A}}}{A_{\\mathrm{Cd}}+A_{\\overline{\\mathrm{S}}}} \\\\\n& =\\frac{\\left(4.82 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(5.82 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}\\left(6.022 \\times 10^{23} \\text { formula units } / \\mathrm{mol}\\right)}{\\left(112.41 \\mathrm{~g} / \\mathrm{mol}+32.06 \\mathrm{~g} / \\mathrm{mol}\\right)} \\\\\n& =3.96 \\text { or almost } 4\n\\end{aligned}\n\\]\nTherefore, there are four \\(\\mathrm{Cd}^{2+}\\) and four \\(\\mathrm{S}^{2-}\\) per unit cell."
},
{
"idx": 512,
"question": "A hypothetical \\(A X\\) type of ceramic material is known to have a density of \\(2.65 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and a unit cell of cubic symmetry with a cell edge length of \\(0.43 \\mathrm{~nm}\\). The atomic weights of the A and \\(X\\) elements are 86.6 and \\(40.3 \\mathrm{~g} / \\mathrm{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": "We are asked to specify possible crystal structures for an AX type of ceramic material given its density \\(\\left(2.65 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\), that the unit cell has cubic symmetry with edge length of \\(0.43 \\mathrm{~nm}\\left(4.3 \\times 10^{-8} \\mathrm{~cm}\\right)\\), and the atomic weights of the A and X elements ( 86.6 and \\(40.3 \\mathrm{~g} / \\mu \\mathrm{mol}\\), respectively). Using Equation and solving for \\(n^{\\prime}\\) yields\n\\[\n\\begin{aligned}\nn^{\\prime} & =\\frac{\\rho V_{C} N_{\\mathrm{A}}}{\\sum A_{\\mathrm{C}}+\\sum A_{\\mathrm{A}}} \\\\\n& =\\frac{\\left(2.65 \\mathrm{~g} / \\mathrm{cm}^{2}\\right)\\left[\\left(4.30 \\times 10^{-8} \\mathrm{~cm}\\right)^{3} /\\right. \\text { unit cell }\\left[\\left(6.022 \\times 10^{23} \\text { formula units/mol }\\right.\\right.}{ } \\\\\n& (86.6+40.3) \\mathrm{g} / \\mathrm{mol} \\\\\n& =1.00 \\text { formula units/unit cell }\n\\end{aligned}\n\\]\nOf the three possible crystal structures, only cesium chloride has one formula unit per unit cell, and therefore, is the only possibility."
},
{
"idx": 513,
"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": 514,
"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": 515,
"question": "Calculate the fraction of lattice sites that are Schottky defects for sodium chloride at its melting temperature \\(\\left(801^{\\circ} \\mathrm{C}\\right)\\). Assume an energy for defect formation of \\(2.3 \\mathrm{eV}\\).",
"answer": "We are asked in this problem to calculate the fraction of lattice sites that are Schottky defects for \\(\\mathrm{NaCl}\\) at its melting temperature \\(\\left(801^{\\circ} \\mathrm{C}_{\\mathrm{}}\\right)\\), assuming that the energy for defect formation is \\(2.3 \\mathrm{eV}\\). In order to solve this problem it is necessary to use Equation and solve for the \\(N_{s} / N\\) ratio. Rearrangement of this expression and substituting values for the several parameters leads to\n\\[\n\\begin{aligned}\n\\frac{N_{s}}{N} & =\\exp \\left(-\\frac{Q_{S}}{2 k T}\\right) \\\\\n& =\\exp \\left\\lvert\\,-\\frac{2.3 \\mathrm{eV}}{\\left(2\\right)\\left(8.62 \\times 10^{-5} \\mathrm{eV} / \\mathrm{K}\\right)\\left(801+273 \\mathrm{~K}\\right)}\\right\\rfloor \\\\\n& =4.03 \\times 10^{-6}"
},
{
"idx": 516,
"question": "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": "For Li+ substituting for Ca2+ in CaO, oxygen vacancies would be created. For each Li+ substituting for Ca2+, one positive charge is removed; in order to maintain charge neutrality, a single negative charge may be removed. Negative charges are eliminated by creating oxygen vacancies, and for every two Li+ ions added, a single oxygen vacancy is formed."
},
{
"idx": 516,
"question": "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": "For Cl- substituting for O2- in CaO, calcium vacancies would be created. For each Cl- substituting for an O2-, one negative charge is removed; in order to maintain charge neutrality, two Cl- ions will lead to the formation of one calcium vacancy."
},
{
"idx": 517,
"question": "What point defects are possible for Al2O3 as an impurity in MgO?",
"answer": "The possible point defects are Al3+ substitution for Mg2+ and the formation of Mg2+ vacancies to maintain charge neutrality."
},
{
"idx": 517,
"question": "How many Al3+ ions must be added to form each of these defects in MgO?",
"answer": "For every Al3+ ion that substitutes for Mg2+, a single positive charge is added. To maintain charge neutrality, one Mg2+ vacancy is formed for every two Al3+ ions added."
},
{
"idx": 518,
"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 chemical formula for kaolinite clay may also be written as Al2O3·2SiO2·2H2O. Thus, if we remove the chemical water, the formula becomes Al2O3·2SiO2. The formula weight for Al2O3 is just (2)(26.98 g/mol)+(3)(16.00 g/mol)=101.96 g/mol; and for SiO2 the formula weight is 28.09 g/mol+(2)(16.00 g/mol)=60.09 g/mol. Thus, the composition of this product, in terms of the concentration of Al2O3, CAl2O3, in weight percent is just CAl2O3=(101.96 g/mol)/(101.96 g/mol+(2)(60.09 g/mol))×100=45.9 wt%."
},
{
"idx": 518,
"question": "What are the liquidus and solidus temperatures of the material remaining after heating kaolinite clay [Al2(Si2O5)(OH)4] to drive off chemical water?",
"answer": "The liquidus and solidus temperatures for this material as determined from the SiO2-Al2O3 phase diagram, Figure 12.27, are 1825°C and 1587°C, respectively."
},
{
"idx": 519,
"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": 519,
"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": 520,
"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": "This problem asks that we compute the crack tip radius ratio before and after etching. Let\n\\[\n\\begin{array}{c}\n\\rho_{t}=\\text { original crack tip radius, and } \\\\\n\\rho_{t}=\\text { etched crack tip radius }\n\\end{array}\n\\]\nAlso,\n\\[\n\\begin{array}{l}\n\\sigma_{t}^{\\star}=\\sigma_{f} \\\\\na^{\\star}=\\frac{a}{3} \\\\\n\\sigma_{0}^{\\star}=8 \\sigma_{0}\n\\end{array}\n\\]\nSolving for \\(\\frac{\\rho_{t}^{\\star}}{\\rho_{t}}\\) from the following\n\\[\n\\sigma_{f}=2 \\sigma_{0}\\left(\\frac{a}{\\rho_{f}}\\right)^{1 / 2}=\\sigma_{f}^{\\star}=2 \\sigma_{0}^{\\star}\\left(\\frac{a^{\\star}}{\\rho_{t}^{\\star}}\\right)^{1 / 2}\n\\]\nyields\n\\[\n\\frac{\\rho_{t}^{\\star}}{\\rho_{t}}=\\left(\\frac{\\sigma_{0}^{\\star}}{\\sigma_{0}}\\right)^{2}\\left(\\frac{a^{\\star}}{a}\\right)=\\left(\\frac{8 \\sigma_{0}}{\\sigma_{0}}\\right)^{2}\\left(\\frac{a / 3}{a}\\right)=21.3"
},
{
"idx": 521,
"question": "A circular specimen of \\(\\mathrm{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 \\(425 \\mathrm{~N}\\left(95.5 \\mathrm{lb} \\mathrm{b}_{1}\\right)\\), the flexural strength is \\(105 \\mathrm{MPa}(15,000 \\mathrm{psi})\\), and the separation between load points is \\(50 \\mathrm{~mm}(2.0 \\mathrm{in})\\).",
"answer": "We are asked to calculate the maximum radius of a circular specimen of \\(\\mathrm{MgO}\\) that is loaded using threepoint bending. Solving for \\(R\\) from Equation\n\\[\nR=\\left[\\frac{F_{f} L}{\\sigma_{f s} \\pi}\\right]^{1 / 3}\n\\]\nwhich, when substituting the parameters stipulated in the problem statement, yields\n\\[\n\\begin{aligned}\nR & =\\left[\\frac{(425 \\mathrm{~N})\\left(50 \\times 10^{-3} \\mathrm{~m}\\right)}{\\left(105 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}\\right)(\\pi)}\\right]^{1 / 3} \\\\\n& =4.0 \\times 10^{-3} \\mathrm{~m}=4.0 \\mathrm{~mm} \\quad(0.16 \\mathrm{in})\n\\end{aligned}\n\\]\nExcerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1376 United States Copyright Act without the permission of the copyright owner is unlawful."
},
{
"idx": 522,
"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 2.5 mm (0.10 in.) and the support point separation distance is 30 mm (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": "This portion of the problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of 390 MPa (56,600 psi) and a radius of 2.5 mm will fracture when subjected to a load of 620 N in a three-point bending test; the support point separation is given as 30 mm. Using Equation we will calculate the value of σ; if this value is greater than σ_fs (390 MPa), then fracture is expected to occur. Employment of Equation yields σ = (FL)/(πR^3) = (620 N)(30 × 10^-3 m)/(π(2.5 × 10^-3 m)^3) = 379 × 10^6 N/m^2 = 379 MPa (53,500 psi). Since this value is less than the given value of σ_fs (390 MPa), then fracture is not predicted."
},
{
"idx": 522,
"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, and since this value of σ (379 MPa) is relatively close to σ_fs (390 MPa) then there is some chance that fracture will occur."
},
{
"idx": 523,
"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": 524,
"question": "Compute the modulus of elasticity for nonporous beryllium oxide (BeO) given that the modulus of elasticity for BeO with 5 vol% porosity is 310 GPa (45 x 10^6 psi).",
"answer": "To compute the modulus of elasticity for nonporous BeO (E0), we use the equation E0 = E / (1 - 1.9P + 0.9P^2), where E = 310 GPa and P = 0.05. Substituting these values, we get E0 = 310 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 342 GPa (49.6 x 10^6 psi)."
},
{
"idx": 524,
"question": "Compute the modulus of elasticity for beryllium oxide (BeO) with 10 vol% porosity, given that the modulus of elasticity for nonporous BeO is 342 GPa (49.6 x 10^6 psi).",
"answer": "To compute the modulus of elasticity for BeO with 10 vol% porosity (E), we use the equation E = E0 (1 - 1.9P + 0.9P^2), where E0 = 342 GPa and P = 0.10. Substituting these values, we get E = 342 GPa [1 - (1.9)(0.10) + (0.9)(0.10)^2] = 280 GPa (40.6 x 10^6 psi)."
},
{
"idx": 525,
"question": "Compute the modulus of elasticity for the nonporous material given that the modulus of elasticity for boron carbide (B4C) having 5 vol% porosity is 290 GPa (42 x 10^6 psi).",
"answer": "To compute the modulus of elasticity for nonporous B4C, we solve the equation E0 = E / (1 - 1.9P + 0.9P^2) using P = 0.05. This gives E0 = 290 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 320 GPa (46.3 x 10^6 psi)."
},
{
"idx": 525,
"question": "At what volume percent porosity will the modulus of elasticity be 235 GPa (34 x 10^6 psi) for boron carbide (B4C), given that the modulus of elasticity for the nonporous material is 320 GPa?",
"answer": "To find the volume percent porosity at which the elastic modulus is 235 GPa, we use the equation E/E0 = 1 - 1.9P + 0.9P^2, where E0 = 320 GPa. This gives 235 GPa / 320 GPa = 0.734 = 1 - 1.9P + 0.9P^2. Rearranging, we get 0.9P^2 - 1.9P + 0.266 = 0. Solving the quadratic equation yields P = (1.9 ± sqrt((-1.9)^2 - (4)(0.9)(0.266))) / (2)(0.9). The physically meaningful root is P = 0.151, or 15.1 vol%."
},
{
"idx": 526,
"question": "Cite the two desirable characteristics of glasses.",
"answer": "Two desirable characteristics of glasses are optical transparency and ease of fabrication."
},
{
"idx": 527,
"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": 527,
"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": 528,
"question": "For refractory ceramic materials, what are three characteristics that improve with increasing porosity?",
"answer": "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": 528,
"question": "For refractory ceramic materials, what are two characteristics that are adversely affected by increasing porosity?",
"answer": "Two characteristics that are adversely affected are (1) load-bearing capacity and (2) resistance to attack by corrosive materials."
},
{
"idx": 529,
"question": "How do 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": 529,
"question": "How do 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": 530,
"question": "What is the distinction between glass transition temperature and melting temperature?",
"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.\nThe 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": 531,
"question": "Compare the temperatures at which soda-lime, borosilicate, \\(96 \\%\\) silica, and fused silica may be annealed.",
"answer": "The annealing point is that temperature at which the viscosity of the glass is \\(10^{12} \\mathrm{~Pa} \\cdot \\mathrm{s}\\left(10^{13} \\mathrm{P}\\right)\\). From Figure 13.7, these temperatures for the several glasses are as follows:\n\\begin{tabular}{lc} \nGlass & Annealing Temperature \\\\\nSoda-lime & \\(500^{\\circ} \\mathrm{C}\\left(930^{\\circ} \\mathrm{F}\\right)\\) \\\\\nBorosilicate & \\(565^{\\circ} \\mathrm{C}\\left(1050^{\\circ} \\mathrm{F}\\right)\\) \\\\\n\\(96 \\%\\) Silica & \\(930^{\\circ} \\mathrm{C}\\left(1705^{\\circ} \\mathrm{F}\\right)\\) \\\\\nFused silica & \\(1170^{\\circ} \\mathrm{C}\\left(2140^{\\circ} \\mathrm{F}\\right)\\)\n\\end{tabular}"
},
{
"idx": 532,
"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 \\times 10^{6} \\mathrm{~Pa} \\cdot \\mathrm{s}\\); from Figure 13.7, these temperatures for the \\(96 \\%\\) silica, borosilicate, and soda-lime glasses are \\(1540^{\\circ} \\mathrm{C}\\left(2800^{\\circ} \\mathrm{F}\\right), 830^{\\circ} \\mathrm{C}\\left(1525^{\\circ} \\mathrm{F}\\right)\\), and \\(700^{\\circ} \\mathrm{C}\\left(1290^{\\circ} \\mathrm{F}\\right)\\), respectively."
},
{
"idx": 533,
"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": 533,
"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": 534,
"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": 535,
"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 \\(\\mathrm{K}^{+}\\)and \\(\\mathrm{Cs}^{+}\\)fill these criteria, having ionic radii of 0.138 and 0.170 \\(\\mathrm{nm}\\), respectively, which are larger than the ionic radius of \\(\\mathrm{Na}^{+}(0.102 \\mathrm{~nm})\\). In fact, soda-lime glasses are tempered by a \\(\\mathrm{K}^{+}-\\mathrm{Na}^{+}\\)ion exchange."
},
{
"idx": 536,
"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": 537,
"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": 538,
"question": "What are the three main components of a whiteware ceramic such as porcelain?",
"answer": "The three components of a whiteware ceramic are clay, quartz, and a flux."
},
{
"idx": 538,
"question": "What role does each component play in the forming and firing procedures of a whiteware ceramic such as porcelain?",
"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": 539,
"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": 539,
"question": "Cite three factors that influence the rate of drying of a ceramic body, 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": 540,
"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": 541,
"question": "(a) Name three factors that influence the degree to which vitrification occurs in clay-based ceramic wares.",
"answer": "Three factors that influence the degree to which vitrification occurs in clay-based ceramic wares are: (1) composition (especially the concentration of flux present); (2) the temperature of firing; and (3) the time at the firing temperature."
},
{
"idx": 541,
"question": "(b) 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. Firing distortion will increase with degree of vitrification since more liquid phase will be present at the firing temperature. 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": 542,
"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": 543,
"question": "Compute repeat unit molecular weight for poly(vinyl chloride), where each repeat unit consists of two carbons, three hydrogens, and one chlorine. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), and chlorine (35.45 g/mol).",
"answer": "For poly(vinyl chloride), the molecular weight is calculated as: m = 2(12.01 g/mol) + 3(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol."
},
{
"idx": 543,
"question": "Compute repeat unit molecular weight for poly(ethylene terephthalate), where each repeat unit has ten carbons, eight hydrogens, and four oxygens. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), and oxygen (16.00 g/mol).",
"answer": "For poly(ethylene terephthalate), the molecular weight is calculated as: m = 10(12.01 g/mol) + 8(1.008 g/mol) + 4(16.00 g/mol) = 192.16 g/mol."
},
{
"idx": 543,
"question": "Compute repeat unit molecular weight for polycarbonate, where each repeat unit has sixteen carbons, fourteen hydrogens, and three oxygens. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), and oxygen (16.00 g/mol).",
"answer": "For polycarbonate, the molecular weight is calculated as: m = 16(12.01 g/mol) + 14(1.008 g/mol) + 3(16.00 g/mol) = 254.27 g/mol."
},
{
"idx": 543,
"question": "Compute repeat unit molecular weight for polydimethylsiloxane, where each repeat unit has two carbons, six hydrogens, one silicon, and one oxygen. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), silicon (28.09 g/mol), and oxygen (16.00 g/mol).",
"answer": "For polydimethylsiloxane, the molecular weight is calculated as: m = 2(12.01 g/mol) + 6(1.008 g/mol) + 28.09 g/mol + 16.00 g/mol = 74.16 g/mol."
},
{
"idx": 544,
"question": "The number-average molecular weight of a polypropylene is \\(1,000,000 \\mathrm{~g} / \\mathrm{mol}\\). Compute the degree of polymerization.",
"answer": "We are asked to compute the degree of polymerization for polypropylene, given that the number-average molecular weight is \\(1,000,000 \\mathrm{~g} / \\textrm{mol}\\). The repeat unit molecular weight of polypropylene is just\n\\[\n\\begin{array}{c}\nm=3\\left(\\mathrm{~A}_{\\mathrm{C}}\\right)+6\\left(\\mathrm{~A}_{\\mathrm{H}}\\right) \\\\\n=(3)(12.01 \\mathrm{~g} / \\mathrm{mol})+(6)(1.008 \\mathrm{~g} / \\mathrm{mol})=42.08 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nNow it is possible to compute the degree of polymerization using Equation as\n\\[\nD P=\\frac{\\bar{M}_{n}}{m}=\\frac{1,000,000 \\mathrm{~g} / \\mathrm{mol}}{42.08 \\mathrm{~g} / \\mathrm{mol}}=23,760\n\\]"
},
{
"idx": 545,
"question": "Compute the repeat unit molecular weight of polystyrene.",
"answer": "The repeat unit molecular weight of polystyrene is called for in this portion of the problem. For polystyrene, each repeat unit has eight carbons and eight hydrogens. Thus, m = 8(A_C) + 8(A_H) = (8)(12.01 g/mol) + (8)(1.008 g/mol) = 104.14 g/mol"
},
{
"idx": 545,
"question": "Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000.",
"answer": "We are now asked to compute the number-average molecular weight. Since the degree of polymerization is 25,000, using the equation M_n = (DP)m = (25,000)(104.14 g/mol) = 2.60 × 10^6 g/mol"
},
{
"idx": 546,
"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": "For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for 5% Cl substitution of all original hydrogen atoms. Consider 50 carbon atoms; there are 100 possible side-bonding sites. Ninety-five are occupied by hydrogen and five are occupied by Cl. Thus, the mass of these 50 carbon atoms, m_C, is just m_C=50(A_C)=(50)(12.01 g/mol)=600.5 g. Likewise, for hydrogen and chlorine, m_H=95(A_H)=(95)(1.008 g/mol)=95.76 g and m_Cl=5(A_Cl)=(5)(35.45 g/mol)=177.25 g. Thus, the concentration of chlorine, C_Cl, is determined as C_Cl = (m_Cl)/(m_C + m_H + m_Cl) × 100 = (177.25 g)/(600.5 g + 95.76 g + 177.25 g) × 100 = 20.3 wt%."
},
{
"idx": 546,
"question": "In what ways does 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": 547,
"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": 547,
"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": 548,
"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": 548,
"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": 549,
"question": "The number-average molecular weight of a poly(styrene-butadiene) alternating copolymer is \\(1,350,000 \\mathrm{~g} / \\mathrm{mol}\\); determine the average number of styrene and butadiene repeat units per molecule.",
"answer": "Since it is an alternating copolymer, the number of both types of repeat units will be the same. Therefore, consider them as a single repeat unit, and determine the number-average degree of polymerization. For the styrene repeat unit, there are eight carbon atoms and eight hydrogen atoms, while the butadiene repeat consists of four carbon atoms and six hydrogen atoms. Therefore, the styrene-butadiene combined repeat unit weight is just\n\\[\n\\begin{aligned}\nm & =12\\left(\\mathrm{~A}_{\\mathrm{C}}\\right)+14\\left(\\mathrm{~A}_{\\mathrm{H}}\\right) \\\\\n& =(12)(12.01 \\mathrm{~g} / \\mathrm{mol})+(14)(1.008 \\mathrm{~g} / \\mathrm{mol})=158.23 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]\nFrom Equation, the degree of polymerization is just\n\\[\nD P=\\frac{\\bar{M}_{n}}{m}=\\frac{1,350,000 \\mathrm{~g} / \\mathrm{mol}}{158.23 \\mathrm{~g} / \\mathrm{mol}}=8530\n\\]\nThus, there is an average of 8530 of both repeat unit types per molecule."
},
{
"idx": 550,
"question": "Calculate the repeat unit molecular weight of acrylonitrile in poly(acrylonitrile-butadiene) copolymer, given that it consists of three carbon, one nitrogen, and three hydrogen atoms.",
"answer": "The repeat unit molecular weight of acrylonitrile is calculated as: (3)(12.01 g/mol) + 14.01 g/mol + (3)(1.008 g/mol) = 53.06 g/mol."
},
{
"idx": 550,
"question": "Calculate the repeat unit molecular weight of butadiene in poly(acrylonitrile-butadiene) copolymer, given that it consists of four carbon and six hydrogen atoms.",
"answer": "The repeat unit molecular weight of butadiene is calculated as: (4)(12.01 g/mol) + (6)(1.008 g/mol) = 54.09 g/mol."
},
{
"idx": 550,
"question": "Calculate the average repeat unit molecular weight of the poly(acrylonitrile-butadiene) copolymer, given that the fraction of butadiene repeat units is 0.30 and the fraction of acrylonitrile repeat units is 0.70.",
"answer": "The average repeat unit molecular weight is calculated as: (0.70)(53.06 g/mol) + (0.30)(54.09 g/mol) = 53.37 g/mol."
},
{
"idx": 550,
"question": "Calculate the number-average molecular weight of the poly(acrylonitrile-butadiene) copolymer, given that the degree of polymerization is 2000 and the average repeat unit molecular weight is 53.37 g/mol.",
"answer": "The number-average molecular weight is calculated as: (53.37 g/mol)(2000) = 106740 g/mol."
},
{
"idx": 551,
"question": "What is the average molecular weight per repeat unit (m̄) for an alternating copolymer with a number-average molecular weight of 250,000 g/mol and a degree of polymerization of 3420?",
"answer": "The average molecular weight per repeat unit (m̄) is calculated as m̄ = M̄n / DP = 250,000 g/mol / 3420 = 73.10 g/mol."
},
{
"idx": 551,
"question": "What is the molecular weight of the styrene repeat unit (ms) in the alternating copolymer?",
"answer": "The molecular weight of the styrene repeat unit (ms) is calculated as ms = 8(AC) + 8(AH) = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol."
},
{
"idx": 551,
"question": "What is the molecular weight of the unknown repeat unit (mx) in the alternating copolymer, given that the chain fractions of styrene and the unknown repeat unit are both 0.5?",
"answer": "The molecular weight of the unknown repeat unit (mx) is calculated as mx = (m̄ - fs * ms) / fx = (73.10 g/mol - (0.5)(104.14 g/mol)) / 0.5 = 42.06 g/mol."
},
{
"idx": 551,
"question": "What are the molecular weights of the possible repeat units: ethylene, propylene, tetrafluoroethylene, and vinyl chloride?",
"answer": "The molecular weights are calculated as follows: m_ethylene = 2(AC) + 4(AH) = 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol; m_propylene = 3(AC) + 6(AH) = 3(12.01 g/mol) + 6(1.008 g/mol) = 42.08 g/mol; m_TFE = 2(AC) + 4(AF) = 2(12.01 g/mol) + 4(19.00 g/mol) = 100.02 g/mol; m_VC = 2(AC) + 3(AH) + (ACl) = 2(12.01 g/mol) + 3(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol."
},
{
"idx": 551,
"question": "Which of the possible repeat units (ethylene, propylene, tetrafluoroethylene, or vinyl chloride) matches the calculated molecular weight of the unknown repeat unit (42.06 g/mol)?",
"answer": "Propylene is the other repeat unit type since its molecular weight (42.08 g/mol) is almost the same as the calculated mx (42.06 g/mol)."
},
{
"idx": 552,
"question": "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": "To determine the ratio of butadiene to styrene repeat units, first calculate the average repeat unit molecular weight of the copolymer, m̄, using the formula: m̄ = M̄_n / DP = 350,000 g/mol / 4425 = 79.10 g/mol. Let f_b be the chain fraction of butadiene repeat units, then the chain fraction of styrene repeat units f_s is 1 - f_b. The repeat unit molecular weights are: m_b = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol for butadiene and m_s = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol for styrene. Solving for f_b: f_b = (m̄ - m_s) / (m_b - m_s) = (79.10 g/mol - 104.14 g/mol) / (54.09 g/mol - 104.14 g/mol) = 0.50. Thus, f_s = 1 - 0.50 = 0.50, and the ratio is f_b / f_s = 0.50 / 0.50 = 1.0."
},
{
"idx": 552,
"question": "Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?",
"answer": "The copolymer can be alternating, random, graft, or block. The alternating copolymer requires a 1:1 ratio of repeat units, which matches the result from part (a). However, random, graft, and block copolymers can also have a 1:1 ratio, so all four types are possible."
},
{
"idx": 553,
"question": "Crosslinked copolymers consisting of \\(60 \\mathrm{wt} \\%\\) ethylene and \\(40 \\mathrm{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": "For a copolymer consisting of \\(60 \\mathrm{wt} \\%\\) ethylene and \\(40 \\operatorname{wt} \\%\\) propylene, we are asked to determine the fraction of both repeat unit types.\nIn \\(100 \\mathrm{~g}\\) of this material, there are \\(60 \\mathrm{~g}\\) of ethylene and \\(40 \\mathrm{~g}\\) of propylene. The ethylene \\(\\left(\\mathrm{C}_{2} \\mathrm{H}_{4}\\right)\\) molecular weight is\n\\[\n\\begin{array}{l}\nm\\left(\\text { ethylene }\\right)=2\\left(A_{\\mathrm{C}}\\right)+4\\left(A_{\\mathrm{H}}\\right) \\\\\n=(2)(12.01 \\mathrm{~g} / \\mathrm{mol})+(4)(1.008 \\mathrm{~g} / \\mathrm{mol})=28.05 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nThe propylene \\(\\left(\\mathrm{C}_{2} \\mathrm{H}_{6}\\right)\\) molecular weight is\n\\[\n\\begin{array}{l}\nm\\text { (propylene) }=3\\left(A_{\\mathrm{C}}\\right)+6\\left(A_{\\mathrm{H}}\\right) \\\\\n=(3)(12.01 \\mathrm{~g} / \\mathrm{mol})+(6)(1.008 \\mathrm{~g} / \\mathrm{mol})=42.08 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nTherefore, in \\(100 \\mathrm{~g}\\) of this material, there are\n\\[\n\\frac{60 \\mathrm{~g}}{28.05 \\mathrm{~g} / \\mathrm{mol}}=2.14 \\mathrm{~mol} \\text { of ethylene }\n\\]\nand\n\\[\n\\frac{40 \\mathrm{~g}}{42.08 \\mathrm{~g} / \\mathrm{mol}}=0.95 \\mathrm{~mol} \\text { of propylene }\n\\]\nThus, the fraction of the ethylene repeat unit, \\(f\\) (ethylene), is just\n\\[\nf\\left(\\text { ethylene }\\right)=\\frac{2.14 \\mathrm{~mol}}{2.14 \\mathrm{~mol}+0.95 \\mathrm{~mol}}=0.69\n\\]\nLikewise,\n\n\\[\nf \\text { (propylene) }=\\frac{0.95 \\mathrm{~mol}}{2.14 \\mathrm{~mol}+0.95 \\mathrm{~mol}}=0.31\n\\]"
},
{
"idx": 554,
"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": 555,
"question": "For the pair of linear and syndiotactic poly(vinyl chloride) and linear and isotactic polystyrene, (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.",
"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": 555,
"question": "For the pair of network phenol-formaldehyde and linear and heavily crosslinked cis-isoprene, (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.",
"answer": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize."
},
{
"idx": 555,
"question": "For the pair of linear polyethylene and lightly branched isotactic polypropylene, (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.",
"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": 555,
"question": "For the pair of alternating poly(styrene-ethylene) copolymer and random poly(vinyl chloride-tetrafluoroethylene) copolymer, (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.",
"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": 556,
"question": "Consider the diffusion of water vapor through a polypropylene (PP) sheet \\(2 \\mathrm{~mm}\\) thick. The pressures of \\(\\mathrm{H}_{2} \\mathrm{O}\\) at the two faces are \\(1 \\mathrm{kPa}\\) and \\(10 \\mathrm{kPa}\\), which are maintained constant. Assuming conditions of steady state, what is the diffusion flux [in \\(\\left[\\mathrm{cm}^{3} \\mathrm{STP}\\right] / \\mathrm{cm}^{2}-\\mathrm{s}\\) ] at \\(298 \\mathrm{~K}\\) ?",
"answer": "This is a permeability problem in which we are asked to compute the diffusion flux of water vapor through a 2 -mm thick sheet of polypropylene. In order to solve this problem it is necessary to employ Equation. The permeability coefficient of \\(\\mathrm{H}_{2} \\mathrm{O}\\) through \\(\\mathrm{PP}\\) is given in Table 14.6 as \\(38 \\times 10^{-13}\\left(\\mathrm{~cm}^{3} \\mathrm{STP}\\right)-\\mathrm{cm} / \\mathrm{cm}^{2}-\\mathrm{s}-\\mathrm{Pa}\\). Thus, from Equation\n\\[\nJ=P_{M} \\frac{\\Delta P}{\\Delta x}=P_{M} \\frac{P_{2}-P_{1}}{\\Delta x}\n\\]\nand taking \\(P_{1}=1 \\mathrm{kPa}(1,000 \\mathrm{~Pa})\\) and \\(P_{2}=10 \\mathrm{kPa}(10,000 \\mathrm{~Pa})\\) we get\n\\[\n\\begin{aligned}\n& =\\left\\lfloor 38 \\times 10^{-13}\\left(\\mathrm{~cm}^{2} \\mathrm{STP}\\right)(\\mathrm{cm})\\right\\lfloor\\left(\\frac{10,000 \\mathrm{~Pa}-1,000 \\mathrm{~Pa}}{0.2 \\mathrm{~cm}}\\right) \\\\\n& =1.71 \\times 10^{-7} \\frac{\\left(\\mathrm{~cm}^{3} \\mathrm{STP}\\right)}{\\mathrm{cm}^{2}-\\mathrm{s}}\n\\end{aligned}\n\\]"
},
{
"idx": 557,
"question": "Argon diffuses through a high density polyethylene (HDPE) sheet \\(40 \\mathrm{~mm}\\) thick at a rate of \\(4.0 \\times 10^{-7}\\) \\(\\left(\\mathrm{cm}^{3} \\mathrm{STP} / \\mathrm{cm}^{2}-\\mathrm{s}\\right.\\) at \\(325 \\mathrm{~K}\\). The pressures of argon at the two faces are \\(5000 \\mathrm{kPa}\\) and \\(1500 \\mathrm{kPa}\\), which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at \\(325 \\mathrm{~K}\\) ?",
"answer": "This problem asks us to compute the permeability coefficient for argon through high density polyethylene at \\(325 \\mathrm{~K}\\) given a steady-state permeability situation. It is necessary for us to Equation in order to solve this problem. Rearranging this expression and solving for the permeability coefficient gives\n\\[\nP_{M}=\\frac{J \\Delta x}{\\Delta P}=\\frac{J \\Delta x}{P_{2}-P_{1}}\n\\]\nTaking \\(P_{1}=1500 \\mathrm{kPa}(1,500,000 \\mathrm{~Pa})\\) and \\(P_{2}=5000 \\mathrm{kPa}(5,000,000 \\mathrm{~Pa})\\), the permeability coefficient of Ar through HDPE is equal to\n\\[\nP_{M}=\\frac{\\left[4.0 \\times 10^{-7} \\frac{\\left(\\mathrm{cm}^{3} \\mathrm{STP}\\right)}{\\mathrm{cm}^{2}-\\mathrm{s}}\\right](4 \\mathrm{~cm})}{(5,000,000 \\mathrm{~Pa}-1,500,000 \\mathrm{~Pa})}\n\\]\n\\[\n\\begin{aligned}\n& =4.57 \\times 10^{-13} \\frac{\\left(\\mathrm{cm}^{3} \\mathrm{STP}\\left(\\mathrm{cm}\\right)\\right.}{\\mathrm{cm}^{2}-\\mathrm{s}-\\mathrm{Pa}}\n\\end{aligned}\n\\]"
},
{
"idx": 558,
"question": "Compute the permeability coefficient (P_M) at 350 K for hydrogen in poly(dimethyl siloxane) (PDMSO) using the given equation P_M = P_M0 exp(-Q_p / R T), where P_M0 = 1.45 × 10^-8 (cm³ STP)(cm) / (cm²-s-Pa), Q_p = 13.7 kJ/mol, and R = 8.31 J/mol-K.",
"answer": "P_M = [1.45 × 10^-8 (cm³ STP)(cm) / (cm²-s-Pa)] exp[-13,700 J/mol / (8.31 J/mol-K)(350 K)] = 1.31 × 10^-10 (cm³ STP)(cm) / (cm²-s-Pa)"
},
{
"idx": 558,
"question": "Compute the diffusion flux (J) for hydrogen through a PDMSO sheet 20 mm thick with hydrogen pressures of 10 kPa and 1 kPa at the two faces, using the permeability coefficient P_M = 1.31 × 10^-10 (cm³ STP)(cm) / (cm²-s-Pa) and the equation J = P_M (ΔP / Δx), where ΔP = P2 - P1 and Δx is the thickness.",
"answer": "J = 1.31 × 10^-10 (cm³ STP)(cm) / (cm²-s-Pa) × (10,000 Pa - 1,000 Pa) / 2.0 cm = 5.90 × 10^-7 (cm³ STP) / (cm²-s)"
},
{
"idx": 559,
"question": "Determine the relaxation time (τ) for the viscoelastic polymer given the initial stress σ(0) = 2.76 MPa, stress after 60 s σ(60) = 1.72 MPa, and elapsed time t = 60 s.",
"answer": "The relaxation time τ is calculated using the formula τ = -t / ln[σ(t)/σ(0)]. Substituting the given values, τ = -60 s / ln[1.72 MPa / 2.76 MPa] = 127 s."
},
{
"idx": 559,
"question": "Calculate the stress σ(10) at t = 10 s for the viscoelastic polymer using the relaxation time τ = 127 s and initial stress σ(0) = 2.76 MPa.",
"answer": "The stress at t = 10 s is calculated using σ(t) = σ(0) exp(-t/τ). Substituting the values, σ(10) = 2.76 MPa exp(-10 s / 127 s) = 2.55 MPa."
},
{
"idx": 559,
"question": "Determine the relaxation modulus E_r(10) for the viscoelastic polymer given the stress σ(10) = 2.55 MPa and the constant strain ε0 = 0.6.",
"answer": "The relaxation modulus is calculated using E_r(t) = σ(t)/ε0. Substituting the values, E_r(10) = 2.55 MPa / 0.6 = 4.25 MPa (616 psi)."
},
{
"idx": 560,
"question": "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": 560,
"question": "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": 561,
"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": 562,
"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, (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": 562,
"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, (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. The likelihood of crystallization for both syndiotactic and isotactic polypropylene is about the same, and, therefore, degree is crystallization is not a factor. Furthermore, tensile modulus is relatively insensitive to degree of polymerization (i.e., molecular weight)-the fact that branched PP has the higher molecular weight is not important."
},
{
"idx": 562,
"question": "For the pair of polymers: branched polyethylene with a number-average molecular weight of 250,000 g/mol; linear and isotactic poly(vinyl chloride) with a number-average molecular weight of 200,000 g/mol, (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": 563,
"question": "For syndiotactic polystyrene having a number-average molecular weight of 600,000 g/mol and atactic polystyrene having a number-average molecular weight of 500,000 g/mol, state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; if possible, note which has the higher tensile strength and cite the reason(s) for your choice; if not possible, state why.",
"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": 563,
"question": "For random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked and block acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked, state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; if possible, note which has the higher tensile strength and cite the reason(s) for your choice; if not possible, state why.",
"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": 563,
"question": "For network polyester and lightly branched polypropylene, state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; if possible, note which has the higher tensile strength and cite the reason(s) for your choice; if not possible, state why.",
"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": 564,
"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": 565,
"question": "Name the polymer(s) from the following list that would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate.",
"answer": "Of the polymers listed, polycarbonate would be suitable for the fabrication of cups to contain hot coffee."
},
{
"idx": 565,
"question": "Why is polycarbonate suitable for the fabrication of cups to contain hot coffee?",
"answer": "Polycarbonate has a glass transition temperature of 100 degrees Celsius or above, which means it does not soften at the maximum temperature of hot coffee (slightly below 100 degrees Celsius)."
},
{
"idx": 566,
"question": "For the pair of polymers: 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, (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.",
"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": 566,
"question": "For the pair of polymers: linear polyethylene that has a degree of polymerization of 5,000; linear and isotactic polypropylene that has a degree of polymerization of 6,500, (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.",
"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": 566,
"question": "For the pair of polymers: 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, (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.",
"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": 567,
"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": 568,
"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": 569,
"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": 570,
"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": 571,
"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 calculated as k_max = k_m V_m + k_p V_p = k_Co V_Co + k_TiC V_TiC = (69 W/m-K)(0.15) + (27 W/m-K)(0.85) = 33.3 W/m-K."
},
{
"idx": 571,
"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 calculated as k_min = (k_Co k_TiC) / (V_Co k_TiC + V_TiC k_Co) = (69 W/m-K)(27 W/m-K) / [(0.15)(27 W/m-K) + (0.85)(69 W/m-K)] = 29.7 W/m-K."
},
{
"idx": 572,
"question": "What is the upper limit for the elastic modulus of the composite given the volume fractions of tungsten (0.60) and copper (0.40), and the moduli of elasticity for copper (110 GPa) and tungsten (407 GPa)?",
"answer": "The upper limit for the elastic modulus of the composite is calculated as follows: E_c(u) = E_Cu V_Cu + E_W V_W = (110 GPa)(0.40) + (407 GPa)(0.60) = 288 GPa."
},
{
"idx": 572,
"question": "What is the specific gravity of the composite given the volume fractions of tungsten (0.60) and copper (0.40), and the specific gravities for copper (8.9) and tungsten (19.3)?",
"answer": "The specific gravity of the composite is calculated as follows: ρ_c = ρ_Cu V_Cu + ρ_W V_W = (8.9)(0.40) + (19.3)(0.60) = 15.14."
},
{
"idx": 572,
"question": "What is the specific stiffness of the composite using the upper limit of elastic modulus (288 GPa) and the specific gravity (15.14)?",
"answer": "The specific stiffness of the composite is calculated as follows: Specific Stiffness = E_c(u) / ρ_c = 288 GPa / 15.14 = 19.0 GPa."
},
{
"idx": 572,
"question": "What is the specific stiffness of the composite using the specific stiffness-volume fraction product for both copper and tungsten, given the volume fractions of tungsten (0.60) and copper (0.40), the moduli of elasticity for copper (110 GPa) and tungsten (407 GPa), and the specific gravities for copper (8.9) and tungsten (19.3)?",
"answer": "The specific stiffness of the composite is calculated as follows: Specific Stiffness = (E_Cu / ρ_Cu) V_Cu + (E_W / ρ_W) V_W = (110 GPa / 8.9)(0.40) + (407 GPa / 19.3)(0.60) = 17.6 GPa."
},
{
"idx": 573,
"question": "What is the distinction between cement and concrete?",
"answer": "Concrete consists of an aggregate of particles that are bonded together by a cement."
},
{
"idx": 573,
"question": "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": 573,
"question": "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": 574,
"question": "What is 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": 574,
"question": "What is one difference between precipitation hardening and dispersion strengthening?",
"answer": "The hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening."
},
{
"idx": 574,
"question": "What is another difference between precipitation hardening and dispersion strengthening?",
"answer": "The strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening."
},
{
"idx": 575,
"question": "For a continuous and aligned fiber-reinforced composite consisting of 30 vol% aramid fibers and 70 vol% polycarbonate matrix, with the following mechanical characteristics: Aramid fiber modulus of elasticity = 131 GPa (19x10^6 psi), tensile strength = 3600 MPa (520,000 psi); Polycarbonate modulus of elasticity = 2.4 GPa (3.5x10^5 psi), tensile strength = 65 MPa (9425 psi); and stress on the polycarbonate matrix when aramid fibers fail = 45 MPa (6500 psi), compute the longitudinal tensile strength.",
"answer": "The longitudinal tensile strength is determined as: sigma_cl* = sigma_m*(1-V_f) + sigma_f* V_f = (45 MPa)(0.70) + (3600 MPa)(0.30) = 1100 MPa (160,000 psi)."
},
{
"idx": 575,
"question": "For a continuous and aligned fiber-reinforced composite consisting of 30 vol% aramid fibers and 70 vol% polycarbonate matrix, with the following mechanical characteristics: Aramid fiber modulus of elasticity = 131 GPa (19x10^6 psi), tensile strength = 3600 MPa (520,000 psi); Polycarbonate modulus of elasticity = 2.4 GPa (3.5x10^5 psi), tensile strength = 65 MPa (9425 psi); and stress on the polycarbonate matrix when aramid fibers fail = 45 MPa (6500 psi), compute the longitudinal modulus of elasticity.",
"answer": "The longitudinal elastic modulus is computed as: E_cl = E_m V_m + E_f V_f = (2.4 GPa)(0.70) + (131 GPa)(0.30) = 41 GPa (5.95x10^6 psi)."
},
{
"idx": 576,
"question": "For a continuous and oriented fiber-reinforced composite with a volume fraction of fibers of 0.25, determine the modulus of elasticity of the matrix phase given the modulus of elasticity in the longitudinal direction is 19.7 GPa (2.8×10^5 psi).",
"answer": "Using the equation for the longitudinal elastic modulus: E_cl = E_m(1 - V_f) + E_f V_f, where E_cl = 19.7 GPa and V_f = 0.25, we can write: 19.7 GPa = E_m(1 - 0.25) + E_f(0.25). Solving this equation simultaneously with the transverse elastic modulus equation leads to E_m = 2.79 GPa (4.04×10^5 psi)."
},
{
"idx": 576,
"question": "For a continuous and oriented fiber-reinforced composite with a volume fraction of fibers of 0.25, determine the modulus of elasticity of the fiber phase given the modulus of elasticity in the transverse direction is 3.66 GPa (5.3×10^5 psi).",
"answer": "Using the equation for the transverse elastic modulus: E_ct = (E_m E_f) / [(1 - V_f) E_f + V_f E_m], where E_ct = 3.66 GPa and V_f = 0.25, we can write: 3.66 GPa = (E_m E_f) / [(1 - 0.25) E_f + 0.25 E_m]. Solving this equation simultaneously with the longitudinal elastic modulus equation leads to E_f = 70.4 GPa (10.2×10^6 psi)."
},
{
"idx": 577,
"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. The modulus of elasticity for glass fiber is 72.5 GPa (10.5 x 10^6 psi) and for nylon 6,6 is 3.0 GPa (4.35 x 10^6 psi).",
"answer": "Given that the fibers carry 94% of the load, the ratio of the load carried by the fibers to that carried by the matrix is F_f/F_m = 0.94/0.06 = 15.67. Using the equation F_f/F_m = (E_f V_f)/(E_m (1-V_f)), where E_f is the modulus of elasticity of the fibers (72.5 GPa), E_m is the modulus of elasticity of the matrix (3.0 GPa), and V_f is the volume fraction of fibers, we substitute the known values to get 15.67 = (72.5 GPa V_f)/(3.0 GPa (1-V_f)). Solving for V_f yields V_f = 0.393."
},
{
"idx": 577,
"question": "What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 MPa (4350 psi). The tensile strength of glass fiber is 3400 MPa (490,000 psi) and of nylon 6,6 is 76 MPa (11,000 psi). The volume fraction of fibers is 0.393.",
"answer": "The tensile strength of the composite is calculated using the equation σ_cl* = σ_m* (1-V_f) + σ_f* V_f, where σ_m* is the matrix stress at fiber failure (30 MPa), σ_f* is the tensile strength of the fibers (3400 MPa), and V_f is the volume fraction of fibers (0.393). Substituting the values gives σ_cl* = (30 MPa)(1-0.393) + (3400 MPa)(0.393) = 1354 MPa (196,400 psi)."
},
{
"idx": 578,
"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": 578,
"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": 578,
"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": 579,
"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": 579,
"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": 580,
"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": 580,
"question": "Which reaction occurs at the anode and which at the cathode?",
"answer": "Oxidation occurs at the anode; reduction at the cathode."
},
{
"idx": 581,
"question": "(a)(i) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in HCl solution.",
"answer": "In HCl, possible reactions are\nMg -> Mg2+ + 2 e- (oxidation)\n2 H+ + 2 e- -> H2 (reduction)"
},
{
"idx": 581,
"question": "(a)(ii) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in an HCl solution containing dissolved oxygen.",
"answer": "In an HCl solution containing dissolved oxygen, possible reactions are\nMg -> Mg2+ + 2 e- (oxidation)\n4 H+ + O2 + 4 e- -> 2 H2O (reduction)"
},
{
"idx": 581,
"question": "(a)(iii) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in an HCl solution containing dissolved oxygen and Fe2+ ions.",
"answer": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are\nMg -> Mg2+ + 2 e- (oxidation)\n4 H+ + O2 + 4 e- -> 2 H2O (reduction)\nFe2+ + 2 e- -> Fe (reduction)"
},
{
"idx": 581,
"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": 582,
"question": "Demonstrate that the value of F in Equation 17.19 is 96,500 C/mol.",
"answer": "The Faraday constant F is just the product of the charge per electron and Avogadro's number; that is F = e N_A = (1.602 x 10^-19 C/electron)(6.022 x 10^23 electrons/mol) = 96,472 C/mol."
},
{
"idx": 582,
"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 = [(8.31 J/mol-K)(298 K)]/[(n)(96,472 C/mol)] (2.303) log x = (0.0592/n) log x. This gives units in volts since a volt is a J/C."
},
{
"idx": 583,
"question": "Is a voltage generated between the two cell halves of a Zn/Zn2+ concentration cell where one half has a Zn2+ concentration of 1.0 M and the other has 10^-2 M?",
"answer": "Yes, a voltage is generated between the two cell halves."
},
{
"idx": 583,
"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 generated is 0.0592 V."
},
{
"idx": 583,
"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": 584,
"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 \\mathrm{in}^{2}\\) and that approximately \\(2.6 \\mathrm{~kg}\\) 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 \\(7.9 \\mathrm{~g} / \\mathrm{cm}^{3}\\).",
"answer": "This problem calls for us to compute the time of submersion of a steel plate. In order to solve this problem, we must first rearrange Equation, as\n\\[\nt=\\frac{K W}{\\rho A(\\mathrm{CPR})}\n\\]\nThus, using values for the various parameters given in the problem statement\n\\[\n\\begin{aligned}\nt & =\\frac{(534)\\left(2.6 \\times 10^{6} \\mathrm{mg}\\right)}{\\left(7.9 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(10 \\mathrm{in}^{2}\\right)(200 \\mathrm{mpy})} \\\\\n& =8.8 \\times 10^{4} \\mathrm{~h}=10 \\mathrm{yr}\n\\end{aligned}\n\\]"
},
{
"idx": 585,
"question": "A thick steel sheet of area 400 cm² is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of 375 g due to corrosion. To what rate of corrosion, in mm/yr, does this correspond?",
"answer": "This problem asks for us to calculate the CPR in mm/yr for a thick steel sheet of area 400 cm² which experiences a weight loss of 375 g after one year. Employment of Equation leads to CPR(mm/yr) = (K W)/(ρ A t) = (87.6)(375 g)(10³ mg/g)/(7.9 g/cm³)(400 cm²)(24 h/day)(365 day/yr)(1 yr) = 1.2 mm/yr"
},
{
"idx": 585,
"question": "A thick steel sheet of area 400 cm² is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of 375 g due to corrosion. To what rate of corrosion, in mpy, does this correspond?",
"answer": "This problem asks for us to calculate the CPR in mpy for a thick steel sheet of area 400 cm² which experiences a weight loss of 375 g after one year. Employment of Equation leads to CPR(mpy) = (534)(375 g)(10³ mg/g)/(7.9 g/cm³)(400 in.²)(1 in/2.54 cm)²(24 h/day)(365 day/yr)(1 yr) = 46.7 mpy"
},
{
"idx": 586,
"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": 586,
"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": 586,
"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": 587,
"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 data: V(M/M2+) = -0.47 V, V(H+/H2) = 0 V, i0 for M = 5 × 10^-10 A/cm2, i0 for H = 2 × 10^-9 A/cm2, β for M = +0.15, β for H = -0.12.",
"answer": "To compute the rate of oxidation for metal M, we establish potential expressions for both oxidation and reduction reactions. For hydrogen reduction: VH = V(H+/H2) + βH log(i/i0H). For M oxidation: VM = V(M/M2+) + βM log(i/i0M). Setting VH = VM and solving for log(iC) gives: log(iC) = [1/(βM - βH)][V(H+/H2) - V(M/M2+) - βH log(i0H) + βM log(i0M)]. Substituting the given values: log(iC) = [1/(0.15 - (-0.12))][0 - (-0.47) - (-0.12)log(2 × 10^-9) + 0.15log(5 × 10^-10)] = -7.293. Thus, iC = 10^-7.293 = 5.09 × 10^-8 A/cm2. The corrosion rate is then: r = iC/(nF) = (5.09 × 10^-8 C/s-cm2)/(2 × 96,500 C/mol) = 2.64 × 10^-13 mol/cm2-s."
},
{
"idx": 587,
"question": "Compute the corrosion potential for the reaction involving metal M given the following data: V(H+/H2) = 0 V, i0 for H = 2 × 10^-9 A/cm2, β for H = -0.12, and the corrosion current density iC = 5.09 × 10^-8 A/cm2.",
"answer": "The corrosion potential VC can be calculated using the potential expression for hydrogen reduction: VC = V(H+/H2) + βH log(iC/i0H). Substituting the given values: VC = 0 + (-0.12 V) log[(5.09 × 10^-8 A/cm2)/(2 × 10^-9 A/cm2)] = -0.169 V."
},
{
"idx": 588,
"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": 588,
"question": "Name two common types of alloy that passivate.",
"answer": "Stainless steels and aluminum alloys often passivate."
},
{
"idx": 589,
"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": 590,
"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": 591,
"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": 592,
"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": 592,
"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": 593,
"question": "What is 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": 593,
"question": "What is 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": 594,
"question": "Compute the electrical conductivity of a 5.1-mm (0.2-in.) diameter cylindrical silicon specimen 51 mm (2 in.) long in which a current of 0.1 A passes in an axial direction. A voltage of 12.5 V is measured across two probes that are separated by 38 mm (1.5 in.).",
"answer": "We use Equations for the conductivity, as sigma = 1/rho = (I l)/(V A) = (I l)/(V pi (d/2)^2). And, incorporating values for the several parameters provided in the problem statement, leads to sigma = (0.1 A)(38 × 10^-3 m)/(12.5 V)(pi)((5.1 × 10^-3 m)/2)^2 = 14.9 (Ω·m)^-1."
},
{
"idx": 594,
"question": "Compute the resistance over the entire 51 mm (2 in.) of the specimen.",
"answer": "The resistance, R, may be computed using Equations, as R = rho l/A = l/(sigma A) = l/(sigma pi (d/2)^2) = (51 × 10^-3 m)/([14.9 (Ω·m)^-1](pi)((5.1 × 10^-3 m)/2)^2) = 168 Ω."
},
{
"idx": 595,
"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 Ge may be determined using Equation. Since the room temperature mobility of electrons is 0.38 m2/V-s, and the electric field is 1000 V/m (as stipulated in the problem statement), vd = μe E = (0.38 m2/V-s)(1000 V/m) = 380 m/s."
},
{
"idx": 595,
"question": "Under these circumstances, how long does it take an electron to traverse a 25-mm (1-in.) length of crystal?",
"answer": "The time, t, required to traverse a given length, l(=25 mm), is just t = l / vd = (25 x 10-3 m) / (380 m/s) = 6.6 x 10-5 s."
},
{
"idx": 596,
"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 may be computed using the equation n = σ / (|e| μ_e), where σ is the electrical conductivity, |e| is the electron charge, and μ_e is the electron mobility. Substituting the given values: n = (6.0 x 10^7 (Ω·m)^-1) / ((1.602 x 10^-19 C)(0.0030 m^2/V·s)) = 1.25 x 10^29 m^-3."
},
{
"idx": 596,
"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, and the density of copper is 8.9 g/cm^3. What is the number of free electrons per copper atom?",
"answer": "To calculate the number of free electrons per copper atom, first determine the number of copper atoms per cubic meter, N_Cu, using the equation N_Cu = (N_A ρ') / A_Cu, where N_A is Avogadro's number, ρ' is the density, and A_Cu is the atomic weight of copper. Substituting the given values: N_Cu = ((6.022 x 10^23 atoms/mol)(8.9 g/cm^3)(10^6 cm^3/m^3)) / 63.55 g/mol = 8.43 x 10^28 m^-3. Then, the number of free electrons per copper atom is n / N_Cu = (1.25 x 10^29 m^-3) / (8.43 x 10^28 m^-3) = 1.48."
},
{
"idx": 597,
"question": "At room temperature the electrical conductivity of \\(\\mathrm{PbTe}\\) is \\(500(\\Omega \\cdot \\mathrm{m})^{-1}\\), whereas the electron and hole mobilities are 0.16 and \\(0.075 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\\), respectively. Compute the intrinsic carrier concentration for \\(\\mathrm{PbTe}\\) at room temperature.",
"answer": "In this problem we are asked to compute the intrinsic carrier concentration for PbTe at room temperature. Since the conductivity and both electron and hole mobilities are provided in the problem statement, all we need do is solve for \\(n\\) and \\(p\\) (i.e., \\(\\left.n_{j}\\right)\\) using Equation. Thus,\n\\[\n\\begin{aligned}\nn_{i} & =\\frac{\\sigma}{\\left|e\\left[\\left(\\mu_{e}+\\mu_{h}\\right)\\right.\\right.} \\\\\n& =\\frac{500(\\Omega \\cdot \\mathrm{m})^{-1}}{\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)(0.16+0.075) \\mathrm{m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}} \\\\\n& =1.33 \\times 10^{22} \\mathrm{~m}^{-3}\n\\end{aligned}\n\\]"
},
{
"idx": 598,
"question": "Define the term 'intrinsic' as it pertains to semiconducting materials and provide an example.",
"answer": "Intrinsic refers to pure, undoped semiconducting materials. Example: high purity (undoped) Si, GaAs, CdS, etc."
},
{
"idx": 598,
"question": "Define the term 'extrinsic' as it pertains to semiconducting materials and provide an example.",
"answer": "Extrinsic refers to doped semiconducting materials. Example: P-doped Ge, B-doped Si, S-doped GaP, etc."
},
{
"idx": 598,
"question": "Define the term 'compound' as it pertains to semiconducting materials and provide an example.",
"answer": "Compound refers to semiconducting materials composed of two or more elements. Example: GaAs, InP, CdS, etc."
},
{
"idx": 598,
"question": "Define the term 'elemental' as it pertains to semiconducting materials and provide an example.",
"answer": "Elemental refers to semiconducting materials composed of a single element. Example: Ge and Si."
},
{
"idx": 599,
"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": 599,
"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": 600,
"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. (Antimony is from group VA of the periodic table--Ge is from group IVA.)"
},
{
"idx": 600,
"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": "Since this material is n-type extrinsic, each Sb will donate a single electron, or the electron concentration is equal to the Sb concentration since all of the Sb atoms are ionized at room temperature; that is n = 5 × 10^22 m^-3, and, as given in the problem statement, μe = 0.1 m^2/V-s. Thus σ = n|e|μe = (5 × 10^22 m^-3)(1.602 × 10^-19 C)(0.1 m^2/V-s) = 800 (Ω·m)^-1."
},
{
"idx": 601,
"question": "Calculate the electron and hole mobilities for intrinsic InP given the following electrical characteristics at room temperature: conductivity (sigma) = 2.5 x 10^-6 (Omega.m)^-1, electron concentration (n) = 3.0 x 10^13 m^-3, hole concentration (p) = 3.0 x 10^13 m^-3.",
"answer": "For intrinsic InP, the conductivity expression is: sigma = n|e| mu_e + p|e| mu_h. Substituting the given values: 2.5 x 10^-6 (Omega.m)^-1 = (3.0 x 10^13 m^-3)(1.602 x 10^-19 C) mu_e + (3.0 x 10^13 m^-3)(1.602 x 10^-19 C) mu_h. This simplifies to 0.52 = mu_e + mu_h."
},
{
"idx": 601,
"question": "Calculate the electron and hole mobilities for extrinsic n-type InP given the following electrical characteristics at room temperature: conductivity (sigma) = 3.6 x 10^-5 (Omega.m)^-1, electron concentration (n) = 4.5 x 10^14 m^-3, hole concentration (p) = 2.0 x 10^12 m^-3.",
"answer": "For extrinsic n-type InP, the conductivity expression is: sigma = n|e| mu_e + p|e| mu_h. Substituting the given values: 3.6 x 10^-5 (Omega.m)^-1 = (4.5 x 10^14 m^-3)(1.602 x 10^-19 C) mu_e + (2.0 x 10^12 m^-3)(1.602 x 10^-19 C) mu_h. This simplifies to 112.4 = 225 mu_e + mu_h."
},
{
"idx": 601,
"question": "Solve the simultaneous equations 0.52 = mu_e + mu_h and 112.4 = 225 mu_e + mu_h to find the electron mobility (mu_e) and hole mobility (mu_h).",
"answer": "Solving the simultaneous equations: 0.52 = mu_e + mu_h and 112.4 = 225 mu_e + mu_h, we subtract the first equation from the second to get 111.88 = 224 mu_e. Solving for mu_e gives mu_e = 0.50 m^2/V-s. Substituting mu_e back into the first equation gives mu_h = 0.02 m^2/V-s."
},
{
"idx": 602,
"question": "Some metal alloy is known to have electrical conductivity and electron mobility values of \\(1.5 \\times 10^{7}\\) \\((\\Omega \\cdot \\mathrm{m})^{-1}\\) and \\(0.0020 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\\), respectively. Through a specimen of this alloy that is \\(35 \\mathrm{~mm}\\) thick is passed a current of 45 A. What magnetic field would need to be imposed to yield a Hall voltage of \\(-1.0 \\times 10^{-7} \\mathrm{~V}\\) ?",
"answer": "In this problem we are asked to determine the magnetic field required to produce a Hall voltage of \\(-1.0 \\times\\) \\(10^{-7} \\mathrm{~V}\\), given that \\(\\sigma=1.5 \\times 10^{7}(\\Omega \\cdot \\mathrm{m})^{-1}, \\mu_{\\mathrm{e}}=0.0020 \\mathrm{~m}^{2} / \\mathrm{v} \\cdot \\mathrm{s}, I_{\\mathrm{x}}=45 \\mathrm{~A}\\), and \\(d=35 \\mathrm{~mm}\\). Combining Equations, and after solving for \\(B_{Z}\\), we get\n\\[\n\\begin{aligned}\nB_{Z} & =\\frac{\\left|V_{\\mathrm{H}}\\right| d}{I_{X} R_{\\mathrm{H}}}=\\frac{\\left|V_{\\mathrm{H}}\\right| \\sigma d}{I_{X} \\mu_{\\mathrm{e}}} \\\\\n& =\\frac{\\left(-1.0 \\times 10^{-7} \\mathrm{~V}\\right)\\left[1.5 \\times 10^{7}(\\Omega-\\mathrm{m})^{-1}\\right](35 \\times 10^{-3} \\mathrm{~m})}{(45 \\mathrm{~A})(0.0020 \\mathrm{~m}^{2} / \\mathrm{W} \\cdot \\mathrm{s})} \\\\\n& =0.58 \\mathrm{tesla}\n\\end{aligned}\n\\]"
},
{
"idx": 603,
"question": "What are the two functions that a transistor may perform in an electronic circuit?",
"answer": "In an electronic circuit, a transistor may be used to (1) amplify an electrical signal, and (2) act as a switching device in computers."
},
{
"idx": 604,
"question": "At temperatures between \\(775^{\\circ} \\mathrm{C}\\left(1048 \\mathrm{~K}\\right)\\) and \\(1100^{\\circ} \\mathrm{C}\\left(1373 \\mathrm{~K}\\right)\\), the activation energy and preexponential for the diffusion coefficient of \\(\\mathrm{Fe}^{2+}\\) in \\(\\mathrm{FeO}\\) are \\(102,000 \\mathrm{~J} / \\mathrm{mol}\\) and \\(7.3 \\times 10^{-8} \\mathrm{~m}^{2} / \\mathrm{s}\\), respectively. Compute the mobility for an \\(\\mathrm{Fe}^{2+}\\) ion at \\(1000^{\\circ} \\mathrm{C}(1273 \\mathrm{~K})\\).",
"answer": "For this problem, we are given, for \\(\\mathrm{FeO}\\), the activation energy \\((102,000 \\mathrm{~J} / \\mathrm{mol})\\) and preexponential \\((7.3 \\times\\) \\(10^{-8} \\mathrm{~m}^{2} / \\mathrm{s}\\) ) for the diffusion coefficient of \\(\\mathrm{Fe}^{2+}\\) and are asked to compute the mobility for a \\(\\mathrm{Fe}^{2+}\\) ion at \\(1273 \\mathrm{~K}\\). The mobility, \\(\\mathrm{H}_{\\mathrm{Fe}} 2+\\), may be computed using Equation; however, this expression also includes the diffusion coefficient \\(D_{\\mathrm{Fe} 2^{+}}\\), which is determined using Equation as\n\\[\n\\begin{aligned}\nD_{\\mathrm{Fe} 2^{+}} & =D_{0} \\exp \\left(-\\frac{Q_{d}}{R T}\\right) \\\\\n& =\\left(7.3 \\times 10^{-8} \\mathrm{~m}^{2 / \\mathrm{s}}\\right) \\exp \\left[-\\frac{102,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(1273 \\mathrm{~K})}\\right] \\\\\n& =4.74 \\times 10^{-12} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{aligned}\n\\]\nNow solving for \\(\\mu_{\\mathrm{Fe} 2^{+}}\\)yields\n\\[\n\\begin{aligned}\n\\mu_{\\mathrm{Fe} 2^{+}} & =\\frac{n_{\\mathrm{Fe} 2^{+}} e D_{\\mathrm{Fe} 2^{+}}}{k T} \\\\\n& =\\frac{(2)\\left(1.602 \\times 10^{-19} \\mathrm{C} / \\mathrm{atom}\\right)\\left(4.74 \\times 10^{-12} \\mathrm{~m} \\mathrm{~m}^{2} / \\mathrm{s}\\right)}{(1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{atom}-\\mathrm{K})(1273 \\mathrm{~K})} \\\\\n& =8.64 \\times 10^{-11} \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\n\\end{aligned}\n\\]\n(Note: the value of \\(n_{\\mathrm{Fe}} 2^{+}\\)is two, inasmuch as two electrons are involved in the ionization of \\(\\mathrm{Fe}\\) to \\(\\mathrm{Fe}^{2+}\\).)"
},
{
"idx": 605,
"question": "A parallel-plate capacitor using a dielectric material having an \\(\\varepsilon_{r}\\) of 2.5 has a plate spacing of 1 \\(\\mathrm{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": "We want to compute the plate spacing of a parallel-plate capacitor as the dielectric constant is increased form 2.5 to 4.0 , while maintaining the capacitance constant. Combining Equations yields\n\\[\nC=\\frac{\\varepsilon A}{l}=\\frac{\\varepsilon_{r} \\varepsilon_{0} A}{l}\n\\]\nNow, let us use the subscripts 1 and 2 to denote the initial and final states, respectively. Since \\(C_{1}=C_{2}\\), then\n\\[\n\\frac{\\varepsilon_{r 1} e_{0} A}{l_{1}}=\\frac{\\varepsilon_{r 2} e_{0} A}{l_{2}}\n\\]\nAnd, solving for \\(l_{2}\\)\n\\[\nl_{2}=\\frac{\\varepsilon_{r 2} l_{1}}{\\varepsilon_{r 1}}=\\frac{(4.0)(1 \\mathrm{~mm})}{2.5}=1.6 \\mathrm{~mm}\n\\]"
},
{
"idx": 606,
"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": 606,
"question": "For solid lead titanate (PbTiO3), gaseous neon, diamond, solid KCl, and liquid NH3, what kind(s) of polarization is (are) possible? Why?",
"answer": "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": 607,
"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": "For aluminum, Cv at 50 K may be approximated by Equation, since this temperature is significantly below the Debye temperature (375 K). The value of Cv at 30 K is used to determine the constant A as A = Cv/T^3 = 0.81 J/mol-K / (30 K)^3 = 3.00 × 10^-5 J/mol-K^4. Therefore, at 50 K, Cv = A T^3 = (3.00 × 10^-5 J/mol-K^4)(50 K)^3 = 3.75 J/mol-K. The specific heat cv is then calculated as cv = (3.75 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 139 J/kg-K."
},
{
"idx": 607,
"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": "Since 425 K is above the Debye temperature, a good approximation for Cv is Cv = 3 R = (3)(8.31 J/mol-K) = 24.9 J/mol-K. The specific heat cv is then calculated as cv = (24.9 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 923 J/kg-K."
},
{
"idx": 608,
"question": "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": 608,
"question": "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": 609,
"question": "A \\(0.1 \\mathrm{~m}(3.9 \\mathrm{in}\\).) rod of a metal elongates \\(0.2 \\mathrm{~mm}(0.0079 \\mathrm{in}\\).) on heating from 20 to \\(100^{\\circ} \\mathrm{C}(68 \\mathrm{to}\\) \\(212^{\\circ} \\mathrm{F}\\) ). Determine the value of the linear coefficient of thermal expansion for this material.",
"answer": "The linear coefficient of thermal expansion for this material may be determined using a rearranged form of Equation as\n\\[\n\\begin{aligned}\n\\alpha_{l}= & \\frac{\\Delta l}{l_{0} \\Delta T}=\\frac{\\Delta l}{l_{0}\\left(T_{f}-T_{0}\\right)}=\\frac{0.2 \\times 10^{-3} \\mathrm{~m}}{\\left(0.1 \\mathrm{~m}\\right)\\left(100^{\\circ} \\mathrm{C}-20^{\\circ} \\mathrm{C}\\right)} \\\\\n& =25.0 \\times 10^{-6}\\left({ }^{\\circ} \\mathrm{C}\\right)^{-1}\n\\end{aligned}\n\\]"
},
{
"idx": 610,
"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": 611,
"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": 612,
"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": 612,
"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": 613,
"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": 614,
"question": "For the pair of materials pure copper and aluminum bronze (95 wt% Cu-5 wt% Al), which has the larger thermal conductivity? Justify your choice.",
"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": 614,
"question": "For the pair of materials fused silica and quartz, which has the larger thermal conductivity? Justify your choice.",
"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": 614,
"question": "For the pair of materials linear polyethylene and branched polyethylene, which has the larger thermal conductivity? Justify your choice.",
"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": 614,
"question": "For the pair of materials random poly(styrene-butadiene) copolymer and alternating poly(styrene-butadiene) copolymer, which has the larger thermal conductivity? Justify your choice.",
"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": 615,
"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": 615,
"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": 615,
"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": 616,
"question": "Confirm that there are 2.2 Bohr magnetons associated with each iron atom, given that the saturation magnetization is \\(1.70 \\times 10^{6} \\mathrm{~A} / \\mathrm{m}\\), that iron has a BCC crystal structure, and that the unit cell edge length is 0.2866 \\(\\mathrm{nm}\\).",
"answer": "We want to confirm that there are 2.2 Bohr magnetons associated with each iron atom. Therefore, let \\(n_{\\mathrm{B}}^{\\prime}\\) be the number of Bohr magnetons per atom, which we will calculate. This is possible using a modified and rearranged form of Equation-that is\n\\[\nn_{\\mathrm{B}}^{\\prime}=\\frac{M_{\\mathrm{s}}}{\\mu_{\\mathrm{B}} N}\n\\]\nNow, \\(N\\) is just the number of atoms per cubic meter, which is the number of atoms per unit cell (two for BCC) divided by the unit cell volume-- that is,\n\\[\nN=\\frac{2}{V_{C}}=\\frac{2}{a^{3}}\n\\]\n\\(a\\) being the BCC unit cell edge length. Thus\n\\[\n\\begin{aligned}\nn_{\\mathrm{B}}^{\\prime} & =\\frac{M_{\\mathrm{s}}}{N \\mu_{\\mathrm{B}}}=\\frac{M_{\\mathrm{s}} a^{3}}{2 \\mu_{\\mathrm{B}}} \\\\\n& =\\frac{\\left(1.70 \\times 10^{6} \\mathrm{~A} / m\\right)\\left[0.2866 \\times 10^{-9} \\mathrm{~m}\\right)^{3} / \\mathrm{unit} \\mathrm{cell}\\right] \\\\\n& \\left(2 \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell}\\right)\\left(9.27 \\times 10^{-24} \\mathrm{~A} \\cdot \\mathrm{m}^{2} / \\mathrm{BM}\\right)\n\\end{aligned}\n\\]\n\\[\n=2.16 \\text { Bohr magnetons/atom }\n\\]"
},
{
"idx": 617,
"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": 618,
"question": "Calculate the number of Bohr magnetons per unit cell (n_B) for yttrium iron garnet given the saturation magnetization (1.0 x 10^4 A/m), cubic unit cell edge length (1.2376 nm), and Bohr magneton value (9.27 x 10^-24 A·m^2/BM).",
"answer": "n_B = (M_s * a^3) / μ_B = (1.0 x 10^4 A/m) * (1.2376 x 10^-9 m)^3 / (9.27 x 10^-24 A·m^2/BM) = 2.04 Bohr magnetons/unit cell."
},
{
"idx": 618,
"question": "Determine the number of Bohr magnetons per formula unit for yttrium iron garnet, given there are 8 formula units per unit cell and the previously calculated n_B value (2.04 Bohr magnetons/unit cell).",
"answer": "Bohr magnetons per formula unit = n_B / 8 = 2.04 / 8 = 0.255 Bohr magnetons/formula unit."
},
{
"idx": 618,
"question": "Calculate the net magnetic moment contribution per formula unit from the Fe^3+ ions in yttrium iron garnet, given there are 2 Fe^3+ ions on a sites and 3 Fe^3+ ions on d sites per formula unit, each with 5 Bohr magnetons, and their magnetic moments are aligned antiparallel.",
"answer": "Net magnetic moment from Fe^3+ ions = (2 * 5 BM) - (3 * 5 BM) = 10 BM - 15 BM = -5 Bohr magnetons/formula unit (antiparallel to Y^3+ contribution)."
},
{
"idx": 618,
"question": "Compute the number of Bohr magnetons associated with each Y^3+ ion in yttrium iron garnet, given the Bohr magnetons per formula unit (0.255 BM), the net magnetic moment from Fe^3+ ions (-5 BM), and there are 3 Y^3+ ions per formula unit.",
"answer": "Number of Bohr magnetons/Y^3+ = (0.255 BM + 5 BM) / 3 = 5.255 BM / 3 = 1.75 Bohr magnetons/Y^3+ ion."
},
{
"idx": 619,
"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": 619,
"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": 620,
"question": "An iron bar magnet having a coercivity of \\(4000 \\mathrm{~A} / \\mathrm{m}\\) is to be demagnetized. If the bar is inserted within a cylindrical wire coil \\(0.15 \\mathrm{~m}\\) long and having 100 turns, what electric current is required to generate the necessary magnetic field?",
"answer": "In order to demagnetize a magnet having a coercivity of \\(4000 \\mathrm{~A} / \\mathbf{m}\\), an \\(H\\) field of \\(4000 \\mathrm{~A} / \\mathrm{m}\\) must be applied in a direction opposite to that of magnetization. According to Equation\n\\[\n\\begin{aligned}\nI & =\\frac{H I}{N} \\\\\n& =\\frac{(4000 \\mathrm{~A} / \\mathrm{m})(0.15 \\mathrm{~m})}{100 \\text { turns }}=6 \\mathrm{~A}\n\\end{aligned}\n\\]"
},
{
"idx": 621,
"question": "Visible light having a wavelength of 6 × 10^-7 m appears orange. Compute the frequency of a photon of this light.",
"answer": "v = c / λ = (3 × 10^8 m/s) / (6 × 10^-7 m) = 5 × 10^14 s^-1"
},
{
"idx": 621,
"question": "Visible light having a wavelength of 6 × 10^-7 m appears orange. Compute the energy of a photon of this light.",
"answer": "E = hc / λ = (6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (6 × 10^-7 m) = 3.31 × 10^-19 J (2.07 eV)"
},
{
"idx": 622,
"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": 622,
"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": 622,
"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": 623,
"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": 624,
"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": 625,
"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": 626,
"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": 627,
"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": "This problem calls for a calculation of the reflectivity between two corundum grains having different orientations and indices of refraction (1.757 and 1.779) in the direction of light propagation, when the light is at normal incidence to the grain boundary. We must employ Equation since the beam is normal to the grain boundary. Thus,\n\\[\n\\begin{aligned}\nR & =\\left(\\frac{n_{2}-n_{1}}{n_{2}+n_{1}}\\right)^{2} \\\\\n& =\\left(\\frac{1.779-1.757}{1.779+1.757}\\right)^{2}=3.87 \\times 10^{-5}\n\\end{aligned}\n\\]"
},
{
"idx": 628,
"question": "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": 628,
"question": "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": 629,
"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": 630,
"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": 631,
"question": "Zinc has five naturally occurring isotopes: \\(48.63 \\%\\) of \\({ }^{64} \\mathrm{Zn}\\) with an atomic weight of 63.929 amu; \\(27.90 \\%\\) of \\({ }^{66} \\mathrm{Zn}\\) with an atomic weight of \\(65.926 \\mathrm{amu}^{\\text {; }} 4.10 \\%\\) of \\({ }^{67} \\mathrm{Zn}\\) with an atomic weight of \\(66.927 \\mathrm{amu} ; 18.75 \\%\\) of \\({ }^{68} \\mathrm{Zn}\\) with an atomic weight of \\(67.925 \\mathrm{amu} ;\\) and \\(0.62 \\%\\) of \\({ }^{70} \\mathrm{Zn}\\) with an atomic weight of \\(69.925 \\mathrm{amu}\\). Calculate the average atomic weight of \\(\\mathrm{Zn}\\).",
"answer": "The average atomic weight of zinc \\(\\bar{A}_{\\mathrm{Zn}}\\) is computed by adding fraction-of-occurrence atomic weight products for the five isotopes - i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent of occurrence divided by 100 .) Thus\n\\[\n\\bar{A}_{\\mathrm{Zn}}=f_{64 \\mathrm{Zn}} A_{66 \\mathrm{Zn}}^{*}+f_{\\phi \\mathrm{Zn}} A_{6 \\mathrm{Zn}}^{*}+f_{\\phi \\mathrm{Zm}} A_{6 \\mathrm{Zn}}^{*} A_{\\mathrm{Zn}}^{*}+f_{\\phi \\mathrm{Zn}}^{*} A_{68 \\mathrm{Zn}}^{*}+f_{70 \\mathrm{Zn}}^{*} A_{70 \\mathrm{Zn}}^{*}\n\\]\nIncluding data provided in the problem statement we solve for \\(\\bar{A}_{\\mathrm{Zn}}\\) as\n\\[\n\\begin{array}{l}\n\\bar{A}_{\\mathrm{Zn}}=(0.4863)(63.929 \\mathrm{amu})+(0.2790)(65.926 \\mathrm{amu}) \\\\\n\\quad+(0.0410)(66.927 \\mathrm{amu})+(0.1875)(67.925 \\mathrm{amu})+(0.0062)(69.925 \\mathrm{amu})\n\\end{array}\n\\]\n\\[\n=65.400 \\mathrm{amu}\n\\]"
},
{
"idx": 632,
"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 average atomic weight of indium \\(\\left(\\bar{A}_{\\mathrm{In}}\\right)\\) is computed by adding fraction-ofoccurrence-atomic weight products for the two isotopes-i.e., using Equation, or\n\\[\n\\bar{A}_{\\mathrm{In}}=f_{11 \\mathrm{In}} A_{113 \\mathrm{In}}+f_{115 \\mathrm{In}} A_{115 \\mathrm{In}}\n\\]\nBecause there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000 ; or\n\\[\nf_{113 \\mathrm{In}}+f_{115 \\mathrm{In}}\n\\]\nwhich means that\n\\[\nf_{113 \\mathrm{In}}=1.000 \\quad f_{115 \\mathrm{In}}\n\\]\nSubstituting into this expression the one noted above for \\(f_{113 \\mathrm{In}}\\), and incorporating the atomic weight values provided in the problem statement yields\n\\[\n\\begin{array}{c}\n114.818 \\mathrm{amu}=f_{113 \\mathrm{In}} A_{113 \\mathrm{In}}+f_{135 \\mathrm{In}} A_{115 \\mathrm{In}} \\\\\n114.818 \\mathrm{amu}=(1.000 \\quad f_{115 \\mathrm{In}}) A_{113 \\mathrm{In}}+f_{115 \\mathrm{~m}} A_{115 \\mathrm{~m}} \\\\\n114.818 \\mathrm{amu}=(1.00 \\quad f_{115 \\mathrm{~m}})(112.904 \\mathrm{amu})+f_{115 \\mathrm{~m}}(114.904 \\mathrm{amu})\n\\end{array}\n\\]Solving this expression for \\(f_{115_{\\text {In }}}\\) yields \\(f_{115_{\\text {In }}}=0.957\\). Furthermore, because\n\\[\n\\begin{array}{rll}\nf_{113_{\\text {In }}} & =1.000 & f_{115_{\\text {In }}}\n\\end{array}\n\\]\nthen\n\\[\nf_{113_{\\text {In }}}=1.000 \\ \\ 0.957=0.043\n\\]"
},
{
"idx": 633,
"question": "The K+ ion has an electron structure that is identical to which inert gas?",
"answer": "The K+ ion has an electron configuration the same as argon."
},
{
"idx": 633,
"question": "The I- ion has an electron structure that is identical to which inert gas?",
"answer": "The I- ion has an electron configuration the same as xenon."
},
{
"idx": 634,
"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": 635,
"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)."
},
{
"idx": 636,
"question": "Without consulting Figure or Table, determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.\n(a) \\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}\\)\n(b) \\(1 s^{2} 2 s^{2} 2 p^{6} s 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}\\)\n(c) \\(1 s^{2} 2 s^{2} 2 p^{6} \\cdot 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}\\)\n(d) \\(1 s^{2} 2 s^{2} 2 p^{6} .3 s^{2} 3 p^{6} 4 s^{1}\\)\n(e) \\(1 s^{2} 2 s^{2} 2 p^{6} x^{2} 3 p^{6} 3 d^{10} 4 s 2 4 p^{6} 4 d^{5} 5 s^{2}\\)\n(f) \\(1 s^{2} 2 s^{2} 2 p^{6} t^{3} s^{2}\\)",
"answer": "(a) The \\(1 s^{2} 2 s^{2} 2 p^{6} 63 s^{2} 3 p^{5}\\) electron configuration is that of a halogen because it is one electron deficient from having a filled \\(p\\) subshell.\n(b) The \\(1 s^{2} 2 s^{2} 2 p^{63 s^{2} 3 p^{6} 3 d^{7} 4 \\mathrm{~s}^{2}\\) electron configuration is that of a transition metal because of an incomplete \\(d\\) subshell.\n(c) The \\(1 s^{2} 2 s^{2} 2 p^{6}{ }_{3} s^{2} 3 p^{6} 3 d^{10} 4 \\mathrm{~s}^{2} 4 p^{6}\\) electron configuration is that of an inert gas because of filled \\(4 s\\) and \\(4 p\\) subshells.\n(d) 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.\n(e) 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.\n(f) The \\(1 s^{2} 2 s^{2} 2 p^{61} 3 s^{2}\\) electron configuration is that of an alkaline earth metal because of two \\(s\\) electrons."
},
{
"idx": 637,
"question": "The net potential energy between two adjacent ions, \\(E_{N}\\), may be represented by the sum of Equations 2.9 and 2.11; that is,\n\\[\nE_{N}=\\frac{A}{r}+\\frac{B}{r^{n}}\n\\]\nCalculate the bonding energy \\(E_{0}\\) in terms of the parameters \\(A, B\\), and \\(n\\) using the following procedure:\n1. 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}\\).\n2. Solve for \\(r\\) in terms of \\(A, B\\), and \\(n\\), which yields \\(r_{0}\\), the equilibrium interionic spacing.\n3. Determine the expression for \\(E_{0}\\) by substitution of \\(r_{0}\\) into Equation.",
"answer": "Differentiation of Equation yields\n\\[\n\\begin{aligned}\n\\frac{d E_{N}}{d r} & =\\frac{d\\left(-A\\right)}{d r}+\\frac{d\\left(\\frac{B}{r^{n}}\\right)}{d r} \\\\\n& =\\frac{A}{r^{(1+1)}}-\\frac{n B}{r^{(n+1)}}=0\n\\end{aligned}\n\\]\nNow, solving for \\(r\\left(=r_{0}\\right)\\)\nor\n\\[\n\\frac{A}{r_{0}^{2}}=\\frac{n B}{r_{0}^{(n+1)}}\n\\]\\[\n\\begin{array}{c}\nE_{0}=-\\frac{A}{r_{0}^{2}}+\\frac{B}{r_{0}^{2}} \\\\\n\\\\=-\\frac{A}{\\left(\\frac{A}{nB}\\right)^{1 /(1-n)}}+\\frac{B}{\\left(\\frac{A}{nB}\\right)^{n /(1-n)}}\n\\end{array}\n\\]"
},
{
"idx": 638,
"question": "Explain why hydrogen fluoride (HF) has a higher boiling temperature than hydrogen \\\\ chloride \\((\\mathrm{HCl})\\left(19.4 \\mathrm{vs} .-85^{\\circ} \\mathrm{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": 639,
"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": 639,
"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": 639,
"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": 639,
"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": 639,
"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": 639,
"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": 640,
"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} \\cdot 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": 641,
"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": 642,
"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": 643,
"question": "If the atomic radius of lead is \\(0.175 \\mathrm{~nm}\\), calculate the volume of its unit cell in cubic meters.",
"answer": "Lead has an FCC crystal structure and, as given in the problem statement, an atomic radius of \\(0.1750 \\mathrm{~nm}\\) . The FCC unit cell volume may be computed from Equation as\n\\[\n\\begin{array}{c}\nV_{C}=16 R^{3} \\sqrt{2}=(16)(0.175 \\times 10^{-9} \\mathrm{~m})^{3} \\sqrt{2} \\\\\n\\quad=1.213 \\times 10^{-28} \\mathrm{~m}^{3}\n\\end{array}\n\\]"
},
{
"idx": 644,
"question": "Calculate the radius of a palladium \\((P d)\\) atom, given that \\(P d\\) has an FCC crystal structure, a density of \\(12.0 \\mathrm{~g} / \\mathrm{cm}^{3}\\), and an atomic weight of \\(106.4 \\mathrm{~g} / \\mathrm{mol}\\).",
"answer": "We are asked to determine the radius of a palladium atom, given that \\(\\mathrm{Pd}\\) has an FCC crystal structure. For FCC, \\(n=4\\) atoms/unit cell, and \\(V_{C}=16 R^{3} \\sqrt{2}\\). Now, the density of \\(\\mathrm{Pd}\\) may be expressed using a form of Equation as follows:\n\\[\n\\begin{array}{l}\n=\\frac{n A_{\\mathrm{Pd}}}{V_{C} N_{\\mathrm{A}}} \\\\\n=\\frac{n A_{\\mathrm{Pd}}}{\\left(16 R^{3} \\sqrt{2}\\right) N_{\\mathrm{A}}} \\\\\n\\text { Solving for } R \\text { from the above expression yields } \\\\\nR=\\left\\langle\\frac{n A_{\\mathrm{Pd}}}{16 N_{\\mathrm{A}} \\sqrt{2}}\\right\\rangle^{1 / 3}\n\\end{array}\n\\]\nIncorporation into this expression values for \\(A_{\\mathrm{Pd}}, n\\), and \\(\\rho\\) leads to the following value of \\(R\\) :\n\\[\n\\begin{array}{l}\n=\\left[\\frac{(4 \\text { atoms/unit cell })(106.4 \\mathrm{~g} / \\mathrm{mol})}{(16)\\left(12.0 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol})(\\sqrt{2})}\\right]^{1 / 3} \\\\\n\\quad=1.38 \\times 10^{-8} \\mathrm{~cm}=0.138 \\mathrm{~nm}\n\\end{array}\n\\]"
},
{
"idx": 645,
"question": "Calculate the radius of a tantalum (Ta) atom, given that Ta has a BCC crystal structure, \\(a\\) density of \\(16.6 \\mathrm{~g} / \\mathrm{cm}^{3}\\), and an atomic weight of \\(180.9 \\mathrm{~g} / \\mathrm{mol}\\).",
"answer": "It is possible to compute the radius of a Ta atom using a rearranged form of Equation.\nFor \\(\\mathrm{BCC}, n=2\\) atoms/unit cell. Furthermore, because \\(V_{C}=\\alpha^{3}\\) and \\(a=\\frac{4 R}{\\sqrt{3}}\\) an expression for the \\(\\mathrm{BCC}\\) unit cell volume is as follows:\n\\[\nV_{C}=\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}=\\frac{64 R^{3}}{3 \\sqrt{3}}\n\\]\nFor Ta, Equation takes the form\n\\[\n\\begin{aligned}\n& =\\frac{n A_{\\mathrm{Ta}}}{V_{C} N_{\\mathrm{A}}} \\\\\n& =\\frac{n A_{\\mathrm{Ta}}}{\\left(\\frac{64 R^{3}}{3 \\sqrt{3}}\\right) N_{\\mathrm{A}}} \\\\\n\\text { Solving for } R \\text { in this equation yields } \\\\\nR & =\\left(\\frac{3 n A_{\\mathrm{Ta}} \\sqrt{3}}{64} \\frac{1 / 3}{N_{\\mathrm{A}}}\\right)^{1 / 3}\n\\end{aligned}\n\\]\nUpon incorporation of values of \\(n, A_{\\mathrm{Ta}}\\), and \\(\\rho\\) into this equation leads to the atomic radius of \\(\\mathrm{Ta}\\) as follows:\n\\[\nR=\\left[\\frac{(3)(2 \\text { atoms/unit cell})(180.9 \\mathrm{~g} / \\mathrm{mol})(\\sqrt{3})}{(64)(16.6 \\mathrm{~g} / \\mathrm{cm}^{3})(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol})}\\right]^{1 / 3}\n\\]\\[\n\\hspace{-0.5cm}+1.43\\times 10^{-8}\\,{\\rm cm} = 0.143\\,{\\rm nm}\n\\]"
},
{
"idx": 646,
"question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm³. What is the volume of its unit cell in cubic meters?",
"answer": "The volume of the Ti unit cell may be computed using a rearranged form of Equation as V_C = (n A_Ti) / (ρ N_A). For HCP, n = 6 atoms/unit cell, and the atomic weight for Ti, A_Ti = 47.87 g/mol. Thus, V_C = (6 atoms/unit cell)(47.87 g/mol) / (4.51 g/cm³)(6.022 × 10^23 atoms/mol) = 1.058 × 10^-22 cm³/unit cell = 1.058 × 10^-28 m³/unit cell."
},
{
"idx": 646,
"question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm³. If the c/a ratio is 1.58, compute the values of c and a.",
"answer": "From Equation, for HCP the unit cell volume is V_C = (3 a² c √3) / 2. Since, for Ti, c = 1.58 a, then V_C = (3(1.58) a³ √3) / 2 = 1.058 × 10^-22 cm³/unit cell. Now, solving for a: a = [(2)(1.058 × 10^-22 cm³) / (3)(1.58)(√3)]^(1/3) = 2.96 × 10^-8 cm = 0.296 nm. And finally, the value of c is c = 1.58 a = (1.58)(0.296 nm) = 0.468 nm."
},
{
"idx": 647,
"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 Mg unit cell may be computed using a rearranged form of the equation as V_C = (n A_Mg) / (ρ N_A). For HCP, n = 6 atoms/unit cell, and the atomic weight for Mg, A_Mg = 24.31 g/mol. Thus, V_C = (6 atoms/unit cell)(24.31 g/mol) / (1.74 g/cm³)(6.022 × 10²³ atoms/mol) = 1.39 × 10⁻²² cm³/unit cell."
},
{
"idx": 647,
"question": "Magnesium (Mg) has an HCP crystal structure, a density of 1.74 g/cm³, and a c/a ratio of 1.624. Compute the values of c and a.",
"answer": "From the equation for HCP, the unit cell volume is V_C = (3 a² c √3) / 2. Given c = 1.624 a and V_C = 1.39 × 10⁻²² cm³/unit cell, solving for a gives a = [(2)(1.39 × 10⁻²² cm³) / (3)(1.624)(√3)]^(1/3) = 3.21 × 10⁻⁸ cm = 0.321 nm. The value of c is c = 1.624 a = (1.624)(0.321 nm) = 0.521 nm."
},
{
"idx": 648,
"question": "Given the orthorhombic unit cell parameters for uranium (U) as a = 0.286 nm, b = 0.587 nm, and c = 0.495 nm, calculate the unit cell volume (V_C).",
"answer": "The unit cell volume (V_C) is calculated as the product of the three lattice parameters: V_C = a * b * c = (0.286 nm) * (0.587 nm) * (0.495 nm) = 0.286 * 0.587 * 0.495 * 10^-21 cm^3 = 8.31 * 10^-23 cm^3."
},
{
"idx": 648,
"question": "Using the density (19.05 g/cm^3), atomic weight (238.03 g/mol), and the calculated unit cell volume (V_C = 8.31 * 10^-23 cm^3) for uranium (U), determine the number of atoms per unit cell (n).",
"answer": "The number of atoms per unit cell (n) is calculated using the formula: n = (density * V_C * N_A) / atomic weight = (19.05 g/cm^3 * 8.31 * 10^-23 cm^3 * 6.022 * 10^23 atoms/mol) / 238.03 g/mol = 4.01 atoms/unit cell."
},
{
"idx": 648,
"question": "Given the atomic radius of uranium (U) as 0.1385 nm, calculate the volume of a single uranium atom (V_atom).",
"answer": "The volume of a single uranium atom (V_atom) is calculated using the formula for the volume of a sphere: V_atom = (4/3) * π * R^3 = (4/3) * π * (0.1385 nm)^3 = (4/3) * π * (1.385 * 10^-8 cm)^3 = 1.11 * 10^-23 cm^3."
},
{
"idx": 648,
"question": "Using the number of atoms per unit cell (n = 4.01) and the volume of a single uranium atom (V_atom = 1.11 * 10^-23 cm^3), calculate the total sphere volume (V_S) in the unit cell.",
"answer": "The total sphere volume (V_S) is calculated as the product of the number of atoms per unit cell and the volume of a single atom: V_S = n * V_atom = 4.01 * 1.11 * 10^-23 cm^3 = 4.45 * 10^-23 cm^3."
},
{
"idx": 648,
"question": "Using the total sphere volume (V_S = 4.45 * 10^-23 cm^3) and the unit cell volume (V_C = 8.31 * 10^-23 cm^3), compute the atomic packing factor (APF) for uranium (U).",
"answer": "The atomic packing factor (APF) is calculated as the ratio of the total sphere volume to the unit cell volume: APF = V_S / V_C = (4.45 * 10^-23 cm^3) / (8.31 * 10^-23 cm^3) = 0.536."
},
{
"idx": 649,
"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": "For indium, and from the definition of the APF is APF = VS / VC = n(4/3 πR3) / a2c we may solve for the number of atoms per unit cell, n, as n = (APF) a2 c / (4/3 πR3) = (0.693)(4.59)2(4.95)(10^-24 cm3) / (4/3 π(1.625 × 10^-8 cm)3) = 4.0 atoms/unit cell"
},
{
"idx": 649,
"question": "The atomic weight of indium is 114.82 g/mol; compute its theoretical density using the tetragonal unit cell parameters a = 0.459 nm and c = 0.495 nm, and the number of atoms per unit cell determined to be 4.0.",
"answer": "In order to compute the density, we just employ the equation ρ = n AIn / (a2 c NA) = (4 atoms/unit cell)(114.82 g/mol) / [(4.59 × 10^-8 cm)2(4.95 × 10^-8 cm)/unit cell](6.022 × 10^23 atoms/mol) = 7.31 g/cm3"
},
{
"idx": 650,
"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": "The volume of an HCP unit cell is given by V_C = 6 R^2 c sqrt(3). Given c = 1.568 a and a = 2 R, then c = (1.568)(2 R) = 3.136 R. Substituting this expression for c into the volume equation leads to V_C = (6)(3.14) R^3 sqrt(3) = (6)(3.14)(sqrt(3))[0.1143 × 10^-7 cm]^3 = 4.87 × 10^-23 cm^3/unit cell."
},
{
"idx": 650,
"question": "Using the unit cell volume calculated previously (4.87 × 10^-23 cm^3/unit cell) and given that Be has an atomic weight of 9.01 g/mol, calculate the theoretical density of Be and compare it with the literature value.",
"answer": "The theoretical density of Be is determined using the formula rho = (n A_Be)/(V_C N_A). For HCP, n = 6 atoms/unit cell, and for Be, A_Be = 9.01 g/mol. Using the value of V_C = 4.87 × 10^-23 cm^3/unit cell, the theoretical density of Be is rho = (6 atoms/unit cell)(9.01 g/mol)/(4.87 × 10^-23 cm^3/unit cell)(6.022 × 10^23 atoms/mol) = 1.84 g/cm^3. The literature value is 1.85 g/cm^3."
},
{
"idx": 651,
"question": "Magnesium ( \\(\\mathrm{Mg}\\) ) has an HCP crystal structure, a c/a ratio of 1.624 , and a density of 1.74 \\(\\mathrm{g} / \\mathrm{cm}^{3}\\). Compute the atomic radius for \\(\\mathrm{Mg}\\).",
"answer": "This problem calls for us to compute the atomic radius for \\(\\mathrm{Mg}\\). In order to do this we must use Equation, as well as the expression that relates the atomic radius to the unit cell volume for HCP—Equation 3.7b - that is\n\\[\nV_{C}=6 R^{2} c \\sqrt{3}\n\\]\nIn this case \\(c=1.624 a\\), but, for HCP, \\(a=2 R\\), which means that\n\\[\nV_{C}=6 R^{2}(1.624)(2 R) \\sqrt{3}=(1.624)(12 \\sqrt{3}) R^{3}\n\\]\nAnd from Equation, the density is equal to\n\\[\n=\\frac{n A_{\\mathrm{Mg}}}{V_{C} N_{\\mathrm{A}}}=\\frac{n A_{\\mathrm{Mg}}}{(1.624)(12 \\sqrt{3}) R^{3} N_{\\mathrm{A}}}\n\\]\nAnd, solving for \\(R\\) from the above equation leads to the following:\n\\[\nR=\\left[\\frac{n A_{\\mathrm{Mg}}}{(1.624)(10 \\sqrt{3}) N_{\\mathrm{A}}}\\right]^{1 / 3}\n\\]\nFor the HCP crystal structure, \\(n=6\\) and from the inside cover, the atomic weight of \\(\\mathrm{Mg}\\) is 24.31 \\(\\mathrm{g} / \\mathrm{mol}\\). Upon incorporation of these values as well as the density of \\(\\mathrm{Mg}\\) provided in the problem statement, we compute magnesium's atomic radius as follows:\\[\nR=\\left[\\frac{\\left(6\\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell}\\right)\\left(24.31\\mathrm{\\ g} / \\mathrm{mol}\\right)}{\\left(1.624\\right)\\left(12 \\sqrt{3}\\right)\\left(1.74\\mathrm{\\ g} / \\mathrm{cm}^{3}\\right)\\left(6.022\\mathrm{\\ \\times \\ }10^{23} \\mathrm{\\ atoms} / \\mathrm{mol}\\right)}\\right]^{-1/3}\n\\]\n\\[\n\\begin{array}{ll}\n& =1.60\\quad 10^{\\;8}\\;\\mathrm{cm}=0.160\\;\\mathrm{nm} \\\\\n\\end{array}\n\\]"
},
{
"idx": 652,
"question": "Cobalt (Co) has an HCP crystal structure, an atomic radius of \\(0.1253 \\mathrm{~nm}\\), and a c/a ratio of 1.623. Compute the volume of the unit cell for Co.",
"answer": "This problem asks that we calculate the unit cell volume for Co, which has an HCP crystal structure. In order to do this, it is necessary to use Equation---an expression for the volume of an HCP unit cell in terms of the atomic radius \\(R\\) and the lattice parameter - that is\n\\[\nV_{C}=6 R^{2} c \\sqrt{3}\n\\]\nThe problem states that \\(c=1.623 a\\); also, it is the case for HCP that \\(a=2 R\\). Making these substitutions into the previous equation yields\n\\[\nV_{C}=(1.623)\\left(12 \\sqrt{3}\\right) R^{3}\n\\]\nAnd incorporation of the value of \\(R\\) provided in the problem statement leads to the following value for the unit cell volume:\n\\[\n\\begin{aligned}\nV_{C} & =(1.623)\\left(12 \\sqrt{3}\\right)(1.253 \\quad 10^{8} \\mathrm{~cm})^{3} \\\\\n& =6.64 \\times 10^{-23} \\mathrm{~cm}^{3}=6.64 \\times 10^{-2} \\mathrm{~nm}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 653,
"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 \\(0.700 \\mathrm{~nm}\\). If the measured density is \\(2.32 \\mathrm{~g} / \\mathrm{cm}^{3}\\), how many \\(\\mathrm{Si}^{4+}\\) and \\(\\mathrm{O}^{2}\\)-ions are there per unit cell?",
"answer": "We are asked to determine the number of \\(\\mathrm{Si}^{4+}\\) and \\(\\mathrm{O}^{ 2-}\\) ions per unit cell for a crystalline form of silica \\(\\left(\\mathrm{SiO}_{2}\\right)\\). For this material, \\(a=0.700 \\mathrm{~nm}\\) and \\(\\rho=2.32 \\mathrm{~g} / \\mathrm{cm}^{3}\\). Solving for \\(n^{\\prime}\\) from Equation, we obtain the following:\n\\[\n\\begin{aligned}\n\\mathrm{r}^{n} & =\\frac{\\rho V_{\\mathrm{C}} N_{\\mathrm{A}}}{\\sum A_{\\mathrm{Si}}+\\sum A_{\\mathrm{O}}}\n\\end{aligned}\n\\]\nNow if a formula \\(\\left(\\mathrm{SiO}_{2}\\right)\\) unit there is \\(1 \\mathrm{Si}^{4+}\\) and \\(2 \\mathrm{O}^{2-}\\) ions. Therefore\n\\[\n\\begin{array}{c}\nA_{\\mathrm{Si}}=A_{\\mathrm{Si}}=28.09 \\mathrm{~g} / \\mathrm{mol} \\\\\nA_{\\mathrm{O}}=2 A_{\\mathrm{O}}=2(16.00 \\mathrm{~g} / \\mathrm{mol})=32.00 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\n(The atomic weight values for \\(\\mathrm{Si}\\) and \\(\\mathrm{O}\\) came from inside the front cover.)\nAlso, because the unit cell has cubic symmetry, it is the case that\n\\[\nV_{C}=a^{3}\n\\]\nHere, \\(a\\) is the unit cell edge length-viz., \\(0.700 \\mathrm{~nm}\\) (or \\(7.00 \\quad 10^{8} \\mathrm{~cm}\\) ). Now, realizing that the density, \\(\\rho\\), has a value of \\(2.32 \\mathrm{~g} / \\mathrm{cm}^{3}{ }_{\\text {, }}\\), the value of \\(n^{\\prime}\\) is determined from Equation as follows\"\n\\[\n\\begin{aligned}\n\\mathrm{r}^{n} & =\\frac{ \\rho V_{C} N_{\\mathrm{A}}}{\\sum A_{\\mathrm{Si}}+\\sum \\mathrm{A}_{\\mathrm{O}}} \\\\\n& =\\frac{\\left(2.32 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(7.00 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}\\left(6.022 \\times 10^{23} \\text { formula units/mol }\\right)}{28.09 \\mathrm{~g} / \\mathrm{mol}+32.00 \\mathrm{~g} / \\mathrm{mol}}\n\\end{aligned}\n\\]\\[\n = 7.97 \\text { or almost } 8\n\\]\nTherefore, there are \\(8 \\mathrm{Si}^{4+}\\) ions and \\(16 \\mathrm{O}^{2-}\\) ions per \\(\\mathrm{SiO}_{2}\\) unit cell."
},
{
"idx": 654,
"question": "Given a hypothetical AX type of ceramic material with a density of 2.10 g/cm³, a cubic unit cell edge length of 0.57 nm, and atomic weights of A and X elements as 28.5 g/mol and 30.0 g/mol respectively, calculate the number of formula units per unit cell (n').",
"answer": "The number of formula units per unit cell (n') is calculated as follows: n' = (ρ * V_C * N_A) / (ΣA_C + ΣA_A) = (2.10 g/cm³ * (5.70 × 10⁻⁸ cm)³/unit cell * 6.022 × 10²³ formula units/mol) / (30.0 g/mol + 28.5 g/mol) = 4.00 formula units/unit cell."
},
{
"idx": 654,
"question": "Based on the calculated number of formula units per unit cell (n' = 4.00), which of the following crystal structures are possible for the AX type of ceramic material: sodium chloride, cesium chloride, or zinc blende?",
"answer": "For n' = 4.00 formula units/unit cell, the possible crystal structures are sodium chloride and zinc blende, as cesium chloride has n' = 1."
},
{
"idx": 655,
"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 r_Fe2+/r_O2- = 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": 655,
"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 r_Ti4+/r_O2- = 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": 655,
"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": 655,
"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."
},
{
"idx": 656,
"question": "The interplanar spacing \\(d_{\\text {hkl }}\\) for planes in a unit cell having orthorhombic geometry is \\\\ given by \\\\ \\(\\frac{1}{a_{h k l}^{2}}=\\frac{h^{2}}{a^{2}}+\\frac{k^{2}}{b^{2}}+\\frac{l^{2}}{c^{2}}\\) \\\\ \\\\ where \\(a, b\\), and \\(c\\) are the lattice parameters. \\\\ (a) To what equation does this expression reduce for crystals having cubic symmetry? \\\\ (b) For crystals having tetragonal symmetry? \\\\",
"answer": " (a) For the crystals having cubic symmetry, \\(a=b=c\\). Making this substitution into the \\\\ above 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}}\\) \\\\ \\\\ (b) For crystals having tetragonal symmetry, \\(a=b \\neq c\\). Replacing \\(b\\) with \\(a\\) in the \\\\ equation found in the problem statement leads to \\\\ \\(\\frac{1}{a_{h k l}^{2}}=\\frac{b^{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": 657,
"question": "The number-average molecular weight of a polystyrene is \\(500,000 \\mathrm{~g} / \\mathrm{mol}\\). Compute the degree of polymerization.\n\\title{",
"answer": "We are asked to compute the degree of polymerization for polystyrene, given that the number-average molecular weight is \\(500,000 \\mathrm{~g} / \\mathrm{mol}\\). The repeat unit for polystyrene has eight carbon atoms and eight hydrogen atoms. Therefore, the repeat unit molecular weight of polystyrene is just\n\\[\n\\begin{array}{c}\nm=8\\left(A_{\\mathrm{C}}\\right)+8\\left(A_{\\mathrm{H}}\\right) \\\\\n=(8)(12.01 \\mathrm{~g} / \\mathrm{mol})+(8)(1.008 \\mathrm{~g} / \\mathrm{mol})=104.14 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nNow it is possible to compute the degree of polymerization using Equation as follows:\n\\[\nD_{\\bar{P}}=\\frac{\\bar{M}_{n}}{m}=\\frac{500,000 \\mathrm{~g} / \\mathrm{mol}}{104.14 \\mathrm{~g} / \\mathrm{mol}}=4800\n\\]"
},
{
"idx": 658,
"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": "For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for 8% Cl substitution of all original hydrogen atoms. Consider 50 carbon atoms; for polyethylene there are 100 possible side-bonding sites. Ninety-two are occupied by hydrogen and eight are occupied by Cl. Thus, the mass of these 50 carbon atoms, m_C, is just m_C=50(A_C)=(50)(12.01 g/mol)=600.5 g. Likewise, for hydrogen and chlorine, m_H=92(A_H)=(92)(1.008 g/mol)=92.74 g, m_Cl=8(A_Cl)=(8)(35.45 g/mol)=283.60 g. Thus, the concentration of chlorine, C_Cl, is determined as follows: C_Cl=(m_Cl)/(m_C+m_H+m_Cl)×100=(283.60 g)/(600.5 g+92.74 g+283.60 g)×100=29.0 wt%."
},
{
"idx": 658,
"question": "In what ways does 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": 659,
"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": 659,
"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": 660,
"question": "The number-average molecular weight of a poly(acrylonitrile-butadiene) alternating copolymer is \\(1,000,000 \\mathrm{~g} / \\mathrm{mol}\\); determine the average number of acrylonitrile and butadiene repeat units per molecule.",
"answer": "For a poly(acrylonitrile-butadiene) alternating copolymer with a number-average molecular weight of \\(1,000,000 \\mathrm{~g} / \\mathrmmol}\\), we are asked to determine the average number of acrylonitrile and butadiene repeat units per molecule.\nSince it is an alternating copolymer, the number of both types of repeat units will be the same. Therefore, consider them as a single repeat unit, and determine the number-average degree of polymerization. For the acrylonitrile repeat unit, there are three carbon atoms, three hydrogen atoms, and one nitrogen atom, while the butadiene repeat consists of four carbon atoms and six hydrogen atoms. Therefore, the acrylonitrile-butadiene combined repeat unit weight is just\n\\[\n\\begin{aligned}\nm & =7\\left(A_{\\mathrm{C}}\\right)+9\\left(A_{\\mathrm{H}}\\right)+1\\left(A_{\\mathrm{N}}\\right) \\\\\n& =(7)(12.01 \\mathrm{~g} / \\mathrm{mol})+(9)(1.008 \\mathrm{~g} / \\mathrm{mol})+(14.01 \\mathrm{~g} / \\mathrm{mol})=107.15 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]\nFrom Equation, the degree of polymerization is just\n\\[\nD P=\\frac{\\bar{M}_{n}}{m}=\\frac{1,000,000 \\mathrm{~g} / \\mathrm{mol}}{107.15 \\mathrm{~g} / \\mathrm{mol}}=9333\n\\]\nThus, there is an average of 9333 of both repeat unit types per molecule."
},
{
"idx": 661,
"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": "This problem asks for us to calculate the number-average molecular weight of a random poly(isobutylene-isoprene) copolyner. The isobutylene repeat unit consists of four carbon atoms and eight hydrogen atoms . Therefore, its repeat unit molecular weight \\(\\left(m_{\\mathrm{Ib}}\\right)\\) is\n\\[\n\\begin{aligned}\nm_{\\mathrm{Ib}} & =4\\left(A_{\\mathrm{C}}\\right)+8\\left(A_{\\mathrm{H}}\\right) \\\\\n& =(4)(12.01 \\mathrm{~g} / \\mathrm{mol})+(8)(1.008 \\mathrm{~g} / \\mathrm{mol})=56.10 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]\nThe isoprene repeat unit is composed of five carbon and eight hydrogen atoms . Thus, its repeat unit molecular weight \\(\\left(m_{\\mathrm{Ip}}\\right)\\) is\n\\[\n\\begin{aligned}\nm_{\\mathrm{IP}} & =5\\left(A_{\\mathrm{C}}\\right)+8\\left(A_{\\mathrm H}\\right) \\\\\n& =(5)(12.01 \\mathrm{~g} / \\mathrm{mol})+(9)(1.008 \\mathrm{~g} / \\mathrm{mol})=68.11 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]\nFrom Equation, the average repeat unit molecular weight is just\n\\[\n\\begin{aligned}\n\\bar{m} & =f_{\\mathrm{Ib}} m_{\\mathrm{Ib}}+f_{\\mathrm{Ib}} m_{\\mathrm{Ib}} \\\\\n& =(0.25)(56.10 \\mathrm{~g} / \\mathrm{mol})+(1-0.25)(68.11 \\mathrm{~g} / \\mathrm{mol})=65.11 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}%\n\\]\nSince \\(D P=1500\\) (as given in the problem statement), \\(\\widetilde{M}_{h}\\) may be computed using a rearranged form of Equation as follows:\\[\n\\bar{M}_{n}=\\bar{m}(DP)=(65.11\\mathrm{\\ g/mol})(1500)=97,700\\mathrm{\\ g/mol}\n\\]"
},
{
"idx": 662,
"question": "An alternating copolymer is known to have a number-average molecular weight of \\(100,000 \\mathrm{~g} / \\mathrm{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": "For an alternating copolymer that has a number-average molecular weight of 100,000 \\(\\mathrm{g} / \\mathrm{mol}\\) and a degree of polymerization of 22 10 , we are to determine one of the repeat unit types if the other type is ethylene. It is first necessary to calculate \\(\\bar{m}\\) using a rearranged form of Equation as follows:\n\\[\n\\bar{m}=\\frac{\\bar{M}_{n}}{D P}=\\frac{100,000 \\mathrm{~g} / \\mathrm{mol}}{2210}=42.25 \\mathrm{~g} / \\mathrm{mol}\n\\]\nSince this is an alternating copolymer we know that chain fraction of each repeat unit type is 0.5 ; that is \\(f_{e}=f_{x}=0.5, f_{e}\\) and \\(f_{x}\\) being, respectively, the chain fractions of the ethylene and unknown repeat units. Also, the repeat unit for ethylene contains 2 carbon atoms and 4 hydrogen atoms; hence, the molecular weight for the ethylene repeat unit is\n\\[\n\\begin{aligned}\nm_{e} & =2\\left(A_{C}\\right)+4\\left(A_{\\mathrm{H}}\\right) \\\\\n& =2(12.01 \\mathrm{~g} / \\mathrm{mol})+4(1.008 \\mathrm{~g} / \\mathrm{mol})=28.05 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]\nThe modified form of Equation for this problem reads as follows:\n\\[\n\\bar{m}=f_{e} m_{e}+f_{x} m_{x}\n\\]\nin which \\(m_{x}\\) represents the repeat unit weight of the unknown repeat unit type. Rearranging this expression such that \\(m_{x}\\) is the dependent variable leads to the following:\\[\nm_{x}=\\frac{\\bar{m}-f_{e}m_{e}}{f_{x}}\n\\]\nFrom above we know that\n\\[\n\\begin{array}{l}\nf_{e}=f_{x}=0.5 \\\\\nm_{e}=28.05 \\mathrm{~g} / \\mathrm{mol} \\\\\n\\bar{m}=42.25 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nInsertion of these values into the above equation leads to the following value for \\(m_{X}\\) :\n\\[\n\\begin{array}{l}\nm_{x}=\\frac{\\bar{m}-f_{e} m_{e}}{f_{x}} \\\\\n\\quad=\\frac{45.25 \\mathrm{~g} / \\mathrm{mol}-\\left(0.5\\right)\\left(28.05 \\mathrm{~g} / \\mathrm{mol}\\right)}{0.5}=62.45 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\n\nFinally, it is necessary to calculate the repeat unit molecular weights for each of the possible other repeat unit types. These are calculated below:\n\n\\[\n\\begin{array}{l}\nm_{\\text {styrene }}=8\\left(A_{\\mathrm{C}}\\right)+8\\left(A_{\\mathrm{H}}\\right)=8(12.01 \\mathrm{~g} / \\mathrm{mol})+8(1.008 \\mathrm{~g} / \\mathrm{mol})=104.16 \\mathrm{~g} / \\mathrm{mol} \\\\\nm_{\\text {propylene }}=3\\left(A_{\\mathrm{C}}\\right)+6\\left(A_{\\mathrm{H}}\\right)=3(12.01 \\mathrm{~g} / \\mathrm{mol})+6(1.008 \\mathrm{~g} / \\mathrm{mol})=42.08 \\mathrm{~g} / \\mathrm{mol} \\\\\nm_{\\text {TFE }}=2\\left(A_{\\mathrm{C}}\\right)+4\\left(A_{\\mathrm{F}}\\right)=2(12.01 \\mathrm{~g} / \\mathrm{mol})+4(19.00 \\mathrm{~g} / \\mathrm{mol})=100.02 \\mathrm{~g} / \\mathrm{mol} \\\\\nm_{\\mathrm{VC}}=2\\left(A_{\\mathrm{C}}\\right)+3\\left(A_{\\mathrm{H}}\\right)+\\left(A_{\\mathrm{C}}\\right)=2(12.01 \\mathrm{~g} / \\textrm{mol})+3(1.008 \\mathrm{~g} / \\mathrm{mol})+35.45 \\mathrm{~g} / \\mathrm{mol}=62.49\n\\end{array}\n\\]\n\nTherefore, vinyl chloride is the other repeat unit type since its \\(m\\) value is almost the same as the calculated \\(m_{X}\\)."
},
{
"idx": 663,
"question": "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": "To determine the ratio of butadiene to acrylonitrile repeat units, first calculate the average repeat unit molecular weight of the copolymer, m_bar, using the formula: m_bar = M_n / DP = 250,000 g/mol / 4640 = 53.88 g/mol. Designate f_b as the chain fraction of butadiene repeat units, so the chain fraction of acrylonitrile repeat units f_a is 1 - f_b. The repeat unit molecular weights are: m_b = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol for butadiene and m_a = 3(12.01 g/mol) + 3(1.008 g/mol) + 14.01 g/mol = 53.06 g/mol for acrylonitrile. Solve for f_b using the equation: f_b = (m_bar - m_a) / (m_b - m_a) = (53.88 g/mol - 53.06 g/mol) / (54.09 g/mol - 53.06 g/mol) = 0.80. Then, f_a = 1 - 0.80 = 0.20. The ratio is f_b / f_a = 0.80 / 0.20 = 4.0."
},
{
"idx": 663,
"question": "Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?",
"answer": "The possible copolymers are random, graft, and block. The alternating copolymer is excluded because it requires a 1:1 ratio of repeat unit types, which is not the case here (the ratio is 4:1)."
},
{
"idx": 664,
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the molecular weight of the ethylene repeat unit (C2H4).",
"answer": "The molecular weight of ethylene (C2H4) is calculated as follows: m(ethylene) = 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol."
},
{
"idx": 664,
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the molecular weight of the propylene repeat unit (C3H6).",
"answer": "The molecular weight of propylene (C3H6) is calculated as follows: m(propylene) = 3(12.01 g/mol) + 6(1.008 g/mol) = 42.08 g/mol."
},
{
"idx": 664,
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the number of moles of ethylene repeat units in 100 g of the material.",
"answer": "The number of moles of ethylene in 100 g of the material is calculated as follows: n_m(ethylene) = 35 g / 28.05 g/mol = 1.25 mol."
},
{
"idx": 664,
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the number of moles of propylene repeat units in 100 g of the material.",
"answer": "The number of moles of propylene in 100 g of the material is calculated as follows: n_m(propylene) = 65 g / 42.08 g/mol = 1.54 mol."
},
{
"idx": 664,
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the fraction of ethylene repeat units.",
"answer": "The fraction of ethylene repeat units is calculated as follows: f(ethylene) = 1.25 mol / (1.25 mol + 1.54 mol) = 0.45."
},
{
"idx": 664,
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the fraction of propylene repeat units.",
"answer": "The fraction of propylene repeat units is calculated as follows: f(propylene) = 1.54 mol / (1.25 mol + 1.54 mol) = 0.55."
},
{
"idx": 665,
"question": "For the pair of polymers: linear and atactic poly(vinyl chloride); linear and isotactic polypropylene, (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.",
"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": 665,
"question": "For the pair of polymers: linear and syndiotactic polypropylene; crosslinked cis-polyisoprene, (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.",
"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": 665,
"question": "For the pair of polymers: network phenol-formaldehyde; linear and isotactic polystyrene, (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.",
"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": 665,
"question": "For the pair of polymers: block poly(acrylonitrile-isoprene) copolymer; graft poly(chloroprene-isobutylene) copolymer, (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.",
"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": 666,
"question": "For some hypothetical metal, the equilibrium number of vacancies at \\(900^{\\circ} \\mathrm{C}\\) is \\(2.3 \\times 10^{25}\\) \\(\\mathrm{m}^{-3}\\). If the density and atomic weight of this metal are \\(7.40 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and \\(85.5 \\mathrm{~g} / \\mathrm{mol}\\), respectively, calculate the fraction of vacancies for this metal at \\(900^{\\circ} \\mathrm{C}\\).",
"answer": "This problem is solved using two steps: (1) calculate the total number of lattice sites in silver, \\(N\\), using Equation, and (2) take the ratio of the equilibrium number of vacancies given in the problem statement \\(\\left(N_{v}=2.3 \\times 10^{25} \\mathrm{~m}^{-3}\\right)\\) and this value of \\(N\\). From Equation\n\\[\n\\begin{array}{c}\nN=\\frac{N_{\\mathrm{A}} \\rho}{A} \\\\\n=\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(7.40 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(10^{6} \\mathrm{~cm}^{3} / \\mathrm{m}^{3}\\right)}{85.85 \\mathrm{~g} / \\mathrm{mol}} \\\\\n=5.21 \\times 10^{28} \\mathrm{atoms} / \\mathrm{m}^{3}\n\\end{array}\n\\]\nThe fraction of vacancies is equal to the \\(N_{\\mathrm{v}} / N\\) ratio, which is computed as follows:\n\\[\n\\begin{aligned}\n\\frac{N_{\\mathrm{v}}}{N} & =\\frac{2.3 \\times 10^{25} \\mathrm{~m}^{-2}}{5.21 \\times 10^{28} \\mathrm{~a} \\mathrm{~m}^{3}} \\\\\n& =4.41 \\times 10^{-4}\n\\end{aligned}\n\\]"
},
{
"idx": 667,
"question": "Calculate the fraction of atom sites that are vacant for copper (Cu) at its melting temperature of 1084 degrees Celsius (1357 K). Assume an energy for vacancy formation of 0.90 eV/atom.",
"answer": "To compute the fraction of atom sites that are vacant in copper at 1357 K, we use the equation: NV/N = exp(-QV/kT). Given QV = 0.90 eV/atom, the calculation is as follows: NV/N = exp[-0.90 eV/atom / (8.62 × 10^-5 eV/atom-K × 1357 K)] = 4.56 × 10^-4."
},
{
"idx": 667,
"question": "Calculate the fraction of atom sites that are vacant for copper (Cu) at room temperature (298 K). Assume an energy for vacancy formation of 0.90 eV/atom.",
"answer": "To compute the fraction of atom sites that are vacant in copper at 298 K, we use the equation: NV/N = exp(-QV/kT). Given QV = 0.90 eV/atom, the calculation is as follows: NV/N = exp[-0.90 eV/atom / (8.62 × 10^-5 eV/atom-K × 298 K)] = 6.08 × 10^-16."
},
{
"idx": 667,
"question": "What is the ratio of NV/N(1357 K) and NV/N(298 K) for copper (Cu)?",
"answer": "The ratio of NV/N(1357 K) and NV/N(298 K) is calculated as follows: (4.56 × 10^-4) / (6.08 × 10^-16) = 7.5 × 10^11."
},
{
"idx": 668,
"question": "Calculate the number of vacancies per cubic meter in gold \\((\\mathrm{Au})\\) at \\(900^{\\circ} \\mathrm{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} / \\mathrm{cm}^{3}\\) (at \\(900^{\\circ} \\mathrm{C}\\) ) and \\(196.9 \\mathrm{~g} / \\mathrm{mol}\\), respectively.",
"answer": "Determination of the number of vacancies per cubic meter in gold at \\(900^{\\circ} \\mathrm{C}(1173 \\mathrm{~K})\\) requires the utilization of Equations as follows:\n\\[\nN_{\\mathrm{v}}=N \\exp \\left(-\\frac{Q_{\\mathrm{v}}}{k T}\\right)=-\\frac{N_{\\mathrm{A}} \\rho_{\\mathrm{Au}}}{A_{\\mathrm{Au}}} \\exp \\left(-\\frac{Q_{\\mathrm{v}}}{k T}\\middle)\n\\]\nInserting into this expression he density and atomic weight values for gold leads to the following:\n\\[\nN_{\\mathrm{v}}=\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(18.63 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)}{196.9 \\mathrm{~g} / \\mathrm{mol}} \\exp \\left[-\\frac{0.98 \\mathrm{eV} / \\mathrm{atom}}{\\left(8.62 \\times 10^{-5} \\mathrm{eV} / \\mathrm{atom} \\cdot \\mathrm{K}\\right)(1173 \\mathrm{~K})}\\right]\n\\]\n\\[\n\\begin{array}{l}\n=3.52 \\times 10^{18} \\mathrm{~cm}^{-3}=(3.52 \\times 10^{18} \\mathrm{~cm} ^{-3})(10^{6} \\mathrm{~m}^{-3} / \\mathrm{cm}^{-3})=3.52 \\times 10^{24} \\mathrm{~m}^{-3}\n\\end{array}\n\\]"
},
{
"idx": 669,
"question": "Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at its melting temperature \\(\\left(645^{\\circ} \\mathrm{C}\\right)\\). Assume an energy for defect formation of \\(1.86 \\mathrm{eV}\\).",
"answer": "In order to solve this problem it is necessary to use Equation and solve for the \\(N_{S} / N\\) ratio. Rearrangement of Equation and substituting values for temperature and the energy of defect formation \\(\\left(Q_{s}\\right)\\) given in the problem statement leads to\n\\[\n\\begin{array}{l}\n\\frac{N_{s}}{N}=\\exp \\left(-\\frac{Q_{s}}{2 k T}\\right) \\\\\n\\quad =\\exp \\left[-\\frac{1.86 \\mathrm{eV}}{(2)\\left(8.62 \\times 10^{-5} \\mathrm{eV} / \\mathrm{K}\\right)(645+273 \\mathrm{~K})}\\right] \\\\\n\\quad =7.87 \\times 10^{-6}\n\\end{array}\n\\]"
},
{
"idx": 670,
"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": 671,
"question": "If cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. Under these conditions, 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": 671,
"question": "If 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": 671,
"question": "If 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": 672,
"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": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystal structures of the two compounds must be the same or similar. 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, Δr%, is calculated as follows: Δr% = (0.077 nm - 0.072 nm) / 0.077 nm × 100 = 6.5%, which is within the acceptable range for a high degree of solubility. Additionally, the crystal structures of both MgO and FeO are rock salt. Therefore, FeO and MgO are expected to form high degrees of solid solubility, and experimentally, these two ceramics exhibit 100% solubility."
},
{
"idx": 672,
"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, Δr%, is calculated as follows: Δr% = (0.136 nm - 0.072 nm) / 0.136 nm × 100 = 47%. This Δr% value is much larger than the ±15% range, and therefore, BaO is not expected to experience any appreciable solubility in MgO. Experimentally, the solubility of BaO in MgO is very small; this is also the case for the solubility of MgO in BaO."
},
{
"idx": 672,
"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, Δr%, is calculated as follows: Δr% = (0.120 nm - 0.072 nm) / 0.120 nm × 100 = 40%. This Δr% value is much larger than the ±15% range, and therefore, PbO is not expected to experience any appreciable solubility in MgO."
},
{
"idx": 672,
"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, Δr%, is calculated as follows: Δr% = (0.072 nm - 0.072 nm) / 0.072 nm × 100 = 0%, which is within the acceptable range for a high degree (probably 100%) of solubility."
},
{
"idx": 673,
"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": 673,
"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": 674,
"question": "Which of these elements would you expect to form a substitutional solid solution having complete solubility with nickel? The Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni 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. The elements and their properties are given in the table.",
"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": 674,
"question": "Which of these elements would you expect to form a substitutional solid solution of incomplete solubility with nickel? The Hume-Rothery rules must be considered: (1) the difference in atomic radii between Ni 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. The elements and their properties are given in the table.",
"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": 674,
"question": "Which of these elements would you expect to form an interstitial solid solution with nickel? The elements and their atomic radii are given in the table.",
"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": 675,
"question": "What is the composition, in atom percent, of silver (Ag) in an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu, given the atomic weights of Ag (107.87 g/mol) and Cu (63.55 g/mol)?",
"answer": "The composition of silver in atom percent is calculated as follows: C_Ag = (C_Ag * A_Cu) / (C_Ag * A_Cu + C_Cu * A_Ag) * 100 = (92.5 * 63.55 g/mol) / (92.5 * 63.55 g/mol + 7.5 * 107.87 g/mol) * 100 = 87.9 at%."
},
{
"idx": 675,
"question": "What is the composition, in atom percent, of copper (Cu) in an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu, given the atomic weights of Ag (107.87 g/mol) and Cu (63.55 g/mol)?",
"answer": "The composition of copper in atom percent is calculated as follows: C_Cu = (C_Cu * A_Ag) / (C_Ag * A_Cu + C_Cu * A_Ag) * 100 = (7.5 * 107.87 g/mol) / (92.5 * 63.55 g/mol + 7.5 * 107.87 g/mol) * 100 = 12.1 at%."
},
{
"idx": 676,
"question": "What is the atomic percent of Pb in an alloy that consists of 5.5 wt% Pb and 94.5 wt% Sn, given the atomic weights of Pb (207.2 g/mol) and Sn (118.71 g/mol)?",
"answer": "The atomic percent of Pb is calculated as follows: C_Pb = (5.5 * 118.71) / (5.5 * 118.71 + 94.5 * 207.2) * 100 = 3.23 at%."
},
{
"idx": 676,
"question": "What is the atomic percent of Sn in an alloy that consists of 5.5 wt% Pb and 94.5 wt% Sn, given the atomic weights of Pb (207.2 g/mol) and Sn (118.71 g/mol)?",
"answer": "The atomic percent of Sn is calculated as follows: C_Sn = (94.5 * 207.2) / (5.5 * 118.71 + 94.5 * 207.2) * 100 = 96.77 at%."
},
{
"idx": 677,
"question": "What is the composition, in weight percent, of an alloy that consists of 5 at\\% Cu and 95 at\\% Pt?",
"answer": "In order to compute composition, in weight percent, of a 5 at\\% Cu-95 at\\% Pt alloy, we employ Equation given the atomic weights of copper and platinum (found on the inside of the book's cover):\n\\[\n\\begin{array}{l}\nA_{\\mathrm{Cu}}=63.55 \\mathrm{~g} / \\mathrm{mol} \\\\\nA_{\\mathrm{Pt}}=195.08 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nThese compositions in weight percent are determined as follows:\n\\[\n\\begin{array}{l}\nC_{\\mathrm{Cu}}=\\frac{C_{\\mathrm{Cu}}^{\\prime} A_{\\mathrm{Cu}}}{C_{\\mathrm{Cu}}^{\\prime} A_{\\mathrm{Cu}}+C_{\\mathrm{Pt}}^{\\prime} A_{\\mathrm{Pt}}} \\times 100 \\\\\n=\\frac{(5)(63.55 \\mathrm{~g} / \\mathrm{mol})}{(5)(63.55 \\mathrm{~g} / \\mathrm{mol})+(95)(195.08 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n=1.68 \\mathrm{wt} \\% \\\\\nC_{\\mathrm{Pt}}=\\frac{C_{\\mathrm{Pt}}^{\\prime} A_{\\mathrm{Pt}}}{C_{\\mathrm{Cu}}^{\\prime} A_{\\mathrm{Cu}^{\\prime}}+C_{\\mathrm{Pt}}^{\\prime} A_{\\mathrm{Pt}}} \\times 100 \\\\\n=\\frac{(95)(195.08 \\mathrm{~g} / \\mathrm{mol})}{(5)(63.55 \\mathrm{~g} / \\mathrmmol)+(95)(195.08 \\mathrm{~g} / \\mathrm{moI})} \\times 100 \\\\\n=98.32 \\mathrm{wt} \\%\n\\end{array}\n\\]"
},
{
"idx": 678,
"question": "Calculate the weight percent of iron in an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium.",
"answer": "The concentration of iron (C_Fe) is calculated as follows: C_Fe = (m_Fe / (m_Fe + m_C + m_Cr)) * 100 = (105 kg / (105 kg + 0.2 kg + 1.0 kg)) * 100 = 98.87 wt%."
},
{
"idx": 678,
"question": "Calculate the weight percent of carbon in an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium.",
"answer": "The concentration of carbon (C_C) is calculated as follows: C_C = (m_C / (m_Fe + m_C + m_Cr)) * 100 = (0.2 kg / (105 kg + 0.2 kg + 1.0 kg)) * 100 = 0.19 wt%."
},
{
"idx": 678,
"question": "Calculate the weight percent of chromium in an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium.",
"answer": "The concentration of chromium (C_Cr) is calculated as follows: C_Cr = (m_Cr / (m_Fe + m_C + m_Cr)) * 100 = (1.0 kg / (105 kg + 0.2 kg + 1.0 kg)) * 100 = 0.94 wt%."
},
{
"idx": 679,
"question": "What is the composition, in atom percent, of an alloy that contains \\(44.5 \\mathrm{lb} \\mathrm{m}\\) of \\(\\mathrm{Ag}, 83.7\\) \\(\\mathrm{lb}_{\\mathrm{m}}\\) of \\(\\mathrm{Au}\\), and \\(5.3 \\mathrm{lb}_{\\mathrm{m}}\\) of \\(\\mathrm{Cu}\\) ?",
"answer": "In this problem we are asked to determine the concentrations, in atom percent, of the Ag\\(\\mathrm{Au}-\\mathrm{Cu}\\) alloy. It is first necessary to convert the amounts of \\(\\mathrm{Ag}, \\mathrm{Au}\\), and \\(\\mathrm{Cu}\\) into grams, by multiplying the mass (in pound-mass) for each component by the conversion factor for poundmass to grams - i.e., \\(453.6 \\mathrm{~g} / \\mathrm{lb} \\mathrm{m}^{\\text {. }}\\). Therefore,\n\\[\n\\begin{array}{l}\nm_{\\mathrm{Ag}}^{\\prime}=(44.5 \\mathrm{lbm})(453.6 \\mathrm{~g} / \\mathrm{lbm})=20,185 \\mathrm{~g} \\\\\nm_{\\mathrm{Au}}^{\\prime}=(83.7 \\mathrm{lbm})(453.6 \\mathrm{~g} / \\mu \\mathrm{b} \\mathrm{m})=37,966 \\mathrm{~g} \\\\\nm_{\\mathrm{Cu}}^{\\prime}=(5.3 \\mathrm{lbm})(453.6 \\mathrm{~g} / \\boldsymbol{l b m})=2,404 \\mathrm{~g}\n\\end{array}\n\\]\nThese masses must next be converted into moles , using atomic weight values for \\(\\mathrm{Ag}, \\mathrm{Au}\\), and \\(\\mathrm{Cu}\\), found inside the front cover of the books. Thus,\n\\[\n\\begin{aligned}\nn_{m_{\\mathrm{Ag}}} & =\\frac{m_{\\mathrm{Ag}}^{\\prime}}{A_{\\mathrm{Ag}}}=\\frac{20,185 \\mathrm{~g}}{107.87 \\mathrm{~g} / \\mathrm{mol}}=187.1 \\mathrm{~mol} \\\\\nn_{m_{\\mathrm{Au}}}^{\\prime} & =\\frac{37,966 \\mathrm{~g}}{196.97 \\mathrm{~g} / \\mathrm{mol}}=192.8 \\mathrm{~mol} \\\\\nn_{m_{\\mathrm{Cu}}}^{\\prime} & =\\frac{2,404 \\mathrm{~g}}{63.55 \\mathrm{~g} / \\mathrm{mol}}=37.8 \\mathrm{~mol}\n\\end{aligned}\n\\]\nNow, employment of Equation, gives\\[\n\\begin{array}{l}\n\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\;C_{\\mathrm{Ag}}^{\\prime}\\;=\\;\\frac{n_{m_{\\mathrm{Ag}}}}{n_{m_{\\mathrm{Ag}}}+n_{m_{\\mathrm{Au}}}+n_{m_{\\mathrm{Cu}}}}\\; \\times\\;100\\\\\n\\\\=\\frac{187.1\\mathrm{mol}}{187.1\\mathrm{mol}+192.8\\mathrm{mol}+37.8\\mathrm{mol}}\\; \\times\\;100=44.8\\mathrm{at} \\%\\\\\n\\\\\n\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\;\\;C_{\\mathrm{Au}}^{\\prime}=\\frac{192.8\\mathrm{mol}}{187.1\\mathrm{mol}+191.2\\mathrm{~mol}+37.8\\mathrm{mol}}\\; \\times\\;1\\;00=46.2\\mathrm{at} \\%\\\\\n\\\\\n\\quad\\quad\\quad\\quad\\;\\;C_{\\mathrm{Cu}}^{\\prime}=\\frac{37.8\\mathrm{mol}}{187.1\\mathrm{mol}+{192.8\\mathrm{mol}+37.8\\mathrm{mol}}} \\times\\;100=9.0\\mathrm{at} \\%\n\\end{array}\n\\]"
},
{
"idx": 680,
"question": "Calculate the number of atoms per cubic meter in \\(\\mathrm{Pb}\\).",
"answer": "This problem calls for a determination of the number of atoms per cubic meter for lead. In order to solve this problem, one must employ Equation,\n\\[\nN=\\frac{N_{\\mathrm{A}} / \\rho_{\\mathrm{Pb}}}{A_{\\mathrm{Pb}}}\n\\]\nThe density of \\(\\mathrm{Pb}\\) is \\(11.35 \\mathrm{~g} / \\mathrm{cm}^{3}\\), while its atomic weight is \\(207.2 \\mathrm{~g} / \\mathrm{mol}\\). Thus,\n\\[\n\\begin{aligned}\nN= & \\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(11.35 \\mathrm{~g} / \\mathrm{cm}^{3)}\\right)}{207.2 \\mathrm{~g} / \\mathrm{mol}} \\\\\n& =3.30 \\times 10^{22} \\mathrm{atoms} / \\mathrm{cm}^{3}=3.30 \\times 10^{28} \\mathrm{atoms} / \\mathrm{m}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 681,
"question": "Calculate the number of atoms per cubic meter in chromium.",
"answer": "This problem calls for a determination of the number of atoms per cubic meter for chromium. In order to solve this problem, one must employ Equation,\n\\[\nN=\\frac{N_{\\mathrm{A}} \\rho_{\\mathrm{Cr}}}{A_{\\mathrm{Cr}}}\n\\]\nThe density of \\(\\mathrm{Cr}\\) is \\(7.19 \\mathrm{~g} / \\mathrm{cm}^{3}\\), while its atomic weight is \\(52.00 \\mathrm{~g} / \\mathrm{mol}\\). Thus,\n\\[\n\\begin{aligned}\nN= & \\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(7.19 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)}{52.00 \\mathrm{~g} / \\mathrm{mol}} \\\\\n= & 8.33 \\times 10^{22} \\mathrm{atoms} / \\mathrm{cm}^{3}=8.33 \\times 10^{28} \\mathrm{atoms} / \\mathrm{m}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 682,
"question": "The concentration of \\(\\mathrm{Si}\\) in an \\(\\mathrm{Fe}-\\mathrm{Si}\\) alloy is \\(0.25 \\mathrm{wt} \\%\\). What is the concentration in kilograms of Si per cubic meter of alloy?",
"answer": "In order to compute the concentration in \\(\\mathrm{kg} / \\mathrm{m}^{3}\\) of \\(\\mathrm{Si}\\) in a \\(0.25 \\mathrm{wt} \\% \\mathrm{Si}-99.75 \\mathrm{wt} \\% \\mathrm{Fe}\\) alloy we must employ Equation as\n\\[\nC_{\\mathrm{Si}}^{\\mathrm{s}}=\\frac{C_{\\mathrm{Si}}}{\\frac{C_{\\mathrm{Si}}}{\\rho_{\\mathrm{Si}}}+\\frac{C_{\\mathrm{Fe}}}{\\rho_{\\mathrm{Fe}}}} \\times 10^{3}\n\\]\nFrom inside the front cover, densities for silicon and iron are 2.33 and \\(7.87 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively; therefore\n\\[\n\\begin{aligned}\nC_{\\mathrm{Si}}^{\\mathrm{s}}= & \\frac{0.25}{\\frac{0.25}{2.33 \\mathrm{~g} / \\mathrm{cm}^{3}}+\\frac{99.75}{7.87 \\mathrm{~g} / \\mathrm{cm}^{3}}} \\times 10^{3} \\\\\n& =19.6 \\mathrm{~kg} / \\mathrm{m}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 683,
"question": "Determine the approximate density of a Ti-6Al-4V titanium alloy that has a composition of \\(90 \\mathrm{wt} \\% \\mathrm{Ti}, 6 \\mathrm{wt} \\% \\mathrm{Al}\\), and \\(4 \\mathrm{wt} \\% \\mathrm{~V}\\).",
"answer": "In order to solve this problem, Equation, which is modified to take the following form:\n\\[\n\\rho_{\\mathrm{ave}}=\\frac{100}{\\frac{C_{\\mathrm{Ti}}}{\\rho_{\\mathrm{Ti}}}+\\frac{C_{\\mathrm{Al}}}{\\rho_{\\mathrm{Al}}}+\\frac{C_{\\mathrm{V}}}{\\rho_{\\mathrm{V}}}}\n\\]\nAnd, using the density values for \\(\\mathrm{Ti}, \\mathrm{Al}\\), and \\(\\mathrm{V}\\)-i.e., \\(4.51 \\mathrm{~g} / \\mathrm{cm}^{3}, 2.71 \\mathrm{~g} / \\mathrm{cm}^{3}\\), and \\(6.10 \\mathrm{~g} / \\mathrm{cm}^{3}-\\) (as taken from inside the front cover of the text), the density is computed as follows:\n\\[\n\\begin{aligned}\n\\rho_{\\text {ave }} & =\\frac{100}{\\frac{90 \\mathrm{wt} \\%}{4.51 \\mathrm{~g} / \\mathrm{cm}^{3}}+\\frac{6 \\mathrm{wt} \\%}{2.71 \\mathrm{~g} / \\mathrm{cm}^{3}}+\\frac{4 \\mathrm{wt} \\%}{6.10 \\mathrm{~g} / \\mathrm{cm}^{3}}}\n\\end{aligned}\n\\]\n\\[\n\\begin{aligned}\n& =4.38 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 684,
"question": "What is the average atomic weight (A_ave) for an 80 wt% Ag-20 wt% Pd alloy, given the atomic weights of Ag (107.87 g/mol) and Pd (106.4 g/mol)?",
"answer": "The average atomic weight (A_ave) is calculated using the formula: A_ave = 100 / (C_Ag/A_Ag + C_Pd/A_Pd), where C_Ag = 80 wt%, C_Pd = 20 wt%, A_Ag = 107.87 g/mol, and A_Pd = 106.4 g/mol. Substituting these values gives A_ave = 100 / (80/107.87 + 20/106.4) = 107.44 g/mol."
},
{
"idx": 684,
"question": "What is the average density (ρ_ave) for an 80 wt% Ag-20 wt% Pd alloy, given the densities of Ag (10.49 g/cm³) and Pd (12.02 g/cm³)?",
"answer": "The average density (ρ_ave) is calculated using the formula: ρ_ave = 100 / (C_Ag/ρ_Ag + C_Pd/ρ_Pd), where C_Ag = 80 wt%, C_Pd = 20 wt%, ρ_Ag = 10.49 g/cm³, and ρ_Pd = 12.02 g/cm³. Substituting these values gives ρ_ave = 100 / (80/10.49 + 20/12.02) = 10.76 g/cm³."
},
{
"idx": 684,
"question": "Calculate the unit cell edge length (a) for an FCC 80 wt% Ag-20 wt% Pd alloy, given the average atomic weight (A_ave = 107.44 g/mol), average density (ρ_ave = 10.76 g/cm³), and Avogadro's number (N_A = 6.022 × 10²³ atoms/mol).",
"answer": "The unit cell edge length (a) for an FCC structure is calculated using the formula: a = ( (n × A_ave) / (ρ_ave × N_A) )^(1/3), where n = 4 atoms/unit cell for FCC, A_ave = 107.44 g/mol, ρ_ave = 10.76 g/cm³, and N_A = 6.022 × 10²³ atoms/mol. Substituting these values gives a = ( (4 × 107.44) / (10.76 × 6.022 × 10²³) )^(1/3) = 4.050 × 10⁻⁸ cm = 0.4050 nm."
},
{
"idx": 685,
"question": "Given an alloy composed of 25 wt% of metal A and 75 wt% of metal B, with densities of 6.17 g/cm³ and 8.00 g/cm³, and atomic weights of 171.3 g/mol and 162.0 g/mol, respectively, calculate the average density of the alloy.",
"answer": "The average density (ρ_ave) of the alloy is calculated using the formula for the reciprocal of the average density based on weight percentages: ρ_ave = 100 / ( (C_A / ρ_A) + (C_B / ρ_B) ), where C_A = 25 wt%, C_B = 75 wt%, ρ_A = 6.17 g/cm³, and ρ_B = 8.00 g/cm³. Substituting these values gives: ρ_ave = 100 / ( (25 / 6.17) + (75 / 8.00) ) = 7.33 g/cm³."
},
{
"idx": 685,
"question": "Given the same alloy composition, calculate the average atomic weight of the alloy.",
"answer": "The average atomic weight (A_ave) of the alloy is calculated using the formula: A_ave = 100 / ( (C_A / A_A) + (C_B / A_B) ), where C_A = 25 wt%, C_B = 75 wt%, A_A = 171.3 g/mol, and A_B = 162.0 g/mol. Substituting these values gives: A_ave = 100 / ( (25 / 171.3) + (75 / 162.0) ) = 164.1 g/mol."
},
{
"idx": 685,
"question": "Using the average density (7.33 g/cm³), average atomic weight (164.1 g/mol), and given unit cell edge length (0.332 nm), determine the number of atoms per unit cell (n) for the alloy.",
"answer": "The number of atoms per unit cell (n) is calculated using the formula: n = (ρ_ave * a³ * N_A) / A_ave, where ρ_ave = 7.33 g/cm³, a = 0.332 nm = 3.32 × 10⁻⁸ cm, N_A = 6.022 × 10²³ atoms/mol, and A_ave = 164.1 g/mol. Substituting these values gives: n = (7.33 * (3.32 × 10⁻⁸)³ * 6.022 × 10²³) / 164.1 = 1.00 atom/unit cell."
},
{
"idx": 685,
"question": "Based on the calculated number of atoms per unit cell (n = 1.00), determine the crystal structure of the alloy (simple cubic, face-centered cubic, or body-centered cubic).",
"answer": "For n = 1.00 atom/unit cell, the crystal structure is simple cubic."
},
{
"idx": 686,
"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 \\mathrm{wt} \\%\\) Mo and \\(83.6 \\mathrm{wt} \\%\\) W. The densities of pure molybdenum and tungsten are 10.22 and \\(19.30 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively.",
"answer": "This problem asks us to determine the number of molybdenum atoms per cubic centimeter for a \\(16.4 \\mathrm{wt} \\% \\mathrm{Mo}-83.6 \\mathrm{wt} \\% \\mathrm{~W}\\) solid solution. To solve this problem, employment of Equation is necessary, using the following values:\n\\[\n\\begin{array}{l}\nC_{1}=C_{\\mathrm{Mo}}=16.4 \\mathrm{wt} \\% \\\\\n\\rho_{1}=\\rho_{\\mathrm{Mo}}=10.22 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n\\rho_{2}=\\rho_{\\mathrm{W}}=19.30 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\nA_{1}=A_{\\mathrm{Mo}}=95.94 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nThus\n\\[\n\\begin{aligned}\nN_{\\mathrm{Mo}} & =\\frac{N_{\\mathrm{A}} C_{\\mathrm{Mo}}}{\\frac{C_{\\mathrm{Mo}} A_{\\mathrm{Mo}}}{\\rho_{\\mathrm{Mo}}}+\\frac{A_{\\mathrm{Mo}}}{\\rho_{\\mathrm{W}}}\\left(100-C_{\\mathrm{Mo}}\\right)} \\\\\n& =\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)(16.4 \\mathrm{wt} \\%)}{(16.4 \\mathrm{wt} \\%)(95.94 \\mathrm{~g} / \\mathrm{mol})+95.94 \\mathrm{~g} / \\mathrm{mol}}(100-16.4 \\mathrm{wt} \\%)} \\\\\n& =1.73 \\times 10^{22} \\mathrm{atoms} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]"
},
{
"idx": 687,
"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 \\mathrm{wt} \\% \\mathrm{Nb}\\) and \\(76 \\mathrm{wt} \\% \\mathrm{~V}\\). The densities of pure niobium and vanadium are 8.57 and \\(6.10 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively.",
"answer": "This problem asks us to determine the number of niobium atoms per cubic centimeter for a \\(24 \\mathrm{wt} \\% \\mathrm{Nb}-76 \\mathrm{wt} \\% \\mathrm{~V}\\) solid solution. To solve this problem, employment of Equation is necessary, using the following values:\n\\[\n\\begin{array}{l}\nC_{1}=C_{\\mathrm{Nb}}=24 \\mathrm{wt} \\% \\\\\n\\rho_{1}=\\rho_{\\mathrm{Nb}}=8.57 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n\\rho_{2}=\\rho_{\\mathrm{V}}=6.10 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\nA_{1}=A_{\\mathrm{Nb}}=92.91 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nThus\n\\[\n\\begin{array}{l}\nN_{\\mathrm{Nb}}=\\frac{N_{\\mathrm{A}} C_{\\mathrm{Nb}}}{\\frac{C_{\\mathrm{Nb}} A_{\\mathrm{Nb}}}{\\rho_{\\mathrm{Nb}}}+\\frac{A_{\\mathrm{Nb}}}{\\rho_{\\mathrm{V}}}\\left(100-C_{\\mathrm{Nb}}\\right)} \\\\\n=\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)(24 \\mathrm{wt} \\%)}{\\left(24 \\mathrm{wt} \\%\\right)\\left(92.91 \\mathrm{~g} / \\mathrm{mol}\\right)+\\frac{92.91 \\mathrm{~g} / \\mathrm{mol}}{6.10 \\mathrm{~g} / \\mathrm{cm}^{3}}(100-24 \\mathrm{wt} \\%)} \\\\\n=1.02 \\times 10^{22} \\mathrm{atoms} / \\mathrm{cm}^{3}\n\\end{array}\n\\]"
},
{
"idx": 688,
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the number of carbon atoms per cubic centimeter of alloy.",
"answer": "To compute the number of carbon atoms per cubic centimeter of alloy, we use the following parameters: A_C = 12.011 g/mol, C_C = 0.1 wt%, ρ_C = 2.25 g/cm³, ρ_Fe = 7.87 g/cm³. The calculation is as follows: N_C = (6.022 × 10²³ atoms/mol)(0.1 wt%) / [(0.1 wt%)(12.011 g/mol) / 2.25 g/cm³ + (12.011 g/mol / 7.87 g/cm³)(100 - 0.1 wt%)] = 3.94 × 10²⁰ C atoms/cm³."
},
{
"idx": 688,
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the number of iron atoms per cubic centimeter of alloy.",
"answer": "To compute the number of iron atoms per cubic centimeter of alloy, we use the following parameters: A_Fe = 55.85 g/mol, C_Fe = 99.9 wt%, ρ_Fe = 7.87 g/cm³, ρ_C = 2.25 g/cm³. The calculation is as follows: N_Fe = (6.022 × 10²³ atoms/mol)(99.9 wt%) / [(99.9 wt%)(55.85 g/mol) / 7.87 g/cm³ + (55.85 g/mol / 2.25 g/cm³)(100 - 99.9 wt%)] = 8.46 × 10²² Fe atoms/cm³."
},
{
"idx": 688,
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the number of unit cells per cubic centimeter of alloy.",
"answer": "From Section 3.4, there are two Fe atoms associated with each unit cell. Therefore, the number of unit cells per cubic centimeter is: n_i = (8.46 × 10²² atoms/cm³) / (2 atoms/unit cell) = 4.23 × 10²² unit cells/cm³."
},
{
"idx": 688,
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the fraction of unit cells that contain carbon atoms.",
"answer": "The fraction of unit cells that contain carbon atoms, n_f, is the number of carbon atoms per cubic centimeter divided by the number of unit cells per cubic centimeter: n_f = (3.94 × 10²⁰ C atoms/cm³) / (4.23 × 10²² unit cells/cm³) = 9.31 × 10⁻³ atoms/unit cell. The reciprocal of n_f is 107.5 unit cells/atom, meaning there is one carbon atom per 107.5 unit cells."
},
{
"idx": 689,
"question": "What is the weight percent of gold (Au) that must be added to silver (Ag) to yield an alloy containing 5.5 × 10^21 Au atoms per cubic centimeter, given the densities of pure Au and Ag are 19.32 g/cm³ and 10.49 g/cm³, respectively, and the atomic weights of Au and Ag are 196.97 g/mol and 107.87 g/mol?",
"answer": "To compute the weight percent of gold (Au) in the alloy, use the following equation with the given values: N_Au = 5.5 × 10^21 atoms/cm³, ρ_Au = 19.32 g/cm³, ρ_Ag = 10.49 g/cm³, A_Au = 196.97 g/mol, A_Ag = 107.87 g/mol, and Avogadro's number N_A = 6.022 × 10^23 atoms/mol. The calculation is as follows: C_Au = 100 / [1 + (N_A × ρ_Ag) / (N_Au × A_Au) - (ρ_Ag / ρ_Au)] = 100 / [1 + (6.022 × 10^23 × 10.49) / (5.5 × 10^21 × 196.97) - (10.49 / 19.32)] = 15.9 wt%."
},
{
"idx": 690,
"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 \\times\\) \\(10^{21}\\) Ge atoms per cubic centimeter. The densities of pure \\(\\mathrm{Ge}\\) and \\(\\mathrm{Si}\\) are 5.32 and \\(2.33 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively.",
"answer": "To solve this problem, employment of Equation is necessary, using the following values:\n\\[\n\\begin{array}{l}\nN_{1}=N_{\\mathrm{Ge}}=2.43 \\times 10^{21} \\mathrm{atoms} / \\mathrm{cm}^{3} \\\\\n\\rho_{1}=\\rho_{\\mathrm{Ge}}=5.32 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n\\rho_{2}=\\rho_{\\mathrm{Si}}=2.33 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\nA_{1}=A_{\\mathrm{Ge}}=72.64 \\mathrm{~g} / \\mathrm{mol} \\text { (found inside front cover) } \\\\\nA_{2}=A_{\\mathrm{Si}}=28.09 \\mathrm{~g} / \\mathrm{mol} \\text { (found inside front cove) }\n\\end{array}\n\\]\nThus\n\\[\n\\begin{aligned}\nC_{\\mathrm{Ge}} & =\\frac{100}{1+\\frac{N_{\\mathrm{A}} \\rho_{\\mathrm{Si}}}{N_{\\mathrm{Ge}} A_{\\mathrm{Ge}}}-\\frac{\\rho_{\\mathrm{Si}}}{\\rho_{\\mathrm{Ge}}}} \\\\\n& =\\frac{100}{1+\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(2.33 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)}{\\left(2.43 \\times 10^{21} \\mathrm{atoms} / c m^{3}\\right)\\left(72.64 \\mathrm{~g} / \\mathrm{mol}\\right)}-\\left(\\frac{2.33 \\mathrm{~g} / \\mathrm{cm}^{3}}{5.32 \\mathrm{~g} / \\mathrm{cm}^{3}}\\right)} \\\\\n& =11.7 \\mathrm{wt} \\%\n\\end{aligned}\n\\]"
},
{
"idx": 691,
"question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch at a magnification of 100x?",
"answer": "This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 6 at a magnification of 100x. All we need do is solve for the parameter n in Equation, inasmuch as G=6. Thus n = 2^(G-1) = 2^(6-1) = 32 grains/in.^2"
},
{
"idx": 691,
"question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch without any magnification?",
"answer": "Now it is necessary to compute the value of n for no magnification. In order to solve this problem it is necessary to use Equation: n_M (M/100)^2 = 2^(G-1) where n_M = the number of grains per square inch at magnification M, and G is the ASTM grain size number. Without any magnification, M in the above equation is 1, and therefore, n_1 (1/100)^2 = 2^(6-1) = 32. And, solving for n_1, n_1 = 320,000 grains/in.^2."
},
{
"idx": 692,
"question": "Determine the ASTM grain size number if 30 grains per square inch are measured at a magnification of \\(250 \\times\\).",
"answer": "In order to solve this problem we make use of Equation:\n\\[\nn_{M}\\left(\\frac{M}{100}\\right)^{2}=2^{G-1}\n\\]\nwhere \\(n_{M}=\\) the number of grains per square inch at magnification \\(M\\), and \\(G\\) is the ASTM grain size number. Solving the above equation for \\(G\\), and realizing that \\(n_{M}=30\\), while \\(M=250\\). Some algebraic manipulation of Equation is necessary. Taking logarithms of both sides of this equation leads to\n\\[\n\\log \\left[ n_{M}\\left(\\frac{M}{100}\\right)^{2}\\right]=\\log \\left(2^{G-1}\\right)\n\\]\nOr\n\\[\n\\log n_{M}+2\\log \\left( \\frac{M}{100}\\right)=(G-1)\\log 2\n\\]\nAnd solving this equation for \\(G\\)\n\\[\n\\frac{\\log n_{M}+2 \\log \\left( \\frac{M}{100}\\right)}{\\log 2}=G-1\n\\]\nor\n\\[\nG=\\frac{\\log n_{M}+2 \\log \\left( \\frac{M}{\\mathbf{1 0 0}}\\right)}{\\log 2}+1\n\\]\\[\n\\begin{array}{l}\nn_{M}=30\\\\\nM=250\\\\\n\\mbox{And, therefore}\n\\end{array}\n\\]\n\\[\nG=\\frac{\\log 30\\,+\\,2\\,\\log \\left(\\frac{250}{100}\\right)}{\\log 2}\\,+\\,1=8.6\n\\]"
},
{
"idx": 693,
"question": "Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of \\(75 \\times\\).",
"answer": "In order to solve this problem we make use of Equation :\n\\[\nn_{M}\\left(\\frac{M}{100}\\right)^{2}=2^{G-1}\n\\]\nwhere \\(n_{M}=\\) the number of grains per square inch at magnification \\(M\\), and \\(G\\) is the ASTM grain size number. Solving the above equation for \\(G\\), and realizing that \\(n_{M}=25\\), while \\(M=75\\). Some algebraic manipulation of Equation is necessary. Taking logarithms of both sides of this equation leads to\n\\[\n\\log \\left[ n_{M}\\left(\\frac{M}{100}\\right)^{2}\\right]=\\log \\left(2^{G-1}\\right)\n\\]\nOr\n\\[\n\\log n_{M}+2 \\log \\left( \\frac{M}{100} \\right)=(G-1) \\log 2\n\\]\nAnd solving this equation for \\(G\\)\n\\[\n\\frac{\\log n_{M}+2 \\log \\left(\\frac{M}{100}\\right)}{\\log 2}=G-1\n\\]\nor\n\\[\nG=\\frac{\\log n_{M}+2 \\log \\left(\\frac{M}{120}\\right)}{\\log 2}+1\n\\]\\[\n\\begin{array}{l}\nn_{M}=25\\\\\nM=75\\\\\n\\mbox{And, therefore}\n\\end{array}\n\\]\n\\[\nG=\\frac{\\log 25\\,+\\,2\\,\\log \\left(\\begin{array}{c}\n75\\\\\n100\n\\end{array}\\right)}{\\log 2}+1=4.8\n\\]"
},
{
"idx": 694,
"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": 695,
"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": 696,
"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": 696,
"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": 697,
"question": "A sheet of steel 5.0-mm thick has nitrogen atmospheres on both sides at \\(900^{\\circ} \\mathrm{C}\\) and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is \\(1.85 \\times 10^{-10} \\mathrm{~m}^{2} / \\mathrm{s}\\), and the diffusion flux is found to be \\(1.0 \\times 10^{-7}\\) \\(\\mathrm{kg} / \\mathrm{m}^{2} \\cdot \\mathrm{s}\\). Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is \\(2 \\mathrm{~kg} / \\mathrm{m}^{3}\\). How far into the sheet from this high-pressure side will the concentration be \\(0.5 \\mathrm{~kg} / \\mathrm{m}^{3}\\) ? Assume a linear concentration profile.",
"answer": "This problem is solved by using Equation in the form\n\\[\nJ=-D \\frac{C_{\\mathrm{A}}-C_{\\mathrm{B}}}{x_{\\mathrm{A}}-x_{\\mathrm{B}}}\n\\]\nIf we take \\(C_{\\mathrm{A}}\\) to be the point at which the concentration of nitrogen is \\(2 \\mathrm{~kg} / \\mathrm{m}^{3}\\), then it becomes necessary to solve the above equation for \\(x_{\\mathrm{B}}\\), as\n\\[\nx_{\\mathrm{B}}=x_{\\mathrm{A}}+D\\left[\\frac{C_{\\mathrm{A}}-C_{\\mathrm{B}}}{J}\\right]\n\\]\nAssume \\(x_{\\mathrm{A}}\\) is zero at the surface, in which case\n\\[\n\\begin{aligned}\nx_{\\mathrm{B}}=0+\\left(1.85 \\times 10^{-10} \\mathrm{~m} \\mathrm{~m}^{2} / \\mathrm{s}\\right) & {\\left[\\frac{2 \\mathrm{~kg} / \\mathrm{m}^{3}-0.5 \\mathrm{~kg} / \\mathrm{m}^{3}}{1.0 \\times 10^{-7} \\mathrm{~kg} / \\mathrm{m}^{2} \\cdot \\mathrm{s}}\\right] } \\\\\n& =2.78 \\times 10^{-3} \\mathrm{~m}=2.78 \\mathrm{~mm}\n\\end{aligned}\n\\]"
},
{
"idx": 698,
"question": "For a steel alloy it has been determined that a carburizing heat treatment of \\(15 \\mathrm{~h}\\) duration will raise the carbon concentration to \\(0.35 \\mathrm{wt} \\%\\) at a point \\(2.0 \\mathrm{~mm}\\) 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": "This problem calls for an estimate of the time necessary to achieve a carbon concentration of \\(0.35 \\mathrm{wt} \\%\\) at a point \\(6.0 \\mathrm{~mm}\\) from the surface. From Equation,\n\\[\n\\frac{x^{2}}{D t}=\\text { constant }\n\\]\nBut since the temperature is constant, so also is \\(D\\) constant, which means that\n\\[\n\\frac{x^{2}}{t}=\\text { constant }\n\\]\nor\n\\[\n\\frac{\\frac{x_{1}^{2}}{t}}{t_{1}}=\\frac{x_{2}^{2}}{t_{2}}\n\\]\nThus, if we assign \\(x_{1}=0.2 \\mathrm{~mm}, x_{2}=0.6 \\mathrm{~mm}\\), and \\(t_{1} 15 \\mathrm{~h}\\), then\n\\[\n\\frac{(2.0 \\mathrm{~mm})^{2}}{15 h}=\\frac{(6.0 \\mathrm{~mm})^{2}}{t_{2}}\n\\]\nfrom which\n\\[\nt_{2}=135 \\mathrm{~h}\n\\]"
},
{
"idx": 699,
"question": "The preexponential and activation energy for the diffusion of chromium in nickel are 1.1 \\(\\times 10^{-4} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(272,000 \\mathrm{~J} / \\mathrm{mol}\\), respectively. At what temperature will the diffusion coefficient have a value of \\(1.2 \\times 10^{-14} \\mathrm{~m}^{2} / \\mathrm{s}\\) ?",
"answer": "We are asked to calculate the temperature at which the diffusion coefficient for the diffusion of \\(\\mathrm{Cr}\\) in \\(\\mathrm{Ni}\\) has a value of \\(1.2 \\times 10^{-14} \\mathrm{m}^{2} / \\mathrm{s}\\). Solving for \\(T\\) from Equation leads to\n\\[\nT=-\\frac{Q_{d}}{R\\left(\\ln D-\\ln D_{0}\\right)}\n\\]\nand using the data the problem statement (i.e., \\(D_{0}=1.1 \\times 10^{-4} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(Q_{d}=272,000 \\mathrm{~J} / \\mathrm{mol}\\) ), we get\n\\[\n\\begin{aligned}\nT=-\\frac{272,000 \\mathrm{~J} / \\mathrm{mol}}{\\left(8.31 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}\\right)\\left[\\ln \\left(1.2 \\times 10^{-14} \\mathrm{~m}^{3} / \\mathrm{s}\\right)-\\ln \\left(1.1 \\times 10^{-4} \\mathrm{~m}^{2 / \\mathrm{s}}\\right)\\right]} \\\\\n=1427 \\mathrm{~K}=1154^{\\circ} \\mathrm{C}\n\\end{aligned}\n\\]\nNote: This problem may also be solved using the \"Diffusion\" module in the VMSE software. Open the \"Diffusion\" module, click on the \"D vs \\(1 / T\\) Plot\" submodule, and then do the following:\n1. In the left-hand window that appears, there is a preset set of data, but none for the diffusion of \\(\\mathrm{Cr}\\) in \\(\\mathrm{Ni}\\). This requires us specify our settings by clicking on the \"Custom1\" box.\n2. In the column on the right-hand side of this window enter the data for this problem, which are given in the problem statement. In the window under \"D0\" enter \"1.1e-4\". Next just below in the \"Qd\" window enter the activation energy value (in KJ/mol), in this case \"272\". It is now necessary to specify a temperature range over which the data is to be plotted-let usarbitrarily pick \\(1000^{\\circ} \\mathrm{C}\\) to be the minimum, which is entered in the \"T Min\" under \"Temperature Range\" window; we will also select \\(1500^{\\circ} \\mathrm{C}\\) to be the maximum temperature, which is entered in the \"T Max\" window.\n3. Next, at the bottom of this window, click the \"Add Curve\" button.\n4. A \\(\\log \\mathrm{D}\\) versus \\(1 / \\mathrm{T}\\) plot then appears, with a line for the temperature dependence of the diffusion coefficient for \\(\\mathrm{Cr}\\) in \\(\\mathrm{Ni}\\). At the top of this curve is a diamond-shaped cursor. We next click-and-drag this cursor down the line to the point at which the entry under the \"Diff Coeff (D):\" label reads \\(1.2 \\times 10^{-14} \\mathrm{~m}^{2} / \\mathrm{s}\\). The temperature that appears under the \"Temperature (T)\" label is \\(1430 \\mathrm{~K}\\)."
},
{
"idx": 700,
"question": "Given the activation energy for the diffusion of copper in silver is 193,000 J/mol and the diffusion coefficient D at 1000 K is 1.0 × 10^-14 m²/s, calculate the pre-exponential factor D₀.",
"answer": "To solve for D₀, use the rearranged equation: D₀ = D exp(Qd / (R T)). Substituting the given values: D₀ = (1.0 × 10^-14 m²/s) exp[193,000 J/mol / (8.31 J/mol-K × 1000 K)] = 1.22 × 10^-4 m²/s."
},
{
"idx": 700,
"question": "Using the pre-exponential factor D₀ = 1.22 × 10^-4 m²/s and the activation energy Qd = 193,000 J/mol, calculate the diffusion coefficient D at 1200 K.",
"answer": "To solve for D at 1200 K, use the equation: D = D₀ exp(-Qd / (R T)). Substituting the values: D = (1.22 × 10^-4 m²/s) exp[-193,000 J/mol / (8.31 J/mol-K × 1200 K)] = 4.8 × 10^-13 m²/s."
},
{
"idx": 701,
"question": "Carbon is allowed to diffuse through a steel plate \\(10-\\mathrm{mm}\\) thick. The concentrations of carbon at the two faces are 0.85 and \\(0.40 \\mathrm{~kg} \\mathrm{C} / \\mathrm{cm}^{3}\\) Fe, which are maintained constant. If the preexponential and activation energy are \\(5.0 \\times 10^{-7} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(77,000 \\mathrm{~J} / \\mathrm{mol}\\), respectively, compute the temperature at which the diffusion flux is \\(6.3 \\times 10^{-10} \\mathrm{~kg} / \\mathrm{m}^{2} \\cdot \\mathrm{s}\\).",
"answer": "This problem asks that we compute the temperature at which the diffusion flux is \\(6.3 \\times\\) \\(10^{-10} \\mathrm{~kg} / \\mathrm{m}^{ 2 - \\mathrm{s}}\\). Combining Equations yields\n\\[\n\\begin{aligned}\nJ & =-D \\frac{\\Delta C}{\\Delta x} \\\\\n& =-D_{0} \\frac{\\Delta C}{\\Delta x} \\exp \\left(-\\frac{Q_{d}}{R T}\\right)\n\\end{aligned}\n\\]\nTaking natural logarithms of both sides of this equation yields the following expression:\n\\[\n\\ln J=\\ln \\left(-D_{0} \\frac{\\Delta C}{\\Delta x}\\right)-\\frac{Q_{d}}{R T}\n\\]\nOr, upon rearrangement\n\\[\n\\frac{Q_{d}}{R T}=\\ln \\left(-D_{0} \\frac{\\Delta C}{\\Delta x}\\left)-\\ln J=\\ln \\left(-\\frac{D_{0} \\Delta C}{J \\Delta x}\\right)\\right.\n\\]\nAnd solving for \\(T\\) from this expression leads to\n\\[\nT=\\left(\\begin{array}{c}\nQ_{d} \\\\\nR\n\\end{array}\\right) \\frac{1}{\\ln \\left(-\\frac{D_{0} \\Delta C}{J \\Delta x} \\right)}\n\\]\nNow, incorporating values of the parameters in this equation provided in the problem statement leads to the following:\\[\n\\begin{array}{l}\nT=\\left(\\begin{array}{c}\n77,000\\mathrm{~J} / \\mathrm{mol} \\\\\n8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\n\\end{array}\\right) \\frac{1}{\\left[\\begin{array}{c}\n\\left(5.0 \\times 10^{-7} \\mathrm{~m}^{2} / \\mathrm{s}\\right)\\left(0.40 \\mathrm{~kg} / \\mathrm{m}^{3}-0.85 \\mathrm{~kg} / \\mathrm{m}^{3}\\right) \\\\\n\\left(6.3 \\times 10^{-10} \\mathrm{~kg} / \\mathrm{m}^{2}-\\mathrm{s}\\right)\\left(10 \\times 10^{-3} \\mathrm{~m}\\right)\n\\end{array}\\right]} \\\\\n\\quad=884 \\mathrm{~K}=611^{\\circ} \\mathrm{C}\n\\end{array}\n\\]"
},
{
"idx": 702,
"question": "Given the steady-state diffusion flux through a metal plate is 7.8 × 10^-8 kg/m²·s at a temperature of 1200°C (1473 K) and a concentration gradient of -500 kg/m⁴, calculate the value of D₀ assuming an activation energy for diffusion of 145,000 J/mol.",
"answer": "To compute D₀, use the expression: D₀ = - (J / (ΔC/Δx)) * exp(Qd / (R T)). Substituting the given values: D₀ = - (7.8 × 10^-8 kg/m²·s / -500 kg/m⁴) * exp(145,000 J/mol / (8.31 J/mol·K * 1473 K)) = 2.18 × 10^-5 m²/s."
},
{
"idx": 702,
"question": "Using the calculated D₀ value of 2.18 × 10^-5 m²/s and the same concentration gradient of -500 kg/m⁴, compute the diffusion flux at 1000°C (1273 K) with an activation energy for diffusion of 145,000 J/mol.",
"answer": "The diffusion flux at 1273 K is calculated using: J = -D₀ (ΔC/Δx) exp(-Qd / (R T)). Substituting the values: J = - (2.18 × 10^-5 m²/s) (-500 kg/m⁴) exp(-145,000 J/mol / (8.31 J/mol·K * 1273 K)) = 1.21 × 10^-8 kg/m²·s."
},
{
"idx": 703,
"question": "Consider the diffusion of some hypothetical metal \\(Y\\) into another hypothetical metal \\(Z\\) at \\(950^{\\circ} \\mathrm{C}\\); after \\(10 \\mathrm{~h}\\) the concentration at the \\(0.5 \\mathrm{~mm}\\) position (in metal Z) is \\(2.0 \\mathrm{wt} \\% Y\\). At what position will the concentration also be \\(2.0 \\mathrm{wt} \\% Y\\) after a \\(17.5 \\mathrm{~h}\\) heat treatment again at \\(950^{\\circ} \\mathrm{C}\\) ? Assume preexponential and activation energy values of \\(4.3 \\times 10^{-4} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(180,000 \\mathrm{~J} / \\mathrm{mol}\\), respectively, for this diffusion system.",
"answer": "In order to determine the position within this diffusion system at which the concentration of \\(\\mathrm{X}\\) in \\(\\mathrm{Y}\\) is \\(2.0 \\mathrm{wt} \\%\\) after \\(17.5 \\mathrm{~h}\\), we must employ Equation with \\(D\\) constant (since the temperature is constant). (Inasmuch as \\(D\\) is constant it is not necessary to employ preexponential and activation energy values cited in the problem statement.) Under these circumstances, Equation takes the form\n\\[\n\\frac{x^{2}}{t}=\\text { constant }\n\\]\nOr\n\\[\n\\frac{x_{1}^{2}}{t_{1}}=\\frac{x_{2}^{2}}{t_{2}}\n\\]\nIf we assume the following:\n\\[\n\\begin{aligned}\nx_{1} & =0.5 \\mathrm{~mm} \\\\\nt_{1} & =10 \\mathrm{~h} \\\\\nt_{2} & =17.5 \\mathrm{~h}\n\\end{aligned}\n\\]\nNow upon solving the above expression for \\(x_{2}\\) and incorporation of these values we have\n\\[\n\\begin{aligned}\nx_{2} & =\\sqrt{\\frac{x_{1}^{2} t_{2}}{t_{1}}} \\\\\n& =\\sqrt{\\frac{(0.5 \\mathrm{~mm})^{2}(17.5 \\mathrm{~h})}{10 \\mathrm{~h}}}\n\\end{aligned}\n\\]\\[=0.66\\mathrm{\\ mm}\n\\]"
},
{
"idx": 704,
"question": "Carbon dioxide diffuses through a high-density polyethylene (HDPE) sheet \\(50 \\mathrm{~mm}\\) thick at a rate of \\(2.2 \\times 10^{-8}\\left(\\mathrm{~cm}^{3} \\mathrm{STP} / \\mathrm{cm}^{2}\\right.\\)-s at \\(325 \\mathrm{~K}\\). The pressures of carbon dioxide at the two faces are \\(4000 \\mathrm{kPa}\\) and \\(2500 \\mathrm{kPa}\\), which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at \\(325 \\mathrm{~K}\\) ?",
"answer": "This problem asks us to compute the permeability coefficient for carbon dioxide through high-density polyethylene at \\(325 \\mathrm{~K}\\) given a steady-state permeability situation. It is necessary for us to use Equation in to solve this problem. Rearranging this expression and solving for the permeability coefficient gives the following:\n\\[\nP_{M}=\\frac{-J}{\\Delta P} \\frac{\\Delta x}{P_{2}-P_{1}}=\\frac{-J}{P_{2}-P_{1}} \\frac{\\Delta x}{P_{1}}\n\\]\nTaking \\(P_{1}=4000 \\mathrm{kPa}(4,000,000 \\mathrm{~Pa})\\) and \\(P_{2}=2500 \\mathrm{kPa}(2,500,000 \\mathrm{~Pa})\\), we compute the permeability coefficient of \\(\\mathrm{CO}_{2}\\) through HDPE as follows:\n\\[\nP_{M}=\\frac{\\left[-2.2 \\times 10^{-8} \\frac{(\\mathrm{cm}^{3} \\mathrm{STP})}{\\mathrm{cm}^{2} \\cdot \\mathrm{s}}\\right](5 \\mathrm{~cm})}{(2,500,000 \\mathrm{~Pa}-4,000,000 \\mathrm{~Pa})}\n\\]\n\\[\n\\begin{array}{c}\n=0.73 \\times 10^{-13}(\\mathrm{~cm}^{3} \\mathrm{STP})(\\mathrm{cm}) \\\\\n\\mathrm{cm}^{2} \\cdot \\mathrm{s}-\\mathrm{Pa}\n\\end{array}\n\\]"
},
{
"idx": 705,
"question": "A cylindrical specimen of a nickel alloy having an elastic modulus of \\(207 \\mathrm{GPa}\\left(30 \\times 10^{6} \\mathrm{psi}\\right)\\) and an original diameter of \\(10.2 \\mathrm{~mm}(0.40 \\mathrm{in}\\).) experiences only elastic deformation when a tensile load of \\(8900 \\mathrm{~N}\\left(2000 \\mathrm{lb} f_{\\mathrm{f}}\\right)\\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \\(0.25 \\mathrm{~mm}(0.010 \\mathrm{in}\\).).",
"answer": "We are asked to compute the maximum length of a cylindrical nickel specimen (before deformation) that is deformed elastically in tension. For a cylindrical specimen the original cross-sectional area \\(A_{0}\\) is equal to\n\\[\nA_{0}=\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}\n\\]\nwhere \\(d_{0}\\) is the original diameter. Combining Equations and solving for \\(l_{0}\\) leads to\n\\[\nl_{0}=\\frac{\\Delta l}{\\varepsilon}=\\frac{\\Delta l}{\\frac{\\sigma}{E}}=\\frac{\\Delta l}{\\frac{F}{A_{0}}}=\\frac{\\Delta l E \\pi}{\\frac{F}{F}}\\left(\\frac{d_{0}}{2}\\right)^{2}=\\frac{\\Delta l \\, E \\pi d_{0}^{2}}{4 F}\n\\]\nSubstitution into this expression, values of \\(d_{0}, \\Delta l, E\\), and \\(F\\) given in the problem statement yields the following\n\\[\n\\begin{aligned}\nl_{0}= & \\frac{\\left(0.25 \\times 10^{-3} \\mathrm{~m}\\right)\\left(207 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)\\left(\\pi\\right)\\left(10.2 \\times 10^{-3} \\mathrm{~m}\\right)^{2}}{(4)(8900 \\mathrm{~N})} \\\\\n& =0.475 \\mathrm{~m}=475 \\mathrm{~mm} \\quad(18.7 \\mathrm{in} .)\n\\end{aligned}\n\\]"
},
{
"idx": 706,
"question": "An aluminum bar \\(125 \\mathrm{~mm}\\) (5.0 in.) long and having a square cross section \\(16.5 \\mathrm{~mm}\\) (0.65 in.) on an edge is pulled in tension with a load of \\(66,700 \\mathrm{~N}(15,000 \\mathrm{lb})\\) and experiences an elongation of \\(0.43 \\mathrm{~mm}(1.7 \\times 10^{-2}\\) in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.",
"answer": "This problem asks us to compute the elastic modulus of aluminum. For a square crosssection, \\(A_{0}=b_{0}^{2}\\), where \\(b_{0}\\) is the edge length. Combining Equations and solving for the modulus of elasticity, \\(E\\), leads to\n\\[\nE=\\frac{F}{\\varepsilon}=\\frac{A_{0}}{\\frac{\\Delta l}{l_{0}}}=\\frac{F l_{0}}{b_{0}^{2} \\Delta l}\n\\]\nSubstitution into this expression, values of \\(l_{0}, b_{0} \\Delta l\\), and \\(F\\) given in the problem statement yields the following\n\\[\n\\begin{aligned}\nE & =\\frac{(67,700 \\mathrm{~N})\\left(125 \\times 10^{-3} \\mathrm{~m}\\right)}{\\left(16.5 \\times 10^{-3} \\mathrm{~m}\\right)^{2}\\left(0.43 \\times 10^{-3} \\mathrm{~m}\\right)} \\\\\n& =71.2 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}=71.2 \\mathrm{GPa} \\quad\\left(10.4 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]"
},
{
"idx": 707,
"question": "Consider a cylindrical nickel wire \\(2.0 \\mathrm{~mm}(0.08 \\mathrm{in}\\).) in diameter and \\(3 \\times 10^{4} \\mathrm{~mm}\\) (1200 in.) long. Calculate its elongation when a load of \\(300 \\mathrm{~N}\\left(67 \\mathrm{lb}^{2}\\right)\\) is applied. Assume that the deformation is totally elastic.",
"answer": "In order to compute the elongation of the \\(\\mathrm{Ni}\\) wire when the \\(300 \\mathrm{~N}\\) load is applied we must employ Equations. Combining these equations and solving for \\(\\Delta l\\) leads to the following\n\\[\n\\Delta l=l_{0} \\varepsilon=l_{0} \\frac{\\sigma}{E}=\\frac{l_{0} F}{E A_{0}}=\\frac{l_{0} F}{E \\pi\\left(\\frac{d_{0}}{2}\\right)^{2}}=\\frac{4 l_{0} F}{E \\pi d_{0}^{2}}\n\\]\nsince the cross-sectional area \\(A_{0}\\) of a cylinder of diameter \\(d_{0}\\) is equal to\n\\[\nA_{0}=\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}\n\\]\nIncorporating into this expression values for \\(l_{0}, F\\), and \\(d_{0}\\) in the problem statement, and realizing that for \\(\\mathrm{Ni}, E=207 \\mathrm{GPa}\\left(30 \\times 10^{6} \\mathrm{psi}\\right)\\) , and that \\(3 \\times 10^{4} \\mathrm{~mm}=30 \\mathrm{~m}\\), the wire elongation is\n\\[\n\\Delta l=\\frac{(4)(30 \\mathrm{~m})(300 \\mathrm{~N})}{\\left(207 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)\\left(\\pi\\right)\\left(2 \\times 10^{-3} \\mathrm{~m}\\right)^{2}}=0.0138 \\mathrm{~m}=13.8 \\mathrm{~mm}(0.53 \\mathrm{in} .)\n\\]"
},
{
"idx": 708,
"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": "This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation (F_y). Taking the yield strength to be 345 MPa, and employment of Equation leads to F_y = sigma_y A_0 = (345 x 10^6 N/m^2)(133 mm^2)(1 m/10^3 mm)^2 = 44,850 N (10,000 lbf)."
},
{
"idx": 708,
"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 76 mm (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 sample may be deformed without plastic deformation is determined by combining Equations as l_i = z l_0 + l_0 = l_0(z + 1) l_i = l_0(sigma/E + 1). Incorporating values of l_0, sigma, and E provided in the problem statement leads to the computation of l_i as follows: l_i = (76 mm)[(345 MPa)/(103 x 10^3 MPa) + 1] = 76.25 mm (3.01 in.)."
},
{
"idx": 709,
"question": "A cylindrical rod of steel \\((E=207 \\mathrm{GPa}, 30 \\times 10^{6} \\mathrm{psi})\\) having a yield strength of 310 \\(\\mathrm{MPa}(45,000 \\mathrm{psi})\\) is to be subjected to a load of \\(11,100 \\mathrm{~N}(2500 \\mathrm{lb})\\). If the length of the rod is \\(500 \\mathrm{~mm}(20.0 \\mathrm{in}\\).), what must be the diameter to allow an elongation of \\(0.38 \\mathrm{~mm}(0.015 \\mathrm{in}\\).)?",
"answer": "This problem asks us to compute the diameter of a cylindrical specimen of steel in order to allow an elongation of \\(0.38 \\mathrm{~mm}\\). Employing Equations, assuming that deformation is entirely elastic, we may write the following expression:\n\\[\n\\begin{aligned}\n\\sigma & =\\frac{F}{A_{0}}=\\frac{F}{\\pi\\left(\\frac{d_{0}^{2}}{4}\\right)} \\\\\n& =E \\varepsilon=E\\left(\\frac{\\Delta l}{l_{0}}\\right)\n\\end{aligned}\n\\]\nAnd upon simplification\n\\[\n\\frac{F}{\\pi\\left(\\frac{d_{0}^{2}}{2}\\right)}=E\\left(\\frac{\\Delta l}{l_{0}}\\right)\n\\]\nSolving for the original cross-sectional area, \\(d_{0}\\), leads to\n\\[\nd_{0}=\\sqrt{\\frac{4 l_{0} F}{\\pi E \\Delta l}}\n\\]\nIncorporation of values for \\(l_{0}, F, E\\), and \\(\\Delta l\\) provided in the problem statement yields the following:\n\\[\nd_{0}=\\sqrt{\\frac{(4)\\left(500 \\times 10^{-3} \\mathrm{~m}\\right)(11,100 \\mathrm{~N})}{\\left(\\pi\\right)\\left(207 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)\\left(0.38 \\times 10^{-3} \\mathrm{~m}\\right)}}\n\\]\\[\n = 9.5 \\times 10^{-3} \\mathrm{\\ m} = 9.5 \\mathrm{\\ mm} \\mathrm{\\ (0.376\\ in.)} \n\\]\\title{"
},
{
"idx": 710,
"question": "A cylindrical specimen of a metal alloy \\(10 \\mathrm{~mm}(0.4 \\mathrm{in.})\\) in diameter is stressed elastically in tension. A force of \\(15,000 \\mathrm{~N}(3,370 \\mathrm{lb} r)\\) produces a reduction in specimen diameter of \\(7 \\times 10^{-3}\\) \\(\\mathrm{mm}(2.8 \\times 10^{-4}\\) in.). Compute Poisson's ratio for this material if its elastic modulus is \\(100 \\mathrm{GPa}\\) \\(\\left(14.5 \\times 10^{6} \\mathrm{psi}\\right)\\).",
"answer": "This problem asks that we compute Poisson's ratio for the metal alloy. From Equations\n\\[\n\\varepsilon_{z}=\\frac{\\sigma}{E}=\\frac{F}{A_{0}}\\left(\\frac{1}{E}\\right)=\\frac{F}{A_{0} E}=\\frac{F}{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}} E=\\frac{4 F}{\\pi d_{0}^{2} E}\n\\]\nSince the transverse strain \\(\\varepsilon_{X}\\) is equal to\n\\[\n\\varepsilon_{x}=\\frac{\\Delta d}{d_{0}}\n\\]\nand Poisson's ratio is defined by Equation, then\n\\[\nv=-\\frac{\\varepsilon_{x}}{\\varepsilon_{z}}=-\\frac{\\Delta d / d_{0}}{\\left(\\frac{4 F}{\\pi d_{0}^{2} E}\\right)}=-\\frac{d_{0} \\Delta d \\pi E}{4 F}\n\\]\nNow, incorporating values of \\(d_{0}, \\Delta d, E\\) and \\(F\\) from the problem statement yields the following value for Poisson's ratio\n\\[\nv=-\\frac{\\left(10 \\times 10^{-3} \\mathrm{~m}\\right)\\left(-7 \\times 10^{-6} \\mathrm{~m}\\right)\\left(\\pi\\right)\\left(100 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}\\right)}{(4)(15,000 \\mathrm{~N})}=0.367\n\\]"
},
{
"idx": 711,
"question": "What is the formula to calculate the modulus of elasticity (E) for a cylindrical specimen under tensile force, given the force (F), initial diameter (d0), transverse strain (Δd/d0), and Poisson's ratio (ν)?",
"answer": "The modulus of elasticity (E) can be calculated using the formula: E = (4 * F * ν) / (π * d0 * Δd), where F is the tensile force, ν is Poisson's ratio, d0 is the initial diameter, and Δd is the change in diameter."
},
{
"idx": 711,
"question": "Given a cylindrical specimen with an initial diameter of 10.0 mm (0.39 in.), a tensile force of 1500 N (340 lb), a reduction in diameter of 6.7 × 10^-4 mm (2.64 × 10^-5 in.), and Poisson's ratio of 0.35, compute the modulus of elasticity (E).",
"answer": "Using the formula E = (4 * F * ν) / (π * d0 * Δd), substitute the given values: F = 1500 N, ν = 0.35, d0 = 10.0 × 10^-3 m, Δd = -6.7 × 10^-7 m. The calculation is E = (4 * 1500 * 0.35) / (π * 10.0 × 10^-3 * -6.7 × 10^-7) = 10^11 Pa = 100 GPa (14.7 × 10^6 psi)."
},
{
"idx": 712,
"question": "For a cylindrical rod 500 mm long with a diameter of 12.7 mm subjected to a tensile load of 29,000 N, which of the four metals or alloys (aluminum alloy, brass alloy, copper, steel alloy) will not experience plastic deformation? Use the given yield strength values from the table.",
"answer": "First, compute the stress: sigma = F / A0 = 29000 N / (pi * (12.7e-3 m / 2)^2) = 230 MPa. Materials with yield strength greater than 230 MPa are: aluminum alloy (255 MPa), brass alloy (345 MPa), and steel alloy (450 MPa). Copper (210 MPa) is excluded."
},
{
"idx": 712,
"question": "For the same cylindrical rod under the same load, which of the candidate materials (aluminum alloy, brass alloy, steel alloy) will not elongate more than 1.3 mm? Use the given modulus of elasticity values from the table.",
"answer": "Calculate elongation for each candidate: For aluminum alloy (E = 70 GPa): delta_l = (230 MPa * 500 mm) / 70e3 MPa = 1.64 mm (exceeds limit). For brass alloy (E = 100 GPa): delta_l = (230 MPa * 500 mm) / 100e3 MPa = 1.15 mm (acceptable). For steel alloy (E = 207 GPa): delta_l = (230 MPa * 500 mm) / 207e3 MPa = 0.56 mm (acceptable). Final candidates: brass alloy and steel alloy."
},
{
"idx": 713,
"question": "A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm. Calculate the ductility in terms of percent reduction in area.",
"answer": "Percent reduction in area is computed using the formula: %RA = ((A0 - Af)/A0) * 100 = ((π*(d0/2)^2 - π*(df/2)^2)/(π*(d0/2)^2)) * 100. Substituting the given values: %RA = ((π*(12.8 mm/2)^2 - π*(8.13 mm/2)^2)/(π*(12.8 mm/2)^2)) * 100 = 60%."
},
{
"idx": 713,
"question": "A cylindrical metal specimen having an original gauge length of 50.80 mm is pulled in tension until fracture occurs. The fractured gauge length is 74.17 mm. Calculate the ductility in terms of percent elongation.",
"answer": "Percent elongation is computed using the formula: %EL = ((lf - l0)/l0) * 100. Substituting the given values: %EL = ((74.17 mm - 50.80 mm)/50.80 mm) * 100 = 46%."
},
{
"idx": 714,
"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": "For this problem, we are given two values of \\(\\varepsilon_{T}\\) and \\(\\sigma_{T}\\), from which we are asked to calculate the true stress that produces a true plastic strain of 0.21 . From Equation, we want to set up two simultaneous equations with two unknowns (the unknowns being \\(K\\) and \\(n\\) ). Taking logarithms of both sides of Equation leads to\n\\[\n\\log \\sigma_{T}=\\log K+n \\log \\varepsilon_{T}\n\\]\nThe two simultaneous equations using data provided in the problem statement are as follows:\n\\[\n\\begin{array}{c}\n\\log (60,000 \\mathrm{psi})=\\log K+n \\log (0.15) \\\\\n\\log (70,000 \\mathrm{psi})=\\log K+n \\log (0.25)\n\\end{array}\n\\]\nSolving for \\(n\\) from these two expressions yields\n\\[\nn=\\frac{\\log (60,000)-\\log (70,000)}{\\log (0.15)-\\log (0.25)}=0.302\n\\]\nWe solve for \\(K\\) by substitution of this value of \\(n\\) into the first simultaneous equation as\\[\n\\begin{array}{c}\n\\log K=\\log \\sigma_{T}-n\\log \\varepsilon_{T} \\\\\n\\\\= \\log (60,000) \\quad (0.302)[\\log (0.15)]=5.027\n\\end{array}\n\\]\nThus, the value of \\(K\\) is equal to\n\\[\nK=10^{5.027}=106,400 \\mathrm{psi}\n\\]\nThe true stress required to produce a true strain of 0.21 is determined using Equation as follows:\n\\[\nT=K\\left(\\quad T\\right)^{n}=(106,400 \\mathrm{psi})(0.21)^{0.302}=66,400 \\mathrm{psi}(460 \\mathrm{MPa})\n\\]"
},
{
"idx": 715,
"question": "A circular specimen of \\(\\mathrm{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 \\(5560 \\mathrm{~N}\\) \\((1250 \\mathrm{lb})\\), the flexural strength is \\(105 \\mathrm{MPa}(15,000 \\mathrm{psi})\\), and the separation between load points is \\(45 \\mathrm{~mm}\\) (1.75 in.).",
"answer": "We are asked to calculate the maximum radius of a circular specimen of \\(\\mathrm{MgO}\\) that is loaded using three-point bending. Solving for the specimen radius, \\(R\\), from Equation leads to the following:\n\\[\nR=\\left[\\frac{F_{f} L}{\\sigma_{f s} \\pi}\\right]^{1 / 3}\n\\]\nwhich, when substituting the parameters stipulated in the problem statement, yields\n\\[\n\\begin{aligned}\nR & =\\left[\\frac{(5560 \\mathrm{~N})\\left(45 \\times 10^{-3} \\mathrm{~m}\\right)}{\\left(105 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}\\right)(\\pi)}\\right]^{1 / 3} \\\\\n& =9.1 \\times 10^{-3} \\mathrm{~m}=9.1 \\mathrm{~mm} \\quad(0.36 \\mathrm{in} .)\n\\end{aligned}\n\\]\nThus, the minimum allowable specimen radius is \\(9.1 \\mathrm{~mm}(0.36 \\mathrm{in}\\).)"
},
{
"idx": 716,
"question": "What is the flexural strength of the aluminum oxide specimen with a circular cross section of radius 5.0 mm (0.20 in.) that fractured at a load of 3000 N (675 lbf) when the distance between the support points was 40 mm (1.6 in.)?",
"answer": "The flexural strength is calculated as follows: sigma_fs = (F_f * L) / (pi * R^3) = (3000 N * 40 x 10^-3 m) / (pi * (5.0 x 10^-3 m)^3) = 306 x 10^6 N/m^2 = 306 MPa (42,970 psi)."
},
{
"idx": 716,
"question": "At what load would a specimen of the same material with a square cross section of 15 mm (0.6 in.) length on each edge fracture if the support point separation is maintained at 40 mm (1.6 in.), given the flexural strength is 306 MPa (42,970 psi)?",
"answer": "The fracture load is calculated as follows: F_f = (2 * sigma_fs * d^3) / (3 * L) = (2 * 306 x 10^6 N/m^2 * (15 x 10^-3 m)^3) / (3 * 40 x 10^-3 m) = 17,200 N (3870 lbf)."
},
{
"idx": 717,
"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 5.0 mm (0.20 in.) and the support point separation distance is 15.0 mm (0.61 in.), would you expect the specimen to fracture when a load of 7500 N (1690 lbf) is applied? Justify your answer.",
"answer": "This portion of the problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of 300 MPa and a radius of 5.0 mm will fracture when subjected to a load of 7500 N in a three-point bending test; the support point separation is given as 15.0 mm. Using Equation we will calculate the value of σ, in this case the flexural stress; if this value is greater than the flexural strength, σ_fs (300 MPa), then fracture is expected to occur. Employment of Equation to compute the flexural stress yields σ = (FL)/(πR^3) = (7500 N)(15 x 10^-3 m)/(π)(5 x 10^-3 m)^3 = 286.5 x 10^6 N/m^2 = 286.5 MPa (40,300 psi). Since this value (286.5 MPa) is less than the given value of σ_fs (300 MPa), then fracture is not predicted."
},
{
"idx": 717,
"question": "Would you be 100% certain of the answer in part (a)? Why or why not?",
"answer": "However, the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and since this value of the flexural stress, σ, is relatively close to σ_fs there is some chance that fracture will occur."
},
{
"idx": 718,
"question": "Compute the modulus of elasticity for nonporous spinel (MgAl2O4) given that the modulus of elasticity for spinel having 5 vol% porosity is 240 GPa (35 x 10^6 psi).",
"answer": "To compute the modulus of elasticity for the nonporous material (E0), we use the equation: E0 = E / (1 - 1.9P + 0.9P^2), where E = 240 GPa and P = 0.05. Substituting the values: E0 = 240 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 265 GPa (38.6 x 10^6 psi)."
},
{
"idx": 718,
"question": "Compute the modulus of elasticity for spinel (MgAl2O4) having 15 vol% porosity, given that the modulus of elasticity for nonporous spinel is 265 GPa (38.6 x 10^6 psi).",
"answer": "To compute the modulus of elasticity for the material with 15 vol% porosity (E), we use the equation: E = E0 (1 - 1.9P + 0.9P^2), where E0 = 265 GPa and P = 0.15. Substituting the values: E = (265 GPa) [1 - (1.9)(0.15) + (0.9)(0.15)^2] = 195 GPa (28.4 x 10^6 psi)."
},
{
"idx": 719,
"question": "Compute the modulus of elasticity for nonporous titanium carbide (TiC) given that the modulus of elasticity for TiC with 5 vol% porosity is 310 GPa (45 x 10^6 psi).",
"answer": "To compute the modulus of elasticity for nonporous TiC (E0), we use the given modulus of elasticity (E) of 310 GPa for a material with 5 vol% porosity (P = 0.05). The equation is as follows: E0 = E / (1 - 1.9P + 0.9P^2). Substituting the values, we get E0 = 310 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 342 GPa (49.6 x 10^6 psi)."
},
{
"idx": 719,
"question": "Determine the volume percent porosity at which the modulus of elasticity for titanium carbide (TiC) becomes 240 GPa (35 x 10^6 psi), given that the modulus of elasticity for nonporous TiC is 342 GPa.",
"answer": "To find the volume percent porosity (P) that results in a modulus of elasticity (E) of 240 GPa, we use the equation E/E0 = 1 - 1.9P + 0.9P^2, where E0 is 342 GPa. Substituting the values, we get 240 GPa / 342 GPa = 0.702 = 1 - 1.9P + 0.9P^2. Rearranging gives 0.9P^2 - 1.9P + 0.298 = 0. Solving the quadratic equation yields roots P+ = 1.94 and P- = 0.171. Only the negative root (0.171) is physically meaningful, corresponding to 17.1 vol% porosity."
},
{
"idx": 720,
"question": "Compute the flexural strength for a completely nonporous specimen of this material given the flexural strength and associated volume fraction porosity for two specimens: (70 MPa, 0.10) and (60 MPa, 0.15).",
"answer": "Given the flexural strengths at two different volume fraction porosities, we determine the flexural strength for a nonporous material. Taking the natural logarithm of both sides of Equation 7.22 yields the following expression: ln σ_fs = ln σ_0 - nP. Using the data provided, two simultaneous equations may be written as: ln(70 MPa) = ln σ_0 - (0.10)n and ln(60 MPa) = ln σ_0 - (0.15)n. Solving for n and σ_0 leads to the values: n = 3.08 and σ_0 = 95.3 MPa. For the nonporous material, P = 0, and from Equation 7.22, σ_0 = σ_fs. Thus, σ_fs for P = 0 is 95.3 MPa."
},
{
"idx": 720,
"question": "Compute the flexural strength for a 0.20 volume fraction porosity given the flexural strength and associated volume fraction porosity for two specimens: (70 MPa, 0.10) and (60 MPa, 0.15), and the previously determined values n = 3.08 and σ_0 = 95.3 MPa.",
"answer": "We are asked for σ_fs at P = 0.20 for this material. From Equation 7.22: σ_fs = σ_0 exp(-nP) = (95.3 MPa) exp[-(3.08)(0.20)] = 51.5 MPa."
},
{
"idx": 721,
"question": "Determine the relaxation time (τ) for the viscoelastic polymer given the initial stress σ(0) = 3.5 MPa, the stress at t = 30 s σ(30) = 0.5 MPa, and the elapsed time t = 30 s.",
"answer": "To determine τ, we use the equation: τ = -t / ln[σ(t)/σ(0)]. Substituting the given values: τ = -30 s / ln[0.5 MPa / 3.5 MPa] = 15.4 s."
},
{
"idx": 721,
"question": "Calculate the stress σ(10) at t = 10 s for the viscoelastic polymer using the relaxation time τ = 15.4 s and the initial stress σ(0) = 3.5 MPa.",
"answer": "Using the equation σ(t) = σ(0) exp(-t/τ), we find: σ(10) = 3.5 MPa exp(-10 s / 15.4 s) = 1.83 MPa."
},
{
"idx": 721,
"question": "Determine the flexural modulus E_f(10) for the viscoelastic polymer at t = 10 s given the strain ε_0 = 0.5 and the stress σ(10) = 1.83 MPa.",
"answer": "The flexural modulus is calculated as: E_f(10) = σ(10) / ε_0 = 1.83 MPa / 0.5 = 3.66 MPa (522 psi)."
},
{
"idx": 722,
"question": "Compute the average hardness value from the given Rockwell G hardness measurements: 47.3, 48.7, 47.1, 52.1, 50.0, 50.4, 45.6, 46.2, 45.9, 49.9, 48.3, 46.4, 47.6, 51.1, 48.5, 50.4, 46.7, 49.7",
"answer": "The average hardness value is calculated by summing all the hardness values and dividing by the number of measurements. Sum = 47.3 + 48.7 + 47.1 + 52.1 + 50.0 + 50.4 + 45.6 + 46.2 + 45.9 + 49.9 + 48.3 + 46.4 + 47.6 + 51.1 + 48.5 + 50.4 + 46.7 + 49.7 = 871.9. Number of measurements = 18. Average = 871.9 / 18 = 48.4."
},
{
"idx": 722,
"question": "Compute the standard deviation of hardness values from the given Rockwell G hardness measurements: 47.3, 48.7, 47.1, 52.1, 50.0, 50.4, 45.6, 46.2, 45.9, 49.9, 48.3, 46.4, 47.6, 51.1, 48.5, 50.4, 46.7, 49.7, with an average hardness of 48.4",
"answer": "The standard deviation is calculated by taking the square root of the average of the squared differences from the mean. Squared differences: (47.3-48.4)^2 = 1.21, (48.7-48.4)^2 = 0.09, (47.1-48.4)^2 = 1.69, (52.1-48.4)^2 = 13.69, (50.0-48.4)^2 = 2.56, (50.4-48.4)^2 = 4.00, (45.6-48.4)^2 = 7.84, (46.2-48.4)^2 = 4.84, (45.9-48.4)^2 = 6.25, (49.9-48.4)^2 = 2.25, (48.3-48.4)^2 = 0.01, (46.4-48.4)^2 = 4.00, (47.6-48.4)^2 = 0.64, (51.1-48.4)^2 = 7.29, (48.5-48.4)^2 = 0.01, (50.4-48.4)^2 = 4.00, (46.7-48.4)^2 = 2.89, (49.7-48.4)^2 = 1.69. Sum of squared differences = 64.95. Variance = 64.95 / 17 = 3.82. Standard deviation = sqrt(3.82) = 1.95."
},
{
"idx": 723,
"question": "Compute the average yield strength (in MPa) from the given values: 274.3, 277.1, 263.8, 267.5, 258.6, 271.2, 255.4, 266.9, 257.6, 270.8, 260.1, 264.3, 261.7, 279.4, 260.5",
"answer": "The average yield strength is calculated as (274.3 + 277.1 + 263.8 + 267.5 + 258.6 + 271.2 + 255.4 + 266.9 + 257.6 + 270.8 + 260.1 + 264.3 + 261.7 + 279.4 + 260.5) / 15 = 3988.2 / 15 = 265.9 MPa"
},
{
"idx": 723,
"question": "Compute the standard deviation of yield strength (in MPa) from the given values: 274.3, 277.1, 263.8, 267.5, 258.6, 271.2, 255.4, 266.9, 257.6, 270.8, 260.1, 264.3, 261.7, 279.4, 260.5, with an average of 265.9 MPa",
"answer": "The standard deviation is calculated as sqrt([(274.3 - 265.9)^2 + (277.1 - 265.9)^2 + (263.8 - 265.9)^2 + (267.5 - 265.9)^2 + (258.6 - 265.9)^2 + (271.2 - 265.9)^2 + (255.4 - 265.9)^2 + (266.9 - 265.9)^2 + (257.6 - 265.9)^2 + (270.8 - 265.9)^2 + (260.1 - 265.9)^2 + (264.3 - 265.9)^2 + (261.7 - 265.9)^2 + (279.4 - 265.9)^2 + (260.5 - 265.9)^2] / 14) = sqrt(753.75 / 14) = 7.34 MPa"
},
{
"idx": 724,
"question": "Upon what three criteria are factors of safety based?",
"answer": "The criteria upon which factors of safety are based are (1) consequences of failure, (2) previous experience, (3) accuracy of measurement of mechanical forces and/or material properties, and (4) economics."
},
{
"idx": 725,
"question": "Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 degrees and 35 degrees, 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?",
"answer": "The resolved shear stress is calculated using the equation: tau_R = sigma * cos(phi) * cos(lambda) = (12 MPa) * (cos 60 degrees) * (cos 35 degrees) = 4.91 MPa (717 psi). Since the resolved shear stress (4.91 MPa) is less than the critical resolved shear stress (6.2 MPa), the single crystal will not yield."
},
{
"idx": 725,
"question": "What stress will be necessary to cause the metal single crystal to yield, given the normal to the slip plane and the slip direction are at angles of 60 degrees and 35 degrees, respectively, with the tensile axis, and the critical resolved shear stress is 6.2 MPa (900 psi)?",
"answer": "The stress at which yielding occurs is calculated using the equation: sigma_y = tau_crss / (cos(phi) * cos(lambda)) = 6.2 MPa / (cos 60 degrees * cos 35 degrees) = 15.1 MPa (2200 psi)."
},
{
"idx": 726,
"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": "Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the largest cos λ. Cosines for the possible λ values are given below. cos(30 degrees) = 0.87, cos(48 degrees) = 0.67, cos(78 degrees) = 0.21. Thus, the slip direction is at an angle of 30 degrees with the tensile axis."
},
{
"idx": 726,
"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 tensile axis. If plastic deformation begins at a tensile stress of 2.5 MPa, determine the critical resolved shear stress for zinc.",
"answer": "The critical resolved shear stress is just crss = σy(cos φ cos λ)max = (2.5 MPa)[cos(65 degrees) cos(30 degrees)] = 0.91 MPa (130 psi)."
},
{
"idx": 727,
"question": "The critical resolved shear stress for copper \\((\\mathrm{Cu})\\) is \\(0.48 \\mathrm{MPa}(70 \\mathrm{psi})\\). Determine the maximum possible yield strength for a single crystal of Cu pulled in tension.\n\\title{",
"answer": "In order to determine the maximum possible yield strength for a single crystal of \\(\\mathrm{Cu}\\) pulled in tension, we simply employ Equation 8.5 as\n\\[\n\\sigma_{y}=2 \\tau_{\\text {crss }}=(2)(0.48 \\mathrm{MPa})=0.96 \\mathrm{MPa} \\quad \\text { (140 psi) }\n\\]"
},
{
"idx": 728,
"question": "What are the values of σ0 and ky for an iron with given yield points and grain diameters? (Given: σy = 230 MPa at d = 1 × 10−2 mm, σy = 275 MPa at d = 6 × 10−3 mm)",
"answer": "The values are σ0 = 75.4 MPa and ky = 15.46 MPa(mm)1/2. These are obtained by solving the simultaneous equations: 230 MPa = σ0 + (10.0) ky and 275 MPa = σ0 + (12.91) ky, where d−1/2 values are 10.0 mm−1/2 and 12.91 mm−1/2 respectively."
},
{
"idx": 728,
"question": "What grain diameter (d) will result in a lower yield point of 310 MPa for the same iron? (Given: σ0 = 75.4 MPa, ky = 15.46 MPa(mm)1/2)",
"answer": "The grain diameter will be 4.34 × 10−3 mm. This is calculated using the equation 310 MPa = 75.4 MPa + [15.46 MPa(mm)1/2] d−1/2, solving for d−1/2 = (310 MPa − 75.4 MPa) / 15.46 MPa(mm)1/2, then d = [15.46 MPa(mm)1/2 / (310 MPa − 75.4 MPa)]2."
},
{
"idx": 729,
"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, \\(\\mathrm{mm}\\) ) & Rectangular \\((\\mathrm{mm})\\) \\\\\n\\hline Original dimensions & 18.0 & \\(20 \\times 50\\) \\\\\nDeformed dimensions & 15.9 & \\(13.7 \\times 55.1\\) \\\\\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 one that has experienced the greatest degree of cold work. Therefore, all we need do is to compute the \\%CW for each specimen using Equation. For the circular one\n\\[\n\\begin{aligned}\n\\% \\mathrm{CW} & =\\left(\\frac{A_{0}-A_{d}}{A_{0}}\\right) \\times 100 \\\\\n& =\\left(\\frac{\\pi r_{0}^{2}-\\pi r_{d}^{2}}{\\pi r_{0}^{2}}\\right) \\times 100 \\\\\n& =\\left[\\frac{\\pi \\frac{\\left(18.0 \\mathrm{~mm}\\right)^{2}}{2}-\\pi \\frac{\\left(15.9 \\mathrm{~mm}\\right)^{2}}{2}}{\\pi \\frac{\\left(18.0 \\mathrm{~mm}\\right)^2}{2}}\\right] \\times 100=22.0 \\% \\mathrm{CW}\n\\end{aligned}\n\\]\nWhile, for the rectangular one (using \\(l\\) and \\(w\\) to designate specimen length and width, respectively:\\[\n\\begin{array}{l}\n\\% \\mathrm{CW} = \\left( \\frac{w_{0} l_{0}-w_{d} l_{d}}{w_{0} l_{0}} \\right) \\times 100 \\\\\n\\quad = \\left[ \\frac{(20 \\mathrm{~mm})(50 \\mathrm{~mm}) \\;-\\; (13.7 \\mathrm{~mm})(55.1 \\mathrm{~mm})}{(20 \\mathrm{~mm})(50 \\mathrm{~mm})} \\right] \\times 100=24.5 \\% \\mathrm{CW}\n\\end{array}\n\\]\nTherefore, the deformed rectangular specimen will be harder since it has the greater \\(\\% \\mathrm{CW}\\)."
},
{
"idx": 730,
"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_{\\mathrm{D}}\\) as\n\\[\n\\tau_{\\text {crss }}=\\tau_{0}+A \\sqrt{\\rho_{\\mathrm{D}}}\n\\]\nwhere \\(\\tau_{0}\\) and \\(A\\) are constants. For copper, the critical resolved shear stress is \\(0.69 \\mathrm{MPa}(100 \\mathrm{psi})\\) at a dislocation density of \\(10^{4} \\mathrm{~mm}^{-2}\\). If it is known that the value of \\(\\tau_{0}\\) for copper is \\(0.069 \\mathrm{MPa}\\) (10 psi), compute \\(\\tau_{\\text {crss }}\\) at a dislocation density of \\(10^{6} \\mathrm{~mm}^{-2}\\).",
"answer": "We are asked in this problem to compute the critical resolved shear stress at a dislocation density of \\(10^{6} \\mathrm{~mm}^{-} .2\\). It is first necessary to compute the value of the constant \\(A\\) from the one set of data as follows:\n\\[\nA=\\frac{\\tau_{\\text {crss }}-\\tau_{0}}{\\sqrt{\\rho_{D}}}=\\frac{0.69 \\mathrm{MPa}-0.069 \\mathrm{MPa}}{\\sqrt{10^{4} \\mathrm{~mm}^{-2}}}=6.21 \\times 10^{-3} \\mathrm{MPa} \\cdot \\mathrm{mm} \\quad(0.90 \\mathrm{psi} \\cdot \\mathrm{mm})\n\\]\nNow, since we know the values for both \\(A\\) and \\(\\tau_{0}\\), it is possible to compute the critical resolved shear stress at a dislocation density of \\(10^{6} 8 \\mathrm{~mm}^{-2}\\) as follows:\n\\[\n\\begin{array}{c}\n\\tau_{\\text {crss }}=\\tau_{0}+A \\sqrt{\\text { \\(\\rho_{D}\\) }} \\\\\n=(0.069 \\mathrm{MPa})+(6.21 \\times 10^{-3} \\mathrm{MPa} \\mathrm{mm}) \\sqrt{10^{6} 8 \\mathrm{~mm}^{-2}}=6.28 \\mathrm{MPa} \\quad(910 \\mathrm{psi})\n\\end{array}\n\\]"
},
{
"idx": 731,
"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": 732,
"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": 732,
"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": 733,
"question": "What was the original grain diameter for a brass material measured as a function of time at 650 degrees Celsius, given the following data: Time (min) = 40, Grain Diameter (mm) = 5.6 x 10^-2; Time (min) = 100, Grain Diameter (mm) = 8.0 x 10^-2?",
"answer": "Using the data given and Equation (for n=2)—that is d^2 - d0^2 = K t, we may set up two simultaneous equations with d0 and K as unknowns, as follows: (5.6 x 10^-2 mm)^2 - d0^2 = (40 min) K; (8.0 x 10^-2 mm)^2 - d0^2 = (100 min) K. Solution of these expressions yields a value for d0, the original grain diameter, of d0 = 0.031 mm."
},
{
"idx": 733,
"question": "What grain diameter would you predict after 200 min at 650 degrees Celsius for the brass material, given the original grain diameter d0 = 0.031 mm and K = 5.44 x 10^-5 mm^2/min?",
"answer": "At 200 min, the diameter d is computed using a rearranged form of Equation (incorporating values of d0 and K that were just determined) as follows: d = sqrt(d0^2 + K t) = sqrt((0.031 mm)^2 + (5.44 x 10^-5 mm^2/min)(200 min)) = 0.109 mm."
},
{
"idx": 734,
"question": "An undeformed specimen of some alloy has an average grain diameter of \\(0.050 \\mathrm{~mm}\\). You are asked to reduce its average grain diameter to \\(0.020 \\mathrm{~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 \\(0.050 \\mathrm{~mm}\\) to \\(0.020 \\mathrm{~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 \\mathrm{~mm}\\)."
},
{
"idx": 735,
"question": "A non-cold-worked brass specimen of average grain size \\(0.01 \\mathrm{~mm}\\) has a yield strength of \\(150 \\mathrm{MPa}(21,750 \\mathrm{psi})\\). Estimate the yield strength of this alloy after it has been heated to \\(500^{\\circ} \\mathrm{C}\\) for \\(1000 \\mathrm{~s}\\), if it is known that the value of \\(\\sigma_{0}\\) is \\(25 \\mathrm{MPa}(3625 \\mathrm{psi})\\).",
"answer": "This problem calls for us to calculate the yield strength of a brass specimen after it has been heated to an elevated temperature at which grain growth was allowed to occur; the yield strength (150 MPa) was given at a grain size of \\(0.01 \\mathrm{~mm}\\). It is first necessary to calculate the constant \\(k_{y}\\) in Equation, using values for \\(\\sigma_{y}(150 \\mathrm{MPa}), \\sigma_{0}(25 \\mathrm{MPa})\\), and \\(d(0.01 \\mathrm{~mm})\\), and as follows:\n\\[\n\\begin{aligned}\nk_{y} & =\\frac{\\sigma_{y}-\\sigma_{0}}{d^{-1 / 2}} \\\\\n& =\\frac{150 \\mathrm{MPa}-25 \\mathrm{MPa}}{(0.01 \\mathrm{~mm})^{-1 / 2}} \\\\\n& =12.5 \\mathrm{MPa} \\cdot \\mathrm{mm}^{1 / 2}\n\\end{aligned}\n\\]\nNext, we must determine the average grain size after the heat treatment. From Figure 8.25 at \\(500^{\\circ} \\mathrm{C}\\) after \\(1000 \\mathrm{~s}\\) (16.7 min) the average grain size of a brass material is about \\(0.016 \\mathrm{~mm}\\). Therefore, calculating \\(\\sigma_{y}\\) at this new grain size using Equation we get\n\\[\n\\begin{aligned}\n\\sigma_{y} & =\\sigma_{0}+k_{y} d^{-1 / 2} \\\\\n& =25 \\mathrm{MPa}+(12.5 \\mathrm{MPa} \\cdot \\mathrm{mm}^{1/2})(0.016 \\mathrm{~mm})^{-1 / 2} \\\\\n& =124 \\mathrm{MPa}(18,000 \\mathrm{psi})\n\\end{aligned}\n\\]"
},
{
"idx": 736,
"question": "Normal butane and isobutane have boiling temperatures of \\(-0.5^{\\circ} \\mathrm{C}\\) and \\(-12.3^{\\circ} \\mathrm{C}\\left(31.1^{\\circ} \\mathrm{F}\\right.\\) and \\(9.9^{\\circ} \\mathrm{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": 737,
"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": "This problem gives us the tensile strengths and associated number-average molecular weights for two poly(methyl methacrylate) materials and then asks that we estimate the tensile strength for \\(\\bar{M}_{n}=40,000 \\mathrm{~g} / \\mathrm{mol}\\). Equation cites the dependence of the tensile strength on \\(\\bar{M}_{n} \\sim\\) that is\n\\[\nT S=T S_{\\infty}-\\frac{A}{\\bar{M}_{n}}\n\\]\nUsing data provided in the problem statement, we may set up two simultaneous equations from which it is possible to solve for the two constants \\(T S_{\\infty}\\) and \\(A\\). These equations are as follows:\n\\[\n\\begin{array}{l}\n50 \\mathrm{MPa}=T S_{\\infty}-\\frac{A}{30,000 \\mathrm{~g} / \\mathrm{mol}} \\\\\n150 \\mathrm{MPa}=T S_{\\infty}-\\frac{A} {50,000 \\mathrm{~g} / \\mathrm{mol}}\n\\end{array}\n\\]\nValues of the two constants are: \\(T S_{\\infty}=300 \\mathrm{MPa}\\) and \\(A=7.50 \\times 10^{6} \\mathrm{MPa}-\\mathrm{g} / \\mathrm{mol}\\). Substituting these values into Equation for \\(\\bar{M}_{n}=40,000 \\mathrm{~ g} / \\mathrm{mol}\\) leads to\\[\n\\begin{array}{l}\n\\begin{array}{l}\nT S = T S_{\\infty} - \\displaystyle\\frac{A}{40,000\\, \\mathrm{g/mol}} \\\\\n\\end{array} \\\\\n\\begin{array}{l}\n\\begin{array}{l}\n\\begin{array}{l}\n= 300\\,\\,\\mathrm{MPa} - \\displaystyle\\frac{7.50\\,\\times 10^{6}\\,\\mathrm{MPa-g/mol}}{40,000\\mathrm{g/mol}} \\\\\n\\end{array}\n\\end{array}\n\\end{array} \\\\\n\\begin{array}{l}\n\\begin{array}{l} = 112.5\\,\\mathrm{MPa} \\end{array}\n\\end{array}\n\\end{array}\n\\]"
},
{
"idx": 738,
"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 \\(\\mathrm{MPa}\\).",
"answer": "This problem gives us the tensile strengths and associated number-average molecular weights for two polyethylene materials and then asks that we estimate the \\(\\bar{M}_{n}\\) required for a tensile strength of \\(140 \\mathrm{MPa}\\). Equation cites the dependence of the tensile strength on \\(\\bar{M}_{n}-\\) that is\n\\[\nT S=T S_{\\infty}-\\frac{A}{\\bar{M}_{n}}\n\\]\nUsing data provided in the problem statement, we may set up two simultaneous equations from which it is possible to solve for the two constants \\(T S_{\\infty}\\) and \\(A\\). These equations are as follows:\n\\[\n\\begin{array}{c}\n90 \\mathrm{MPa}=T S_{\\infty}-\\frac{A}{20,000 \\mathrm{~g} / \\mathrm{mol}} \\\\\n180 \\mathrm{MPa}=T S_{\\infty}-\\frac{A}{40,000 \\mathrm{~g} / \\mathrm{mol}}\n\\end{array}\n\\]Values of the two constants are: \\(T S_{\\infty}=270 \\mathrm{MPa}\\) and \\(A=3.6 \\times 10^{6} \\mathrm{MPa}-\\mathrm{g} / \\mathrm{mol}\\). Solving for \\(\\bar{M}_{n}\\) in Equation and substituting \\(T S=140 \\mathrm{MPa}\\) as well as the above values for \\(T S_{\\infty}\\) and \\(A\\) leads to\n\\[\n\\begin{array}{l}\n\\bar{M}_{n}=\\frac{A}{T S_{\\infty}-T S} \\\\\n\\\\=\\frac{3.6 \\times 10^{6} \\mathrm{MPa}-\\mathrm {g} / \\mathrm{mol}}{270 \\mathrm{MPa}-140 \\mathrm{MPa}}=27,700 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]"
},
{
"idx": 739,
"question": "For the pair of polymers: 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, (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.",
"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": 739,
"question": "For the pair of polymers: Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene copolymer with 10% of possible sites crosslinked, (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.",
"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": 739,
"question": "For the pair of polymers: 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, (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.",
"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": 740,
"question": "For the pair of linear and isotactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol and branched and atactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol, (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.",
"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": 740,
"question": "For the pair of graft acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked and alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked, (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.",
"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": 740,
"question": "For the pair of network polyester and lightly branched polytetrafluoroethylene, (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.",
"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": 741,
"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 \\times 10^{-4} \\mathrm{~mm}\\left(7.5 \\times 10^{-6} \\mathrm{in.}\\right)\\) and a crack length of \\(3.8 \\times 10^{-2}\\) \\(\\mathrm{mm}\\left(1.5 \\times 10^{-3}\\right.\\) in.) when a tensile stress of \\(140 \\mathrm{MPa}(20,000 \\mathrm{psi})\\) is applied?",
"answer": "Equation is employed to solve this problem-i.e.,\n\\[\n\\sigma_{m}=2 \\sigma_{0}\\left(\\frac{a}{\\rho_{t}}\\right)^{1 / 2}\n\\]\nValues for \\(\\sigma_{0}\\left(140 \\mathrm{MPa}\\right), 2 a\\left(3.8 \\times 10^{-2} \\mathrm{~mm}\\right)\\), and \\(\\rho_{t}\\left(1.9 \\times 10^{-4} \\mathrm{~mm}\\right)\\) are provided in the problem statement. Therefore, we solve for \\(\\sigma_{m}\\) as follows:\n\\[\n\\sigma_{m}=(2)(140 \\mathrm{MPa})\\left(\\frac{3.8 \\times 10^{-2} \\mathrm{~mm}}{2}\\right)^{1 / 2}=2800 \\mathrm{MPa}(400,000 \\mathrm{psi})\n\\]"
},
{
"idx": 742,
"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 \\mathrm{~mm}(0.02 \\mathrm{in}\\).) and a tip radius of curvature of \\(5 \\times 10^{-3} \\mathrm{~mm}\\left(2 \\times 10^{-4} \\mathrm{in}\\right)\\), when a stress of \\(1035 \\mathrm{MPa}(150,000 \\mathrm{psi})\\) is applied.",
"answer": "In order to estimate the theoretical fracture strength of this material it is necessary to calculate \\(\\sigma_{m}\\) using Equation given that \\(\\sigma_{0}=1035 \\mathrm{MPa}, a=0.5 \\mathrm{~mm}\\), and \\(\\rho_{t}=5 \\times 10^{-3} \\mathrm{~mm}\\). Thus,\n\\[\n\\begin{aligned}\n\\sigma_{m} & =2 \\sigma_{0}\\left(\\frac{a}{\\rho_{t}}\\right)^{1 / 2} \\\\\n& =(2)(1035 \\mathrm{MPa})\\left(\\frac{0.5 \\mathrm{~mm}}{5 \\times 10^{-3} \\mathrm{~mm}}\\right)^{1 / 2} \\\\\n& =2.07 \\times 10^{4} \\mathrm{MPa}=20.7 \\mathrm{GPa} \\quad\\left(3 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]"
},
{
"idx": 743,
"question": "A specimen of a 4340 steel alloy with a plane strain fracture toughness of \\(54.8 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) (50 ksi \\(\\sqrt{\\mathrm{m}}\\).) is exposed to a stress of \\(1030 \\mathrm{MPa}(150,000 \\mathrm{psi})\\). Will this specimen experience fracture if the largest surface crack is \\(0.5 \\mathrm{~mm}\\) ( 0.02 in.) long? Why or why not? Assume that the parameter \\(Y\\) has a value of 1.0 .",
"answer": "This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of \\(1030 \\mathrm{MPa}\\). This requires that we solve for \\(\\sigma_{c}\\) from Equation, given the values of \\(K_{I c}(54.8 \\mathrm{MPa} \\sqrt{\\mathrm{m}})\\), the largest value of \\(a(0.5 \\mathrm{~mm})\\), and \\(Y(1.0)\\). Thus, the critical stress for fracture is equal to\n\\[\n\\begin{aligned}\n\\sigma_{c} & =\\frac{K_{I c}}{Y \\sqrt{\\pi a}} \\\\\n& =\\frac{54.8 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(1.0) \\sqrt{(\\pi)(0.5 \\times 10^{-3} \\mathrm{~m})}} \\\\\n& =1380 \\mathrm{MPa}(200,000 \\mathrm{psi})\n\\end{aligned}\n\\]\nTherefore, fracture will not occur because this specimen will tolerate a stress of \\(1380 \\mathrm{MPa}\\) (200,000 psi) before fracture, which is greater than the applied stress of \\(1030 \\mathrm{MPa}(150,00\\) psi \\()\\)."
},
{
"idx": 744,
"question": "For an aluminum alloy with a plane strain fracture toughness of 40 MPa sqrt(m), determine the parameter Y when fracture occurs at a stress of 300 MPa and a maximum internal crack length of 4.0 mm.",
"answer": "Y is calculated using the formula Y = K_IC / (σ sqrt(π a)), where K_IC = 40 MPa sqrt(m), σ = 300 MPa, and a = 2.0 mm (half of the maximum internal crack length). Substituting these values, Y = 40 MPa sqrt(m) / (300 MPa sqrt(π × 2.0 × 10^-3 m)) = 1.68."
},
{
"idx": 744,
"question": "For the same component and alloy, calculate the product Y σ sqrt(π a) when the stress level is 260 MPa and the maximum internal crack length is 6.0 mm, using the previously determined Y value of 1.68.",
"answer": "The product Y σ sqrt(π a) is calculated as (1.68)(260 MPa) sqrt(π × 3.0 × 10^-3 m) = 42.4 MPa sqrt(m)."
},
{
"idx": 744,
"question": "Determine if fracture will occur at a stress level of 260 MPa and a maximum internal crack length of 6.0 mm, given that the plane strain fracture toughness (K_IC) of the alloy is 40 MPa sqrt(m).",
"answer": "Fracture will occur because the calculated Y σ sqrt(π a) value of 42.4 MPa sqrt(m) is greater than the K_IC of the material, which is 40 MPa sqrt(m)."
},
{
"idx": 745,
"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 \\mathrm{MPa} \\sqrt{\\mathrm{m}}(23.7 \\mathrm{ksi} \\sqrt{\\mathrm{m}}\\).). It has been determined that fracture results at a stress of \\(112 \\mathrm{MPa}(16,240 \\mathrm{psi})\\) when the maximum internal crack length is \\(8.6 \\mathrm{~mm}(0.34 \\mathrm{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 \\mathrm{~mm}(0.24 \\mathrm{in.})\\).",
"answer": "This problem asks us to determine the stress level at which an a wing component on an aircraft will fracture for a given fracture toughness \\((26 \\mathrm{MPa} \\sqrt{\\mathrm{m}})\\) and maximum internal crack length \\((6.0 \\mathrm{~mm})\\), given that fracture occurs for the same component using the same alloy at one stress level (112 MPa) and another internal crack length (8.6 mm). (Note: Because the cracks are internal, their lengths are equal to \\(2 a\\).) It first becomes necessary to solve for the parameter \\(Y\\) for the conditions under which fracture occurred using Equation. Therefore,\n\\[\nY=\\frac{K_{I c}}{\\sigma \\sqrt{\\pi a}}=\\frac{26 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{\\left(112 \\mathrm{MPa}\\right) \\sqrt{(\\pi)\\left(\\frac{8.6 \\times 10^{-3} \\mathrm{~m}}{2}\\right)}=2.0}\n\\]\nNow we will solve for \\(\\sigma_{c}\\) (for a crack length of \\(6 \\mathrm{~mm}\\) ) using Equation as\n\\[\n\\sigma_{c}=\\frac{K_{I c}}{Y \\sqrt{\\pi a}}=\\frac{26 \\mathrm{MPa} \\sqrt{m}}{(2.0) \\sqrt{(\\pi)\\left(\\frac{6 \\times 10^{-3} \\mathrm{~m}}{2}\\right)}}=134 \\mathrm{MPa}(19,300 \\mathrm{psi})\n\\]"
},
{
"idx": 746,
"question": "What is the value of the design parameter Y for a structural component made from an alloy with a plane-strain fracture toughness of 62 MPa sqrt(m), failing at a stress of 250 MPa when the maximum length of a surface crack is 1.6 mm?",
"answer": "The design parameter Y is calculated using the formula Y = K_IC / (σ sqrt(π a_C)), where K_IC = 62 MPa sqrt(m), σ = 250 MPa, and a_C = 1.6 x 10^-3 m. Substituting these values, Y = 62 / (250 sqrt(π x 1.6 x 10^-3)) = 3.50."
},
{
"idx": 746,
"question": "What is the maximum allowable surface crack length (in mm) without fracture for the 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), using the previously calculated Y value of 3.50?",
"answer": "The maximum crack length a_C is calculated using the formula a_C = 1/π (K_IC / (σ Y))^2, where K_IC = 51 MPa sqrt(m), σ = 250 MPa, and Y = 3.50. Substituting these values, a_C = 1/π (51 / (250 x 3.50))^2 = 0.00108 m = 1.08 mm."
},
{
"idx": 747,
"question": "A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of \\(82.4 \\mathrm{MPa} \\sqrt{\\mathrm{m}}(75.0 \\mathrm{ksi} \\sqrt{\\mathrm{m}}\\).). If the plate is exposed to a tensile stress of \\(345 \\mathrm{MPa}(50,000 \\mathrm{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": "For this problem, we are given values of \\(K_{I_{C}}\\left(82.4 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right), \\sigma\\) (345 MPa), and \\(Y(1.0)\\) for a large plate and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is to solve for \\(a_{c}\\) using Equation; therefore\n\\[\n\\begin{aligned}\na_{c} & =\\frac{1}{\\pi}\\left(\\frac{K_{I_{C}}}{Y \\sigma}\\right)^{2} \\\\\n& =\\frac{1}{\\pi}\\left[\\frac{82.4 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(1.0)(345 \\mathrm{MPa})}\\right]^{2} \\\\\n& =0.0182 \\mathrm{~m}=18.2 \\mathrm{~mm} \\quad(0.72 \\mathrm{in} .)\n\\end{aligned}\n\\]"
},
{
"idx": 748,
"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 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) ( \\(90 \\mathrm{ksi} \\sqrt{\\mathrm{in}}\\).) and a yield strength of 860 \\(\\mathrm{MPa}\\) (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is \\(3.0 \\mathrm{~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": "This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus \\((3.0 \\mathrm{~mm})\\), the value of \\(K_{I_{C}}\\) \\(\\left(98.9 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right)\\), the design stress \\(\\left(\\sigma_{y} / 2\\right.\\) in which \\(\\sigma_{y}=860 \\mathrm{MPa}\\) ), and \\(Y=1.0\\). We first need to compute the value of \\(a_{C}\\) using Equation; thus\n\\[\n\\begin{aligned}\na_{c} & =\\frac{1}{\\pi}\\left(\\frac{K_{I_{C}}}{\\sigma Y}\\right)^{2} \\\\\n& =\\frac{1}{\\pi}\\left[\\frac{98.9 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(1.0)\\left(\\frac{860 \\mathrm{MPa}}{2}\\right)}\\right] \\\\\n& =0.0168 \\mathrm{~m}=16.8 \\mathrm{~mm} \\quad(0.66 \\mathrm{in} .)\n\\end{aligned}\n\\]\nTherefore, the critical flaw is subject to detection since this value of \\(a_{c}(16.8 \\mathrm{~mm})\\) is greater than the \\(3.0 \\mathrm{~mm}\\) resolution limit."
},
{
"idx": 749,
"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": 749,
"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 probability of the existence of a flaw that is capable of initiating a crack diminishes."
},
{
"idx": 750,
"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": "This problem asks that we compute the crack tip radius ratio before and after etching. Let\n\\[\n\\begin{array}{c}\nt=\\text { original crack tip radius, and } \\\\\nt=\\text { etched crack tip radius }\n\\end{array}\n\\]\nAlso, as given in the problem statement\n\\[\n\\begin{array}{c}\n\\sigma_{f}^{\\prime}=\\sigma_{f} \\\\\n\\\\\na^{\\prime}=\\frac{a}{2} \\\\\n\\\\\n\\sigma_{0}^{\\prime}=4 \\sigma_{0}\n\\end{array}\n\\]\nWhen we incorporate the above relationships into two expressions of Equation as follows:\n\\[\n\\sigma_{f}=2 \\sigma_{0}\\left(\\frac{a}{\\rho_{t}}\\right)^{1 / 2}=\\sigma_{f}^{\\prime}=2 \\sigma_{0}^{\\prime}\\left(\\frac{a^{\\prime}}{\\rho_{t}^{\\prime}}\\right)^{1 / 2}\n\\]\nWe now solve for the \\(\\stackrel{t}{t}\\) ratio, which yields the following:\\[\n\\frac{\\rho_{t}^{'}}{\\rho_{t}}=\\left(\\frac{\\sigma_{0}^{'}}{\\sigma_{0}}\\right)^{2}\\left(\\frac{a^{'}}{a}\\right)=\\left(\\frac{4\\sigma_{0}}{\\sigma_{0}}\\right)^{2}\\left(\\frac{a/2}{a}\\right)=8\n\\]\nwhich is the solution requested in the problem statement."
},
{
"idx": 751,
"question": "What is the maximum carbon content possible for a plain carbon steel that must have an impact energy of at least \\(200 \\mathrm{~J}\\) at \\(-50^{\\circ} \\mathrm{C}\\) ?",
"answer": "From the curves in Figure 9.23, for only the \\(0.11 \\mathrm{wt} \\% \\mathrm{C}\\) and \\(0.01 \\mathrm{wt} \\% \\mathrm{C}\\) steels are impact energies (and therefore, ductile-to-brittle temperatures) greater than \\(200 \\mathrm{~J}\\). Therefore, \\(0.11 \\mathrm{wt} \\%\\) is the maximum carbon concentration."
},
{
"idx": 752,
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for tin (Sn), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
"answer": "For Sn, 0.4 Tm = (0.4)(232 + 273) = 202 K or -71 C (-96 F)"
},
{
"idx": 752,
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for molybdenum (Mo), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
"answer": "For Mo, 0.4 Tm = (0.4)(2617 + 273) = 1049 K or 776 C (1429 F)"
},
{
"idx": 752,
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for iron (Fe), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
"answer": "For Fe, 0.4 Tm = (0.4)(1538 + 273) = 724 K or 451 C (845 F)"
},
{
"idx": 752,
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for gold (Au), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
"answer": "For Au, 0.4 Tm = (0.4)(1064 + 273) = 535 K or 262 C (504 F)"
},
{
"idx": 752,
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for zinc (Zn), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
"answer": "For Zn, 0.4 Tm = (0.4)(420 + 273) = 277 K or 4 C (39 F)"
},
{
"idx": 752,
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for chromium (Cr), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
"answer": "For Cr, 0.4 Tm = (0.4)(1875 + 273) = 859 K or 586 C (1087 F)"
},
{
"idx": 753,
"question": "What is the stress imposed on a cylindrical specimen of S-590 alloy with a diameter of 13.2 mm when exposed to a tensile load of 27,000 N?",
"answer": "The stress is calculated using the formula sigma = F / A0, where F is the force and A0 is the cross-sectional area. Given d0 = 13.2 mm (13.2 x 10^-3 m) and F = 27,000 N, the stress is equal to sigma = 27,000 N / (pi * (13.2 x 10^-3 m / 2)^2) = 197 x 10^6 N/m^2 = 197 MPa, or approximately 200 MPa."
},
{
"idx": 753,
"question": "At approximately what temperature will the steady-state creep be 10^-3 h^-1 for the S-590 alloy under a stress of 200 MPa?",
"answer": "From Figure 9.40, the point corresponding to 10^-3 h^-1 and 200 MPa lies approximately midway between 730°C and 815°C lines; therefore, the temperature would be approximately 775°C."
},
{
"idx": 754,
"question": "Steady-state creep rate data are given in the following table for a nickel alloy at \\(538^{\\circ} \\mathrm{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 \\\\\n\\(10^{-6}\\) & 36.1 \\\\\n\\hline\n\\end{tabular}\nCompute the stress at which the steady-state creep is \\(10^{-5} \\mathrm{~h}^{-1}\\) (also at \\(538^{\\circ} \\mathrm{C}\\) ).",
"answer": "This problem gives \\(\\dot{\\varepsilon}_{s}\\) values at two different stress levels and \\(538^{\\circ} \\mathrm{C}\\), and asks that we determine the stress at which the steady-state creep rate is \\(10^{-5} \\mathrm{~h}^{-1}\\) (alsoat \\(538^{\\circ} \\mathrm{C}\\) ). It is possible to solve this problem using Equation. First of all, it is necessary to determine values of \\(K_{1}\\) and \\(n\\). Taking the logarithms of both sides of Equation leads to the following expression:\n\\[\n\\log \\dot{\\varepsilon}_{s}=\\log K_{1}+n \\log \\sigma\n\\]\nWe can now generate two simultaneous equations from the data given in the problem statement in which the two unknowns are \\(K_{1}\\) and \\(n\\). These two equations are as follows:\n\\[\n\\begin{array}{l}\n\\log \\left(10^{-7}\\right)=\\log K_{1}+n \\log (22.0 \\mathrm{MPa}) \\\\\n\\log \\left(10^{-6}\\right)=\\log K_{1}+n \\log (36.1 \\mathrm{MPa})\n\\end{array}\n\\]\nSimultaneous solution to these expressions leads to the following:\n\\[\nK_{1}=6.12 \\times 10^{-14}\n\\]\\[\nn=4.63\n\\]\nUsing these values we may determine the stress at which \\(\\dot{\\varepsilon_{s}}=10^{-5} \\mathrm{~h}^{-1}\\). Let us rearrange Equation above such that \\(\\log \\sigma\\) is the dependent variable:\n\\[\n\\log \\sigma=\\frac{\\log \\dot{\\varepsilon_{s}}-\\log K_{1}}{n}\n\\]\nIncorporation of values of \\(\\dot{\\varepsilon_{s}}, K_{1}\\), and \\(n\\) yields the following:\n\\[\n\\begin{array}{c}\n\\log \\sigma=\\frac{\\log \\left(10^{-5}\\right)-\\log \\left(6.12 \\times 10^{-14}\\right)}{4.63} \\\\\n=1.774\n\\end{array}\n\\]\nFrom which we determine \\(\\sigma\\) as follows:\n\\[\n\\sigma=10^{1.774}=59.4 \\mathrm{MPa}\n\\]"
},
{
"idx": 755,
"question": "Given the steady-state creep rate data at 200°C (473 K) for an alloy with activation energy for creep of 140,000 J/mol, determine the stress exponent (n) and constant (K2) using the following data: 2.5×10−3 h−1 at 55 MPa and 2.4×10−2 h−1 at 69 MPa.",
"answer": "Using the equation ln(˙εs) = ln(K2) + n ln(σ) − (Qc/RT), set up two equations with the given data: ln(2.5×10−3 h−1) = ln(K2) + n ln(55 MPa) − (140,000 J/mol)/(8.31 J/mol·K)(473 K) and ln(2.4×10−2 h−1) = ln(K2) + n ln(69 MPa) − (140,000 J/mol)/(8.31 J/mol·K)(473 K). Solving these simultaneously yields n = 9.97 and K2 = 3.27×10−5 h−1."
},
{
"idx": 755,
"question": "Using the determined stress exponent (n = 9.97) and constant (K2 = 3.27×10−5 h−1), compute the steady-state creep rate at 250°C (523 K) and a stress level of 48 MPa (7000 psi).",
"answer": "Using the equation ˙εs = K2 σ^n exp(−Qc/RT), substitute the values: ˙εs = (3.27×10−5 h−1)(48 MPa)^9.97 exp[−(140,000 J/mol)/(8.31 J/mol·K)(523 K)]. The calculation results in a steady-state creep rate of 1.94×10−2 h−1."
},
{
"idx": 756,
"question": "Given the steady-state creep data for an iron at a stress level of 140 MPa (20,000 psi) with strain rates of 6.6×10⁻⁴ h⁻¹ at 1090 K and 8.8×10⁻² h⁻¹ at 1200 K, and a stress exponent n=8.5, determine the values of K₂ and Qₐ.",
"answer": "Using the equation ln(ε̇ₛ) = ln(K₂) + n ln(σ) - (Qₐ)/(RT), we set up two equations:\n1. ln(6.6×10⁻⁴ h⁻¹) = ln(K₂) + 8.5 ln(140 MPa) - (Qₐ)/(8.31 J/mol·K × 1090 K)\n2. ln(8.8×10⁻² h⁻¹) = ln(K₂) + 8.5 ln(140 MPa) - (Qₐ)/(8.31 J/mol·K × 1200 K)\nSolving these simultaneously yields K₂ = 57.5 h⁻¹ and Qₐ = 483,500 J/mol."
},
{
"idx": 756,
"question": "Using the determined values of K₂=57.5 h⁻¹ and Qₐ=483,500 J/mol, compute the steady-state creep rate at a stress level of 83 MPa (12,000 psi) and temperature of 1300 K.",
"answer": "Applying the equation ε̇ₛ = K₂ σⁿ exp(-Qₐ/(RT)):\nε̇ₛ = (57.5 h⁻¹)(83 MPa)^8.5 exp[-483,500 J/mol / (8.31 J/mol·K × 1300 K)]\nThe calculated steady-state creep rate is 4.31×10⁻² h⁻¹."
},
{
"idx": 757,
"question": "If ice homogeneously nucleates at \\(-40^{\\circ} \\mathrm{C}\\), calculate the critical radius given values of \\(-\\) \\(3.1 \\times 10^{8} \\mathrm{~J} / \\mathrm{m}^{3}\\) and \\(25 \\times 10^{-3} \\mathrm{~J} / \\mathrm{m}^{2}\\), respectively, for the latent heat of fusion and the surface free energy.",
"answer": "This problem states that ice homogeneously nucleates at \\(-40^{\\circ} \\mathrm{C}^{\\text {, and we are to calculate }}\\) the critical radius given the latent heat of fusion \\(\\left(-3.1 \\times 10^{8} \\mathrm{~J} / \\mathbf{m}^{3}\\right)\\) and the surface free energy \\(\\left(25 \\times 10^{-3} \\mathrm{~J} / \\mathrm {m}^{2}\\right)\\). Solution to this problem requires the utilization of Equation as\n\\[\n\\begin{aligned}\nr * & =\\left(-\\frac{2 \\gamma T_{m}}{\\Delta H_{f}}\\right)\\left(\\frac{1}{T_{m}-T}\\right) \\\\\n& =\\left[-\\frac{(2)\\left(25 \\times 10^{-3} \\mathrm{~J} / m^{2}\\right)(273 \\mathrm{~K})}{-3.1 \\times 10^{8} \\mathrm{~J} / \\max ^{3}}\\right]\\left(\\frac{1}{40 \\mathrm{~K}}\\right) \\\\\n& =1.10 \\quad 10^{9} \\mathrm{~m}=1.10 \\mathrm{~nm}\n\\end{aligned}\n\\]"
},
{
"idx": 758,
"question": "Given the Avrami equation for recrystallization kinetics with n=5.0, and a fraction recrystallized y=0.30 after t=100 min, calculate the value of the rate constant k.",
"answer": "To calculate k, rearrange the Avrami equation: exp(-k t^n)=1-y. Take natural logarithms: -k t^n=ln(1-y). Solve for k: k=-ln(1-y)/t^n. Substituting y=0.30, t=100 min, n=5.0: k=-ln(1-0.30)/(100 min)^5=3.57×10^-11."
},
{
"idx": 758,
"question": "Using the calculated k=3.57×10^-11 and n=5.0, determine the time t_0.5 for the fraction recrystallized to reach y=0.5.",
"answer": "Rearrange the Avrami equation to solve for t: t^n=-ln(1-y)/k. For t_0.5: t_0.5=[-ln(1-0.5)/k]^(1/n). Substituting k=3.57×10^-11 and n=5.0: t_0.5=[-ln(1-0.5)/3.57×10^-11]^(1/5)=114.2 min."
},
{
"idx": 758,
"question": "Calculate the rate of recrystallization using the determined t_0.5=114.2 min.",
"answer": "The rate of recrystallization is the reciprocal of t_0.5: rate=1/t_0.5=1/114.2 min=8.76×10^-3 min^-1."
},
{
"idx": 759,
"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": "he first thing necessary is to set up two expressions of the form of Equation, and then to solve simultaneously for the values of \\(n\\) and \\(k\\). In order to expedite this process, we will rearrange and do some algebraic manipulation of Equation. First of all, we rearrange as follows:\n\\[\n1 \\quad y=\\exp \\left(k t^{n}\\right)\n\\]\nNow taking natural logarithms\n\\[\n\\ln (1 \\quad y)=k t^{n}\n\\]\nOr\n\\[\n\\ln (1 \\quad y)=k t^{n}\n\\]\nWhich may also be expressed as\n\\[\n\\ln \\left(\\frac{1}{1-y}\\right)=k t^{n}\n\\]\nNow taking natural logarithms again of both sides of this equation, leads to\n\\[\n\\ln \\left[\\ln \\left(\\frac{1}{1-y}\\right)\\right]=\\ln k+n \\ln t\n\\]Which is the form of the equation that we will now use. Using values cited in the problem statement, the two equations are thus\n\\[\n\\begin{array}{l}\n\\ln \\left\\{\\ln \\left[\\frac{1}{1} \\frac{1}{0.2}\\right]\\right\\}=\\ln k+n \\ln (280 \\mathrm{~s}) \\\\\n\\ln \\left\\{\\ln \\left[\\frac{1}{1} \\frac{2}{0.6}\\right]\\right\\}=\\ln k+n \\ln (425 \\mathrm{~s})\n\\end{array}\n\\]\nSolving these two expressions simultaneously for \\(n\\) and \\(k\\) yields \\(n=3.385\\) and \\(k=1.162 \\times 10^{-9}\\) for time in seconds.\nNow it becomes necessary to solve for the value of \\(t\\) at which \\(y=0.95\\). Equation given above may be rewritten as follows:\n\\[\nt^{n}=\\frac{\\ln (1 \\quad y)}{k}\n\\]\nAnd solving for \\(t\\) leads to\n\\[\nt=\\left[-\\frac{\\ln (1-y)}{k}\\right]^{1 / n}\n\\]\nNow incorporating into this expression values for \\(n\\) and \\(k\\) determined above, the time required for \\(95 \\%\\) austenite transformation is equal to\n\\[\nt=\\left[-\\frac{\\ln (1-0.95)}{1.162 \\times 10^{-9}}\\right]^{1 / 3.385}=603 \\mathrm{~s}\n\\]"
},
{
"idx": 760,
"question": "The fraction recrystallized-time data for the recrystallization at \\(350^{\\circ} \\mathrm{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 \\(116.8 \\mathrm{~min}\\).\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 first thing necessary is to set up two expressions of the form of Equation, and then to solve simultaneously for the values of \\(n\\) and \\(k\\). In order to expedite this process, we will rearrange and do some algebraic manipulation of Equation. First of all, we rearrange as follows:\n\\[\n1 \\quad y=\\exp \\left(k t^{n}\\right)\n\\]\nNow taking natural logarithms\n\\[\n\\ln (1 \\quad y)=k t^{n}\n\\]\nOr\n\\[\n\\ln (1 \\quad y)=k t^{n}\n\\]\nWhich may also be expressed as\n\\[\n\\ln \\left(\\frac{1}{1-y}\\right)=k t^{n}\n\\]\nNow taking natural logarithms again of both sides of the above equation, leads to\\[\n\\ln \\left\\{\\ln \\left(\\frac{1}{1} \\quad \\right) \\right\\} = \\ln k + n \\ln t\n\\]\nWhich is the form of the equation that we will now use. The two equations are as follows:\n\\[\n\\begin{array}{l}\n\\ln \\left\\{\\ln \\left[\\frac{1}{1}-\\frac{1}{0.30}\\right]\\right\\}=\\ln k+n \\ln (95.2 \\,\\min ) \\\\\n\\ln \\left\\{\\ln \\left[\\frac{1}{1}-\\frac{2}{0.80}\\right]\\right\\}=\\ln k+n \\ln (126.6 \\,\\min )\n\\end{array}\n\\]\nSolving these two expressions simultaneously for \\(n\\) and \\(k\\) yields \\(n=5.286\\) and \\(k=1.239 \\times 10^{-11}\\) for time in minutes.\nNow it becomes necessary to solve for \\(y\\) when \\(t=116.8 \\mathrm{~min}\\). Application of Equation leads to\n\\[\n\\begin{array}{c}\ny=1 \\quad \\exp \\left(k t^{n}\\right) \\\\\n=1-\\exp \\left[-\\left(1.239 \\times 10^{-11}\\right)(116.8 \\mathrm{\\ min })^{5.286}\\right]=0.65\n\\end{array}\n\\]"
},
{
"idx": 761,
"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": 761,
"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": 762,
"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": 763,
"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} \\mathrm{C}\\left(212^{\\circ} \\mathrm{F}\\right)\\). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of \\(100^{\\circ} \\mathrm{C}\\) or above, and would be suitable for this application."
},
{
"idx": 764,
"question": "For the pair of polymers: branched polyethylene having a number-average molecular weight of 850,000 g/mol and linear polyethylene having a number-average molecular weight of 850,000 g/mol, (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.",
"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": 764,
"question": "For the pair of polymers: polytetrafluoroethylene having a density of 2.14 g/cm³ and a weight-average molecular weight of 600,000 g/mol and PTFE having a density of 2.20 g/cm³ and a weight-average molecular weight of 600,000 g/mol, (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.",
"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": 764,
"question": "For the pair of polymers: linear and syndiotactic poly(vinyl chloride) having a number-average molecular weight of 500,000 g/mol and linear polyethylene having a number-average molecular weight of 225,000 g/mol, (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.",
"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": 764,
"question": "For the pair of polymers: linear and syndiotactic polypropylene having a weight-average molecular weight of 500,000 g/mol and linear and atactic polypropylene having a weight-average molecular weight of 750,000 g/mol, (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.",
"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."
},
{
"idx": 765,
"question": "Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length in which a current of 0.25 A passes in an axial direction. A voltage of 24 V is measured across two probes that are separated by 45 mm (1.75 in.).",
"answer": "This problem calls for us to compute the electrical conductivity of a silicon specimen. We use Equations for the conductivity, as follows: sigma = 1/rho = (I l)/(V A) = (I l)/(V pi (d/2)^2). Since the cross-sectional area of a cylinder having a diameter d is A = pi (d/2)^2. Now, incorporation of values for the several parameters provided in the problem statement, leads to the following: sigma = (I l)/(V pi (d/2)^2) = (0.25 A)(45 x 10^-3 m)/(24 V)(pi)((7.0 x 10^-3 m)/2)^2 = 12.2 (Ohm m)^-1."
},
{
"idx": 765,
"question": "Compute the resistance over the entire 57 mm (2.25 in.) of the specimen with diameter 7.0 mm (0.28 in.), given the electrical conductivity sigma = 12.2 (Ohm m)^-1.",
"answer": "The resistance, R, over the entire specimen length may be computed by combining and rearranging Equations as follows: R = (l rho)/A = (l (1/sigma))/A = l/(sigma A) = l/(sigma pi (d/2)^2). The value of R may be computed upon entering values given in the problem statement into the preceding expression: R = l/(sigma pi (d/2)^2) = 57 x 10^-3 m/[12.2 (Ohm m)^-1](pi)((7.0 x 10^-3 m)/2)^2 = 121.4 Ohm."
},
{
"idx": 766,
"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": 767,
"question": "In terms of electron energy band structure, why do metals have high electrical conductivity?",
"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."
},
{
"idx": 767,
"question": "In terms of electron energy band structure, why do semiconductors and insulators have lower electrical conductivity compared to metals?",
"answer": "There 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."
},
{
"idx": 767,
"question": "In terms of electron energy band structure, why do semiconductors have higher electrical conductivity than insulators?",
"answer": "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": 768,
"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": 769,
"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": 769,
"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": 770,
"question": "Calculate the drift velocity of electrons in silicon at room temperature when the magnitude of the electric field is 500 V/m. The room temperature mobility of electrons is 0.145 m2/V-s.",
"answer": "The drift velocity of electrons in Si may be determined using the equation vd = μeE. Since the room temperature mobility of electrons is 0.145 m2/V-s, and the electric field is 500 V/m, the drift velocity is equal to vd = (0.145 m2/V-s)(500 V/m) = 72.5 m/s."
},
{
"idx": 770,
"question": "Under the circumstances where the drift velocity of electrons in silicon is 72.5 m/s, how long does it take an electron to traverse a 25-mm (1 in.) length of crystal?",
"answer": "The time, t, required to traverse a given length, l (=25 mm), is just the length divided by the drift velocity, as follows: t = l/vd = (25 x 10-3 m)/(72.5 m/s) = 3.45 x 10-4 s."
},
{
"idx": 771,
"question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 × 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 may be computed using rearranged form of Equation as follows: n = σ / |e| μ_e = 3.8 × 10^7 (Ω·m)^-1 / (1.602 × 10^-19 C)(0.0012 m^2/V·s) = 1.98 × 10^29 m^-3"
},
{
"idx": 771,
"question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 × 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": "In order to calculate the number of free electrons per aluminum atom, we must first determine the number of aluminum atoms per cubic meter, N_Al. From Equation (and using the atomic weight for Al found inside the front cover-viz. 26.98 g/mol) then the values of N_Al is determined as follows: N_Al = (N_A ρ') / A_Al = (6.022 × 10^23 atoms/mol)(2.7 g/cm^3)(10^6 cm^3/m^3) / 26.98 g/mol = 6.03 × 10^28 m^-3. And, finally, the number of free electrons per aluminum atom (n/N_Al) is equal to the following: n/N_Al = 1.98 × 10^29 m^-3 / 6.03 × 10^28 m^-3 = 3.28 electrons/Al atom"
},
{
"idx": 772,
"question": "At room temperature the electrical conductivity of \\(P b S\\) is \\(25\\left(\\Omega \\cdot m\\right)^{-1}\\), whereas the electron and hole mobilities are 0.06 and \\(0.02 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\\), respectively. Compute the intrinsic carrier concentration for \\(\\mathrm{PbS}\\) at room temperature.",
"answer": "In this problem we are asked to compute the intrinsic carrier concentration for \\(\\mathrm{PbS}\\) at room temperature. Since the conductivity and both electron and hole mobilities are provided in the problem statement, all we need do is solve for \\(n\\) or \\(p\\) (i.e., \\(n_{i}\\) ) using Equation. Making the appropriate rearrangement of Equation such that \\(n_{i}\\) is the dependent variable leads to the following:\n\\[\n\\begin{array}{l}\nn_{i}=\\frac{\\sigma}{\\left|e\\right|\\left(\\mu_{e}+\\mu_{h}\\right)} \\\\\n=\\frac{25(\\Omega \\cdot \\mathrm{m})^{-1}}{\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)(0.06+0.02) \\mathrm{m}^{2} \\mathrm{~V} \\cdot \\mathrm{s}} \\\\\n=1.95 \\times 10^{21} \\mathrm{~m}^{-3}\n\\end{array}\n\\]"
},
{
"idx": 773,
"question": "An \\(n\\)-type semiconductor is known to have an electron concentration of \\(5 \\times 10^{17} \\mathrm{~m}^{-3}\\). If the electron drift velocity is \\(350 \\mathrm{~m} / \\mathrm{s}\\) in an electric field of \\(1000 \\mathrm{~V} / \\mathrm{m}\\), calculate the conductivity of this material.",
"answer": "For this problem we are to determine the electrical conductivity of and \\(n\\)-type semiconductor, given that \\(n=5 \\times 10^{17} \\mathrm{~m}^{-3}\\) and the electron drift velocity is \\(350 \\mathrm{~m} / \\mathrm{s}^{\\text {in }}\\) an electric field of \\(1000 \\mathrm{~V} / \\mathrm{ m}\\). The conductivity of this material may be computed using Equation . But before this is possible, it is necessary to calculate the value of \\(\\mu_{e}\\) using Equation, as follows:\n\\[\n\\begin{aligned}\n\\mu_{e} & =\\frac{v_{d}}{E} \\\\\n& =\\frac{350 \\mathrm{~m} / \\mathrm{s}}{1000 \\mathrm{~V} / \\mathrm{m}}=0.35 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\n\\end{aligned}\n\\]\nIt is now possible to compute the conductivity using Equation,\n\\[\n\\begin{aligned}\n\\sigma & =n|e| \\mu_{e} \\\\\n& =\\left(5 \\times 10^{17} \\mathrm{~m}^{-3}\\right)\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)\\left(0.35 \\mathrm{~m}^{2} / \\mathrm{V}^{-} \\mathrm{s}\\right) \\\\\n& =0.028(\\Omega \\cdot \\mathrm{m})^{-1}\n\\end{aligned}\n\\]"
},
{
"idx": 774,
"question": "Predict whether nitrogen (N) will act as a donor or an acceptor when added to silicon (Si) as a substitutional impurity.",
"answer": "Nitrogen will act as a donor in Si. Since N is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
},
{
"idx": 774,
"question": "Predict whether boron (B) will act as a donor or an acceptor when added to germanium (Ge) as a substitutional impurity.",
"answer": "Boron will act as an acceptor in Ge. Since B is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom."
},
{
"idx": 774,
"question": "Predict whether sulfur (S) will act as a donor or an acceptor when added to indium antimonide (InSb) as a substitutional impurity.",
"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": 774,
"question": "Predict whether indium (In) will act as a donor or an acceptor when added to cadmium sulfide (CdS) as a substitutional impurity.",
"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": 774,
"question": "Predict whether arsenic (As) will act as a donor or an acceptor when added to zinc telluride (ZnTe) as a substitutional impurity.",
"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": 775,
"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. (Arsenic is from group VA of the periodic table--Ge is from group IVA.)"
},
{
"idx": 775,
"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": "Since this material is n-type extrinsic, Equation is valid. Furthermore, each As atom will donate a single electron, or the electron concentration is equal to the As concentration since all of the As atoms are ionized at room temperature; that is n=10^24 m^-3, and, as given in the problem statement, μe=0.1 m^2/V-s. Thus, the conductivity is equal to σ=n|e| μe =(10^24 m^-3)(1.602×10^-19 C)(0.1 m^2/V-s) =1.60×10^4 (Ω·m)^-1"
},
{
"idx": 776,
"question": "A metal alloy is known to have electrical conductivity and electron mobility values of 1.2 \\(\\times 10^{7}(\\Omega \\cdot \\mathrm{m})^{-1}\\) and \\(0.0050 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\\), respectively. A current of \\(40 \\mathrm{~A}\\) is passed through a specimen of this alloy that is \\(35 \\mathrm{~mm}\\) thick. What magnetic field would need to be imposed to yield a Hall voltage of \\(-3.5 \\times 10^{-7} \\mathrm{~V}\\) ?",
"answer": "In this problem we are asked to determine the magnetic field required to produce a Hall voltage of \\(-3.5 \\times 10^{-7} \\mathrm {~V}\\), given that \\(\\sigma=1.2 \\times 10^{7}(\\Omega \\cdot \\mathrm{m})^{-1}, \\mu_{e}=0.0050 \\mathrm{~m}^{2} / \\mathrm{v} \\cdot \\mathrm{s}, I_{x}=40 \\mathrm{~A}\\), and \\(d=35 \\mathrm{~mm}\\). We now take a rearranged form of Equation\n\\[\n\\left|R_{\\mathrm{H}}\\right|=\\frac{\\mu_{e}}{\\sigma}\n\\]\nand incorporate into it Equation as follows:\n\\[\n\\begin{aligned}\nV_{\\mathrm{H}} & =\\frac{R_{\\mathrm{H}} I_{x} B_{z}}{d} \\\\\n& =\\frac{e^{I_{x} B_{z}}}{d}\n\\end{aligned}\n\\]\nWe now rearrange this expression such that \\(B_{z}\\) is the dependent variable, and after inserting values for the remaining parameters that were given in the problem statement, the value of \\(B_{z}\\) is determined as follows:\n\\[\n\\begin{aligned}\nB_{z} & =\\frac{\\left|V_{H}\\right| \\sigma d}{I_{x} \\mu_{e}} \\\\\n& =\\frac{\\left(-3.5 \\times 10^{-7} \\mathrm{~V}\\right)\\left[1.2 \\times 10^{7}(\\Omega \\cdot \\mathrm{m})^{-1}\\right](35 \\times 10^{-3} \\mathrm{~m})}{(40 \\mathrm{~A})(0.0050 \\mathrm{~m}^{2} / \\mathrm{Vs})}\n\\end{aligned}\n\\]\\[\n = 0.735 \\, {\\rm tesla}\n\\]"
},
{
"idx": 777,
"question": "Given the activation energy (173,000 J/mol) and preexponential factor (4.0 x 10^-4 m^2/s) for the diffusion coefficient of Na+ in NaCl, compute the diffusion coefficient D_Na+ at 873 K.",
"answer": "D_Na+ = D0 exp(-Qd / RT) = (4.0 x 10^-4 m^2/s) exp[-173,000 J/mol / (8.31 J/mol-K)(873 K)] = 1.76 x 10^-14 m^2/s"
},
{
"idx": 777,
"question": "Using the computed diffusion coefficient D_Na+ (1.76 x 10^-14 m^2/s) at 873 K, calculate the mobility μ_Na+ for a Na+ ion, assuming the valence n_Na+ = 1.",
"answer": "μ_Na+ = (n_Na+ * e * D_Na+) / (k T) = (1)(1.602 x 10^-19 C/atom)(1.76 x 10^-14 m^2/s) / (1.38 x 10^-23 J/atom-K)(873 K) = 2.34 x 10^-13 m^2/V-s"
},
{
"idx": 778,
"question": "A parallel-plate capacitor using a dielectric material having an \\(\\varepsilon_{\\mathrm{r}}\\) of 2.2 has a plate spacing of \\(2 \\mathrm{~mm}\\) ( 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": "We want to compute the plate spacing of a parallel-plate capacitor as the dielectric constant is increased form 2.2 to 3.7, while maintaining the capacitance constant. Combining Equations yields the following expression:\n\\[\nC=\\varepsilon \\frac{A}{l}=\\varepsilon_{r} \\varepsilon_{0} \\frac{A}{l}\n\\]\nNow, let us use the subscripts 1 and 2 to denote the initial and final states, respectively. Since \\(C_{1}\\) \\(=C_{2}\\), then we can write the following:\n\\[\n\\frac{\\varepsilon_{r 1} \\varepsilon_{0} A}{l_{1}}=\\frac{\\varepsilon_{r 2} \\varepsilon_{0} A}{l_{2}}\n\\]\nAnd, solving for \\(l_{2}\\) leads to the following:\n\\[\n\\begin{array}{c}\nl_{2}=\\frac{\\varepsilon_{r 2} l_{1}}{\\varepsilon_{r 1}}=\\frac{(3.7)(2 \\mathrm{~mm})}{2.2}=3.36 \\mathrm{~mm}\n\\end{array}\n\\]"
},
{
"idx": 779,
"question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1 mm (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": "We are first asked to compute the capacitance. Combining Equations leads to the following expression: C = ε A / l = εr ε0 A / l. Incorporation of values given in the problem statement for parameters on the right-hand side of this expression, and solving for C yields C = (3.5)(8.85 × 10-12 F/m)(3225 mm2)(1 m2 / 106 mm2) / 10-3 m = 10-10 F = 100 pF."
},
{
"idx": 779,
"question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1 mm (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": "Now we are asked to compute the electric field that must be applied in order for 2 × 10-8 C to be stored on each plate. First we need to solve for V in Equation as V = Q / C = (2 × 10-8 C) / (10-10 F) = 200 V. The electric field E may now be determined using Equation, as follows: E = V / l = (200 V) / (10-3 m) = 2.0 × 105 V/m."
},
{
"idx": 780,
"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": "We solve this portion of the problem by computing the value of εr using Equation. Rearranging this equation such that εr is the dependent parameter gives the following: εr = P / (ε0 E) + 1. When values of the parameters on the right-hand side of this equation are inserted the value of εr equal to εr = (4.0 × 10^-6 C/m^2) / ((8.85 × 10^-12 F/m)(1 × 10^5 V/m)) + 1 = 5.52"
},
{
"idx": 780,
"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 when an electric field of 10^5 V/m is applied?",
"answer": "The dielectric displacement D may be determined using Equation, as follows: D = ε0 E + P = (8.85 × 10^-12 F/m)(1 × 10^5 V/m) + 4.0 × 10^-6 C/m^2 = 4.89 × 10^-6 C/m^2"
},
{
"idx": 781,
"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": "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": 781,
"question": "(b) 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": 782,
"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": 782,
"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": 782,
"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": 782,
"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": 782,
"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": 783,
"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 degrees C for an extended time period."
},
{
"idx": 783,
"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": 783,
"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": 784,
"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": 784,
"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": 784,
"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": 784,
"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": 785,
"question": "For the \\(\\mathrm{MgO}-\\mathrm{Al}_{2} \\mathrm{O}_{3}\\) system, what is the maximum temperature that is possible without the formation of a liquid phase? At what composition or over what range of compositions will this maximum temperature be achieved?",
"answer": "This problem asks that we specify, for the \\(\\mathrm{MgO}-\\mathrm{Al}_{2} \\mathrm{~O}_{3}\\) system, the maximum temperature without the formation of a liquid phase; it is approximately \\(2800^{\\circ} \\mathrm{C}\\), which is possible for pure \\(\\mathrm{MgO}\\)."
},
{
"idx": 786,
"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": 786,
"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": 786,
"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": 786,
"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": 786,
"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": 787,
"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": 788,
"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": 789,
"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": 789,
"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": 789,
"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": 790,
"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": 790,
"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": 791,
"question": "What is the maximum thermal conductivity for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix, given thermal conductivities of 27 W/m·K for TiC and 67 W/m·K for Ni?",
"answer": "The maximum thermal conductivity k_max is calculated as follows: k_max = k_m V_m + k_p V_p = k_Ni V_Ni + k_TiC V_TiC = (67 W/m·K)(0.10) + (27 W/m·K)(0.90) = 31.0 W/m·K"
},
{
"idx": 791,
"question": "What is the minimum thermal conductivity for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix, given thermal conductivities of 27 W/m·K for TiC and 67 W/m·K for Ni?",
"answer": "The minimum thermal conductivity k_min is calculated as follows: k_min = (k_Ni k_TiC) / (V_Ni k_TiC + V_TiC k_Ni) = (67 W/m·K)(27 W/m·K) / [(0.10)(27 W/m·K) + (0.90)(67 W/m·K)] = 28.7 W/m·K"
},
{
"idx": 792,
"question": "Estimate the upper limit for the elastic modulus of the composite consisting of tungsten particles within a copper matrix with volume fractions of 0.70 for tungsten and 0.30 for copper, given the modulus of elasticity for copper is 110 GPa and for tungsten is 407 GPa.",
"answer": "The upper limit for the elastic modulus of the composite is calculated as follows: E_c(u) = E_Cu V_Cu + E_W V_W = (110 GPa)(0.30) + (407 GPa)(0.70) = 318 GPa."
},
{
"idx": 792,
"question": "Estimate the upper limit for the specific gravity of the composite consisting of tungsten particles within a copper matrix with volume fractions of 0.70 for tungsten and 0.30 for copper, given the specific gravity for copper is 8.9 and for tungsten is 19.3.",
"answer": "The upper limit for the specific gravity of the composite is calculated as follows: ρ_c = ρ_Cu V_Cu + ρ_W V_W = (8.9)(0.30) + (19.3)(0.70) = 16.18."
},
{
"idx": 792,
"question": "Calculate the specific stiffness of the composite using the upper limits for the elastic modulus (318 GPa) and specific gravity (16.18) of the composite.",
"answer": "The specific stiffness is calculated as follows: Specific Stiffness = E_c(u) / ρ_c = 318 GPa / 16.18 = 19.65 GPa."
},
{
"idx": 792,
"question": "Calculate the specific stiffness of the composite using the specific stiffness-volume fraction product for both copper and tungsten, given the modulus of elasticity for copper is 110 GPa and for tungsten is 407 GPa, and the specific gravity for copper is 8.9 and for tungsten is 19.3, with volume fractions of 0.30 for copper and 0.70 for tungsten.",
"answer": "The specific stiffness is calculated as follows: Specific Stiffness = (E_Cu / ρ_Cu) V_Cu + (E_W / ρ_W) V_W = (110 GPa / 8.9)(0.30) + (407 GPa / 19.3)(0.70) = 18.47 GPa."
},
{
"idx": 793,
"question": "For a continuous and oriented fiber-reinforced composite with moduli of elasticity in the longitudinal and transverse directions of 33.1 GPa and 3.66 GPa respectively, and a fiber volume fraction of 0.30, determine the modulus of elasticity of the fiber phase.",
"answer": "Using the longitudinal modulus equation: E_cl = E_m(1 - V_f) + E_f V_f, and the transverse modulus equation: E_ct = (E_m E_f) / [(1 - V_f) E_f + V_f E_m], we solve these simultaneously for E_f. The solution yields E_f = 104 GPa (15 x 10^6 psi)."
},
{
"idx": 793,
"question": "For a continuous and oriented fiber-reinforced composite with moduli of elasticity in the longitudinal and transverse directions of 33.1 GPa and 3.66 GPa respectively, and a fiber volume fraction of 0.30, determine the modulus of elasticity of the matrix phase.",
"answer": "Using the longitudinal modulus equation: E_cl = E_m(1 - V_f) + E_f V_f, and the transverse modulus equation: E_ct = (E_m E_f) / [(1 - V_f) E_f + V_f E_m], we solve these simultaneously for E_m. The solution yields E_m = 2.6 GPa (3.77 x 10^5 psi)."
},
{
"idx": 794,
"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. The modulus of elasticity for carbon fiber is 260 GPa (37x10^6 psi) and for nylon 6,6 is 2.8 GPa (4.0x10^5 psi).",
"answer": "To compute the volume fraction of fibers (V_f) required for the fibers to carry 97% of the load, we use the equation: F_f/F_m = (E_f V_f)/(E_m (1-V_f)). Given F_f/F_m = 0.97/(1-0.97) = 32.3, E_f = 260 GPa, and E_m = 2.8 GPa, we substitute these values into the equation: 32.3 = (260 GPa V_f)/(2.8 GPa (1-V_f)). Solving for V_f yields V_f = 0.258."
},
{
"idx": 794,
"question": "What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 50 MPa (7250 psi). The tensile strength of carbon fiber is 4000 MPa (580,000 psi) and of nylon 6,6 is 76 MPa (11,000 psi). The volume fraction of fibers (V_f) is 0.258.",
"answer": "The tensile strength of the composite (σ_cl*) is calculated using the equation: σ_cl* = σ_m* (1-V_f) + σ_f* V_f. Given σ_m* = 50 MPa, V_f = 0.258, and σ_f* = 4000 MPa, we substitute these values into the equation: σ_cl* = (50 MPa)(1-0.258) + (4000 MPa)(0.258) = 1070 MPa (155,000 psi)."
},
{
"idx": 795,
"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": 795,
"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": 796,
"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": 796,
"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": 797,
"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": "For a hybrid composite having all fibers aligned in the same direction, the longitudinal modulus of elasticity may be determined using a modified form of Equation as follows: E_cl = E_m V_m + E_f1 V_f1 + E_f2 V_f2 in which the subscripts f1 and f2 refer to the two types of fibers."
},
{
"idx": 797,
"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)]. The elastic moduli of aramid and glass fibers are, respectively, 131 GPa (19x10^6 psi) and 72.5 GPa (10.5x10^6 psi).",
"answer": "Now we are asked to compute the longitudinal elastic modulus for a glass- and aramid-fiber hybrid composite. Thus, from the previous expression E_cl = (4 GPa)(1.0 - 0.25 - 0.35) + (131 GPa)(0.25) + (72.5 GPa)(0.35) = 59.7 GPa (8.67x10^6 psi)."
},
{
"idx": 798,
"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": 798,
"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": 799,
"question": "Is a voltage generated between the two cell halves of an Fe/Fe2+ concentration cell where one half has an Fe2+ concentration of 0.5 M and the other has 2 x 10^-2 M?",
"answer": "Yes, a voltage is generated between the two cell halves."
},
{
"idx": 799,
"question": "What is the magnitude of the voltage generated in the Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 x 10^-2 M?",
"answer": "The magnitude of the voltage generated is 0.0414 V."
},
{
"idx": 799,
"question": "Which electrode will be oxidized in the Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 x 10^-2 M?",
"answer": "The electrode in the cell half with the lower Fe2+ concentration (2 x 10^-2 M) will be oxidized."
},
{
"idx": 800,
"question": "For the aluminum and cast iron alloys 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": 800,
"question": "For the Inconel and nickel alloys 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": 800,
"question": "For the cadmium and zinc alloys 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": 800,
"question": "For the brass and titanium alloys 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": 800,
"question": "For the low-carbon steel and copper alloys 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": 801,
"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 \\mathrm{~cm}^{2}\\) and that approximately \\(7.6 \\mathrm{~kg}\\) had corroded away during the submersion. Assuming a corrosion penetration rate of \\(4 \\mathrm{~mm} / \\mathrm{yr}\\) for this alloy in seawater, estimate the time of submersion in years. The density of the alloy is \\(4.5 \\mathrm{~g} / \\mathrm{cm}^{3}\\).",
"answer": "This problem calls for us to compute the time of submersion of a metal plate. In order to solve this problem, we must first rearrange Equation such that time, \\(t\\), is the dependent variable:\n\\[\nt=\\frac{K W}{\\rho A(\\mathrm{CPR})}\n\\]\nThus, using values for the various parameters given in the problem statement yields the following value for \\(t\\) :\n\\[\n\\begin{aligned}\nt= & \\frac{(87.6)(7.6 \\mathrm{~kg})(10^{6} \\mathrm{mg} / \\mathrm{kg})}{\\left(4.5 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(800 \\mathrm{~cm}^{2}\\right)(4 \\mathrm{~mm} / \\mathrm{yr})} \\\\\n& =4.62 \\times 10^{4} \\mathrm{~h}=5.27 \\mathrm{yr}\n\\end{aligned}\n\\]"
},
{
"idx": 802,
"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 data: For Metal M: V(M/M2+) = -0.90 V, i0 = 10-12 A/cm2, β = +0.10; For Hydrogen: V(H+/H2) = 0 V, i0 = 10-10 A/cm2, β = -0.15.",
"answer": "The rate of oxidation for metal M is calculated by first establishing the potential expressions for both oxidation and reduction reactions. For hydrogen reduction: VH = V(H+/H2) + βH log(i/i0H). For M oxidation: VM = V(M/M2+) + βM log(i/i0M). Setting VH = VM and solving for log iC gives: log iC = [1/(βM - βH)][V(H+/H2) - V(M/M2+) - βH log i0H + βM log i0M]. Substituting the given values: log iC = [1/(0.10 - (-0.15))][0 - (-0.90) - (-0.15){log(10-10)} + (0.10){log(10-12)}] = -7.20. Thus, iC = 10-7.20 = 6.31 × 10-8 A/cm2 = 6.31 × 10-8 C/s·cm2. The oxidation rate is then: r = iC / (nF) = (6.31 × 10-8 C/s·cm2) / (2 × 96,500 C/mol) = 3.27 × 10-13 mol/cm2·s."
},
{
"idx": 802,
"question": "Compute the corrosion potential for the reaction involving metal M given the following data: For Metal M: V(M/M2+) = -0.90 V, i0 = 10-12 A/cm2, β = +0.10; For Hydrogen: V(H+/H2) = 0 V, i0 = 10-10 A/cm2, β = -0.15, and the corrosion current density iC = 6.31 × 10-8 A/cm2.",
"answer": "The corrosion potential Vc is calculated using the expression for hydrogen reduction: Vc = V(H+/H2) + βH log(iC/i0H). Substituting the given values: Vc = 0 + (-0.15 V) log[(6.31 × 10-8 A/cm2) / (10-10 A/cm2)] = -0.420 V."
},
{
"idx": 803,
"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": 804,
"question": "For a concentration cell, briefly explain why corrosion occurs at the region having the lower concentration.",
"answer": "For a concentration cell, corrosion occurs at that region having the lower concentration. In order to explain this phenomenon let us consider an electrochemical cell consisting of two divalent metal \\(\\mathrm{M}\\) electrodes each of which is immersed in a solution containing a different concentration of its \\(\\mathrm{M}^{2+}\\) ion; let us designate the low and high concentrations of \\(\\mathrm{M}^{2+}\\) as \\(\\left[\\mathrm{M}_{\\mathrm{L}}^{2+}\\right]\\) and \\(\\left[\\mathrm{M}_{\\mathrm{H}}^{2+}\\right]\\), respectively. Now assuming that reduction and oxidation reactions occur in the highand low-concentration solutions, respectively, let us determine the cell potential in terms of the two \\(\\left[\\mathrm{M}^{2+}\\right] \\mathrm{s}\\); if this potential is positive then we have chosen the solutions in which the reduction and oxidation reactions appropriately.\nThus, the two half-reactions in the form of Equations are as follows:\n\\[\n\\begin{array}{l}\n\\mathrm{M}_{\\mathrm{H}}^{2+}+2 e^{-} \\rightarrow \\mathrm{M} \\quad V_{\\mathrm{M}}^{0} \\\\\n\\mathrm{M} \\rightarrow \\mathrm{M}_{\\mathrm{L}}^{2+}+2 e^{-} \\quad-V_{\\mathrm{M}}^{0}\n\\end{array}\n\\]\nWhereas the overall cell reaction is the following:\n\\[\n\\mathrm{M}_{\\mathrm{H}}^{2+}+\\mathrm{M} \\rightarrow \\mathrm{M}+\\mathrm{M}_{\\mathrm{L}}^{2+}\n\\]\nFrom Equation, this yields a cell potential of\n\\[\n\\Delta V=V_{\\mathrm{M}}^{b}-V_{\\mathrm{M}}^{b}-\\frac{R T}{n F} \\ln \\left(\\frac{\\left[M_{\\mathrm{L}}^{2+}\\right]}{\\left[\\mathrm{M}_{\\mathrm{H}}^{2+}\\right]}\\right)\n\\]\\[\n\\hspace{1.5cm} = -\\frac{RT}{nF} \\ln \\left( \\frac{\\left[ \\mathrm{M}_{2}^{2+} \\right] }{\\left[ \\mathrm{M}_{H}^{2+} \\right] } \\right)\n\\]\n(Note: we denote the Faraday constant in the above equations by an \" \\(F\\) \" rather than with the symbol used in the textbook.)\nInasmuch as \\(\\left[\\mathrm{M}_{\\mathrm{L}}^{2+}\\right] \\prec\\left[\\mathrm{M}_{\\mathrm{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": 805,
"question": "For copper, the heat capacity at constant volume \\(C_{V}\\) at \\(20 \\mathrm{~K}\\) is \\(0.38 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\\) and the Debye temperature is \\(340 \\mathrm{~K}\\). Estimate the specific heat for the following:\n(a) at \\(40 \\mathrm{~K}\\)\n(b) at \\(400 \\mathrm{~K}\\)\n\\(\\underline{\\text {",
"answer": "(a) For copper, \\(C_{V}\\) at \\(20 \\mathrm{~K}\\) may be approximated by Equation, since this temperature is significantly below the Debye temperature \\((340 \\mathrm{~K})\\). The value of \\(C_{V}\\) at \\(20 \\mathrm{~K}\\) is given, and thus, we may compute the constant \\(A\\) by rearranging Equation as follows:\n\\[\nA=\\frac{C_{V}}{T^{3}}=\\frac{0.38 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{N}}{(20 \\mathrm{~K})^{3}}=4.75 \\times 10^{-5} \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}^{4}\n\\]\nTherefore, at \\(40 \\mathrm{~K}\\) the value of \\(C_{V}\\) is equal to\n\\[\n\\begin{aligned}\nC_{V} & =A T^{3} \\\\\n& =(4.75 \\quad 10^{5} \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}^{-4})(40 \\mathrm{~K})^{3} \\\\\n& =3.04 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\n\\end{aligned}\n\\]\nand the specific heat is computed using the following expression:\n\\[\nc_{V}=\\left(\\frac{C_{V}}{A^{\\prime}}\\right)\\left(\\begin{array}{c}\n1000 \\mathrm{~g} \\\\\n1 \\mathrm{~kg}\n\\end{array}\\right)\n\\]\nIn this equation, the parameter \\(A^{\\prime}\\) represents the atomic weight of the material. Therefore, the specific heat for copper at \\(40 \\mathrm{~K}\\) is equal to\\[\n\\begin{array}{c}\nc_{V}=\\left(\\begin{array}{c}\n3.04\\mathrm{~J} / \\mathrm{mol}-\\mathrm{K} \\\\\n63.55\\mathrm{~g} / \\mathrm{mol}\n\\end{array}\\right)\\left(\\begin{array}{c}\n1000\\mathrm{~g} \\\\\n1 \\mathrm{~kg}\n\\end{array}\\right) \\\\\n=47.8\\mathrm{~J} / \\mathrm{kg}-\\mathrm{K}\n\\end{array}\n\\]\n(b) Because \\(400 \\mathrm{~K}\\) is above the Debye temperature, a good approximation for \\(C_{V}\\) is as follows:\n\\[\n\\begin{array}{c}\nC_{V}=3 R \\\\\n= (3)(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})=24.9 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\n\\end{array}\n\\]\nWe convert this \\(C_{V}\\) to specific heat using Equation as follows:\n\\[\n\\begin{aligned}\nc_{V} & =\\left(\\begin{array}{c}\nC_{V} \\\\\nA^{\\prime}\n\\end{array}\\right)\\left(\\begin{array}{c}\n100 \\mathrm{~g} \\\\\n1 \\mathrm{~kg}\n\\end{array}\\right. \\\\\n& =\\left(\\begin{array}{c}\n24.9 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{KI} \\\\\n63.55 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\\right)\\left(\n\\begin{array}{c}\n1000 \\mathrm{~g} \\\\\n1 \\mathrm{~kg}\n\\end{aligned}\n\\right) \\\\\n& =392 \\mathrm{~J} / \\mathrm{kg}-\\mathrm{K}\n\\end{aligned}\n\\]"
},
{
"idx": 806,
"question": "A 0.4-m (15.7-in.) rod of a metal elongates \\(0.48 \\mathrm{~mm}\\) (0.019 in.) on heating from \\(20^{\\circ} \\mathrm{C}\\) to \\(100^{\\circ} \\mathrm{C}\\left(68^{\\circ} \\mathrm{F}\\right.\\) to \\(\\left.212^{\\circ} \\mathrm{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 may be determined using a rearranged form of Equation as follows:\n\\[\n\\begin{aligned}\n\\alpha_{l} & =\\frac{\\Delta l}{l_{0} \\Delta T}=\\frac{\\Delta l}{l_{0}\\left(T_{f}-T_{0}\\right)} \\\\\n& =\\frac{0.48 \\times 10^{-3} \\mathrm{~m}}{(0.4 \\mathrm{~m})\\left(100^{\\circ} \\mathrm{C}-20^{\\circ} \\mathrm{C}\\right)} \\\\\n& =15.0 \\times 10^{-6}\\left({ }^{\\circ} \\mathrm{C}\\right)^{-1} \\quad\\left[8.40 \\times 10^{-6}\\left({ }^{\\circ} \\mathrm{F}\\right)^{-1}\\right]\n\\end{aligned}\n\\]"
},
{
"idx": 807,
"question": "For the pair of materials pure silver and sterling silver (92.5 wt% Ag-7.5 wt% Cu), which has the larger thermal conductivity? Justify your choice.",
"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": 807,
"question": "For the pair of materials fused silica and polycrystalline silica, which has the larger thermal conductivity? Justify your choice.",
"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": 807,
"question": "For the pair of materials linear and syndiotactic poly(vinyl chloride) (DP =1000) and linear and syndiotactic polystyrene (DP =1000), which has the larger thermal conductivity? Justify your choice.",
"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": 807,
"question": "For the pair of materials atactic polypropylene (Mw=10^6 g/mol) and isotactic polypropylene (Mw=10^5 g/mol), which has the larger thermal conductivity? Justify your choice.",
"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": 808,
"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": 808,
"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": 808,
"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": 809,
"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": "The magnetic permeability of this material may be determined as follows: mu = B / H = 0.630 tesla / 5 x 10^5 A/m = 1.260 x 10^-6 H/m"
},
{
"idx": 809,
"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, and its magnetic permeability is 1.260 x 10^-6 H/m. Compute the magnetic susceptibility for this material.",
"answer": "The magnetic susceptibility is calculated as follows: chi_m = mu_r - 1 = mu / mu_0 - 1 = (1.260 x 10^-6 H/m) / (1.257 x 10^-6 H/m) - 1 = 2.387 x 10^-3"
},
{
"idx": 809,
"question": "The magnetic susceptibility of a material is 2.387 x 10^-3. 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 chi_m is positive and on the order of 10^-3, there would also be a paramagnetic contribution."
},
{
"idx": 810,
"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.\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": 811,
"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 \\times 10^{5}\\) \\(\\mathrm{A} / \\mathrm{m}\\) and a density of \\(5.40 \\mathrm{~g} / \\mathrm{cm}^{3}\\), estimate the number of Bohr magnetons associated with each \\(\\mathrm{Cu}^{2+}\\) ion.",
"answer": "We want to compute the number of Bohr magnetons per \\(\\mathrm{Cu}^{2+}\\) ion in \\(\\left(\\mathrm{CuFe}_{2} \\mathrm{O}_{4}\\right)\\) 8. Let \\(n_{\\mathrm{B}}\\) represent the number of Bohr magnetons per \\(\\mathrm{Cu}^{2+}\\) 1on; then, using Equation, we write\n\\[\nM_{s}=n_{\\mathrm{B}} \\mu_{\\mathrm{B}} N\n\\]\nin which \\(N\\) is the number of \\(\\mathrm{Cu}^{2+}\\) ions per cubic meter of material. But, from Equation\n\\[\nN=\\frac{\\rho N_{\\mathrm{A}}}{A}\n\\]\nHere \\(A\\) is the molecular weight of \\(\\mathrm{CuFe}_{2} \\mathrm{O}_{4}(239.25 \\mathrm{~g} / \\mathrm{mol})\\). Thus, combining the previous two equations leads to the following:\n\\[\nM_{s}=\\frac{n_{\\mathrm{B}} \\mu_{\\mathrm{B}} \\rho N_{\\mathrm{A}}}{A}\n\\]\nor, upon rearrangement (and expressing the density in units of grams per meter cubed), the number of Bohr magnetons per \\(\\mathrm{Cu}^{2+}\\) on \\(\\left(n_{\\mathrm{B}}\\right)\\) is computed as follows:\n\\[\n\\begin{array}{c}\nn_{\\mathrm{B}}=\\frac{M_{s} A}{\\mu_{\\mathrm{B}} \\rho N_{\\mathrm{A}}} \\\\\n=\\frac{\\left(1.35 \\times 10^{5} \\mathrm{~A} / \\mathrm{m}\\right)\\left(239.25 \\mathrm{~g} / \\mathrm{mol}\\right)}{\\left(9.27 \\times 10^{-24} \\mathrm{~A} \\cdot \\mathrm{m}^{2} / \\mathrm{BM}\\right)\\left(5.40 \\times 10^{6} \\mathrm{~g} / \\mathrm{m}^{3}\\right)\\left(6.022 \\times 10^{23} \\mathrm{ions} / \\mathrm{mol}\\right)}\n\\end{array}\n\\]\\[\n = 1.07 \\mbox{ Bohr magnetons/Cu\\(^{2+}\\) ion }\n\\]"
},
{
"idx": 812,
"question": "Calculate the number of Bohr magnetons per unit cell (n_B) given the saturation magnetization (1.35 x 10^5 A/m), the cubic unit cell edge length (1.2529 nm), and the Bohr magneton value (9.27 x 10^-24 A·m^2/BM).",
"answer": "n_B = (1.35 x 10^5 A/m) * (1.2529 x 10^-9 m)^3 / (9.27 x 10^-24 A·m^2/BM) = 28.64 Bohr magnetons/unit cell."
},
{
"idx": 812,
"question": "Determine the number of Bohr magnetons per formula unit given there are 8 formula units per unit cell and the previously calculated n_B (28.64 Bohr magnetons/unit cell).",
"answer": "Bohr magnetons per formula unit = 28.64 Bohr magnetons/unit cell / 8 formula units/unit cell = 3.58 Bohr magnetons/formula unit."
},
{
"idx": 812,
"question": "Calculate the net magnetic moment contribution per formula unit from the Fe^3+ ions, given there are 2 Fe^3+ ions on a sites and 3 Fe^3+ ions on d sites, each with 5 Bohr magnetons, and their magnetic moments are aligned antiparallel.",
"answer": "Net magnetic moment from Fe^3+ ions = (2 * 5 Bohr magnetons) - (3 * 5 Bohr magnetons) = -5 Bohr magnetons/formula unit."
},
{
"idx": 812,
"question": "Compute the number of Bohr magnetons associated with each Sm^3+ ion, given the net magnetic moment per formula unit (3.58 Bohr magnetons), the contribution from Fe^3+ ions (-5 Bohr magnetons), and there are 3 Sm^3+ ions per formula unit.",
"answer": "Number of Bohr magnetons/Sm^3+ = (3.58 Bohr magnetons + 5 Bohr magnetons) / 3 = 2.86 Bohr magnetons/Sm^3+ ion."
},
{
"idx": 813,
"question": "An iron bar magnet having a coercivity of \\(7000 \\mathrm{~A} / \\mathrm{m}\\) is to be demagnetized. If the bar is inserted within a cylindrical wire coil \\(0.25 \\mathrm{~m}\\) long and having 150 turns, what electric current is required to generate the necessary magnetic field?",
"answer": "In order to demagnetize a magnet having a coercivity of \\(7000 \\mathrm{~A} / \\mathbf{m}\\), an \\(H\\) field of 7000 \\(\\mathrm{A} / \\mathrm{m}\\) must be applied in a direction opposite to that of magnetization. According to Equation\n\\[\n\\begin{aligned}\nI & =\\frac{H l}{N} \\\\\n& =\\frac{(7000 \\mathrm{~A} / \\mathrm{m})(0.25 \\mathrm{~m})}{150 \\mathrm{turns}}=11.7 \\mathrm{~A}\n\\end{aligned}\n\\]"
},
{
"idx": 814,
"question": "Cite the differences between hard and soft magnetic materials in terms of hysteresis behavior.",
"answer": "Relative to hysteresis behavior, a hard magnetic material has a high remanence, a high coercivity, a high saturation flux density, high hysteresis energy losses, and a low initial permeability; a soft magnetic material, on the other hand, has a high initial permeability, a low coercivity, and low hysteresis energy losses."
},
{
"idx": 814,
"question": "Cite the differences between hard and soft magnetic materials in terms of typical applications.",
"answer": "Soft magnetic materials are used in devices that experience alternating and when energy losses must be low-such as transformer cores, generators, motors (electromagnet), and switching circuits. Hard magnetic materials are used in applications that require permanent magnets-such as motors (permanent magnet motors), audio and video recorders, hearing aids, and computer peripherals."
},
{
"idx": 815,
"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": 816,
"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": 817,
"question": "Visible light having a wavelength of 5 × 10^-7 m appears green. Compute the frequency of a photon of this light.",
"answer": "v = c / λ = (3 × 10^8 m/s) / (5 × 10^-7 m) = 6 × 10^14 s^-1"
},
{
"idx": 817,
"question": "Visible light having a wavelength of 5 × 10^-7 m appears green. Compute the energy of a photon of this light.",
"answer": "E = h c / λ = (6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (5 × 10^-7 m) = 3.98 × 10^-19 J (2.48 eV)"
},
{
"idx": 818,
"question": "Distinguish among materials that are opaque, translucent, and transparent in terms of \\\\ their appearance and light transmittance.",
"answer": "Opaque materials are impervious to light transmission; it is not possible to see through them.\nLight is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material.\nVirtually all of the incident light is transmitted through transparent materials, and one can see clearly through them."
},
{
"idx": 819,
"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": 820,
"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": "This problem calls for a calculation of the reflectivity between two quartz grains having different orientations and indices of refraction (1.544 and 1.553) in the direction of light propagation, when the light is at normal incidence to the grain boundary. We must employ Equation since the beam is normal to the grain boundary. Thus,\n\\[\n\\begin{aligned}\nR & =\\frac{\\left(n_{2}-n_{1}\\right)^{2}}{\\left(n_{2}+n_{1}\\right)^{2}} \\\\\n& =\\frac{\\left(1.553-1.544\\right)^{2}}{\\left(1.553+1.544\\right)^{2}} \\\\\n& =8.45 \\quad 10^{6}\n\\end{aligned}\n\\]"
},
{
"idx": 821,
"question": "Zinc selenide has a band gap of \\(2.58 \\mathrm{eV}\\). Over what range of wavelengths of visible light is it transparent?",
"answer": "This problem asks us to determine the range of visible light wavelengths over which \\(\\mathrm{ZnSe}\\left(E_{g}=2.58 \\mathrm{eV}\\right)\\) is transparent. Only photons having energies of \\(2.58 \\mathrm{eV}\\) or greater are absorbed by valence-band-to-conduction-band electron transitions. Thus, photons having energies less than \\(2.58 \\mathrm{eV}\\) are not absorbed; the minimum photon energy for visible light is 1.8 \\(\\mathrm{eV}\\) , which corresponds to a wavelength of \\(0.7 \\mu \\mathrm{m}\\). From a rearranged form of Equation, the wavelength of a photon having an energy of \\(2.58 \\mathrm{eV}\\) (i.e., the band-gap energy) is just\n\\[\n\\begin{aligned}\n\\lambda & =\\frac{h c}{E} \\\\\n& =\\frac{\\left(4.13 \\times 10^{-15} \\mathrm{eV} \\cdot \\mathrm{s}\\right)\\left(3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right)}{2.58 \\mathrm{eV}} \\\\\n& =4.80 \\times 10^{-7} \\mathrm{~m}=0.48 \\mu \\mathrm{m}\n\\end{aligned}\n\\]\nThus, pure \\(\\mathrm{ZnSe}\\) is transparent to visible light having wavelengths between 0.48 and \\(0.7 \\mu \\mathrm{m}\\)."
}
]
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