4929 lines
872 KiB
JSON
4929 lines
872 KiB
JSON
[
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{
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"idx": 1,
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"question": "Temperature indicators are sometimes produced from a coiled metal strip that unc oils a specific amount when the temperature increases. How does this work; from what kind of material would the indicator be made; and what are the important properties that the material in the indicator must possess?",
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"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. 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.",
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"is_select": 0
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},
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{
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"idx": 2,
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"question": "What properties should the head of a carpenter's hammer possess? How would you manufacture a hammer head?",
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"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.\nThe 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.",
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"is_select": 0
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},
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{
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"idx": 3,
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"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?",
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"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\\]",
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"is_select": 1
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},
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{
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"idx": 4,
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"question": "(b) Using the densities and atomic weights, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium",
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"answer": "(i) In lead:\n\\[\n\\begin{array}{c}\n\\left(11.36 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(1 \\mathrm{~cm}^{3}\\right)\\left(6.02 \\times 10^{23} \\mathrm{~atms} / \\mathrm{mol}\\right)=3.3 \\times 10^{22} \\mathrm{atoms} / \\mathrm{cm}^{3} \\\\\n207.19 \\mathrm{~g} / \\mathrm{mol} \\\\\n\\text { (ii) In lithium: } \\\\\n\\left(\\frac{0.534 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(} 1 \\mathrm{~cm}^{3}\\right)\\left(6.02 \\mathrm{~g} \\times 10^{23} \\text { atoms } / \\mathrm{mol} \\right)=4.63 \\times 10^{22} \\mathrm{atoms} / \\mathrm{ cm}^{3} \\\\\n6.94 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]",
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"is_select": 1
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},
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{
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"idx": 5,
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"question": "(a)Using data, calculate the number of iron atoms in one ton (2000 pounds).",
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"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}\\)",
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"is_select": 1
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},
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{
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"idx": 6,
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"question": "(b) Using data, calculate the volume in cubic centimeters occupied by one mole of boron.",
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"answer": "(\\frac{(1 \\mathrm{~mol})(10.81 \\mathrm{~g} / \\mathrm{mol})}{2.3 \\mathrm{~g} / \\mathrm{cm}^{3}}=4.7 \\mathrm{~cm}^{3}\\)",
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"is_select": 1
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},
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{
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"idx": 7,
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"question": "In order to plate a steel part having a surface area of \\(200 \\mathrm{in} .^{2}\\) with a 0.002 in. thick layer of nickel, (a) how many atoms of nickel are required and (b) how many moles of nickel are required?",
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"answer": "Volume \\(=(200 \\mathrm{in} .{ }^{2})(0.002 \\mathrm{in} .)(2.54 \\mathrm{~cm} / \\mathrm{in} .{ }^{3})=6.555 \\mathrm{~cm}^{3}\\)\n(a) \\((6.555 \\mathrm{~cm}^{3})(8.902 \\mathrm{~g} / \\mathrm{cm}^{3})(6.02 \\times 10^{23}\\) atoms \\(/ \\mathrm{mol})\\)\n\\(=5.98 \\times 10^{23}\\) atoms\n(b) \\((6.555 \\mathrm{~cm}^{3})(8.002 \\mathrm{~g} / \\mathrm{cm}^{3})\\)\n\\(=0.994 \\mathrm{~mol} \\mathrm{Ni}\\) required\n\\(58.71 \\mathrm{~g} / \\mathrm{mol}\\)",
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"is_select": 1
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},
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{
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"idx": 8,
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"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?",
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"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.",
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"is_select": 1
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},
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{
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"idx": 9,
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"question": "Calculate the fraction of bonding of \\(\\mathrm{MgO}\\) that is ionic.",
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"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",
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"is_select": 1
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},
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{
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"idx": 10,
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"question": "Would you expect \\(\\mathrm{MgO}\\) or magnesium to have the higher modulus of elasticity?\nExplain.",
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"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}\\).",
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"is_select": 0
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},
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{
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"idx": 11,
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"question": "Calculate the atomic radius in \\(\\mathrm{cm}\\) for the following: (a) BCC metal with \\(a_{0}=0.3294 \\mathrm{~nm}\\) and one atom per lattice point; and (b) FCC metal with \\(a_{0}=4.0862 \\AA\\) and one atom per lattice point.\n\\end{abstract}",
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"answer": "(a) For BCC metals,\n\\[\nr=\\frac{(\\sqrt{3}) a_{0}}{4}=\\frac{(\\sqrt{3})(0.3294 \\mathrm{~nm})}{4}=0.1426 \\mathrm{~nm}=1.426 \\times 10^{-8} \\mathrm{~cm}\n\\]\n(b) For FCC metals,\n\\[\nr=\\frac{(\\sqrt{2}) a_{0}}{4}=\\frac{(\\sqrt{2})(4.0862 \\AA)}{4}=1.4447 \\AA=1.4447 \\times 10^{-8} \\mathrm{~cm}\n\\]",
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"is_select": 1
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},
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{
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"idx": 12,
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"question": "Determine the crystal structure for the following: (a) a metal with \\(a_{0}=4.9489 \\AA\\), \\(r=1.75 \\AA\\) and one atom per lattice point; and (b) a metal with \\(a_{0}=0.42906 \\mathrm{~nm}\\), \\(r=0.1858 \\mathrm{~nm}\\) and one atom per lattice point.",
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"answer": "We want to determine if \" \\(x\\) \" in the calculations below equals \\(\\sqrt{2}\\) (for FCC) or \\(\\sqrt{3}\\) (for BCC):\n(a) \\((x)(4.9489 \\AA)=(4)(1.75 \\AA)\\)\n\\(x=\\sqrt{2}\\), therefore FCC\n(b) \\((x)(0.42906 \\mathrm{~nm})=(4)(0.1858 \\mathrm{~nm})\\)\n\\(x=\\sqrt{3}\\), therefore BCC",
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"is_select": 1
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},
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{
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"idx": 13,
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"question": "The density of potassium, which has the BCC structure and one atom per lattice point, is \\(0.855 \\mathrm{~g} / \\mathrm{cm}^{3}\\). The atomic weight of potassium is \\(39.09 \\mathrm{~g} / \\mathrm{mol}\\). Calculate (a) the lattice parameter; and (b) the atomic radius of potassium.",
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"answer": "(a) Using Equation 3-5:\n\\[\n\\begin{array}{l}\n0.855 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{\\left(2 \\text { atoms } / \\mathrm{cell}\\right)(39.09 \\mathrm{~g} / \\mathrm{mol})}{\\left(a_{0}\\right)^{3}\\left(6.02 \\times 10^{-33} \\text { atoms } / \\mathrm{mol}\\right)} \\\\\na_{3}^{3}=1.5189 \\times 10^{-22} \\mathrm{~cm}^{3} \\text { or } a_{0}=5.3355 \\times 10^{-8} \\mathrm{~cm}\n\\end{array}\n\\]\n(b) From the relationship between atomic radius and lattice parameter:\n\\[\nr=\\frac{(\\sqrt{3})(5.3355 \\times 10^{-8} \\mathrm{~cm})}{4}=2.3103 \\times 10^{-8} \\mathrm{~cm}\n\\]",
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"is_select": 1
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},
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{
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"idx": 14,
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"question": "The density of thorium, which has the FCC structure and one atom per lattice point, is \\(11.72 \\mathrm{~g} / \\mathrm{cm}^{3}\\). The atomic weight of thorium is \\(232 \\mathrm{~g} / \\mathrm{mol}\\). Calculate (a) the lattice parameter and (b) the atomic radius of thorium.",
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"answer": "(a) From Equation 3-5:\n\\[\n\\begin{array}{l}\n11.72 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{(4 \\text { atoms } / \\mathrm{cell})(232 \\mathrm{~g} / \\mathrm{mol})}{\\left(a_{0}\\left)^{3}\\left(6.02 \\times 10^{-23} \\text { atoms } / \\mathrm{mol}\\right)\\right.} \\\\\na_{0}^{3}=1.315297 \\times 10^{-22} \\mathrm{~cm}^{3} \\mathrm{or} a_{0}=5.0856 \\times 10^{-8} \\mathrm{~cm} \\\\\n\\text { (b) From the relationship between atomic radius and lattice parameter: } \\\\\nr=\\frac{(\\sqrt{2})(5.0856 \\times 10^{-8} \\mathrm{~cm})}{4}=1.7980 \\times 10^{-8} \\mathrm{~cm}\n\\",
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"is_select": 1
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},
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{
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"idx": 15,
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"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.",
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"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",
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"is_select": 1
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},
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{
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"idx": 16,
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"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}\\).",
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"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",
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"is_select": 1
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},
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{
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"idx": 17,
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"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?",
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"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\\]",
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"is_select": 1
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},
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{
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"idx": 18,
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"question": "Bismuth has a hexagonal structure, with \\(a_{0}=0.4546 \\mathrm{~nm}\\) and \\(c_{0}=1.186 \\mathrm{~nm}\\). The density is \\(9.808 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and the atomic weight is \\(208.98 \\mathrm{~g} / \\mathrm{mol}\\). Determine (a) the volume of the unit cell and (b) the number of atoms in each unit cell.",
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"answer": "(a) The volume of the unit cell is \\(V=a_{1}^{2} c_{\\text {vcos30 }}\\).\n\\[\n\\begin{aligned}\nV & =\\left(0.4546 \\mathrm{~nm}\\right)^{2}(1.186 \\mathrm{~nm})(\\cos 30)=0.21226 \\mathrm{~nm}^{3} \\\\\n& =2.1226 \\times 10^{-22} \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]\n(b) If \" \\(x\\) \" is the number of atoms per unit cell, then:\n\\[\n\\begin{aligned}\n9.808 \\mathrm{~g} / \\mathrm{cm}^{3} & =\\frac{(x \\text { atoms } / \\mathrm{cell})(208.98 \\mathrm{~g} / \\mathrm{mol})}{\\left(2.1226 \\times 10^{-22} \\mathrm{~ cm}^{3}\\right)\\left(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol}\\right)} \\\\\nx & =6 \\text { atoms } / \\mathrm{cell}\n\\end{aligned}\n\\]",
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"is_select": 1
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},
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{
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"idx": 19,
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"question": "Gallium has an orthorhombic structure, with \\(a_{0}=0.45258 \\mathrm{~nm}, b_{0}=0.45186 \\mathrm{~nm}\\), and \\(c_{0}=0.76570 \\mathrm{~nm}\\). The atomic radius is \\(0.1218 \\mathrm{~nm}\\). The density is \\(5.904 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and the atomic weight is \\(69.72 \\mathrm{~g} / \\mathrm{mol}\\). Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell.",
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"answer": "The volume of the unit cell is \\(V=a_{0} b_{0} c_{0}\\) or\n\\[\n\\begin{aligned}\nV & =\\left(0.45258 \\mathrm{~nm}\\right)(0.45186 \\mathrm{~nm})(0.76570 \\mathrm{~nm})=0.1566 \\mathrm{~nm}^{3} \\\\\n& =1.566 \\times 10^{-22} \\mathrm{~cm}^{3}\nend\n\\end{aligned}\n\\]\n(a) From the density equation:\n\\[\n\\begin{aligned}\n5.904 \\mathrm{~g} / \\mathrm{cm}^{3} & =\\frac{(\\text { x atoms } / \\mathrm{cell})(69.72 \\mathrm{~g} / \\mathrm{mol})}{\\left(1.566 \\times 10^{-22} \\mathrm{~cm3}\\right)\\left(6.02 \\times 10^{23} a_{0} \\text { atoms } / \\mathrm{mol}\\right)} \\\\\nx & =8 \\text { atoms } / \\mathrm{cell}\n\\end{aligned}\n\\]\n(b) From the packing factor (PF) equation:\n\\[\n\\mathrm{PF}=\\frac{(8 \\text { atoms } / \\mathrm{cell})(4 \\mathrm{rt} / 3)(0.1218 \\mathrm{~nm})^{3}}{0.1566 \\mathrm{~nm}^{3}}=0.387\n\\]",
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"is_select": 1
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},
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{
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"idx": 20,
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"question": "Beryllium has a hexagonal crystal structure, with \\(a_{0}=0.22858 \\mathrm{~nm}\\) and \\(c_{0}=0.35842 \\mathrm{~nm}\\). The atomic radius is \\(0.1143 \\mathrm{~nm}\\), the density is \\(1.848 \\mathrm{~g} / \\mathrm{cm}^{3}\\), and the atomic weight is \\(9.01 \\mathrm{~g} / \\mathrm{mol}\\). Determine (a) the numberof atoms in each unit cell and (b) the packing factor in the unit cell.",
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"answer": "(\\quad V=(0.22858 \\mathrm{~nm})^{2}(0.35842 \\mathrm{~nm}) \\cos 30=0.01622 \\mathrm{~nm}^{3}=16.22 \\times 10^{-24} \\mathrm{~cm}^{3}\\)\n(a) From the density equation:\n\\[\n\\begin{aligned}\n1.848 \\mathrm{~g} / \\mathrm{cm}^{3} & =\\overbrace{(x \\text { atoms } / \\mathrm{cell})(9.01 \\mathrm{~g} / \\mathrm{mol})}_{(16.22 \\times 10^{-24} \\mathrm{~cm3})(6.02 \\times 10^{23} \\text { atoms } / 0.3)} \\\\\nx & =2 \\text { atoms } / \\mathrm{cell} \\\\\n\\text { (b) The packing factor (PF) is: } \\\\\n\\mathrm{PF} & =\\frac{(2 \\text { atoms } / \\mathrm{cell})(4 \\mathrm{rt} / 23)(0.1143 \\mathrm{~nm})^{3}}{0.01622 \\mathrm{~nm}^{3}}=0.77\n\\end{aligned}\n\\]",
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"is_select": 1
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},
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{
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"idx": 21,
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"question": "A typical paper clip weighs \\(0.59 \\mathrm{~g}\\) and consists of BCC iron. Calculate (a) the number of unit cells and (b) the number of iron atoms in the paper clip.",
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"answer": "The lattice parameter for BCC iron is \\(2.866 \\times 10^{-8} \\mathrm{~cm}\\). Therefore\n\\[\nV_{\\text {unit cell }}=\\left(2.866 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}=2.354 \\times 10^{-23} \\mathrm{~cm}^{3}\n\\]\n(a) The density is \\(7.87 \\mathrm{~g} / \\mathrm{cm}^{3}\\). The number of unit cells is:\n\\[\n\\text { number }=\\frac{0.59 \\mathrm{~g}}{\\left(7.87 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(2.354 \\times 10^{-23} \\mathrm{~cm}{ }^{3} / \\mathrm{cell}\\right)}=3.185 \\times 10^{21} \\text { cells }\n\\]\n(b) There are 2 atoms/cell in BCC iron. The number of atoms is:\n\\[\n\\text { number }=\\left(3.185 \\times 10^{21} \\text { cells }\\right)(2 \\text { atoms } / \\text { cell })=6.37 \\times 10^{21} \\text { atoms }\n\\]",
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"is_select": 1
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},
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{
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"idx": 22,
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"question": "Determine the planar density and packing fraction for BCC lithium in the (100), (110), and (111) planes. Which, if any, of these planes is close packed?",
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"answer": "(\\quad a_{\\mathrm{o}}=3.5089 \\AA\\)\nFor (100):\n\\[\n\\begin{array}{l}\n\\text { planar density }=\\frac{1}{\\left(3.5089 \\times 10^{-8} \\mathrm{~cm}\\right)^{2}}=0.0812 \\times 10^{16} \\text { points } / \\mathrm{cm}{ }^{2} \\\\\n\\text { packing fraction }=\\frac{\\pi\\left[\\sqrt{3} a_{\\mathrm{o}} / 4\\right]^{2}}{a_{\\mathrm{o}}^{2}}=0.589\n\\end{array}\n\\]24 The Science and Engineering of Materials Instructor's Solutions Manual\nFor (110):\nplanar density \\(=\\frac{2}{\\sqrt{2}\\left(3.5089 \\times 10^{-8} \\mathrm{~cm}\\right)^{2}}=0.1149 \\times 10^{16}\\) points \\(/ \\mathrm{cm}^{2}\\)\npacking fraction \\(=\\frac{2 \\sqrt{3} a_{0} / 4}{\\sqrt{2} a_{0}^{2}}=0.833\\)\nFor (111):\nThere are only \\((3)\\left(\\lambda_{0}\\right)=\\frac{1 / 2}{\\lambda_{0}}\\) points in the plane, which has an area of \\(0.866 a_{0}^{2}\\).\nplanar density \\(=\\frac{1 / 2}{0.866\\left(3.5089 \\times 10^{-8} \\mathrm{cm}\\right)^{2}}=0.0469 \\times 10^{16}\\) points \\(/ \\mathrm{cm}^{\\mathbf{2}}\\)\npacking fraction \\(=\\frac{1 / 2}{0.866 a_{0}^{2}} \\sqrt{3} a_{0} / 4\\)\nThere is no close-packed plane in BCC structures.",
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"is_select": 1
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},
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{
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"idx": 23,
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"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.",
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"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",
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"is_select": 1
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},
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{
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"idx": 24,
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"question": "In a FCC unit cell, how many \\(d_{111}\\) are present between the \\(0,0,0\\) point and the 1,1,1 point?",
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"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",
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"is_select": 1
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},
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{
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"idx": 25,
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"question": "Determine the minimum radius of an atom that will just fit into (a) the tetrahedral interstitial site in FCC nickel and (b) the octahedral interstitial site in BCC lithium.",
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"answer": "(a) For the tetrahedral site in FCC nickel \\(\\left(\\sigma_{\\mathrm{c}}=3.5167 \\AA\\right)\\) :\n\\[\n\\begin{array}{l}\nr_{\\mathrm{Ni}}=\\frac{\\sqrt{2}(3.5167 \\AA)}{4}=1.243 \\AA \\\\\nr / r_{\\mathrm{Ni}}=0.225 \\text { for a tetrahedral site. Therefore: } \\\\\nr=\\langle 1.243 \\AA\\rangle(0.225)=0.2797 \\AA \\\\\n\\text { (b) For the octahedral site in BCC lithium }\\left(a_{\\mathrm{o}}=3.5089 \\AA\\right) \\\\\nr_{\\mathrm{Li}}=\\frac{\\sqrt{3}(3.5089)}{4}=1.519 \\AA \\\\\nr / r_{\\mathrm{Li}}=0.414 \\text { for an octahedral site. Therefore: } \\\\\nr=(1.519 \\AA)(0.414)=0.629 \\AA\n\\end{array}\n\\]",
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"is_select": 1
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},
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{
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"idx": 26,
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"question": "What is the radius of an atom that will just fit into the octahedral site in FCC copper without disturbing the crystal structure?",
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"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\\]",
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"is_select": 1
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},
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{
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"idx": 27,
|
||
"question": "Using the ionic radii, determine the coordination number expected for the following compounds.\n(a) \\(\\mathrm{Y}_{2} \\mathrm{O}_{3}\\)\n(b) \\(\\mathrm{UO}_{2}\\)\n(c) \\(\\mathrm{BaO}\\)\n(d) \\(\\mathrm{Si}_{4} \\mathrm{~N}_{4}\\)\n(e) \\(\\mathrm{GeO}_{2}\\)\n(f) \\(\\mathrm{MnO}\\)\n(g) \\(\\mathrm{MgS}\\)\n(h) \\(\\mathrm{KBr}\\)",
|
||
"answer": "(a) \\(r_{\\mathrm{Y}}{ }^{3 / 2} / r_{\\mathrm{O}}{ }^{-2}=\\frac{0.89}{1.32}=0.67 \\quad \\mathrm{CN}=6\\) (e) \\(r_{\\mathrm{ce}}{ }^{4 / 4} / r_{\\mathrm{O}}{ }^{-2}=\\frac{0.53}{1.32}=0.40 \\quad \\mathrm{CN}=4\\)\n(b) \\(r_{\\mathrm{U}}{ }^{1 / 4} / r_{\\mathrm{O}}{ }^{-2}=\\frac{\\frac{0.97}{1.32}=0.73}{} \\mathrm{CN}=6\\) (f) \\(r_{\\mathrm{XBa}}{ }^{2 / 2} / r_{\\mathrm{O}}{ }^{-2}=\\frac{\\frac{1.08}{1.32}=0.61}{} \\mathrm{CN}=6\\)\n(c) \\(r_{\\mathrm{O}}{ }^{-2} / r_{\\mathrm{Ba}}{ }^{2}=\\frac{1.32}{1.34}=0.99 \\quad \\mathrm{CN}=8\\) (g) \\(r_{\\mathrm{Mg}}{ }^{2} / r_{\\mathrm{Si}}{ }^{2}=\\frac{0.66}{1.32}=0.50 \\quad \\mathrm{CN}=6\\)\n(d) \\(r_{\\mathrm{N}}{ }^{3} / r_{\\mathrm{Si}}{ }^{4}=0.15 \\quad 0.42 \\quad \\mathrm{CN}=4\\) (h) \\(r_{\\mathrm{K}}{ }^{1} / r_{\\mathrm{B}}{ }^{-1}=\\frac{1.33}{1.96}=0.68 \\quad \\mathrm{CN}=6\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 28,
|
||
"question": "Would you expect \\(\\mathrm{NiO}\\) to have the cesium chloride, sodium chloride, or zinc blende structure? Based on your answer, determine (a) the lattice parameter, (b) the density, and (c) the packing factor.",
|
||
"answer": "(\\quad r_{\\mathrm{N}+2}^{\\prime \\prime}=0.69 \\AA \\quad r_{\\mathrm{O}^{-2}}^{\\prime \\prime}=1.32 \\AA \\quad r_{\\mathrm{N}+2}^{\\prime \\prime}=0.52 \\quad \\mathrm{CN}=6\\)\nA coordination number of 8 is expected for the \\(\\mathrm{CsCl}\\) structure, and a coordination number of 4 is expected for \\(\\mathrm{ZnS}\\). But a coordination number of 6 is consistent with the \\(\\mathrm{NaCl}\\) structure.\n(a) \\(a_{0}=2(0.69)+2(1.32)=4.02 \\AA\\)\n(b) \\(\\rho=\\frac{(4 \\text { of each ion }(\\text { cell })(58.71+16 \\mathrm{~g} / \\mathrm{mol})}{(4.02 \\times 10^{-8} \\mathrm{~cm})^{3}(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol})}=7.64 \\mathrm{~g} / \\mathrm{cm}^{3}}=7.64 \\mathrm{~g} / \\mathrm{cm}^{3}\\)\n(c) \\(\\mathrm{PF}=\\frac{(4 \\pi / 3)(4 \\text { ions } / \\mathrm{cell})\\left[(0.69)^{3}+(1.32)^{3}\\right]}{(4.02)^{3}}=0.678\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 29,
|
||
"question": "Would you expect \\(\\mathrm{UO}_{2}\\) to have the sodium chloride, zinc blende, or fluoride structure? Based on your answer, determine (a) the lattice parameter, (b) the density, and (c) the packing factor.",
|
||
"answer": "(\\quad r_{\\mathrm{U}^{+4}}=0.97 \\AA \\quad r_{\\mathrm{O}^{-2}}=1.32 \\AA \\quad r_{\\mathrm{U}^{+4}}=0.97 / 1.32=0.735\\)\nvalence of \\(\\mathrm{U}=+4\\), valence of \\(\\mathrm{O}=-2\\)\nThe radius ratio predicts a coordination number of 8 ; however there must be twice as many oxygen ions as uranium ions in order to balance the charge. The fluorite structure will satisfy these requirements, with:\n\\(\\mathrm{U}=\\mathrm{FCC}\\) position (4) \\(\\mathrm{O}=\\) tetrahedral position (8)\n(a) \\(\\sqrt{3} a_{0}=4 r_{0}+4 r_{0}=4(0.97+1.32)=9.16\\) or \\(a_{0}=5.2885 \\AA\\)\n(b) \\(\\rho=\\frac{4(238.03 \\mathrm{~g} / \\mathrm{mol})+8(16 \\mathrm{~g} / \\mathrm{mol})}{(5.2885 \\times 10^{-3} \\mathrm{~cm})^{3}(6.02 \\times 10^{\\circ} 23 \\text { atoms } / \\mathrm{mol})}=12.13 \\mathrm{~g} / \\mathrm{cm}^{3}\\)\n(c) \\({\\mathrm{PF}=\\frac{(4 \\pi / 3)(4(0.97)^{3}+8(1.32)^{3}}{(5.2885)^{3}}}=0.624\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 30,
|
||
"question": "Would you expect \\(\\mathrm{BeO}\\) to have the sodium chloride, zinc blende, or fluorite structure? Based on your answer, determine (a) the lattice parameter, (b) the density, and (c) the packing factor.",
|
||
"answer": "(\\quad r_{\\mathrm{He}^{+}}^{2}=0.35 \\AA \\quad r_{\\mathrm{O}^{-2}}^{2}=1.32 \\AA\\)\n\\[\n\\begin{array}{l}\nr_{\\mathrm{He} / r_{\\mathrm{O}}}=0.265 \\quad \\mathrm{CN}=4 \\quad \\therefore \\text { Zinc Blende } \\\\\n\\text { (a) } \\sqrt{3} a_{0}=4 r_{\\mathrm{He}^{+}}^{2}+4 r_{\\mathrm{O}^{-2}}^{2}=4(0.35+1.32)=6.68 \\text { or } a_{0}=3.8567 \\AA \\\\\n\\text { (b) } \\rho=\\frac{4(9.01+16 \\mathrm{~g} / \\mathrm{mol})}{(3.8567 \\times 10^{-8} \\mathrm{~cm})^{3}\\left(6.02 \\times 10^{23} \\text { atoms } / m o l\\right)}=2.897 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n\\text { (c) } \\mathrm{PF}=\\frac{(4 \\pi / 3)(4)(0.35)^{3}+8(1.32)^{3}}{(3.8567)^{3}}=0.684\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 31,
|
||
"question": "Would you expect CsBr to have the sodium chloride, zinc blende, fluorite, or cesium chloride structure? Based on your answer, determine (a) the lattice parameter, (b) the density, and (c) the packing factor.",
|
||
"answer": "(\\quad r_{\\mathrm{Cs}}{ }^{+1}=1.67 \\AA \\quad r_{\\mathrm{Br}}{ }^{-1}=1.96 \\AA\\)\n\\[\n\\begin{array}{l}\nr_{\\mathrm{Cs}}{ }^{-1}=0.852 \\quad \\mathrm{CN}=8 \\quad \\therefore \\mathrm{CsCl} \\\\\nr_{\\mathrm{Br}}{ }^{-1} \\\\\n\\text { (a) } \\sqrt{3} a_{0}=2 r_{\\mathrm{Cs}}{ }^{+1}+2 r_{\\mathrm{Br}}{ }^{-1}=2(1.96+1.67)=7.26 \\text { or } a_{0}=4.1916 \\AA \\\\\n\\text { (b) } \\rho=\\frac{(4.1916 \\times 10^{-8} \\mathrm{~cm})^{3}\\left(6.02 \\times 10^{23} \\mathrm{~atoms} / \\mathrm{mol}\\right)}{\\left(4 \\pi / 3\\right)\\left[(1.96)^{3}+\\left(1.67\\right)^{3}\\right]}=4.8 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n\\text { (c) } \\mathrm{PF}=\\frac{(4.1916)^{3}}{(4.1916)^{3}}=0.693\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 32,
|
||
"question": "Sketch the ion arrangement on the (110) plane of \\(\\mathrm{ZnS}\\) (with the zinc blende structure) and compare this arrangement to that on the (110) plane of \\(\\mathrm{CaF}_{2}\\) (with the flourite structure). Compare the planar packing fraction on the (110) planes for these two materials.",
|
||
"answer": "(\\quad \\mathrm{ZnS}\\) :\n\\[\n\\begin{aligned}\n\\sqrt{3} a_{0} & =4 r_{\\mathrm{Zn}}{ }^{+2}+4 r_{\\mathrm{S}}{ }^{-2} \\\\\n\\sqrt{3} a_{0} & =4(0.074 \\mathrm{~nm})+4(0.184 \\mathrm{~nm}) \\\\\na_{0} & =0.596 \\mathrm{~nm} \\\\\n\\mathrm{PPF} & =\\frac{\\left(2\\left(\\pi r_{\\mathrm{Zn}}{ }^{+}\\right)+\\left(2\\right)\\left(\\pi r_{\\mathrm{S}}{ }^{+}\\right)}{\\left(\\sqrt{2} a_{0}\\right) a_{0}}=\\frac{2 \\pi\\left(0.074\\right)^{2}+2 \\pi\\left(0.184\\right)^{2}}{\\sqrt{2}(0.596 \\mathrm{~nm})^{2}}=0.492\n\\end{aligned}\n\\]\n\n\\[\n\\begin{array}{l}\n\\mathrm{CaF}_{2} \\text { s } \\\\\n\\sqrt{3} a_{0}=4 r_{\\mathrm{Cs}}{ }^{+2}+4 r_{\\mathrm{F}}{ }^{-1} \\\\\n\\sqrt{3} a_{0}=4(0.099 \\mathrm{~nm})+4(0.133 \\mathrm{~nm}) \\\\\na_{0}=0.536 \\mathrm{~nm} \\\\\n\\mathrm{PPF}=\\frac{\\left(2\\left(\\pi r_{\\mathrm{C}}{ }^{+}\\right)+\\left(4\\right)\\left(\\pi r_{\\mathrm{F}}^{+}\\right)}{\\left(\\sqrt{2} a_{0}\\right) a{ }_{0}}=\\frac{2 \\pi(0.099)^{2}+4 \\pi(0.133)^{2}}{\\sqrt{2}(0.536 \\mathrm{~nm})^{2}}=0.699\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 33,
|
||
"question": "\\(\\mathrm{MgO}\\), which has the sodium chloride structure, has a lattice parameter of \\(0.396 \\mathrm{~nm}\\). Determine the planar density and the planar packing fraction for the (111) and (222) planes of \\(\\mathrm{MgO}\\). What ions are present on each plane?",
|
||
"answer": "As described in the answer to Problem 3-57, the area of the (111) plane is \\(0.866 a_{\\mathrm{a}}^{2}\\).\n\\[\na_{\\mathrm{o}}=2 r_{\\mathrm{Mg}}^{2-2}+2 r_{\\mathrm{O}}^{-2}=2(0.66+1.32)=3.96 \\AA\n\\]\n(111): P.D. \\(=\\frac{2 \\mathrm{Mg}}{(0.866)(3.96 \\times 10^{-8} \\mathrm{~cm})^{2}}=0.1473 \\times 10^{16}\\) points \\(/ \\mathrm{cm}^{2}\\)\n\\[\n\\begin{aligned}\n\\mathrm{PPF}= & \\frac{2 \\pi(0.66)^{2}}{(0.866)(3.96)^{2}}=0.202 \\\\\n& \\text { (222): P.D. }=0.1473 \\times 10^{16} \\text { points } / \\mathrm{cm}^{2} \\\\\n& \\mathrm{PPF}= & \\frac{2 \\pi(1.32)^{2}}{(0.866)(3.96)^{2}} =0.806\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 37,
|
||
"question": "The density of a sample of HCP beryllium is \\(1.844 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and the lattice parameters are \\(a_{0}=0.22858 \\mathrm{~nm}\\) and \\(c_{0}=0.35842 \\mathrm{~nm}\\). Calculate (a) the fraction of the lattice points that contain vacancies and (b) the total number of vacancies in a cubic centimeter:",
|
||
"answer": "(\\quad V_{\\mathrm{u}, \\mathrm{c}}=(0.22858 \\mathrm{~nm})^{2}(0.35842 \\mathrm{~nm}) \\cos 30=0.01622 \\mathrm{~nm}^{3}\\)\n\\[\n\\begin{array}{l}\n=1.622 \\times 10^{-23} \\mathrm{~cm}^{3} \\\\\n=1.844 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{(x)(9.01 \\mathrm{~g} / \\mathrm{mol})}{\\left(1.622 \\times 10^{-23} \\mathrm{~cm^{3}}\\right)(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol})} \\quad x=1.9984 \\\\\n\\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad\n\\end{array}\n\\]\n\\[\n\\begin{array}{l}\n\\text { fraction }=\\frac{2-1.9984}{2}=0.0008 \\\\\n\\text { (b) number }=\\frac{0.0016 \\text { vacancies/uc }}{1.622 \\times 10^{-23} \\mathrm{~cm}{ }^{3}}=0.986 \\times 10^{20} \\text { vacancies } / \\mathrm{cm}^{3}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 38,
|
||
"question": "BCC lithium has a lattice parameter of \\(3.5089 \\times 10^{-8} \\mathrm{~cm}\\) and contains one vacancy per 200 unit cells. Calculate (a) the number of vacancies per cubic centimeter and (b) the density of Li.",
|
||
"answer": "(a) \\(\\frac{1 \\text { vacancy }}{(200)(3.5089 \\times 10^{-8} \\mathrm{~cm})^{3}}=1.157 \\times 10^{20} \\text { vacancies } / \\mathrm{cm}{ }^{3}\\)\n(b) In 200 unit cells, there are \\(399 \\mathrm{Li}\\) atoms. The atoms/cell are \\(399 / 200\\) :\n\\[\n\\rho=\\frac{(399 / 200)(6.94 \\mathrm{~g} / \\mathrm{mol})}{\\left(3.5089 \\times 10^{-8} \\mathrm{~cm}^{3}\\right)(6.02 \\times 10^{23} \\text { atoms/ mol })}=0.532 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 39,
|
||
"question": "FCC lead has a lattice parameter of \\(0.4949 \\mathrm{~nm}\\) and contains one vacancy per \\(500 \\mathrm{~Pb}\\) atoms. Calculate (a) the density and (b) the number of vacancies per gram of \\(\\mathrm{Pb}\\).",
|
||
"answer": "(a) The number of atoms/cell \\(=(499 / 500)(4\\) sites/cell)\n\\[\n\\rho=\\frac{(499 / 500)(4)(207.19 \\mathrm{~g} / \\mathrm{mol})}{\\left(4.949 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}(6.02 \\times 10^{23} \\text { atoms } / m \\mathrm{~mol})}=11.335 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\]\n(b) The \\(500 \\mathrm{~Pb}\\) atoms occupy \\(500 / 4=125\\) unit cells:\n\\[\n\\frac{1 \\text { vacancy }}{\\left(125 \\text { cells }\\right)} \\times\\left[(1 / 11.335 \\mathrm{~g} / \\mathrm{cm}^{2})\\right]=5.82 \\times 10^{18} \\text { vacancies } / \\mathrm{g}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 43,
|
||
"question": "Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of \\(0.2867 \\mathrm{~nm}\\). For the \\(\\mathrm{Fe}-\\mathrm{C}\\) alloy, find (a) the density and (b) the packing factor.",
|
||
"answer": "There is one carbon atom per 100 iron atoms, or \\(1 \\mathrm{C} / 50\\) unit cells, or \\(1 / 50 \\mathrm{C}\\) per unit cell:\n(a) \\(\\rho=\\frac{(2)(55.847 \\mathrm{~g} / \\mathrm{mol})+(1 / 50)(12 \\mathrm{~g} / \\mathrm{mol})}{(2.867 \\times 10^{-8} \\mathrm{~cm})^{3}(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol})}=7.89 \\mathrm{~g} / \\mathrm{cm}^{3}\\)\n\\[\n\\text { (b) }\n\\]\n\\[\n\\text { Packing Factor }=\\frac{2(4 \\pi t / 3)(1.241)^{2}+(1 / 50)(4 \\pi t / 3)(0.77)^{3}}{(2.867)^{3}}=0.681\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 44,
|
||
"question": "The density of BCC iron is \\(7.882 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and the lattice parameter is \\(0.2886 \\mathrm{~nm}\\) when hydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms and (b) the number of unit cells required on average that contain hydrogen atoms.",
|
||
"answer": "(a) \\(7.882 \\mathrm{~g} / \\mathrm{cm}^{2}=\\frac{2(55.847 \\mathrm{~g} / \\mathrm{mol})+x(1.00797 \\mathrm{~g} / \\mathrm{mol})}{(2.866 \\times 10^{-8} \\mathrm{~cm})^{3}(6. \\\\\n\\left. \\times 0.0081 \\mathrm{H} \\text { atoms } / \\mathrm{cell}\\right.\\)\nThe total atoms per cell include \\(2 \\mathrm{Fe}\\) atoms and \\(0.0081 \\mathrm{H}\\) atoms. Thus:\n\\[\nf_{\\mathrm{H}}=\\frac{0.0081}{2.0081}=0.004\n\\]\n(b) Since there is \\(0.0081 \\mathrm{H} / \\mathrm{cell}\\), then the number of cells containing \\(\\mathrm{H}\\) atoms is:\n\\[\n\\text { cells }=1 / 0.0081=123.5 \\text { or } 1 \\mathrm{H} \\text { in } 123.5 \\text { cells }\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 45,
|
||
"question": "Suppose one Schottky defect is present in every tenth unit cell of \\(\\mathrm{MgO}\\). \\(\\mathrm{MgO}\\) has the sodium chloride crystal structure and a lattice parameter of \\(0.396 \\mathrm{~nm}\\). Calculate (a) the number of anion vacancies per \\(\\mathrm{cm}^{3}\\) and (b) the density of the ceramic.",
|
||
"answer": "In 10 unit cells, we expect \\(40 \\mathrm{Mg}+40 \\mathrm{O}\\) ions, but due to the defect:\n\\[\n\\begin{array}{l}\n40 \\mathrm{Mg}-1=39 \\\\\n40 \\mathrm{O}-1=39\n\\end{array}\n\\]\n(a) 1 vacancy \\(/(10\\) cells \\()(3.96 \\times 10^{-8} \\mathrm{~cm})^{3}=1.61 \\times 10^{21}\\) vacancies \\(/ \\mathrm{cm}^{3}\\)\n(b) \\(\\quad(39 / 40)(4)(24.312 \\mathrm{~g} / \\mathrm{mol})+(39 / 40)(4)(16 \\mathrm{~g} / \\mathrm{mol})\\)\n\\[\n\\text { (3.96 } \\times 10^{-8} \\mathrm{~cm})^{3}(6.02 \\times 10^{23} \\text { atoms } / \\mathrm{mol})=4.205 \\mathrm{~g} / \\mathrm{cm}^{3}\\)\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 46,
|
||
"question": "\\(\\mathrm{ZnS}\\) has the zinc blende structure. If the density is \\(3.02 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and the lattice parameter is \\(0.59583 \\mathrm{~nm}\\), determine the number of Schottky defects (a) per unit cell and (b) per cubic centimeter.",
|
||
"answer": "Let \\(x\\) be the number of each type of ion in the unit cell. There normally are 4 of each type.\n(a) \\(3.02 \\mathrm{~g} / \\mathrm{cm}^{3}=\\frac{x(65.38 \\mathrm{~g} / \\mathrm{mol})+x(32.064 \\mathrm{~g} / \\mathrm{mol})}{\\left(5.9583 \\times 10^{-8} \\mathrm{~cm}\\right)^{6}(0.02 \\times 10^{23} \\text { ions } / \\mathrm{mol})} \\quad x=3.9465\\)\n\\[\n\\begin{array}{l}\n4-3.9465=0.0535 \\text { defects } / \\mathrm{u.c} \\\\\n\\text { (b) \\# of unit cells } / \\mathrm{cm}^{3}=1 /\\left(5.9683 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}=4.704 \\times 10^{21} \\\\\n\\text { Schottky defects per } \\mathrm{cm}^{3}=\\left(4.704 \\times 10^{21}\\right)(0.0535)=2.517 \\times 10^{20}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 47,
|
||
"question": "Calculate the length of the Burgers vector in the following materials:\n(a) BCC niobium (b) FCC silver (c) diamond cubic silicon",
|
||
"answer": "(a) The repeat distance, or Burgers vector, is half the body diagonal, or:\n\\[\nb=\\text { repeat distance }=\\left(\\sqrt{2}\\right) \\cdot\\left(\\sqrt{3}\\right)(3.294 \\AA)=2.853 \\AA\n\\]\n(b) The repeat distance, or Burgers vector, is half of the face diagonal, or:\n\\[\nb=\\left(\\sqrt{2}\\right) \\cdot\\left(\\sqrt{2} d_{\\mathrm{e}}\\right)=\\left(\\sqrt{2}\\right) \\cdot\\left(\\sqrt{2}\\right)(4.0862 \\AA)=2.889 \\AA\n\\]\n(c) The slip direction is , where the repeat distance is half of the face diagonal:\n\\[\nb=\\left(\\sqrt{2}\\right) \\cdot\\left(\\sqrt{\\Sigma}\\right)(5.4307 \\AA)=3.840 \\AA\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 48,
|
||
"question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of \\(5000 \\mathrm{psi}\\). Calculate the resolved shear stress acting on the (111) slip plane in the [T, and [T] slip directions. Which slip system(s) will become active first?",
|
||
"answer": "(\\quad \\phi=54.76^{\\circ} \\quad \\tau=5000 \\cos 54.76 \\cos \\lambda\\)\n\\[\n\\begin{array}{l}\n\\lambda_{110}=90^{\\circ} \\quad \\tau=0 \\\\\n\\lambda_{011}=45^{\\circ} \\quad \\tau=2040 \\text { psi active } \\\\\n\\lambda_{101}=45^{\\circ} \\quad \\tau=2040 \\text {\npsi active }\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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), (011), and (101) slip planes.",
|
||
"answer": "(\\quad \\mathrm{CRSS}=12,000 \\mathrm{psi}=\\sigma \\cos \\phi \\cos \\lambda\\)\n\\[\n\\begin{array}{l}\n\\lambda=54.76^{\\alpha} \\quad \\frac{12,000 \\mathrm{psi}}{\\cos \\phi \\cos \\lambda}=\\sigma \\\\\n\\phi_{110}=90^{\\circ} \\quad \\sigma=\\infty \\\\\n\\phi_{011}=45^{\\circ} \\quad \\sigma=29,412 \\mathrm{psi} \\\\\n\\phi_{101}=45^{\\circ} \\quad \\sigma=29,412\\) psi\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 50,
|
||
"question": "The strength of titanium is found to be 65,000 psi when the grain size is \\(17 \\times\\) \\(10^{-6} \\mathrm{~m}\\) and 82,000 psi when the grain size is \\(0.8 \\times 10^{-6} \\mathrm{~m}\\). Determine (a) the constants in the Hall-Petch equation and (b) the strength of the titanium when the grain size is reduced to \\(0.2 \\times 10^{-6} \\mathrm{~m}\\).",
|
||
"answer": "(\\quad 65000=\\sigma_{\\mathrm{o}}+\\mathrm{K} \\frac{1}{\\sqrt{17} \\times 10^{-5}}=\\sigma_{\\mathrm{o}}+242.5 \\mathrm{~K}\\)\n\\[\n82000=\\sigma_{\\mathrm{o}}+\\mathrm{K} \\frac{I}{\\sqrt{0.8 \\times 10^{-6}}}=\\sigma_{\\mathrm{o}}+1118.0 \\mathrm{~K}\n\\]\n(a) By solving the two simultaneous equations:\n\\[\n\\mathrm{K}=19.4 \\mathrm{psi} / \\sqrt{d} \\quad \\sigma_{\\mathrm{o}}=60,290 \\mathrm{psi}\n\\]\n(b) \\(\\sigma=60,290+19.4 / \\sqrt{0.2 \\times 10^{-6}}=103670 \\mathrm{psi}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 51,
|
||
"question": "For an ASTM grain size number of 8, calculate the number of grains per square inch\n(a) at a magnification of 100 and (b) with no magnification.",
|
||
"answer": "(\\quad(a) N=2^{n-1} \\quad N=2^{n-1}=2^{7}=128\\) grains/in. \\({ }^{2}\\)\n(b) No magnification means that the magnification is \" 1 \":\n\\[\n(27)(100 / 1)^{2}=1.28 \\times 10^{6} \\text { grains/in. }{ }^{2}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 57,
|
||
"question": "The diffusion coefficient for \\(\\mathrm{Cr}^{+3}\\) in \\(\\mathrm{Cr}_{2} \\mathrm{O}_{3}\\) is \\(6 \\times 10^{-15} \\mathrm{~cm}^{2} / \\mathrm{s}\\) at \\(727^{\\circ} \\mathrm{C}\\) and is \\(1 \\times 10^{-9} \\mathrm{~cm}^{2} / \\mathrm{s}\\) at \\(1400^{\\circ} \\mathrm{C}\\). Calculate (a) the activation energy and (b) the constant \\(D_{y}\\) :\n\\[\n\\begin{array}{l}\n\\text {",
|
||
"answer": "\\quad \\text { (a) } \\frac{6 \\times 10^{-15}}{1 \\times 10^{-9}}=\\frac{D_{0} \\exp \\left[-Q /(1.987)(1000)\\right]}{D_{0} \\exp \\left[-Q /(1.987)(1673)\\right]} \\\\\n\\quad 6 \\times 10^{-6}=\\exp \\left[-Q(0.000503-0.00030)\\right]=\\exp \\left[-0.000203 Q\\right] \\\\\n-12.024=-0.000203 Q \\quad \\text { or } \\quad Q=59,230 \\mathrm{cal} / \\mathrm{mol} \\\\\n\\text { (b) } 1 \\times 10^{-9} \\mathrm{D}_{0} \\exp \\left[-59,230 /(1.987)(1673)\\right]=D_{0} \\exp \\left(-17.818\\right) \\\\\n1 \\times 10^{-9}=1.828 \\times 10^{-8} D_{0} \\quad \\text { or } \\quad D_{0}=0.055 \\mathrm{~cm}^{2} / \\mathrm{s}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 58,
|
||
"question": "A \\(0.2-\\mathrm{mm}\\) thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains \\(1 \\mathrm{~Sb}\\) atom per \\(10^{8} \\mathrm{Si}\\) atoms and the other surface contains \\(500 \\mathrm{Sb}\\) atoms per \\(10^{8} \\mathrm{Si}\\) atoms. The lattice parameter for \\(\\mathrm{Si}\\) is \\(5.407 \\AA\\) (Appendix A). Calculate the concentration gradient in (a) atomic percent \\(\\mathrm{Sb}\\) per \\(\\mathrm{cm}\\) and (b) \\(\\mathrm{Sb}\\) atoms \\(/ \\mathrm{cm}^{3} \\cdot \\mathrm{cm}\\).",
|
||
"answer": "(\\quad \\Delta c / \\Delta x=\\frac{(1 / 10^{8}-500 / 10^{6}}{0.02 \\mathrm{~cm}} \\times 100 \\%=-0.02495\\) at \\(/ \\mathrm{cm}\\)\n\\[\n\\begin{array}{l}\na_{0}=5.4307 \\AA \\quad V_{\\text {unit cell }}=160.16 \\times 10^{-24} \\mathrm{~cm}^{3} \\\\\nc_{1}=\\frac{(8 \\mathrm{Si}\\right.\\) atoms \\(/ \\mu \\mathrm{c}.)\\left(1 \\mathrm{~Sb} / 10^{8} \\mathrm{Si}\\right)}{160.16 \\times 10^{-24} \\mathrm{~c} m^{3} / \\mathrm{u} . c}=-0.04995 \\times 10^{16} \\mathrm{~Sb} \\text { atoms } / \\mathrm{cm}^{3} \\\\\nc_{2}=\\frac{(8 \\mathrm{Si}\\right.\\) atoms \\(/ \\mu \\cdot \\mathrm{c}) \\cdot(500 \\mathrm{Sb} / 10^{8} \\mathrm{Si})}{160.16 \\times 10^{-24} \\mathrm{~m}^{3} / \\mathrm{u} . c}=-24.975 \\times 10^{16} \\mathrm{~Sb} \\text {- atoms } / \\mathrm{cm}^{3} \\\\\n\\Delta c / \\Delta x=\\frac{(0.04995-24.975) \\times 10^{16}}{0.02 \\mathrm{~cm}}=-1.246 \\times 10^{-19} \\mathrm{~Sb} \\text { atoms } / \\mathrm{cm}^{-3} \\cdot \\mathrm{cm}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 59,
|
||
"question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion \\(0.025 \\mathrm{~mm}\\) away contains 20 atomic percent zinc. The lattice parameter for the FCC alloy is \\(3.63 \\times 10^{-8} \\mathrm{~cm}\\). Determine the concentration gradient in (a) atomic percent \\(\\mathrm{Zn}\\) per \\(\\mathrm{cm}\\), (b) weight percent \\(\\mathrm{Zn}\\) per \\(\\mathrm{cm}\\).",
|
||
"answer": "(a) \\(\\Delta c / \\Delta x=\\frac{20 \\%-25 \\%}{(0.025 \\mathrm{~mm})(0.1 \\mathrm{~cm} / \\mathrm{mm})}=-2000\\) at \\(\\% \\mathrm{Zn} / \\mathrm{cm}\\)\n(b) We now need to determine the wt\\% of zinc in each portion:\n\\[\n\\begin{array}{l}\n\\text { wt } \\% \\text { Zn }=\\frac{(20)(65.38 \\mathrm{~g} / \\mathrm{mol})}{(20)(65.38)+(80)(63.54)} \\times 100=20.46 \\\\\n\\text { wt } \\% \\text { Zn }=\\frac{(25)(65.38 \\mathrm{~g} / \\mathrm{mol})}{\\left(25)(65.38)+(75)(63.54)} \\times 100=25.54 \\\\\n\\Delta c / \\Delta x=\\frac{20.46 \\%-25.54 \\%}{0.0025 \\mathrm{~cm}}=-2032 \\mathrm{wt} \\% \\mathrm{Zn} / \\mathrm{cm} \\\\\n\\",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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^{\\circ} \\mathrm{C} .5 \\times 10^{8} \\mathrm{H}\\) atoms \\(/ \\mathrm{cm}^{3}\\) are in equilibrium with the hot side of the foil, while \\(2 \\times 10^{3} \\mathrm{H}\\) atoms \\(/ \\mathrm{cm}^{3}\\) are in equilibrium with the cold side Determine (a) the concentration gradient of hydrogen and (b) the flux of hydrogen through the foil.",
|
||
"answer": "(a) \\(\\Delta c / \\Delta x=\\frac{2 \\times 10^{3}-5 \\times 10^{8}}{(0.001 \\mathrm{in})(2.54 \\mathrm{~cm} / \\mathrm{in})}\\) \\(=-1969 \\times 10^{8} \\mathrm{H}\\) atoms \\(/ \\mathrm{\\mathrm{cm}^{3} \\cdot \\mathrm{cm}}\\)\n(b) \\(J=-D(\\Delta c / \\Delta x)=-0.0012 \\exp \\left[-3600 /(1.987)(923)\\right]\\left(-1969 \\times 10^{8}\\right)\\)\n\\(J=0.33 \\times 10^{8} \\mathrm{H}\\) atoms \\(/ \\mathrm { cm}^{2} \\cdot \\mathrm{s}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 61,
|
||
"question": "A \\(1-\\mathrm{mm}\\) sheet of FCC iron is used to contain nitrogen in a heat exchanger at \\(1200^{\\circ} \\mathrm{C}\\). The concentration of \\(\\mathrm{N}\\) at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. Determine the flux of nitrogen through the foil in \\(\\mathrm{N}\\) atoms \\(/ \\mathrm{cm}^{2} \\cdot \\mathrm{s}\\).",
|
||
"answer": "(\\quad(a) \\Delta c / \\Delta x=\\frac{(0.00005-0.0004)(4 \\text { atoms per cell }) /(3.589 \\times 10^{-8} \\mathrm{~cm})^{3}}{(1 \\mathrm{~mm})(0.1 \\mathrm{~cm} / \\mathrm{mm})}\\)\n\\[\n\\begin{aligned}\n& =-3.03 \\times 10^{20} \\mathrm{~N} \\text { atoms } / \\mathrm{cm}^{3} \\cdot \\mathrm{cm} \\\\\n& (b) J=-D(\\Delta c / \\Delta x)=-0.0034 \\exp \\left[-34.600 /(1.987)(1.473) \\cdot(3.03 \\times 10^{20}\\right) \\\\\n& =7.57 \\times 10^{12} \\mathrm{~N} \\text { atoms } / \\mathrm{cm}^{2} \\cdot \\mathrm{s}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 63,
|
||
"question": "A BCC iron structure is to be manufactured that will allow no more than \\(50 \\mathrm{~g}\\) of hydrogen to be lost per year through each square centimeter of the iron at \\(400^{\\circ} \\mathrm{C}\\). If the concentration of hydrogen at one surface is \\(0.05 \\mathrm{H}\\) atom per unit cell and is 0.001 \\(\\mathrm{H}\\) atom per unit cell at the second surface, determine the minimum thickness of the iron.",
|
||
"answer": "(\\quad c_{1}=0.05 \\mathrm{H} /\\left(2.866 \\times 10^{-8} \\mathrm{em}\\right)^{3}=212.4 \\times 10^{19} \\mathrm{H}\\) atoms \\(/ \\mathrm{cm}^{3}\\)\n\\[\n\\begin{array}{l}\nc_{2}=0.001 \\mathrm{H} /\\left(2.866 \\times 10^{-6} \\mathrm{~cm}\\right)^{3}=4.25 \\times 10^{19} \\mathrm{H} \\text { atoms } / \\mathrm{cm}^{3} \\\\\n\\Delta c / \\Delta x=\\frac{4.25 \\times 10^{19}-212.4 \\times 10^{19}}{\\Delta x}=-\\frac{2.08 \\times 10^{21}}{\\Delta x}\n\\end{array}\n\\]\n\\[\n\\begin{array}{l}\nJ=\\frac{\\left(50 \\mathrm{~g} / \\mathrm{cm}^{2} \\mathrm{y}\\right)\\left(6.02 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)}{\\left(1.00797 \\mathrm{~g} / \\mathrm{mol}\\right)\\left(31.536 \\times 10^{6} \\mathrm{~s} / \\mathrm{y}\\right)}=9.47 \\times 10^{17} \\mathrm{H} \\mathrm{atoms} / \\mathrm{cm}^{2} \\cdot \\mathrm{s} \\\\\nJ=9.47 \\times 10^{17} \\mathrm{H}^{\\text {atoms }} / \\mathrm{cm}^{2} \\cdot \\mathrm{s} \\\\\n=(-2.08 \\times 10^{23} / \\mathrm{A} \\times)(0.0012) \\exp [-3600 \\mathrm{~L} / \\mathrm{(1.987)(673)})] \\\\\n\\Delta x=0.179 \\mathrm{~cm}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 65,
|
||
"question": "Compare the rate at which oxygen ions diffuse in \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\) with the rate at which aluminum ions diffuse in \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\) at \\(1500^{\\circ} \\mathrm{C}\\). Explain the difference.",
|
||
"answer": "(\\quad D_{\\mathrm{O}}{ }^{-2}=\\) 1900 exp[-152,000/(1.987)(1773)] = 3.47 \\times 10^{-16} \\mathrm{~cm}^{2} / \\mathrm{s}\\)\n\\[\nD_{\\mathrm{Al}}{ }^{+3}=28 \\exp [-114,000 /(1.987)(1773)]=2.48 \\times 10^{-13} \\mathrm{~cm}^{2} / \\mathrm{s}\n\\]\nThe ionic radius of the oxygen ion is \\(1.32 \\AA\\), compared with the aluminum ionic radius of \\(0.51 \\AA\\); consequently it is much easier for the smaller aluminum ion to diffuse in the ceramic.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 66,
|
||
"question": "Compare the diffusion coefficients of carbon in BCC iron at the allotropic transformation temperature of \\(912^{\\circ} \\mathrm{C}\\) and explain the difference.",
|
||
"answer": "(\\quad D_{\\mathrm{BCC}}=0.011 \\exp [-20.900 /(1.987)(1185)]=1.51 \\times 10^{-6} \\mathrm{~cm}^{2} / \\mathrm{s}\\)\n\\[\n\\begin{array}{l}\nD_{\\mathrm{FCC}}=0.23 \\exp [-32,900 /(1.987)(1185)]=1.92 \\times 10^{-7} \\mathrm{~cm}^{2} / \\mathrm{s}\n\\end{array}\n\\]\nPacking 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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 67,
|
||
"question": "Iron containing \\(0.05 \\% \\mathrm{C}\\) is heated to \\(912^{\\circ} \\mathrm{C}\\) in an atmosphere that produces \\(1.20 \\% \\mathrm{C}\\) at the surface and is held for \\(24 \\mathrm{~h}\\). Calculate the carbon content at \\(0.05 \\mathrm{~cm}\\) beneath the surface if (a) the iron is BCC and (b) the iron is FCC. Explain the difference.\n}",
|
||
"answer": "(\\quad t=(24 \\mathrm{~h})(3600 \\mathrm{~s} / \\mathrm{h})=86,400 \\mathrm{~s}\\)\n\\[\n\\begin{array}{l}\nD_{\\mathrm{BCC}}=0.011 \\exp [-20,900 /(1.987)(1185)]=1.54 \\times 10^{-6} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\nD_{\\mathrm{FCC}}=0.23 \\exp [-32,900 /(1.987)(1185)]=1.97 \\times 10^{-7} \\mathrm{~cm}^{2} / \\mathrm{s}\n\\end{array}\n\\]\n\\[\n\\begin{array}{l}\n\\mathrm{BCC}: \\frac{1.2-c_{\\mathrm{c}}}{1.2-0.05}=\\mathrm{erf}[0.05 /(2 \\sqrt{(1.54 \\times 10^{-6})(86,400)})]=\\mathrm{erf}[0.0685]=0.077 \\\\\n\\mathrm{FCC}: \\frac{1.2-c_{\\mathrm{c}}}{1.2 -0.05}=\\mathrm{erf}[0.05 /(2 \\sqrt{1.97 \\times 10^{-7})(86,400)})]=\\mathrm{erf}[0.192]=0.2139 \\\\\nc_{\\mathrm{y}}=0.95 \\% \\mathrm{C}\n\\end{array}\n\\]\nFaster diffusion occurs in the looser packed BCC structure, leading to the higher carbon content at point \" \\(\\mathrm{X}\\) \".",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 69,
|
||
"question": "A \\(0.15 \\% \\mathrm{C}\\) steel is to be carburized at \\(1100^{\\circ} \\mathrm{C}\\), giving \\(0.35 \\% \\mathrm{C}\\) at a distance of \\(1 \\mathrm{~mm}\\) beneath the surface. If the surface composition is maintained at \\(0.90 \\% \\mathrm{C}\\), what time is required?",
|
||
"answer": "(\\frac{0.9-0.35}{0.9-0.15}=0.733=\\operatorname{erf}[0.1 / 2 \\sqrt{Dt}]\\)\n\\[\n\\begin{array}{l}\n0.1 / 2 \\sqrt{D t}=0.786 \\quad \\text { or } \\quad \\sqrt{D t}=0.0636 \\quad \\text { or } \\quad D t=0.00405 \\\\\nD=0.23 \\exp \\left[-32.900 /(1.987)(1373)\\right]=1.332 \\times 10^{-6} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\nt=0.00405 / 1.332 \\times 10^{-6}=3040 \\mathrm{~s}=51 \\mathrm{~min}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 71,
|
||
"question": "A \\(1.2 \\% \\mathrm{C}\\) tool steel held at \\(1150^{\\circ} \\mathrm{C}\\) is exposed to oxygen for \\(48 \\mathrm{~h}\\). The carbon content at the steel surface is zero. To what depth will the steel be decarburized to less than \\(0.20 \\% \\mathrm{C}\\) ?",
|
||
"answer": "(\\frac{0-0.2}{0-1.2}=0.1667 \\therefore x / 2 \\sqrt{D t}=0.149\\)\n\\[\n\\begin{array}{l}\nD=0.23 \\exp \\frac{-32.900 /(1.987)(1423)]}{t} \\approx 0.234 \\times 10^{-6} \\mathrm{~cm}^{2} / \\mathbf{s} \\\\\nt=(48 \\mathrm{~h})(3600 \\mathrm{~s} / \\mathrm{h})=17.28 \\times 10^{4} \\mathrm{~s} \\\\\n\\sqrt{D t}=0.5929 \\\\\n\\text { Then from above, } x=(0.149)(2)(0.5929)=0.177 \\mathrm{~cm}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 77,
|
||
"question": "A ceramic part made of \\(\\mathrm{MgO}\\) is sintered successfully at \\(1700^{\\circ} \\mathrm{C}\\) in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to\\(1500^{\\circ} \\mathrm{C}\\). Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions? What time will be required at the lower temperature?",
|
||
"answer": "Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.\n\\[\n\\begin{array}{l}\nD_{1760}=0.000043 \\mathrm{exp}\\left[-82,100 /(1.987)(1973)\\right]=3.455 \\times 10^{-14} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\nD_{15600}=0.000043 \\mathrm{exp}\\left[-82.100 /(1.987)(1773)\\right]=3.255 \\times 10^{-15} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\n\\quad I_{1560}=D_{1760} I_{1760} / D_{1560}=\\frac{(3.455 \\times 10^{-14}) \\mathrm{g} 0)}{3.255 \\times 10^{-15}}=955 \\mathrm{~min}=15.9 \\mathrm{~h}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 78,
|
||
"question": "A 850-1b force is applied to a 0.15 -in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55,000 psi. Determine (a) whether the wire will plastically deform and (b) whether the wire will experience necking.",
|
||
"answer": "(a) First determine the stress acting on the wire:\n\\[\n\\sigma=F / A=850 \\mathrm{lb} /(\\sigma / 4)(0.15 \\mathrm{in})^{2}=48,100 \\mathrm{psi}\n\\]\nBecause \\(\\sigma\\) is greater than the yield strength of 45,000 psi, the wire will plastically deform.\n(b) Because \\(\\sigma\\) is less than the tensile strength of \\(55,000 \\mathrm{psi}\\), no necking will occur.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 79,
|
||
"question": "A force of \\(100,000 \\mathrm{~N}\\) is applied to a \\(10 \\mathrm{~mm} \\times 20 \\mathrm{~mm}\\) iron bar having a yield strength of \\(400 \\mathrm{MPa}\\) and a tensile strength of \\(480 \\mathrm{MPa}\\). Determine whether the bar will plastically deform and whether the bar will experience necking.",
|
||
"answer": "First determine the stress acting on the wire:\n\\[\n\\sigma=F / A=100,000 \\mathrm{~N} /(10 \\mathrm{~mm})(20 \\mathrm{~mm})=500 \\mathrm{~N} / \\mathrm{mm}^{2}=500 \\mathrm{MPa}\n\\]\nBecause \\(\\sigma\\) is greater than the yield strength of \\(400 \\mathrm{MPa}\\), the wire will plastically deform.\nBecause \\(\\sigma\\) is greater than the tensile strength of \\(480 \\mathrm{MPa}\\), the wire will also neck.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 81,
|
||
"question": "A force of 20,000 \\(\\mathrm{N}\\) will cause a \\(1 \\mathrm{~cm} \\times 1 \\mathrm{~cm}\\) bar of magnesium to stretch from \\(10 \\mathrm{~cm}\\) to \\(10.045 \\mathrm{~cm}\\). Calculate the modulus of elasticity, both in \\(\\mathrm{GPa}\\) and psi.",
|
||
"answer": "The strain \\(e\\) is \\(\\varepsilon=(10.045 \\mathrm{~cm}-10 \\mathrm{~cm}) / 10 \\mathrm{~cm}=0.0045 \\mathrm{~cm} / \\mathrm{cm}\\)\nThe stress \\(\\sigma\\) is \\(\\sigma=20.000 \\mathrm{~N} /(10 \\mathrm{~mm}) / 10 \\mathrm{~mm})=200 \\mathrm{~N} / \\mathrm{mm}^{2}\\) \\(=200 \\mathrm{MPa}\\)\n\\(E=\\sigma / \\varepsilon=200 \\mathrm{MPa} / 0.0045 \\mathrm{~cm} / \\mathrm{cm}=44,444 \\mathrm{MPa}=44.4 \\mathrm{GPa}\\)\n\\(E=(44,444 \\mathrm{MPa})(145 \\mathrm{psi} / \\mathrm{MPa})=6.44 \\times 10^{6} \\mathrm{psi}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 85,
|
||
"question": "(a) A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of \\(16 \\times 10^{6}\\) psi, and Poisson's ratio of 0.30 . Determine the length and diameter of the bar when a \\(500-\\mathrm{lb}\\) load is applied.",
|
||
"answer": "The stress is \\(\\sigma=F / A=500 \\mathrm{lb} /(\\pi / 4)(0.4 \\mathrm{in}.)^{2}=3,979 \\mathrm{psi}\\)The applied stress is much less than the yield strength; therefore Hooke's law can be used.\nThe strain is \\(\\varepsilon=\\sigma / E=3,979 \\mathrm{psi} /\\left(16 \\times 10^{6} \\mathrm{psi}\\right)=0.00024868 \\mathrm{in} . / \\mathrm{in}\\).\n\\[\n\\begin{array}{l}\n\\frac{\\ell_{f}-\\ell_{0}}{\\ell_{0}}=\\frac{\\ell_{f}-12 \\mathrm{in} .}{12 \\mathrm{in} .}=0.00024868 \\mathrm{in} . / \\mathrm{n} . \\\\\n\\ell_{f}=12.00298 \\mathrm{in} .\n\\end{array}\n\\]\nFrom Poisson's ratio, \\(\\mu=-\\varepsilon_{\\text {lat }} / \\varepsilon_{\\text {long }}=0.3\\)\n\\[\n\\begin{array}{l}\n\\varepsilon_{\\text {lab }}=-(0.3)(0.00024868)=-0.0000746 \\mathrm{n} / \\mathrm{in} . \\\\\n\\frac{\\ell_{f}-\\ell_{0}}{\\ell_{p}}=\\frac{d_{f}-0.4 \\mathrm{n}}{0.4}=-0.0000746 \\mathrm{n} / \\mathrm{in}\\). \\\\\n\\frac{\\ell_{f}=0.39997 \\mathrm{n} \\mathrm{n} .}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 86,
|
||
"question": "A three-point bend test is performed on a block of \\(\\mathrm{ZrO}_{2}\\) 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 \\mathrm{lb}\\) is applied, the specimen deflects 0.037 in. and breaks. Calculate (a) the flexural strength and (b) the flexural modulus, assuming that no plastic deformation occurs.",
|
||
"answer": "(a) flexural strength \\(=3 F L / 2 w h^{2}=\\frac{(3)(400 \\mathrm{lb})(4 \\mathrm{in})}{(2)(0.5 \\mathrm{in})(0.25 \\mathrm{in})^{2}}=76,800 \\mathrm{psi}\\)\n(b) flexural modulus \\(=F L^{3} / 4 w h^{3} \\delta\\)\n\\[\n\\begin{aligned}\n& =\\frac{(400 \\mathrm{lb})(4 \\mathrm{in})^{3}}{(4)(0.5 \\mathrm{in})(0.25 \\mathrm{in})^{3}(0.037 \\mathrm{in})} \\\\\n& =22.14 \\times 10^{6} \\mathrm{psi}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 87,
|
||
"question": "A three-point bend test is performed on a block of silicon carbide that is \\(10 \\mathrm{~cm}\\) long, \\(1.5 \\mathrm{~cm}\\) wide, and \\(0.6 \\mathrm{~cm}\\) thick and is resting on two supports \\(7.5 \\mathrm{~cm}\\) apart. The sample breaks when a deflection of \\(0.09 \\mathrm{~mm}\\) is recorded. Calculate (a) the force that caused the fracture and (b) the flexural strength. The flexural modulus for silicon carbide is \\(480 \\mathrm{GPa}\\). Assume that no plastic deformation occurs.",
|
||
"answer": "(a) The force \\(F\\) required to produce a deflection of \\(0.09 \\mathrm{~mm}\\) is\n\\[\n\\begin{array}{l}\nF=(\\text { flexural modulus })\\left(4 w h^{3} \\delta\\right) / L^{3} \\\\\nF=(480,000 \\mathrm{MPa})(4)(15 \\mathrm{~mm})(6 \\mathrm{~mm})^{3}(0.09 \\mathrm{~mm}) /(75 \\mathrm{~mm})^{3} \\\\\nF=1327 \\mathrm{~N}\n\\end{array}\n\\](b) flexural strength \\(=3 F L / 2 w h^{2}=(3)(1327 \\mathrm{~N})(75 \\mathrm{~mm}) /(2)(15 \\mathrm{~mm})(6 \\mathrm{~mm})^{2}\\) \\(=276 \\mathrm{MPa}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 88,
|
||
"question": "(a) A thermosetting polymer containing glass beads is required to deflect \\(0.5 \\mathrm{~mm}\\) when a force of \\(500 \\mathrm{~N}\\) is applied. The polymer part is \\(2 \\mathrm{~cm}\\) wide, \\(0.5 \\mathrm{~cm}\\) thick, and \\(10 \\mathrm{~cm}\\) long. If the flexural modulus is \\(6.9 \\mathrm{GPa}\\), determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is \\(85 \\mathrm{MPa}\\) ? Assume that no plastic deformation occurs.",
|
||
"answer": "(a) The minimum distance \\(L\\) between the supports can be calculated from the flexural modulus.\n\\[\n\\begin{aligned}\nL^{3} & =4 w h^{2} \\delta(\\text { flexural modulus }) / F \\\\\nL^{3} & =(4)(20 \\mathrm{~mm})(5 \\mathrm{~mm})^{3}(0.5 \\mathrm{~mm})(6.9 \\mathrm{GPA})(1000 \\mathrm{MPa} / \\mathrm{GPa}) / \\\\\n& 500 \\mathrm{~N} \\\\\nL^{3} & =69,000 \\mathrm{~mm}^{3} \\quad \\text { or } \\quad L=41 \\mathrm{~mm}\n\\end{aligned}\n\\]\nThe stress acting on the bar when a deflection of \\(0.5 \\mathrm{~mm}\\) is obtained is\n\\[\n\\begin{aligned}\n\\sigma & =3 F L / 2 w h^{2}=(3)(500 \\mathrm{~N})(41 \\mathrm{~mm}) /(2)(20 \\mathrm{~mm})(5 \\mathrm{~mm})^{2} \\\\\n& =61.5 \\mathrm{MPa}\n\\end{aligned}\n\\]\nThe applied stress is less than the flexural strength of \\(85 \\mathrm{MPa}\\); the polymer is not expected to fracture.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 89,
|
||
"question": "(b) The flexural modulus of alumina is \\(45 \\times 10^{6} \\mathrm{psi}\\) and its flexural strength is \\(46,000 \\mathrm{psi}\\). A bar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs.",
|
||
"answer": "(b) The force required to break the bar is\n\\[\n\\begin{array}{l}\nF=2 w h^{2}(\\text { flexural strength } / 3 L \\\\\nF=(2)(1 \\text { in. })(0.3 \\text { in. })^{2}(46,000 \\mathrm{psi} /(3)(7 \\text { in. })=394 \\mathrm{lb}\n\\end{array}\n\\]\nThe deflection just prior to fracture is\n\\[\n\\begin{array}{l}\n\\delta=F L^{3} / 4 \\mathrm{wh}^{3}(\\text { flexural modulus }) \\\\\n\\delta=(394 \\mathrm{lb})(7 \\text { in. })^{3} /(4)(1 \\text { in. })(0.3 \\mathrm{in. })^{3}(45 \\times 10^{6} \\mathrm{psi})=0.0278 \\text { in. }\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 95,
|
||
"question": "A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain fracture toughness of 5,000 psi \\(/ \\mathrm{in}\\). To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05 in. long. Assuming that \\(f=1.4\\), does our nondestructive test have the required sensitivity? Explain.",
|
||
"answer": "The applied stress is \\(\\sigma=(\\%) /(75,000 \\mathrm{psi})=25,000 \\mathrm{psi}\\)\n\\[\n\\begin{array}{l}\na=(1 / \\pi /\\left[K_{k_{c}} / f \\sigma \\right]^{2}=(1 / \\pi)[5,000 \\mathrm{psi} \\sqrt{\\mathrm{m}}, f /(1.4)(25,000 \\mathrm{psi})]^{2} \\\\\na=0.0065 \\mathrm{in} .\n\\end{array}\n\\]\nThe length of internal flaws is \\(2 a=0.013 \\mathrm{in}\\).\nOur nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size required for failure. Thus our NDT test is not satisfactory.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 96,
|
||
"question": "to survive for one million cycles under conditions that provide for equal compressive and tensile stresses. What is the fatigue strength, or maximum stress amplitude, required? What are the maximum stress, the minimum stress, and the mean stress on the part during its use? What effect would the frequency of the stress application have on your answers? Explain.",
|
||
"answer": "From the figure, the fatigue strength at one million cycles is \\(22 \\mathrm{MPa}\\).\nThe maximum stress is \\(+22 \\mathrm{MPa}\\), the minimum stress is \\(-22 \\mathrm{MPa}\\), and the mean stress is \\(0 \\mathrm{MPa}\\).\nA 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.\n7-27 The high-strength steel in Figure 7-21 is subjected to a stress alternating at 200 revolutions per minute between \\(600 \\mathrm{MPa}\\) and \\(200 \\mathrm{MPa}\\) (both tension). Calculate the growth rate of a surface crack when it reaches a length of \\(0.2 \\mathrm{~mm}\\) in both m/cycle and \\(\\mathrm{m} / \\mathrm{s}\\). Assume that \\(f=1.0\\).\nSolution: For the steel, \\(C=1.62 \\times 10^{-12}\\) and \\(\\mathrm{n}=3.2\\). The change in the stress intensity factor \\(\\Delta K\\) is\n\\[\n\\Delta K-\\sqrt{A S} \\sqrt{\\alpha u}=(1.2)(600 \\mathrm{MPa}-200 \\mathrm{MPa}) \\sqrt{\\alpha}(0.0002 \\mathrm{~m})=12.03 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\n\\]\nThe crack growth rate is\n\\[\n\\begin{array}{l}\n\\operatorname{dual} N=1.62 \\times 10^{-12}(\\Delta K)^{-2} \\\\\n\\operatorname{dual} N=1.62 \\times 10^{-\\mathrm{t} 2}(12.03)^{\\mathrm{t} 2}=4.638 \\times 10^{-9} \\mathrm{~m} / \\mathrm{cycle}\n\\end{array}\n\\]daldt \\(=(4.638 \\times 10^{-9} \\mathrm{~m} /\\) cycle \\()\\) (200 cycles \\(/ \\mathrm{min}) / 60 \\mathrm{~s} / \\mathrm{min}\\) daldt \\(=1.55 \\times 10^{-8} \\mathrm{~m} / \\mathrm{s}\\)\n7-28 The high-strength steel in Figure 7-21, which has a critical fracture toughness of \\(80 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\), is subjected to an alternating stress varying from \\(-900 \\mathrm{MPa}\\) (compression) to \\(+900 \\mathrm{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\\).\nSolution: (a) Only the tensile portion of the applied stress is considered in \\(\\Delta r\\). Based on the applied stress of \\(900 \\mathrm{MPa}\\) and the fracture toughness of \\(80 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) , the size of a surface crack required for failure to occur is\n\\[\n\\begin{array}{l}\nK=f \\sigma \\sqrt{\\pi a_{c}} \\quad \\text { or } \\quad a_{c}=(1 / \\pi)\\left[K / \\text { far }\\right]^{2} \\\\\na_{c}=(1 / \\pi)\\left[80 \\mathrm{MPa} \\sqrt{\\mathrm{m}} /(1)(900 \\mathrm{MPa})^{2}=0.0025 \\mathrm{~m}=2.5 \\mathrm{~mm}\n\\end{array}\n\\]\n(b) The largest initial surface crack tolerable to prevent failure within \\(10^{5}\\) cycles is\n\\[\n\\begin{array}{l}\n\\mathrm{N}=10^{5} \\text { cycles }= \\frac{2\\left[(0.0025 \\mathrm{~m})^{(2-3.2) / 2}-a_{c}^{(2-3.2) / 2}\\right]}{(2-3.2)(1.62 \\times 10^{-12})(1)^{3-2}(900)^{3-2}(7)^{3-2 / 2}} \\\\\n10^{5}=\\frac{2[36.41-(a)^{-0.60}]}{(-1.2)(1.62 \\times 10^{-12})(1)(2.84 \\times 10^{9})(6.244)} \\\\\n\\left(a_{f}\\right)^{-0.6}=1760 \\\\\na_{t}=3.9 \\times 10^{-6} \\mathrm{~m}=0.0039 \\mathrm{~mm}\n\\end{array}\n\\]\n7-29 The acrylic polymer from which Figure",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 99,
|
||
"question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the activation energy \\(Q_{r}\\) and the constant \" \\(m\\) \" for rupture in the temperature range 980 to \\(1090^{\\circ} \\mathrm{C}\\).",
|
||
"answer": "The appropriate equation is \\(t_{r}=K \\sigma^{\\prime \\prime} \\exp (Q_{r} / R T)\\).\nFrom Figure 7-27(a), we can determine the rupture time versus tempera ture for a fixed stress, say \\(\\sigma=1000 \\mathrm{psi}\\) :\n\\[\n\\begin{array}{lll}\nt_{r}= & 2,400 \\mathrm{~h} & \\text { at } 1090^{\\circ} \\mathrm{C} \\\\\nt_{r}= & 14,000 \\mathrm{~h} & \\text { at } 1040^{\\circ} \\mathrm{C} \\\\\nt_{r}= & 100,000 \\mathrm{~h} & \\text { at } 980^{\\circ} \\mathrm{C} \\\\\n\\end{array}\n\\]\nFrom this data, the equation becomes \\(t_{r}=K^{\\prime} \\exp \\left(Q_{r} / R T\\right)\\) and we can find \\(Q_{r}\\) by simultaneous equations or graphically.\n\\(Q_{r}=117,000 \\mathrm{cal} / \\mathrm{mol}\\)We can also determine the rupture time versus applied stress for a constant temperature, say \\(1090^{\\circ} \\mathrm{C}\\) :\n\\[\n\\begin{array}{ll}\nt_{r}=10^{5} \\mathrm{~h} & \\text { for } \\sigma=450 \\mathrm{psi} \\\\\nt_{r}=10^{4} \\mathrm{~h} & \\text { for } \\sigma=800 \\mathrm{psi} \\\\\nt_{r}=10^{3} \\mathrm{~h} & \\text { for } \\sigma=1200 \\mathrm{psi} \\\\\nt_{r}=10^{2} \\mathrm{~h} & \\text { for } \\sigma=2100 \\mathrm{psi}\n\\end{array}\n\\]\nWith this approach, the equation becomes \\(t_{r}=K^{\\prime \\prime} \\sigma^{m}\\), where \" \\(m\\) \" is obtained graphically or by simultaneous equations:\n\\[\nm=3.9\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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 \\mathrm{in}\\). In a second case, the 2-in.-diameter rod is reduced in one step from 2 in. to a \\(1 \\mathrm{in}\\). diameter. Calculate the \\(\\% \\mathrm{CW}\\) for both cases.",
|
||
"answer": "(\\quad \\% \\mathrm{CW}=\\frac{(2)^{2}-(1)^{2}}{(2)^{2}} \\times 100=75 \\%\\) in both cases",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 103,
|
||
"question": "Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required, and (b) the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is \\(0.356 \\mathrm{~nm}\\).",
|
||
"answer": "From Table \\(9-1, \\Delta T_{\\max }=480^{\\circ} \\mathrm{C}\\)\n\\[\n\\begin{aligned}\nr^{\\text {s }}= & \\frac{\\left(2\\right)\\left(255 \\times 10^{-7} \\mathrm{~J} / \\mathrm{cm}^{2}\\right)(1453+273)}{\\left(2756 \\mathrm{~J} / \\mathrm{cm}^{3}\\right)(480)}=6.65 \\times 10^{-8} \\mathrm{~cm} \\\\\na_{0}= & 3.56 \\AA \\quad V=45.118 \\times 10^{-24} \\mathrm{~cm}^{3} \\\\\n& V_{\\text {nucleus }}=(4 \\pi / 3)(6.65 \\times 10^{-8} \\mathrm{~cm})^{3}=1232 \\times 10^{-24} \\mathrm{~cm}^{3}\n\\end{aligned}\n\\]\nnumber of unit cells \\(=1232 / 45.118=27.3\\)\natoms per nucleus \\(=(4\\) atoms/cell \\()(27.3\\) cells \\()=109\\) atoms",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 104,
|
||
"question": "Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required, and (b) the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is \\(2.92 \\AA\\).",
|
||
"answer": "(\\quad r^{\\text {s }}=\\frac{\\left(2\\right)\\left(204 \\times 10^{-7} \\mathrm{~J} / \\mathrm{cm}{ }^{2}\\right)(1538+273)}{\\left(1737 \\mathrm{~J} / \\mathrm{cm}^{3}\\right)(420)}=10.128 \\times 10^{-8} \\mathrm{~cm}\\)\n\\[\nV=(4 \\pi / 3)(10.128)^{3}=4352 \\AA^{3}=4352 \\times 10^{-24} \\mathrm{~cm}^{3}\nV_{\\text {lc }}=(2.92 \\AA)^{3}=24.897 \\AA^{3}=24.897 \\times 10^{-24} \\mathrm{~cm}^{3}\n\\]\nnumber of unit cells \\(=4352 / 24.897=175\\)\natoms per nucleus \\(=(175\\) cells)(2 atoms/cell) \\(=350\\) atoms",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 107,
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates (a) at \\(10^{\\circ} \\mathrm{C}\\) undercooling, (b) at \\(100^{\\circ} \\mathrm{C}\\) undercooling, and (c) homogeneously. The specific heat of iron is \\(5.78 \\mathrm{~J} / \\mathrm{cm}^{3} \\cdot{ }^{\\circ} \\mathrm{C}\\).",
|
||
"answer": "(\\quad f=\\frac{c \\Delta T}{\\Delta H_{f}}=\\frac{(5.78 \\mathrm{~J} / \\mathrm{cm}^{3} .{ }^{\\circ} \\mathrm{C})(10^{\\circ} \\mathrm{C})}{1737 \\mathrm{~J} / \\mathrm{cm}^{3}}=0.0333\\)\n\\[\n\\begin{array}{l}\n\\frac{c \\Delta T}{\\Delta H_{f}}=\\frac{(5.7^{2} \\mathrm{~J} / \\mathrm{cm}^{3} \\cdot{ }^{*} \\mathrm{C})(100^{\\circ} \\mathrm{C})}{1737 \\mathrm{~kJ} / \\mathrm{cm}^{3}}=0.333 \\\\\n\\frac{c \\Delta T}{\\Delta H_{f}}=\\frac{(5.87 \\mathrm{~J} / \\mathrm{cm}^{3} \\cdot{ }_{\\mathrm{C}}{ }^{\\circ} \\mathrm{C}(420^{\\circ} \\mathrm{C})}{1737 \\mathrm{~L} / \\mathrm{cm}^{3}} \\text {, therefore, all dendritically }\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 108,
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates (a) at \\(10^{\\circ} \\mathrm{C}\\) undercooling, (b) at \\(100^{\\circ} \\mathrm{C} \\text { undercooling, and (c) homogeneously. The specific heat of silver is \\(3.25 \\mathrm{~J} / \\mathrm{cm}^{3} \\cdot{ }{ }^{\\circ} \\mathrm{C}\\).",
|
||
"answer": "(\\quad f=\\underbrace{c \\Delta T}_{\\Delta H_{f}}=\\frac{(3.25 \\mathrm{~J} / \\mathrm{cm}^{3} .{ }^{\"} \\mathrm{C})(10^{\\circ} \\mathrm{C})}{965 \\mathrm{~J} / \\mathrm{cm}^{3}}=0.0237\\)\n\\[\n\\begin{array}{l}\n\\frac{c \\Delta T}{ \\Delta H_{f}}=\\frac{(3.25 \\mathrm{~J} / c^{2} \\mathrm{~J} / \\mathrm{cm}^{3} \\times \\mathrm{C})(100^{\\circ} \\mathrm{C})}{965 \\mathrm{~J}\\left(\\mathrm{~cm}^{3}\\right)}=0.337 \\\\\n\\frac{c \\Delta T}{ \\Delta H_{f}}=\\frac{(3.35 \\mathrm{~J} / \\mathrm{cm}^{3} \\cdot{ }}{965 \\mathrm{~J} / \\mathrm{cm}^{3}}=0.842\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 109,
|
||
"question": "A 2-in. cube solidifies in \\(4.6 \\mathrm{~min}\\). Calculate (a) the mold constant in Chvorinov's rule and (b) the solidification time for a \\(0.5 \\mathrm{in} . \\times 0.5 \\mathrm{in} . \\times 6 \\mathrm{in}\\). bar cast under the same conditions. Assume that \\(n=2\\).",
|
||
"answer": "(a) We can find the volume and surface area of the cube:\n\\[\n\\begin{array}{l}\nV=(2)^{3}=8 \\mathrm{in} \\cdot{ }^{3} \\quad A=6(2)^{2}=24 \\mathrm{in} \\cdot{ }^{2} \\quad t=4.6=B(8 / 24)^{2} \\\\\nB=4.6 /(0.333)^{2}=41.48 \\mathrm{min} / \\mathrm{in}^{2} \\\\\n\\text { (b) For the bar, assuming that } B=41.48 \\mathrm{min} / \\mathrm{in}^{2} \\text {. } \\\\\nV=(0.5)(0.5)(6)=1.5 \\mathrm{in} \\cdot{ }^{2} \\\\\nA=2(0.5)(0.5)+4(0.5)(6)=12.5 \\mathrm{in} \\cdot{ }^{2} \\\\\nt=(41.48)(1.5 / 12.5)^{2}=0.60 \\mathrm{~min}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 111,
|
||
"question": "Calculate the volume, diameter, and height of the cylindrical riser required to prevent shrinkage in a 4 in. \\(\\times 10\\) in. \\(\\times 20\\) in. casting if the \\(H / D\\) of the riser is 1.5 .",
|
||
"answer": "(\\quad(V A)_{C}=\\frac{(4)(10)(20)}{2(4)(10)+2(4)(20)+2(10)(20)}=\\underline{800 / 640}=1.25\\)\n\\[\n\\begin{array}{c}\n(V A)_{F}=\\frac{(\\pi / 4) D^{2} H}{2(\\pi / 4) D^{2}+\\pi D H}=\\frac{(\\pi / 4)(3 / 2) D^{3}}{(\\pi / 2) D^{2}+(3 \\pi / 2) D^{2}}=\\frac{3 D / 8}{2}=3 D / 6 \\approx 1.25 \\\\\nD \\geq 6.67 \\text { in. } \\quad H \\geq 10 \\text { in. } \\quad V \\geq 349 \\text { in. }^{3}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 112,
|
||
"question": "Calculate the volume, diameter, and height of the cylindrical riser required to prevent shrinkage in \\(\\mathrm{a} 1 \\mathrm{in} . \\times 6\\) in. \\(\\times 6\\) in. casting if the \\(H / D\\) of the riser is 1.0 .",
|
||
"answer": "(\\quad V=(1)(6)(6)=36 \\mathrm{in}^{3} \\quad A=2(1)(6)+2(1)(6)+2(6)(6)=96\\) in. \\({ }^{2}\\)\n\\[\n\\begin{array}{c}\n(V A)_{C}=36 / 96=0.375 \\\\\n(V A)_{F}=\\frac{(\\pi / 4) D^{2} \\mathrm{H}}{2(\\pi / 4) D^{2}+\\pi D H}=\\frac{+(\\pi / 4) D^{3}}{(3 \\pi / 2) D^{2}}=D / 6 \\geq 0.375 \\\\\nD \\geq 2.25 \\text { in. } \\quad H \\geq 2.25 \\text { in. } \\quad V \\geq 8.95 \\text { in. }^{3}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 113,
|
||
"question": "A 4-in-diameter sphere of liquid copper is allowed to solidify, producing a spherical shrinkage cavity in the center of the casting. Compare the volume and diameter of the shrinkage cavity in the copper casting to that obtained when a 4-in. sphere of liquid iron is allowed to solidify.",
|
||
"answer": "Cr: \\(5.1 \\% \\quad\\) Fe: \\(3.4 \\% \\quad r_{\\text {sphere }}=4 / 2=2\\) in.\n\\(\\mathrm{Cr}: \\mathrm{V}_{\\text {shrinkage }}=(4 \\pi / 3)(2)^{2}(0.051)=1.709\\) in. \\(^{3}\\)\n\\(\\mathrm{dr}_{\\text {sphere }}=1.48\\) in.\n\\(\\mathrm{dr}_{\\text {cavity }}=1.30\\) in.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 115,
|
||
"question": "A \\(2 \\mathrm{~cm} \\times 4 \\mathrm{~cm} \\times 6 \\mathrm{~cm}\\) magnesium casting is produced. After cooling to room tempera-\n\nage cavity at the center of the casting and (b) the percent shrinkage that must have occurred during solidification.",
|
||
"answer": "The density of the magnesium is \\(1.738 \\mathrm{~g} / \\mathrm{cm}^{3}\\)\n(a) \\(V_{\\text {initial }}=(2)(4)(6)=48 \\mathrm{~cm}^{3}\\)\n\\(V_{\\text {initial }}=80 \\mathrm{~g} / 1.738 \\mathrm{~g} / \\mathrm{cm}^{3}=46.03 \\mathrm{~cm}^{3}\\)\n(b) \\(\\%\\) shrinkage \\(=\\frac{48-46.03}{48} \\times 100 \\%=4.1 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 116,
|
||
"question": "A \\(2 \\mathrm{in} . \\times 8 \\mathrm{in} . \\times 10 \\mathrm{in}\\). iron casting is produced and, after cooling to room temperature, is found to weigh \\(43.9 \\mathrm{lb}\\). Determine (a) the percent shrinkage that must have occurred during solidification and (b) the number of shrinkage pores in the casting if all of the shrinkage occurs as pores with a diameter of 0.05 in.",
|
||
"answer": "The density of the iron is \\(7.87 \\mathrm{~g} / \\mathrm{cm}^{3}\\)\n(a) \\(v_{\\text {actual }}=\\frac{(43.9 \\mathrm{lb})(454 \\mathrm{~g})}{7.87 \\mathrm{~g} / \\mathrm{cm}^{3}}=2532.5 \\mathrm{~cm}^{3}\\)\n\\(V_{\\text {intended }}=(2)(8)(10)=160 \\mathrm{in}^{3} \\times(2.54 \\mathrm{~cm} / \\mathrm{in})^{3}=2621.9 \\mathrm{~cm}^{3}\\)\n\\[\n\\text { shrinkage }=\\frac{2621.9-2532.5}{2621.9} \\times 100 \\%=3.4 \\%\n\\]\n(b) \\(V_{\\text {process }}=2621.9-2532.5=89.4 \\mathrm{~cm}^{3}\\)\n\\(V_{\\text {process }}=(0.05 \\mathrm{in} . / 2)(2.54 \\mathrm{~cm} / \\mathrm{in})=0.0635 \\mathrm{~cm}\\)\n\\[\n\\# \\text { pores }=\\frac{89.4 \\mathrm{~cm}^{3}}{(4 \\pi / 3)(0.0635 \\mathrm{~cm})^{3}}=83,354 \\text { pores }\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 118,
|
||
"question": "A liquid cast iron has a density of \\(7.65 \\mathrm{~g} / \\mathrm{cm}^{3}\\). Immediately after solidification, the density of the solid cast iron is found to be \\(7.71 \\mathrm{~g} / \\mathrm{cm}^{3}\\). Determine the percent volume change that occurs during solidification. Does the cast iron expand or contract during solidification?",
|
||
"answer": "(\\quad 1 / 7.65-1 / 7.71\\)\n\\(1 / 7.65 \\times 100 \\%=\\frac{0.1307 \\mathrm{~cm}^{3}-0.1297 \\mathrm{~cm}^{3}}{0.1307 \\mathrm{~cm}^{3}} \\times 100 \\%=0.77 \\%\\)\nThe casting contracts.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 120,
|
||
"question": "The unary phase diagram for \\(\\mathrm{SiO}_{2}\\) is 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. What do the other \"triple\" points indicate?",
|
||
"answer": "(a) The solid-liquid-vapor triple point occurs at \\(1713^{\\circ} \\mathrm{C}\\); the solid phase present at this point is \\(\\beta\\)-cristobalite.\n(b) The other triple points describe the equilibrium between two solids and a vapor phase.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 121,
|
||
"question": "Based on Hume-Rothery's conditions, which of the following systems would be expected to display unlimited solid solubility? Explain.\n(a) \\(\\mathrm{Au}-\\mathrm{Ag}\\)\n(b) \\(\\mathrm{Al}-\\mathrm{Cu}\\)\n(c) \\(\\mathrm{Al}-\\mathrm{Au}\\)\n(d) \\(\\mathrm{U}-\\mathrm{W}\\)\n(e) \\(\\mathrm{Mo}-\\mathrm{Ta}\\)\n(f) \\(\\mathrm{Nb}-\\mathrm{W}\\)\n(g) \\(\\mathrm{Mg}-\\mathrm{Zn}\\)\n(h) \\(\\mathrm{Mg}-\\mathrm{Cd}\\)",
|
||
"answer": "(a) \\(r_{\\mathrm{Au}}=1.442 \\quad v=+1 \\quad\\) FCC\n\\[\n\\begin{array}{l}\nr_{\\mathrm{Au}}=1.445 \\quad v=+1 \\quad \\text { FCC } \\\\\n\\phi r=0.2 \\%\n\\end{array}\n\\]\n\\begin{tabular}{lll} \n(b) \\(r_{\\mathrm{Al}}=1.432 \\quad v=+3\\) & FCC \\\\\n\\(r_{\\mathrm{Cu}}=1.278 \\quad v=+1\\) & FCC \\\\\n\\(\\phi r=10.7 \\%\\) & No \\\\\n\\hline\n\\end{tabular}\n(c) \\(r_{\\mathrm{Al}}=1.432 \\quad v=+\\quad 3 \\quad\\) FCC\n\\[\n\\begin{array}{l}\nr_{\\mathrm{Al}}=1.442 \\quad v=+1 \\quad \\text { FCC } \\\\\n\\phi r=0 \\quad \\text { V }\n\\end{array}\n\\]\n(d) \\(r_{\\mathrm{F}}=1.38 \\quad v=+4 \\quad\\) Ortho\n\\[\n\\begin{array}{ll}\nr_{\\mathrm{W}}=1.371 \\quad v=+4 \\quad \\text { FCC } \\\\\n\\phi r=0.7 \\%\n\\end{array}\n\\](e) \\(r_{\\mathrm{Mo}}=1.363 \\quad v=+4\\)\n\\(\\begin{aligned} r_{\\mathrm{Ta}} & =1.43 \\quad v=+5 \\\\ \\phi r & =4.7 \\%\\end{aligned}\\)\n(f) \\(r_{\\mathrm{RG}}=1.426 \\quad v=+4\\)\n\\(\\begin{aligned} r_{\\mathrm{W}} & =1.371 \\quad v=+4 \\\\ \\phi r & =3.9 \\%\\end{aligned}\\)\n(g) \\(r_{\\mathrm{RG}}=1.604 \\quad v=+2\\)\n\\(r_{\\mathrm{Rr}}=1.332 \\quad v=+2\\)\n\\(\\phi r=17 \\%\\)\n(h) \\(r_{\\mathrm{RG}}=1.604 \\quad v=+\\quad 2\\)\n\\(r_{\\mathrm{Cd}}=1.490 \\quad v=+2\\)\n\\(\\phi r=7.1 \\%\\)\nThe \\(\\mathrm{Au}-\\mathrm{Ag}, \\mathrm{Mo}-\\mathrm{Ta}\\), and \\(\\mathrm{Mg}-\\mathrm{Cd}\\) systems have the required radius ratio, the same crystal structures, and the same valences. Each of these might be expected to display complete solid solubility. [The \\(\\mathrm{Au}-\\mathrm{Ag}\\) and \\(\\mathrm{Mo}-\\mathrm{Ta}\\) do have isomorphous phase diagrams. In addition, the \\(\\mathrm{Mg}-\\mathrm{Cd}\\) alloys all solidify like isomorphous alloys; however a number of solid-state phase transformations complicate the diagram.]",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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? Is any of the alloying elements expected to have unlimited solid solubility in copper?\n(a) \\(\\mathrm{Au}\\)\n(b) \\(\\mathrm{Mn}\\)\n(c) \\(\\mathrm{Sr}\\)\n(d) \\(\\mathrm{Si}\\)\n(e) \\(\\mathrm{Co}\\)",
|
||
"answer": "For copper: \\(r_{\\mathrm{Cu}}=1.278 \\AA\\)\n(a) \\(\\mathrm{Au}: r=1.442\\)\n\\(\\phi r=\\frac{r_{\\mathrm{Au}}-r_{\\mathrm{Cu}}}{r_{\\mathrm{Cu}}}=+12.8 \\%\\)\n(b) \\(\\mathrm{Mn}: r=1.12\\)\n\\(\\phi r=-12.4 \\%\\)\n(c) \\(\\mathrm{Sr}: r=2.151\\)\n\\(\\phi r=+68.3 \\%\\)\n(d) \\(\\mathrm{Si}: r=1.176\\)\n\\(\\phi r=-8.0 \\%\\)\n(e) \\(\\mathrm{Co}: r=1.253\\)\n\\(\\phi r=-2.0 \\%\\)\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline \\begin{tabular}{l} \nMay be Un \\\\\nsolubility\n\\end{tabular} & & & & & \\\\\n\\hline Different structure & & & & & \\\\\n\\hline Highest Strength & & & & & \\\\\n\\hline Different structure & & & & & \\\\\n\\hular}{\\begin{tabular}{l} \nThe \\(\\mathrm{Cu}-\\mathrm{Sr}\\) alloy would be expected to be strongest (largest size difference). \\\\\nThe \\(\\mathrm{Cu}-\\mathrm{Au}\\) alloy satisfies Hume-Rothery's conditions and might be expected \\\\\nto display complete solid solubility-in fact it freezes like an isomorphous \\\\\nseries of alloys, but a number of solid-state transformations occur at lower \\\\\ntemperatures.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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 vould be expected to give the least reduction in electrical conductivity? Is any of the alloy elements expected to have unlimited solid solubility in aluminum?\n(a) \\(\\mathrm{Li}\\)\n(b) \\(\\mathrm{Ba}\\)\n(c) \\(\\mathrm{Be}\\)\n(d) \\(\\mathrm{Cd}\\)\n(e) \\(\\mathrm{Ga}\\)",
|
||
"answer": "For aluminum: \\(r=1.432 \\AA\\) (FCC structure with valence of 3)\n(a) \\(\\mathrm{Li}: r=1.519 \\quad \\phi r=6.1 \\% \\quad \\mathrm{BCC}\\) valence \\(=1\\)\n(b) \\(\\mathrm{Ba}: r=2.176 \\quad \\phi r=-52.0 \\% \\quad \\mathrm{BCC}\\) valence \\(=2\\)\n(c) \\(\\mathrm{Be}: r=1.143 \\quad \\phi r=-20.2 \\% \\quad \\mathrm{HCP}\\) valence \\(=2\\)\n(d) \\(\\mathrm{Cd}: r=1.49 \\quad \\phi r=4.1 \\% \\quad \\mathrm{HCP}\\) valence \\(=2\\)\n(e) \\(\\mathrm{Ga}: r=1.218 \\quad \\phi r=14.9 \\% \\quad\\) Orthorhombic valence \\(=3\\)\nThe cadmium would be expected to give the smallest reduction in electrical conductivity, since the \\(\\mathrm{Cd}\\) atoms are most similar in size to the aluminum atoms.\nNone are expected to have unlimited solid solubility, due either to difference in valence, atomic radius, or crystal structure.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 125,
|
||
"question": "How many grams of nickel must be added to 500 grams of copper to produce an\nalloy that contains \\(50 \\mathrm{wt} \\% \\alpha\\) at \\(1300^{\\circ} \\mathrm{C}\\) ?",
|
||
"answer": "At \\(1300^{\\circ} \\mathrm{C}\\), the composition of the two phases in equilibrium are\n\\[\n\\begin{array}{l}\nL: 46 \\mathrm{wt} \\% \\mathrm{Ni} \\text { and } \\alpha: 58 \\mathrm{wt} \\% \\mathrm{Ni} \\\\\n\\text { The alloy required to give } 50 \\% \\alpha \\text { is then } \\\\\n\\frac{x-46}{58-46} \\times 100=50 \\% \\alpha \\text { or } x=52 \\mathrm{wt} \\% \\mathrm{Ni}\n\\end{array}\n\\]The number of grams of \\(\\mathrm{Ni}\\) must be:\n\\[\n\\frac{x}{x+500} \\times 100 \\%=52 \\text { or } x=541.7 \\mathrm{~g} \\mathrm{Ni}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 126,
|
||
"question": "How many grams of \\(\\mathrm{MgO}\\) must be added to \\(1 \\mathrm{~kg}\\) of \\(\\mathrm{NiO}\\) to produce a ceramic that has a solidus temperature of \\(2200^{\\circ} \\mathrm{C}\\) ?",
|
||
"answer": "\\(\\mathrm{MW}_{\\mathrm{MgO}}=40.312 \\mathrm{~g} / \\mathrm{mol} \\quad \\mathrm{MW}_{\\mathrm{NO}}=74.71 \\mathrm{~g} / \\mathrm{mol}\\)\n\\(38 \\mathrm{~mol} \\% \\mathrm{MgO}\\) is needed to obtain the correct solidus temperature.\n\\[\n\\begin{array}{l}\n\\text { wt } \\% \\mathrm{MgO}=\\frac{1828 \\mathrm{~m} 0.3122}{1882 \\mathrm{~m} 0.3122+\\mathrm{K} 22 \\mathrm{~F} 4.712} \\times 100 \\%=24.9 \\% \\\\\n\\text { The number of grams required is: } \\\\\n\\frac{x}{x+1000} \\times 100 \\%=24.9 \\% \\text { or } x=332 \\mathrm{~g} \\text { of } \\mathrm{MgO}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 128,
|
||
"question": "A Nb-W alloy held at \\(2800^{\\circ} \\mathrm{C}\\) is partly liquid and partly solid. (a) If possible, determine the composition of each phase in the alloy; and (b) if possible, determine the amount of each phase in the alloy.",
|
||
"answer": "(a) L: \\(49 \\mathrm{wt} \\% \\mathrm{~W} \\quad \\alpha: 70 \\mathrm{wt} \\% \\mathrm{~W}\\)\n(b) Not possible unless we know the original composition of the alloy.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 129,
|
||
"question": "A Nb-W alloy contains \\(55 \\% \\alpha\\) at \\(2600^{\\circ} \\mathrm{C}\\). Determine (a) the composition of each phase; and (b) the original composition of the alloy.",
|
||
"answer": "(a) L \\(: 22 \\mathrm{wt} \\% \\mathrm{~W} \\quad \\alpha: 42 \\mathrm{wt} \\% \\mathrm{~W}\\)\n(b) \\(0.55=\\frac{x-22}{42-22} \\quad\\) or \\(\\quad x=33 \\mathrm{wt} \\% \\mathrm{~W}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 130,
|
||
"question": "Suppose a \\(1200-\\mathrm{lb}\\) bath of a \\(\\mathrm{Nb}-40 \\mathrm{wt} \\% \\mathrm{~W}\\) alloy is held at \\(2800^{\\circ} \\mathrm{C}\\). How many pounds of tungsten can be added to the bath before any solid forms? How many pounds of tungsten must be added to cause the entire bath to be solid?",
|
||
"answer": "Solid starts to form at \\(2800^{\\circ} \\mathrm{C}\\) when \\(49 \\mathrm{wt} \\% \\mathrm{~W}\\) is in the alloy. In \\(1200 \\mathrm{lb}\\) of the original \\(\\mathrm{Nb}-40 \\% \\mathrm{~W}\\) alloy, there are \\((0.4)(1200)=480 \\mathrm{lb} \\mathrm{W}\\) and \\(720 \\mathrm{lb} \\mathrm{Nb}\\). The total amount of tungsten that must be in the final alloy is:\n\\[\n\\begin{array}{lll}\n0.49 & =\\frac{x}{x+720} & \\text { or } \\quad x=692 \\mathrm{lb} \\mathrm{W} \\text { total } \\\\\n& & \\\\\n\\text { or } & 692-480=212 \\text { additional pounds of } \\mathrm{W} \\text { must be added }\n\\end{array}\n\\]\nTo be completely solid at \\(2800^{\\circ} \\mathrm{C}\\), the alloy must contain \\(70 \\mathrm{wt} \\% \\mathrm{~W}\\). The total amount of tungsten required in the final alloy is:\n\\[\n\\begin{array}{lll}\n0.70 & =\\frac{x}{x+720} & \\text { or } \\begin{array}{l}\nx=1680 \\mathrm{lb} \\mathrm{W} \\text { total }\n\\end{array} \\\\\n\\text { or } & 1680-480=1200 \\text { additional pounds of } \\mathrm{W} \\text { must be added }\n%\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 132,
|
||
"question": "A NiO-60 mol\\% MgO ceramic is allowed to solidify. Determine (a) the composition of the first solid to form and (b) the composition of the last liquid to solidify under equilibrium conditions.",
|
||
"answer": "(a) \\(1 \\mathrm{st} \\alpha: 80 \\% \\mathrm{MgO} \\quad\\) (b) Last L: \\(35 \\% \\mathrm{MgO}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 133,
|
||
"question": "A Nb-35\\% W alloy is allowed to solidify. Determine (a) the composition of the first solid to form and \\((\\mathrm{b})\\) the composition of the last liquid to solidify under equilibrium conditions.",
|
||
"answer": "(a) \\(1 \\mathrm{st} \\alpha: 55 \\% \\mathrm{~W} \\quad\\) (b) Last L: \\(18 \\% \\mathrm{~W}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 135,
|
||
"question": "Consider a Pb-15\\% Sn alloy. During solidification, determine (a) the composition of the first solid to form. (b) the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy, (c) the amounts and compositions of each phase at \\(260^{\\circ} \\mathrm{C}\\), (d) the amounts and compositions of each phase at \\(183^{\\circ} \\mathrm{C}\\), and (e) the amounts and compositions of each phase at \\(25^{\\circ} \\mathrm{C}\\).",
|
||
"answer": "(a) \\(8 \\% \\mathrm{Sn}\\)\n(b) liquidus \\(=290^{\\circ} \\mathrm{C}\\), solidus \\(=240^{\\circ} \\mathrm{C}\\), solvus \\(=170^{\\circ} \\mathrm{C}\\), freezing range \\(=50^{\\circ} \\mathrm{C}\\)(c) \\(L: 30 \\% \\mathrm{Sn} \\quad \\alpha: 12 \\% \\mathrm{Sn}\\);\n\\(\\% L=\\frac{15-12}{30-12} \\times 100 \\%=17 \\% \\quad \\% \\alpha=83 \\%\\)\n(d) \\(\\alpha: 15 \\% \\mathrm{Sn} \\quad 100 \\% \\alpha\\)\n(e) \\(\\alpha: 2 \\% \\mathrm{~Pb} \\quad \\beta: 100 \\% \\mathrm{Sn}\\)\n\\(\\% \\alpha=\\frac{100-15}{100-2} \\times 100=87 \\% \\quad \\% \\beta=13 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 136,
|
||
"question": "Consider a \\(\\mathrm{Pb}-35 \\% \\mathrm{Sn}\\) alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at \\(184^{\\circ} \\mathrm{C}\\), (d) the amounts and compositions of each base at \\(182^{\\circ} \\mathrm{C}\\), (e) the amounts and compositions of each microconstituent at \\(182^{\\circ} \\mathrm{C}\\), and (f) the amounts and compositions of each phase at \\(25^{\\circ} \\mathrm{С}\\).",
|
||
"answer": "(a) hypoeutectic\n(b) \\(14 \\% \\mathrm{Sn}\\)\n(c) \\(\\alpha: 19 \\% \\mathrm{Sn} \\quad L: 61.9 \\% \\mathrm{Sn}\\)\n\\(\\% \\alpha=\\frac{61.9-35}{61.9-19} \\times 100 \\%=63 \\% \\quad \\% L=37 \\%\\)\n(d) \\(\\alpha: 19 \\% \\mathrm{Sn} \\quad \\beta: 97.5 \\% \\mathrm{Sn}\\)\n\\(\\% \\alpha=\\frac{97.5-35}{97.5-19} \\times 100 \\%=80 \\% \\quad \\% \\beta=20 \\%\\)\n(e) primary \\(\\alpha: 19 \\% \\mathrm{Sn} \\quad \\%\\) primary \\(\\alpha=63 \\%\\)\neutectic: \\(61.9 \\% \\mathrm{Sn} \\quad \\%\\) eutectic \\(=37 \\%\\)\n(f) \\(\\alpha: 2 \\% \\mathrm{Sn} \\quad \\beta: 100 \\% \\mathrm{Sn}\\)\n\\(\\%\\left.\\alpha=\\frac{100-35}{100-2} \\times 100 \\%=66 \\% \\quad \\beta=34 \\%\\right]\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 137,
|
||
"question": "Consider a \\(\\mathrm{Pb}-70 \\% \\mathrm{Sn}\\) alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at \\(184^{\\circ} \\mathrm{C}\\), (d) the amounts and compositions of each phase at \\(182^{\\circ} \\mathrm{C}\\), (e) the amounts and compositions of each microconstituent at \\(182^{\\circ} \\mathrm{C}\\), and (f) the amounts and compositions of each phase at \\(25^{\\circ} \\mathrm{C}\\).",
|
||
"answer": "(a) hypereutectic\n(b) \\(98 \\% \\mathrm{Sn}\\)\n(c) \\(\\beta: 97.5 \\% \\mathrm{Sn} \\quad L: 61.9 \\% \\mathrm{Sn}\\)\n\\(\\% \\beta=\\frac{70-61.9}{97.5-61.9} \\times 100 \\%=22.8 \\% \\quad \\% L=77.2 \\%\\)\n(d) \\(\\alpha: 19 \\% \\mathrm{Sn} \\quad \\beta: 97.5 \\% \\mathrm{Sn}\\)\n\\(\\% \\alpha=\\frac{97.5-70}{97.5-19} \\times 100 \\%=35 \\% \\quad \\% \\beta=65 \\%\\)\n(e) primary \\(\\beta: 97.5 \\% \\mathrm{Sn} \\%\\) primary \\(\\beta=22.8 \\%\\)\neutectic: \\(61.9 \\% \\mathrm{Sn} \\%\\) eutectic \\(=77.2 \\%\\)\n(f) \\(\\alpha: 2 \\% \\mathrm{Sn} \\quad \\beta: 100 \\% \\mathrm{Sn}\\)\n\\(\\% \\alpha=\\frac{100-70}{100-2} \\times 100 \\%=30 \\% \\quad \\% \\beta=70 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 138,
|
||
"question": "Consider an Al-4\\% Si alloy. Determine (a) if the alloy is hypo Connecticut or hyper euretic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at \\(578^{\\circ} \\mathrm{C}\\), (d) the amounts and compositions of each phase at \\(576^{\\circ} \\mathrm{C}\\), the amounts and compositions of each microconstituent at \\(576^{\\circ} \\mathrm{C}\\), and (e) the amounts and compositions of each phase at \\(25^{\\circ} \\mathrm{C}\\).",
|
||
"answer": "(a) hypo eutectic\n(b) \\(1 \\% \\mathrm{Si}\\)\n(c) \\(\\alpha: 1.65 \\% \\mathrm{Si} \\quad L: 12.6 \\% \\mathrm{Si}\\)\n\\[\n\\begin{array}{l}\n\\% \\alpha=\\frac{12.6-4}{12.6-1.65}=78.5 \\% \\quad \\% L=21.5 \\% \\\\\n\\text { (d) } \\alpha: 1.65 \\% \\mathrm{Si} \\quad \\beta: 99.83 \\% \\mathrm{Si} \\\\\n\\% \\alpha=\\frac{99.83-4}{99.83-1.65}=97.6 \\% \\quad \\beta=2.4 \\%\n\\end{array}\n\\]\nprimary \\(\\alpha: 1.65 \\% \\mathrm{Si} \\quad \\%\\) primary \\(\\alpha=78.5 \\%\\)\neutectic: \\(12.6 \\% \\mathrm{Si} \\quad \\%\\) eutectic \\(=21.5 \\%\\)\n(e) \\(\\alpha: 0 \\% \\mathrm{Si} \\quad \\beta: 100 \\% \\mathrm{Si} \\% \\alpha=100-4=96 \\% \\quad \\% \\beta=4 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 139,
|
||
"question": "Consider a Al-25\\% Si alloy. Determine (a)if the alloy is hypo eutectic or hyper euretic, (b) the composition of the first solid to form during solidified, (c) the amounts and compositions of each phase at \\(578^{\\circ \\mathrm{C}}\\), (d) the amounts and compositions of each phase at \\(576^{\\circ \\mathrm{C}}\\), (e) the amounts and compositions of each microconstituent at \\(576^{\\circ} C\\), and (f) the amounts and compositions of each phase at \\(25^{\\circ} \\mathrm{CC}\\).",
|
||
"answer": "(a) hyper euretic\n(b) \\(100 \\% \\mathrm{Si}\\)(c) \\(\\beta: 99.83 \\%\\) Si \\(\\quad L: 12.6 \\%\\) Si\n\\[\n\\begin{array}{l}\n\\% L=\\frac{99.83-25}{99.83-12.6}=85.8 \\% \\quad \\% \\beta=14.2 \\% \\\\\n\\text { (d) } \\alpha: 1.65 \\% \\text { Si } \\quad \\beta: 99.83 \\% \\text { Si } \\\\\n\\% \\alpha=\\frac{99.83-25}{99.83-1.65}=76.2 \\% \\quad{ }^{\\prime} \\% \\beta=23.8 \\%\n\\end{array}\n\\]\n(c) primary \\(\\beta: 99.83 \\%\\) Si \\(\\quad \\%\\) primary \\(\\beta=14.2 \\%\\) utectic: \\(12.6 \\% \\mathrm{Si} \\quad \\%\\) utectic \\(=85.8 \\%\\)\n(f) \\(\\alpha: 0 \\% \\mathrm{Si} \\quad \\beta: 100 \\% \\mathrm{Si} \\quad \\% \\alpha=\\frac{100-25}{100-0}=75 \\% \\quad \\% \\beta=25 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 140,
|
||
"question": "A Pb-Sn alloy contains \\(45 \\% \\alpha\\) and \\(55 \\% \\beta\\) at \\(100^{\\circ} \\mathrm{C}\\). Determine the composition of the alloy. Is the alloy hypoeutectic or hypereutectic?",
|
||
"answer": "(\\quad \\% \\alpha=45=\\frac{98.0-x}{98.0-5} \\times 100 \\quad\\) or \\(\\quad x=56.15 \\% \\mathrm{Sn} \\quad\\) Hypocutectic",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 141,
|
||
"question": "An Al-Si alloy contains \\(85 \\% \\alpha\\) and \\(15 \\% \\beta\\) at \\(500^{\\circ} \\mathrm{C}\\). Determine the composition of the alloy. \\(\\mathrm{Is}\\) the alloy hypoeutectic or hypereutectic?",
|
||
"answer": "(\\quad \\% \\quad \\alpha=85=\\frac{100-x}{100-1} \\times 100 \\quad\\) or \\(\\quad x=15.85 \\% \\mathrm{Si} \\quad\\) Hypereutectic",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 144,
|
||
"question": "Observation of a microstructure shows that there is \\(28 \\%\\) eutectic and \\(72 \\%\\) primary \\(\\beta\\) in an Al-Li alloy . (a) Determine the composition of the alloy and whether it is hypoeutectic or hypereutectic. (b) How much \\(\\alpha\\) and \\(\\beta\\) are in the eutectic microconstituent?",
|
||
"answer": "(a) \\(28=\\frac{20.4-x}{20.4-9.9} \\times 100 \\quad\\) or \\(\\quad x=17.46 \\% \\mathrm{Li} \\quad\\) Hypereutectic\n(b) \\(\\% \\alpha_{\\text {Est }}=\\frac{20.4-9.9}{20.4-4} \\times 100 \\%=64 \\% \\quad\\) and \\(\\% \\beta_{\\text {Est }}=36 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 145,
|
||
"question": "Calculate the total amount of \\(\\alpha\\) and \\(\\beta\\) and the amount of each microconstituent in a \\(\\mathrm{Pb}-50 \\% \\mathrm{Sn}\\) alloy at \\(182^{\\circ} \\mathrm{C}\\). What fraction of the total \\(\\alpha\\) in the alloy is contained in the eutectic microconstituent?",
|
||
"answer": "(\\quad \\alpha_{\\text {total }}=\\frac{97.5-50}{97.5-19} \\times 100 \\%=60.5 \\% \\quad \\beta_{\\text {Total }}=39.5 \\%\\)\n\\[\n\\begin{array}{l}\n\\alpha_{\\text {Primary }}=\\frac{61.9-50}{61.9-19} \\times 100 \\%=27.7 \\% \\quad \\text { Eutectic }=72.3 \\% \\\\\n\\alpha_{\\text {in }} \\text { eutectic }=60.5-27.7=32.8 \\% \\\\\nf=32.8 / 60.5=0.54\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 146,
|
||
"question": "(a) Recommend an artificial age-hardening heat treatment for a \\(\\mathrm{Cu}-1.2 \\%\\) Be alloy (see Figure 12-34). Include appropriate temperatures. (b) Compare the amount of the \\(\\gamma_{2}\\) precipitate that forms by artificial aging at \\(400^{\\circ} \\mathrm{C}\\) with the amount of the precipitate that forms by natural aging.",
|
||
"answer": "(a) For the \\(\\mathrm{Cu}-1.2 \\%\\) Be alloy, the peritectite temperature is \\(870^{\\circ} \\mathrm{C}\\); above this temperature, liquid may form. The solvus temperature is about \\(530^{\\circ} \\mathrm{C}\\). Therefore:\n1) Solution treat between \\(530^{\\circ} \\mathrm{C}\\) and \\(870^{\\circ} \\mathrm{C}\\left(780^{\\circ} \\mathrm{C}\\right.\\) is typical for beryllium copper alloys)\n2) Querci\n3) Age below \\(530^{\\circ} \\mathrm{C}\\left(330^{\\circ} \\mathrm{C}\\right.\\) is typical for these alloys)\n(b) We can perform lever law calculations at \\(400^{\\circ} \\mathrm{C}\\) and at room temperature. The solubility of \\(\\mathrm{Be}\\) in \\(\\mathrm{Cu}\\) at \\(400^{\\circ} \\mathrm{C}\\) is about \\(0.6 \\% \\mathrm{Be}\\) and that at room temperature is about \\(0.2 \\% \\mathrm{Be}\\) :\n\\[\n\\begin{array}{l}\n\\gamma_{2} \\text { at } 400^{\\circ} \\mathrm{C} =\\frac{1.2-0.6}{11.7-0.6} \\times 100=5.4 \\% \\\\\n\\gamma_{2} \\text { room } T =\\frac{1.2-0.2}{12-0.2} \\times 100=8.5 \\%\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 147,
|
||
"question": "For an \\(\\mathrm{Fe}-0.35 \\% \\mathrm{C}\\) alloy, determine (a) the temperature at which austenite first begins to transform on cooling, (b) the primary microconstituent that forms, (c) the composition and amount of each phase present at \\(728^{\\circ} \\mathrm{C}\\), (d) the composition and amount of each phase present at \\(726^{\\circ} \\mathrm{C}\\), and (e) the composition and amount of each microconstituent present at \\(726^{\\circ} \\mathrm{C}\\).",
|
||
"answer": "(a) \\(795^{\\circ} \\mathrm{C} \\quad\\) (b) primary \\(\\alpha\\)-ferrite\n(c) \\(\\alpha: 0.0218 \\% \\mathrm{C} \\quad \\% \\alpha=\\frac{0.77-0.35}{0.77-0.0218} \\times 100=56.1 \\%\\) \\(\\gamma: 0.77 \\% \\mathrm{C} \\quad \\% \\gamma=43.9 \\%\\)\n(d) \\(\\alpha: 0.0218 \\% \\mathrm{C} \\%=6.67-0.35\\)\n\\(\\mathrm{Fe}_{3} \\mathrm{C}: 6.67 \\% \\mathrm{C} \\% \\mathrm{Fe}_{3} \\mathrm{C}=4.9 \\%\\)\n(e) primary \\(\\alpha: 0.0218 \\% \\mathrm{C} \\%\\) primary \\(\\alpha=56.1 \\%\\) pearlite: \\(0.77 \\% \\mathrm{C} \\%\\) Pearlite \\(=43.9 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 148,
|
||
"question": "For an \\(\\mathrm{Fe}-1.15 \\% \\mathrm{C}\\) alloy, determine (a) the temperature at which austENe first begins to transform on cooling, (b) the primary microconstituent that forms, \\((c)\\) the composition and amount of each phase present at \\(728^{\\circ} \\mathbf{C}\\), (d) the composition and amount of each phase present at \\(720^{\\circ} \\mathrm{C}\\), and (e) the composition and amount of \\(\\mathrm{each}\\) microconstituent present at \\(726^{\\circ} \\mathrm{C}\\)",
|
||
"answer": "(a) \\(880^{\\circ} \\mathrm{C} \\quad\\) (b) primary \\(\\mathrm{Fe}_{3} \\mathrm{C}\\)\n(c) \\(\\mathrm{Fe}_{3} \\mathrm{C}: 6.67 \\% \\quad \\mathrm{C} \\% \\mathrm{Fe}_{3} \\mathrm{C}=\\frac{1.15-0.77}{6.67-0.77} \\times 100=6.4 \\%\\)\n\\(\\gamma: 0.77 \\% \\mathrm{C} \\quad \\% \\gamma=\\mathbf{9 3 . 6 \\%}\\)\n(d) \\(\\alpha: 0.0218 \\% \\mathrm{C}\\) \\(\\% \\alpha=\\frac{6.67-1.15}{6.67-0.0218} \\times 100=83 \\%\\) \\(\\mathrm{Fe}_{3} \\mathrm{C}: 6.67 \\%\\) C \\% \\(\\mathrm{Fe}_{3} \\mathrm{C}=17 \\%\\)\n(e) primary \\(\\mathrm{Fe}_{3} \\mathrm{C}: 6.67 \\% 0 \\%\\) primary \\(\\mathrm{Fe}_{3} \\mathrm{C}=6.4 \\%\\) pearlite: \\(0.77 \\% \\mathrm{C} \\quad \\%\\) Pearlite \\(=93.6 \\%\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 149,
|
||
"question": "A steel contains \\(8 \\%\\) cementite and \\(92 \\%\\) ferrite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectoid or hypereutectoid?",
|
||
"answer": "(\\quad \\alpha=0.92=\\frac{6.67-x}{6.67-0} \\quad x=0.53 \\% \\mathrm{C}, \\therefore\\) Hypoeutectoid",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 150,
|
||
"question": "A steel contains \\(18 \\%\\) cementite and \\(82 \\%\\) ferrite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectoid or hypereutectoid?",
|
||
"answer": "(\\quad \\alpha=0.82=\\frac{6.67-x}{6.67-0} \\quad x=1.20 \\% \\mathrm{C}, \\therefore\\) Hypereutectoid",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 151,
|
||
"question": "A steel contains \\(18 \\%\\) pearlite and \\(82 \\%\\) primary ferrite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectoroid or hypereutectoid?",
|
||
"answer": "primary \\(\\alpha=0.82=\\frac{0.77-x}{0.77-0.0218}\\)\n\\[\nx=0.156 \\% \\mathrm{C}, \\therefore \\text { Hypoeutectoid }\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 152,
|
||
"question": "A steel contains \\(94 \\%\\) pearlite and \\(6 \\%\\) primary cementite at room temperature. Estimate the carbon content of the steel. Is the steel hypoeutectod or hypereutectoid?",
|
||
"answer": "Pearlite \\(=0.94=\\frac{6.67-x}{6.67-0.77} \\quad x=1.124 \\% \\mathrm{C}, \\therefore\\) Hypereutectoid",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 155,
|
||
"question": "A steel is heated until \\(40 \\%\\) austenite, with a carbon content of \\(0.5 \\%\\), forms. Estimate the temperature and the overall carbon content of the steel.",
|
||
"answer": "In order for \\(\\gamma\\) to contain \\(0.5 \\% \\mathrm{C}\\), the austenitizing temperature must be about \\(760^{\\circ} \\mathrm{C}\\) (from the tie line). At this temperature:\n\\[\n0.4=\\frac{x-0.02}{0.5-0.02} \\quad x=0.212 \\% \\mathrm{C}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 156,
|
||
"question": "An isothermally transformed eutectoid steel is found to have a yield strength of \\(410 \\mathrm{MPa}\\). Estimate (a) the transformation temperature and (b) the interlamellar spacing in the pearlite.",
|
||
"answer": "We can first find the interlamellar spacing from Figure 12-18; then using this interlamellar spacing, we can find the transformation temperature from Figure 12-19.\n(a) transformation temperature \\(=615^{\\circ} \\mathrm{C}\\)\n(b) \\(1 / \\lambda=60,000\\) or \\(\\lambda=1.67 \\times 10^{-5} \\mathrm{~cm}\\)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 157,
|
||
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the following hardness values:\n(a) HRC 38\n(b) HRC 42\n(c) HRC 48\n(d) HRC 52",
|
||
"answer": "(a) \\(600^{\\circ} \\mathrm{C}\\)\n(b) \\(400^{\\circ} \\mathrm{C}\\)\n(c) \\(340^{\\circ} \\mathrm{C}\\)\n(d) \\(300^{\\circ} \\mathrm{C}\\)\npenrile\nbanite\nbanite\nbanite",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 158,
|
||
"question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to \\(800^{\\circ} \\mathrm{C}\\) for \\(1 \\mathrm{~h}\\), quenched to \\(350^{\\circ} \\mathrm{C}\\) and held for \\(750 \\mathrm{~s}\\), and finally quenched to room temperature.",
|
||
"answer": "HRC \\(=47\\) and the microstructure is all baninite.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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 .",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 181,
|
||
"question": "(a) Calculate \\(\\%\\) IC of the interatomic bonds for the intermetallic compound \\(\\mathrm{TiAl}_{3}\\).\n(b) On the basis of this result what type of interatomic bonding would you expect to be found in \\(\\mathrm{TiAl}_{3}\\) ?\n- \\(\\quad\\) van der Waals\n- ionic\n- metallic\n- covalent",
|
||
"answer": "(a) The percent ionic character is a function of the electron negativities of the ions \\(X_{\\mathrm{A}}\\) and \\(X_{\\mathrm{B}}\\) according to Equation. The electronegativities for \\(\\mathrm{Al}\\) and \\(\\mathrm{Ti}\\) are both 1.5 and. Therefore, the percent ionic character is determined using Equation as follows:\n\\[\n\\% \\mathrm{IC}=\\left[1-\\exp (-0.25)(1.5-1.5)^{2}\\right] \\times 100=0 \\%\n\\]\n(b) Because the percent ionic character is zero and this intermetallic compound is composed of two metals, the bonding is completely metallic.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 187,
|
||
"question": "Rhodium has an atomic radius of \\(0.1345 \\mathrm{~nm}\\), a density of \\(12.41 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and an atomic weight of 102.91 \\(\\mathrm{g} / \\mathrm{mol}\\). What is rhodium's crystal structure?\na) Simple cubic\nb) \\(\\mathrm{BCC}\\)\nc) \\(\\mathrm{FCC}\\)",
|
||
"answer": "The number of atoms associated with the rhodium unit cell is a function of the unit cell volume, the density, and the atomic weight according to a rearranged form of Equation-that is\n\\[\nn=\\frac{\\rho V_{C} N_{\\mathrm{A}}}{A}\n\\]\nSubstitution into this expression values for rand \\(A\\) provided in the problem statement (as well as the value for \\(N A\\) ) yields the following:\n\\[\n\\begin{aligned}\nn= & \\frac{\\left(12.41 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(V_{C}\\right)}{102.91 \\mathrm{~g} / \\mathrm{mol}} \\\\\n= & 7.26 \\times 10^{22}\\left(V_{C}\\right) \\mathrm{atom} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]\nFor the three possible crystal structures (SC, FCC, BCC), the dependence of unit cell volume on the atomic radii are according to the following equations:\n\\[\n\\begin{aligned}\n\\mathrm{SC} & V_{C}=a^{3}=(2 R)^{3}=8 R^{3} \\\\\n\\mathrm{BCC} & V_{\\mathrm{C}}=a^{3}=\\left(\\frac{4 R}{\\sqrt{3}}\\right)^{3}=\\frac{64 R^{3}}{3 \\sqrt{3}} \\\\\n\\mathrm{FCC} & V_{\\mathrm{C}}=a^{3}=(2 R \\sqrt{2})^{3}=16 R^{3} \\sqrt{2}\n\\end{aligned}\n\\]\nValues of \\(a\\) for \\(\\mathrm{BCC}\\) and \\(\\mathrm{FCC}\\) crystal structures are noted in Equations, respectively. We now insert each of these expressions for \\(V C\\) into the above expression for \\(n\\) and incorporate the value of \\(R\\) provided in the problem statement \\((0.1345 \\mathrm{~nm}=1.345^{\\circ} 10-8 \\mathrm{~cm})\\). If this value of \\(n\\) is 1.0 , the crystal structure is \\(\\mathrm{SC}\\); if the value is 2.0 the crystal structure is \\(\\mathrm{BCC}\\); and if the value is 4.0 it is FCC. These computations are as follows:\nFor SC:\n\\[\n\\begin{aligned}\nn & =\\left(7.26 \\times 10^{22} \\mathrm{atom} / \\mathrm{cm}^{3}\\right)\\left(8 R^{3}\\right) \\\\\n& =\\left(7.26 \\times 10^{22} \\mathrm { atoms } / \\mathrm{cm}^{3}\\right)\\left(8\\right)\\left(1.345 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}=1.41 \\mathrm{atoms}\n\\end{aligned}\n\\]\nFor BCC\n\\[\n\\begin{aligned}\nn & =\\left(7.26 \\times \\mathrm{10}^{22} \\mathrm{atom} / \\mathrm{cm}^{3}\\right) \\frac{64 R^{3}}{3 \\sqrt{3}} \\\\\n& =\\left(7.26 \\times 10^{22} a t o m / \\mathrm{cm}^{3}\\right)\\left[\\frac{(64)(1.345 \\times 10^{-8} \\mathrm{~cm})^{3}}{3 \\sqrt{3}}\\right]=2.18 \\mathrm{atoms}\n\\end{aligned}\n\\]\\[\n\\begin{array}{c}\nn=\\left(7.26 \\times 10^{22} \\mathrm{atom} / \\mathrm{cm}^{3}\\right)\\left(16 R^{3} \\sqrt{3}\\right) \\\\\n=\\left(7.26 \\times 10^{22} \\mathrm{atom} / \\mathrm{cm}^{3}\\right)(16)\\left(1.345 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}(\\sqrt{2})=4.0\n\\end{array}\n\\]\nTherefore, the crystal structure for rhodium is FCC; for FCC only, is this computed value of \\(n\\) the same as the theoretical value.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 199,
|
||
"question": "Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen.For this polymer, determine the following:\n(a) The concentration of \\(\\mathrm{F}\\) (in wt \\%) that must be added if this substitution occurs for \\(18.6 \\%\\) of all of the original hydrogen atoms.\n(b) The concentration of \\(\\mathrm{F}\\) (in wt \\%) that must be addied to completely fluorinate the material, i.e. to produce polytetrafluoroethylene (PTFE).\nAtomic weights for several elements are included in the following table:\nCarbon \\(\\quad 12.01 \\mathrm{~g} / \\mathrm{mol}\\)\nChlorine \\(\\quad 35.45 \\mathrm{~g} / \\mathrm{mol}\\)\nFluorine \\(\\quad 19.00 \\mathrm{~g} / \\mathrm{mol}\\)\nHydrogen \\(1.008 \\mathrm{~g} / \\mathrm{mol}\\)\nOxygen \\(\\quad 16.00 \\mathrm{~g} / \\mathrm{mol}\\)",
|
||
"answer": "(a) For fluorinated polyethylene, we are asked to determine the weight percent of fluorine added for \\(8 \\% \\mathrm{~F}\\) 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 \\(\\mathrm{F}\\). Thus, the mass of these 50 carbon atoms, \\(m \\mathrm{C}\\), is just\n\\(m_{\\mathrm{C}}=50\\left(A_{\\mathrm{C}}\\right)=(50)(12.01 \\mathrm{~g} / \\mathrm{mol})=600.5 \\mathrm{~g}\\)\nLikewise, for hydrogen and fluorine,\n\\(m_{\\mathrm{H}}=92\\left(A_{\\mathrm{H}}\\right)=(92)(1.008 \\mathrm{~g} / \\mathrm{mol})=92.74 \\mathrm{~g}\\)\n\\(m_{\\mathrm{F}}=8\\left(A_{\\mathrm{F}}\\right)=(8)(19.00 \\mathrm{~g} / \\mathrm{mol})=152.00 \\mathrm{~g}\\)\nThus, the concentration of fluorine, \\(C_{\\mathrm{F}}\\), is determined using Equation athrm{C}}+m_{\\mathrm{H}}+m_{\\mathrm{F}}} \\times 100 \\\\\n& =\\frac{152.00 \\mathrm{~g}}{600.5 \\mathrm{~g}+92.74 \\mathrm{~g}+152.00 \\mathrm{~g}} \\times 100=18.0 \\mathrm{wt} \\%\n\\end{aligned}\n\\]\n(b) In order to completely fluorinate the polyethylene all of the hydrogen sites are replaced with fluorine. As in part (a), we will consider 50 carbon atoms with 100 side-bonding sites. Since one atom of fluorine is bonded to each of these sites, there will be 100 fluorine atoms. As above, the mass of the 50 carbon atoms is \\(600.5 \\mathrm{~g}\\). Whereas the mass of fluorine, \\(m_{\\mathrm{F}}\\), is computed as follows:\n\\[\nm_{\\mathrm{F}}=100\\left(A_{\\mathrm{F}}\\right)=(100)(19.00 \\mathrm{~g} / \\mathrm{mol})=_{1900.00 \\mathrm{~g}}\n\\]In order to compute the concentration of fluorine, we use Equation, as follows:\n\\[\n\\begin{aligned}\nC_{\\mathrm{F}} & =\\frac{m_{\\mathrm{F}}}{m_{\\mathrm{C}}+m_{\\mathrm{F}}} \\times 100 \\\\\n& =\\frac{1900.00 \\mathrm{~g}}{600.5 \\mathrm{~g}+1900.00 \\mathrm{~g}} \\times 100=76.0 \\mathrm{wt} \\%\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 205,
|
||
"question": "The number of vacancies present in some metal at \\(864^{\\circ} \\mathrm{C}\\) is \\(1.1 \\times 10^{24} \\mathrm{~m}^{-3}\\). Calculate the number of vacancies at \\(463^{\\circ} \\mathrm{C}\\) given that the energy for vacancy formation is \\(1.25 \\mathrm{eV} /\\) atom; assume that the density at both temperatures is the same",
|
||
"answer": "The number of vacancies per cubic meter may be calculated using Equation, that is\n\\[\nN_{v}=N \\exp \\left(-\\frac{Q_{v}}{k T}\\right)\n\\]\nOr, solving for \\(N\\), the total number of atomic sites per cubic meter\n\\[\nN=\\frac{N_{v}}{\\exp \\left(-\\frac{Q_{v}}{k T}\\right)}=N_{v} \\exp \\left(\\frac{Q_{v}}{k T}\\right)\n\\]\nBecause the densities are assumed to be the same at both temperatures, so also will the value of \\(N\\) be the same at both \\(T^{s}\\). If we use the subscripts \" 1 \" and \" 2 \" to denote the lower and higher temperatures respectively, then it is the case that\n\\[\nN_{v 1} \\exp \\left(\\frac{Q_{v}}{k T_{1}}\\right)=N_{v 2} \\exp \\left(\\frac{Q_{v}}{k T_{2}}\\right)\n\\]\nIn this problem we are asked to determine the number of vacancies at the lower temperature \\(N_{v 1}\\). Solving for \\(N_{v 1}\\) from the above expression leads to the following\n\\[\nN_{v 1}=N_{v 2} \\frac{\\exp \\left(\\frac{Q_{v}}{k T_{2}}\\right)}{\\exp \\left(\\frac{Q_{v}}{k T_{1}}\\right)}=N_{v 2} \\exp \\left(\\frac{Q_{v}}{k}\\right)\\left(\\frac{1}{T_{2}}-\\frac{1}{T_{1}}\\right)\n\\]\nGiven the following values:\n\\[\n\\begin{array}{l}\nN_{12}=1.4 \\times 10^{24} \\mathrm{~m}^{-3} \\\\\nT_{2}=881^{\\circ} \\mathrm{C} \\\\\nQ_{1}=1.15 \\mathrm{eV} / \\mathrm{atom} \\\\\nk=8.62 \\times 10^{-5} \\mathrm{eV} / \\mathrm{atom}-\\mathrm{K}\n\\end{array}\n\\]\\[\n\\begin{array}{l}\nN_{v 1}=\\left(1.4 \\times 10^{24} \\mathrm{~m}^{-3}\\right) \\exp \\left(\\frac{1.15 \\mathrm{eV} / \\mathrm{atom}}{8.62 \\times 10^{-5} \\mathrm{eV} / \\mathrm{atom}-\\mathrm{K}}\\right)\\left(\\frac{1}{881+273 \\mathrm{~K}}-\\frac{1}{628+273 \\mathrm{~K}}\\right) \\\\\n\\quad=5.45 \\times 10^{22} \\mathrm{~m}^{-3}\\left(5.45 \\mathrm{E}+22 \\mathrm{~m}^{-3}\\right)\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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 answers.\n(a) \\(\\mathrm{MgO}\\)\n(b) \\(\\mathrm{CaO}\\)\n(c) \\(\\mathrm{BeO}\\)\n(d) \\(\\mathrm{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 \\((<\\) \\(\\pm 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. Inasmuch as crystal structure information for the compounds in this problem, for the most part, is not provided in the text, we will base our criteria only on the relative ion sizes and charges on the anions and cations. Because all of these ceramics are oxides and the charge on all of the cations is +2 , in making solubility determinations, only the relative sizes of the cations will be considered.\n(a) For \\(\\mathrm{MgO}\\), the ionic radii of the \\(\\mathrm{Mn}^{2+}\\) and \\(\\mathrm{Mg}^{2+}\\) (found inside the front cover of the book) are \\(0.067 \\mathrm{~nm}\\) and \\(0.072 \\mathrm{~nm}\\), respectively. Therefore the percentage difference in ionic radii, \\(\\Delta r \\%\\) is determined as follows:\n\\[\n\\begin{aligned}\n\\Delta r \\% & =\\frac{r_{\\mathrm{Mg} 2^{+}}-r_{\\mathrm{Mn} 2^{+}}}{r_{\\mathrm{Mg} 2^{+}}} \\times 100 \\\\\n& =\\frac{0.072 \\mathrm{~nm}-0.067 \\mathrm{~nm}}{0.072 \\mathrm{~nm}} \\times 100=6.9 \\%\n\\end{aligned}\n\\]\nwhich value is with the acceptable range for a high degree of solubility.(b) For \\(\\mathrm{CaO}\\), the ionic radii of the \\(\\mathrm{Mn}^{2+}\\) and \\(\\mathrm{Ca}^{2+}\\) (found inside the front cover of the book) are 0.067 \\(\\mathrm{nm}\\) and \\(0.100 \\mathrm{~nm}\\), respectively. Therefore, the percentage difference in ionic radii, \\(\\Delta r \\%\\) is determined as follows:\n\\[\n\\begin{aligned}\n\\Delta r \\% & =\\frac{r_{\\mathrm{Ca} 2^{+}}-r_{\\mathrm{Mn} 2^{+}}}{r_{\\mathrm{Ca} 2^{+}}} \\times 100 \\\\\n& =\\frac{0.100 \\mathrm{~nm}-0.067 \\mathrm{~nm}}{0.100 \\mathrm{~nm}} \\times 100=33 \\%\n\\end{aligned}\n\\]\nThis \\(\\Delta r \\%\\) value is much larger than the \\(\\pm 15 \\%\\) range, and, therefore, \\(\\mathrm{CaO}\\) is not expected to experience any appreciable solubility in MnO.\n(c) For \\(\\mathrm{BeO}\\), the ionic radii of the \\(\\mathrm{Mn}^{24}\\) and \\(\\mathrm{Be}^{2+}\\) (found inside the front cover of the book) are 0,067 \\(\\mathrm{nm}\\) and \\(0.035 \\mathrm{~nm}\\), respectively. Therefore, the percentage difference in ionic radii,\\(\\Delta r \\%\\) is determined as follows:\n\\[\n\\begin{aligned}\n\\Delta \\mathrm{r} \\% & =\\frac{r_{\\mathrm{Mn} 2^{+}}-r_{\\mathrm{Be} 2^{+}}}{r_{\\mathrm{Mn} 2^{+}}} \\times 100 \\\\\n& =\\frac{0 . 067 \\mathrm{~nm}-0.035 \\mathrm{~nm}}{0.067 \\mathrm{~nm}} \\times 100=48 \\%\n\\end{aligned}\n\\]\nThis \\(\\Delta r \\%\\) value is much larger than the \\(\\pm 15 \\%\\) range, and, therefore, \\(\\mathrm{\\mathrm{BeO}}\\) is not expected to experience any appreciable solubility in \\(\\mathrm{MnO}\\).\n(d) For \\(\\mathrm{NiO}\\), the ionic radii of the \\(\\mathrm{Mn}^{2}\\) + and \\(\\mathrm{Ni}^{2+}\\) (found inside the front cover of the book) are 0 . 067 \\(\\mathrm{nm}\\) and \\(0.069 \\mathrm{~nm}\\), respectively. Therefore, the percentage difference in ionic radii \\(\\Delta r \\%\\) is determined as follows:\n\\[\n\\begin{aligned}\n\\quad \\Delta r \\% & =\\frac{r_{\\mathrm{Mn} 2^{+}}-\\mathrm{r}_{\\mathrm{Ni} 2^{+}} \\times 100}{r_{\\mathrm{Mn} 2^{+}}} \\\\\n& =\\frac{0.069 \\mathrm{~nm}-0.067 \\mathrm{~nm}}{0 . 069 \\mathrm{~nm}} \\times 100=3 \\%\n\\end{aligned}\n\\]\nwhich value is, of course, within the acceptable range for a high degree of solubility.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 208,
|
||
"question": "Atomic radius, crystal structure, electronegativity, and the most common valence are given in the following table for several elements; for those that are nonmetals, only atomic radii are indicated.\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline Element & \\begin{tabular}{c} \nAtomic \\\\\nRadius \\\\\n\\((\\mathrm{nm})\\)\n\\end{tabular} & \\begin{tabular}{c} \nCrystal \\\\\nStructure\n\\end{tabular} & \\begin{tabular}{c} \nElectro- \\\\\nnegativity\n\\end{tabular} & Valence \\\\\n\\hline \\(\\mathrm{Cu}\\) & 0.1278 & FCC & 1.9 & +2 \\\\\n\\hline \\(\\mathrm{C}\\) & 0.071 & & & \\\\\n\\hline \\(\\mathrm{H}\\) & 0.046 & & & \\\\\n\\hline O & 0.060 & & & \\\\\n\\hline \\(\\mathrm{Ag}\\) & 0.1445 & FCC & 1.9 & +1 \\\\\n\\hline \\(\\mathrm{Al}\\) & 0.1431 & FCC & 1.5 & +3 \\\\\n\\hline \\(\\mathrm{Co}\\) & 0.1253 & HCP & 1.8 & +2 \\\\\n\\hline \\(\\mathrm{Cr}\\) & 0.1249 & BCC & 1.6 & +3 \\\\\n\\hline \\(\\mathrm{Fe}\\) & 0.1241 & BCC & 1.8 & +2 \\\\\n\\hline \\(\\mathrm{Ni}\\) & 0.1246 & FCC & 1.8 & +2 \\\\\n\\hline \\(\\mathrm{Pd}\\) & 0.1376 & FCC & 2.2 & +2 \\\\\n\\hline \\(\\mathrm{Pt}\\) & 0.1387 & FCC & 2.2 & +2 \\\\\n\\hline \\(\\mathrm{Zn}\\) & 0.1332 & HCP & 1.6 & +2 \\\\\n\\hline\n\\end{tabular}\nChoose which of these elements you would expect to form the following with copper:A substitutional solid solution having complete solubility\n- \\(\\mathrm{Ni}\\)\n- \\(\\mathrm{O}\\)\n- \\(\\mathrm{H}\\)\n- \\(\\mathrm{Pt}\\)\n- \\(\\mathrm{Pd}\\)\n- \\(\\mathrm{Co}\\)\n- \\(\\mathrm{C}\\)\n- \\(\\mathrm{Zn}\\)\n- \\(\\mathrm{Ag}\\)\n- \\(\\mathrm{Al}\\)\n- \\(\\mathrm{Cr}\\)\n- \\(\\mathrm{Fe}\\)\nA substitutional solid solution of incomplete solubility\n- \\(\\mathrm{Pd}\\)\n- \\(\\mathrm{Al}\\)\n- \\(\\mathrm{Cr}\\)\n- Fe\n- \\(\\mathrm{H}\\)\n- \\(\\mathrm{Pt}\\)\n- \\(\\mathbf{N i}\\)\n- \\(\\mathrm{Zn}\\)\n- \\(\\mathrm{C}\\)\n- \\(\\mathrm{Ag}\\)\n- \\(\\mathrm{O}\\)\n- \\(\\mathrm{Co}\\)\nAn interstitial solid solution\n- \\(\\mathrm{C}\\)\n- \\(\\mathrm{Pd}\\)\n- \\(\\mathrm{Pt}\\)\n- \\(\\mathrm{H}\\)\n- \\(\\mathrm{Cr}\\)\n- \\(\\mathrm{O}\\)\n- \\(\\mathrm{Zn}\\)\n- \\(\\mathrm{Ag}\\)\n\\(\\cdot \\mathrm{Al}\\)\n- \\(\\mathrm{Co}\\)\n- \\(\\mathrm{Fe}\\)\n- \\(\\mathrm{Ni}\\)",
|
||
"answer": "In this problem we are asked to cite which of the elements listed form with \\(\\mathrm{Cu}\\) the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between \\(\\mathrm{Cu}\\) and the other element \\((\\Delta \\mathrm{R} \\%)\\) must be less than \\(\\pm 15 \\%, 2)\\) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are tabulated, for the various elements, these criteria.\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline & & Crystal & \\(\\Delta\\) Electro- & \\\\\n\\hline Element & \\(\\Delta \\mathrm{R} \\%\\) & Structure & negativity & Valence \\\\\n\\hline \\(\\mathbf{C u}\\) & & FCC & & \\(2+\\) \\\\\n\\hline \\(\\mathrm{C}\\) & -44 & & & \\\\\n\\hline \\(\\mathrm{H}\\) & -64 & & & \\\\\n\\hline \\(\\mathrm{O}\\) & -53 & & & \\\\\n\\hline \\(\\mathrm{Ag}\\) & +13 & FCC & 0 & \\(1+\\) \\\\\n\\hline \\(\\mathrm{Al}\\) & +12 & FCC & -0.4 & \\(3+\\) \\\\\n\\hline \\(\\mathrm{Co}\\) & -2 & HCP & -0.1 & \\(2+\\) \\\\\n\\hline \\(\\mathrm{Cr}\\) & -2 & BCC & -0.3 & \\(3+\\) \\\\\n\\hline \\(\\mathrm{Fe}\\) & -3 & BCC & -0.1 & \\(2+\\) \\\\\n\\hline \\(\\mathrm{Ni}\\) & -3 & FCC & -0.1 & \\(2+\\) \\\\\n\\hline \\(\\mathrm{Pd}\\) & +8 & FCC & +0.3 & \\(2+\\) \\\\\n\\hline \\(\\mathrm{Pt}\\) & +9 & FCC & +0.3 & \\(2+\\) \\\\\n\\hline \\(\\mathrm\tZ n\\) & +4 & HCP & -0.3 & \\(2+\\) \\\\\n\\hline\n\\end{tabular}\n(a) \\(\\mathrm{Ni}, \\mathrm{Pd}\\), and \\(\\mathrm{Pt}\\) meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures \\(\\mathrm{Co}\\) and \\(\\mathrm{Fe}\\) experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures.\n(b) \\(\\mathrm{Ag}, \\mathrm{Al}, \\mathrm{Co}, \\mathrm{Cr}, \\mathrm{Fe}\\), and \\(\\mathrm{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 \\(\\mathrm{Cu}\\) are greater than \\(\\pm 15 \\%\\), and/or have a valence different than \\(2+\\).\n(c) \\(\\mathrm{C}, \\mathrm{H}\\), and \\(\\mathrm{O}\\) form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of \\(\\mathrm{Cu}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 210,
|
||
"question": "13}\nSome hypothetical alloy is composed of \\(12.5 \\mathrm{wt} \\%\\) of metal A and 87.5 wt\\% of metal B. If the densities of metals \\(\\mathrm{A}\\) and \\(\\mathrm{B}\\) are 4.27 and \\(6.35 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively, whereas their respective atomic weights are 61.4 and \\(125.7 \\mathrm{~g} / \\mathrm{mol}\\), determine whether the crystal structure for this alloy is simple cubic, facercentered cubic, or body-centered cubic. Assume a unit cell edge length of \\(0.395 \\mathrm{~nm}\\).",
|
||
"answer": "In order to solve this problem it is necessary to employ Equation; in this expression density and atomic weight will be averages for the alloy-that is\n\\[\n\\rho_{\\mathrm{ave}}=\\frac{n A_{\\mathrm{ave}}}{V_{C} N_{\\mathrm{A}}}\n\\]\nInasmuch as for each of the possible crystal structures, the unit cell is cubic, then \\(\\mathrm{V}_{\\mathrm{C}}=a^{3}\\), or\n\\[\n\\rho_{\\mathrm{ave}}=\\frac{n A_{\\mathrm{we}}}{a^{3} N_{\\mathrm{A}}}\n\\]\nAnd, in order to determine the crystal structure it is necessary to solve for \\(\\mathrm{n}\\), the number of atoms per unit cell. For \\(\\mathrm{n}=1\\), the crystal structure is simple cubic, whereas for \\(\\mathrm{n}\\) values of 2 and 4 , the crystal structure will be either BCC or FCC, respectively. When we solve the above expression for \\(\\mathrm{n}\\) the result is as follows:\n\\[\nn=\\frac{\\rho_{\\mathrm{ave}} a^{3} N_{\\mathrm{A}}}{A_{\\mathrm{ave}}}\n\\]\nExpressions for Aave and \\(\\rho_{\\text {ave }}\\) are found in Equations, respectively, which, when incorporated into the above expression yields\n\\[\nn=\\frac{\\left(\\begin{array}{c}\n100 \\\\\n\\frac{C_{\\mathrm{A}}}{\\rho_{\\mathrm{A}}}+\\frac{C_{\\mathrm{B}}}{\\rho_{\\mathrm{B}}}\n\\end{array}\\right) a^{3} N_{\\mathrm{A}}}{\\left(\\begin{array}{c}\n100 \\\\\n\\frac{C_{\\mathrm{\\mathrm{A}}}}{A_{\\mathrm{A}}}+\\frac{C_{\\mathrm{B}}}{A_{\\mathrm{B}}}\n\\end{array}\\right)}\n\\]\nSubstitution of the concentration values (i.e., \\(\\mathrm{C}_{\\mathrm{A}}=12.5 \\mathrm{wt} \\%\\) and \\(\\mathrm{C}_{\\mathrm{B}}=87.5 \\mathrm{wt} \\%\\) ) as well as values for the other parameters given in the problem statement, into the above equation gives\n\\[\nn=\\frac{\\left(\\begin{array}{c}\n100 \\\\\n \\frac{12.5 \\mathrm{wt} \\%}{4.27 \\mathrm{~g} / \\mathrm{cm}^{3}}+\\frac{87.5 \\mathrm{wt} \\%}{6.35 \\mathrm{~g} / \\mathrm{cm}^{3}}\n\\end{array}\\right)\\left(3.95 \\times 10^{-8} \\mathrm{~nm}\\right)^{3}\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)}{=2.00 \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell}}\n\\]\nTherefore, on the basis of this value, the crystal structure is body-centered cubic.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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} \\%\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 218,
|
||
"question": "The cornea is the transparent outer layer of the human eye. Because it must be transparent to light, it does not normally contain blood vessels. Therefore, it must receive its nutrients via diffusion. Oxygen from the surrounding air diffuses to the cornea through the surface tears whereas other nutrients diffuse to the cornea from the inner parts of the eye, such as the vitreous humor and lens.\nDuring operation, the cornea produces waste in the form of \\(\\mathrm{CO}_{2}\\) gas that must be expelled to keep the eye healthy and functioning. This is accomplished by the simultaneous diffusion of \\(\\mathrm{CO}_{2}\\) from the cornea to the surrounding atmosphere, which generally features a low \\(\\mathrm{CO}_{2}\\) concentration.\nIt is therefore critical that modern contact lens materials allow sufficient diffusion rates of oxygen and carbon dioxide. Without oxygen, the cornea will warp, loose transparency, and become susceptible to scarring. The body may also react by growing additional blood vessels into the eye, which can damage the cornea.\nIf an increased steady-state flow rate of \\(\\mathrm{O}_{2}\\) (oxygen molecules per second) to the cornea is desired, which of the following contact lens / ambient condition modifications is not likely to be useful?\nNote: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.\n(a) Increase the contact lens thickness\n(b) Increase the diffusivity of oxygen gas by decreasing the contact lens porosity\n(c) Increase the ambient temperature\n(d) Increase the ambient partial pressure of oxygen gas\n(e) All of the suggestions (a-d) are useful for increasing the flow rate of oxygen",
|
||
"answer": "Note: Since the area of the contact lens \\((A)\\) doesn't change, the flow rate of oxygen molecules \\(\\left(J^{*} A\\right)\\) increases whenever the diffusion flux of oxygen \\((J)\\) increases.\n(a) If the contact lens thickness is increased and all other factors are preserved, the concentration gradient decrease and the oxygen diffusion flux should decrease.\n(b) 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.(c) Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate.\n(d) Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 219,
|
||
"question": "For a steel alloy it has been determined that a carburizing heat treatment of \\(16 \\mathrm{~h}\\) duration at \\(757^{\\circ} \\mathrm{C}\\) will raise the carbon concentration to \\(0.5 \\mathrm{wt} \\%\\) at a point \\(2.3 \\mathrm{~mm}\\) from the surface. Estimate the time necessary to achieve the same concentration at a \\(8 \\mathrm{~mm}\\) position for an identical steel and at a carburizing temperature of \\(1130^{\\circ} \\mathrm{C}\\). Assume that \\(D_{0}\\) is \\(4.6 \\times 10^{-5} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(Q_{d}\\) is \\(104 \\mathrm{~kJ} / \\mathrm{mol}\\).",
|
||
"answer": "For this problem the solution to Fick's second law is according to Equation, that is\n\\[\n\\frac{x^{2}}{D t}=\\text { constant }\n\\]\nFor two temperatures, \\(T_{1}\\) and \\(T_{2}\\), we may write the following expression:\n\\[\n\\frac{x_{1}^{2}}{D_{1} t_{1}}=\\frac{x_{2}^{2}}{D_{2} t_{2}}\n\\]\nFurthermore, the dependence of \\(D\\) depends on \\(T\\) is represented by Equation, which means that\n\\[\n\\begin{array}{l}\nD_{1}=D_{0} \\exp \\left(-\\frac{Q_{d}}{R T_{1}}\\right) \\\\\nD_{2}=D_{0} \\exp \\left(-\\frac{Q_{d}}{R_{T_{2}}}\\right)\n\\end{array}\n\\]\nCombining these two expressions into the previous equation, leads to the following:\n\\[\n\\frac{x_{1}^{2}}{t_{1} \\exp \\left(-\\frac{Q_{d}}{R T_{1}} \\right)}=\\frac{x_{2}^{2}}{t_{2} \\exp \\left(-\\frac{Q_{d}}{R T_{2}} \\right)}\n\\]Assuming that \\(t_{2}\\) is unknown time to be determined, the above equation takes the form\n\\[\nt_{2}=\\frac{t_{1} x_{2}^{2} D_{0} \\exp \\left(-\\frac{Q_{d}}{R T_{1}}\\right)}{x_{1}^{2} D_{0} \\exp \\left(-\\frac{Q_{d} \\quad}{R T_{2}}\\right)}\n\\]\nwhich reduces to the following\n\\[\nt_{2}=\\left(\\frac{t_{1} x_{2}^{2}}{x_{1}^{2}}\\right) \\exp \\left[-\\frac{Q_{d}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\\right]\n\\]\nFrom our problem statement:\n\\[\n\\begin{array}{l}\nt_{1}=17 \\mathrm{~h} \\\\\nT_{1}=898^{\\circ} \\mathrm{C}\\left(=1171 \\mathrm{~K}\\right) \\\\\nx_{1}=3.1 \\mathrm{~mm}\\left(3.1 \\times 10^{-3} \\mathrm{~m}\\right) \\\\\nT_{2}=1010^{\\circ} \\mathrm{C}\\left(=1283 \\mathrm{~K}\\right) \\\\\nx_{2}=5.4 \\mathrm{~mm}\\left(5.4 \\times 10^{3} \\mathrm{~m}\\right) \\\\\nQ_{d}=118,000 \\mathrm{~J} / \\mathrm{mol} \\\\\nD_{0}=2.5 \\times 10^{-5} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{array}\n\\]\nWhen these values are included into the above equation, the value of \\(t_{2}\\) is computed as follows:\n\\[\n\\begin{aligned}\nt_{2}= & {\\left[\\frac{(17 \\mathrm{~h})\\left(5.4 \\times 10^{-3} \\mathrm{~m}\\right)^{2}}{(3.1 \\times 10^{-3} \\mathrm{~m})^{2}}\\right] \\exp \\left[-\\frac{(118,000 \\mathrm{~J} / \\mathrm{mol})}{8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}}\\left(\\frac{1}{1171 \\mathrm{~K}}-\\frac{1}{1283 \\mathrm{~K}}\\right)\\right] } \\\\\n& =17.9 \\mathrm{~h}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 222,
|
||
"question": "For a bronze alloy, the stress at which plastic deformation begins is \\(277 \\mathrm{MPa}\\) and the modulus of elasticity is \\(117 \\mathrm{GPa}\\).\n(a) What is the maximum load that may be applied to a specimen having a cross-sectional area of \\(327 \\mathrm{~mm}^{2}\\) without plastic deformation?\n(b) If the original specimen length is \\(148 \\mathrm{~mm}\\), what is the maximum length to which it may be stretched without causing plastic deformation?",
|
||
"answer": "(a) This part of the problem asks for us to compute the maximum load that may be applied without yielding. From the definition of stress, Equation, the yield strength and load at yielding \\(F_{y}\\) are related as follows:\n\\[\nF_{y}=\\sigma_{y} A_{0}\n\\]\nSince, for this problem, \\(\\sigma_{y}=271 \\mathrm{MPa}=271 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}\\) and \\(A_{0}=320 \\mathrm{~mm}^{2}=320 \\times 10^{-6} \\mathrm{~m}^{2}\\) then\n\\[\nF_{y}=\\left(271 \\times 10^{6} \\mathrm{~N} / \\mathbf{m}^{2}\\right)\\left(320 \\times 10^{-6} \\mathrm{~m}^{2)}\\right)=86,700 \\mathrm{~N}\n\\]\n(b) Now we are asked to compute the maximum length to which the specimen may be stretched without plastic deformation. At the point of yielding, Equation takes on the following rearranged form:\n\\[\n\\varepsilon_{y}=\\frac{\\sigma_{y}}{E}\n\\]\nHere \\(\\varepsilon_{y}\\) is the strain at yielding. The definition of strain from Equation is\n\\[\n\\varepsilon=\\frac{l_{t}-l_{0}}{l_{0}}\n\\]Or, making this substitution for \\(\\varepsilon\\) into the previous equation leads to\n\\[\n\\frac{l_{i}-l_{0}}{l_{0}}=\\frac{\\sigma_{y}}{E}\n\\]\nAnd solving for \\(l_{i}\\), in this case the length to which the specimen may be elongated without plastic deformation gives\n\\[\n\\begin{aligned}\nl_{i} & =l_{0}\\left(\\frac{\\sigma_{y}}{E}+1\\right) \\\\\n& =(131 \\times 10^{-3} \\mathrm{~m})\\left(\\frac{271 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}}{119 \\times 10^{9} \\mathrm{~N} / \\mathrm{m}^{2}}+1\\right) \\\\\n0.13130 \\mathrm{~m} & =131.30 \\mathrm{~mm}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 225,
|
||
"question": "A cylindrical rod \\(380 \\mathrm{~mm}\\) (15.0 in.) long and having a diameter of \\(10.0 \\mathrm{~mm}(0.40 \\mathrm{in}\\).), is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than \\(0.9 \\mathrm{~mm}\\) ( 0.035 in.) when the applied load is \\(24,500 \\mathrm{~N}\\left(5500 \\mathrm{lb}_{\\mathrm{f}}\\right)\\), which of the four metals or alloys listed below are possible candidates?\n\\begin{tabular}{lccc}\n\\hline & \\begin{tabular}{c} \nModulus of Elasticity \\\\\nMaterial\n\\end{tabular} & \\begin{tabular}{c} \nYield Strength \\\\\n\\((\\mathbf{G P a})\\)\n\\end{tabular} & \\begin{tabular}{c} \nTensile Strength \\\\\n\\((\\mathbf{M P a})\\)\n\\end{tabular} \\\\\n\\hline Aluminum alloy & 70 & 255 & 420 \\\\\nBrass alloy & 100 & 345 & 420 \\\\\nCopper & 110 & 250 & 290 \\\\\nSteel alloy & 207 & 450 & 550 \\\\\n\\hline\n\\end{tabular}\n- Aluminum alloy\n- Brass alloy\n- Copper\n- Steel alloy",
|
||
"answer": "This problem asks that we ascertain which of four metal alloys will not (1) experience plastic deformation, and (2) elongate more than \\(0.9 \\mathrm{~mm}\\) when a tensile load of \\(24,500 \\mathrm{~N}\\) is applied. It is first necessary to compute the stress using Equation; a material to be used for this application must necessarily have a yield strength greater than this value. Thus,\n\\[\n\\sigma=\\frac{F}{A_{0}}=\\frac{24,500 \\mathrm{~N}}{\\pi\\left(\\frac{10.0 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}}=312 \\mathrm{MPa}\n\\]Of the metal alloys listed, only brass and steel have yield strengths greater than this stress.\nNext, we must compute the elongation produced in both brass and steel using Equations in order to determine whether or not this elongation is less than \\(0.9 \\mathrm{~mm}\\). For brass\n\\[\n\\Delta l=\\frac{\\sigma l_{0}}{E}=\\frac{(312 \\mathrm{MPa})(380 \\mathrm{~mm})}{100 \\times 10^{3} \\mathrm{MPa}}=1.19 \\mathrm{~mm}\n\\]\nThus, brass is not a candidate. However, for steel\n\\[\n\\Delta l=\\frac{\\sigma l_{0}}{E}=\\frac{\\left(312 \\mathrm{MPa}\\right)(380 \\mathrm{~mm})}{207 \\times 10^{3} \\mathrm{MPa}}=0.57 \\mathrm{~mm}\n\\]\nTherefore, of these four alloys, only steel satisfies the stipulated criteria.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 226,
|
||
"question": "A tensile test is performed on a specimen of some metal alloy, and it is found that a true plastic strain of 0.12 is produced when a true stress of \\(280 \\mathrm{MPa}\\) is applied. For this alloy, the value of the strain hardening exponent is 0.3 . On the basis of these data what true plastic strain would be expected for a total true plastic stress of \\(330 \\mathrm{MPa}\\) ?",
|
||
"answer": "We are asked to compute the true strain that results from the application of a true stress of \\(340 \\mathrm{MPa}\\); other true stress-strain data are also given. It first becomes necessary to solve for \\(n\\) in Equation. Taking logarithms of this expression leads to\n\\[\n\\log \\sigma_{T}=\\log K+n \\log \\varepsilon_{T}\n\\]\nNext we rearrange this equation such that \\(\\log K\\) is the dependent variable:\n\\[\n\\log K=\\log \\sigma_{T}-n \\log \\varepsilon_{T}\n\\]\nWe now solve for \\(n\\) using the following data given in the problem statement:\n\\[\n\\begin{aligned}\ns_{T} & =260 \\mathrm{MPa} \\\\\ne T & =0.10 \\\\\nn & =0.2\n\\end{aligned}\n\\]\nThus\n\\[\n\\log K=\\log (260 \\mathrm{MPa})-(0.2)[\\log (0.10)]=2.615\n\\]\nOr\n\\[\nK=10^{2.615}=412 \\mathrm{MPa}\n\\]\nWe now rearrange Equation such that \\(\\mathcal{E}_{T}\\) is the dependent variable; we first divide both sides of the Equation by \\(K\\), which leads to the following expression:\n\\[\n\\varepsilon_{T}^{n}=\\frac{\\sigma_{T}}{K}\n\\]\n\\(\\mathcal{E}_{T}\\) becomes the dependent variable by taking the \\(1 / n\\) root of both sides of this expression, as\n\\[\n\\varepsilon_{T}=\\left(\\frac{\\sigma_{T}}{K}\\right)^{1 / n}\n\\]\nFinally, using values of \\(K\\) and \\(n\\), we solve for the true strain at a true stress of \\(340 \\mathrm{MPa}\\) :\n\\[\n\\varepsilon_{T}=\\left(\\frac{\\sigma_{T}}{K} \\right)^{1 / n}=\\left(\\frac{340 \\mathrm{MPa}}{412 \\mathrm{MPa}}\\right)^{1 / 0.2}=0.383\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 227,
|
||
"question": "A cylindrical specimen of a metal alloy \\(48.8 \\mathrm{~mm}\\) long and \\(9.09 \\mathrm{~mm}\\) in diameter is stressed in tension. A true stress of \\(327 \\mathrm{MPa}\\) causes the specimen to plastically elongate to a length of \\(55 \\mathrm{~mm}\\). If it is known that the strain-hardening exponent for this alloy is 0.3 , calculate the true stress (in \\(\\mathrm{MPa}\\) )necessary to plastically elongate a specimen of this same material from a length of \\(48.8 \\mathrm{~mm}\\) to a length of \\(57.6 \\mathrm{~mm}\\).",
|
||
"answer": "We are asked to compute the true stress that is required to elongate a cylindrical specimen from 54.8 \\(\\mathrm{mm}\\) to \\(64.7 \\mathrm{~mm}\\); other true stress-strain data are also given. We first want to compute the value of \\(K\\) in Equation-the expression that relates true stress and true strain:\n\\[\n\\sigma_{T}=K \\varepsilon_{T}^{n}\n\\]\nFurthermore, substitution of the definition of true strain—namely Equation\n\\[\n\\varepsilon_{T}=\\ln \\left(\\frac{l_{i}}{l_{0}}\\right)\n\\]\ninto Equation leads to\n\\[\n\\sigma_{T}=K\\left[\\ln \\left(\\frac{l_{i}}{l_{0}}\\right)\\right]^{n}\n\\]\nAnd solving for \\(K\\) yields\n\\[\nK=\\frac{\\sigma_{T}}{\\left[\\ln \\left(\\frac{l_{i}}{l_{0}}\\right)\\frac{1}{n}\\right]}\n\\]\nWe now solve for the value of \\(K\\) using the following data given in the problem statement:\n\\[\n\\begin{array}{l}\ns_{T}=365 \\mathrm{MPa} \\\\\nl_{0}=54.8 \\mathrm{~mm} \\\\\nl_{i}=61.8 \\mathrm{~mm} \\\\\nn=0.2\n\\end{array}\n\\]\nThus\n\\[\nK=\\frac{365 \\mathrm{MPa}}{\\left[\\ln \\left(\\frac{61.8 \\mathrm{~mm}}{54.8 \\mathrm{~mm}}\\right)\\right]^{0.2}}=558 \\mathrm{MPa}\n\\]\nIt is now possible to compute the stress required to elongate the specimen from \\(54.8 \\mathrm{~mm}\\) to \\(64.7 \\mathrm{~mm}\\) using the modified Equation cited above. Therefore,\\[\n\\begin{array}{c}\n\\sigma_{T}=K\\left[\\ln \\left(\\frac{l_{t}}{l_{0}}\\right)\\right]^{n} \\\\\n=(558 \\mathrm{MPa})\\left[\\ln \\left(\\frac{64.7 \\mathrm{~mm}}{54.8 \\mathrm{~mm}}\\right)\\right]^{0.2}=390 \\mathrm{MPa}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 228,
|
||
"question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of \\(390 \\mathrm{MPa}(56,600 \\mathrm{psi})\\). If the specimen radius is \\(2.5 \\mathrm{~mm}(0.10\\) in.) and the support point separation distance is \\(30 \\mathrm{~mm}(1.2 \\mathrm{in}\\).), predict whether or not you would expect the specimen to fracture when a load of \\(620 \\mathrm{~N}\\left(140 \\mathrm{lb}_{\\mathrm{f}}\\right)\\) is applied. Calculate the value of flexural strength for this test.",
|
||
"answer": "The problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of \\(390 \\mathrm{MPa}(56,60\\) psi \\()\\) and a radius of \\(2.5 \\mathrm{~mm}\\) will fracture when subjected to a load of \\(620 \\mathrm{~N}\\) in a three-point bending test; the support point separation is given as 30 \\(\\mathrm{mm}\\). Using Equation we will calculate the value of \\(\\sigma\\); if this value is greater than \\(\\sigma_{f s}(390 \\mathrm{MPa})\\), then fracture is expected to occur. Employment of Equation yields\n\\[\n\\sigma=\\frac{F L}{\\pi R^{3}}\n\\]\n\\[\n=\\frac{(620 \\mathrm{~N})\\left(30 \\times 10^{-3} \\mathrm{~m}\\right)}{(\\pi)\\left(2.5 \\times 10^{-3} \\mathrm{~m}\\right)^{3}}=379 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=379 \\mathrm{MPa}(53,500 \\mathrm{psi})\n\\]\nSince this value is less than the given value of \\(\\sigma_{f s}(390 \\mathrm{MPa})\\), then fractionure is not predicted.\nThe 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 \\(\\sigma(379 \\mathrm{MPa})\\) is relatively close to \\(\\sigma_{f s}(390 \\mathrm{MPa})\\) then there is some chance that fracture will occur.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 229,
|
||
"question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows:\n\\begin{tabular}{cc}\n\\hline \\(\\sigma_{\\mathrm{fs}}\\) (MPa) & \\(\\mathbf{P}\\) \\\\\n\\hline 100 & 0.05 \\\\\n50 & 0.20 \\\\\n\\hline\n\\end{tabular}\n(a) Compute the flexural strength for a completely nonporous specimen of this material.\n(b) Compute the flexural strength for a 0.10 volume fraction porosity.",
|
||
"answer": "(a) Given the flexural strengths at two different volume fraction porosities, we are asked to determine the flexural strength for a nonporous material. If the natural logarithm is taken of both sides of Equation, then\n\\[\n\\ln \\sigma_{f s}=\\ln \\sigma_{0}-n P\n\\]\nUsing the data provided in the problem statement, two simultaneous equations may be written as\n\\[\n\\begin{array}{l}\n\\ln (100 \\mathrm{MPa})=\\ln \\sigma_{0}-(0.05) n \\\\\n\\ln (50 \\mathrm{MPa})=\\ln \\sigma_{0}-(0.20) n\n\\end{array}\n\\]\nSolving for \\(n\\) and \\(\\sigma_{0}\\) leads to \\(n=4.62\\) and \\(\\sigma_{0}=126 \\mathrm{MPa}\\). For the nonporous material, \\(P=0\\), and, from Equation, \\(\\sigma_{0}=\\sigma_{f s}\\). Thus, \\(\\sigma_{f s}\\) for \\(P=0\\) is \\(126 \\mathrm{MPa}\\).\n(b) Now, we are asked for \\(\\sigma_{f s}\\) at \\(P=0.10\\) for this same material. Utilizing Equation yields\n\\[\n\\begin{aligned}\n\\sigma_{f s} & =\\sigma_{0} \\exp (-n P) \\\\\n& =(126 \\mathrm{MPa}) \\exp \\left[-(4.62)(0.10)\\right] \\\\\n& =79.4 \\mathrm{MPa}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 236,
|
||
"question": "Most metals strain harden at room temperature.\na) True\nb) False",
|
||
"answer": "True. Most metals strain harden at room temperature.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 237,
|
||
"question": "During the recovery of a cold-worked material, which of the following statement(s) is (are) true?\na) Some of the internal strain energy is relieved.\nb) All of the internal strain energy is relieved.\nc) There is some reduction in the number of dislocations.d) There is a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state.\ne) The electrical conductivity is recovered to its precold-worked state.\nf) The thermal conductivity is recovered to its precold-worked state.\ng) The metal becomes more ductile, as in its precold-worked state.\nh) Grains with high strains are replaced with new, unstrained grains.",
|
||
"answer": "During the recovery process the following take place:\n- Some of the internal strain energy is relieved.\n- There is some reduction in the number of dislocations.\n- The electrical conductivity is recovered to its precold-worked state.\n- The thermal conductivity is recovered to its precold-worked state.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 238,
|
||
"question": "During the recrystallization of a cold-worked material, which of the following statement(s) is (are) true?\na) Some of the internal strain energy is relieved.\nb) All of the internal strain energy is relieved.\nc) There is some reduction in the number of dislocations.\nd) There is a significant reduction in the number of dislocations, to approximately the number found In the precold-worked state.\ne) The electrical conductivity is recovered to its precold-worke state.\nf) The thermal conductivity is recovered to its precold-worked state.\ng) \\(\\quad\\) The metal becomes more ductile, as in its precold-worked state.\nh) Grain with high strains are replaced with new, unstrained grains.",
|
||
"answer": "During th e recovery process the following take place:\n- All of the internal strain energy is relieved.\n- There is significant reduction in the number of dislocations.- The metal becomes more ductile, as in its precold-worked state.\n- Grains with high strains are replaced with new, unstrained grains.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 240,
|
||
"question": "A hypothetical metal alloy has a grain diameter of \\(2.4 \\times 10^{-2} \\mathrm{~mm}\\). After a heat treatment at \\(575^{\\circ} \\mathrm{C}\\) for \\(500 \\mathrm{~min}\\), the grain diameter has increased to \\(7.3 \\times 10^{-2} \\mathrm{~mm}\\). Compute the time required for a specimen of this same material (i.e., \\(\\mathrm{d}_{0}=2.4 \\times 10^{-2} \\mathrm{~mm}\\) ) to achieve a grain diameter of \\(5.5 \\times 10^{-2} \\mathrm{~mm}\\) while being heated at \\(575^{\\circ} \\mathrm{C}\\). Assume the \\(\\mathrm{n}\\) grain diameter exponent has a value of 2.2.",
|
||
"answer": "To solve this problem requires that we use Equation with \\(n=2.2\\). It is first necessary to solve for the parameter \\(K\\) in this equation, using values given in the problem statement of \\(d_{0}\\left(2.4 \\times 10^{-2} \\mathrm{~mm}\\right)\\) and the grain diameter after the 500 -min heat treatment \\(\\left(7.3 \\times 10^{-2} \\mathrm{~mm}\\right)\\). The computation for \\(K\\) using a rearranged form of Equation in which \\(K\\) becomes the dependent parameter is as follows:\n\\[\n\\begin{aligned}\nK & =\\frac{d^{2.2}-d_{0}^{2.2}}{t} \\\\\n& =\\frac{\\left(7.3 \\times 10^{-2} \\mathrm{~mm}\\end{array}\\left)^{2.2}-\\left(2.4 \\times 10^{-2} \\mathrm{~mm}\\frac{1}{2}\\right)^{2.2}}{500 \\mathrm{~min}} \\\\\n& =5.77 \\times 10^{-6} \\mathrm{~mm}^{2.2} / \\mathrm{min}\n\\end{aligned}\n\\]\nIt is now possible to solve for the time required to yield a value of \\(5.5 \\times 10^{-2} \\mathrm{~mm}\\), for \\(d\\) using a rearranged form of Equation :\n\\[\n\\begin{aligned}\nt & =\\frac{d^{2.2}-d_{0}^{2.2}}{{K}} \\\\\n& =\\frac{\\left(5.5 \\times 10^{-2} \\mathrm{~mm}\\right)^{2.2}-\\left(2.4 \\times 10^{-2}\\right.\\) mm \\()^{2.2}}{5.77 \\times 10^{-6} \\mathrm{~mm}^{12} / \\mathrm{min}} \\\\\n& =246 \\mathrm{~min}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 248,
|
||
"question": "Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?\n- Intergranular\n- Transgranular",
|
||
"answer": "Both intergranular and transgranular fractures are brittle. For transgranualar, crack propagation is through the grains, whereas for intergranular, crack propagation is along grain boundaries.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 250,
|
||
"question": "A structural component in the shape of a flat plate \\(27.3 \\mathrm{~mm}\\) thick is to be fabricated from a metal alloy for which the yield strength and plane strain fracture toughness values are \\(535 \\mathrm{MPa}\\) and \\(31.3 \\mathrm{MPa} \\cdot \\mathrm{m}^{-1 / 2}\\), respectively. For this particular geometry, the value of \\(Y\\) is 1.8 . Assuming a design stress of 0.5 times the yield strength, calculate the critical length of a surface flaw.",
|
||
"answer": "In order to solve this problem it is necessary to use Equation, with \\(\\sigma=(0.4) \\sigma_{y}=(0.4)(533 \\mathrm{MPa})=213 \\mathrm{MPa}\\). Therefore, the value of \\(a_{c}\\) is computed as follows:\n\\[\na_{c}=\\frac{1}{\\pi}\\left(\\frac{K_{I c}}{\\sigma Y}\\right)^{2}=\\frac{1}{\\pi}\\left[\\frac{22.0 \\mathrm{MPa} \\cdot \\mathrm{m}^{-1 / 2}}{(213 \\mathrm{MPa})(1.3)}\\right]^{2}=2.0 \\times 10^{-3} \\mathrm{~m}=2.0 \\mathrm{~mm}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 252,
|
||
"question": "Which of the following factor(s) favor(s) brittle fracture in polymers?\na) Increasing in temperature.\nb) Increasing in strain rate.\nc) The presence of a sharp notch.\nd) Decreasing specimen thickness.",
|
||
"answer": "Increasing strain rate, the presence of a sharp notch, and decreasing temperature all favor brittle fracture of polymers.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 255,
|
||
"question": "A structural component is fabricated from an alloy that has a plane strain fracture toughness of 45 \\(\\mathrm{MPa}\\). It has been determined that this component fails at a stress of \\(300 \\mathrm{MPa}\\) when the maximum length of a surface crack is \\(0.95 \\mathrm{~mm}\\). What is the maximum allowable surface crack length (in \\(\\mathrm{mm}\\) ) without fracture for this same component exposed to a stress of \\(300 \\mathrm{MPa}\\) and made from another alloy with a plane strain fracture toughness of \\(100.0 \\mathrm{MPa}\\) ? The geometry factor \\(Y\\) is the same in both cases.",
|
||
"answer": "This problem asks us to determine the maximum allowable surface crack length without fracture for a structural component for a specified fracture toughness \\((57.5 \\mathrm{MPa} \\sqrt{\\mathrm{m}})\\), given that fracture occurs for the same component using the same stress level ( \\(300 \\mathrm{MPa}\\) ) and another fracture toughness ( \\(45 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) ). (Note: Because the cracks surface cracks, their lengths are equal to \\(a\\).) The maximum crack length without fracture \\(a_{\\mathrm{c}}\\) may be calculated using Equation-i.e.,\n\\[\na_{\\mathrm{c}}=\\frac{1}{\\pi\\left(\\frac{K_{I c}}{\\sigma Y}\\right)^{2}}\n\\]\nwhere \\(K_{I c}\\) is the plane strain fracture toughness, \\(\\sigma\\) is the applied stress, and \\(Y\\) is a design parameter. Using data for the first alloy \\(\\left[K_{I c}=45 \\mathrm{MPa} \\sqrt{\\mathrm{m}}, a_{\\mathrm{c}}=0.95 \\mathrm{~mm}\\left(9.5 \\times 10^{-4} \\mathrm{~m}\\right), \\sigma=300 \\mathrm{MPa}\\right]\\), it is possible to calculate the value of \\(Y\\) using a rearranged form of the above equation as follows:\n\\[\n\\begin{aligned}\nY & =\\frac{K_{I c}}{\\sigma \\sqrt{\\pi a_{c}}} \\\\\n& =\\frac{45 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(300 \\mathrm{MPa}) \\sqrt{(\\pi)(9.5 \\times 10^{-4} \\mathrm{~m})}} \\\\\n& =2.75\n\\end{aligned}\n\\]Using this value of \\(Y\\), the maximum crack length for the second alloy is determined, again using Equation as\n\\[\n\\begin{aligned}\na_{c} & =\\frac{1}{\\pi\\left(\\frac{K_{I c}}{\\sigma Y}\\right)^{2}} \\\\\n& =\\frac{1}{\\pi}\\left[\\frac{57.5 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(300 \\mathrm{MPa})(2.75)}\\right] \\\\\n& =0.00155 \\mathrm{~m}=1.55 \\mathrm{~mm}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 259,
|
||
"question": "Select all of the two-phase material systems from the list below. Select all that apply.\n(a) Solid ductile cast iron (ferrite solid solution + embedded graphite spheres)\n(b) Solid sodium chloride (salt, \\(\\mathrm{NaCl}\\) )\n(c) Liquid bronze \\((\\mathrm{Cu}+\\mathrm{Sn}\\) liquid solution)\n(d) Solid gray cast iron (ferrite solid solution + embedded graphite flakes)\n(e) Solid aluminum featuring dissolved silicon\n(f) Solid lead-tin solder (a mixture of \\(\\mathrm{Pb}\\)-rich and \\(\\mathrm{Sn}\\)-rich solid solutions)\n(g) Partially melted aluminum\n(h) Frozen water with trapped air bubbles\n(i) Epoxy embedded with carbon fibers",
|
||
"answer": "(a), (d), (f), (h), (i) are two-phase systems because there is a physical boundary beyond the particle level (atom, ions, molecules, etc.) that separates chemically \\& structurally distinct volumes. In the case of \\((\\mathrm{g})\\), the physical boundary merely separates volumes that are structurally distinct (crystalline and liquid).\n(b) is a single-phase compound, not a solution. The ratio of the two ions is in a fixed stoichiometry and they adopt an orderly arrangement in the crystal rather than random substitution or random placement in interstitial sites.\n(c) is a liquid solution, and solutions - themselves - are single-phase by definition.\n(e) is a solid solution since silicon is implied to substitute for aluminum. Solutions are single-phase by definition.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 260,
|
||
"question": "Complete the following statements regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility.\nThe solute and host species must have a very [w] sizes.\nThe solute and host species must attempt to pack with \\([\\mathrm{x}]\\) crystal structure.\nThe solute and host species must feature [y] valence electron configuration.\nThe solute and host species must feature \\([z]\\) ability to attract electrons (electronegativity).\n\\(\\mathrm{w}=\\) similar, different\n\\(\\mathrm{x}=\\) a similar (or the same), a different\n\\(\\mathrm{y}=\\) a similar (or the same), a different\n\\(\\mathrm{z}=\\) a similar, a different",
|
||
"answer": "The Hume-Rothery rules state that two species must have similar size, the same (or similar) crystal structure, the same (or similar) valence, and a similar electronegativity in order to have extensive solid-solubility.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 264,
|
||
"question": "Which of the following kinds of information may be determined with the aid of a phase diagram?\na) The phase(s) present at a specified temperature and composition.\nb) The composition(s) of phase(s) present at a specified temperature and composition.\nc) 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:\n- The phase(s) present at a specified temperature and composition.\n- The composition(s) of phase(s) present at a specified temperature and composition.\n- The fraction(s) of phase(s) present at a specified temperature and composition.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 272,
|
||
"question": "Compute the number of electrons that each aluminum atom donates, on average, to a bulk piece of aluminum metal. Room temperature data for aluminum:\nThe resistivity of aluminum is \\(2.63 \\times 10^{-8} \\Omega \\cdot \\mathrm{m}\\)\nThe electron mobility of aluminum is \\(0.0012 \\mathrm{~m}^{2} /(\\mathrm{V} \\cdot \\mathrm{s})\\)\nThe mass density of aluminum is \\(2.7 \\mathrm{~g} / \\mathrm{cm}^{3}\\)\nThe atomic weight of aluminum is \\(27 \\mathrm{~g} / \\mathrm{mol}\\)",
|
||
"answer": "The atom concentration of aluminum at room temperature is:\n\\[\n\\frac{n}{V}=\\frac{\\rho N_{A}}{A}=\\frac{(2.7 \\mathrm{~g} / \\mathrm{cm}^{3})(6.022 \\times 10^{23} \\mathrm{atom} / \\mathrm{mol})}{27 \\mathrm{~g} / \\mathrm{mol}}=6.022 \\times 10^{22} \\mathrm{atom} / \\mathrm{cm}^{3}\n\\]\nThe free electron concentration at room temperature is:\n\\[\nn=\\frac{\\sigma}{e \\mu_{e}}=\\frac{1}{\\rho e \\mu_{e}}=\\frac{1}{(2.63 \\times 10^{-8} \\Omega \\mathrm{m})(1.6 \\times 10^{-19} \\mathrm{C})(0.0012 \\mathrm{~m}^{2}) /(\\mathrm{V} \\mathrm{s}))}=1.98 \\times 10^{29} \\mathrm{electron} / \\mathrm{m}^{3}\n\\]\nTherefore, the number of free electrons donated by each atom, on average, is:\n\\[\n\\frac{1.98 \\times 10^{29} \\text { electron } / \\mathrm{m}^{3}}{6.022 \\times 10^{28} \\text { atom } / \\mathrm{m}^{3}}=3.29 \\text { electron } / \\mathrm{atom}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 291,
|
||
"question": "Select T/F for each of the following statements regarding aluminum / aluminum alloys:\n[a] Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.\n[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.\n[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.\n[d] Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.\n[e] The relatively low melting point of aluminum is often considered a significant limitation for structural applications.",
|
||
"answer": "[a] \\(\\mathrm{F}\\)\n[b] \\(\\mathrm{T}\\)\n[c] \\(\\mathrm{F}\\)\n[d] \\(\\mathrm{F}\\)\n[e] \\(\\mathrm{T}\\)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 292,
|
||
"question": "Select T/F for each of the following statements regarding aluminum / aluminum alloys:\n[a] Aluminum alloys are generally viable as lightweight structural materials in humid environments because they are not very susceptible to corrosion by water vapor.\n[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.\n[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.\n[d] Compared to other metals, like steel, pure aluminum is very resistant to failure via fatigue.\n[e] Aluminum exhibits one of the highest melting points of all metals, which makes it difficult and expensive to cast.",
|
||
"answer": "These questions could be scrambled to create additional versions.\n[a] \\(\\mathrm{T}\\)\n[b] \\(\\mathrm{F}\\)\n[c] \\(\\mathrm{T}\\)\n[d] \\(\\mathrm{F}\\)\n[e] \\(\\mathrm{F}\\)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 293,
|
||
"question": "48a (question pool)}\nSelect T/F for each of the following statements regarding copper \\& copper alloys:\n[a] Copper has a higher elastic modulus than aluminum.\n[b] The density of copper is closer to that of aluminum than it is to iron.\n[c] Bronze is an alloy of copper and zinc.\n[d] Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates.\n[e] Copper is relatively resistant to corrosion by neutral and even mildly basic water, making it useful for freshwater plumbing applications.",
|
||
"answer": "([a] T\\)\n\\([b] F\\)\n[c] \\(\\mathrm{F}\\)\n[d] \\(\\mathrm{T}\\)\n[e] \\(\\mathrm{T}\\)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 294,
|
||
"question": "49a (question pool)}\nSelect \\(\\mathrm{T} / \\mathrm{F}\\) for each of the following statements regarding copper \\& copper alloys:\n[a] Copper is much more abundant in the earth's crust compared to iron or aluminum.\n[b] Copper is one of just a few metals that can be found in metallic form in nature.\n[c] Pure and/or annealed copper is more difficult to machine compared to its work-hardened form or its alloys.\n[d] Copper is a minor component (by weight) of most brass \\& bronze alloys.\n[e] Amongst metals and alloys copper is one of the best conductors of heat.",
|
||
"answer": "([a] \\mathrm{F}\\)\n[b] \\(\\mathrm{T}\\)\n[c] \\(\\mathrm{T}\\)\n[d] \\(\\mathrm{F}\\)\n[e] \\(\\mathrm{T}\\)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 295,
|
||
"question": "50a (question pool)}\nSelect \\(\\mathrm{T} / \\mathrm{F}\\) for each of the following statements regarding various metals \\& alloys:\n[a] Nickel is majority component (by mass) in certain superalloys such as Waspaloy \\({ }^{\\mathrm{TM}}\\).\n[b] Tungsten is the lowest density metal that has structural use.\n[c] Tantalum offers extremely good corrosion resistance, especially at low temperatures.[d] Magnesium metal is very similar to aluminum, in terms of its physical and mechanical properties.\n[e] Beryllium metal is commonly used as an alloying agent in copper metal.",
|
||
"answer": "([a] \\mathrm{T}\\)\n\\([b] \\mathrm{F}\\)\n[c] \\(\\mathrm{T}\\)\n[d] \\(\\mathrm{T}\\)\n[e] \\(\\mathrm{T}\\)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 298,
|
||
"question": "As the porosity of refractory ceramic bricks increases,\na) Strength decreases.\nb) Strength increases.\nc) Chemical resistance decreases.\nd) Chemical resistance increases.\ne) Thermal insulation decreases.\nf) Thermal insulation increases.",
|
||
"answer": "As the porosity of refractory ceramic bricks increases, strength decreases, chemical resistance decreases, and thermal insulation increases.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 302,
|
||
"question": "Such that the bond strength across the fiber-epoxy interface is [s] MPa, and the shear yield strength of the epoxy is [y] MPa, compute the minimum fiber length, in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. The tensile strength of these carbon fibers is \\([f] \\mathrm{MPa}\\).",
|
||
"answer": ":\n\\[\n\\begin{array}{c}\n\\tau_{\\text {fiber-matrix }}<\\tau_{y}, \\text { matrix } \\mathrm{SO} \\tau_{c}=\\tau_{\\text {fiber-matrix }} \\\\\nl=\\frac{\\sigma_{f}^{\\star} d}{2 \\tau_{c}}=\\frac{\\sigma_{f}^{\\star} d}{2 \\tau_{\\text {fiber-matric }}}=[f][d] \\\\\n2[s] 1000\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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 .",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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 .",
|
||
"is_select": 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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 314,
|
||
"question": "Match each fiber type with its description.\nWhiskers\n- Polycrystalline or amorphous materials with small diameters\n- Single crystals with extremely large length-to-diameter ratios\n- Metals having relatively large diameters\nFibers\n- Single crystals with extremely large length-to-diameter ratios\n- Metals having relatively large diameters\n- Polycrystalline or amorphous materials with small diameters\nWires\n- Metals having relatively large diameters\n- Polycrystalline or amorphous materials with small diameters\n- \\(\\quad\\) Single crystals with extremely large length-to-diameter ratios",
|
||
"answer": "Whiskers are single crystals with extremely large length-to-diameter ratios. Fibers are polycrystalline or amorphous materials with small diameters. Wires are large-diameter metals.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 319,
|
||
"question": "Carbon fiber-reinforced composites have which of the following properties?\na) Relatively high strengths\nb) Relatively high stiffnesses\nc) High service temperatures \\(\\left(>200^{\\circ} \\mathrm{C}\\right)\\)",
|
||
"answer": "Carbon fiber-reinforced composites have relatively high strengths, relatively high stiffnesses, and high service temperatures \\(\\left(>200^{\\circ} \\mathrm{C}\\). \\()\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 324,
|
||
"question": "Carbon-carbon composites exhibit which of the following properties/characteristics?\na) High tensile moduli at elevated temperatures\nb) High tensile strengths at elevated temperatures\nc) Resistance to creep\nd) Large fracture toughness values\ne) High thermal conductivities\nf) Low coefficients of thermal expansion\ng) Resistance to oxidation at elevated temperatures\nh) Low cost",
|
||
"answer": "Carbon-carbon composites have high tensile moduli at elevated temperatures, high tensile strengths at elevated temperatures, large fracture toughness values, high thermal conductivities, and low coefficients of thermal expansion, and they are highly resistant to creep.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 336,
|
||
"question": "The formula for yttrium iron garnet \\(\\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{3} \\mathrm{O}_{12}\\right)\\) may be written in the form \\(\\mathrm{Y}_{3}^{*} \\mathrm{Fe}_{2}^{*} \\mathrm{Fe}_{3}^{d} \\mathrm{O}_{12}^{2}\\), where the superscripts a, c, and d represent different sites on which the \\(\\mathrm{Y}^{3+}\\) and \\(\\mathrm{Fe}^{3+}\\) ions are located. The spin magnetic moments for the \\(\\mathrm{Y}^{3+}\\) and \\(\\mathrm{Fe}^{\\text {out }}\\) ions positioned in the \\(\\mathrm{a}\\) and \\(\\mathrm{c}\\) sites are oriented parallel to one another and antiparallel to the \\(\\mathrm{Fe}^{3+}\\) ions in \\(\\mathrm{d}\\) sites. Compute the number of Bohr magnetons associated with each \\(\\mathrm{Y}^{3+}\\) ion, given the following information: (1) each unit cell consists of eight formula \\(\\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right)\\) units; (2) the unit cell is cubic with an edge length of \\(1.2376 \\mathrm{~nm}\\); (3) the saturation magnetization for this material is \\(1.0 \\times 10^{4} \\mathrm{~A} / \\mathrm{m}\\); and (4) assume that there are 5 Bohr magnetons associated with each \\(\\mathrm{Fe}^{3+}\\) ion.",
|
||
"answer": "For this problem, we are given that yttrium iron garnet may be written in the form \\(\\mathrm{Y}_{3}^{c} \\mathrm{Fe}_{2}^{a} \\mathrm{Fe}_{3}^{d} \\mathrm{O}_{12}\\) where the superscripts \\(a, c\\), and \\(d\\) represent different sites on which the \\(\\mathrm{Y}^{3+}\\) and \\(_{\\text {ions }}\\) are located, and that the spin magnetic moments for the ions on \\(a\\) and \\(c\\) sites are oriented parallel to one another and antiparallel to the \\(\\mathrm{Fe}{ }^{3+}\\) ions on the \\(d\\) sites. We are to determine the number of Bohr magnetons associated with each \\(\\mathrm{Y}^{3+} \\mathrm{ion}\\) given that each unit cell consists of eight formula units, the unit cell is cubic with an edge length of \\(1.2376 \\mathrm{\\mathrm{nm}}\\), the saturation magnetization for the material is \\(1.0 \\times 10^{4} \\mathrm{~A}-\\mathrm{m}\\), and that there are 5 Bohr magnetons for each \\(\\mathrm{Fe}^{3+}\\) ion.\nThe first thing to do is to calculate the number of Bohr magnetons per unit cell, which we will denote \\(n_{\\mathrm{B}}\\).Solving for \\(n_{\\mathrm{B}}\\) using Equation, we get\n\\[\n\\begin{aligned}\nn_{\\mathrm{B}} & =\\frac{M_{s} \\alpha^{3}}{\\mu_{\\mathrm{B}}} \\\\\n& =\\frac{\\left(1.0 \\times 10^{4} \\mathrm{~A} / \\mathrm{m}\\right)\\left(1.2376 \\times 10^{-9} \\mathrm{~m}\\right)^{3}}{9.27 \\times 10^{-24} \\mathrm{~A} \\cdot \\mathrm{m}^{2} / \\mathrm{BM}}=2.04 \\mathrm{Bohr} \\text { magnetons/unit cell }\n\\end{aligned}\n\\]\nNow, there are 8 formula units per unit cell or \\(\\frac{2.04}{8}=0.255\\) Bohr magnetons per formula unit.\nFurthermore, for each formula unit there are two \\(\\mathrm{Fe}^{3+}\\) ions on \\(a\\) sites and three \\(\\mathrm{Fe}^{3+}\\) on \\(d\\) sites which magnetic moments are aligned antiparallel. Since there are 5 Bohr magnetons associated with each \\(\\mathrm{Fe}^{3+}\\) ion, the net magnetic moment contribution per formula unit from the \\(\\mathrm{Fe}^{3+}\\) ions is 5 Bohr magnetons. This contribution is antiparallel to the contribution from the \\(\\mathrm{Y}^{3+}\\) ions, and since there are three \\(\\mathrm{Y}^{3+}\\) ions per formula unit, then\n\\[\n\\text { No. of Bohr magnetons } / \\mathrm{Y}^{3+}=\\frac{0.255 \\mathrm{BM}+5 \\mathrm{BM}}{3}=1.75 \\mathrm{BM}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 346,
|
||
"question": "Match the type of light transmission with its description,\nTransmits light with relative little absorption.\n- Translucent\n- Transparent\n- Opaque\nTransmits light diffusely\n- Opaque\n- Translucent\n- Transparent\nIs impervious to light transmission.\n- Transparent\n- Opaque\n- Translucent",
|
||
"answer": "A transparent material transmits light with relatively little absorption.\nA translucent material transmits light diffusely.\nAn opaque material is impervious to light transmission.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 350,
|
||
"question": "Match the following material types with their light transmission characteristics.\nSingle crystal electrical insulators\n- Opaque\n- Translucent\n- Transparent\nPolycrystalline and nonporous electrical insulators\n- Translucent\n- Transparent\n- Opaque\nPorous electrical insulators\n- Transparent\n- Opaque\n- Translucent",
|
||
"answer": "Single crystal electrical insulators are transparent. Polycrystalline and nonporous electrical insulators are translucent. And, porous electrical insulators are opaque to visible light.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 356,
|
||
"question": "Only pure materials luminesce.\na) True\nb) False",
|
||
"answer": "False. Luminescent materials contain impurities.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 359,
|
||
"question": "(a) How many grams are there in one amu of a material?\n(b) 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": "(a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, \\(\\mathrm{g} / \\mathrm{mol}\\), and atom/mol relationships is all that is necessary, as\n\\[\n\\begin{aligned}\n\\# \\mathrm{~g} / \\mathrm{amu}= & \\left(\\frac{1 \\mathrm{~mol}}{6.022 \\times 10^{23} \\mathrm{atoms}}\\right)\\left(\\frac{1 \\mathrm{~g} / \\mathrm{mol}}{1 \\mathrm{amu} / \\mathrm{atom}}\\right) \\\\\n= & 1.66 \\times 10^{-24} \\mathrm{~g} / \\mathrm{amu}\n\\end{aligned}\n\\]\n(b) Since there are \\(453.6 \\mathrm{~g} / \\mathrm{lb}_{\\mathrm{m}}\\),\n\\[\n\\begin{aligned}\n1 \\mathrm{lb}-\\mathrm{mol}= & \\left(453.6 \\mathrm{~g} / \\mathrm{lb}_{\\mathrm {m }}\\right)\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{g}-\\mathrm{mol}\\right) \\\\\n= & 2.73 \\times 10^{26} \\mathrm{atoms} / \\mathrm{lb}-\\mathrm{mol}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 360,
|
||
"question": "(a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.\n(b) Cite two important additional refinements that resulted from the wave-mechanical atomic model.",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 361,
|
||
"question": "Relative to electrons and electron states, what does each of the four quantum numbers specify?\n}",
|
||
"answer": "The \\(n\\) quantum number designates the electron shell.\nThe \\(l\\) quantum number designates the electron subshell.\nThe \\(m_{j}\\) quantum number designates the number of electron states in each electron subshell.\nThe \\(m_{\\mathrm{s}}\\) quantum number designates the spin moment on each electron.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 362,
|
||
"question": "Allowed values for the quantum numbers of electrons are as follows:\n\\[\n\\begin{aligned}\nn & =1,2,3, \\ldots \\\\\nl & =0,1,2,3, \\ldots, n-1 \\\\\nm_{l} & =0, \\pm 1, \\pm 2, \\pm 3, \\ldots, \\pm l \\\\\nm_{\\mathrm{S}} & =\\pm \\frac{1}{2}\n\\end{aligned}\n\\]\nThe relationships between \\(n\\) and the shell designations are noted. Relative to the subshells,\n\\[\n\\begin{array}{l}\nl=0 \\text { corresponds to an } s \\text { subshell } \\\\\nl=1 \\text { corresponds to a } p \\text { subshell } \\\\\nl=2 \\text { corresponds to } a d \\text { subshell } \\\\\nl=3 \\text { corresponds to an } f \\text { subshell }\n\\end{array}\n\\]\nFor the \\(K\\) shell, the four quantum numbers for each of the two electrons in the \\(1 \\mathrm{~s}\\) state, in the order of \\(\\operatorname{nlm}_{i} m_{s}\\), are \\(100\\left(\\frac{1}{2}\\right)\\) and \\(100\\left(-\\frac{1}{2}\\right)\\). Write the four quantum numbers for all of the electrons in the \\(L\\) and \\(M\\) shells, and note which correspond to the \\(s, p\\), and \\(d\\) subshells.",
|
||
"answer": "For the \\(L\\) state, \\(n=2\\), and eight electron states are possible. Possible \\(l\\) values are 0 and 1 , while possible \\(m_{l}\\) values are 0 and \\pm 1 ; and possible \\(m_{s}\\) values are \\(\\pm \\frac{1}{2}\\). Therefore, for the \\(s\\) states, the quantum numbers are \\(200\\left(\\frac{1}{2}\\right)\\) and \\(200\\left(-\\frac{1}{2}\\right)\\). For the \\(p\\) states, the quantum numbers are \\(210\\left(\\frac{1}{2}\\right), 210\\left(-\\frac{1}{2}\\right), 211\\left(\\frac{1}{2}\\right), 211\\left(-\\frac{1}{2}\\right), 21(-1)\\left(\\frac{1}{2}\\right)\\), and \\(21(-1)\\left(-\\frac{1}{2}\\right)\\).\nFor the \\(M\\) state, \\(n=3\\), and 18 states are possible. Possible \\(l\\) values are 0,1 , and 2 ; possible \\(m_{l}\\) values are \\(0, \\pm 1\\), and \\pm 2 ; and possible \\(m_{s}\\) values are \\(\\pm \\frac{1}{\\mathrm{~L}}\\). Therefore, for the \\(s\\) states, the quantum numbers are \\(300\\left(\\frac{1}{2}\\right)\\), \\(300\\left(-\\frac{1}{2}\\right)\\), for the \\(p\\) states they are \\(310\\left(\\frac{1}{2}\\right), 310\\left(-\\frac{1}{2}\\right), 311\\left(\\frac{1}{2}\\right), 311\\left(-\\frac{1}{2}\\right), 31(-1)\\left(\\frac{1}{2}\\right)\\), and \\(31(-1)\\left(-\\frac{1}{2}\\right)\\); for the \\(d\\) states they are \\(320\\left(\\frac{1}{2}\\right), 320\\left(-\\frac{1}{2}\\right), 321\\left(\\frac{1}{2}\\right), 321\\left(-\\frac{1}{2}\\right), 32(-1)\\left(\\frac{1}{2}\\right), 32(-1)\\left(-\\frac{1}{2}\\right), 322\\left(\\frac{1}{2}\\right), 322\\left(-\\frac{1}{2}\\right), 32(-2)\\left(\\frac{1}{2}\\right)\\), and \\(32(-2)\\left(-\\frac{1}{2}\\right)\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 363,
|
||
"question": "Give the electron configurations for the following ions: \\(\\mathrm{Fe}^{2+}, \\mathrm{Al}^{2+}, \\mathrm{Cu}^{+}, \\mathrm{Ba}^{2+}, \\mathrm{Br}^{-}\\), and \\(\\mathrm{O}^{2-}\\).",
|
||
"answer": "(\\mathrm{Fe}^{2+}\\) : From Table 2.2, the electron configuration for an atom of iron is \\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}\\). 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 \\(\\mathrm{Fe}^{2+}\\) ion is \\(1 s^{2} 2 s^{2} 2 p^{6} \\mathrm{~s}^{2} 3 p^{6} 3 d^{6}\\).\n\\(\\mathrm{Al}^{3+}\\) : From Table 2.2, the electron configuration for an atom of aluminum is \\(1 s^{2} 2 s^{2} 2 p^{6} s^{2} 32 p^{1}\\). In order to become an ion with a plus three charge, it must lose three electrons-in this case two \\(3 \\mathrm{~s}\\) and the one \\(3 p\\). Thus, the electron configuration for an \\(\\mathrm{Al}^{3+}\\) ion is \\(1 s^{2} 2 s^{2} 2\\) p \\(^{6}\\).\n\\(\\mathrm{Cu}^{+}\\)From Table 2.2, the electron configuration for an atom of copper is \\(1 s^{2} 2 s^{2} 2 p^{6} x s^{2} 3 p^{6} 3 d^{10} 4 s^{1}\\). 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 \\(\\mathrm{Cu}^{+}\\)ion is \\(1 s^{2} 2 s^{2} 2 p^{6} p^{6} 3 s^{2} 3 p^{6} 3 d 10\\).\n\\(\\mathrm{Ba}^{2+}\\) : 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 \\(1 s^{2} 2 s^{2} 2 p^{6} v s^{2} 3^{2} 3 p^{6} 3 d^{10} 4 s^{\\prime} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 6 s^{2}\\). In order to become an ion with a plus two charge, it must close two electrons-in this case two the 6s. Thus, the electron configuration for a \\(\\mathrm{Ba}^{2+}\\) ion is \\(1 s^{2} 2 s^{2}{ }^{2} p^{6} 3 s^{2} 3 p^{6} 3 \\mathrm{~d}^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2}{ }^{2} 5 p^{6}\\).\n\\(\\mathrm{Br}^{+}\\)From Table 2.2, the electron configuration for an atom of bromine is \\(1 s^{2} 2 s^{2} 2 p^{6} r^{3} 3 s^{2} 3 p^{6} 3 d^{10} 4 \\mathrm{~s}^{2} 4 p^{5}\\). In order to become an ion with a minus one charge, it must acquire one electron-in this case another \\(4 p\\). Thus, the electron configuration for a \\(\\mathrm{Br}^{-}\\)ion is \\(1 s^{2} 2 s^{2} 2 p^{\\text {6 }} 3 s^{2} 3 p^{6} 3 d^{10} \\mathrm{~A} 2^{2} 4 p^{6}\\).\n\\(\\mathrm{O}^{2-}\\) : From Table 2.2, the electron configuration for an atom of oxygen is \\(1 s^{2} 2 s^{2} 2 p^{4}\\). In order to become an ion with a minus two charge, it must acquire two electrons-in this case another two \\(2 p\\). Thus, the electron configuration for an \\(\\mathrm{O}^{2-}\\) ion is \\(1 s^{2} 2 s^{2} 2 p{ }^{6}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 368,
|
||
"question": "(a) What electron subshell is being filled for the rare earth series of elements on the periodic table?\n(b) What electron subshell is being filled for the actinide series?",
|
||
"answer": "(a) The \\(4 f\\) subshell is being filled for the rare earth series of elements.\n(b) The \\(5 f\\) subshell is being filled for the actinide series of elements.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 370,
|
||
"question": "(a) Briefly cite the main differences between ionic, covalent, and metallic bonding.\n(b) State the Pauli exclusion principle.",
|
||
"answer": "(a) The main differences between the various forms of primary bonding are:\nIonic--there is electrostatic attraction between oppositely charged ions.\nCovalent--there is electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration.\nMetallic--the positively charged ion cores are shielded from one another, and also \"glued\" together by the sea of valence electrons.\n(b) The Pauli exclusion principle states that each electron state can hold no more than two electrons, which must have opposite spins.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 371,
|
||
"question": "What type(s) of bonding would be expected for each of the following materials: brass (a copper-zinc alloy), rubber, barium sulfide (BaS), solid xenon, bronze, nylon, and aluminum phosphide (AlP)?",
|
||
"answer": "For brass, the bonding is metallic since it is a metal alloy.\nFor rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)\nFor BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of \\(\\mathrm{Ba}\\) and \\(\\mathrm{S}\\) in the periodic table.\nFor solid xenon, the bonding is van der Waals since xenon is an inert gas.\nFor bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).\nFor nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)\nFor AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of \\(\\mathrm{Al}\\) and \\(\\mathrm{P}\\) in the periodic table.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 378,
|
||
"question": "Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. \n\\begin{tabular}{cccc}\n\\hline Alloy & \\begin{tabular}{c} \nAtomic Weight \\\\\n\\((\\mathbf{g} / \\mathbf{m o l})\\)\n\\end{tabular} & \\begin{tabular}{c} \nDensity \\\\\n\\(\\mathbf{( g / c m}^{2})\\)\n\\end{tabular} & \\begin{tabular}{c} \nAtomic Radius \\\\\n\\((\\mathbf{m m})\\)\n\\end{tabular} \\\\\n\\hline A & 77.4 & 8.22 & 0.125 \\\\\nB & 107.6 & 13.42 & 0.133 \\\\\nC & 127.3 & 9.23 & 0.142 \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "For each of these three alloys we need, by trial and error, to calculate the density using Equation, and compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, the respective values of \\(n\\) are 1,2 , and 4 , whereas the expressions for \\(a\\) (since \\(V_{C}=a^{3}\\) ) are \\(2 R, 2 R \\sqrt{2}\\), and \\(\\frac{4 R}{\\sqrt{3}}\\).\nFor alloy A, let us calculate \\(\\rho\\) assuming a simple cubic crystal structure.\n\\[\n\\begin{aligned}\n\\rho & =\\frac{n A_{\\mathrm{A}}}{V_{C} N_{\\mathrm{A}}} \\\\\n& =\\frac{n A_{\\mathrm{A}}}{(2 R)^{3} N_{\\mathrm{A}}} \\\\\n& =\\frac{(1 \\text { atom/unit cell })(77.4 \\mathrm{~g} / \\mathrm{mol})}{\\left\\{\\left[(2)(1.25 \\times 10^{-8})\\right]^{\\frac{1}{3}} /(\\text { unit cell })\\right\\}(6.022 \\times 10^{23} \\text { atoms/mol })} \\\\\n& =8.22 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]\nTherefore, its crystal structure is simple cubic.\nFor alloy B, let us calculate \\(\\rho\\) assuming an FCC crystal structure.\n\\[\n\\rho=\\frac{n A_{\\mathrm{B}}}{(2 R \\sqrt{2})^{3} N_{\\mathrm{A}}}\n\\]\n\n\n\\[\n\\begin{array}{l}\n=\\frac{(4 \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell})(107.6 \\mathrm{~g} / \\mathrm{mol})}{\\left[\\left[(2 \\sqrt{2})\\left(1.33 \\times 10^{-8} \\mathrm{~cm}\\right)\\right]^{\\frac{3}{2}} /(\\mathrm{unit} \\mathrm{cell})\\right]}\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right) \\\\\n=13.42 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{array}\n\\]\nTherefore, its crystal structure is FCC.\nFor alloy \\(\\mathrm{C}\\), let us calculate \\(\\rho\\) assuming a simple cubic crystal structure.\n\\[\n\\begin{aligned}\n& =\\frac{n A_{\\mathrm{c}}}{(2 R)^{3} N_{\\mathrm{A}}} \\\\\n& =\\frac{(1 \\mathrm{atom} / \\mathrm{unit} \\mathrm{cell})(127.3 \\mathrm{~g} / \\mathrm{mol})}{\\left[\\left[(2)(1.42 \\times 10^{-8} \\mathrm{~cm})\\right]^{\\frac{3}{2}} /(\\mathrm{unit} \\mathrm{cell)}\\right]}\\left(6.022 \\times 10^{23}\\right. \\mathrm{atoms} / \\mathrm{mol}\\right) \\\\\n& =9.23 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]\nTherefore, its crystal structure is simple cubic.",
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"is_select": 1
|
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},
|
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{
|
||
"idx": 379,
|
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"question": "The unit cell for tin has tetragonal symmetry, with \\(a\\) and \\(b\\) lattice parameters of 0.583 and \\(0.318 \\mathrm{~nm}\\), respectively. If its density, atomic weight, and atomic radius are \\(7.30 \\mathrm{~g} / \\mathrm{cm}^{3}, 118.69 \\mathrm{~g} / \\mathrm{mol}\\), and \\(0.151 \\mathrm{~nm}\\), respectively, compute the atomic packing factor.",
|
||
"answer": "In order to determine the APF for \\(\\mathrm{Sn}\\), we need to compute both the unit cell volume \\(\\left(V_{C}\\right)\\) which is just the \\(a^{2} c\\) product, as well as the total sphere volume \\(\\left(V_{S}\\right)\\) which is just the product of the volume of a single sphere and the number of spheres in the unit cell \\((n)\\). The value of \\(n\\) may be calculated from Equation as\n\\[\n\\begin{aligned}\nn & =\\frac{\\rho V_{C} N_{\\mathrm{A}}}{A_{\\mathrm{Sn}}} \\\\\n& =\\frac{\\left(7.30 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)(5.83)^{2}(3.18)\\left(\\times 10^{-24} \\mathrm{~cm}^{3}\\right)\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)}{118.69 \\mathrm{~g} / \\mathrm{mol}} \\\\\n& =4.00 \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell}\n\\end{aligned}\n\\]\nTherefore\n\\[\n\\begin{aligned}\n\\mathrm{APF} & =\\frac{V_{\\mathrm{S}}}{V_{C}}=\\frac{\\left(4\\left(\\frac{4}{3} \\pi R^{3}\\right)\\right.}{(a)^{2}(c)} \\\\\n& =\\frac{\\left(4\\left[\\frac{4}{3}(\\pi)(1.51 \\times 10^{-8} \\mathrm{~cm})^{3}\\right]\\right.}{(5.83 \\times 10^{-8} \\mathrm{~cm})^{2}(3.18 \\times 10^{-8} \\mathrm{~cm})} \\\\\n& =0.534\n\\end{aligned}\n\\]",
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"is_select": 1
|
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},
|
||
{
|
||
"idx": 380,
|
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"question": "List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure.",
|
||
"answer": " the barium ions are situated at all corner positions. The point coordinates for these ions are as follows: \\(000,100,110,010,001,101,111\\), and 011 .\nThe oxygen ions are located at all face-centered positions; therefore, their coordinates are \\(\\frac{1}{2} \\frac{1}{2} 0, \\frac{1}{2} \\frac{1}{2} 1\\), \\(\\frac{1}{2} \\frac{1}{2}, \\frac{0}{2} \\frac{1}{2}, \\frac{1}{2} 0 \\frac{1}{2}\\), and \\(\\frac{1}{2} 1 \\frac{1}{2}\\).\nAnd, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates \\(\\frac{1}{2} \\frac{1}{2} \\frac{1}{2}\\).",
|
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"is_select": 0
|
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},
|
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{
|
||
"idx": 381,
|
||
"question": "List the point coordinates of all atoms that are associated with the diamond cubic unit cell.",
|
||
"answer": "First of all, one set of carbon atoms occupy all corner positions of the cubic unit cell; the coordinates of these atoms are as follows: \\(000,100,110,010,001,101,111\\), and 011 .\nAnother set of atoms reside on all of the face-centered positions, with the following coordinates: \\(\\frac{1}{2} \\frac{1}{2} 0\\), \\(\\frac{1}{2} \\frac{1}{2} 1, \\frac{1}{2} \\frac{1}{2} \\cdot 0, \\frac{1}{2} \\frac{1}{2} \\cdot \\frac{1}{2} 0 \\frac{1}{2}\\), and \\(\\frac{1}{2} \\frac{1}{2} \\cdot\\)\nThe third set of carbon atoms are positioned within the interior of the unit cell. the coordinates of the atom that lies toward the lower-left-front of the unit cell has the coordinates \\(\\frac{3}{4} \\frac{1}{4} \\frac{1}{4}\\), whereas the atom situated toward the lower-right-back of the unit cell has coordinates of \\(\\frac{1}{3} \\frac{1}{3} 1\\). Also, the carbon atom that resides toward the upper-left-back of the unit cell has the \\(\\frac{1}{4} \\frac{1}{4} \\frac{3}{4}\\) coordinates. And, the coordinates of the final atom, located toward the upper-right-front of the unit cell, are \\(\\frac{3 \\cdot 3}{4 \\cdot 4} \\cdot\\)",
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"is_select": 0
|
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},
|
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{
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"idx": 382,
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"question": "Using the Molecule Definition Utility found in both \"Metallic Crystal Structures and Crystallography\" and \"Ceramic Crystal Structures\" modules of VMSE, located on the book's web site [www.wiley.com/college/Callister (Student Companion Site)], generate a three-dimensional unit cell for the intermetallic compound \\(\\mathrm{AuCu}_{3}\\) given the following: (1) the unit cell is cubic with an edge length of \\(0.374 \\mathrm{~nm}\\), (2) gold atoms are situated at all cube corners, and (3) copper atoms are positioned at the centers of all unit cell faces.",
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||
"answer": "First of all, open the \"Molecular Definition Utility\"; it may be found in either of \"Metallic Crystal Structures and Crystallography\" or \"Ceramic Crystal Structures\" modules.\nIn the \"Step 1\" window, it is necessary to define the atom types, colors for the spheres (atoms), and specify atom sizes. Let us enter \"Au\" as the name for the gold atoms (since \"Au\" the symbol for gold), and \"Cu\" as the name for the copper atoms. Next it is necessary to choose a color for each atom type from the selections that appear in the pull-down menu-for example, \"Yellow\" for Au and \"Red\" for \\(\\mathrm{Cu}\\). In the \"Atom Size\" window, it is necessary to enter an atom/ion size. In the instructions for this step, it is suggested that the atom/ion diameter in nanometers be used. From the table found inside the front cover of the textbook, the atomic radii for gold and copper are \\(0.144 \\mathrm{~nm}\\) and \\(0.128 \\mathrm{~nm}\\), respectively, and, therefore, their ionic diameters are twice these values (i.e., \\(0.288 \\mathrm{~nm}\\) and \\(0.256 \\mathrm{~nm}\\) ); therefore, we enter the values \" 0.288 \" and \" 0.256 \" for the two atom types. Now click on the \"Register\" button, followed by clicking on the \"Go to Step 2\" button.\nIn the \"Step 2\" window we specify positions for all of the atoms within the unit cell; their point coordinates are specified in the problem statement. Let's begin with gold. Click on the yellow sphere that is located to the right of the \"Molecule Definition Utility\" box. Again, Au atoms are situated at all eight corners of the cubic unit cell. One Au will be positioned at the origin of the coordinate system-i.e., its point coordinates are 000, and, therefore, we enter a \"0\" (zero) in each of the \" \\(\\mathrm{x}\\) \", \" \\(\\mathrm{y}\\) \", and \" \\(\\mathrm{z}\\) \" atom position boxes. Next we click on the \"Register Atom Position\" button. Now we enter the coordinates of another gold atom; let us arbitrarily select the one that resides at the corner of the unit cell that is one unit-cell length along the \\(x\\)-axis (i.e., at the 100 point coordinate). Inasmuch as it is located a distance of \\(a\\) units along the \\(x\\)-axis the value of \" 0.374 \" is entered in the \" \\(\\mathrm{x}\\) \" atom position box (since this is the value of \\(a\\) given in the problem statement); zeros are entered in each of the \" \\(\\mathrm{y}\\) \" and \" \\(z\\) \" position boxes. We repeat this procedure for the remaining six Au atoms.\nAfter this step has been completed, it is necessary to specify positions for the copper atoms, which are located at all six face-centered sites. To begin, we click on the red sphere that is located next to the \"Molecule Definition Utility\" box. The point coordinates for some of the \\(\\mathrm{Cu}\\) atoms are fractional ones; in these instances, the \\(a\\) unit cell length (i.e., 0.374) is multiplied by the fraction. For example, one \\(\\mathrm{Cu}\\) atom is located \\(1 \\frac{1}{2} \\frac{1}{2}\\) coordinate. Therefore, the \\(\\mathrm{x}, \\mathrm{y}\\), and \\(\\mathrm{z}\\) atoms positions are (1)(0.374) \\(=0.374, \\frac{1}{2}(0.374)=0.187\\), and \\(\\frac{1}{2}(0.374)=0.187\\), respectively.\nFor the gold atoms, the \\(\\mathrm{x}, \\mathrm{y}\\), and \\(\\mathrm{z}^{\\prime}\\) atom position entries for all 8 sets of point coordinates are as follows:\n\n\\[\n\\begin{array}{l}\n0,0, \\text { and } 0 \\\\\n0.374,0, \\text { and } 0 \\\\\n0,0.374, \\text { and } 0 \\\\\n0,0, \\text { and } 0.374 \\\\\n0,0.374,0.374 \\\\\n0.374,0.374 \\\\\n0.374,0 .374,0 \\\\\n0.374,0.374,0.374\n\\end{array}\n\\]\nNow, for the copper atoms, the \\(\\mathrm{x}, \\mathrm{y}\\), and \\(\\mathrm{z}\\) atom position entries for all 6 sets of point coordinates are as follows:\n\\[\n\\begin{array}{l}\n0.187,0.187,0 \\\\\n0.187,0,0.187 \\\\\n0,0.187,0.187 \\\\\n0.374,0.187,0.187 \\\\\n0.187,0.374,0.187 \\\\\n0.187,0.187,0.374\n\\end{array}\n\\]\nIn Step 3, we may specify which atoms are to be represented as being bonded to one another, and which type of bond(s) to use (single solid, single dashed, double, and triple are possibilities), or we may elect to not represent any bonds at all (in which case we are finished). If it is decided to show bonds, probably the best thing to do is to represent unit cell edges as bonds. This image may be rotated by using mouse click-and-drag\nYour image should appear as the following screen shot. Here the gold atoms appear lighter than the copper atoms.\n\n\n\n[Note: Unfortunately, with this version of the Molecular Definition Utility, it is not possible to save either the data or the image that you have generated. You may use screen capture (or screen shot) software to record and store your image.]",
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"is_select": 0
|
||
},
|
||
{
|
||
"idx": 383,
|
||
"question": "Convert the (010) and (101) planes into the four-index Miller-Bravais scheme for hexagonal unit cells.\n}",
|
||
"answer": "For (010), \\(h=0, k=1\\), and \\(l=0\\), and, from Equation, the value of \\(i\\) is equal to \\\\ \\(\\quad i=-(h+k)=-(0+1)=-1\\) \\\\ Therefore, the (010) plane becomes \\(\\left(01 \\overline{-1} 0\\right)\\). \\\\ Now for the (101) plane, \\(h=1, k=0\\), and \\(l=1\\), and computation of \\(i\\) using Equation leads to \\\\ \\(\\quad i=-(h+k)=-[1+0]=-1\\)\n}\nsuch that (101) becomes \\(\\left(10 \\overline{-1} 1\\right)\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 384,
|
||
"question": "(a) Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius \\(R\\).\n(b) Compute the planar density value for this same plane for magnesium.",
|
||
"answer": "(a) A (0001) plane for an HCP unit cell is show below.\n\nEach 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.\n\nIn 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\n\n\\[\n\\begin{aligned}\n\\mathrm{PD}_{0001} & =\\frac{\\text { number of atoms centered on (0001) plane }}{\\text { area of (0001) plane }} \\\\\n& =\\frac{3 \\text { atoms }}{6 R^{2} \\sqrt{3}}=\\frac{1}{2 R^{2} \\sqrt{3}}\n\\end{aligned}\n\\]\n\n(b) From the table inside the front cover, the atomic radius for magnesium is \\(0.160 \\mathrm{~nm}\\). Therefore, the planar density for the (0001) plane is\n\n\\[\n\\mathrm{PD}_{0001}(\\mathrm{Mg})=\\frac{1}{2 R^{2} \\sqrt{3}}=\\frac{1}{2(0.160 \\mathrm{~nm})^{2} \\sqrt{3}}=11.28 \\mathrm{~nm}^{-2}=1.128 \\times 10^{19} \\mathrm{~m}^{-2}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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^{\\circ}\\) (first-order reflection) when monochromatic \\(x\\)-radiation having a wavelength of \\(0.1542 \\mathrm{~nm}\\) is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for an iridium atom.",
|
||
"answer": "(a) From the data given in the problem, and realizing that \\(69.22^{\\circ}=2 \\theta\\), the interplanar spacing for the (220) set of planes for iridium may be computed using Equation as\n\\[\nd_{220}=\\frac{n \\lambda}{2 \\sin \\theta}=\\frac{(1)(0.1542 \\mathrm{~nm})}{(2)\\left(\\sin \\frac{69.22^{\\circ}}{2}\\right)}=0.1357 \\mathrm{~nm}\n\\]\n(b) In order to compute the atomic radius we must first determine the lattice parameter, \\(a\\), using Equation 3.14, and then \\(R\\) from Equation since Ir has an FCC crystal structure. Therefore,\n\\[\na=d_{220} \\sqrt{(2)^{2}+(2)^{2}+(0)^{2}}=(0.1357 \\mathrm{~nm})(\\sqrt{8})=0.3838 \\mathrm{~nm}\n\\]\nAnd, from Equation \n\\[\nR=\\frac{a}{2 \\sqrt{2}}=\\frac{0.3838 \\mathrm{~nm}}{2 \\sqrt{2}}=0.1357 \\mathrm{~nm}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 387,
|
||
"question": "Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.\n\\begin{tabular}{lcccc}\n\\hline Element & \\begin{tabular}{c} \nAtomic Radius \\\\\n\\((\\mathbf{m m})\\)\n\\end{tabular} & Crystal Structure & Electronegativity & Valence \\\\\n\\hline \\(\\mathrm{Cu}\\) & 0.1278 & FCC & 1.9 & +2 \\\\\n\\(\\mathrm{C}\\) & 0.071 & & & \\\\\n\\(\\mathrm{H}\\) & 0.046 & & & \\\\\n\\(\\mathrm{O}\\) & 0.060 & & & \\\\\n\\(\\mathrm{Ag}\\) & 0.1445 & FCC & 1.9 & +1 \\\\\n\\(\\mathrm{Al}\\) & 0.1431 & FCC & 1.5 & +3 \\\\\n\\(\\mathrm{Co}\\) & 0.1253 & HCP & 1.8 & +2 \\\\\n\\(\\mathrm{Cr}\\) & 0.1249 & BCC & 1.6 & +3 \\\\\n\\(\\mathrm{Fe}\\) & 0.1241 & BCC & 1.8 & +2 \\\\\n\\(\\mathrm{Ni}\\) & 0.1246 & FCC & 1.8 & +2 \\\\\n\\(\\mathrm{Pd}\\) & 0.1376 & FCC & 2.2 & +2 \\\\\n\\(\\mathrm{Pt}\\) & 0.1387 & FCC & 2.2 & +2 \\\\\n\\(\\mathrm{Zn}\\) & 0.1332 & HCP & 1.6 & +2 \\\\\n\\hline\n\\end{tabular}\nWhich of these elements would you expect to form the following with copper:\n(a) A substitutional solid solution having complete solubility\n(b) A substitutional solid solution of incomplete solubility\n(c) An interstitial solid solution",
|
||
"answer": "In this problem we are asked to cite which of the elements listed form with \\(\\mathrm{Cu}\\) the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between \\(\\mathrm{Cu}\\) and the other element \\((\\Delta \\mathrm{R} \\%)\\) must be less than \\(\\pm 15 \\%, 2)\\) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are tabulated, for the various elements, these criteria.\n\\begin{tabular}{lccc} \nElement & \\(\\Delta \\mathrm{R} \\%\\) & \\begin{tabular}{c} \nCrystal \\\\\nStructure\n\\end{tabular} & \\begin{tabular}{c}\n\\(\\Delta\\) Electro- \\\\\nnegativity\n\\end{tabular} \\\\\n\\hline \\(\\mathbf{C u}\\) & & FCC & \\(\\mathbf{2}+\\)\n\\end{tabular}\n\n\n\\begin{tabular}{lllll} \nH & -64 & & & \\\\\nO & -53 & & & \\\\\nAg & +13 & FCC & 0 & \\(1+\\) \\\\\nAl & +12 & FCC & -0.4 & \\(3+\\) \\\\\nCo & -2 & HCP & -0.1 & \\(2+\\) \\\\\nCr & -2 & BCC & -0.3 & \\(3+\\) \\\\\nFe & -3 & BCC & -0.1 & \\(2+\\) \\\\\nNi & -3 & FCC & -0.1 & \\(2+\\) \\\\\nPd & +8 & FCC & +0.3 & \\(2+\\) \\\\\nPt & +9 & FCC & +0.3 & \\(2+\\) \\\\\nZn & +4 & HCP & -0.3 & \\(2+\\)\n\\end{tabular}\n(a) \\(\\mathrm{Ni}, \\mathrm{Pd}\\), and \\(\\mathrm{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.\n(b) \\(\\mathrm{Ag}, \\mathrm{Al}, \\mathrm{Co}, \\mathrm{Cr}, \\mathrm{Fe}\\), and \\(\\mathrm{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 \\(\\mathrm{Cu}\\) are greater than \\(\\pm 15 \\%\\), and/or have a valence different than \\(2+\\).\n(c) \\(\\mathrm{C}, \\mathrm{H}\\), and \\(\\mathrm{O}\\) form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of \\(\\mathrm{Cu}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 390,
|
||
"question": "What is the composition, in atom percent, of an alloy that contains \\(99.7 \\mathrm{lb}_{\\mathrm{m}}\\) copper, \\(102 \\mathrm{lb}_{\\mathrm{m}}\\) zinc, and \\(2.1 \\mathrm{lb}_{\\mathrm{m}}\\) lead?\n",
|
||
"answer": "In this problem we are asked to determine the concentrations, in atom percent, of the \\(\\mathrm{Cu}-\\mathrm{Zn}-\\mathrm{Pb}\\) alloy. It is first necessary to convert the amounts of \\(\\mathrm{Cu}, \\mathrm{Zn}\\), and \\(\\mathrm{Pb}\\) into grams.\n\n\\[\n\\begin{aligned}\nm_{\\mathrm{Cu}}^{\\prime} & =\\left(99.7 \\mathrm{lb}_{\\mathrm{m}}\\right)\\left(453.6 \\mathrm{~g} / \\mathrm{lb}_{\\mathrm{m}}\\right)=45,224 \\mathrm{~g} \\\\\nm_{\\mathrm{Zn}}^{\\prime} & =\\left(102 \\mathrm{lb}_{\\mathrm{m}}\\right)\\left(453.1 \\mathrm{~g} / \\mathrm{lb}_{\\mathrm{m}}\\right)=(46,267 \\mathrm{~g} \\\\\nm_{\\mathrm{Pb}}^{\\prime} & =\\left(2.1 \\mathrm{lb}_{\\mathrm{m}}\\right)\\left(453.3 \\mathrm{~g} / \\mathrm{lb}_{\\mathrm{m}}\\right) \\text { = } 953 \\mathrm{~g}\n\\end{aligned}\n\\]\n\nThese masses must next be converted into moles , as\n\n\\[\n\\begin{aligned}\nn_{m_{\\mathrm{Cu}}} & =\\frac{m_{\\mathrm{Cu}}^{\\prime}}{A_{\\mathrm{Cu}}}=\\frac{45,224 \\mathrm{~g}}{63.55 \\mathrm{~g} / \\mathrm{mol}}=711.6 \\mathrm{~mol} \\\\\nn_{m_{\\mathrm{Zn}}} & =\\frac{46,267 \\mathrm{~g}}{65.41 \\mathrm{~g} / \\mathrm{mol}}=707.3 \\mathrm{~mol} \\\\\nn_{m_{\\mathrm{Pb}}} & =\\frac{953 \\mathrm{~g}}{207.2 \\mathrm{~g} / \\mathrm{mol}}=4.6 \\mathrm{~mol}\n\\end{aligned}\n\\]\n\nNow, employment of a modified form of Equation, gives\n\n\\[\n\\begin{aligned}\nC_{\\mathrm{Cu}}^{\\prime} & =\\frac{n_{m_{\\mathrm{Cu}}}}{n_{m_{\\mathrm{Cu}}}+n_{m_{\\mathrm{Zn}}}+n_{m_{\\mathrm{Pb}}} \\times 100 \\\\\n& =\\frac{711.6 \\mathrm{~mol}}{711.6 \\mathrm{~mol}+707.3 \\mathrm{~mol}+4.6 \\mathrm{~mol}} \\times 100=50.0 \\mathrm{at} \\% \\\\\nC_{\\mathrm{Zn}}^{\\prime} & =\\frac{707.3 \\mathrm{~mol}}{711.6 \\mathrm{~mol}+\\frac{707.3 \\mathrm{~mol}+4.6 \\mathrm{~m}}{6}} \\times 100=49.7 \\mathrm{at} \\%\n\\end{aligned}\n\\]\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 1976 United States Copyright Act without the permission of the copyright owner is unlovelul.\n\n\\[\nC_{\\mathrm{Pb}}^{\\star}=\\frac{4.6 \\mathrm{~mol}}{711.6 \\mathrm{~mol}+707.3 \\mathrm{~mol}+4.6 \\mathrm{~mol}} \\times 100=0.3 \\mathrm{at} \\%\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 1976 United States Copyright Act without the permission of the copyright owner is unlovelful.",
|
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"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
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"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
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"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
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"is_select": 1
|
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},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 397,
|
||
"question": "(a) 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?\n(b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 398,
|
||
"question": "(a) Briefly describe a twin and a twin boundary.\n(b) Cite the difference between mechanical and annealing twins.",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 400,
|
||
"question": "(a) Briefly explain the concept of a driving force.\n(b) What is the driving force for steady-state diffusion?",
|
||
"answer": "(a) The driving force is that which compels a reaction to occur.\n(b) The driving force for steady-state diffusion is the concentration gradient.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 401,
|
||
"question": "For a steel alloy it has been determined that a carburizing heat treatment of 10-h duration will raise the carbon concentration to \\(0.45 \\mathrm{wt} \\%\\) at a point \\(2.5 \\mathrm{~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 calls for an estimate of the time necessary to achieve a carbon concentration of \\(0.45 \\mathrm{wt} \\%\\) at a point \\(5.0 \\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, and\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\\]\nThus,\n\\[\n\\frac{(2.5 \\mathrm{~mm})^{2}}{10 \\mathrm{~h}}=\\frac{(5.0 \\mathrm{~mm})^{2}}{t_{2}}\n\\]\nfrom which\n\\[\nt_{2}=40 \\mathrm{~h}\n\\]\n}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 403,
|
||
"question": "The activation energy for the diffusion of carbon in chromium is \\(111,000 \\mathrm{~J} / \\mathrm{mol}\\). Calculate the diffusion coefficient at \\(1100 \\mathrm{~K}\\left(827^{\\circ} \\mathrm{C}\\right)\\), given that \\(\\mathrm{D}\\) at \\(1400 \\mathrm{~K}\\left(1127^{\\circ} \\mathrm{C}\\right)\\) is \\(6.25 \\times 10^{-11} \\mathrm{~m}^{2} / \\mathrm{s}\\).",
|
||
"answer": "To solve this problem it first becomes necessary to solve for \\(D_{0}\\) from Equation as\n\\[\n\\begin{aligned}\nD_{0} & =D \\exp \\left(\\frac{Q_{d}}{R T}\\right) \\\\\n& =\\left(6.25 \\times 10^{-11} \\mathrm{~m} \\mathrm{~s}^{2} / \\mathrm{s}\\right) \\exp \\left[\\frac{111,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(1400 \\mathrm{~K})}\\right] \\\\\n& =8.7 \\times 10^{-7} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{aligned}\n\\]\nNow, solving for \\(D\\) at \\(1100 \\mathrm{~K}\\) (again using Equation) gives\n\\[\n\\begin{aligned}\nD & =\\left(8.7 \\times 10^{-7} \\mathrm{~m}^{2 / \\mathrm{s}}\\right) \\exp \\left[-\\frac{111,000 \\mathrm{~J} / \\mathrm{m} \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm {mol}-\\mathrm{K})(1100 \\mathrm{~K})}\\right] \\\\\n& =4.6 \\times 10^{-12} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{aligned}%\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 404,
|
||
"question": "The diffusion coefficients for iron in nickel are given at two temperatures:\n\\begin{tabular}{cc}\n\\hline\\(T(\\mathrm{~K})\\) & \\(D\\left(m^{2} / \\mathrm{s}\\right)\\) \\\\\n\\hline 1273 & \\(9.4 \\times 10^{-16}\\) \\\\\n1473 & \\(2.4 \\times 10^{-14}\\) \\\\\n\\hline\n\\end{tabular}\n(a) Determine the values of \\(D_{0}\\) and the activation energy \\(Q_{d}\\).\n(b) What is the magnitude of \\(D\\) at \\(1100^{\\circ} \\mathrm{C}(1373 \\mathrm{~K})\\) ?",
|
||
"answer": "(a) Using Equation, we set up two simultaneous equations with \\(Q_{d}\\) and \\(D_{0}\\) as unknowns as follows:\n\\[\n\\begin{array}{l}\n\\ln D_{1}=\\ln D_{0}-\\frac{Q_{d}}{R}\\left(\\frac{1}{T_{1}}\\right) \\\\\n\\ln D_{2}=\\ln D_{0}-\\frac{Q_{d}}{R}\\left(\\begin{array}{c}\n1 \\\\\nT_{2}\n\\end{array}\\right)\n\\end{array}\n\\]\nNow, solving for \\(Q_{d}\\) in terms of temperatures \\(T_{1}\\) and \\(T_{2}\\left(1273 \\mathrm{~K}\\right.\\) and \\(1473 \\mathrm{~K}\\) ) and \\(D_{1}\\) and \\(D_{2}\\left(9.4 \\times 10^{-16}\\right.\\) and \\(2.4 \\times\\) \\(10^{-14} \\mathrm{~m}^{2} / \\mathrm{s}\\) ), we get\n\\[\n\\begin{aligned}\nQ_{d} & =-R \\frac{\\ln D_{1}-\\ln D_{2}}{\\frac{1}{T_{1}}-\\frac{1}{T_{2}}} \\\\\n& =-\\left(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\\right) \\frac{\\left[\\ln \\left(9.4 \\times 10^{-16}\\right)-\\ln \\left(2.4 \\times 10^{-14}\\right)\\right]}{\\frac{1}{1273 \\mathrm{~K}}-\\frac{1}{1473 \\mathrm{~K}}} \\\\\n& =252,400 \\mathrm{~J} / \\mathrm{mol}\n\\end{aligned}\n\\]\nNow, solving for \\(D_{0}\\) from Equation (and using the \\(1273 \\mathrm{~K}\\) value of \\(D\\) )\n\\[\nD_{0}=D_{1} \\exp \\left(\\frac{Q_{d}}{R T_{1}}\\right)\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 1976 United States Copyright Act without the permission of the copyright owner is unlovelul.\n\n\\[\n\\begin{array}{c}\n=\\left(9.4 \\times 10^{-16} \\mathrm{~m}^{2} / \\mathrm{s}\\right) \\exp \\left[\\frac{252,400 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(1273 \\mathrm{~K})}\\right] \\\\\n=2.2 \\times 10^{-5} \\mathrm{~m}^{2} / \\mathrm{s} \\\\\n\\text { (b) Using these values of } D_{0} \\text { and } Q_{d}, D \\text { at } 1373 \\mathrm{~K} \\text { is just } \\\\\nD=\\left(2.2 \\times 10^{-5} \\mathrm{~m}^{2 / \\mathrm{s}}\\right) \\exp \\left[-\\frac{252,400 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\left.\\mathrm{K}\\right)(1373 \\mathrm{~K})}\\right] \\\\\n=5.4 \\times 10^{-15} \\mathrm{~m}^{2} / \\mathrm{s}\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 \"Do and Qd from Experimental Data\" submodule, and then do the following:\n1. In the left-hand window that appears, enter the two temperatures from the table in the book (viz. \"1273\" and \"1473\", in the first two boxes under the column labeled \"T (K)\". Next, enter the corresponding diffusion coefficient values (viz. \"9.4e-16\" and \"2.4e-14\").\n3. Next, at the bottom of this window, click the \"Plot data\" button.\n4. A \\(\\log \\mathrm{D}\\) versus \\(1 / \\mathrm{T}\\) plot then appears, with a line for the temperature dependence for this diffusion system. At the top of this window are give values for \\(D_{0}\\) and \\(Q_{d}\\) : for this specific problem these values are \\(2.17 \\times\\) \\(10^{-5} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(252 \\mathrm{~kJ} / \\mathrm{mol}\\), respectively\n5. To solve the (b) part of the problem we utilize the diamond-shaped cursor that is located at the top of the line on this plot. Click-and-drag this cursor down the line to the point at which the entry under the \"Temperature (T):\" label reads \"1373\". The value of the diffusion coefficient at this temperature is given under the label \"Diff Coeff (D):\". For our problem, this value is \\(5.4 \\times 10^{-15} \\mathrm{~m}^{3} / \\mathrm{s}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 405,
|
||
"question": "The diffusion coefficients for silver in copper are given at two temperatures:\n\\begin{tabular}{cc}\n\\hline\\(T\\left({ }^{\\circ} \\mathrm{C}\\right)\\) & \\(D\\left(m^{2} / \\mathrm{s}\\right)\\) \\\\\n\\hline 650 & \\(5.5 \\times 10^{-16}\\) \\\\\n900 & \\(1.3 \\times 10^{-13}\\) \\\\\n\\hline\n\\end{tabular}\n(a) Determine the values of \\(D_{0}\\) and \\(Q_{d}\\).\n(b) What is the magnitude of \\(D\\) at \\(875^{\\circ} \\mathrm{C}\\) ?",
|
||
"answer": "(a) Using Equation, we set up two simultaneous equations with \\(Q_{d}\\) and \\(D_{0}\\) as unknowns as follows:\n\\[\n\\begin{array}{l}\n\\ln D_{1}=\\ln D_{0}-\\frac{Q_{d}}{R}\\left(\\frac{1}{T_{1}}\\right) \\\\\n\\ln D_{2}=\\ln D_{0}-\\frac{Q_{d}}{R}\\left(\\left.\\frac{1}{T_{2}}\\right)\\right.\n\\end{array}\n\\]\nSolving for \\(Q_{d}\\) in terms of temperatures \\(T_{1}\\) and \\(T_{2}\\left(923 \\mathrm{~K}\\left[650^{\\circ} \\mathrm{C}\\right]\\right.\\) and \\(\\left.1173 \\mathrm{~K}\\left[900^{\\circ} \\mathrm{C}\\right]\\right)\\) and \\(D_{1}\\) and \\(D_{2}\\left(5.5 \\times 10^{-}\\right.\\) \\({ }^{16}\\) and \\(1.3 \\times 10^{-13} \\mathrm{~m}^{2} / \\mathrm{s}\\) ), we get\n\\[\n\\begin{aligned}\nQ_{d} & =-R \\frac{\\ln D_{1}-\\ln D_{2}}{\\frac{1}{T_{1}}-\\frac{1}{T_{2}}} \\\\\n& =-\\frac{\\left(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\\right)\\left[\\ln \\left(5.5 \\times 10^{-16}\\right)-\\ln \\left(1.3 \\times 10^{-13}\\right)\\right]}{\\frac{1}{923 \\mathrm{~K}}-\\frac{1}{1173 \\mathrm{~K}}} \\\\\n& =196,700 \\mathrm{~J} / \\mathrm{mol}\n\\end{aligned}\n\\]\nNow, solving for \\(D_{0}\\) from Equation (and using the \\(650^{\\circ} \\mathrm{C}\\) value of \\(D\\) )\n\\[\nD_{0}=D_{1} \\exp \\left(\\frac{Q_{d}}{R T_{1}}\\right)\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 1976 United States Copyright Act without the permission of the copyright owner is unlovelul.\n\n\\[\n\\begin{array}{c}\n=\\left(5.5 \\times 10^{-16} \\mathrm{~m}^{2} / \\mathrm{s}\\right) \\exp \\left\\lfloor\\frac{196,700 \\mathrm{~J} / \\mathrm{mol}}{\\left(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\\right)\\left(923 \\mathrm{~K}\\right)}\\right\\rfloor \\\\\n=7.5 \\times 10^{-5} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{array}\n\\]\n(b) Using these values of \\(D_{0}\\) and \\(Q_{d}, D\\) at \\(1148 \\mathrm{~K}\\left(875^{\\circ} \\mathrm{C}\\right)\\) is just\n\\[\n\\begin{aligned}\nD & =\\left(7.5 \\times 10^{-5} \\mathrm{~m}^{2 / \\mathrm{s}}\\right) \\exp \\left[-\\frac{196,700 \\mathrm{~J} / \\mathrm{s} / \\mathrm{mol}}{\\left(8.31 \\mathrm{~J}_{\\mathrm{m} / \\mathrm{ol}}-\\mathrm{K}\\right)(1148 \\mathrm{~K})}\\right] \\\\\n& =8.3 \\times 10^{-14} \\mathrm{~m}^{2} / \\mathrm{s}\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 \"Do and Qd from Experimental Data\" submodule, and then do the following:\n1. In the left-hand window that appears, enter the two temperatures from the table in the book (converted from degrees Celsius to Kelvins) (viz. \"923\" \\(\\left(650^{\\circ} \\mathrm{C}\\right)\\) and \"1173\" \\(\\left(900^{\\circ} \\mathrm{C}\\right)\\), in the first two boxes under the column labeled \"T (K)\". Next, enter the corresponding diffusion coefficient values (viz. \"5.5e-16\" and \"1.3e-13\").\n3. Next, at the bottom of this window, click the \"Plot data\" button.\n4. A \\(\\log \\mathrm{D}\\) versus \\(1 / \\mathrm{T}\\) plot then appears, with a line for the temperature dependence for this diffusion system. At the top of this window are give values for \\(D_{0}\\) and \\(Q_{d}\\) for this specific problem these values are \\(7.55 \\times\\) \\(10^{-5} \\mathrm{~m}^{2} / \\mathrm{s}\\) and \\(196 \\mathrm{~kJ} / \\mathrm{mol}\\), respectively\n5. To solve the (b) part of the problem we utilize the diamond-shaped cursor that is located at the top of the line on this plot. Click-and-drag this cursor down the line to the point at which the entry under the \"Temperature (T):\" label reads \"1148\" (i.e., \\(875^{\\circ} \\mathrm{C}\\) ). The value of the diffusion coefficient at this temperature is given under the label \"Diff Coeff (D):\". For our problem, this value is \\(8.9 \\times 10^{-14} \\mathrm{~m}^{2} / s\\).",
|
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"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 407,
|
||
"question": "The steady-state diffusion flux through a metal plate is \\(5.4 \\times 10^{-10} \\mathrm{~kg} / \\mathrm{m}^{2}-\\mathrm{s}\\) at a temperature of \\(727^{\\circ} \\mathrm{C}\\) \\((1000 \\mathrm{~K})\\) and when the concentration gradient is \\(-350 \\mathrm{~kg} / \\mathrm{m}^{4}\\). Calculate the diffusion flux at \\(1027^{\\circ} \\mathrm{C}(1300 \\mathrm{~K})\\) for the same concentration gradient and assuming an activation energy for diffusion of \\(125,000 \\mathrm{~J} / \\mathrm{mol}\\).",
|
||
"answer": "In order to solve this problem, we must first compute the value of \\(D_{0}\\) from the data given at \\(727^{\\circ} \\mathrm{C}(1000\\) \\(\\mathrm{K})\\); this requires the combining of both Equations as\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 \\(D_{0}\\) from the above expression gives\n\\[\n\\begin{aligned}\nD_{0} & =-\\frac{J}{\\Delta C} \\exp \\left(\\frac{Q_{d}}{R T}\\right) \\\\\n& =-\\left(\\frac{5.4 \\times 10^{-10} \\mathrm{~kg} / m^{2}-\\mathrm{s}}{-350 \\mathrm{~kg} / \\mathrm{m}^{4}}\\right) \\exp \\left[\\frac{125,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(1000 \\mathrm{~K})}\\right] \\\\\n& =5.26 \\times 10^{-6} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{aligned}\n\\]\nThe value of the diffusion flux at \\(1300 \\mathrm{~K}\\) may be computed using these same two equations as follows:\n\\[\n\\begin{aligned}\nJ & =-D_{0}\\left(\\frac{\\Delta C}{\\Delta x}\\right) \\exp \\left(-\\frac{Q_{d}}{R T}\\right) \\\\\n& =-\\left(5.26 \\times 10^{-6} \\mathrm{~m}^{3} / \\mathrm{s}\\right)\\left(-350 \\mathrm{~kg} / \\mathrm{m}^{4}\\right) \\exp \\left[-\\frac{125,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathbf{K})(1300 \\mathrm{~K})}\\right] \\\\\n& =1.74 \\times 10^{-8} \\mathrm{~kg} / \\mathrm{m}^{2}-\\mathrm{s}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 408,
|
||
"question": "For a bronze alloy, the stress at which plastic deformation begins is \\(275 \\mathrm{MPa}(40,000 \\mathrm{psi})\\), and the modulus of elasticity is \\(115 \\mathrm{GPa}\\left(16.7 \\times 10^{6} \\mathrm{psi}\\right)\\).\n(a) What is the maximum load that may be applied to a specimen with a cross-sectional area of \\(325 \\mathrm{~mm}^{2}\\) (0.5 in. \\({ }^{2}\\) ) without plastic deformation?\n(b) If the original specimen length is \\(115 \\mathrm{~mm}\\) (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation?",
|
||
"answer": "(a) This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation \\(\\left(F_{y}\\right)\\). Taking the yield strength to be \\(275 \\mathrm{MPa}\\), and employment of Equation leads to\n\\[\n\\begin{aligned}\nF_{y}= & \\sigma_{y} A_{0}=\\left(275 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}\\right)\\left(325 \\times 10^{-6} \\mathrm{~m}^{2}\\right) \\\\\n= & 89,375 \\mathrm{~N}\\left(20,000 \\mathrm{lb}_{\\mathrm{f}}\\right)\n\\end{aligned}\n\\]\n(b) The maximum length to which the sample may be deformed without plastic deformation is determined from Equations as\n\\[\n\\begin{aligned}\nl_{i} & =l_{0}\\left(1+\\frac{\\sigma}{E}\\right) \\\\\n& =\\left(115 \\mathrm{~mm}\\right)\\left[1+\\frac{275 \\mathrm{MPa}}{115 \\times 10^{3} \\mathrm{MPa}}\\right]=115.28 \\mathrm{~mm}\\left(4.51 \\mathrm{in} .\\right)\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 410,
|
||
"question": "Compute the elastic moduli for the following metal alloys, whose stress-strain behaviors may be observed in the \"Tensile Tests\" module of Virtual Materials Science and Engineering (VMSE): (a) titanium, (b) tempered steel, (c) aluminum, and (d) carbon steel. How do these values compare with those presented in Table 6.1 for the same metals?",
|
||
"answer": "The elastic modulus is the slope in the linear elastic region as\n\\[\nE=\\frac{\\Delta \\sigma}{\\Delta \\varepsilon}=\\frac{\\sigma_{2}-\\sigma_{1}}{\\varepsilon_{2}-\\varepsilon_{1}}\n\\]\nSince stress-strain curves for all of the metals/alloys pass through the origin, we make take \\(\\sigma_{1}=0\\) and \\(\\varepsilon_{1}=0\\).\nDeterminations of \\(\\sigma_{2}\\) and \\(\\varepsilon_{2}\\) are possible by moving the cursor to some arbitrary point in the linear region of the curve and then reading corresponding values in the \"Stress\" and \"Strain\" windows that are located below the plot.\n(a) For the titanium alloy, we selected \\(\\sigma_{2}=404.2 \\mathrm{MPa}\\) with its corresponding \\(\\varepsilon_{2}=0.0038\\). Therefore,\n\\[\nE=\\frac{\\sigma_{2}-\\sigma_{1}}{\\varepsilon_{2}-v_{1}}=\\frac{404.2 \\mathrm{MPa}-0 \\mathrm{MPa}}{0.0038-0}=106,400 \\mathrm{MPa}=106.4 \\mathrm{GPa}\n\\]\nThe elastic modulus for titanium given in Table 6.1 is \\(107 \\mathrm{GPa}\\), which is in very good agreement with this value,\n(b) For the tempered steel, we selected \\(\\sigma_{2}=962.2 \\mathrm{MPa}\\) with its corresponding \\(\\varepsilon_{2}=\\mathbf{0 . 0 0 4 7}\\). Therefore,\n\\[\nE=\\frac{\\sigma_{2}-\\sigma_{1}}{\\overline{\\varepsilon_{2}-\\varepsilon_{1}}}=\\frac{962.2 \\mathrm{MPa}-0 \\mathrm{MPa}}{0.047-0}=204,700 \\mathrm{MPa}=204.7 \\mathrm{GPa}\n\\]\n\n\nThe elastic modulus for steel given in Table 6.1 is \\(207 \\mathrm{GPa}\\), which is in reasonably good agreement with this value.\n(c) For the aluminum, we selected \\(\\sigma_{2}=145.1 \\mathrm{MPa}\\) with its corresponding \\(\\varepsilon_{2}=0.0021\\). Therefore,\n\\[\nE=\\frac{\\sigma_{2}-\\sigma_{1}}{\\varepsilon_{2}-\\varepsilon_{1}}=\\frac{145.1 \\mathrm{MPa}-0 \\mathrm{MPa}}{0.0021-0}=69,100 \\mathrm{MPa}=69.1 \\mathrm{GPa}\n\\]\nThe elastic modulus for aluminum given in Table 6.1 is \\(69 \\mathrm{GPa}\\), which is in excellent agreement with this value.\n(d) For the carbon steel, we selected \\(\\sigma_{2}=129 \\mathrm{MPa}\\) with its corresponding \\(\\varepsilon_{2}=0\\).0006. Therefore,\n\\[\nE=\\frac{\\sigma_{2}-\\sigma_{1}}{\\vfrac{\\varepsilon_{2}-\\varepsilon_{1}}{1}}=\\frac{129 \\mathrm{MPa}-0 \\mathrm{MPa}}{0.0006-0}=215,000 \\mathrm{MPa}=215 \\mathrm{GPa}\n\\]\nThe elastic modulus for steel given in Table 6.1 is \\(207 \\mathrm{TPa}\\), which is in reasonable agreement with this value.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 411,
|
||
"question": "Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \\(8.0 \\mathrm{~mm}(0.31\\) in.). A tensile force of \\(1000 \\mathrm{~N}\\left(225 \\mathrm{lb}_{\\mathrm{f}}\\right)\\) produces an elastic reduction in diameter of \\(2.8 \\times 10^{-4} \\mathrm{~mm}\\left(1.10 \\times 10^{-5} \\mathrm{in}\\right.\\).). Compute the modulus of elasticity for this alloy, given that Poisson's ratio is 0.30 .",
|
||
"answer": "This problem asks that we calculate the modulus of elasticity of a metal that is stressed in tension. Combining Equations leads to\n\\[\nE=\\frac{\\sigma}{\\varepsilon_{z}}=\\frac{F}{A_{0} \\varepsilon_{z}}=\\frac{F}{\\varepsilon_{z} \\pi\\left(\\frac{d_{0}}{2}\\right)^{2}}=\\frac{.4 F}{\\varepsilon_{z} \\pi \\frac{d_{0}^{2}}{2}}\n\\]\nFrom the definition of Poisson's ratio, and realizing that for the transverse strain, \\(\\varepsilon_{\\lambda}=\\frac{\\Delta d}{d_{0}}\\)\n\\[\n\\varepsilon_{\\mathrm{Z}}=-\\frac{\\varepsilon_{\\lambda}}{\\mathrm{v}}=-\\frac{\\Delta d}{d_{0} \\mathrm{v}}\n\\]\nTherefore, substitution of this expression for \\(\\varepsilon_{\\mathrm{z}}\\) into the above equation yields\n\\[\n\\begin{aligned}\nE & =\\frac{4 F}{\\varepsilon_{\\mathrm{z}} \\pi \\frac{d_{0}^{2}}{2}}=\\frac{4 F \\mathrm{v}}{\\pi d_{0} \\Delta d} \\\\\n& =\\frac{(4)(1000 \\mathrm{~N})(0.30)}{\\pi\\left(8 \\times 10^{-3} \\mathrm{~m}\\right)\\left(2.8 \\times 10^{-7} \\mathrm{~m}\\right)}=1.705 \\times 10^{11} \\mathrm{~Pa}=170.5 \\mathrm{GPa}\\left(24.7 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 412,
|
||
"question": "A cylindrical rod \\(100 \\mathrm{~mm}\\) long and having a diameter of \\(10.0 \\mathrm{~mm}\\) is to be deformed using a tensile load of \\(27,500 \\mathrm{~N}\\). It must not experience either plastic deformation or a diameter reduction of more than \\(7.5 \\times 10^{-3} \\mathrm{~mm}\\). Of the materials listed as follows, which are possible candidates? Justify your choice(s).\n\\begin{tabular}{lccc}\n\\hline Material & \\begin{tabular}{c} \nModulus of Elasticity \\\\\n\\((\\mathbf{G P a})\\)\n\\end{tabular} & \\begin{tabular}{c} \nYield Strength \\\\\n\\((\\mathbf{M P a})\\)\n\\end{tabular} & Poisson's Ratio \\\\\n\\hline Aluminum alloy & 70 & 200 & 0.33 \\\\\nBrass alloy & 101 & 300 & 0.34 \\\\\nSteel alloy & 207 & 400 & 0.30 \\\\\nTitanium alloy & 107 & 650 & 0.34 \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of \\(27,500 \\mathrm{~N}\\) is applied; this means that the stress corresponding to this load not exceed the yield strength of the material. Upon computing the stress\n\\[\n\\sigma=\\frac{F}{A_{0}}=\\frac{F}{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}}=\\frac{27,500}{\\pi\\left(\\frac{10 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}}=350 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=350 \\mathrm{MPa}\n\\]\nOf the alloys listed, the Ti and steel alloys have yield strengths greater than \\(350 \\mathrm{MPa}\\).\nRelative to the second criterion (i.e., that \\(\\Delta d\\) be less than \\(7.5 \\times 10^{-3} \\mathrm{~mm}\\) ), it is necessary to calculate the change in diameter \\(\\Delta d\\) for these three alloys. From Equation\n\\[\n\\mathrm{v}=-\\frac{\\dot{\\varepsilon}_{\\mathrm{x}}}{\\varepsilon_{\\mathrm{z}}}=-\\frac{\\frac{\\Delta d}{d_{0}}}{\\frac{\\sigma}{E}}=-\\frac{E \\Delta d}{\\sigma d_{0}}\n\\]\nNow, solving for \\(\\Delta d\\) from this expression,\n\\[\n\\Delta d=-\\frac{v \\sigma d_{0}}{E}\n\\]For the steel alloy}\n\n\n\\[\n\\Delta d=-\\frac{(0.30)(350 \\text { MPa })(10 \\text { mm })}{207 \\times 10^{3} \\text { MPa }}=-5.1 \\times 10^{-3} \\mathrm{~mm}\n\\]\nTherefore, the steel is a candidate.\nFor the Ti alloy\n\\[\n\\Delta d=-\\frac{(0.34)(350 \\text { MPa })(10 \\text { mm })}{107 \\times 10^{3} \\text { MPa }}=-11.1 \\times 10^{-3} \\mathrm{~mm}\n\\]\nHence, the titanium alloy is not a candidate.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 413,
|
||
"question": "A cylindrical metal specimen having an original diameter of \\(12.8 \\mathrm{~mm}\\) ( 0.505 in.) and gauge length of \\(50.80 \\mathrm{~mm}\\) (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is \\(6.60 \\mathrm{~mm}\\) (0.260 in.), and the fractured gauge length is \\(72.14 \\mathrm{~mm}\\) (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.",
|
||
"answer": "This problem calls for the computation of ductility in both percent reduction in area and percent elongation. Percent reduction in area is computed using Equation as\n\\[\n\\% \\mathrm{RA}=\\frac{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}-\\pi\\left(\\frac{d_{f}}{2}\\right)^{2}}{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}} \\times 100\n\\]\nin which \\(d_{0}\\) and \\(d_{f}\\) are, respectively, the original and fracture cross-sectional areas. Thus,\n\\[\n\\% \\mathrm{RA}=\\frac{\\pi\\left(\\frac{12.8 \\mathrm{~mm}}{2}\\right)^{2}-\\pi\\left(\\frac{6.60 \\mathrm{~mm}}{2}\\right)^{2}}{\\pi\\left(\\frac{12.8 \\mathrm{~mm}}{2} \\frac{1}{2}\\right)} \\times 100=73.4 \\%\n\\]\nWhile, for percent elongation, we use Equation as\n\\[\n\\begin{aligned}\n\\% \\mathrm{EL} & =\\left(\\frac{l_{f}-l_{0}}{l_{0}}\\right) \\times 100 \\\\\n& =\\frac{72.14 \\mathrm{~mm}-50.80 \\mathrm{~mm}}{50.80 \\mathrm{~mm}} \\times 100=42 \\%\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 414,
|
||
"question": "For some metal alloy, a true stress of \\(415 \\mathrm{MPa}(60,175 \\mathrm{psi})\\) produces a plastic true strain of 0.475 . How much will a specimen of this material elongate when a true stress of \\(325 \\mathrm{MPa}(46,125 \\mathrm{psi})\\) is applied if the original length is \\(300 \\mathrm{~mm}\\) (11.8 in.)? Assume a value of 0.25 for the strain-hardening exponent \\(n\\).",
|
||
"answer": "Solution of this problem requires that we utilize Equation. It is first necessary to solve for \\(K\\) from the given true stress and strain. Rearrangement of this equation yields\n\\[\nK=\\frac{\\sigma_{T}}{\\left(\\varepsilon_{T}\\right)^{n}}=\\frac{415 \\mathrm{MPa}}{(0.475)^{0.25}}=500 \\mathrm{MPa} \\quad(72,500 \\mathrm{psi})\n\\]\nNext we must solve for the true strain produced when a true stress of \\(325 \\mathrm{MPa}\\) is applied, also using Equation. Thus\n\\[\n\\varepsilon_{T}=\\left(\\frac{\\sigma_{T}}{K}\\right)^{1 / n}=\\left(\\frac{325 \\mathrm{MPa}}{500 \\mathrm{MPa}}\\right)^{1 / 0.25}=0.179=\\ln \\left(\\frac{l_{i}}{l_{0}}\\right)\n\\]\nNow, solving for \\(l_{i}\\) gives\n\\[\nl_{i}=l_{0} e^{0.179}=(300 \\mathrm{~mm}) e^{0.179}=358.8 \\mathrm{~mm} \\quad(14.11 \\mathrm{in} .)\n\\]\nAnd finally, the elongation \\(\\Delta l\\) is just\n\\[\n\\Delta l=l_{i}-l_{0}=358.8 \\mathrm{~mm}-300 \\mathrm{~mm}=58.8 \\mathrm{~mm} \\quad(2.31 \\mathrm{in} .)\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 416,
|
||
"question": "To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of \\(10^{4} \\mathrm{~mm}^{-2}\\). Suppose that all the dislocations in \\(1000 \\mathrm{~mm}^{3}\\left(1 \\mathrm{~cm}^{3}\\right)\\) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to \\(10^{10}\\) \\(\\mathrm{mm}^{-2}\\) by cold working. What would be the chain length of dislocations in \\(1000 \\mathrm{~mm}^{3}\\) of material?",
|
||
"answer": "The dislocation density is just the total dislocation length per unit volume of material (in this case per cubic millimeters). Thus, the total length in \\(1000 \\mathrm{~mm}^{3}\\) of material having a density of \\(10^{4} \\mathrm{~mm}^{-2}\\) is just\n\\[\n\\left(10^{4} \\mathrm{~mm}^{-2}\\right)\\left(1000 \\mathrm{~mm}^{3}\\right)=10^{7} \\mathrm{~mm}=10^{4} \\mathrm{~m}=6.2 \\mathrm{mi}\n\\]\nSimilarly, for a dislocation density of \\(10^{10} \\mathrm{~mm}^{-2}\\), the total length is\n\\[\n\\left(10^{10} \\mathrm{~mm}^{-2}\\right)\\left(1000 \\quad \\mathrm{~mm}^{3}\\right)=10^{13} \\mathrm{~mm}=10^{10} \\mathrm{~m}=6.2 \\times 10^{6} \\mathrm{mi}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 417,
|
||
"question": "For each of edge, screw, and mixed dislocations, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.",
|
||
"answer": "For the various dislocation types, the relationships between the direction of the applied shear stress and the direction of dislocation line motion are as follows:\nedge dislocation--parallel\nscrew dislocation--perpendicular\nmixed dislocation--neither parallel nor perpendicular\n\n\nSlip Systems\n}",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 418,
|
||
"question": "(a) Define a slip system.\n(b) Do all metals have the same slip system? Why or why not?",
|
||
"answer": "(a) A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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^{\\circ}\\) with the tensile axis. Three possible slip directions make angles of \\(62.4^{\\circ}, 72.0^{\\circ}\\), and \\(81.1^{\\circ}\\) with the same tensile axis.\n(a) Which of these three slip directions is most favored?\n(b) If plastic deformation begins at a tensile stress of \\(1.95 \\mathrm{MPa}(280 \\mathrm{psi})\\), determine the critical resolved shear stress for aluminum.",
|
||
"answer": "We are asked to compute the critical resolved shear stress for Al. As stipulated in the problem, \\(\\phi=28.1^{\\circ}\\), while possible values for \\(\\lambda\\) are \\(62.4^{\\circ}, 72.0^{\\circ}\\), , and \\(81.1^{\\circ}\\).\n(a) Slip will occur along that direction for which (cos \\(\\phi \\cos \\lambda\\) ) is a maximum, or, in this case, for the largest \\(\\cos \\lambda\\). Cosines for the possible \\(\\lambda\\) values are given below.\n\\[\n\\begin{array}{l}\n\\cos \\left(62.4^{\\circ}\\right)=0.46 \\\\\n\\cos \\left(72.0^{\\circ}\\right)=0.31 \\\\\n\\cos \\left(81.1^{\\circ}\\right)=0.15\n\\end{array}\n\\]\nThus, the slip direction is at an angle of \\(62.4^{\\circ}\\) with the tensile axis.\n(b) From Equation, the critical resolved shear stress is just\n\\[\n\\begin{aligned}\n\\tau_{\\text {cross }} & =\\sigma_{y}(\\cos \\phi \\cos \\lambda)_{\\max } \\\\\n& =\\left(1.95 \\mathrm{MPa}\\right)\\left[\\cos \\left(28.1^{\\circ}\\right) \\cos \\left(62.4^{\\circ}\\right)\\right]=0.80 \\mathrm{MPa} \\quad \\text { (114 psi) }\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 422,
|
||
"question": "List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result.",
|
||
"answer": "Four major differences between deformation by twinning and deformation by slip are as follows: (1) with slip deformation there is no crystallographic reorientation, whereas with twinning there is a reorientation; (2) for slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples; (3) slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems; and (4) normally slip results in relatively large deformations, whereas only small deformations result for twinning.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 425,
|
||
"question": "Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are \\(16 \\mathrm{~mm}\\) and \\(11 \\mathrm{~mm}\\), respectively. The second specimen, with an initial radius of \\(12 \\mathrm{~mm}\\), must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.",
|
||
"answer": "In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to the same percent cold work. For the first specimen\n\\[\n\\begin{array}{l}\n\\% \\mathrm{CW}=\\frac{A_{0}-A_{d}}{A_{0}} \\times 100=\\frac{\\pi r_{0}^{2}-\\pi r_{d}^{2}}{\\pi r_{0}^{2}} \\times 100 \\\\\n=\\frac{\\pi(16 \\mathrm{~mm})^{2}-\\pi(11 \\mathrm{~mm})^{2}}{\\pi(16 \\mathrm{~mm})^{2}} \\times 100=52.7 \\% \\mathrm{CW}\n\\end{array}\n\\]\nFor the second specimen, the deformed radius is computed using the above equation and solving for \\(r_{d}\\) as\n\\[\n\\begin{aligned}\nr_{d} & =r_{0} \\sqrt{1-\\frac{\\% \\mathrm{CW}}{100}} \\\\\n& =(12 \\mathrm{~mm}) \\sqrt{1-\\frac{52.7 \\% \\mathrm{CW}}{100}}=8.25 \\mathrm{~mm}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 433,
|
||
"question": "A specimen of a 4340 steel alloy having a plane strain fracture toughness of \\(45 \\mathrm{MPa} \\sqrt{\\mathrm{m}}(41 \\mathrm{ksi} \\sqrt{\\mathrm{in}}\\).) is exposed to a stress of \\(1000 \\mathrm{MPa}(145,000 \\mathrm{psi})\\). Will this specimen experience fracture if it is known that the largest surface crack is \\(0.75 \\mathrm{~mm}(0.03 \\mathrm{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 \\(1000 \\mathrm{MPa}\\), given the values of \\(K_{I C}, Y\\), and the largest value of \\(a\\) in the material. This requires that we solve for \\(\\sigma_{c}\\) from Equation. Thus\n\\[\n\\sigma_{c}=\\frac{K_{I C}}{Y \\sqrt{\\pi a}}=\\frac{45 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(1.0) \\sqrt{( \\pi ) ( 0.75 \\times 10^{-3} \\mathrm{~m})}}=927 \\mathrm{MPa} \\quad(133,500 \\mathrm{psi})\n\\]\nTherefore, fracture will most likely occur because this specimen will tolerate a stress of \\(927 \\mathrm{MPa}(133,500 \\mathrm{psi})\\) before fracture, which is less than the applied stress of \\(1000 \\mathrm{MPa}(145,00\\) psi \\()\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 434,
|
||
"question": "Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of \\(35 \\mathrm{MPa} \\sqrt{\\mathrm{m}} \\quad(31.9 \\mathrm{ksi} \\sqrt{\\mathrm{m}}\\). It has been determined that fracture results at a stress of \\(250 \\mathrm{MPa}\\) \\((36,250 \\mathrm{psi})\\) when the maximum (or critical) internal crack length is \\(2.0 \\mathrm{~mm}(0.08 \\mathrm{in}\\).). For this same component and alloy, will fracture occur at a stress level of \\(325 \\mathrm{MPa}(47,125 \\mathrm{psi})\\) when the maximum internal crack length is \\(1.0 \\mathrm{~mm}(0.04 \\mathrm{in})\\).? Why or why not?",
|
||
"answer": "We are asked to determine if an aircraft component will fracture for a given fracture toughness (35 \\(\\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) ), stress level ( \\(325 \\mathrm{MPa}\\) ), and maximum internal crack length \\((1.0 \\mathrm{~mm})\\), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. It first becomes necessary to solve for the parameter \\(Y\\), using Equation, for the conditions under which fracture occurred (i.e., \\(\\sigma=250\\) \\(\\mathrm{MPa}\\) and \\(2 a=2.0 \\mathrm{~mm}\\) ). Therefore,\n\\[\nY=\\frac{K_{I c}}{\\sigma \\sqrt{\\pi a}}=\\frac{35 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{\\left(250 \\mathrm{MPa}\\right) \\sqrt{(\\pi) \\frac{2 \\times 10^{-3} \\mathrm{~m}}{2}}}=2.50\n\\]\nNow we will solve for the product \\(Y \\sigma \\sqrt{\\pi a}\\) for the other set of conditions, so as to ascertain whether or not this value is greater than the \\(K_{I c}\\) for the alloy. Thus,\n\\[\n\\begin{aligned}\nY \\sigma \\sqrt{\\pi a} & =(2.50)(325 \\mathrm{MPa}) \\sqrt{(\\pi) \\frac{1 \\times 10^{-3} \\mathrm{~m}}{2}} \\\\\n& =32.2 \\mathrm{MPa} \\sqrt{\\mathrm{m}} \\quad(29.5 \\mathrm{ksi} \\sqrt{\\mathrm{m}})\n\\end{aligned}\n\\]\nTherefore, fracture will not occur since this value \\((32.3 \\mathrm{MPa} \\sqrt{\\mathrm{m}})\\) is less than the \\(K_{I C}\\) of the material, \\(35 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 435,
|
||
"question": "Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of \\(40 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) (36.4 k \\(\\left.\\mathrm{si} \\sqrt{\\mathrm{m}}\\right)\\). It has been determined that fracture results at a stress of \\(365 \\mathrm{MPa}(53,000 \\mathrm{psi})\\) when the maximum internal crack length is \\(2.5 \\mathrm{~mm}(0.10 \\mathrm{in.})\\). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of \\(4.0 \\mathrm{~mm}(0.16 \\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 \\(\\left(40 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right)\\) and maximum internal crack length \\((4.0 \\mathrm{~mm})\\), given that fracture occurs for the same component using the same alloy at one stress level ( \\(365 \\mathrm{MPa}\\) ) and another internal crack length \\((2.5 \\mathrm{~mm})\\). 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{40 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{\\left(365 \\mathrm{MPa}\\right) \\sqrt{(\\tau) \\left(\\frac{2.5 \\times 10^{-3} \\mathrm{~m}}{2}\\right)}}=1.75\n\\]\nNow we will solve for \\(\\sigma_{c}\\) using Equation as\n\\[\n\\sigma_{c}=\\frac{K_{I c}}{Y \\sqrt{\\pi a}}=\\frac{40 \\mathrm{MPa} \\sqrt{m}}{(1.75) \\sqrt{(\\tau) \\frac{4 \\times 10^{-3} \\mathrm{~m}}{2}}}=288 \\mathrm{MPa} \\quad(41,500 \\mathrm{psi})\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 440,
|
||
"question": "Briefly explain the difference between fatigue striations and beachmarks both in terms of (a) size and (b) origin.",
|
||
"answer": "(a) 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.\n(b) With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue striation is corresponds to the advance of a fatigue crack during a single load cycle.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 442,
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals: nickel, copper, iron, tungsten, lead, and aluminum.",
|
||
"answer": "Creep becomes important at about \\(0.4 T_{m}, T_{m}\\) being the absolute melting temperature of the metal. (The melting temperatures in degrees Celsius are found inside the front cover of the book.)\nFor \\(\\mathrm{Ni}, 0.4 T_{m}=(0.4)(1455+273)=691 \\mathrm{~K}\\) or \\(418^{\\circ} \\mathrm{C}\\left(785^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Cu}, 0.4 T_{m}=(0.4)(1085+273)=543 \\mathrm{~K}\\) or \\(270^{\\circ} \\mathrm{C}\\left(518^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Fe}, 0.4 T_{m}=(0.4)(1538+273)=725 \\mathrm{~K}\\) or \\(450^{\\circ} \\mathrm{C}\\left(845^{\\circ} \\mathrm{F}\\right)\\)\nFor W, \\(0.4 T_{m}=(0.4)(3410+273)=1473 \\mathrm{~K}\\) or \\(1200^{\\circ} \\mathrm{C}\\left(2190^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Pb}, 0.4 T_{m}=(0.4)(327+273)=240 \\mathrm{~K}\\) or \\(-33^{\\circ} \\mathrm{C}\\left(-27^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Al}, 0.4 T_{m}=(0.4)(660+273)=373 \\mathrm{~K}\\) or \\(100^{\\circ} \\mathrm{C}\\left(212^{\\circ} \\mathrm{F}\\right)\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 445,
|
||
"question": "For alloys of two hypothetical metals \\(A\\) and \\(B\\), there exist an \\(\\alpha, A\\)-rich phase and \\(a \\beta, B\\)-rich phase. From the mass fractions of both phases for two different alloys provided in the table below, (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for both \\(\\alpha\\) and \\(\\beta\\) phases at this temperature.\n\\begin{tabular}{ccc}\n\\hline Alloy Composition & \\begin{tabular}{c} \nFraction \\(\\alpha\\) \\\\\nPhase\n\\end{tabular} & \\begin{tabular}{c} \nFraction \\(\\beta\\) \\\\\nPhase\n\\end{tabular} \\\\\n\\hline \\(60 \\mathrm{wt} \\% \\mathrm{~A}-40 \\mathrm{wt} \\% \\mathrm{~B}\\) & 0.57 & 0.43 \\\\\n\\(30 \\mathrm{wt} \\% \\mathrm{~A}-70 \\mathrm{wt} \\% \\mathrm{~B}\\) & 0.14 & 0.86 \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "The problem is to solve for compositions at the phase boundaries for both \\(\\alpha\\) and \\(\\beta\\) phases (i.e., \\(C_{\\alpha}\\) and \\(C_{\\beta}\\) ). We may set up two independent lever rule expressions, one for each composition, in terms of \\(C_{\\alpha}\\) and \\(C_{\\beta}\\) as follows:\n\\[\n\\begin{array}{l}\nW_{\\alpha 1}=0.57=\\frac{C_{\\beta}-C_{01}}{C_{\\beta}-C_{\\alpha}}=\\frac{C_{\\beta}-60}{C_{\\beta}-C_{\\alpha}} \\\\\nW_{\\alpha 2}=0.14=\\frac{C_{\\beta}-C_{02}}{C_{\\beta}-C_{\\alpha}}=\\frac{C_{\\beta} -30}{C_{\\beta}-C_{\\alpha}}\n\\end{array}\n\\]\nIn these expressions, compositions are given in wt\\% of A. Solving for \\(C_{\\alpha}\\) and \\(C_{\\beta}\\) from these equations, yield\n\\[\n\\begin{array}{l}\nC_{\\alpha}=90 \\text { (or } 90 \\text { wt } \\% \\text { A-10 wt } \\% \\text { B) } \\\\\nC_{\\beta}=20.2 \\text { (or } 20.2 \\text { wt } \\% \\text { A-79.8 wt } \\% \\text { B) }\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 446,
|
||
"question": "A hypothetical \\(A-B\\) alloy of composition \\(55 \\mathrm{wt} \\% B-45 \\mathrm{wt} \\% A\\) at some temperature is found to consist of mass fractions of 0.5 for both \\(\\alpha\\) and \\(\\beta\\) phases. If the composition of the \\(\\beta\\) phase is \\(90 \\mathrm{wt} \\% \\mathrm{~B}-10 \\mathrm{wt} \\% \\mathrm{~A}\\), what is the composition of the \\(\\alpha\\) phase?",
|
||
"answer": "For this problem, we are asked to determine the composition of the \\(\\beta\\) phase given that\n\\[\n\\begin{array}{l}\nC_{0}=55(\\text { or } 55 \\text { wt } \\% \\text { B- } 45 \\text { wt } \\% \\text { A }) \\\\\nC_{\\beta}=90(\\text { or } 90 \\text { wt } \\% \\text { B-10 wt } \\% \\text { A }) \\\\\nW_{\\alpha}=W_{\\beta}=0.5\n\\end{array}\n\\]\nIf we set up the lever rule for \\(W_{\\alpha}\\)\n\\[\nW_{\\alpha}=0.5=\\frac{C_{\\beta}-C_{0}}{C_{\\beta}-C_{\\alpha}}=\\frac{90-55}{90-C_{\\alpha}}\n\\]\nAnd solving for \\(C_{\\alpha}\\)\n\\[\nC_{\\alpha}=20(\\text { or } 20 \\text { wt } \\% \\text { B- } 80 \\text { wt } \\% \\text { A })\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 447,
|
||
"question": "For \\(11.20 \\mathrm{~kg}\\) of a magnesium-lead alloy of composition \\(30 \\mathrm{wt} \\% \\mathrm{~Pb}-70 \\mathrm{wt} \\% \\mathrm{Mg}\\), is it possible, at equilibrium, to have \\(\\alpha\\) and \\(\\mathrm{Mg}_{2} \\mathrm{~Pb}\\) phases having respective masses of \\(7.39 \\mathrm{~kg}\\) and \\(3.81 \\mathrm{~kg}\\) ? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, explain why.",
|
||
"answer": "Yes, it is possible to have a \\(30 \\mathrm{wt} \\% \\mathrm{~Pb}-70 \\operatorname{wt} \\% \\mathrm{Mg}\\) alloy which has masses of \\(7.39 \\mathrm{~kg}\\) and \\(3.8 \\mathrm{~kg}\\) for the \\(\\alpha\\) and \\(\\mathrm{Mg}_{2} \\mathrm{~P} \\mathrm{~b}\\) phases, respectively. In order to demonstrate this, it is first necessary to determine the mass fraction of each phase as follows:\n\\[\n\\begin{aligned}\nW_{\\alpha}= & \\frac{m_{\\alpha}}{m_{\\alpha}+m_{\\mathrm{Mg}_{2} \\mathrm{~Pb}}}=\\frac{7.39 \\mathrm{~kg}}{7.39 \\mathrm{~kg}+3.81 \\mathrm{~kg}}=0.66 \\\\\n& W_{\\mathrm{Mg}_{2} \\mathrm{~Pb}}=1.00-0.66=0.34\n\\end{aligned}\n\\]\nNow, if we apply the lever rule expression for \\(W_{\\alpha}\\)\n\\[\nW_{\\alpha}=\\frac{C_{\\mathrm{Mg}_{2} \\mathrm{~Pb}}-C_{0}}{C_{\\mathrm{Mg}_{2} \\mathrm{~Pb}}-c_{\\alpha}}\n\\]\nSince the \\(\\mathrm{Mg}_{2} \\mathrm{~Pb}\\) phase exists only at \\(81 \\mathrm{wt} \\% \\mathrm{~Pb}\\), and \\(C_{0}=30 \\mathrm{wt} \\% \\mathrm{~Pb}\\)\n\\[\nW_{\\alpha}=0.66=\\frac{81-30}{81-C_{\\alpha}}\n\\]\nSolving for \\(C_{\\alpha}\\) from this expression yields \\(C_{\\alpha}=3.7 \\mathrm{wt} \\% \\mathrm{~Pb}\\). The position along the \\(\\alpha-\\left(\\alpha+\\mathrm{Mg}_{2} \\mathrm{~Pb}\\right)\\) phase boundary of Figure 9.20 corresponding to this composition is approximately \\(190^{\\circ} \\mathrm{C}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 448,
|
||
"question": "(a) Briefly describe the phenomenon of coring and why it occurs.\n(b) Cite one undesirable consequence of coring.",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 451,
|
||
"question": "Two intermetallic compounds, \\(A B\\) and \\(A B_{2}\\), exist for elements \\(A\\) and \\(B\\). If the compositions for \\(A B\\) and \\(A B_{2}\\) are \\(34.3 \\mathrm{wt} \\% \\mathrm{~A}-65.7 \\mathrm{wt} \\% \\mathrm{~B}\\) and \\(20.7 \\mathrm{wt} \\% \\mathrm{~A}-79.3 \\mathrm{wt} \\% \\mathrm{~B}\\), respectively, and element \\(A\\) is potassium, identify element \\(B\\).",
|
||
"answer": "First of all, consider the \\(A B\\) intermetallic compound; inasmuch as it contains the same numbers of \\(A\\) and \\(\\mathrm{B}\\) atoms, its composition in atomic percent is 50 at\\% A-50 at\\% B. Equation may be written in the form:\n\\[\nC_{\\mathrm{B}}=\\frac{C_{\\mathrm{B}} A_{\\mathrm{A}}}{C_{\\mathrm{A}} A_{\\mathrm{B}}+C_{\\mathrm{B}} A_{\\mathrm{A}}} \\times 100\n\\]\nwhere \\(A_{\\mathrm{A}}\\) and \\(A_{\\mathrm{B}}\\) are the atomic weights for elements \\(\\mathrm{A}\\) and \\(\\mathrm{B}\\), and \\(C_{\\mathrm{A}}\\) and \\(C_{\\mathrm{B}}\\) are their compositions in weight percent. For this \\(A B\\) compound, and making the appropriate substitutions in the above equation leads to\n\\[\n50 \\text { at } \\% \\mathrm{~B}=\\frac{(65.7 \\text { wt } \\% \\mathrm{~B})\\left(A_{\\mathrm{A}}\\right)}{(34.3 \\mathrm{wt} \\% \\mathrm{~A})\\left(A_{\\mathrm{B}}\\right)+(65.7 \\text { wt } \\% \\mathrm{~B})\\left(A \\mathrm{~A}\\right)} \\times 100\n\\]\nNow, solving this expression yields,\n\\[\nA_{\\mathrm{B}}=1.916 A_{\\mathrm{A}}\n\\]\nSince potassium is element \\(\\mathrm{A}\\) and it has an atomic weight of \\(39.10 \\mathrm{~g} / \\mathrm{mol}\\), the atomic weight of element \\(\\mathrm{B}\\) is just\n\\[\nA_{\\mathrm{B}}=(1.916)(39.10 \\mathrm{~g} / \\mathrm{mol})=74.92 \\mathrm{~g} / \\mathrm{mol}\n\\]\nUpon consultation of the period table of the elements we note the element that has an atomic weight closest to this value is arsenic (74.92 g/mol). Therefore, element B is arsenic, and the two intermetallic compounds are KAs and \\(\\mathrm{KAs}_{2}\\).\n\n\n\nCongruent Phase Transformations \\\\ Eutectoid and Peritectic Reactions\n}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 453,
|
||
"question": "Compute the mass fractions of \\(\\alpha\\) ferrite and cementite in pearlite.",
|
||
"answer": "This problem asks that we compute the mass fractions of \\(\\alpha\\) ferrite and cementite in pearlite. The lever-rule expression for ferrite is\n\\[\nW_{\\alpha}=\\frac{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{0}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\alpha}}\n\\]\nand, since \\(C_{\\mathrm{Fe}_{3} \\mathrm{C}}=6.70 \\mathrm{wt} \\% \\mathrm{C}, C_{0}=0.76 \\mathrm{wt} \\% \\mathrm{C}\\), and \\(C_{\\alpha}=0.022 \\mathrm{wt} \\% \\mathrm{C}\\)\n\\[\nW_{\\alpha}=\\frac{6.70-0.76}{6.70-0.022}=0.89\n\\]\nSimilarly, for cementite\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}}=\\frac{C_{0}-C_{\\alpha}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{a}}=\\frac{0.76-0.022}{6.70-0.022}=0.11\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 454,
|
||
"question": "(a) What is the distinction between hypoeutectoid and hypereutectoid steels?\n(b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each?",
|
||
"answer": "(a) 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.\n(b) 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 \\mathrm{wt} \\% \\mathrm{C}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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? Why?",
|
||
"answer": "In this problem we are given values of \\(W_{\\alpha}\\) and \\(W_{\\mathrm{Fe}_{3} \\mathrm{C}}\\) (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 \\(\\alpha\\) leads to\n\\[\nW_{\\alpha}=0.92=\\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}=0.56 \\mathrm{wt} \\% \\mathrm{C}\\). Therefore, the proeutectoid phase is \\(\\alpha\\) ferrite since \\(C_{0}\\) is less than \\(0.76 \\mathrm{wt} \\% \\mathrm{C}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 457,
|
||
"question": "Consider \\(1.0 \\mathrm{~kg}\\) of austenite containing \\(1.15 \\mathrm{wt} \\% \\mathrm{C}\\), cooled to below \\(727^{\\circ} \\mathrm{C}\\left(1341^{\\circ} \\mathrm{F}\\right)\\).\n(a) What is the proeutectoid phase?\n(b) How many kilograms each of total ferrite and cementite form?\n(c) How many kilograms each of pearlite and the proeutectoid phase form?",
|
||
"answer": "(a) The proeutectoid phase will be \\(\\mathrm{Fe}_{3} \\mathrm{C}\\) since \\(1.15 \\mathrm{wt} \\% \\mathrm{C}\\) is greater than the eutectoid composition ( 0.76 \\(\\mathrm{wt} \\% \\mathrm{C}\\) ).\n(b) For this portion of the problem, we are asked to determine how much total ferrite and cementite form. Application of the appropriate lever rule expression yields\n\\[\nW_{\\alpha}=\\frac{C_{\\mathrm{Fe} 3 \\mathrm{C}}-C_{0}}{C_{\\mathrm{Fe} 3 \\mathrm{C}}-C_{\\alpha}}=\\frac{6.70-1.15}{6.70-0.022}=0.83\n\\]\nwhich, when multiplied by the total mass of the alloy \\((1.0 \\mathrm{~kg})\\), gives \\(0.83 \\mathrm{~kg}\\) of total ferrite.\nSimilarly, for total cementite,\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}}=\\frac{C_{0}-C_{\\alpha}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\alpha}}=\\frac{1.15-0.022}{6.70-0.022}=0.17\n\\]\nAnd the mass of total cementite that forms is \\((0.17)(1.0 \\mathrm{~kg})=0.17 \\mathrm{~kg}\\).\n(c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form. Applying Equation, in which \\(C_{1}^{\\prime}=1.15 \\mathrm{wt} \\% \\mathrm{C}\\)\n\\[\nW_{p}=\\frac{6.70-C_{1}^{\\prime}}{6.70-0.76}=\\frac{6.70-1.15}{6.70-0 .76}=0.93\n\\]\nwhich corresponds to a mass of \\(0.93 \\mathrm{~kg}\\). Likewise, from Equation\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}}=\\frac{c_{1}^{\\prime}-0.76}{5.94}=\\frac{1.15-0.76}{5.94}=0.07\n\\]\nwhich is equivalent to \\(0.07 \\mathrm{~kg}\\) of the total \\(1.0 \\mathrm{~kg}\\) mass.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 458,
|
||
"question": "Consider \\(2.5 \\mathrm{~kg}\\) of austenite containing \\(0.65 \\mathrm{wt} \\% \\mathrm{C}\\), cooled to below \\(727^{\\circ} \\mathrm{C}\\left(1341^{\\circ} \\mathrm{F}\\right)\\).\n(a) What is the proeutectoid phase?\n(b) How many kilograms each of total ferrite and cementite form?\n(c) How many kilograms each of pearlite and the proeutectoid phase form?",
|
||
"answer": "(a) Ferrite is the proeutectoid phase since \\(0.65 \\mathrm{wt} \\% \\mathrm{C}\\) is less than \\(0.76 \\mathrm{wt} \\% \\mathrm{C}\\).\n(b) For this portion of the problem, we are asked to determine how much total ferrite and cementite form.\nFor ferrite, application of the appropriate lever rule expression yields\n\\[\nW_{\\alpha}=\\frac{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{0}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\alpha}}=\\frac{6.70-0.65}{6.70-0.022}=0.91\n\\]\nwhich corresponds to \\((0.91)(2.5 \\mathrm{~kg})=2.27 \\mathrm{~kg}\\) of total ferrite.\nSimilarly, for total cementite,\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}}=\\frac{C_{0}-C_{\\alpha}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{a}}=\\frac{0.65-0.022}{6.70-0.022}=0.09\n\\]\nOr \\((0.09)(2.5 \\mathrm{~kg})=0.23 \\mathrm{~kg}\\) of total cementite form.\n(c) Now consider the amounts of pearlite and proeutectoid ferrite. Using Equation\n\\[\nW_{p}=\\frac{C_{0}^{\\prime}-0.022}{0.74}=\\frac{0.65-0.022}{0.74}=0.85\n\\]\nThis corresponds to \\((0.85)(2.5 \\mathrm{~kg})=2.12 \\mathrm{~kg}\\) of pearlite.\nAlso, from Equation,\n\\[\nW_{\\alpha^{\\prime}}=\\frac{0.76-0.65}{0.74}=0.15\n\\]\nOr, there are \\((0.15)(2.5 \\mathrm{~kg})=0.38 \\mathrm{~kg}\\) of proeutectoid ferrite.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 460,
|
||
"question": "The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these two microconstituents are 0.286 and 0.714 , respectively. Determine the concentration of carbon in this alloy.\n\\[\n\\text {",
|
||
"answer": "This problem asks that we determine the carbon concentration in an iron-carbon alloy, given the mass fractions of proeutectoid ferrite and pearlite. From Equation\n\\[\nW_{p}=0.714=\\frac{C_{0}^{\\star}-0.022}{0.74}\n\\]\nwhich yields \\(C_{0}^{\\prime}=0.55 \\mathrm{wt} \\% \\mathrm{C}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 463,
|
||
"question": "Consider \\(2.0 \\mathrm{~kg}\\) of a \\(99.6 \\mathrm{wt} \\% \\mathrm{Fe}-0.4 \\mathrm{wt} \\% \\mathrm{C}\\) alloy that is cooled to a temperature just below the eutectoid.\n(a) How many kilograms of proeutectoid ferrite form?\n(b) How many kilograms of eutectoid ferrite form?\n(c) How many kilograms of cementite form?",
|
||
"answer": "In this problem we are asked to consider \\(2.0 \\mathrm{~kg}\\) of a \\(99.6 \\operatorname{wt} \\% \\mathrm{Fe}-0.4 \\mathrm{wt} \\% \\quad \\mathrm{C}\\) alloy that is cooled to a temperature below the eutectoid.\n(a) Equation must be used in computing the amount of proeutectoid ferrite that forms. Thus,\n\\[\nW_{\\alpha^{\\prime}}=\\frac{0.76-C_{0}^{\\prime}}{0.74}=\\frac{0.76-0.40}{0.74}=0.49\n\\]\nOr, (0.49)(2.0 kg) \\(=0.98 \\mathrm{~kg}\\) of proeutectoid ferrite forms.\n(b) In order to determine the amount of eutectoid ferrite, it first becomes necessary to compute the amount of total ferrite using the lever rule applied entirely across the \\(\\alpha+\\mathrm{Fe}_{3} \\mathrm{C}\\) phase field, as\n\\[\nW_{\\alpha}=\\frac{C_{\\mathrm{Fe} 3 \\mathrm{C}}-C_{0}^{\\prime}}{C_{\\mathrm{Fe} 3 \\mathrm{C}}-C_{\\alpha}}=\\frac{6.70-0.40}{6.70-0.022}=0.94\n\\]\nwhich corresponds to (0.94)(2.0 kg) \\(=1.88 \\mathrm{~kg}\\). Now, the amount of eutectoid ferrite is just the difference between total and proeutectoid ferrites, or\n\\[\n1.88 \\mathrm{~kg}-0.98 \\mathrm{~kg}=0.90 \\mathrm{~kg}\n\\]\n(c) With regard to the amount of cementite that forms, again application of the lever rule across the entirety of the \\(\\alpha+\\mathrm{Fe}_{3} \\mathrm{C}\\) phase field. leads to\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}}=\\frac{C_{0}^{\\prime}-C_{\\alpha}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\alpha}}=\\frac{0.40-0.022}{6.70-0.022}=0.057\n\\]\nwhich amounts to \\((0.057)(2.0 \\mathrm{~kg})=0.114 \\mathrm{~kg}\\) cementite in the alloy.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 465,
|
||
"question": "Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 and 0.049 , respectively? Why or why not?",
|
||
"answer": "This problem asks if it is possible to have an iron-carbon alloy for which \\(W_{\\alpha}=0.846\\) and \\(W_{\\mathrm{Fe}_{3} \\mathrm{C}^{\\prime}}=0.049\\).\nIn order to make this determination, it is necessary to set up lever rule expressions for these two mass fractions in terms of the alloy composition, then to solve for the alloy composition of each; if both alloy composition values are equal, then such an alloy is possible. The expression for the mass fraction of total ferrite is\n\\[\nW_{\\alpha}=\\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}=0.846\n\\]\nSolving for this \\(C_{0}\\) yields \\(C_{0}=1.05 \\mathrm{wt} \\% \\mathrm{C}\\). Now for \\(W_{\\mathrm{Fe}_{3} \\mathrm{C}^{\\prime}}\\) we utilize Equation as\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}^{\\prime}}=\\frac{c_{1}^{\\prime}-0.76}{5.94}=0.049\n\\]\nThis expression leads to \\(C_{1}^{\\prime}=1.05 \\mathrm{wt} \\% \\mathrm{C}\\). And, since \\(C_{0}=C_{1}^{\\prime}\\), this alloy is possible.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 466,
|
||
"question": "Is it possible to have an iron-carbon alloy for which the mass fractions of total cementite and pearlite are 0.039 and 0.417 , respectively? Why or why not?",
|
||
"answer": "This problem asks if it is possible to have an iron-carbon alloy for which \\(W_{\\mathrm{Fe}_{3} \\mathrm{C}}=0.039\\) and \\(W_{p}=0.417\\).\nIn order to make this determination, it is necessary to set up lever rule expressions for these two mass fractions in terms of the alloy composition, then to solve for the alloy composition of each; if both alloy composition values are equal, then such an alloy is possible. The expression for the mass fraction of total cementite is\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}}=\\frac{C_{0}-C_{a}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{a}}=\\frac{C_{0}-0.022}{6.70-0.022}=0.039\n\\]\nSolving for this \\(C_{0}\\) yields \\(C_{0}=0.28 \\mathrm{wt} \\% \\mathrm{C}\\). Therefore, this alloy is hypoeutectoid since \\(C_{0}\\) is less than the eutectoid composition ( \\(0.76 \\mathrm{wt} \\%)\\). Thus, it is necessary to use Equation for \\(W_{p}\\) as\n\\[\nW_{p}=\\frac{C_{0}^{\\prime}-0.022}{0.74}=0.417\n\\]\nThis expression leads to \\(C_{0}^{\\prime}=0.33 \\mathrm{wt} \\% \\mathrm{C}\\). Since \\(C_{0}\\) and \\(C_{0}^{\\prime}\\) are different, this alloy is not possible.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 467,
|
||
"question": "Compute the mass fraction of eutectoid ferrite in an iron-carbon alloy that contains \\(0.43 \\mathrm{wt} \\% \\mathrm{C}\\).",
|
||
"answer": "In order to solve this problem it is necessary to compute mass fractions of total and proeutectoid ferrites, and then to subtract the latter from the former. To calculate the mass fraction of total ferrite, it is necessary to use the lever rule and a tie line that extends across the entire \\(\\boldsymbol{\\alpha}+\\mathrm{Fe}_{3} \\mathrm{C}\\) phase field as\n\\[\nW_{\\alpha}=\\frac{C_{\\mathrm{Fe} 3 \\mathrm{C}}-C_{0}}{C_{\\mathrm{Fe} 3 \\mathrm{C}}-C_{\\alpha}}=\\frac{6.70-0.43}{6.70-0.022}=0.939\n\\]\nNow, for the mass fraction of proeutectoid ferrite we use Equation as\n\\[\nW_{\\alpha^{\\prime}}=\\frac{0.76-C_{0}^{\\prime}}{0.74}-\\frac{0.76-0.43}{0.74}=0.446\n\\]\nAnd, finally, the mass fraction of eutectoid ferrite \\(W_{\\alpha^{\\prime \\prime}}\\) is just\n\\[\nW_{\\alpha^{\\prime \\prime}}=W_{\\alpha}-W_{\\alpha^{\\prime}}=0.939-0.446=0.493\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 468,
|
||
"question": "The mass fraction of eutectoid cementite in an iron-carbon alloy is 0.104 . On the basis of this information, is it possible to determine the composition of the alloy? If so, what is its composition? If this is not possible, explain why.",
|
||
"answer": "This problem asks whether or not it is possible to determine the composition of an iron-carbon alloy for which the mass fraction of eutectoid cementite is 0.104 ; and if so, to calculate the composition. Yes, it is possible to determine the alloy composition; and, in fact, there are two possible answers. For the first, the eutectoid cementite exists in addition to proeutectoid cementite. For this case the mass fraction of eutectoid cementite \\(\\left(W_{\\mathrm{Fe}_{3} \\mathrm{C}^{*}}\\right)\\) is just the difference between total cementite and proeutectoid cementite mass fractions; that is\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}^{*}}=W_{\\mathrm{Fe}_{3} \\mathrm{C}^{*}}-W_{\\mathrm{Fe}_{3} \\mathrm{C}^{*}}\n\\]\nNow, it is possible to write expressions for \\(W_{\\mathrm{Fe}_{3} \\mathrm{C}}\\) and \\(W_{\\mathrm{Fe}_{3} \\mathrm{C}}\\) in terms of \\(C_{0}\\), the alloy composition. Thus,\n\\[\n\\begin{aligned}\nW_{\\mathrm{Fe}_{3} \\mathrm{C}^{*}} & =\\frac{C_{0}-C_{\\alpha}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}^{*}}-C_{\\alpha}}-\\frac{C_{0}-0.76}{5.94} \\\\\n& =\\frac{C_{0}-0.022}{6.70-0.022}-\\frac{C_{0}-0.76}{5.94}=0.104\n\\end{aligned}\n\\]\nAnd, solving for \\(C_{0}\\) yields \\(C_{0}=1.11 \\mathrm{wt} \\% \\mathrm{C}\\).\nFor the second possibility, we have a hypoeutectoid alloy wherein all of the cementite is eutectoid cementite. Thus, it is necessary to set up a lever rule expression wherein the mass fraction of total cementite is 0.104 . Therefore,\n\\[\nW_{\\mathrm{Fe}_{3} \\mathrm{C}}=\\frac{C_{0}-C_{\\alpha}}{C_{\\mathrm{Fe}_{\\mathrm{3}} \\mathrm{C}}-C_{\\alpha}}=\\frac{C_{0}-0.022}{6.70-0.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 469,
|
||
"question": "The mass fraction of eutectoid ferrite in an iron-carbon alloy is 0.82 . On the basis of this information, is it possible to determine the composition of the alloy? If so, what is its composition? If this is not possible, explain why.",
|
||
"answer": "This problem asks whether or not it is possible to determine the composition of an iron-carbon alloy for which the mass fraction of eutectoid ferrite is 0.82 ; and if so, to calculate the composition. Yes, it is possible to determine the alloy composition; and, in fact, there are two possible answers. For the first, the eutectoid ferrite exists in addition to preoutectoid ferrite. For this case the mass fraction of eutectoid ferrite \\(\\left(W_{\\alpha^{\\prime}}\\right)\\) is just the difference between total ferrite and proeutectoid ferrite mass fractions; that is\n\\[\nW_{\\alpha^{\\prime \\prime}}=W_{\\alpha}-W_{\\alpha^{\\prime}}\n\\]\nNow, it is possible to write expressions for \\(W_{\\alpha}\\) and \\(W_{\\alpha^{\\prime}}\\) in terms of \\(C_{0}\\), the alloy composition. Thus,\n\\[\n\\begin{aligned}\nW_{\\alpha^{\\prime \\prime \\prime}} & =\\frac{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{0}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\alpha}}-\\frac{0.76-C_{0}}{0.74} \\\\\n& =\\frac{6.70-C_{0}}{6.70-0.022}-\\frac{0.76-C_{0}}{0.74}=0.82\n\\end{aligned}\n\\]\nAnd, solving for \\(C_{0}\\) yields \\(C_{0}=0.70 \\mathrm{wt} \\% \\mathrm{C}\\).\nFor the second possibility, we have a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite. Thus, it is necessary to set up a lever rule expression wherein the mass fraction of total ferrite is 0.82 . Therefore,\n\\[\nW_{\\alpha}=\\frac{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\mathrm{0}}}{C_{\\mathrm{Fe}_{3} \\mathrm{C}}-C_{\\alpha}^{\\prime}}=\\frac{6.70-C_{0}}{6.70-0.012}=0.82\n\\]\nAnd, solving for \\(C_{0}\\) yields \\(C_{0}=1.22 \\mathrm{wt} \\% \\mathrm{C}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 470,
|
||
"question": "Name the two stages involved in the formation of particles of a new phase. Briefly describe each.",
|
||
"answer": "The two stages involved in the formation of particles of a new phase are nucleation and growth. The nucleation process involves the formation of normally very small particles of the new phase(s) which are stable and capable of continued growth. The growth stage is simply the increase in size of the new phase particles.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 473,
|
||
"question": "The fraction recrystallized-time data for the recrystallization at \\(600^{\\circ} \\mathrm{C}\\) of a previously deformed steel are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of \\(22.8 \\mathrm{~min}\\).\n\\begin{tabular}{cc}\n\\hline \\begin{tabular}{c} \nFraction \\\\\nRecrystallized\n\\end{tabular} & Time (min) \\\\\n\\hline 0.20 & 13.1 \\\\\n0.70 & 29.1 \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "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\\]\nwhich is the form of the equation that we will now use. The two equations are thus\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 1976 United States Copyright Act without the permission of the copyright owner is unloveful.\n\n\\[\n\\left.\\begin{array}{l}\n\\ln \\left\\{\\ln \\left[\\frac{1}{1-0.20}\\right]\\right\\}=\\ln k+n \\ln(13.1 \\min)\\\\\n\\ln \\left\\{\\ln \\left[\\frac{1}{1-0.70}\\right]\\right\\}=\\ln k+n \\ln(29.1 \\min)\n\\end{array}\\right.\n\\]\nSolving these two expressions simultaneously for \\(n\\) and \\(k\\) yields \\(n=2.112\\) and \\(k=9.75 \\times 10^{-4}\\).\nNow it becomes necessary to solve for \\(y\\) when \\(t=22.8 \\mathrm{~min}\\). Application of Equation leads to\n\\[\n\\begin{aligned}\ny & =1-\\exp \\left(-k t^{n}\\right) \\\\\n& =1-\\exp \\left[-\\left(9.75 \\times 10^{-4}\\right)\\left(22.8 \\mathrm{~min}\\right)^{2.112}\\right]=0.51\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 474,
|
||
"question": "In terms of heat treatment and the development of microstructure, what are two major limitations of the iron-iron carbide phase diagram?",
|
||
"answer": "Two limitations of the iron-iron carbide phase diagram are:\n(1) The nonequilibrium martensite does not appear on the diagram; and\n(2) The diagram provides no indication as to the time-temperature relationships for the formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium ferrite and cementite phases.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 475,
|
||
"question": "(a) Briefly describe the phenomena of superheating and supercooling.\n(b) Why do these phenomena occur?",
|
||
"answer": "(a) Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 476,
|
||
"question": "Briefly cite the differences between pearlite, bainite, and spheroidite relative to microstructure and mechanical properties.",
|
||
"answer": "The microstructures of pearlite, bainite, and spheroidite all consist of \\(\\alpha\\)-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an \\(\\alpha\\)-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.\nBainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 478,
|
||
"question": "Name the microstructural products of eutectoid iron-carbon alloy ( \\(0.76 \\mathrm{wt} \\% \\mathrm{C})\\) specimens that are first completely transformed to austenite, then cooled to room temperature at the following rates:\n(a) \\(200^{\\circ} \\mathrm{C} / \\mathrm{s}\\),\n(b) \\(100^{\\circ} \\mathrm{C} / \\mathrm{s}\\), and\n(c) \\(20^{\\circ} \\mathrm{C} / \\mathrm{s}\\).",
|
||
"answer": "We are called upon to name the microstructural products that form for specimens of an iron-carbon alloy of eutectoid composition that are continuously cooled to room temperature at a variety of rates. \n(a) At a rate of \\(200^{\\circ} \\mathrm{C} / \\mathrm{s}\\), only martensite forms.\n(b) At a rate of \\(100^{\\circ} \\mathrm{C} / \\mathrm{s}\\), both martensite and pearlite form.\n(c) At a rate of \\(20^{\\circ} \\mathrm{C} / \\mathrm{s}\\), only fine pearlite forms.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 479,
|
||
"question": "Cite two important differences between continuous cooling transformation diagrams for plain carbon and alloy steels.",
|
||
"answer": "Two important differences between continuous cooling transformation diagrams for plain carbon and alloy steels are: (1) for an alloy steel, a bainite nose will be present, which nose will be absent for plain carbon alloys; and (2) the pearlite-proeutectoid noses for plain carbon steel alloys are positioned at shorter times than for the alloy steels.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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 the following rates:\n(a) \\(10^{\\circ} \\mathrm{C} / \\mathrm{s}\\),\n(b) \\(1^{\\circ} \\mathrm{C} / \\mathrm{s}\\),\n(c) \\(0.1^{\\circ} \\mathrm{C} / \\mathrm{s}\\), and\n(d) \\(0.01^{\\circ} \\mathrm{C} / \\mathrm{s}\\).",
|
||
"answer": "This problem asks for the microstructural products that form when specimens of a 4340 steel are continuously cooled to room temperature at several rates. \n(a) At a cooling rate of \\(10^{\\circ} \\mathrm{C} / \\mathrm{s}\\), only martensite forms.\n(b) At a cooling rate of \\(1^{\\circ} \\mathrm{C} / \\mathrm{s}\\), both martensite and bainite form.\n(c) At a cooling rate of \\(0.1^{\\circ} \\mathrm{C} / \\mathrm{s}\\), martensite, proeutectoid ferrite, and bainite form.\n(d) At a cooling rate of \\(0.01^{\\circ} \\mathrm{C} / \\mathrm{s}\\), martensite,\\(\\mathrm{proeutectoid}\\) ferrite, pearlite, and bainite form.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 483,
|
||
"question": "Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite.",
|
||
"answer": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\(\\alpha\\)-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 485,
|
||
"question": "Rank the following iron-carbon alloys and associated microstructures from the highest to the lowest tensile strength:\n(a) \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\) with spheroidite,\n(b) \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\) with coarse pearlite,\n(c) \\(0.60 \\mathrm{wt} \\% \\mathrm{C}\\) with fine pearlite, and\n(d) \\(0.60 \\mathrm{wt} \\% \\mathrm{C}\\) with coarse pearlite.\nJustify this ranking.",
|
||
"answer": "This problem asks us to rank four iron-carbon alloys of specified composition and microstructure according to hardness. This ranking is as follows:\n\\[\n\\begin{array}{l}\n0.60 \\mathrm{wt} \\% \\mathrm{C}, \\text { fine pearlite } \\\\\n0.60 \\mathrm{wt} \\% \\mathrm{C}, \\text { coarse pearlite } \\\\\n0.25 \\mathrm{wt} \\% \\mathrm{C}, \\text { coarse pearlite } \\\\\n\\quad 0.25 \\mathrm{wt} \\% \\mathrm{C}, \\text { spheroidite }\n\\end{array}\n\\]\nThe \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\), coarse pearlite is stronger than the \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\), spheroidite since coarse pearlite is stronger than spheroidite; the composition of the alloys is the same. The \\(0.60 \\mathrm{wt} \\% \\mathrm{C}\\), coarse pearlite is stronger than the 0.25 wt \\(\\% \\mathrm{C}\\), coarse pearlite, since increasing the carbon content increases the strength. Finally, the 0.60 wt \\(\\% \\mathrm{C}\\), fine pearlite is stronger than the \\(0.60 \\mathrm{wt} \\% \\mathrm{C}\\), coarse pearite inasmuch as the strength of fine pearlite is greater than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine pearlite.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 486,
|
||
"question": "Briefly explain why the hardness of tempered martensite diminishes with tempering time (at constant temperature) and with increasing temperature (at constant tempering time).",
|
||
"answer": "This question asks for an explanation as to why the hardness of tempered martensite diminishes with tempering time (at constant temperature) and with increasing temperature (at constant tempering time). The hardness of tempered martensite depends on the ferrite-cementite phase boundary area; since these phase boundaries are barriers to dislocation motion, the greater the area the harder the alloy. The microstructure of tempered martensite consists of small sphere-like particles of cementite embedded within a ferrite matrix. As the size of the cementite particles increases, the phase boundary area diminishes, and the alloy becomes softer. Therefore, with increasing tempering time, the cementite particles grow, the phase boundary area decreases, and the hardness diminishes. As the tempering temperature is increased, the rate of cementite particle growth also increases, and the alloy softens, again, because of the decrease in phase boundary area.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 487,
|
||
"question": "(a) Briefly describe the microstructural difference between spheroidite and tempered martensite.\n(b) Explain why tempered martensite is much harder and stronger.",
|
||
"answer": "(a) Both tempered martensite and spheroidite have sphere-like cementite particles within a ferrite matrix; however, these particles are much larger for spheroidite.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 488,
|
||
"question": "(a) List the four classifications of steels. (b) For each, briefly describe the properties and typical applications.",
|
||
"answer": "This question asks that we list four classifications of steels, and, for each, to describe properties and cite typical applications.Low Carbon Steels}\nProperties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable.\nTypical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans.Medium Carbon Steels}\nProperties: heat treatable, relatively large combinations of mechanical characteristics.\nTypical applications: railway wheels and tracks, gears, crankshafts, and machine parts.High Carbon Steels}\nProperties: hard, strong, and relatively brittle.\nTypical applications: chisels, hammers, knives, and hacksaw blades.High Alloy Steels (Stainless and Tool)}\nProperties: hard and wear resistant; resistant to corrosion in a large variety of environments.\nTypical applications: cutting tools, drills, cutlery, food processing, and surgical tools.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 489,
|
||
"question": "(a) Cite three reasons why ferrous alloys are used so extensively. (b) Cite three characteristics of ferrous alloys that limit their utilization.",
|
||
"answer": "(a) Ferrous alloys are used extensively because:\n(1) Iron ores exist in abundant quantities.\n(2) Economical extraction, refining, and fabrication techniques are available.\n(3) The alloys may be tailored to have a wide range of properties.\n(b) Disadvantages of ferrous alloys are:\n(1) They are susceptible to corrosion.\n(2) They have a relatively high density.\n(3) They have relatively low electrical conductivities.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 492,
|
||
"question": "Compare white and nodular cast irons with respect to (a) composition and heat treatment, (b) microstructure, and (c) mechanical characteristics.",
|
||
"answer": "This question asks us to compare white and nodular cast irons.\n(a) With regard to composition and heat treatment:\nWhite iron--2.5 to \\(4.0 \\mathrm{wt} \\% \\mathrm{C}\\) and less than \\(1.0 \\mathrm{wt} \\%\\) Si. No heat treatment; however, cooling is rapid during solidification.\nNodular cast iron--2.5 to \\(4.0 \\mathrm{wt} \\% \\mathrm{~C}, 1.0\\) to \\(3.0 \\mathrm{wt} \\% \\mathrm{Si}\\), and a small amount of Mg or Ce. A heat treatment at about \\(700^{\\circ} \\mathrm{C}\\) may be necessary to produce a ferritic matrix.\n(b) With regard to microstructure:\nWhite iron--There are regions of cementite interspersed within pearlite.\nNodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix.\n(c) With respect to mechanical characteristics:\nWhite iron--Extremely hard and brittle.\nNodular cast iron--Moderate strength and ductility.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 497,
|
||
"question": "Cite advantages and disadvantages of hot working and 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. 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. The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low. The disadvantages of hot working are: (1) A poor surface finish. (2) A variety of mechanical properties is not possible.\n}",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 498,
|
||
"question": "(a) Cite advantages of forming metals by extrusion as opposed to rolling. (b) Cite some disadvantages.",
|
||
"answer": "(a) The advantages of extrusion as opposed to rolling are as follows:\n(1) Pieces having more complicated cross-sectional geometries may be formed.\n(2) Seamless tubing may be produced.\n(b) The disadvantages of extrusion over rolling are as follows:\n(1) Nonuniform deformation over the cross-section.\n(2) A variation in properties may result over a cross-section of an extruded piece.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 500,
|
||
"question": "Compare sand, die, investment, lost foam, and continuous casting techniques.",
|
||
"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.\nFor 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.\nFor 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.\nFor 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.\nFor 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 502,
|
||
"question": "Cite three sources of internal residual stresses in metal components. What are two possible adverse consequences of these stresses?",
|
||
"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.\nTwo adverse consequences of these stresses are distortion (or warpage) and fracture.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 503,
|
||
"question": "Give the approximate minimum temperature at which it is possible to austenitize each of the following iron-carbon alloys during a normalizing heat treatment: (a) \\(0.20 \\mathrm{wt} \\% \\mathrm{C}\\), (b) \\(0.76 \\mathrm{wt} \\% \\mathrm{C}\\), and (c) 0.95 \\(\\mathrm{wt} \\% \\mathrm{C}\\).",
|
||
"answer": "(a) For \\(0.20 \\mathrm{wt} \\% \\mathrm{C}\\), heat to at least \\(905^{\\circ} \\mathrm{C}\\left(1660^{\\circ} \\mathrm{F}\\right)\\) since the \\(A_{3}\\) temperature is \\(850^{\\circ} \\mathrm{C}\\left(1560^{\\circ} \\mathrm{F}\\right)\\).\n(b) For \\(0.76 \\mathrm{wt} \\% \\mathrm{C}\\), heat to at least \\(782^{\\circ} \\mathrm{C}\\left(1440^{\\circ} \\mathrm{F}\\right)\\) since the \\(A_{3}^{\\prime}\\) temperature is \\(727^{\\circ} \\mathrm{C}\\left(1340^{\\circ} \\mathrm{F}\\right)\\).\n(c) For \\(0.95 \\mathrm{wt} \\% \\mathrm{C}\\), heat to at least \\(840^{\\circ} \\mathrm{C}\\left(1545^{\\circ} \\mathrm{F}\\right)\\) since the \\(A_{\\mathrm{cm}}\\) temperature is \\(785^{\\circ} \\mathrm{C}\\left(1445^{\\circ} \\mathrm{F}\\right)\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 504,
|
||
"question": "Give the approximate temperature at which it is desirable to heat each of the following iron-carbon alloys during a full anneal heat treatment: (a) \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\), (b) \\(0.45 \\mathrm{wt} \\% \\mathrm{C}\\), (c) \\(0.85 \\mathrm{wt} \\% \\mathrm{C}\\), and (d) \\(1.10 \\mathrm{wt} \\% \\mathrm{C}\\).",
|
||
"answer": "(a) For \\(0.25 \\mathrm{wt} \\% \\mathrm{C}\\), heat to about \\(880^{\\circ} \\mathrm{C}\\left(1510^{\\circ} \\mathrm{F}\\right)\\) since the \\(A_{3}\\) temperature is \\(830^{\\circ} \\mathrm{C}\\left(1420^{\\circ} \\mathrm{F}\\right)\\).\n(b) For \\(0.45 \\mathrm{wt} \\% \\mathrm{C}\\), heat to about \\(830^{\\circ} \\mathrm{C}\\left(1525^{\\circ} \\mathrm{F}\\right)\\) since the \\(A_{3}\\), temperature is \\(780^{\\circ} \\mathrm{C}\\left(1435^{\\circ} \\mathrm{F}\\right)\\).\n(c) For \\(0.85 \\mathrm{wt} \\% \\mathrm{C}\\), heat to about \\(777^{\\circ} \\mathrm{C}\\left(1430^{\\circ} \\mathrm{F}\\right)\\) since the \\(A_{1}\\) temperature is \\(727^{\\circ} \\mathrm{C}\\left(1340^{\\circ} \\mathrm{F}\\right)\\).\n(d) For \\(1.10 \\mathrm{wt} \\% \\mathrm{C}\\), heat to about \\(777{ }^{\\circ} \\mathrm{C}\\left(1430^{\\circ} F\\right)\\) since the \\(A_{1}\\) temperature is \\(727^{\\mathrm{C}} \\mathrm{C}\\left(1340^{\\circ} \\mathrm{F}^{\\prime}\\right)\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 505,
|
||
"question": "What is the purpose of a spheroidizing heat treatment? On what classes of alloys is it normally used?",
|
||
"answer": "The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure. It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong.\n\n\nHeat Treatment of Steels\n}",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 510,
|
||
"question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions.\n(a) Will the stacking sequence for this structure be FCC or HCP? Why?\n(b) Will cations fill tetrahedral or octahedral positions? Why?\n(c) What fraction of the positions will be occupied?",
|
||
"answer": "This question is concerned with the zinc blende crystal structure in terms of close-packed planes of anions.\n(a) 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 .\n(b) The cations will fill tetrahedral positions since the coordination number for cations is four.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 516,
|
||
"question": "(a) Suppose that \\(\\mathrm{Li}_{2} \\mathrm{O}\\) is added as an impurity to \\(\\mathrm{CaO}\\). If the \\(\\mathrm{Li}^{+}\\)substitutes for \\(\\mathrm{Ca}^{2+}\\), what kind of vacancies would you expect to form? How many of these vacancies are created for every \\(\\mathrm{Li}^{+}\\)added?\n(b) Suppose that \\(\\mathrm{CaCl}_{2}\\) is added as an impurity to \\(\\mathrm{CaO}\\). If the Cl substitutes for \\(\\mathrm{O}^{2-}\\), what kind of vacancies would you expect to form? How many of the vacancies are created for every \\(\\mathrm{Cl}^{-}\\)added?",
|
||
"answer": "(a) For \\(\\mathrm{Li}^{+}\\)substituting for \\(\\mathrm{Ca}^{2+}\\) in \\(\\mathrm{CaO}\\), oxygen vacancies would be created. For each \\(\\mathrm{Li}^{+}\\)substituting for \\(\\mathrm{Ca}{ }^{2+}\\), 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 \\(\\mathrm{Li}^{+}\\)ions added, a single oxygen vacancy is formed.\n(b) For \\(\\mathrm{Cl}^{-}\\)substituting for \\(\\mathrm{O}^{2-}\\) in \\(\\mathrm{CaO}\\), calcium vacancies would be created. For each \\(\\mathrm{Cl}^{-}\\)substituting for an \\(\\mathrm{O}^{2-}\\), one negative charge is removed; in order to maintain charge neutrality, two \\(\\mathrm{Cl}^{-}\\)ions will lead to the formation of one calcium vacancy.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 517,
|
||
"question": "What point defects are possible for \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\) as an impurity in \\(\\mathrm{MgO}\\) ? How many \\(\\mathrm{Al}^{2+}\\) ions must be added to form each of these defects?",
|
||
"answer": "For every \\(\\mathrm{Al}^{3+}\\) ion that substitutes for \\(\\mathrm{Mg}^{2+}\\) in \\(\\mathrm{MgO}\\), a single positive charge is added. Thus, in order to maintain charge neutrality, either a positive charge must be removed or a negative charge must be added.\nNegative charges are added by forming \\(\\mathrm{O}^{2-}\\) interstitials, which are not likely to form.\nPositive charges may be removed by forming \\(\\mathrm{Mg}^{2+}\\) vacancies, and one magnesium vacancy would be formed for every two \\(\\mathrm{Al}^{3+}\\) ions added.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 518,
|
||
"question": "When kaolinite clay \\(\\left[\\mathrm{Al}_{2}\\left(\\mathrm{Si}_{2} \\mathrm{O}_{5}\\right)(\\mathrm{OH})_{4}\\right]\\) is heated to a sufficiently high temperature, chemical water is driven off.\n(a) Under these circumstances, what is the composition of the remaining product (in weight percent \\(\\left.\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right) ?\\)\n(b) What are the liquidus and solidus temperatures of this material",
|
||
"answer": "(a) The chemical formula for kaolinite clay may also be written as \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}{ }^{2-2} \\mathrm{SiO}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}\\). Thus, if we remove the chemical water, the formula becomes \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}{ }^{2-} 2 \\mathrm{SiO}_{2}\\). The formula weight for \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\) is just (2)(26.98 \\(\\mathrm{g} / \\mathrm{mol})+(3)(16.00 \\mathrm{~g} / \\mathrm{mol})=101.96 \\mathrm{~g} / \\mathrm{mol}\\); and for \\(\\mathrm{SiO}_{2}\\) the formula weight is \\(28.09 \\mathrm{~g} / \\mathrm{mol}+(2)(16.00 \\mathrm{~g} / \\mathrm{mol})=\\) \\(60.09 \\mathrm{~g} / \\mathrm{mol}\\). Thus, the composition of this product, in terms of the concentration of \\(\\mathrm{Al}_{2} \\mathrm{O}_{3}, C_{\\mathrm{Al}_{2} \\mathrm{O}_{3}}{ }^{2}\\) in weight percent is just\n\\[\nC_{\\mathrm{Al}_{2} \\mathrm{O}_{3}}=\\frac{101.96 \\mathrm{~g} / \\mathrm{mol}}{101.96 \\mathrm{~g} / \\mathrm{mol}+(2)(60.09 \\mathrm{~g} / \\mathrm{mol})} \\times 100=45.9 \\mathrm{wt} \\%\n\\]\n(b) The liquidus and solidus temperatures for this material as determined from the \\(\\mathrm{SiO}_{2}-\\mathrm{Al}_{2} \\mathrm{O}_{3}\\) phase diagram, Figure 12.27, are \\(1825^{\\circ} \\mathrm{C}\\) and \\(1587^{\\circ} \\mathrm{C}\\), respectively.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 519,
|
||
"question": "Briefly explain\n(a) why there may be significant scatter in the fracture strength for some given ceramic material, and\n(b) why fracture strength increases with decreasing specimen size.",
|
||
"answer": "(a) 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.\n(b) The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 522,
|
||
"question": "(a) A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of \\(390 \\mathrm{MPa}(56,600 \\mathrm{psi})\\). If the specimen radius is \\(2.5 \\mathrm{~mm}(0.10 \\mathrm{in}\\).) and the support point separation distance is \\(30 \\mathrm{~mm}(1.2 \\mathrm{in}\\).), predict whether or not you would expect the specimen to fracture when a load of \\(620 \\mathrm{~N}\\left(140 \\mathrm{lb}_{\\mathrm{j}}\\right)\\) is applied. Justify your prediction.\n(b) Would you be \\(100 \\%\\) certain of the prediction in part (a)? Why or why not?",
|
||
"answer": "(a) This portion of the problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of \\(390 \\mathrm{MPa}(56,60\\) mp \\(\\mathrm{si})\\) and a radius of \\(2.5 \\mathrm{~mm}\\) will fracture when subjected to a load of \\(620 \\mathrm{~N}\\) in a three-point bending test; the support point separation is given as \\(30 \\mathrm{~mm}\\). Using Equation we will calculate the value of \\(\\sigma\\); if this value is greater than \\(\\sigma_{f s}(390 \\mathrm{MPa})\\), then fracture is expected to occur. Employment of Equation yields\n\\[\n\\begin{aligned}\n\\sigma & =\\frac{F L}{\\pi R^{3}} \\\\\n& =\\frac{(620 \\mathrm{~N})\\left(30 \\times 10^{-3} \\mathrm{~m}\\right)}{(\\pi)\\left(2.5 \\times 10^{-3} \\mathrm{~m}\\right)^{3}}=379 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=379 \\mathrm{MPa} \\quad(53,500 \\mathrm{psi})\n\\end{aligned}\n\\]\nSince this value is less than the given value of \\(\\sigma_{f s}(390 \\mathrm{MPa})\\), then traçree is not predicted.\n(b) 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 \\(\\sigma(379 \\mathrm{MPa})\\) is relatively close to \\(\\sigma_{f s}(390 \\mathrm{MPa})\\) then there is some chance that fracture will occur.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 524,
|
||
"question": "The modulus of elasticity for beryllium oxide (BeO) having 5 vol\\% porosity is \\(310 \\mathrm{GPa}\\left(45 \\times 10^{6}\\right.\\) psi).\n(a) Compute the modulus of elasticity for the nonporous material.\n(b) Compute the modulus of elasticity for 10 vol\\% porosity.",
|
||
"answer": "(a) This portion of the problem requests that we compute the modulus of elasticity for nonporous BeO given that \\(E=310 \\mathrm{GPa}\\) for a material having \\(5 \\mathrm{vol} \\%\\) porosity. Thus, we solve Equation for \\(E_{0}\\), using \\(P=0.05\\), which gives\n\\[\n\\begin{aligned}\nE_{0} & =\\frac{E}{1-1.9 P+0.9 P^{2}} \\\\\n& =\\frac{310 \\mathrm{GPa}}{1-(1.9)(0.05)+(0.9)(0.05)^{2}}=342 \\mathrm{GPa}(49.6 \\times 10^{6} \\mathrm{psi})\n\\end{aligned}\n\\]\n(b) Now we are asked to determine the value of \\(E\\) at \\(P=10 \\mathrm{vol} \\%\\) (i.e., 0.10). Using Equation we get\n\\[\n\\begin{aligned}\nE & =E_{0}\\left(1-1.9 P+0.9 P^{2}\\right) \\\\\n& =(342 \\mathrm{GPa})\\left[1-(1.9)(0.10)+(0.9)(0.10)^{2}\\right]=280 \\mathrm{GPa}(40.6 \\times 10^{6} \\mathrm{psi})\n\\end{aligned}%\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 525,
|
||
"question": "The modulus of elasticity for boron carbide \\(\\left(B_{4} C\\right)\\) having 5 vol\\% porosity is \\(290 \\mathrm{GPa}\\left(42 \\times 10^{6} \\mathrm{psi}\\right)\\).\n(a) Compute the modulus of elasticity for the nonporous material.\n(b) At what volume percent porosity will the modulus of elasticity be \\(235 \\mathrm{GPa}\\left(34 \\times 10^{6} \\mathrm{psi}\\right)\\) ?",
|
||
"answer": "(a) This portion of the problem requests that we compute the modulus of elasticity for nonporous \\(\\mathrm{B}_{4} \\mathrm{C}\\) given that \\(E=290 \\mathrm{GPa}\\left(42 \\times 10^{5} \\mathrm{psi}\\right)\\) for a material having \\(5 \\mathrm{vol} \\%\\) porosity. Thus, we solve Equation for \\(E_{0}\\). using \\(P=0.05\\), which gives\n\\[\n\\begin{aligned}\nE_{0} & =\\frac{E}{1-1.9 P+0.9 P^{2}} \\\\\n& =\\frac{290 \\mathrm{GPa}}{1-(1.9)(0.05)+(0.9)(0.05)^{2}}=320 \\mathrm{GPa} \\quad\\left(46.3 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]\n(b) Now we are asked to compute the volume percent porosity at which the elastic modulus of \\(\\mathrm{B}_{4} \\mathrm{C}\\) is 235 \\(\\mathrm{MPa}\\left(34 \\times 10^{6} \\mathrm{psi}\\right.\\). Since from part (a), \\(E_{0}=320 \\mathrm{GPa}\\), and using Equation we get\n\\[\n\\frac{E}{E_{0}}=\\frac{235 \\mathrm{MPa}}{320 \\mathrm{MPa}}=0.734=1-1.9 P+0.9 P^{2}\n\\]\nOr\n\\[\n0.9 P^{2}-1.9 P+0.266=0\n\\]\nNow, solving for the value of \\(P\\) using the quadratic equation solution yields\n\\[\nP=\\frac{1.9 \\pm \\sqrt{(-1.9)^{2}-(4)(0.9)(0.266)}}{(2)(0.9)}\n\\]\nThe positive and negative roots are\n\\[\n\\begin{array}{l}\np^{+}=1.96 \\\\\nP^{+}=0.151\n\\end{array}\n\\]\nObviously, only the negative root is physically meaningful, and therefore the value of the porosity to give the desired modulus of elasticity is \\(15.1 \\mathrm{vol} \\%\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 526,
|
||
"question": "Cite the two desirable characteristics of glasses.",
|
||
"answer": "Two desirable characteristics of glasses are optical transparency and ease of fabrication.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 527,
|
||
"question": "(a) What is crystallization?\n(b) Cite two properties that may be improved by crystallization.",
|
||
"answer": "(a) Crystallization is the process whereby a glass material is caused to transform to a crystalline solid, usually as a result of a heat treatment.\n(b) Two properties that may be improved by crystallization are (1) a lower coefficient of thermal expansion, and (2) higher strength.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 528,
|
||
"question": "For refractory ceramic materials, cite three characteristics that improve with and two characteristics that are adversely affected by increasing porosity.",
|
||
"answer": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock. Two characteristics that are adversely affected are (1) load-bearing capacity and (2) resistance to attack by corrosive materials.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 529,
|
||
"question": "Compare the manner in which the aggregate particles become bonded together in clay-based mixtures during firing and in cements during setting.",
|
||
"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.\nWith 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 533,
|
||
"question": "(a) Explain why residual thermal stresses are introduced into a glass piece when it is cooled.\n(b) Are thermal stresses introduced upon heating? Why or why not?",
|
||
"answer": "(a) 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.\n(b) Yes, thermal stresses will be introduced because of thermal expansion upon heating for the same reason as for thermal contraction upon cooling.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 538,
|
||
"question": "(a) What are the three main components of a whiteware ceramic such as porcelain?\n(b) What role does each component play in the forming and firing procedures?",
|
||
"answer": "(a) The three components of a whiteware ceramic are clay, quartz, and a flux.\n(b) With regard to the role that each component plays:\nQuartz acts as a filler material.\nClay 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.\nThe flux facilitates the formation of a glass having a relatively low melting temperature.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 539,
|
||
"question": "(a) Why is it so important to control the rate of drying of a ceramic body that has been hydroplastically formed or slip cast?\n(b) Cite three factors that influence the rate of drying, and explain how each affects the rate.",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 541,
|
||
"question": "(a) Name three factors that influence the degree to which vitrification occurs in clay-based ceramic wares.\n(b) Explain how density, firing distortion, strength, corrosion resistance, and thermal conductivity are affected by the extent of vitrification.",
|
||
"answer": "(a) 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.\n(b) Density will increase with degree of vitrification since the total remaining pore volume decreases.\nFiring distortion will increase with degree of vitrification since more liquid phase will be present at the firing temperature.\nStrength 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.\nCorrosion 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.\nThermal conductivity will increase with degree of vitrification. The glass phase has a higher conductivity than the pores that it has filled.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 543,
|
||
"question": "Compute repeat unit molecular weights for the following: (a) poly(vinyl chloride), (b) poly(ethylene terephthalate), (c) polycarbonate, and (d) polydimethylsiloxane.",
|
||
"answer": "(a) For poly(vinyl chloride), each repeat unit consists of two carbons, three hydrogens, and one chlorine . If \\(A_{\\mathrm{C}}, A_{\\mathrm{H}}\\) and \\(A_{\\mathrm{Cl}}\\) represent the atomic weights of carbon, hydrogen, and chlorine, respectively, then\n\\[\n\\begin{array}{c}\nm=2\\left(A_{\\mathrm{C}}\\right)+3\\left(A_{\\mathrm{H}}\\right)+\\left(A_{\\mathrm{Cl}}\\right) \\\\\n=(2)(12.01 \\mathrm{~g} / \\mathrm{mol})+(3)(1.008 \\mathrm{~g} / \\mathrm{mol})+35.45 \\mathrm{~g} / \\mathrm{mol}=62.49 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\n(b) For poly(ethylene terephthalate), each repeat unit has ten carbons, eight hydrogens, and four oxygens. Thus,\n\\[\n\\begin{array}{c}\nm=10\\left(A_{\\mathrm{C}}\\right)+8\\left(A_{\\mathrm{H}}\\right)+4\\left(A_{\\mathrm{O}}\\right) \\\\\n=(10)(12.01 \\mathrm{~g} / \\mathrm{mol})+(8)(1.008 \\mathrm{~g} / \\mathrm{mol})+(4)(16.00 \\mathrm{~g} / \\mathrm{mol})=192.16 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\n(c) For polycarbonate, each repeat unit has sixteen carbons, fourteen hydrogens, and three oxygens. Thus,\n\\[\n\\begin{array}{c}\nm=16\\left(A_{\\mathrm{C}}\\right)+14\\left(A_{\\mathrm{H}}\\right)+3\\left(A_{\\mathrm{O}}\\right) \\\\\n=(16)(12.01 \\mathrm{~g} / \\mathrm{mol})+(14)(1.008 \\mathrm{~g} / \\mathrm{mol})+(3)(16.00 \\mathrm{~g} / \\mathrm{mol}) \\\\\n=254.27 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\n(d) For polydimethylsiloxane, from Table 14.5, each repeat unit has two carbons, six hydrogens, one silicon and one oxygen. Thus,\n\\[\n\\begin{array}{c}\nm=2\\left(A_{\\mathrm{c}}\\right)+6\\left(A_{\\mathrm{H}}\\right)+\\left(A_{\\mathrm{Si}}\\right)+\\left(A_{\\mathrm{O}}\\right) \\\\\n=(2)(12.01 \\mathrm{~g} /\\left./ \\mathrm{mol}\\right)+(6)(1.008 \\mathrm{~g} / \\mathrm{mol})+(28.09 \\mathrm{~g} / \\mathrm{mol})+(16.00 \\mathrm{~g} / \\mathrm{mol})=74.16 \\mathrm{~g} / \\mathrm{mol}\n\\end{array} .\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 545,
|
||
"question": "(a) Compute the repeat unit molecular weight of polystyrene.\n(b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000 .",
|
||
"answer": "(a) 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,\n\\[\n\\begin{aligned}\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{aligned}\n\\]\n(b) We are now asked to compute the number-average molecular weight. Since the degree of polymerization is 25,000 , using Equation\n\\[\n\\bar{M}_{n}=(D P) m=(25,000)(104.14 \\mathrm{~g} / \\mathrm{mol})=2.60 \\times 10^{6} \\mathrm{~g} / \\mathrm{mol}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 546,
|
||
"question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen.\n(a) Determine the concentration of \\(\\mathrm{Cl}\\) (in wt\\%) that must be added if this substitution occurs for \\(5 \\%\\) of all the original hydrogen atoms.\n(b) In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)?",
|
||
"answer": "(a) For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for \\(5 \\%\\) \\(\\mathrm{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 \\(\\mathrm{Cl}\\). Thus, the mass of these 50 carbon atoms, \\(m_{\\mathrm{C}}\\), is just\n\\[\nm_{\\mathrm{C}}=50\\left(A_{\\mathrm{C}}\\right)=(50)(12.01 \\mathrm{~g} / \\mathrm{mol})=600.5 \\mathrm{~g}\n\\]\nLikewise, for hydrogen and chlorine,\n\\[\n\\begin{array}{l}\nm_{\\mathrm{H}}=95\\left(A_{\\mathrm{H}}\\right)=(95)(1.008 \\mathrm{~g} / \\mathrm{mol})=95.76 \\mathrm{~g} \\\\\nm_{\\mathrm{Cl}}=5\\left(A_{\\mathrm{Cl}}\\right)=(5)(35.45 \\mathrm{~g} / \\mathrm{mol})=177.25 \\mathrm{~g}\n\\end{array}\n\\]\nThus, the concentration of chlorine, \\(C_{\\mathrm{Cl}}\\), is determined using a modified form of Equation as\n\\[\n\\begin{aligned}\nC_{\\mathrm{Cl}} & =\\frac{m_{\\mathrm{Cl}}}{m_{\\mathrm{C}}+m_{\\mathrm{H}}+m_{\\mathrm{Cl}}} \\times 100 \\\\\n& =\\frac{177.25 \\mathrm{~g}}{600.5 \\mathrm{~g}+95.76 \\mathrm{~g}+177.25 \\mathrm{~g}} \\times 100=20.3 \\mathrm{wt} \\%\n\\end{aligned}\n\\]\n(b) Chlorinated polyethylene differs from poly(vinyl chloride), in that, for PVC, (1) \\(25 \\%\\) of the sidebonding sites are substituted with \\(\\mathrm{Cl}\\), and (2) the substitution is probably much less random.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 547,
|
||
"question": "Make comparisons of thermoplastic and thermosetting polymers (a) on the basis of mechanical characteristics upon heating, and (b) according to possible molecular structures.",
|
||
"answer": "(a) Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers, harden upon heating, while further heating will not lead to softening.\n(b) Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 548,
|
||
"question": "(a) Is it possible to grind up and reuse phenol-formaldehyde? Why or why not?\n(b) Is it possible to grind up and reuse polypropylene? Why or why not?",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 550,
|
||
"question": "Calculate the number-average molecular weight of a random nitrile rubber [poly(acrylonitrilebutadiene) copolymer] in which the fraction of butadiene repeat units is 0.30 ; assume that this concentration corresponds to a degree of polymerization of 2000 .",
|
||
"answer": "This problem asks for us to calculate the number-average molecular weight of a random nitrile rubber copolymer. For the acrylonitrile repeat unit there are three carbon, one nitrogen, and three hydrogen atoms. Thus, its repeat unit molecular weight is\n\\[\n\\begin{aligned}\nm_{\\mathrm{AC}} & =3\\left(A_{\\mathrm{C}}\\right)+\\left(A_{\\mathrm{N}}\\right)+3\\left(A_{\\mathrm{H}}\\right) \\\\\n& =(3)(12.01 \\mathrm{~g} / \\mathrm{mol})+14.01 \\mathrm{~g} / \\mathrm{mol}+(3)(1.008 \\mathrm{~g} / \\mathrm{mol})=53.06 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]\nThe butadiene repeat unit is composed of four carbon and six hydrogen atoms. Thus, its repeat unit molecular weight is\n\\[\n\\begin{aligned}\n& m_{\\mathrm{Bu}}=4\\left(A_{\\mathrm{C}}\\right)+6\\left(A_{\\mathrm{H}}\\right) \\\\\n& =(4)(12.01 \\mathrm{~g} / \\mathrm{mol})+(6)(1.008 \\mathrm{~g} / \\mathrm{mol})= 54.09 \\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{AC}} m_{\\mathrm{AC}}+f_{\\mathrm{Bu}} m_{\\mathrm{Bu}} \\\\\n& =(0.70)(53.06 \\mathrm{~g} / \\mathrm{mol})+(0.30)(54.09 \\mathrm{~g} / \\mathrm{mol})=53.37 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\n\\]\nSince \\(D P=2000\\) (as stated in the problem), \\(\\bar{M}_{n}\\) may be computed using Equation as\n\\[\n\\bar{M}_{n}=\\bar{m}(D P)=(53.37 \\mathrm{~g} / \\mathrm{mol})(2000)=106740 \\mathrm{~g} / \\mathrm{mol}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 551,
|
||
"question": "An alternating copolymer is known to have a number-average molecular weight of \\(250,000 \\mathrm{~g} / \\mathrm{mol}\\) and a degree of polymerization of 3420 . If one of the repeat units is styrene, which of ethylene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit? Why?",
|
||
"answer": "For an alternating copolymer which has a number-average molecular weight of \\(250,000 \\mathrm{~g}\\) /mol and a degree of polymerization of 3420 , we are to determine one of the repeat unit types if the other is styrene. It is first necessary to calculate \\(\\bar{m}\\) using Equation as\n\\[\n\\bar{m}=\\frac{\\bar{M}_{n}}{D P}=\\frac{250,000 \\mathrm{~g} / \\mathrm{mol}}{3420}=73.10 \\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_{s}=f_{x}=\\) \\(0.5, f_{s}\\) and \\(f_{x}\\) being, respectively, the chain fractions of the styrene and unknown repeat units. Also, the repeat unit molecular weight for styrene is\n\\[\n\\begin{aligned}\nm_{s} & =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{aligned}\n\\]\nNow, using Equation, it is possible to calculate the repeat unit weight of the unknown repeat unit type, \\(m_{x}\\). Thus\n\\[\n\\begin{aligned}\nm_{x} & =\\frac{\\bar{m}-f_{s} m_{s}}{f_{x}} \\\\\n& =\\frac{73.10 \\mathrm{~g} / \\mathrm{mol} \\cdot(0.5)(104.14 \\mathrm{~g} / \\mathrm{mol})}{0.5}=42.06 \\mathrm{~g} / \\mathrm{mol}\n\\end{aligned}\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\\begin{array}{l}\nm_{\\text {ethylene }}=2\\left(A_{\\mathrm{C}}\\right)+4\\left(A_{\\mathrm{H}}\\right)=2(12.01 \\mathrm{~g} / \\mathrm{mol})+4(1.008 \\mathrm{~g} / \\mathrm{mol})=28.05 \\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} / \\textrm{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{Cl}}\\right)=2(12.01 \\mathrm{~g} / \\text { mol })+3(1.008 \\mathrm{~g} / \\mathrm{mol})+35.45 \\mathrm{~g} / \\mathrm{mol}=62.49 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nTherefore, propylene is the other repeat unit type since its \\(m\\) value is almost the same as the calculated \\(m_{x}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 552,
|
||
"question": "(a) Determine the ratio of butadiene to styrene repeat units in a copolymer having a numberaverage molecular weight of \\(350,000 \\mathrm{~g} / \\mathrm{mol}\\) and degree of polymerization of 4425 .\n(b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?",
|
||
"answer": "(a) This portion of the problem asks us to determine the ratio of butadiene to styrene repeat units in a copolymer having a weight-average molecular weight of \\(350,000 \\mathrm{~g} / \\mathbf{m o l}\\) and a degree of polymerization of 4425 . It first becomes necessary to calculate the average repeat unit molecular weight of the copolymer, \\(\\bar{m}\\), using Equation as\n\\[\n\\bar{m}=\\frac{\\bar{M}_{B}}{D P}=\\frac{350,000 \\mathrm{~g} / \\mathrm{mol}}{4425}=79.10 \\mathrm{~g} / \\mathrm{mol}\n\\]\nIf we designate \\(f_{b}\\) as the chain fraction of butadiene repeat units, since the copolymer consists of only two repeat unit types, the chain fraction of styrene repeat units \\(f_{s}\\) is just \\(1-f_{b}\\). Now, Equation for this copolymer may be written in the form\n\\[\n\\bar{m}=f_{b} m_{b}+f_{s} m_{s}=f_{b} m_{b}+\\left(1-f_{b}\\right) m_{s}\n\\]\nin which \\(m_{b}\\) and \\(m_{s}\\) are the repeat unit molecular weights for butadiene and styrene, respectively. These values are calculated as follows:\n\\[\n\\begin{array}{l}\nm_{b}=4\\left(A_{\\mathrm{C}}\\right)+6\\left(A_{\\mathrm{H}}\\right)=4(12.01 \\mathrm{~g} / \\mathrm{mol})+6(1.008 \\mathrm{~g} / \\mathrm{mol})=54.09 \\mathrm{~g} / \\mathrm{mol} \\\\\nm_{s}=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})=104.14 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nSolving for \\(f_{b}\\) in the above expression yields\n\\[\nf_{b}=\\frac{\\bar{m}-m_{s}}{m_{b}-m_{s}}=\\frac{79.10 \\mathrm{~g} / \\mathrm{mol}-104.14 \\mathrm{~g} / \\mathrm{mol}}{54.09 \\mathrm{~g} / \\mathrm{mol}-104.14 \\mathbf{g} / \\mathrm{mol}}=0.50\n\\]\nFurthermore, \\(f_{S}=1-f_{b}=1-0.50=0.50\\); or the ratio is just\n\n\n\\[\n\\frac{f_{b}}{f_{s}}=\\frac{0.50}{0.50}=1.0\n\\]\n(b) Of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit types is alternating; the ratio must be 1:1. Therefore, on the basis of the result in part (a), the possibilities for this copolymer are not only alternating, but also random, graft, and block.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 555,
|
||
"question": "For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.\n(a) Linear and syndiotactic poly(vinyl chloride); linear and isotactic polystyrene.\n(b) Network phenol-formaldehyde; linear and heavily crosslinked cis-isoprene.\n(c) Linear polyethylene; lightly branched isotactic polypropylene.\n(d) Alternating poly(styrene-ethylene) copolymer; random poly(vinyl chloride-tetrafluoroethylene) copolymer.",
|
||
"answer": "(a) 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 \\(\\mathrm{Cl}\\) side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize.\n(b) No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize.\n(c) 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.\n(d) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 558,
|
||
"question": "The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the following equation:\n\\[\nP_{M}=P_{M_{0}} \\exp \\left(-\\frac{Q_{p}}{R T}\\right)\n\\]\nwhere \\(P_{M_{0}}\\) and \\(Q_{p}\\) are constants for a given gas-polymer pair. Consider the diffusion of hydrogen through a poly(dimethyl siloxane) (PDMSO) sheet \\(20 \\mathrm{~mm}\\) thick. The hydrogen pressures at the two faces are \\(10 \\mathrm{kPa}\\) and 1 \\(\\mathrm{kPa}\\), which are maintained constant. Compute the diffusion flux [in \\(\\left(\\mathrm{cm}^{3} \\mathrm{STP}\\right) / \\mathrm{cm}^{2}-\\mathrm{s}\\) ] at \\(350 \\mathrm{~K}\\). For this diffusion system\n\\[\n\\begin{array}{l}\nP_{M_{0}}=1.45 \\times 10^{-8}\\left(\\mathrm{~cm}^{3} \\mathrm{STP}\\right)\\left(\\mathrm{cm}^{3} / \\mathrm{cm}^{2}-\\mathrm{s}-\\mathrm{Pa}\\right. \\\\\nQ_{p}=13.7 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{array}\n\\]\nAlso, assume a condition of steady state diffusion",
|
||
"answer": "This problem asks that we compute the diffusion flux at \\(350 \\mathrm{~K}\\) for hydrogen in poly(dimethyl siloxane) (PDMSO). It is first necessary to compute the value of the permeability coefficient at \\(350 \\mathrm{~K}\\). The temperature dependence of \\(P_{M}\\) is given in the problem statement, as follows:\n\\[\nP_{M}=P_{M_{0}} \\exp \\left(-\\left.\\frac{Q_{p}}{R T}\\right)\\right]\n\\]\nAnd, incorporating values provided for the constants \\(P_{M_{0}}\\) and \\(Q_{p}\\), we get\n\\[\n\\begin{aligned}\nP_{M}=\\left[1.45 \\times 10^{-8} \\frac{\\left(\\mathrm{~cm}^{3} \\mathrm{STP}\\right)(\\mathrm{cm})}{\\mathrm{cm}^{2}-\\mathrm{s}-\\mathrm{Pa}}\\right] \\exp & {\\left[-\\frac{13,700 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(350 \\mathrm{~K})}\\right] } \\\\\n& =1.31 \\times 10^{-10} \\frac{\\left(\\mathrm{~cm}^{3} \\mathrm{STP}(\\mathrm{cm})\\right.}{\\mathrm{cm}^{2}-\\mathrm{s}-\\mathrm{Pa}}\n\\end{aligned}\n\\]\nAnd, using Equation, the diffusion flux is equal to\n\\[\n\\begin{aligned}\nJ & =P_{M} \\frac{\\Delta P}{\\Delta x}=P_{M} \\frac{P_{2}-P_{1}}{\\Delta x} \\\\\n& =1.31 \\times 10^{-10} \\frac{\\mathrm{~cm}^{3} \\mathrm{STP}(\\mathrm{cm})}{\\mathrm{cm}^{2}-\\mathrm{s}-\\mathrm{\\text { Pa }}}\\left(\\frac{10,000 \\mathrm{~Pa}-1,000 \\mathrm{~Pa}}{2.0 \\mathrm{~cm}}\\right)\n\\end{aligned}\n\\]\n\\[\n\\begin{array}{l}\n = 5.90 \\times 10^{-7} \\frac{(\\mathrm{cm}^{3} \\mathrm{STP})}{\\mathrm{cm}^{2} \\cdot \\mathrm{s}}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 559,
|
||
"question": "For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to\n\\[\n\\sigma(t)=\\sigma(0) \\exp \\left(-\\frac{t}{\\tau}\\right)\n\\]\nwhere \\(\\sigma(t)\\) and \\(\\sigma(t)\\) represent the time-dependent and initial (i.e., time \\(=0\\) ) stresses, respectively, and \\(t\\) and \\(\\tau\\) denote elapsed time and the relaxation time; \\(\\tau\\) is a time-independent constant characteristic of the material. A specimen of some viscoelastic polymer the stress relaxation of which obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.6 ; the stress necessary to maintain this constant strain was measured as a function of time. Determine \\(E_{\\tau}(10)\\) for this material if the initial stress level was \\(2.76 \\mathrm{MPa}(400 \\mathrm{psi})\\), which dropped to \\(1.72 \\mathrm{MPa}\\) (250 psi) after \\(60 \\mathrm{~s}\\).",
|
||
"answer": "This problem asks for a determination of the relaxation modulus of a viscoelastic material, which behavior is according to Equation 15.10--i.e.,\n\\[\n\\sigma(t)=\\sigma(0) \\exp \\left(-\\frac{1}{\\tau}\\right)\n\\]\nWe want to determine \\(\\sigma(10)\\), but it is first necessary to compute \\(\\tau\\) from the data provided in the problem statement. Thus, solving for \\(\\tau\\) from the above expression,\n\\[\n\\tau=\\frac{-t}{\\ln \\left[\\frac{\\sigma(t)}{\\sigma(0)}\\right]}=\\frac{-60 \\mathrm{~s}}{\\ln \\left[\\frac{1.72}{2.76} \\frac{\\mathrm{MPa}}{\\mathrm{MPa}}\\right]}=127 \\mathrm{~s}\n\\]\nTherefore,\n\\[\n\\sigma(10)=(2.76 \\mathrm{MPa}) \\exp \\left(-\\frac{10 \\mathrm{~s}}{127 \\mathrm{~s}}\\right)=2.55 \\mathrm{MPa}\n\\]\nNow, using Equation 15.1\n\\[\nE_{r}(10)=\\frac{\\sigma(10)}{\\varepsilon_{0}}=\\frac{2.55}{0.6} \\frac{\\mathrm{MPa}}{0.6}=4.25 \\mathrm{MPa} \\quad \\text { (616 psi) }\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 560,
|
||
"question": "(a) Contrast the manner in which stress relaxation and viscoelastic creep tests are conducted.\n(b) For each of these tests, cite the experimental parameter of interest and how it is determined.",
|
||
"answer": "(a) 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.\n(b) 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 \\mathrm{~s}\\) and strain; creep modulus is the ratio of stress and strain taken at a specific time.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 562,
|
||
"question": "For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.\n(a) Random acrylonitrile-butadiene copolymer with \\(10 \\%\\) of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with \\(5 \\%\\) of possible sites crosslinked\n(b) Branched and syndiotactic polypropylene with a degree of polymerization of 5000; linear and isotactic polypropylene with a degree of polymerization of 3000\n(c) Branched polyethylene with a number-average molecular weight of \\(250,000 \\mathrm{~g} / \\mathrm{mol}\\); linear and isotactic poly(vinyl chloride) with a number-average molecular weight of \\(200,000 \\mathrm{~g} / \\mathrm{mol}\\)",
|
||
"answer": "For each of four pairs of polymers, we are to do the following: (1) determine whether or not it is possible to decide which has the higher tensile modulus; (2) if so, note which has the higher tensile modulus and then state the reasons for this choice; and (3) if it is not possible to decide, then state why.\n(a） 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.\n(b) 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.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 563,
|
||
"question": "For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.\n(a) Syndiotactic polystyrene having a number-average molecular weight of \\(600,000 \\mathrm{~g} / \\mathrm{mol}\\); atactic polystyrene having a number-average molecular weight of \\(500,000 \\mathrm{~g} / \\mathrm{mol}\\)\n(b) Random acrylonitrile-butadiene copolymer with \\(10 \\%\\) of possible sites crosslinked; block acrylonitrilebutadiene copolymer with \\(5 \\%\\) of possible sites crosslinked\n(c) Network polyester; lightly branched polypropylene",
|
||
"answer": "For each of three pairs of polymers, we are to do the following: (1) determine whether or not it is possible to decide which has the higher tensile strength; (2) if it is possible, then note which has the higher tensile strength and then state the reasons for this choice; and (3) if it is not possible to decide, to state why.\n(a) 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.\n(b) 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.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 565,
|
||
"question": "Name the following polymer(s) that would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?",
|
||
"answer": "This question asks us to name which, of several polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below \\(100^{\\circ} \\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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 566,
|
||
"question": "For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.\n(a) Isotactic polystyrene that has a density of \\(1.12 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and a weight-average molecular weight of \\(150,000 \\mathrm{~g} / \\mathrm{mol}\\); syndiotactic polystyrene that has a density of \\(1.10 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and a weight-average molecular weight of \\(125,000 \\mathrm{~g} / \\mathrm{mol}\\)\n(b) Linear polyethylene that has a degree of polymerization of 5,000; linear and isotactic polypropylene that has a degree of polymerization of 6,500\n(c) Branched and isotactic polystyrene that has a degree of polymerization of 4,000; linear and isotactic polypropylene that has a degree of polymerizationof 7,500",
|
||
"answer": "(a) Yes, it is possible to determine which of the two polystyrenes has the higher \\(T_{m}\\). 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.\n(b) Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher \\(T_{m}\\) 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.\n(c) 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 \\(T_{m}\\). However, the polystyrene has more branching and a lower degree of polymerization; both of these factors lead to a lowering of the melting temperature.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 571,
|
||
"question": "Estimate the maximum and minimum thermal conductivity values for a cermet that contains \\(85 \\mathrm{vol} \\%\\) titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and \\(69 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K}\\) for TiC and Co, respectively.",
|
||
"answer": "This problem asks for the maximum and minimum thermal conductivity values for a TiC-Co cermet. Using a modified form of Equation the maximum thermal conductivity \\(k_{\\max }\\) is calculated as\n\\[\n\\begin{aligned}\nk_{\\max } & =k_{m} V_{m}+k_{p} V_{p}=k_{\\mathrm{Co}} V_{\\mathrm{Co}}+k_{\\mathrm{TiC}} V_{\\mathrm{TiC}} \\\\\n& =(69 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})(0.15)+(27 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})(0.85)=33.3 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K}\n\\end{aligned}\n\\]\nUsing a modified form of Equation, the minimum thermal conductivity \\(k_{\\min }\\) will be\n\\[\n\\begin{aligned}\nk_{\\min } & =\\frac{k_{\\mathrm{Co}} k_{\\mathrm{TIC}}}{V_{\\mathrm{Co}} k_{\\mathrm{TIC}}+V_{\\mathrm{TIC}} k_{\\mathrm{Co}}} \\\\\n& =\\frac{(69 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})(27 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})}{(0.15)(27 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})+(0.85)(69 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})} \\\\\n& =29.7 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K}\n\\end {aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 572,
|
||
"question": "A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.60 and 0.40 , respectively, estimate the upper limit for the specific stiffness of this composite given the data that follow.\n\\begin{tabular}{lcc}\n\\hline & Specific Gravity & \\begin{tabular}{c} \nModulus of Elasticity \\\\\n(GPa)\n\\end{tabular} \\\\\n\\hline Copper & 8.9 & 110 \\\\\nTungsten & 19.3 & 407 \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "Given the elastic moduli and specific gravities for copper and tungsten we are asked to estimate the upper limit for specific stiffness when the volume fractions of tungsten and copper are 0.60 and 0.4 .0 , respectively. There are two approaches that may be applied to solve this problem. The first is to estimate both the upper limits of elastic modulus \\(\\left[E_{\\mathrm{c}}(u)\\right]\\) and specific gravity \\(\\left(\\rho_{\\mathrm{c}}\\right)\\) for the composite, using expressions of the form of Equation, and then take their ratio. Using this approach\n\\[\n\\begin{aligned}\nE_{\\mathrm{c}}(u) & =E_{\\mathrm{Cu}} V_{\\mathrm{Cu}}+E_{\\mathrm{W}} V_{\\mathrm{W}} \\\\\n& =(110 \\mathrm{GPa})(0.40)+(407 \\mathrm{GPa})(0.60) \\\\\n& =288 \\mathrm{GPa}\n\\end{aligned}\n\\]\nAnd\n\\[\n\\begin{aligned}\n\\rho_{\\mathrm{c}} & =\\rho_{\\mathrm{Cu}} V_{\\mathrm{Cu}}+\\rho_{\\mathrm{W}} V_{\\mathrm{W}} \\\\\n& =(8.9)(0.40)+(19.3)(0.60)=15.14\n\\end{aligned}\n\\]\nTherefore\n\\[\n\\text { Specific Stiffness }=\\frac{E_{\\mathrm{c}}(u)}{\\rho_{\\mathrm{c}}}=\\frac{288 \\mathrm{GPa}}{15.14}=19.0 \\mathrm{GPa}\n\\]\nWith the alternate approach, the specific stiffness is calculated, again employing a modification of Equation, but using the specific stiffness-volume fraction product for both metals, as follows:\n\n\n\\[\n\\begin{array}{l}\n\\text { Specific Stiffness }=\\frac{E_{\\mathrm{Cu}}}{\\rho_{\\mathrm{Cu}}} V_{\\mathrm{Cu}}+\\frac{E_{\\mathrm{W}}}{\\rho_{\\mathrm{W}}} V_{\\mathrm{W}} \\\\\n=\\frac{110 \\mathrm{\\; GPa}}{8.9}(0.40)+\\frac{407 \\mathrm{GPa}}{19.3}(0.60)=17.6 \\mathrm{GPa}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 573,
|
||
"question": "(a) What is the distinction between cement and concrete?\n(b) Cite three important limitations that restrict the use of concrete as a structural material.\n(c) Briefly explain three techniques that are utilized to strengthen concrete by reinforcement.",
|
||
"answer": "(a) Concrete consists of an aggregate of particles that are bonded together by a cement.\n(b) 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.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 574,
|
||
"question": "Cite one similarity and two differences 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.\nThe two differences are: (1) the hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening; and (2) the strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 575,
|
||
"question": "A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol\\% aramid fibers and 70 vol\\% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:\n\\begin{tabular}{lcc}\n\\hline & \\begin{tabular}{c} \nModulus of Elasticity \\\\\n{\\([\\mathrm{GPa}(\\mathrm{psi})]\\)}\n\\end{tabular} & \\begin{tabular}{c} \nTensile Strength [MPa \\\\\n\\((\\mathrm{psi})]\\)\n\\end{tabular} \\\\\n\\hline Aramid fiber & \\(131\\left(19 \\times 10^{6}\\right)\\) & \\(3600(520,000)\\) \\\\\nPolycarbonate & \\(2.4\\left(3.5 \\times 10^{5}\\right)\\) & \\(65(9425)\\) \\\\\n\\hline\n\\end{tabular}\nAlso, the stress on the polycarbonate matrix when the aramid fibers fail is \\(45 \\mathrm{MPa}(6500 \\mathrm{psi})\\).\nFor this composite, compute\n(a) the longitudinal tensile strength, and\n(b) the longitudinal modulus of elasticity",
|
||
"answer": "This problem calls for us to compute the longitudinal tensile strength and elastic modulus of an aramid fiber-reinforced polycarbonate composite.\n(a) The longitudinal tensile strength is determined using Equation as\n\\[\n\\begin{aligned}\n\\sigma_{c l}^{*} & =\\sigma_{m}^{*}\\left(1-V_{f}\\right)+\\sigma_{f}^{*} V_{f} \\\\\n& =(45 \\mathrm{MPa})(0.70)+(3600)(0.30) \\\\\n& =1100 \\mathrm{MPa}(160,000 \\mathrm{psi})\n\\end{aligned}\n\\]\n(b) The longitudinal elastic modulus is computed using Equation as\n\\[\n\\begin{aligned}\nE_{c l} & =E_{m} V_{m}+E_{f} V_{f} \\\\\n& =(2.4 \\mathrm{GPa})(0.70)+(131 \\mathrm{GPa})(0.30) \\\\\n& =41 \\mathrm{GPa}\\left(5.95 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 576,
|
||
"question": "For a continuous and oriented fiber-reinforced composite, the moduli of elasticity in the longitudinal and transverse directions are 19.7 and \\(3.66 \\mathrm{GPa}\\left(2.8 \\times 10^{5}\\right.\\) and \\(\\left.5.3 \\times 10^{5} \\mathrm{psi}\\right)\\), respectively. If the volume fraction of fibers is 0.25 , determine the moduli of elasticity of fiber and matrix phases.",
|
||
"answer": "This problem asks for us to compute the elastic moduli of fiber and matrix phases for a continuous and oriented fiber-reinforced composite. We can write expressions for the longitudinal and transverse elastic moduli using Equations, as\n\\[\n\\begin{array}{c}\nE_{c l}=E_{m}\\left(1-V_{f}\\right)+E_{f} V_{f} \\\\\n19.7 \\mathrm{GPa}=E_{m}(1-0.25)+E_{f}(0.25)\n\\end{array}\n\\]\nAnd\n\\[\nE_{c t}=\\frac{E_{m} E_{f}}{\\left(1-V_{f}\\right) E_{f}+V_{f} E_{m}}\n\\]\n\\[\n3.66 \\mathrm{GPa}=\\frac{E_{m} E_{f}}{(1-0.25) E_{f}+0.25 E_{m}}\n\\]\nSolving these two expressions simultaneously for \\(E_{m}\\) and \\(E_{f}\\) leads to\n\\[\n\\begin{array}{l}\nE_{m}=2.79 \\mathrm{GPa}\\left(4.04 \\times 10^{5} \\mathrm{psi}\\right) \\\\\nE_{f}=70.4 \\mathrm{GPa}\\left(10.2 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.\n(a) Using the data provided, determine the volume fraction of fibers that will be required.\n(b) What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 \\(\\mathrm{MPa}\\) (4350 psi).\n\\begin{tabular}{lcc}\n\\hline & \\begin{tabular}{c} \nModulus of Elasticity \\\\\n{\\(\\left[\\mathrm{GPa}(\\mathrm{psi})\\right]\\)}\n\\end{tabular} & \\begin{tabular}{c} \nTensile Strength \\\\\n{\\(\\left[\\mathrm{MPa}(\\mathrm{psi})\\right]\\)}\n\\end{tabular} \\\\\n\\hline Glass fiber & \\(72.5\\left(10.5 \\times 10^{6}\\right)\\) & \\(3400(490,000)\\) \\\\\nNylon 6,6 & \\(3.0\\left(4.35 \\times 10^{6}\\right)\\) & \\(76(11,000)\\) \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "(a) Given some data for an aligned and continuous glass-fiber-reinforced nylon 6,6 composite, we are asked to compute the volume fraction of fibers that are required such that the fibers carry \\(94 \\%\\) of a load applied in the longitudinal direction. From Equation\n\\[\n\\frac{F_{f}}{F_{m}}=\\frac{E_{f} V_{f}}{E_{m} V_{m}}=\\frac{E_{f} V_{f}}{E_{m}\\left(1-V_{f}\\right)}\n\\]\nNow, using values for \\(F_{f}\\) and \\(F_{m}\\) from the problem statement\n\\[\n\\frac{F_{f}}{F_{m}}=\\frac{0.94}{0.06}=15.67\n\\]\nAnd when we substitute the given values for \\(E_{f}\\) and \\(E_{m}\\) into the first equation leads to\n\\[\n\\frac{F_{f}}{F_{m}}=15.67=\\frac{(72.5 \\mathrm{GPa}) V_{f}}{(3.0 \\mathrm{GPa})\\left(1-V_{f}\\right)}\n\\]\nAnd, solving for \\(V_{f}\\) yields, \\(V_{f}=0.393\\).\n(b) We are now asked for the tensile strength of this composite. From Equation,\n\\[\n\\sigma_{c l}^{*}=\\sigma_{m}^{*}\\left(1-V_{f}\\right)+\\sigma_{f}^{*} V_{f}\n\\]\n\n\n\\[\n\\begin{array}{l}\n=(30 \\mathrm{MPa})(1-0.393)+(3400 \\mathrm{MPa})(0.393) \\\\\n\\quad=1354 \\mathrm{MPa}(196,400 \\mathrm{psi})\n\\end{array}\n\\]\nsince values for \\(\\sigma_{f}^{*}(3400 \\mathrm{MPa})\\) and \\(\\sigma_{m}^{*}(30 \\mathrm{MPa})\\) are given in the problem statement.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 578,
|
||
"question": "For a polymer-matrix fiber-reinforced composite,\n(a) List three functions of the matrix phase.\n(b) Compare the desired mechanical characteristics of matrix and fiber phases.\n(c) Cite two reasons why there must be a strong bond between fiber and matrix at their interface.",
|
||
"answer": "(a) 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.\n(b) The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 579,
|
||
"question": "(a) What is the distinction between matrix and dispersed phases in a composite material?\n(b) Contrast the mechanical characteristics of matrix and dispersed phases for fiber-reinforced composites.",
|
||
"answer": "(a) The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 580,
|
||
"question": "(a) Briefly explain the difference between oxidation and reduction electrochemical reactions.\n(b) Which reaction occurs at the anode and which at the cathode?",
|
||
"answer": "(a) 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.\n(b) Oxidation occurs at the anode; reduction at the cathode.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 581,
|
||
"question": "(a) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in each of the following solutions: (i) \\(\\mathrm{HCl}\\), (ii) an \\(\\mathrm{HCl}\\) solution containing dissolved oxygen, (iii) an \\(\\mathrm{HCl}\\) solution containing dissolved oxygen and, in addition, \\(\\mathrm{Fe}^{2+}\\) ions.\n(b) In which of these solutions would you expect the magnesium to oxidize most rapidly? Why?",
|
||
"answer": "(a) This problem asks that we write possible oxidation and reduction half-reactions for magnesium in various solutions.\n(i) In \\(\\mathrm{HCl}\\), possible reactions are\n\\[\n\\begin{array}{l}\n\\mathrm{Mg} \\rightarrow \\mathrm{Mg}^{2+}+2 e^{-} \\text {(oxidation) } \\\\\n2 \\mathrm{H}^{+}+2 e^{-} \\rightarrow \\mathrm{H}_{2} \\text { (reduction) }\n\\end{array}\n\\]\n(ii) In an \\(\\mathrm{HCl}\\) solution containing dissolved oxygen, possible reactions are\n\\[\n\\begin{array}{l}\n\\mathrm{Mg}^{-} \\rightarrow \\mathrm{Mg}^{2+}+2 e^{-} \\quad \\text { (oxidation) } \\\\\n4 \\mathrm{H}^{+}+\\mathrm{O}_{2}+4 e^{-} \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O} \\quad \\text { (reduction) }\n\\end{array}\n\\]\n(iii) In an \\(\\mathrm{HCl}\\) solution containing dissolved oxygen and \\(\\mathrm{Fe}^{2+}\\) ions, possible reactions are\n\\[\n\\begin{array}{l}\n\\mathrm{Mg}{ }^{-} \\rightarrow \\mathrm{Mg}^{2+}+2 e^{-}(\\text {oxidation) } \\\\\n4 \\mathrm{H}^{+}+\\mathrm{O}_{\\mathrm{2}}+4 e^{-} \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{~O} \\quad \\text { (reduction) } \\\\\n\\mathrm{Fe}^{2+}+2 e^{-} \\rightarrow \\mathrm{Fe} \\quad \\text { (reduction) }\n\\end{array}\n\\]\n(b) The magnesium would probably oxidize most rapidly in the \\(\\mathrm{HCl}\\) solution containing dissolved oxygen and \\(\\mathrm{Fe}^{\\mathbf{2}+}\\) ions because there are two reduction reactions that will consume electrons from the oxidation of magnesium.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 582,
|
||
"question": "Demonstrate that (a) the value of \\(\\mathscr{F}\\) in Equation 17.19 is \\(96,500 \\mathrm{C} / \\mathrm{mol}\\), and (b) at \\(25^{\\circ} \\mathrm{C}(298 \\mathrm{~K})\\),\n\\[\n\\frac{R T}{n F} \\ln x=\\frac{0.0592}{n} \\log x\n\\]",
|
||
"answer": "(a) The Faraday constant \\(\\mathscr{F}\\) (represented here as \" \\(F\\) \") is just the product of the charge per electron and Avogadro's number; that is\n\\[\n\\begin{aligned}\nF= & e N_{\\mathrm{A}}=\\left(1.602 \\times 10^{-19} \\mathrm{C} / \\mathrm{electron}\\right)\\left(6.022 \\times 10^{23} \\mathrm{electrons} / \\mathrm{mol}\\right) \\\\\n& =96,472 \\mathrm{C} / \\mathrm{mol}\n\\end{aligned}\n\\]\n(b) At \\(25^{\\circ} \\mathrm{C}(298 \\mathrm{~N})\\),\n\\[\n\\begin{aligned}\n\\frac{R T}{n F} \\ln (x) & =\\frac{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(298 \\mathrm{~K})}{(n)(96,472 \\mathrm{C} / \\mathrm{mol})}(2.303) \\log (x) \\\\\n& =\\frac{0.0592}{n} \\log (x)\n\\end{aligned}\n\\]\nThis gives units in volts since a volt is a J/C.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 583,
|
||
"question": "A \\(\\mathrm{Zn} / \\mathrm{Zn}^{2+}\\) concentration cell is constructed in which both electrodes are pure zinc. The \\(\\mathrm{Zn}^{2+}\\) concentration for one cell half is \\(1.0 \\mathrm{M}\\), for the other, \\(10^{-2} \\mathrm{M}\\). Is a voltage generated between the two cell halves? If so, what is its magnitude and which electrode will be oxidized? If no voltage is produced, explain this result.",
|
||
"answer": "This problem calls for us to determine whether or not a voltage is generated in a \\(\\mathrm{Zn} / \\mathrm{Zn}^{2+}\\) concentration cell, and, if so, its magnitude. Let us label the \\(\\mathrm{Zn}\\) cell having a \\(1.0 \\mathrm{M} \\mathrm{Zn}^{2+}\\) solution as cell 1 , and the other as cell 2 . Furthermore, assume that oxidation occurs within cell 2 , wherein \\(\\left[\\mathrm{Zn}_{2}^{2+}\\right]=10^{-2} M\\). Hence,\n\\[\n\\mathrm{Zn}_{2}+\\mathrm{Zn}_{1}^{2+} \\rightarrow \\mathrm{Zn}_{2}^{2+}+\\mathrm{Zn}_{1}\n\\]\nand, employing Equation leads to\n\\[\n\\begin{aligned}\n\\Delta V & =-\\frac{0.0592}{2} \\log \\frac{\\left[\\mathrm{Zn}_{2}^{2+}\\right]}{\\left[\\mathrm{Zn}_{1}^{2+}\\right]} \\\\\n& =-\\frac{0.0592}{2} \\log \\left[\\frac{10^{-2} M}{1.0 M}\\right]=+0.0592 \\mathrm{~V}\n\\end{aligned}\n\\]\nTherefore, a voltage of \\(0.0592 \\mathrm{~V}\\) is generated when oxidation occurs in the cell having the \\(\\mathrm{Zn}^{2+}\\) concentration of \\(10^{-2} M\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 585,
|
||
"question": "A thick steel sheet of area \\(400 \\mathrm{~cm}^{2}\\) is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of \\(375 \\mathrm{~g}\\) due to corrosion. To what rate of corrosion, in both mpy and \\(\\mathrm{mm} / \\mathrm{yr}\\), does this correspond?",
|
||
"answer": "This problem asks for us to calculate the CPR in both mpy and \\(\\mathrm{mm} / \\mathrm{yr}\\) for a thick steel sheet of area 400 \\(\\mathrm{cm}^{2}\\) which experiences a weight loss of \\(375 \\mathrm{~g}\\) after one year. Employment of Equation leads to\n\\[\n\\begin{aligned}\n\\operatorname{CPR}(\\operatorname{mm} / \\mathrm{yr}) & =\\frac{K W}{\\rho A t} \\\\\n& =\\frac{(87.6)(375 \\mathrm{~g})\\left(10^{3} \\mathrm{mg} / \\mathrm{g}\\right)}{\\left(7.9 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(400 \\mathrm{~cm}^{2}\\right)(24 \\mathrm{~h} / \\text { day })(365 \\text { day } / \\mathrm{yr})(1 \\text { yr })} \\\\\n& =1.2 \\mathrm{~mm} / \\mathrm{yr}\n\\end{aligned}\n\\]\nAlso\n\\[\n\\begin{aligned}\n\\operatorname{CPR}(\\text { mpy })= & \\frac{(534)(375 \\mathrm{~g})\\left(10^{3} \\text { mg } / \\mathrm{g}\\right)}{\\left(7.9 \\mathrm{~g / \\mathrm{cm}^{3}\\right)(400 \\mathrm{in} .^{2})(1 \\text { in } / 2.54 \\mathrm{~cm})^{2}(24 \\mathrm{~h} / \\text { day })(365 \\text {\nday } / \\mathrm{yr})(1 \\text { yr })} \\\\\n& =46.7 \\mathrm{mpy}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 586,
|
||
"question": "(a) Cite the major differences between activation and concentration polarizations.\n(b) Under what conditions is activation polarization rate controlling?\n(c) Under what conditions is concentration polarization rate controlling?",
|
||
"answer": "(a) 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.\n(b) Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high.\n(c) Concentration polarization is rate controlling when the reaction rate is high and/or the concentration of active species in the liquid solution is low.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 587,
|
||
"question": "The corrosion rate is to be determined for some divalent metal \\(M\\) in a solution containing hydrogen ions. The following corrosion data are known about the metal and solution:\n\\begin{tabular}{cc}\n\\hline For Metal M & For Hydrogen \\\\\n\\hline\\(V_{\\left(\\mathrm{M} / \\mathrm{M}^{2+}\\right)}=-0.47 \\mathrm{~V}\\) & \\(V_{\\left(\\mathrm{H}^{+} / \\mathrm{H}_{2}\\right)}=0 \\mathrm{~V}\\) \\\\\n\\(i_{0}=5 \\times 10^{-10} \\mathrm{~A} / \\mathrm{cm}^{2}\\) & \\(i_{0}=2 \\times 10^{-9} \\mathrm{~A} / \\mathrm{cm}^{2}\\) \\\\\n\\(\\beta=+0.15\\) & \\(\\beta=-0.12\\) \\\\\n\\hline\n\\end{tabular}\n(a) Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal \\(M\\) (in \\(\\mathrm{mol} / \\mathrm{cm}^{2}\\)-s).\n(b) Compute the corrosion potential for this reaction.",
|
||
"answer": "(a) This portion of the problem asks that we compute the rate of oxidation for a divalent metal M given that both the oxidation and reduction reactions are controlled by activation polarization, and also given the polarization data for both \\(\\mathrm{M}\\) oxidation and hydrogen reduction. The first thing necessary is to establish relationships of the form of Equation for the potentials of both oxidation and reduction reactions. Next we will set these expressions equal to one another, and then solve for the value of \\(i\\) which is really the corrosion current density, \\(i_{c}\\). Finally, the corrosion rate may be calculated using Equation. The two potential expressions are as follows:\nFor hydrogen reduction\n\\[\nV_{\\mathrm{H}}=V_{\\left(\\mathrm{H}^{+} / \\mathrm{H}_{2} \\right)}+\\beta_{\\mathrm{H}} \\log \\left(\\frac{i}{i_{\\mathrm{H}}}\\right)\n\\]\nAnd for \\(\\mathrm{M}\\) oxidation\n\\[\nV_{\\mathrm{M}}=V_{\\left(\\mathrm{M} / \\mathrm{M}^{2+}\\cdot\\right)}+\\beta_{\\mathrm{M}} \\log \\left(\\frac{i}{i_{\\mathrm{O}_{\\mathrm{M}}}}\\right)\n\\]\nSetting \\(V_{\\mathrm{H}}=V_{\\mathrm{M}}\\) and solving for \\(\\log i\\left(\\log i_{\\mathrm{C}}\\right)\\) leads to\n\n\n\\[\n\\log \\hat{l}_{\\mathrm{c}}=\\left\\lfloor\\frac{1}{\\beta_{\\mathrm{M}}-\\beta_{\\mathrm{H}}}\\right\\rfloor V_{\\left(\\mathrm{H}^{+} / \\mathrm{H}_{2}\\right)}-V_{\\left(\\mathrm{M} / \\mathrm{M}^{2+}\\right)}-\\beta_{\\mathrm{H}} \\log \\hat{l}_{\\mathrm{h}}+\\beta_{\\mathrm{M}} \\log \\hat{l}_{\\mathrm{h}}\\right\\rfloor\n\\]\nAnd, incorporating values for the various parameters provided in the problem statement leads to\n\\[\n\\begin{array}{l}\n\\log \\hat{l}_{\\mathrm{c}}=\\left\\lfloor\\frac{ 1 }{0.15 -(-0.12)}\\right\\rfloor[0-(-0.47)-(-0.12)\\left\\{ \\log \\left(2 \\times 10^{-9}\\right)\\right\\} + (0.15)\\left\\{ \\log \\left(5 \\times 10^{-10}\\right)\\right\\} ] \\\\\n\\quad=-7.293\n\\end{array}\n\\]\nOr\n\\[\n\\hat{l}_{\\mathrm{c}}=10^{-7.293}=5.09 \\times 10^{-8} \\mathrm{~A} / \\mathrm{cm}^{2}\n\\]\nAnd from Equation\n\\[\n\\begin{aligned}\nr & =\\frac{\\hat{l}_{\\mathrm{c}}}{n F} \\\\\n& =\\frac{5.09 \\times 10^{-8} \\mathrm{C} / \\mathrm{s}-\\mathrm{cm}^{2}}{(2)(96,500 \\mathrm{C} / \\mathrm{mol})}=2.64 \\times 10^{-13} \\mathrm{~mol} / \\mathrm{cm}^{2} \\cdot \\mathrm{s}\n\\end{aligned}\n\\]\n(b) Now it becomes necessary to compute the value of the corrosion potential, \\(V_{\\mathrm{c}}\\). This is possible by using either of the above equations for \\(V_{\\mathrm{H}}\\) or \\(V_{\\mathrm{H}}\\) and substituting for \\(i\\) the value determined above for \\(i_{\\mathrm{c}}\\). Thus\n\\[\n\\begin{aligned}\nV_{c} & =\\mathrm{V}_{\\left(\\mathrm{H}^{+} / \\mathrm{H}_{2}\\right)}\n+\\beta_{\\mathrm{H}} \\log \\left(\\frac{\\hat{l}_{\\mathrm{c}}}{\\hat{l}_{\\mathrm{h}}}\\right) \\\\\n& =0+(-0.12 \\mathrm{~V}) \\log \\left(\\frac{5.09 \\times 10^{-8} \\mathrm{~A} / \\mathrm{cm}^{2}}{2 \\times 10^{-9} \\mathrm{~A} / \\mathrm{cm}^{2}}\\right)=-0.169 \\mathrm{~V}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 588,
|
||
"question": "Briefly describe the phenomenon of passivity. Name two common types of alloy that passivate.",
|
||
"answer": "Passivity is the loss of chemical reactivity, under particular environmental conditions, of normally active metals and alloys. Stainless steels and aluminum alloys often passivate.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 592,
|
||
"question": "(a) What are inhibitors? \\\\ (b) What possible mechanisms account for their effectiveness?",
|
||
"answer": "(a) Inhibitors are substances that, when added to a corrosive environment in relatively low concentrations, \\\\ decrease the environment's corrosiveness. \\\\ (b) 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.\n}",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 593,
|
||
"question": "Briefly describe the two techniques that are used for galvanic protection.",
|
||
"answer": "Descriptions of the two techniques used for galvanic protection are as follows:\n(1) 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.\n(2) An impressed current from an external dc power source provides excess electrons to the metallic structure to be protected.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 594,
|
||
"question": "(a) Compute the electrical conductivity of a 5.1-mm (0.2-in.) diameter cylindrical silicon specimen 51 \\(\\mathrm{mm}\\) (2 in.) long in which a current of 0.1 A passes in an axial direction. A voltage of \\(12.5 \\mathrm{~V}\\) is measured across two probes that are separated by \\(38 \\mathrm{~mm}\\) (1.5 in.).\n(b) Compute the resistance over the entire \\(51 \\mathrm{~mm}\\) (2 in.) of the specimen.",
|
||
"answer": "This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen.\n(a) We use Equations for the conductivity, as\n\\[\n\\sigma=\\frac{1}{\\rho}=\\frac{I I}{\\mathrm{VA}}=\\frac{I I}{V \\pi\\left(\\frac{d}{2}\\right)^{2}}\n\\]\nAnd, incorporating values for the several parameters provided in the problem statement, leads to\n\\[\n\\sigma=\\frac{(0.1 \\mathrm{~A})\\left(38 \\times 10^{-3} \\mathrm{~m}\\right)}{(12.5 \\mathrm{~V})(\\pi)\\left(\\frac{5.1 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}}=14.9(\\Omega \\cdot \\mathrm{m})^{-1}\n\\]\n(b) The resistance, \\(R\\), may be computed using Equations, as\n\\[\nR=\\frac{\\rho l}{A}=\\frac{l}{\\sigma A}=\\frac{l}{\\sigma \\pi\\left(\\frac{d}{2}\\right)^{2}}\n\\]\n\n\\[\n = \\frac{51 \\times 10^{-3} \\mathrm{~m}}{\\left[14.9 \\quad \\left(\\Omega-\\mathrm{m}\\right)^{-1}\\right]\\left(\\pi\\right)\\left(\\frac{5.1 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}}=168 \\Omega\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 595,
|
||
"question": "(a) Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the electric field is \\(1000 \\mathrm{~V} / \\mathrm{m}\\). (b) Under these circumstances, how long does it take an electron to traverse a 25 -mm (1-in.) length of crystal?",
|
||
"answer": "(a) The drift velocity of electrons in Ge may be determined using Equation. Since the room temperature mobility of electrons is \\(0.38 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\) , and the electric field is \\(1000 \\mathrm{~V} / \\mathrm{ m}\\) (as stipulated in the problem statement),\n\\[\n\\begin{aligned}\nv_{d} & =\\mu_{e} E \\\\\n& =\\left(0.38 \\mathrm{~m}^{2} / \\mathrm{V}-s\\right)(1000 \\mathrm{~V} / \\mathrm{m})=380 \\mathrm{~m} / \\mathrm{s}\n\\end{aligned}\n\\]\n(b) The time, \\(t\\), required to traverse a given length, \\(l(=25 \\mathrm{~mm})\\), is just\n\\[\nt=\\frac{l}{v_{d}}=\\frac{25 \\times 10^{-3} \\mathrm{~m}}{380 \\mathrm{~m} / \\mathrm{s}}=6.6 \\times 10^{-5} \\mathrm{~s}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 596,
|
||
"question": "At room temperature the electrical conductivity and the electron mobility for copper are \\(6.0 \\times 10^{7}\\) \\((\\Omega \\cdot \\mathrm{m})^{-1}\\) and \\(0.0030 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\\), respectively. (a) Compute the number of free electrons per cubic meter for copper at room temperature. (b) What is the number of free electrons per copper atom? Assume a density of \\(8.9 \\mathrm{~g} / \\mathrm{cm}^{3}\\).",
|
||
"answer": "(a) The number of free electrons per cubic meter for copper at room temperature may be computed using Equation as\n\\[\n\\begin{aligned}\nn & =\\frac{\\sigma}{\\left|e\\right| \\mu_{e}} \\\\\n& =\\frac{6.0 \\times 10^{7}(\\Omega-m)^{-1}}{\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)\\left(0.003 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\\right)} \\\\\n& =1.25 \\times 10^{29} \\mathrm{~m}^{-3}\n\\end{aligned}\n\\]\n(b) In order to calculate the number of free electrons per copper atom, we must first determine the number of copper atoms per cubic meter, \\(N_{\\mathrm{Cu}}\\). From Equation (and using the atomic weight value for \\(\\mathrm{Cu}\\) found inside the front cover- \\(\\mathrm{viz} .63 .55 \\mathrm{~g} / \\mathrm{mol}\\) )\n\\[\n\\begin{aligned}\nN_{\\mathrm{Cu}} & =\\frac{N_{\\mathrm{A}} \\rho^{\\prime}}{A_{\\mathrm{Cu}}} \\\\\n& =\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(8.9 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(10^{6} \\mathrm{~cm}^{3} / \\mathrm{m}^{3}\\right)}{63.55 \\mathrm{~g} / \\mathrm{mol}} \\\\\n& =8.43 \\times 10^{28} \\mathrm{~m}^{-3}\n\\end{aligned}\n\\]\n(Note: in the above expression, density is represented by \\(\\rho^{\\prime}\\) in order to avoid confusion with resistivity which is designated by \\(\\rho\\).) And, finally, the number of free electrons per aluminum atom is just \\(n / N_{\\mathrm{Cu}}\\)\n\\[\n\\frac{n}{N_{\\mathrm{Cu}}}=\\frac{1.25 \\times 10^{29} \\mathrm{~m} \\mathrm{~m}^{-3}}{8.43 \\times 10^{28} \\mathrm{~m} \\mathrm{~m}^{-3}}=1.48\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 598,
|
||
"question": "Define the following terms as they pertain to semiconducting materials: intrinsic, extrinsic, compound, elemental. Now provide an example of each.",
|
||
"answer": "These semiconductor terms are defined in the Glossary. Examples are as follows: intrinsic--high purity (undoped) Si, GaAs, CdS, etc.; extrinsic--P-doped Ge, B-doped Si, S-doped GaP, etc.; compound--GaAs, InP, CdS, etc.; elemental--Ge and Si.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 599,
|
||
"question": "(a) Explain why no hole is generated by the electron excitation involving a donor impurity atom. (b) Explain why no free electron is generated by the electron excitation involving an acceptor impurity atom.",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 600,
|
||
"question": "Germanium to which \\(5 \\times 10^{22} \\mathrm{~m}^{-3} \\mathrm{~Sb}\\) atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the \\(\\mathrm{Sb}\\) atoms may be thought of as being ionized (i.e., one charge carrier exists for each \\(\\mathrm{Sb}\\) atom). (a) Is this material \\(n\\)-type or p-type? (b) Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and \\(0.05 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\), respectively.",
|
||
"answer": "(a) (a) This germanium material to which has been added \\(5 \\times 10^{22} \\mathrm{~m}^{-3}\\) Sb atoms is \\(n\\)-type since \\(\\mathrm{Sb}\\) is a donor in Ge. (Antimony is from group VA of the periodic table--Ge is from group IVA.)\n(b) Since this material is \\(n\\)-type extrinsic, Equation is valid. Furthermore, each Sb will donate a single electron, or the electron concentration is equal to the \\(\\mathrm{Sb}\\) concentration since all of the \\(\\mathrm{Sb}\\) atoms are ionized at room temperature; that is \\(n=5 \\times 10^{22} \\mathrm{~m}^{-3}\\), and, as given in the problem statement, \\(\\mu_{e}=0.1 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\). Thus\n\\[\n\\begin{aligned}\n\\sigma & =n|e| \\mu_{e} \\\\\n& =\\left(5 \\times 10^{22} \\mathrm{~m}^{-3}\\right)\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)\\left(0.1 \\mathrm{~m}^{2} / \\mathrm{V}-\\right. \\\\\n& =800(\\Omega \\cdot \\mathrm{m})^{-1}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 601,
|
||
"question": "The following electrical characteristics have been determined for both intrinsic and p-type extrinsic indium phosphide ( \\(\\mathrm{InP}\\) ) at room temperature:\n\\begin{tabular}{lccc}\n\\hline & \\(\\boldsymbol{\\sigma}\\left(\\boldsymbol{\\Omega} \\cdot \\mathbf{m}\\right)^{-1}\\) & \\(\\mathbf{n}\\left(\\mathbf{m}^{-3}\\right)\\) & \\(\\mathbf{p}\\left(\\mathbf{m}^{-3}\\right)\\) \\\\\n\\hline Intrinsic & \\(2.5 \\times 10^{-6}\\) & \\(3.0 \\times 10^{13}\\) & \\(3.0 \\times 10^{12}\\) \\\\\nExtrinsic (n-type) & \\(3.6 \\times 10^{-5}\\) & \\(4.5 \\times 10^{14}\\) & \\(2.0 \\times 10^{12}\\) \\\\\n\\hline\n\\end{tabular}\nCalculate electron and hole mobilities.",
|
||
"answer": "In order to solve for the electron and hole mobilities for \\(\\operatorname{InP}\\), we must write conductivity expressions for the two materials, of the form of Equation.,\n\\[\n\\sigma=n|e| \\mu_{e}+p|e| \\mu_{h}\n\\]\nFor the intrinsic material\n\\[\n\\begin{aligned}\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& +\\left(3.0 \\times 10^{13} \\mathrm{~mol}^{-3}\\right)\\left(1.602 \\times 10^{-9} \\mathrm{C}\\right) \\mu_{h}\n\\end{aligned}\n\\]\nwhich reduces to\n\\[\n0.52=\\mu_{e}+\\mu_{h}\n\\]\nWhereas, for the extrinsic \\(\\operatorname{InP}\\)\n\\[\n\\begin{aligned}\n3.6 \\times 10^{-5}(\\Omega \\cdot \\mathrm{m})^{-1} & =\\left(4.5 \\times 10^{14} \\mathrm{~m}^{-3}\\right)\\left(1.602times 10^{-19} \\mathrm{C}\\right) \\mu_{e}\n\\end{aligned}\n\\]\n\\[\n\\begin{aligned}\n+ & \\left(2.0 \\times 10^{12} \\mathrm{~m}^{-3}\\right)\\left(1.602\\times 10^{-19} \\mathrm{C}\\right) \\mu_{h}\n\\end{aligned}\n&\n\\]\nwhich may be simplified to\n\\[\n112.4=225 \\mu_{e}+\\mu_{h}\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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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+}\\).)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 606,
|
||
"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. (b) For solid lead titanate \\(\\left(\\mathrm{PbTiO}_{3}\\right)\\), gaseous neon, diamond, solid \\(\\mathrm{KCl}\\), and liquid \\(\\mathrm{NH}_{3}\\) what kind(s) of polarization is (are) possible? Why?",
|
||
"answer": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.\n(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for \\(\\mathrm{BaTiO}_{3}\\).\nOnly electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments.\nOnly 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.\nBoth electronic and ionic polarizations will be found in solid \\(\\mathrm{KCl}\\), since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material.\nBoth electronic and orientation polarizations are found in liquid \\(\\mathrm{NH}_{3}\\). The \\(\\mathrm{NH}_{3}\\) molecules have permanent dipole moments that are easily oriented in the liquid state.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 607,
|
||
"question": "For aluminum, the heat capacity at constant volume \\(C_{\\mathrm{v}}\\) at \\(30 \\mathrm{~K}\\) is \\(0.81 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\\), and the Debye temperature is \\(375 \\mathrm{~K}\\). Estimate the specific heat (a) at \\(50 \\mathrm{~K}\\) and (b) at \\(425 \\mathrm{~K}\\).",
|
||
"answer": "(a) For aluminum, \\(C_{\\mathrm{v}}\\) at \\(50 \\mathrm{~K}\\) may be approximated by Equation, since this temperature is significantly below the Debye temperature (375 K). The value of \\(C_{\\mathrm{v}}\\) at \\(30 \\mathrm{~K}\\)\nthe constant \\(A\\) as\n\\[\nA=\\frac{C_{v}}{T^{3}}=\\frac{0.81 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}}{(30 \\mathrm{~K})^{3}}=3.00 \\times 10^{-5} \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}^{4}\n\\]\nTherefore, at \\(50 \\mathrm{~K}\\)\n\\[\nC_{\\mathrm{v}}=A T^{3}=\\left(3.00 \\times 10^{-5} \\mathrm{~J} / 0 \\mathrm{~mol}-\\mathrm{K}^{4}\\right)(50 \\mathrm{~K})^{3}=3.75 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\n\\]\nand\n\\[\nc_{\\mathrm{v}}=\\left(3.75 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{L}^{2}\\right)(1 \\mathrm{~mol} / 26.98 \\mathrm{~g})(1000 \\mathrm{~g} / \\mathrm{kg})=139 \\mathrm{~J} / \\mathrm{kg}-\\mathrm{K}\n\\]\n(b) Since \\(425 \\mathrm{~K}\\) is above the Debye temperature, a good approximation for \\(C_{\\mathrm{v}}\\) is\n\\[\n\\begin{aligned}\nC_{\\mathrm{v}} & =3 R \\\\\n& =(3)(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})=24.9 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K}\n\\end{aligned}\n\\]\nAnd, converting this to specific heat\n\\[\nc_{\\mathrm{v}}=(24.9 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{L})(1 \\mathrm{~mol} / 26.98 \\mathrm{~g})(1000 \\mathrm{~g} / \\mathrm{kg})=923 \\mathrm{~J} / \\mathrm{kg}-\\mathrm{K}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 608,
|
||
"question": "(a) Briefly explain why \\(C_{v}\\) rises with increasing temperature at temperatures near \\(0 \\mathrm{~K}\\). (b) Briefly explain why \\(C_{v}\\) becomes virtually independent of temperature at temperatures far removed from \\(0 \\mathrm{~K}\\).",
|
||
"answer": "(a) The reason that \\(C_{v}\\) rises with increasing temperature at temperatures near \\(0 \\mathrm{~N}\\) 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.\n(b) At temperatures far removed from \\(0 \\mathrm{~K}, C_{v}\\) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 612,
|
||
"question": "(a) Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric materials, rendering them more thermally insulative. (b) Briefly explain how the degree of crystallinity affects the thermal conductivity of polymeric materials and why.",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 614,
|
||
"question": "For each of the following pairs of materials, decide which has the larger thermal conductivity. Justify your choices.\n(a) Pure copper; aluminum bronze (95 wt\\% Cu-5 wt\\% Al).\n(b) Fused silica; quartz.\n(c) Linear polyethylene; branched polyethylene.\n(d) Random poly(styrene-butadiene) copolymer; alternating poly(styrene-butadiene) copolymer.",
|
||
"answer": "This question asks for us to decide, for each of several pairs of materials, which has the larger thermal conductivity and why.\n(a) 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.\n(b) 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.\n(c) 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.\n(d) 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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 615,
|
||
"question": "(a) Explain the two sources of magnetic moments for electrons.\n(b) Do all electrons have a net magnetic moment? Why or why not?\n(c) Do all atoms have a net magnetic moment? Why or why not?",
|
||
"answer": "(a) The two sources of magnetic moments for electrons are the electron's orbital motion around the nucleus, and also, its spin.\n(b) Each electron will have a net magnetic moment from spin, and possibly, orbital contributions, which do not cancel for an isolated atom.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 618,
|
||
"question": "The formula for yttrium iron garnet \\(\\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right)\\) may be written in the form \\(\\mathrm{Y}_{3}^{+} \\mathrm{Fe}_{4}^{a} \\mathrm{Fe}_{d}^{a} \\mathrm{O}_{12}\\), where the superscripts \\(a, c\\), and \\(d\\) represent different sites on which the \\(Y^{3+}\\) and \\(\\mathrm{Fe}^{3+}\\) ions are located. The spin magnetic moments for the \\(Y^{3+}\\) and \\(\\mathrm{Fe}^{3+}\\)\n moments for the \\(Y^{3+}\\) and \\(\\mathrm{Fe}^{3-}\\) ions positioned in the \\(a\\) and \\(c\\) sites are oriented parallel to one another and antiparallel to the \\(\\mathrm{Fe}^{3+}\\) ions in \\(d\\) sites. Compute the number of Bohr magnetons associated with each \\(Y^{3+}\\) ion, given the following information: (1) each unit cell consists of eight formula \\(\\left(Y_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right)\\), units; (2) the unit cell is cubic with an edge length of \\(1.2376 \\mathrm{~nm}\\); (3) the saturation magnetization for this material is \\(1.0 \\times 10^{4} \\mathrm{~A} / \\mathrm{m}\\); and (4) assume that there are 5 Bohr magnetons associated with each \\(\\mathrm{Fe}^{3+}\\) ion.",
|
||
"answer": "For this problem we are given that yttrium iron garnet may be written in the form \\(\\mathrm{Y}_{3}^{+} \\mathrm{\\mathrm{Fe}_{4}^{a} \\mathrm{Fe}_{d}^{a} O_{12}\\) where the superscripts \\(a, c\\), and \\(d\\) represent different sites on which by \\(\\mathrm{Y}^{3+}\\) and \\(\\mathrm{Fe}^{3+}\\) ions are located, and that the spin magnetic moments for the ions on \\(a\\) and \\(c\\) sites are oriented parallel to one another and antiparallel to the \\(\\mathrm{Fe}^{3+}\\) ions on the \\(d\\) sites. We are to determine the number of Bohr magnetons associated with each \\(\\mathrm{Y}^{3+}\\) ion given that each unit cell consists of eight formula units, the unit cell is cubic with an edge length of \\(1.2376 \\mathrm {~nm}\\), the saturation magnetization for the material is \\(1.0 \\times 10^{4} \\mathrm{~A}\\) /m, and that there are 5 Bohr magnetons for each \\(\\mathrm{Fe}^{3+}\\) ion.\nThe first thing to do is to calculate the number of Bohr magnetons per unit cell, which we will denote \\(n_{\\mathrm{B}}\\). Solving for \\(n_{\\mathrm{B}}\\) using Equation, we get\n\\[\n\\begin{aligned}\nn_{\\mathrm{B}} & =\\frac{M_{\\mathrm{s}} a^{3}}{\\mu_{\\mathrm{B}}} \\\\\n& =\\frac{\\left(1.0 \\times 10^{4} \\mathrm{~A} / \\mathbf{m}\\right)\\left(1.2376 \\times 10^{-9} \\mathrm{~m}\\right)^{3}}{9.27 \\times 10^{-24} \\mathrm{~A} \\cdot \\mathrm{m}^{2} / \\mathrm{BM}}=2.04 \\mathrm{Bohr} \\mathrm{magnetons} / \\mathrm{unit}\\) cell\n\\end{aligned}\n\\]\nNow, there are 8 formula units per unit cell or \\(\\frac{2.04}{8}=0.255 \\mathrm{Bohr}\\) magnetons per formula unit. Furthermore, for each formula unit there are two \\(\\mathrm{Fe}^{3+}\\) ions on \\(a\\) sites and three \\(\\mathrm{Fe}^{3+}\\) on \\(d\\) sites which magnetic moments are aligned antiparallel. Since there are 5 Bohr magnetons associated with each \\(\\mathrm{Fe}^{3 +}\\) ion, the net magnetic moment contribution per formula unit from the \\(\\mathrm{Fe}^{3+}\\) ions is \\(5 \\mathrm{Bohr}\\) magnetons. This contribution is antiparallel to the contribution from the \\(\\mathrm{Y}^{3+}\\) ions, and since there are three \\(\\mathrm{Y}^{3+}\\) ions per formula unit, then\n\\[\n\\text { No. of Bohr magnetons } / \\mathrm{Y}^{3+}=\\frac{0.255 \\mathrm{BM}+5 \\mathrm{BM}}{3}=1.75 \\mathrm{BM}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 619,
|
||
"question": "Briefly explain why the magnitude of the saturation magnetization decreases with increasing temperature for ferromagnetic materials, and why ferromagnetic behavior ceases above the Curie temperature.",
|
||
"answer": "For ferromagnetic materials, 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. 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 621,
|
||
"question": "Visible light having a wavelength of \\(6 \\times 10^{-7} \\mathrm{~m}\\) appears orange. Compute the frequency and energy of a photon of this light.",
|
||
"answer": "In order to compute the frequency of a photon of orange light, we must use Equation as\n\\[\n\\mathrm{v}=\\frac{c}{\\lambda}=\\frac{3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}}{6 \\times 10^{-7} \\mathrm{~m}}=5 \\times 10^{14} \\mathrm{~s}^{-1}\n\\]\nNow, for the energy computation, we employ Equation as follows:\n\\[\n\\begin{aligned}\nE & =\\frac{h c}{\\lambda}=\\frac{\\left(6.63 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}\\right)\\left(3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right)}{6 \\times 10^{-7} \\mathrm{~m}} \\\\\n& =3.31 \\times 10^{-19} \\mathrm{~J} \\quad(2.07 \\mathrm{eV})\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 622,
|
||
"question": "Distinguish between 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 628,
|
||
"question": "Briefly explain what determines the characteristic color of (a) a metal and (b) a transparent nonmetal.",
|
||
"answer": "(a) The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 633,
|
||
"question": "Potassium iodide (KI) exhibits predominantly ionic bonding. The \\(\\mathrm{K}^{+}\\)and \\(\\Gamma\\) ions have electron structures that are identical to which two inert gases?",
|
||
"answer": "The \\(\\mathrm{K}^{+}\\)ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon.\nThe \\(\\mathrm{I}^{-}\\)ion is an iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 639,
|
||
"question": "What type(s) of bonding would be expected for each of the following materials: solid \\\\ xenon, calcium fluoride \\(\\left(\\mathrm{CaF}_{2}\\right)\\), bronze, cadmium telluride (CdTe), rubber, and tungsten?",
|
||
"answer": "For solid xenon, the bonding is van der Waals since xenon is an inert gas.\nFor \\(\\mathrm{CaF}_{2}\\), the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of \\(\\mathrm{Ca}\\) and \\(\\mathrm{F}\\) in the periodic table.\nFor bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).\nFor CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of \\(\\mathrm{Cd}\\) and \\(\\mathrm{Te}\\) in the periodic table.\nFor rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)\nFor tungsten, the bonding is metallic since it is a metallic element from the periodic table.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.)",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 646,
|
||
"question": "Titanium (Ti) has an HCP crystal structure and a density of \\(4.51 \\mathrm{~g} / \\mathrm{cm}^{3}\\).\n(a) What is the volume of its unit cell in cubic meters?\n(b) If the \\(c / a\\) ratio is 1.58 , compute the values of \\(c\\) and \\(a\\).\n\\(\\underline{\\text {",
|
||
"answer": "(a) The volume of the Ti unit cell may be computed using a rearranged form of Equation as\n\\[\nV_{C}=\\frac{n A_{\\mathrm{T} \\mathrm{i}}}{N_{\\mathrm{A}}}\n\\]\nFor HCP, \\(n=6\\) atoms/unit cell, and the atomic weight for \\(\\mathrm{Ti}, A_{\\mathrm{T} \\mathrm{i}}=47.87 \\mathrm{~g} / \\mathrm{mol}\\) (from inside the front cover of the book). Thus,\n\\[\n\\begin{aligned}\nV_{C} & =\\frac{(6 \\text { atoms/unit cell })(47.87 \\mathrm{~g} / \\mathrm{mol})}{\\left(4.51 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol}\\right)} \\\\\n& =1.058 \\times 10^{-22} \\mathrm{~cm}^{3} / \\mathrm{unit} \\mathrm{cell}=1.058 \\times 10^{-28} \\mathrm{~m}^{3} / \\mathrm{unit} \\mathrm{cell}\n\\end{aligned}\n\\]\n(b) From Equation, for HCP the unit cell volume for is\n\\[\nV_{C}=\\frac{3 a^{2} c \\sqrt{3}}{2}\n\\]\nsince, for \\(\\mathrm{Ti}, c=1.58 a\\), then\n\\[\nV_{C}=\\frac{3(1.58) a^{3} \\sqrt{3}}{2}=1.058 \\quad 10^{22} \\mathrm{~cm}^{3} / \\mathrm{unit} \\text { cell }\n\\]\nNow, solving for \\(a\\)\\[\n\\begin{array}{l}\na=\\left[\\begin{array}{c}\n\\left(2\\right)\\left(1.058\\,\\times\\,10^{\\,\\,22}\\,\\mathrm{cm}^{3}\\right)\\\\\n\\left(3\\right)\\left(1.58\\right)\\left(\\sqrt{3}\\right)\n\\end{array}\\right]^{1/3}\n\\\\\n\\\\=2.96\\quad 10^{\\,\\,8}\\,\\mathrm{cm}=0.296\\,\\mathrm{nm}\n\\end{array}\n\\]\nAnd finally, the value of \\(c\\) is\n\\[\nc=1.58 a=(1.58)(0.296 \\mathrm{~nm})=0.468 \\mathrm{~nm}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 647,
|
||
"question": "Magnesium (Mg) has an HCP crystal structure and a density of \\(1.74 \\mathrm{~g} / \\mathrm{cm}^{3}\\).\n(a) What is the volume of its unit cell in cubic centimeters?\n(b) If the \\(c / a\\) ratio is 1.624 , compute the values of \\(c\\) and \\(a\\).",
|
||
"answer": "(a) The volume of the \\(\\mathrm{Mg}\\) unit cell may be computed using a rearranged form of Equation as\n\\[\nV_{C}=\\frac{n A_{\\mathrm{Mg}}}{N_{\\mathrm{A}}}\n\\]\nNow, for HCP, \\(n=6\\) atoms/unit cell, and the atomic weight for \\(\\mathrm{Mg}, A_{\\mathrm{Mg}}=24.31 \\mathrm{~g} / \\mathrm{mol}\\) (from inside the front cover of the book). Thus,\n\\[\n\\begin{aligned}\nV_{C} & =\\frac{(6 \\text { atoms/unit cell})(24.31 \\mathrm{~g} / \\mathrm{mol})}{\\left(1.74 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol}\\right)} \\\\\n& =1.39 \\times 10^{-22} \\mathrm{~cm}^{3} / \\mathrm{unit} \\mathrm{cell}=1.39 \\times 10^{-28} \\mathrm{~m}^{3} / \\mathrm{unit} \\mathrm{cell}\n\\end{aligned}\n\\]\n(b) From Equation, for HCP the unit cell volume for is\n\\[\nV_{C}=\\frac{3 a^{2} c \\sqrt{3}}{2}\n\\]\nbut, since, for \\(\\mathrm{Mg}, c=1.624 a\\), then\n\\[\nV_{C}=\\frac{3(1.624) a^{3} \\sqrt{3}}{2}=1.39 \\times 10^{-22} \\mathrm{~cm} 3 / \\text { unit cell }\n\\]\nNow, solving for \\(a\\)\\[\n\\begin{array}{l}\na=\\left[\\frac{(2)(1.39\\times10^{\\,\\,22}\\,\\mathrm{cm}^{3})}{(3)(1.624)(\\sqrt{3})}\\right]^{1/3}\\\\\n\\\\=3.21\\quad 10^{\\,\\,8}\\,\\mathrm{cm}=0.321\\,\\mathrm{nm}\\\\\n\\end{array}\n\\]\nAnd finally, the value of \\(c\\) is\n\\[\nc=1.624 a=(1.624)(0.321\\mathrm{\\ nm})=0.521\\mathrm{\\ nm}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 648,
|
||
"question": "The unit cell for uranium \\((U)\\) has orthorhombic symmetry, with \\(a, b\\), and \\(c\\) lattice parameters of \\(0.286,0.587\\), and \\(0.495 \\mathrm{~nm}\\), respectively. If its density, atomic weight, and atomic radius are \\(19.05 \\mathrm{~g} / \\mathrm{cm}^{3}, 238.03 \\mathrm{~g} / \\mathrm{mol}\\), and \\(0.1385 \\mathrm{~nm}\\), respectively, compute the atomic packing factor.\n\\title{",
|
||
"answer": "In order to determine the APF for \\(\\mathrm{U}\\), we need to compute both the unit cell volume \\(\\left(V_{C}\\right)\\) which is just the product of the three unit cell parameters, as well as the total sphere volume \\(\\left(V_{S}\\right)\\) which is just the product of the volume of a single sphere and the number of spheres in the unit cell \\((n)\\). The value of \\(n\\) may be calculated from a rearranged form of Equation as\n\\[\n\\begin{aligned}\nn & =\\frac{V_{C} N_{\\mathrm{A}}}{A_{\\mathrm{U}}} \\\\\n& =\\frac{\\left(19.05 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)(2.86)(5.87)(4.95)\\left(\\begin{array}{cc}\n10 & 24 \\mathrm{~cm}^{3}\n\\end{array}\\right)(6.022 \\quad 10^{23} \\mathrm{atoms} / \\mathrm{mol})}{238.03 \\mathrm{~g} / \\mathrm{mol}} \\\\\n& =4.01 \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell}\n\\end{aligned}\n\\]\nTherefore, we determine the atomic packing factor using Equation as\n\\[\n\\begin{array}{c}\n\\mathrm{APF}=\\frac{V_{S}}{V_{C}}=\\frac{\\left(4\\right)\\left(\\frac{4}{3} \\pi R^{3}\\right)}{(a)(b)(c)} \\\\\n=\\frac{\\left(4\\left[\\frac{4}{3}(\\pi)(1.385 \\times 10^{-8} \\mathrm{~cm})^{3}\\right]\\right.}{\\left(2.86)(5.87)(4.95)(\\times 10^{-24} \\mathrm{~cm}^{3}\\right)} \\\\\n=0.536",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 649,
|
||
"question": "Indium (In) has a tetragonal unit cell for which the \\(a\\) and \\(c\\) lattice parameters are 0.459 and \\(0.495 \\mathrm{~nm}\\), respectively.\n(a) If the atomic packing factor and atomic radius are 0.693 and \\(0.1625 \\mathrm{~nm}\\), respectively, determine the number of atoms in each unit cell.\n(b) The atomic weight of indium is \\(114.82 \\mathrm{~g} / \\mathrm{mol}\\); compute its theoretical density.\n\\title{",
|
||
"answer": "(a) For indium, and from the definition of the APF is\n\\[\n\\mathrm{APF}=\\frac{V_{S}}{V_{C}}=\\frac{n\\left(\\frac{4}{3} \\pi R^{3}\\right)}{a^{2} c}\n\\]\nwe may solve for the number of atoms per unit cell, \\(n\\), as\n\\[\n\\begin{array}{c}\nn=\\frac{(\\mathrm{APF}) a^{2} c}{\\frac{4}{3} \\pi R^{3}} \\\\\n=\\frac{(0.693)(4.59)^{2}(4.95)\\left(10^{-24} \\mathrm{~cm}^{3}\\right)}{\\frac{4}{3} \\pi(1.625 \\times 10^{-8} \\mathrm{~cm})^{3}} \\\\\n=4.0 \\text { atoms/unit cell }\n\\end{array}\n\\]\n(b) In order to compute the density, we just employ Equation as\n\\[\n\\rho=\\frac{n A_{\\mathrm{In}}}{a^{2} c N_{\\mathrm{A}}}\n\\]\\[\n\\begin{array}{c}\n\\left.\\left(\\underline{4\\;atoms} / \\textrm{unit cell}\\right)\\left(\\underline{114.82\\;g / \\textrm{mol}}\\right)\\right.\\\\\n\\left.\\left.\\left[\\underline{\\left(4.59\\times10^{-8}\\; \\textrm{cm}\\right)^{2}\\left(\\underline{4.95\\times10^{-8}\\; \\textrm{cm}}\\right) / \\textrm{unit cell}\\right]\\right.\\left(\\underline{6.022\\times10^{23}\\; \\textrm{atoms} / \\textrm{mol}}\\right)\\right.\\\\\n\\\\=7.31\\; \\textrm{g/cm}^{3}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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 \\mathrm{~nm}\\), (a) determine the unit cell volume, and (b) calculate the theoretical density of Be and compare it with the literature value.",
|
||
"answer": "(a) We are asked to calculate the unit cell volume for Be. The volume of an HCP unit cell is provided by Equation as follows:\n\\[\nV_{C}=6 R^{2} c \\sqrt{3}\n\\]\nBut, \\(c=1.568 a\\), and \\(a=2 R\\), or \\(c=(1.568)(2 R)=3.136 R\\). Substitution of this expression for \\(c\\) into the above equation leads to\n\\[\n\\begin{aligned}\nV_{C} & =(6)(3.14) R^{3} \\sqrt{3} \\\\\n& =(6)(3.14)(\\sqrt{3})\\left[0.1143 \\times 10^{-7} \\mathrm{~cm}\\right]^{3}=4.87 \\times 10^{-23} \\mathrm{~cm}^{3} / \\mathrm{unit} \\mathrm{cell}\n\\end{aligned}\n\\]\n(b) The theoretical density of Be is determined, using Equation, as follows:\n\\[\n=\\frac{n A_{\\mathrm{Be}}}{V_{C} N_{\\mathrm{A}}}\n\\]\nFor HCP, \\(n=6\\) atoms/unit cell, and for \\(\\mathrm{Be}, A_{\\mathrm{Be}}=9.01 \\mathrm{~g} / \\mathrm{mol}\\) (as noted inside the front cover)and the value of \\(V_{C}\\) was determined in part (a). Thus, the theoretical density of Be is\n\\[\n\\rho=\\frac{(6 \\text { atoms/unit cell})(9.01 \\mathrm{~g} / \\mathrm{mol})}{\\left(4.87 \\times 10^{-23} \\mathrm{~cm}^{\\mathrm{3}} / \\text { unit cell }\\right)(6.022 \\times 10^{23} \\text { atoms } / \\mathrm{mol})}\n\\]\\[\n = 1.84\\, \\mathrm{g/cm}^3\n\\]\nThe literature value is \\(1.85 \\mathrm{~g} / \\mathrm{cm}^{3}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 654,
|
||
"question": "A hypothetical AX type of ceramic material is known to have a density of \\(2.10 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and a unit cell of cubic symmetry with a cell edge length of \\(0.57 \\mathrm{~nm}\\). The atomic weights of the \\(A\\) and \\(X\\) elements are 28.5 and \\(30.0 \\mathrm{~g} / \\mathrm{mol}\\), respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: sodium chloride, cesium chloride, or zinc blende? Justify your choice(s).",
|
||
"answer": "We are asked to specify possible crystal structures for an AX type of ceramic material given its density \\(\\left(2.10 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\), that the unit cell has cubic symmetry with edge length of \\(0.57 \\mathrm{~nm}\\), and the atomic weights of the \\(\\mathrm{A}\\) and \\(\\mathrm{X}\\) elements (28.5 and \\(30.0 \\mathrm{~g} / \\mathrm{\\mathrm{mol}}\\), respectively). To determine the possible crystal structure(s), it is necessary to compute the value of \\(n^{\\prime}\\) in Equation; if \\(n^{\\prime}=1\\) the crystal structure is cesium chloride, whereas if \\(n^{\\prime}=4\\), sodium chloride and zinc blende are the possible crystals structures. Using Equation we determine the value of \\(n^{\\prime}\\) as follows:\n\\[\n\\begin{aligned}\nn^{\\prime} & =\\frac{\\rho V_{C} \\cdot N_{\\mathrm{A}}}{\\sum A_{\\mathrm{C}}+\\sum A_{\\mathrm{A}}} \\\\\n& =\\frac{\\left(2.10 \\mathrm{~g} / \\mathrm{cm}^{2}\\right)\\left[\\left(5.70 \\times 10^{-8} \\mathrm{~cm}\\right)^{3} / \\text { unit cell }\\right]\\left(6.022 \\times 10^{23} \\text { formula units } / \\mathrm{mol}\\right)}{(30.0 \\mathrm{~g} / \\mathrm{mol}+28.5 \\mathrm{~g} / \\mathrm{mol})} \\\\\n& =4.00 \\text { formula units } / \\text { unit cell }\n\\end{aligned}\n\\]\nOf the three possible crystal structures, only sodium chloride and zinc blende have four formula units per unit cell, and therefore, are possibilities.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 655,
|
||
"question": "Iron titanate, \\(\\mathrm{FeTiO}_{3}\\), forms in the ilmenite crystal structure that consists of an HCP arrangement of \\(\\mathrm{O}^{2}\\)-ions.\n(a) Which type of interstitial site will the \\(\\mathrm{Fe}^{2+}\\) ions occupy? Why?\n(b) Which type of interstitial site will the \\(\\mathrm{Ti}^{4+}\\) ions occupy? Why?\n(c) What fraction of the total tetrahedral sites will be occupied?\n(d) What fraction of the total octahedral sites will be occupied?",
|
||
"answer": "(a) We are first of all asked to cite, for \\(\\mathrm{FeTiO}_{3}\\), which type of interstitial site the \\(\\mathrm{Fe}^{2+}\\) ions will occupy. From Table, the cation-anion radius ratio is\n\\[\n\\frac{r_{\\mathrm{Fe}^{2+}}}{r_{\\mathrm{O}^{2-}}}=\\frac{0.077 \\mathrm{~nm}}{0.140 \\mathrm{~nm}}=0.550\n\\]\nSince this ratio is between 0.414 and 0.732 , the \\(\\mathrm{Fe}^{2+}\\) ions will occupy octahedral sites .\n(b) Similarly, for the \\(\\mathrm{Ti}^{4+}\\) ions, the titanium-oxygen ionic radius ratio is as follows:\n\\[\n\\frac{r_{\\mathrm{Ti}^{4+}}}{r_{\\mathrm{O}^{2-}}}=\\frac{0.601 \\mathrm{~nm}}{0.140 \\mathrm{~nm}}=n 436\n\\]\nSince this ratio is between 0.414 and 0.723 , the \\(\\mathrm{Ti}^{4+}\\) ions will also occupy octahedral sites.\n(c) Since both \\(\\mathrm{Fe}^{2+}\\) and \\(\\mathrm{Ti}^{4+}\\) ions occupy octahedral sites, no tetrahedral sites will be occupied.\n(d) For every \\(\\mathrm{FeTiO}_{3}\\) formula unit, there are three \\(\\mathrm{O}^{2-}\\) ions, and, therefore, three octahedral sites; since there is one ion each of \\(\\mathrm{Fe}^{2+}\\) and \\(\\mathrm{Ti}^{2+}\\) per formula unit, two-thirds of these octahedral sites will be occupied.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}}\\) \\\\",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 658,
|
||
"question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen.\n(a) Determine the concentration of \\(\\mathrm{Cl}\\) (in wt\\%) that must be added if this substitution occurs for \\(8 \\%\\) of all the original hydrogen atoms.\n(b) In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)?",
|
||
"answer": "(a) For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for \\(8 \\% \\mathrm{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 \\(\\mathrm{Cl}\\). Thus, the mass of these 50 carbon atoms, \\(m_{\\mathrm{C}}\\), is just\n\\[\nm_{\\mathrm{C}}=50\\left(A_{\\mathrm{C}}\\right)=(50)(12.01 \\mathrm{~g} / \\mathrm{mol})=600.5 \\mathrm{~g}\n\\]\nLikewise, for hydrogen and chlorine,\n\\[\n\\begin{array}{l}\nm_{\\mathrm{H}}=92\\left(A_{\\mathrm{H}}\\right)=(92)(1.008 \\mathrm{~g} / \\mathrm{mol})=92.74 \\mathrm{~g} \\\\\nm_{\\mathrm{Cl}}=8\\left(A_{\\mathrm{Cl}}\\right)=(8)(35.45 \\mathrm{~g} / \\mathrm{mol})=283.60 \\mathrm{~g}\n\\end{array}\n\\]\nThus, the concentration of chlorine, \\(C_{\\mathrm{Cl}}\\), is determined as follows:\n\\[\n\\begin{array}{l}\nC_{\\mathrm{Cl}}=\\frac{m_{\\mathrm{Cl}}}{m_{\\mathrm{C}}+m_{\\mathrm{H}}+m_{\\mathrm{Cl}}} \\times 100 \\\\\n=\\frac{283.60 \\mathrm{~g}}{600.5 \\mathrm{~g}+92.74 \\mathrm{~g}+283.60 \\mathrm{~g}} \\times 100=29.0 \\mathrm{wt} \\%\n\\end{array}\n\\](b) Chlorinated polyethylene differs from poly(vinyl chloride), in that, for PVC, (1) \\(25 \\%\\) of the side-bonding sites are substituted with \\(\\mathrm{Cl}\\), and (2) the substitution is probably much less random.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 659,
|
||
"question": "Compare thermoplastic and thermosetting polymers (a) on the basis of mechanical \\\\ characteristics upon heating and (b) according to possible molecular structures.",
|
||
"answer": "(a) Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers, harden upon heating, while further heating will not lead to softening.\n(b) Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 663,
|
||
"question": "(a) Determine the ratio of butadiene to acrylonitrile repeat units in a copolymer having a number-average molecular weight of \\(250,000 \\mathrm{~g} / \\mathrm{mol}\\) and a degree of polymerization of 4640 .\n(b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?",
|
||
"answer": "(a) This portion of the problem asks us to determine the ratio of butadiene to acrylonitrile repeat units in a copolymer having a weight-average molecular weight of 250,000 g/mol and a degree of polymerization of 4640. It first becomes necessary to calculate the average repeat unit molecular weight of the copolymer, \\(\\bar{m}\\), using a rearranged form of Equation as follows:\n\\[\n\\bar{m}=\\frac{\\bar{M}_{n}}{D P}=\\frac{250,000 \\mathrm{~g} / \\mathrm{mol}}{4640}=53.88 \\mathrm{~g} / \\mathrm{mol}\n\\]\nIf we designate \\(f_{b}\\) as the chain fraction of butadiene repeat units, since the copolymer consists of only two repeat unit types, the chain fraction of acrylonitrile repeat units \\(f_{a}\\) is just \\(1-f_{b}\\). Now, Equation for this copolymer may be written in the form\n\\[\n\\bar{m}=f_{b} m_{b}+f_{a} m_{a}=f_{b} m_{b}+\\left(1-f_{b}\\right) m_{a}\n\\]\nin which \\(m_{b}\\) and \\(m_{a}\\) are the repeat unit molecular weights for butadiene and acrylonitrile, respectively. Values of these repeat unit molecular weights are calculated as follows:\n\\[\n\\begin{array}{c}\nm_{b}=4\\left(A_{\\mathrm{C}}\\right)+6\\left(A_{\\mathrm{H}}\\right)=4(12.01 \\mathrm{~g} / \\mathrm{mol})+6(1.008 \\mathrm{~g} / \\mathrm{mol})=54.09 \\mathrm{~g} / \\mathrm{mol} \\\\\nm_{a}=3\\left(A_{\\mathrm{C}}\\right)+3\\left(A_{\\mathrm{H}}\\right)+\\left(A_{\\mathrm{N}}\\right)=3(12.01 \\mathrm{~g} / \\mathrm{mol})+3(1.008 \\mathrm{~g} / \\mathrm{mol})+(14.01 \\mathrm{~g} / \\mathrm{mol}) \\\\\n=53.06 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]Solving for \\(f_{b}\\) in the above expression for \\(\\bar{m}\\) yields\n\\[\nf_{b}=\\frac{\\bar{m}-m_{a}}{m_{b}-m_{a}}\n\\]\nAnd since values for \\(\\bar{m}, m_{a}\\), and \\(m_{b}\\) are noted above, we determine the value of \\(f_{b}\\) as follows:\n\\[\nf_{b}=\\frac{53.88 \\mathrm{~g} / \\mathrm{mol}-53.06 \\mathrm{~g} / \\mathrm{mol}}{54.09 \\mathrm{~g} / \\mathrm{mol}-53.06 \\mathrm {~g} / \\mathrm{mol}}=0.80\n\\]\nFurthermore, \\(f_{a}=1-f_{b}=1-0.80=0.20\\); which means that the ratio is\n\\[\n\\frac{f_{b}}{f_{a}}=\\frac{0.80}{0.20}=4.0\n\\]\n(b) Of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit types is alternating; the ratio must be 1:1. Therefore, on the basis of the result in part (a), the possibilities for this copolymer are random, graft, and block.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 664,
|
||
"question": "Crosslinked copolymers consisting of \\(35 \\mathrm{wt} \\%\\) ethylene and \\(65 \\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.\n\\title{",
|
||
"answer": "For a copolymer consisting of \\(35 \\mathrm{wt} \\%\\) ethylene and \\(65 \\operatorname{wt} \\%\\) propylene, we are asked to determine the fraction of both repeat unit types.\nIn \\(100 \\mathrm{~g}\\) of this material, there are \\(35 \\mathrm{~g}\\) of ethylene and \\(65 \\mathrm{~g}\\) of propylene. The ethylene \\(\\left(\\mathrm{C}_{2} \\mathrm{H}_{4}\\right)\\) molecular weight is\n\\[\n\\begin{aligned}\nm(\\text { ethylene }) & =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{aligned}\n\\]\nThe propylene \\(\\left(\\mathrm{C}_{3} \\mathrm{H}_{6}\\right)\\) molecular weight is\n\\[\n\\begin{aligned}\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{aligned}\n\\]\nWe now determine the number of moles of each repeat unit type. In \\(100 \\mathrm{~g}\\) of this material, the number of moles of ethylene \\(\\left[n_{m}(\\right.\\) ethylene \\(\\left.)\\right]\\) is computed as follows:\n\\[\nn_{m}(\\text { ethylene })=\\frac{35 \\mathrm{~g}}{28.05 \\mathrm{~g} / \\mathrm{mol}}=1.25 \\mathrm{~mol}\n\\]\nFurthermore, for propylene\n\\[\nn_{m}(\\text { propylene })=\\frac{65 \\mathrm{~g}}{42.08 \\mathrm{~g} / \\mathrm{mol}}=1.54 \\mathrm{~mol}\n\\]Thus, the fraction of the ethylene repeat unit, \\(f\\) (ethylene), is determined as follows:\n\\[\n\\begin{aligned}\nf(\\text { ethylene }) & =\\frac{n_{m}(\\text { ethylene })}{n_{m}(\\text { ethylene })+n_{m}(\\text { propylene })} \\\\\n& =\\frac{1.25 \\mathrm{~mol}}{1.25 \\mathrm{~mol}+1.54 \\mathrm{~mol}}=0.45\n\\end{aligned}\n\\]\nLikewise, for propylene\n\\[\n\\begin{aligned}\nf(\\text { propylene }) & =\\frac{n_{m}(\\text { propylene })}{n_{m}(\\text { ethylene })+n_{m}(\\operatorname{propylene})} \\\\\n& =\\frac{1.54 \\mathrm{~mol}}{1.25 \\mathrm{~mol}+1 \\cdot 54 \\mathrm{~mol}}=0.55\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 665,
|
||
"question": "For each of the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.\n(a) Linear and atactic poly(vinyl chloride); linear and isotactic polypropylene\n(b) Linear and syndiotactic polypropylene; crosslinked cis-polyisoprene\n(c) Network phenol-formaldehyde; linear and isotactic polystyrene\n(d) Block poly(acrylonitrile-isoprene) copolymer; graft poly(chloroprene-isobutylene) copolymer",
|
||
"answer": "(a) 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.\n(b) 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.\n(c) 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.\n(d) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 667,
|
||
"question": "(a) Calculate the fraction of atom sites that are vacant for copper \\((\\mathrm{Cu})\\) at its melting temperature of \\(1084^{\\circ} \\mathrm{C}(1357 \\mathrm{~K})\\). Assume an energy for vacancy formation of \\(0.90 \\mathrm{eV} / \\mathrm{atom}\\).\n(b) Repeat this calculation at room temperature (298 K).\n(c) What is ratio of \\(N_{V} / N(1357 \\mathrm{~K})\\) and \\(N_{V} / N(298 \\mathrm{~K})\\) ?",
|
||
"answer": "(a) In order to compute the fraction of atom sites that are vacant in copper at \\(1357 \\mathrm{~K}\\), we must employ Equation. As stated in the problem, \\(Q_{V}=0.90 \\mathrm{eV} /\\) atom. Thus,\n\\[\n\\begin{aligned}\n\\frac{N_{V}}{N}= & \\exp \\left(-\\frac{Q_{V}}{k T}\\right)=\\exp \\left[-\\frac{0.90 \\mathrm{eV} / \\mathrm{atom}}{\\left(8.62 \\times 10^{-5} \\mathrm{eV} / \\mathrm{atom}-\\mathrm{K}\\right)(1357 \\mathrm{~K})}\\right] \\\\\n& =4.56 \\times 10^{-4}=N_{V} / N(1357 \\mathrm{~K})\n\\end{aligned}\n\\]\n(b) We repeat this computation at room temperature (298 K), as follows:\n\\[\n\\begin{aligned}\n\\frac{N_{V}}{N}= & \\left.\\exp \\left(-\\frac{Q_{V}}{k T}\\right)\\right|=\\exp \\left[-\\frac{0.90 \\mathrm{eV} / \\text { atom }}{\\left(8.62 \\times 10^{-5} \\mathrm{e} \\mathrm{V} / \\text { atom }-\\mathrm{K}\\right)(298 \\mathrm{~K})}\\right] \\\\\n& =6.08 \\times 10^{-16}=N_{V} / N(298 \\mathrm{~K})\n\\end{aligned}\n\\]\n(c) And, finally the ratio of \\(N_{V} / N(1357 \\mathrm{~K}\\) ) and \\(N_{V} / N(298 \\mathrm{~K})\\) is equal to the following:\n\\[\n\\frac{N_{V} / N(1357 \\mathrm{~K})}{N_{V} / N(298 \\mathrm{~K})}=\\frac{4.56 \\times 10^{-4}}{6.08 \\times 10^{-16}}=7.5 \\times 10^{11}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 671,
|
||
"question": "If cupric oxide \\((\\mathrm{CuO})\\) is exposed to reducing atmospheres at elevated temperatures, some of the \\(\\mathrm{Cu}^{2+}\\) ions will become \\(\\mathrm{Cu}^{+}\\).\n(a) Under these conditions, name one crystalline defect that you would expect to form in order to maintain charge neutrality.\n(b) How many \\(\\mathrm{Cu}^{+}\\)ions are required for the creation of each defect?\n(c) How would you express the chemical formula for this nonstoichiometric material?",
|
||
"answer": "(a) For a \\(\\mathrm{Cu}^{2+} \\mathrm{O}^{2-}\\) compound in which a small fraction of the copper ions exist as \\(\\mathrm{Cu}^{+}\\), for each \\(\\mathrm{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 \\(\\mathrm{Cu}^{2+}\\) interstitials or \\(\\mathrm{O}^{2-}\\) vacancies.\n(b) There will be two \\(\\mathrm{Cu}^{+}\\)ions required for each of these defects.\n(c) The chemical formula for this nonstoichiometric material is \\(\\mathrm{Cu}_{1+} \\mathrm{O}\\) or \\(\\mathrm{CuO}_{1-x}\\), where \\(x\\) is some small fraction.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 672,
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100\\%) solubility with \\(\\mathrm{MgO}\\) ? Explain your answers.\n(a) \\(\\mathrm{FeO}\\)\n(b) \\(\\mathrm{BaO}\\)\n(c) \\(\\mathrm{PbO}\\)\n(d) \\(\\mathrm{CoO}\\)",
|
||
"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 \\((< \\pm 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. Inasmuch as crystal structure information for the compounds in this problem, for the most part, is not provided in the text, we will base our criteria only on the relative ion sizes and charges on the anions and cations. Because all of these ceramics are oxides and the charge on all of the cations is +2 , in making solubility determinations, only the relative sizes of the cations will be considered.\n(a) For \\(\\mathrm{FeO}\\), the ionic radii of the \\(\\mathrm{Mg}^{2+}\\) and \\(\\mathrm{Fe}^{2+}\\) (found inside the front cover of the book) are \\(0.072 \\mathrm{~nm}\\) and \\(0.077 \\mathrm{~nm}\\), respectively. Therefore the percentage difference in ionic radii, \\(\\Delta r \\%\\) is determined as follows:\n\\[\n\\begin{aligned}\n\\Delta r \\% & =\\frac{\\eta_{\\mathrm{Fe} 2^{+}}-\\eta_{\\mathrm{Mg} 2^{+}}}{\\eta_{\\mathrm{Fe} 2^{+}}} \\times 100 \\\\\n& =\\frac{0.077 \\mathrm{~nm}-0.072 \\mathrm{~nm}}{0.077 \\mathrm{~nm}} \\times 100=6.5 \\%\n\\end{aligned}\n\\]\nwhich value is with the acceptable range for a high degree of solubility. Furthermore, as Table 3.5 notes, the crystal structures of both \\(\\mathrm{MgO}\\) and \\(\\mathrm{FeO}\\) are rock salt; therefore, it is expected thatFeO and \\(\\mathrm{MgO}\\) form high degrees of solid solubility. Experimentally, these two ceramics exhibit \\(100 \\%\\) solubility.\n(b) For \\(\\mathrm{BaO}\\), the ionic radii of the \\(\\mathrm{Mg}^{2+}\\) and \\(\\mathrm{Ba}^{2+}\\) (found inside the front cover of the book) are \\(0.072 \\mathrm{~nm}\\) and \\(0.136 \\mathrm{~nm}\\), respectively. Therefore the percentage difference in ionic radii, \\(\\Delta r \\%\\) is determined as follows:\n\\[\n\\begin{aligned}\n\\Delta r \\% & =\\frac{\\gamma_{\\mathrm{Ba} 2^{+}}-\\gamma_{\\mathrm{Mg} 2^{+}}}{\\gamma_{\\mathrm{Ba} 2^{+}}} \\times 100 \\\\\n& =\\frac{0.136 \\mathrm{~nm}-0.072 \\mathrm{~nm}}{0.136 \\mathrm{~nm}} \\times 100=47 \\%\n\\end{aligned}\n\\]\nThis \\(\\Delta r \\%\\) value is much larger than the \\(\\pm 15 \\%\\) range, and, therefore, \\(\\mathrm{BaO}\\) is not expected to experience any appreciable solubility in \\(\\mathrm{MgO}\\). Experimentally, the solubility of \\(\\mathrm{BaO}\\) in \\(\\mathrm{MgO}\\) is very small; this is also the case for the solubility of \\(\\mathrm{MgO}\\) in \\(\\mathrm{BaO}\\).\n(c) For \\(\\mathrm{PbO}\\), the ionic radii of the \\(\\mathrm{Mg}^{2}+\\) and \\(\\mathrm{Pb}^{2+}\\) (found inside the front cover of the book) are \\(0 . 072 \\mathrm{~nm}\\) and \\(0.120 \\mathrm{~nm}\\), respectively. Therefore the percentage difference in ionic radii, Ad\\% is determined as follows:\n\\[\n\\begin{aligned}\n\\Delta r \\% & =\\overbrace{\\mathrm{Pb}^{2+}-\\gamma_{\\mathrm{Mg}^{2+}}}{\\mathrm{Pb}^{2+}} \\times 100 \\\\\n& =\\frac{0.120 \\mathrm{~nm}-0.072 \\mathrm{~nm}}{0 . 120 \\mathrm{~nm}} \\times 100=40 \\%\n\\end{aligned}\n\\]\nThis Ad\\% value is much larger than the \\(\\pm 15 \\%\\) range, and, therefore,, \\(\\mathrm{PbO}\\) is not expected to experience any appreciable solubility in \\(\\mathrm{ MgO}\\).(d) For CoO, the ionic radii of the \\(\\mathrm{Mg}^{2+}\\) and \\(\\mathrm{Co}^{2+}\\) (found inside the front cover of the book) are \\(0.072 \\mathrm{~nm}\\) and \\(0.072 \\mathrm{~nm}\\), respectively. Therefore the percentage difference in ionic radii, \\(\\Delta r \\%\\) is determined as follows:\n\\[\n\\begin{array}{l}\n\\Delta r \\%=\\frac{r_{\\mathrm{Co} 2^{+}}-r_{\\mathrm{Mg} 2^{+}}}{r_{\\mathrm{Co} 2^{+}}} \\times 100 \\\\\n\\quad=\\frac{0.072 \\mathrm{~nm}-0.072 \\mathrm{~nm}}{0.072 \\mathrm{~nm}} \\times 100=0 \\%\n\\end{array}\n\\]\nwhich value is, of course, within the acceptable range for a high degree (probably 100\\%) of solubility.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 673,
|
||
"question": "(a) Suppose that \\(\\mathrm{CaO}\\) is added as an impurity to \\(\\mathrm{Li}_{2} \\mathrm{O}\\). If the \\(\\mathrm{Ca}^{2+}\\) substitutes for \\(\\mathrm{Li}^{+}\\), what kind of vacancies would you expect to form? How many of these vacancies are created for every \\(\\mathrm{Ca}^{2+}\\) added?\n(b) Suppose that \\(\\mathrm{CaO}\\) is added as an impurity to \\(_{\\mathrm{Cl}} 2\\). If the \\(\\mathrm{O}^{2-}\\) substitutes for \\(\\mathrm{Cl}^{-}\\), what kind of vacancies would you expect to form? How many of these vacancies are createdfor every \\(\\mathrm{O}^{2-}\\) added?",
|
||
"answer": "(a) For \\(\\mathrm{Ca}^{2+}\\) substituting for \\(\\mathrm{Li}^{+}\\)in \\(\\mathrm{Li}_{2} \\mathrm{O}\\), lithium vacancies would be created. For each \\(\\mathrm{Ca}^{2+}\\) substituting for \\(\\mathrm{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 \\(\\mathrm{Ca}^{2+}\\) ion added, a single lithium vacancy is formed.\n(b) For \\(\\mathrm{O}^{2-}\\) substituting for \\(\\mathrm{Cl}^{-}\\)in \\(\\mathrm{CaCl}_{2}\\), chlorine vacancies would be created. For each \\(\\mathrm{O}^{2-}\\) substituting for a \\(\\mathrm{Cl}^{-}\\), one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one \\(\\mathrm{O}^{2-}\\) ion will lead to the formation of one chlorine vacancy.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 674,
|
||
"question": "Atomic radius, crystal structure, electronegativity, and the most common valence are given in the following table for several elements; for those that are nonmetals, only atomic radii are indicated.\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline & Atomic Radius & \\multicolumn{3}{|c|}{ Crystal } \\\\\n\\hline Element & (nm) & Structure & Electronegativity & Valence \\\\\n\\hline \\(\\mathrm{Ni}\\) & 0.1246 & FCC & 1.8 & +2 \\\\\n\\hline\\(C\\) & 0.071 & & & \\\\\n\\hline\\(H\\) & 0.046 & & & \\\\\n\\hline\\(O\\) & 0.060 & & & \\\\\n\\hline \\(\\mathrm{Ag}\\) & 0.1445 & FCC & 1.4 & +1 \\\\\n\\hline \\(\\mathrm{Al}\\) & 0.1431 & FCC & 1.5 & +3 \\\\\n\\hline\\(C o\\) & 0.1253 & \\(H C P\\) & 1.7 & +2 \\\\\n\\hline\\(C r\\) & 0.1249 & \\(B C C\\) & 1.6 & +3 \\\\\n\\hline \\(\\mathrm{Fe}\\) & 0.1241 & \\(B C C\\) & 1.7 & +2 \\\\\n\\hline\\(P t\\) & 0.1387 & \\(F C C\\) & 1.5 & +2 \\\\\n\\hline \\(\\mathrm{Zn}\\) & 0.1332 & \\(H C P\\) & 1.7 & +2 \\\\\n\\hline\n\\end{tabular}\nWhich of these elements would you expect to form the following with nickel:\n(a) a substitutional solid solution having complete solubility\n(b) a substitutional solid solution of incomplete solubility\n(c) an interstitial solid solution",
|
||
"answer": "For complete substitutional solubility the four Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between \\(\\mathrm{Ni}\\) and the other element \\((\\Delta \\mathrm{R} \\%)\\) must be less than \\(\\pm 15 \\%\\);(2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same.\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline \\multirow[b]{2}{*}{ Element } & \\multirow[b]{2}{*}{\\(\\Delta \\mathrm{R} \\%\\)} & \\multirow{2}{*}{\\begin{tabular}{c} \nCrystal \\\\\nStructure\n\\end{tabular}} & \\multicolumn{2}{|c|}{\\(\\Delta\\) Electro- } \\\\\n\\hline & & & \\(\\underline{\\text { negativeity }}\\) & Valence \\\\\n\\hline \\(\\mathrm{Ni}\\) & & FCC & & \\(2+\\) \\\\\n\\hline \\(\\mathrm{C}\\) & -43 & & & \\\\\n\\hline \\(\\mathrm{H}\\) & -63 & & & \\\\\n\\hline \\(\\mathrm{O}\\) & -52 & & & \\\\\n\\hline \\(\\mathrm{Ag}\\) & +16 & FCC & -0.4 & \\(1+\\) \\\\\n\\hline \\(\\mathrm{Al}\\) & +15 & FCC & -0.3 & \\(3+\\) \\\\\n\\hline \\(\\mathrm{Co}\\) & +0.6 & HCP & -0.1 & \\(2+\\) \\\\\n\\hline \\(\\mathrm{Cr}\\) & +0.2 & BCC & -0.2 & \\(3+\\) \\\\\n\\hline \\(\\mathrm{Fe}\\) & -0.4 & BCC & -0.1 & \\(2+\\) \\\\\n\\hline \\(\\mathrm{Pt}\\) & +11 & FCC & -0.3 & \\(2+\\) \\\\\n\\hline \\(\\mathrm{Zn}\\) & +7 & HCP & -0.1 & \\(2+\\) \\\\\n\\hline\n\\end{tabular}\n(a) 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.\n(b) 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 \\(\\mathrm{Ni}\\) are greater than \\(\\pm 15 \\%\\), and/or have a valence different than 2+.\n(c) \\(\\mathrm{C}, \\mathrm{H}\\), and \\(\\mathrm{O}\\) form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of \\(\\mathrm{Ni}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 675,
|
||
"question": "What is the composition, in atom percent, of an alloy that consists of \\(92.5 \\mathrm{wt} \\% \\mathrm{Ag}\\) and \\(7.5 \\mathrm{wt} \\% \\mathrm{Cu}\\) ?",
|
||
"answer": "In order to compute composition, in atom percent, of a \\(92.5 \\mathrm{wt} \\% \\mathrm{Ag}-7.5 \\mathrm{wt} \\% \\mathrm{Cu}\\) alloy, we employ Equation given the atomic weights of silver and copper (found on the inside of the book's cover):\n\\[\n\\begin{array}{l}\nA_{\\mathrm{Ag}}=107.87 \\mathrm{~g} / \\mathrm{mol} \\\\\nA_{\\mathrm{Cu}}=63.55 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nThese compositions in atom percent are determined as follows:\n\\[\n\\begin{aligned}\nC_{\\mathrm{Ag}} & =\\frac{C_{\\mathrm{Ag}} A_{\\mathrm{Cu}}}{C_{\\mathrm{Ag}} A_{\\mathrm{Cu}}+C_{\\mathrm{Cu}} A_{\\mathrm{Ag}}} \\times 100 \\\\\n& =\\frac{(92.5)(63.55 \\mathrm{~g} / \\mathrm{mol})}{(92.5)(63.55 \\mathrm{~g} / \\mathrmmol})+(7.5)(107.87 \\mathrm{~g} / \\mathrm{mol}) \\times 100 \\\\\n& =87.9 \\mathrm{at} \\%\n\\end{aligned}\n\\]\nand\n\\[\n\\begin{array}{l}\nC_{\\mathrm{Cu}}{ }^{\\prime}=\\frac{C_{\\mathrm{Cu}} A_{\\mathrm{Ag}}}{C_{\\mathrm{Ag}} A_{\\mathrm{Cu}}+C_{C \\mathrm{u}} A_{\\mathrm{Ag}}} \\times 100 \\\\\n=\\frac{(7.5)(107.87 \\mathrm{~g} / \\mathrm {mol})}{(92.5)(63.55 \\mathrm{~g/ mol})+(7.5)(107.87 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n=12.1 \\mathrm{at} \\%\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 676,
|
||
"question": "What is the composition, in atom percent, of an alloy that consists of \\(5.5 \\mathrm{wt} \\% \\mathrm{~Pb}\\) and \\(94.5 \\mathrm{wt} \\% \\mathrm{Sn}\\) ?",
|
||
"answer": "In order to compute composition, in atom percent, of a \\(5.5 \\mathrm{wt} \\% \\mathrm{~Pb}-94.5 \\mathrm{wt} \\% \\mathrm{Sn}\\) alloy, we employ Equation given the atomic weights of lead and tin (found on the inside of the book's cover):\n\\[\n\\begin{array}{l}\nA_{\\mathrm{Pb}}=207.2 \\mathrm{~g} / \\mathrm{mol} \\\\\nA_{\\mathrm{Sn}}=118.71 \\mathrm{~g} / \\mathrm{mol}\n\\end{array}\n\\]\nThese compositions in atom percent are determined as follows:\n\\[\n\\begin{aligned}\nC_{\\mathrm{Pb}} & =\\frac{C_{\\mathrm{Pb}} A_{\\mathrm{Sn}}}{C_{\\mathrm{Pb}} A_{\\mathrm{Sn}}+C_{\\mathrm{Sn}} A_{\\mathrm{Pb}}} \\times 100 \\\\\n& =\\frac{(5.5)(118.71 \\mathrm{~g} / \\mathrm{mol})}{(5.5)(118.71 \\mathrm{~g} / \\mathrmmol)+(94.5)(207.2 \\mathrm{~g} / \\mathrm{mol})} \\times 100 \\\\\n& =3.23 \\mathrm{at} \\%\n\\end{aligned}\n\\]\n\\[\n\\begin{array}{l}\nC_{\\mathrm{Sn}}^{\\prime}=\\frac{C_{\\mathrm{Sn}} A_{\\mathrm{Pb}}}{C_{\\mathrm{Pb}} A_{\\mathrm{Sn}}+C \\mathrm{Sn} A_{\\mathrm{Pb}}} \\times 100 \\\\\n=\\frac{(94.5)(207.2 \\mathrm{~g} / \\mathrm {mol})}{(5.5)(118.71 \\mathrm{~g/ mol})+(94.5)(207.2 \\mathrm{~g} / \\mathrm{l} \\mathrm{mol})} \\times 100 \\\\\n=96.77 \\mathrm{at} \\%\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 678,
|
||
"question": "Calculate the composition, in weight percent, of an alloy that contains \\(105 \\mathrm{~kg}\\) of iron, 0.2 \\(\\mathrm{kg}\\) of carbon, and \\(1.0 \\mathrm{~kg}\\) of chromium.\n\\title{",
|
||
"answer": "The concentration, in weight percent, of an element in an alloy may be computed using Equation. For this alloy, the concentration of iron \\(\\left(C_{\\mathrm{Fe}}\\right)\\) is just\n\\[\n\\begin{aligned}\nC_{\\mathrm{Fe}} & =\\frac{m_{\\mathrm{Fe}}}{m_{\\mathrm{Fe}}+m_{\\mathrm{C}}+m_{\\mathrm{Cr}}} \\times 100 \\\\\n& =\\frac{105 \\mathrm{~kg}}{105 \\mathrm{~kg}+0.2 \\mathrm{~kg}+1.0 \\mathrm{~kg}} \\times 100=98.87 \\mathrm{wt} \\%\n\\end{aligned}\n\\]\nSimilarly, for carbon\n\\[\nC_{\\mathrm{C}}=\\frac{0.2 \\mathrm{~kg}}{105 \\mathrm{~kg}+0.1 \\mathrm{~kg}} \\times 100=0.19 \\mathrm{wt} \\%\n\\]\nAnd for chromium\n\\[\nC_{\\mathrm{Cr}}=\\frac{1.0 \\mathrm{~kg}}{105 \\mathrm{~kg}+0.3 \\mathrm{~kg}+1.0 \\mathrm{~kg}} \\times 00=0.94 \\mathrm{wt} \\%\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 684,
|
||
"question": "Calculate the unit cell edge length for an \\(80 \\mathrm{wt} \\% \\mathrm{Ag}-20 \\mathrm{wt} \\% \\mathrm{Pd}\\) alloy. All of the palladium is in solid solution, the crystal structure for this alloy is FCC, and the roomtemperature density of \\(P d\\) is \\(12.02 \\mathrm{~g} / \\mathrm{cm}^{3}\\).",
|
||
"answer": "In order to solve this problem it is necessary to employ Equation; in this expression density and atomic weight will be averages for the alloy - that is\n\\[\n\\rho_{\\mathrm{ave}}=\\frac{n A_{\\mathrm{ave}}}{V_{C} N_{\\mathrm{A}}}\n\\]\nInasmuch as the unit cell is cubic, then \\(V_{C}=a^{3}\\), and\n\\[\n\\rho_{\\mathrm{ave}}=\\frac{n A_{\\mathrm{aev}}}{a^{3} N_{\\mathrm{A}}}\n\\]\nAnd solving this equation for the unit cell edge length, \\(a\\), leads to\n\\[\na=\\left(\\frac{n A_{\\mathrm{ave}}}{\\rho_{\\mathrm{ave}} N_{\\mathrm{A}}}\\right)^{1 / 3}\n\\]\nExpressions for \\(A_{\\text {ave }}\\) and \\(\\rho_{\\text {ave }}\\) are found in Equations , respectively, which, when incorporated into the above expression, yields\\[\na=\\left[\\begin{array}{c}\nn \\\\\n\\left.\\frac{C_{\\mathrm{Ag}}}{A_{\\mathrm{Ag}}}+\\frac{C_{\\mathrm{pd}}}{A_{\\mathrm{pd}}}\\right) \\\\\n\\left.\\frac{100}{C_{\\mathrm{Ag}}}{ }_{+}C_{\\mathrm{pd}}\\right) \\\\\n\\rho_{\\mathrm{Ag}}+\\frac{C_{\\mathrm{pd}}}{\\rho_{\\mathrm{Pd}}}\n\\end{array}\\right]^{1 / 3}\n\\]\nSince the crystal structure is FCC, the value of \\(n\\) in the above expression is 4 atoms per unit cell. The atomic weights for \\(\\mathrm{Ag}\\) and \\(\\mathrm{Pd}\\) are 107.87 and \\(106.4 \\mathrm{~g} / \\mathrm{mol}\\), respectively (Figure 2.8), whereas the densities for the \\(\\mathrm{Ag}\\) and \\(\\mathrm{Pd}\\) are \\(10.49 \\mathrm{~g} / \\mathrm{cm}^{3}\\) (inside front cover) and 12.02 \\(\\mathrm{g} / \\mathrm{cm}^{3}\\). Substitution of these, as well as the concentration values stipulated in the problem statement, into the above equation gives\n\\[\na=\\left[\\begin{array}{c}\n\\left(\\begin{array}{c}\n\\left(\\begin{array}{c}\n100 \\\\\n80 \\mathrm{wt} \\%\n\\end{array}\\right. \\\\\n107.87 \\mathrm{~g} / \\mathrm{mol}^{+} \\\\\n\\left(\\begin{array}{c}\n100 \\\\\n6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol} \\\\\n10.49 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{array}\\right)\n\\end{array}\\right]^{1 / 3}\n\\]\n\\[\n\\begin{array}{l}\n=4.050 \\times 10^{-8} \\mathrm{~cm}=0.4050 \\mathrm{~nm}\n\\end{array}\n\\]\\title{",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 685,
|
||
"question": "Some hypothetical alloy is composed of \\(25 \\mathrm{wt} \\%\\) of metal A and \\(75 \\mathrm{wt} \\%\\) of metal B. If the densities of metals A and B are 6.17 and \\(8.00 \\mathrm{~g} / \\mathrm{cm}^{3}\\), respectively, and their respective atomic weights are 171.3 and \\(162.0 \\mathrm{~g} / \\mathrm{mol}\\), determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic. Assume a unit cell edge length of \\(0.332 \\mathrm{~nm}\\).",
|
||
"answer": "In order to solve this problem it is necessary to employ Equation; in this expression density and atomic weight will be averages for the alloy - that is\n\\[\n\\rho_{\\mathrm{ave}}=\\frac{n A_{\\mathrm{ave}}}{V_{C} N_{\\mathrm{A}}}\n\\]\nInasmuch as for each of the possible crystal structures, the unit cell is cubic, then \\(V_{C}=a^{3}\\), or\n\\[\n\\rho_{\\mathrm{ave}}=\\frac{n A_{\\mathrm{aev}}}{a^{3} N_{\\mathrm{A}}}\n\\]\nNow, in order to determine the crystal structure it is necessary to solve for \\(n\\), the number of atoms per unit cell. For \\(n=1\\), the crystal structure is simple cubic, whereas for \\(n\\) values of 2 and 4 , the crystal structure will be either BCC or FCC, respectively. When we solve the above expression for \\(n\\) the result is as follows:\n\\[\nn=\\frac{\\rho_{\\mathrm{ave}} a^{3} N_{\\mathrm{A}}}{A_{\\mathrm{ave}}}\n\\]\nExpressions for \\(A_{\\text {ave }}\\) and \\(\\rho_{\\text {ave }}\\) are found in Equations, respectively, which, when incorporated into the above expression yields\\[\nn=\\frac{\\left(\\frac{100}{C_{\\mathrm{A}}+C_{\\mathrm{B}}}{\\rho_{\\mathrm{A}}}\\right) a^{3} N_{\\mathrm{A}}}{\\left(\\frac{100}{C_{\\mathrm{A}}+C_{\\mathbf{B}}}{A_{\\mathrm{A}}-A_{\\mathrm{B}}}\\right)}\n\\]\nSubstitution of the concentration values (i.e., \\(C_{\\mathrm{A}}=25 \\mathrm{wt} \\%\\) and \\(C_{\\mathrm{B}}=75 \\mathrm{wt} \\%\\) ) as well as values for the other parameters given in the problem statement, into the above equation gives\n\\[\nn=\\frac{\\left(\\frac{100}{25 \\mathrm{wt} \\%}{\\frac{25 \\mathrm{wt} \\%}{6.17 \\mathrm{~g} / \\mathrm{cm}^{3}}+\\frac{75 \\mathrm{wt} \\%}{8.00 \\mathrm{~g} / \\mathrm{cm}^{3}}}\\right)}{\\left(\\frac{100}{25 \\mathrm{wt} \\%}{ }_{171.3 \\mathrm{~g} / \\mathrm{mol}}+\\frac{75 \\mathrm{wt} \\%}{162.0 \\mathrm{~g} / \\mathrm{mol}}\\right)}\n\\]\n\\[\n\\begin{array}{c}\n=1.00 \\mathrm{atom} / \\mathrm{unit} \\mathrm{cell} \\\\\n\\hline\n\\end{array}\n\\]\nTherefore, on the basis of this value, the crystal structure is simple cubic.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 688,
|
||
"question": "For a BCC iron-carbon alloy that contains \\(0.1 \\mathrm{wt} \\% \\mathrm{C}\\), calculate the fraction of unit cells that contain carbon atoms.\n\\title{",
|
||
"answer": "It is first necessary to compute the number of carbon atoms per cubic centimeter of alloy using Equation. For this problem\n\\[\n\\begin{array}{l}\nA_{1}=A_{\\mathrm{C}}=12.011 \\mathrm{~g} / \\mathrm{mol} \\text { (from inside the front cover) } \\\\\nC_{1}=C_{\\mathrm{C}}=0.1 \\mathrm{wt} \\% \\\\\n\\rho_{1}=\\rho_{\\mathrm{C}}=2.25 \\mathrm{~g} / \\mathrm{cm}^{3} \\text { (from inside front cover) } \\\\\n\\rho_{2}=\\rho_{\\mathrm{Fe}}=7.87 \\mathrm{~g} / \\mathrm{cm}^{3} \\text { ( from inside front cover) }\n\\end{array}\n\\]\nNow using Equation\n\\[\n\\begin{array}{c}\nN_{\\mathrm{C}}=\\frac{N_{\\mathrm{A}} C_{\\mathrm{C}}}{\\frac{C_{\\mathrm{C}} A_{\\mathrm{C}}}{\\rho_{\\mathrm{C}}}+\\frac{A_{\\mathrm{C}}}{\\rho_{\\mathrm{Fe}}}\\left(100-C_{\\mathrm{C}}\\right)} \\\\\n=\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)(0.1 \\mathrm{wt} \\%)}{\\left(0.1 \\mathrm{wt} \\%\\right)\\left(12.011 \\mathrm{~g} / \\mathrm{mol}\\right)}+\\frac{\\left(12.011 \\mathrm{~g} / \\mathrm{mol} \\right)}{7.87 \\mathrm{~g} / \\mathrm{cm}^{3}}\\left(100-0.1 \\mathrm{wt} \\%\\right) \\\\\n=3.94 \\times 10^{20} \\mathrm{C} \\text { atoms } / \\mathrm{cm}^{3}\n\\end{array}\n\\]\nWe now perform the same calculation for iron atoms, where\n\\[\n\\begin{array}{l}\nA_{1}=A_{\\mathrm{Fe}}=55.85 \\mathrm{~g} / \\mathrm{mol} \\text { (from inside the } \\text { front cover) } \\\\\nC_{1}=C_{\\mathrm{Fe}}=100 \\%-0.1 \\mathrm{wt} \\%=99.9 \\mathrm{wt} \\% \\\\\n\\rho_{1}=\\rho_{\\mathrm{Fe}}=7.87 \\mathrm{~g/} \\mathrm{cm}^{3} \\text { (from inside front cover) } \\\\\n\\rho_{2}=\\rho_{\\mathrm{C}}=2.25 \\mathrm{~g/} \\mathrm{cm}^{3} \\text { ( from inside front cover) }\n\\hline\n\\end{array}\n\\]\nTherefore,\\[\n\\begin{aligned}\nN_{\\mathrm{Fe}}= & \\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)(99.9 \\mathrm{wt} \\%)}{\\left(99.9 \\mathrm{wt} \\%\\right)(55.85 \\mathrm{~g} / \\mathrm{mol})}+\\frac{\\left(55.85 \\mathrm{~g} / \\mathrm{mol}\\right)}{2.25 \\mathrm{~g} / \\mathrm{cm}^{3}}\\left(100-99.9 \\mathrm{wt} \\%\\right) \\\\\n& =8.46 \\times 10^{22} \\mathrm{Fe} \\mathrm{atoms} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]\nFrom Section 3.4 there are two Fe atoms associated with each unit cell; hence, the number of unit cells in a centimeter cubed, \\(n_{i}\\), is equal to\n\\[\n\\begin{aligned}\nn_{i}= & \\frac{\\text { No. Fe atoms } / \\mathrm{cm}^{3}}{\\text { No. atoms } / \\text { unit cell }} \\\\\n= & \\frac{8.46 \\times 10^{22} \\mathrm{atoms} / \\mathrm{cm}^{3}}{2 \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell}}=4.23 \\times 10^{22} \\mathrm{unit} \\mathrm{cells} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]\nNow, the fraction of unit cells that contain carbon atoms, \\(n_{f}\\), is just the number of carbon atoms per centimeter cubed \\(\\left(3.94 \\times 10^{20}\\right)\\) divided by the number of unit cells per centimeter cubed \\(\\left(4.23 \\times 10^{22}\\right)\\), or\n\\[\n\\begin{aligned}\nn_{f}= & \\frac{3.94 \\times 10^{20} \\mathrm{C} \\mathrm{atoms} / \\mathrm{cm}^{3}}{4.23 \\times 10^{22} \\mathrm{unit} \\text { cells } / \\mathrm{cm}^{3}} \\\\\n= & 9.31 \\times 10^{-3} \\mathrm{atoms} / \\mathrm{unit} \\mathrm{cell}\n\\end{aligned}\n\\]\nThe reciprocal of \\(n_{f}\\) is the number of unit cells per atom-viz.\n\\[\n\\begin{aligned}\n\\frac{1}{n_{f}} & =\\frac{1}{9.31 \\times 10^{-3} \\mathrm{atoms} / \\text { unit cell }} \\\\\n& =107.5 \\text { unit cells/atom }\n\\end{aligned}\n\\]\nThat is, there is one carbon atom per 107.5 unit cells.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 689,
|
||
"question": "Gold (Au) forms a substitutional solid solution with silver (Ag). Compute the weight percent of gold that must be added to silver to yield an alloy that contains \\(5.5 \\times 10^{21} \\mathrm{Au}\\) atoms per cubic centimeter. The densities of pure \\(\\mathrm{Au}\\) and \\(\\mathrm{Ag}\\) are 19.32 and \\(10.49 \\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{Au}}=5.5 \\times 10^{21} \\mathrm{atoms} / \\mathrm{cm}^{3} \\\\\n\\rho_{1}=\\rho_{\\mathrm{Au}}=19.32 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n\\rho_{2}=\\rho_{\\mathrm{Ag}}=10.49 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\nA_{1}=A_{\\mathrm{Au}}=196.97 \\mathrm{~g} / \\mathrm{mol} \\text { (found inside front cover) } \\\\\nA_{2}=A_{\\mathrm{Ag}}=107.87 \\mathrm{~g} / \\mathrm{mol} \\text { (found inside from cover) }\n\\end{array}\n\\]\nThus\n\\[\n\\begin{aligned}\nC_{\\mathrm{Au}} & =\\frac{100}{1+\\frac{N_{\\mathrm{A}} \\rho_{\\mathrm{Ag}}}{N_{\\mathrm{Au}} A_{\\mathrm{Au}}}-\\frac{\\rho_{\\mathrm{Ag}}}{\\rho_{\\mathrm{Au}}}} \\\\\n& =\\frac{100}{1+\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(10.49 \\mathrm{~g} / \\mathrm{cm}^{3})\\right.}{\\left(5.5 \\times 10^{21} \\mathrm{atoms} / \\textrm{cm}^{3}\\right)(196.97 \\mathrm{~g} / \\mathrm{mol})}-\\left(\\frac{10.49 \\mathrm{~g} / \\mathrm{cm}^{3}}{19.32 \\mathrm{~g} / \\mathrm{cm}^{3}}\\right) \\\\\n& =15.9 \\mathrm{wt} \\%\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 691,
|
||
"question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch under each of the following conditions?\n(a) At a magnification of \\(100 \\times\\)\n(b) Without any magnification?\n\\(\\underline{\\text {",
|
||
"answer": "(a) 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 \\(100 \\times\\). All we need do is solve for the parameter \\(n\\) in Equation, inasmuch as \\(G=6\\). Thus\n\\[\n\\begin{aligned}\nn & =2^{G-1} \\\\\n& =2^{6-1}=32 \\text { grains } / \\mathrm{in} .^{2}\n\\end{aligned}\n\\]\n(b) 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\\[\n{ }_{n_{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. Without any magnification, \\(M\\) in the above equation is 1 , and therefore,\n\\[\nn_{1}\\left(\\frac{1}{100}\\right)^{2}=2^{6-1}=32\n\\]\nAnd, solving for \\(n_{1}, n_{1}=320,000\\) grains/in. \\({ }^{2}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 696,
|
||
"question": "(a) Compare interstitial and vacancy atomic mechanisms for diffusion. \\\\ (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy \\\\ diffusion.",
|
||
"answer": "(a) 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.\n(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 700,
|
||
"question": "The activation energy for the diffusion of copper in silver is \\(193,000 \\mathrm{~J} / \\mathrm{mol}\\). Calculate the diffusion coefficient at \\(1200 \\mathrm{~K}\\left(927^{\\circ} \\mathrm{C}\\right)\\), given that \\(\\mathrm{D}\\) at \\(1000 \\mathrm{~K}\\left(727^{\\circ} \\mathrm{C}\\right)\\) is \\(1.0 \\times 10^{-14} \\mathrm{~m}^{2} / \\mathrm{s}\\).",
|
||
"answer": "To solve this problem it first becomes necessary to solve for \\(D_{0}\\) from a rearranged form of Equation and using \\(1000 \\mathrm{~K}\\) diffusion data:\n\\[\n\\begin{array}{c}\nD_{0}=D \\exp \\left(\\frac{Q_{d}}{R T}\\right) \\\\\n=\\left(1.0 \\times 10^{-14} \\mathrm{~m}^{ 2} / \\mathrm{s}\\right) \\exp \\left[\\frac{193,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(1000 \\mathrm{~K})]}\\right. \\\\\n1.22 \\times 10^{-4} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{array}\n\\]\nNow, solving for \\(D\\) at \\(1200 \\mathrm{~K}\\) gives\n\\[\n\\begin{array}{c}\nD=\\left(1.22 \\times 10^{-4} \\mathrm{~m}^{ 2} / \\mathrm{s}\\right) \\\\\n\\left.\\left.\\left.\\exp \\left[-\\frac{193,000 \\mathrm{~J} / \\mathrm{mo} l}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{L})(1200 \\mathrm{~K})]} \\right.\\right]\\right.\\right. \\\\\n4.8 \\times 10^{-13} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{array} \\quad\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 702,
|
||
"question": "The steady-state diffusion flux through a metal plate is \\(7.8 \\times 10^{-8} \\mathrm{~kg} / \\mathrm{m}^{2} \\cdot \\mathrm{s}\\) at a temperature of \\(1200^{\\circ} \\mathrm{C}(1473 \\mathrm{~K})\\) and when the concentration gradient is \\(-500 \\mathrm{~kg} / \\mathrm{m}^{4}\\). Calculate the diffusion flux at \\(1000^{\\circ} \\mathrm{C}(1273 \\mathrm{~K})\\) for the same concentration gradient and assuming an activation energy for diffusion of \\(145,000 \\mathrm{~J} / \\mathrm{mol}\\).",
|
||
"answer": "In order to solve this problem, we must first compute the value of \\(D_{0}\\) from the data given at \\(1200^{\\circ} \\mathrm{C}(1473 \\textrm{K})\\); this requires the combining of both Equations as follows:\n\\[\n\\begin{array}{l}\nJ=-D \\frac{\\Delta C}{\\Delta x} \\\\\n\\quad=-D_{0} \\frac{\\Delta C}{\\Delta x} \\exp \\left(-\\frac{Q_{d}}{R T}\\right)\n\\end{array}\n\\]\nSolving for \\(D_{0}\\) from the above expression gives\n\\[\nD_{0}=-\\frac{J}{\\frac{\\Delta C}{\\Delta x}} \\exp \\left(\\frac{Q_{d}}{R T}\\right)\n\\]\nAnd incorporating values of the parameters in this expression provided in the problem statement yields the following:\\[\n\\begin{array}{l}\nD_{0}=-\\left(\\frac{7.8 \\times 10^{-8} \\mathrm{~kg} / \\mathrm{m}^{2} \\cdot \\mathrm{s}}{-500 \\mathrm{~kg} / \\mathrm{m}^{4}}\\right) \\exp \\left[\\frac{145,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}) (1200+273 \\mathrm{~K})}\\right] \\\\ \\\\\n\\quad=2.18 \\times 10^{-5} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{array}\n\\]\nThe value of the diffusion flux at \\(1273 \\mathrm{~K}\\) may be computed using Equation S6.33 (above) as follows:\n\\[\n\\begin{array}{l}\nJ=-D_{0}\\left(\\frac{\\Delta C}{\\Delta x}\\right) \\exp \\left(-\\frac{Q_{d}}{R T}\\right) \\\\\n\\quad=-\\left(2.18 \\times 10^{-5} \\mathrm{~m}^{ 2} / \\mathrm{s}\\right)\\left(-500 \\mathrm{~kg} / \\mathrm{m}^{4}\\right) \\exp \\left[-\\frac{145,000 \\mathrm{~J} / \\mathrm{m} 0}{(8.31 \\mathrm{~J} / \\mathrm{mol} \\cdot \\textrm{K}) (1273 \\textrm{K})}\\right] \\\\\n\\quad=1.21 \\times 10^{-8} \\mathrm{~kg} / \\mathrm{m}{ }^{2} \\cdot \\mathrm{s}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 708,
|
||
"question": "For a brass alloy, the stress at which plastic deformation begins is \\(345 \\mathrm{MPa}(50,000 \\mathrm{psi})\\), and the modulus of elasticity is \\(103 \\mathrm{GPa}\\left(15.0 \\times 10^{6} \\mathrm{psi}\\right)\\).\n(a) What is the maximum load that can be applied to a specimen with a cross-sectional area of \\(130 \\mathrm{~mm}^{2}\\) ( 0.2 in. \\({ }^{2}\\) ) without plastic deformation?\n(b) If the original specimen length is \\(76 \\mathrm{~mm}(3.0 \\mathrm{in}\\).), what is the maximum length to which it can be stretched without causing plastic deformation?",
|
||
"answer": "(a) This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation \\(\\left(F_{\\mathrm{y}}\\right)\\). Taking the yield strength to be \\(345 \\mathrm{MPa}\\), and employment of Equation leads to\n\\[\n\\begin{aligned}\nF_{\\mathrm{y}}=\\sigma_{\\mathrm{y}} A_{0} & =\\left(345 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}\\right)\\left(133 \\mathrm{~mm}^{2}\\right)\\left(\\frac{1 \\mathrm{~m}}{10^{3} \\mathrm{~mm}}\\right)^{2} \\\\\n& =44,850 \\mathrm{~N} \\quad\\left(10,000 \\mathrm{lb} \\mathrm{f}\\right)\n\\end{aligned}\n\\]\n(b) The maximum length to which the sample may be deformed without plastic deformation is determined by combining Equations as\n\\[\n\\begin{array}{c}\nl_{i}=z l_{0}+l_{0}=l_{0}(z+1) \\\\\nl_{i}=l_{0}\\left(\\frac{\\sigma}{E}+1\\right)\n\\end{array}\n\\]\nIncorporating values of \\(l_{0}, \\sigma\\), and \\(E\\) provided in the problem statement leads to the computation of \\(l_{i}\\) as follows:\\[\nl_{j}=(76\\,\\,\\mathrm{mm})\\left[\\frac{345\\,\\,\\mathrm{MPa}}{103\\times10^{3}\\,\\mathrm{MPa}}+\\,1\\,\\right]=76.25\\,\\mathrm{mm}\\,\\,\\mathrm{(3.01\\,\\,in.)}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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{",
|
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"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 711,
|
||
"question": "Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \\(10.0 \\mathrm{~mm}(0.39 \\mathrm{in.})\\). A tensile force of \\(1500 \\mathrm{~N}(340 \\mathrm{lb})\\) produces an elastic reduction in diameter of \\(6.7 \\times 10^{-4} \\mathrm{~mm}(2.64 \\times\\)\n\\(10^{-5}\\) in.). Compute the elastic modulus of this alloy, given that Poisson's ratio is 0.35 .",
|
||
"answer": "This problem asks that we calculate the modulus of elasticity of a metal that is stressed in tension. Combining Equations leads to\n\\[\nE=\\frac{\\sigma}{\\varepsilon_{z}}=\\frac{\\frac{F}{A_{0}}}{\\varepsilon_{z}}=\\frac{F}{A_{0} \\varepsilon_{z}}=\\frac{F}{\\varepsilon_{z} \\pi\\left(\\frac{d_{0}}{2}\\right)^{2}}=\\frac{4 F}{\\varepsilon_{z} \\pi d_{0}^{2}}\n\\]\nFrom the definition of Poisson's ratio, and realizing that for the transverse strain,\n\\(x=\\frac{d}{d_{0}}\\) leads to\n\\[\n\\varepsilon_{z}=-\\frac{\\varepsilon_{x}}{\\nu}=-\\frac{\\Delta d / d_{0}}{\\nu}=-\\frac{\\Delta d}{d_{0} v}\n\\]\nTherefore, substitution of this expression for \\(\\varepsilon_{z}\\) into the above equation for \\(E\\) yields\n\\[\nE=\\frac{4 F}{\\varepsilon_{z} \\pi d_{0}^{\\frac{2}{2}}}=\\frac{4 F}{\\left(-\\frac{\\Delta d}{d_{0} v}\\right) \\pi d_{0}^{\\frac{2}{2}}}=-\\frac{4 F \\nu}{\\pi d_{0} \\Delta d}\n\\]\nIncorporation of values for \\(F, v, d_{0}\\), and \\(\\Delta d\\) given in the problem statement (and realizing that \\(\\Delta d\\) is negative) allows us to calculate the modulus of elasticity as follows:\\[\nE=\\frac{(4)(1500\\mathrm{\\ N})(0.35)}{\\pi\\left(10\\times10^{-3}\\mathrm{\\ m}\\right)\\left(-6.7\\times10^{-7}\\mathrm{\\ m}\\right)}=10^{11}\\mathrm{\\ Pa}=100\\mathrm{\\ GPa}\\ \\ (14.7\\times10^{6}\\mathrm{\\ psi})\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 712,
|
||
"question": "A cylindrical rod \\(500 \\mathrm{~mm}\\) (20.0 in.) long and having a diameter of \\(12.7 \\mathrm{~mm}\\) (0.50 in.) is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than \\(1.3 \\mathrm{~mm}(0.05 \\mathrm{in}\\).) when the applied load is \\(29,000 \\mathrm{~N}\\left(6500 \\mathrm{lb}^{2}\\right)\\), which of the four metals or alloys listed in the following table are possible candidates? Justify your choice(s).\n\\begin{tabular}{lcccc}\n\\hline & & Yield Strength & \\multicolumn{1}{c}{ Tensile Strength } \\\\\n\\cline { 3 - 4 } Material & Modulus of Elasticity (GPa) & (MPa) & (MPa) \\\\\n\\hline Aluminum alloy & 70 & 255 & 420 \\\\\nBrass alloy & 100 & 345 & 420 \\\\\nCopper & 110 & 210 & 275 \\\\\nSteel alloy & 207 & 450 & 550 \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "This problem asks that we ascertain which of four metal alloys will not (1) experience plastic deformation, and (2) elongate more than \\(1.3 \\mathrm{~mm}\\) when a tensile load of \\(29,000 \\mathrm{~N}\\) is applied. It is first necessary to compute the stress using Equation; a material to be used for this application must necessarily have a yield strength greater than this value. Thus,\n\\[\n\\sigma=\\frac{F}{A_{0}}=\\frac{F}{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}}=\\frac{29,000 \\mathrm{~N}}{\\pi\\left(\\frac{12.7 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}}=230 \\mathrm{MPa}\n\\]\nOf the metal alloys listed, aluminum, brass and steel have yield strengths greater than this stress.\nNext, we must compute the elongation produced in each of aluminum, brass, and steel by combining Equations in order to determine whether or not this elongation is less than \\(1.3 \\mathrm{~mm}\\). For aluminum \\(\\left(E=70 \\mathrm{GPa}=70 \\times 10^{3} \\mathrm{MPa}\\right)\\)\n\\[\n\\Delta l=\\varepsilon l_{0}=\\frac{\\sigma l_{0}}{E}=\\frac{(230 \\mathrm{MPa})(500 \\mathrm{~mm})}{70 \\times 10^{3} \\mathrm{MPa}}=1.64 \\mathrm{~mm}\n\\]Thus, aluminum is not a candidate because its \\(\\Delta l\\) is greater than \\(1.3 \\mathrm{~mm}\\).\nFor brass \\((E=100 \\mathrm{GPa}=100 \\times 10^{3} \\mathrm{MPa})\\)\n\\[\n\\Delta l=\\frac{\\sigma l_{0}}{E}=\\frac{(230 \\mathrm{MPa})(500 \\mathrm{~mm})}{100 \\times 10^{3} \\mathrm{MPa}}=1.15 \\mathrm{~mm}\n\\]\nThus, brass is a candidate. And, for steel \\((E=207 \\mathrm{GPa}=207 \\times 10^{3} \\mathrm{MPa})\\)\n\\[\n\\begin{array}{l}\n\\Delta l=\\frac{\\sigma l_{0}}{E}=\\frac{(23 \\mathrm{MPa})(500 \\mathrm{~mm})}{207 \\times 10^{3} \\mathrm{MPa}}=0.56 \\mathrm{~mm}\n\\end{array}\n\\]\nTherefore, of these four alloys, only brass and steel satisfy the stipulated criteria.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 713,
|
||
"question": "A cylindrical metal specimen having an original diameter of \\(12.8 \\mathrm{~mm}(0.505 \\mathrm{in}\\).) and gauge length of \\(50.80 \\mathrm{~mm}\\) (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is \\(8.13 \\mathrm{~mm}(0.320 \\mathrm{in}\\).), and the fractured gauge length is \\(74.17 \\mathrm{~mm}\\) (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.",
|
||
"answer": "This problem calls for the computation of ductility in both percent reduction in area and percent elongation. Percent reduction in area, for a cylindrical specimen, is computed using Equation as\n\\[\n\\% \\mathrm{RA}=\\left(\\frac{A_{0}-A_{f}}{A_{0}}\\right) \\times 100=\\frac{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}-\\pi\\left(\\frac{d_{f}}{2}\\right)^{2}}{\\pi\\left(\\frac{d_{0}}{2}\\right)^{2}} \\times 100\n\\]\nin which \\(d_{0}\\) and \\(d_{f}\\) are, respectively, the original and fracture cross-sectional areas. Thus,\n\\[\n\\% \\mathrm{RA}=\\frac{\\pi\\left(\\frac{12.8 \\mathrm{~mm}}{2}\\right)^{2}-\\pi\\left(\\frac{8.13 \\mathrm{~mm}}{2}\\right)^{2}}{\\pi\\left(\\frac{12.8 \\mathrm{~mm}}{2} \\right)^{2}} \\times 100=60 \\%\n\\]\nWhile, for percent elongation, we use Equation as\n\\[\n\\begin{aligned}\n\\% \\mathrm{EL} & =\\left(\\frac{l_{f}-l_{0}}{l_{0}}\\right) \\times 100 \\\\\n& =\\frac{74.17 \\mathrm{~mm}-50.80 \\mathrm{~mm}}{50.80 \\mathrm{~mm}} \\times 100=46 \\%\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 716,
|
||
"question": "A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius \\(5.0 \\mathrm{~mm}(0.20 \\mathrm{in}\\).); the specimen fractured at a load of \\(3000 \\mathrm{~N}\\) \\((675 \\mathrm{lb} f)\\) when the distance between the support points was \\(40 \\mathrm{~mm}\\) (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of \\(15 \\mathrm{~mm}\\) (0.6 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at \\(40 \\mathrm{~mm}\\) (1.6 in.)?",
|
||
"answer": "For this problem, the load is given at which a circular specimen of aluminum oxide fractures when subjected to a three-point bending test; we are then are asked to determine the load at which a specimen of the same material having a square cross-section fractures. It is first necessary to compute the flexural strength of the cylindrical specimen from Equation; then, using this value, calculate the value of \\(F_{f}\\) in Equation (for the specimen having a square cross-section). From Equation, and using the first set of data provided in the problem statement, we compute the flexural strength as follows:\n\\[\n\\begin{aligned}\n\\sigma_{f s} & =\\frac{F_{f} L}{\\pi R^{3}} \\\\\n& =\\frac{(3000 \\mathrm{~N})\\left(40 \\times 10^{-3} \\mathrm{~m}\\right)}{(\\pi)\\left(5.0 \\times 10^{-3} \\mathrm{~m}\\right)^{3}}=306 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=306 \\mathrm{MPa} \\quad(42,970 \\mathrm{psi})\n\\end{aligned}\n\\]\nNow, solving for \\(F_{f}\\) for the second specimen using Equation, realizing that \\(b=d=15 \\mathrm{~mm}\\) \\(\\left(15 \\times 10^{-3} \\mathrm{~m}\\right)\\), yields\n\\[\nF_{f}=\\frac{2 \\sigma_{f s} d^{3}}{3 L}\n\\]\\[\n\\hskip-1cm\n\\mathbf{N}^{(1)}(3) \\equiv \\frac{1}{(3) \\left(306 \\times 10^{2} \\right) \\left(15 \\times 10^{-3} \\mathrm{m} \\right)^{3}} = 17.200 \\mathrm{\\ N} \\hskip1cm \\left(3870 \\mathrm{\\ lbf} \\right)\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 717,
|
||
"question": "(a) A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of \\(300 \\mathrm{MPa}(43,500 \\mathrm{psi})\\). If the specimen radius is \\(5.0 \\mathrm{~mm}(0.20 \\mathrm{in}\\).) and the support point separation distance is \\(15.0 \\mathrm{~mm}(0.61 \\mathrm{in}\\).), would you expect the specimen to fracture when a load of \\(7500 \\mathrm{~N}(1690 \\mathrm{lb} f)\\) is applied? Justify your answer.\n(b) Would you be \\(100 \\%\\) certain of the answer in part (a)? Why or why not?",
|
||
"answer": "(a) This portion of the problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of \\(300 \\mathrm{MPa}\\) and a radius of \\(5.0 \\mathrm{~mm}\\) will fracture when subjected to a load of \\(7500 \\mathrm{~N}\\) in a three-point bending test; the support point separation is given as \\(15.0 \\mathrm{~mm}\\). Using Equation we will calculate the value of \\(\\sigma\\), in this case the flexural stress; if this value is greater than the flexural strength, \\(\\sigma_{f s}(300 \\mathrm{MPa})\\), then fracture is expected to occur. Employment of Equation to compute the flexural stress yields\n\\[\n\\begin{aligned}\n\\sigma & =\\frac{F L}{\\pi R^{3}}=\\frac{(7500 \\mathrm{~N})\\left(15 \\times 10^{-3} \\mathrm{~m}\\right)}{(\\pi)\\left(5 \\times 10^{-3} \\mathrm{~m}\\right)^{3}} \\\\\n& =286.5 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=286.5 \\mathrm{MPa} \\quad(40,300 \\mathrm{psi})\n\\end{aligned}\n\\]\nSince this value (286.5 MPa) is less than the given value of \\(\\sigma_{f s}(300 \\mathrm{MPa})\\), then fraction are is not predicted.\n(b) 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, \\(\\sigma\\), is relatively close to \\(\\sigma_{f s}\\) there is some chance that fracture will occur.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 718,
|
||
"question": "The modulus of elasticity for spinel \\(\\left(\\mathrm{MgAl}_{2} \\mathrm{O}_{4}\\right)\\) having 5 vol\\% porosity is \\(240 \\mathrm{GPa}(35 \\times\\) \\(10^{6} \\mathrm{psi})\\). \\\\ (a) Compute the modulus of elasticity for the nonporous material. \\\\ (b) Compute the modulus of elasticity for \\(15 \\mathrm{vol} \\%\\) porosity.",
|
||
"answer": "(a) This portion of the problem requests that we compute the modulus of elasticity for nonporous spinel given that \\(E=240 \\mathrm{GPa}\\) for a material having 5 vol\\% porosity. Thus, we solve Equation for \\(E_{0}\\) (the modulus of elasticity of the nonporous material) using \\(P=0.05\\), as follows:\n\\[\n\\begin{aligned}\nE_{0} & =\\frac{E}{1-1.9 P+0.9 P^{2}} \\\\\n& =\\frac{240 \\mathrm{GPa}}{1-(1.9)(0.05)+(0.9)(0.05)^{2}}=265 \\mathrm{GPa} \\quad\\left(38.6 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]\n(b) Now we are asked to determine the value of \\(E\\) at \\(P=15 \\mathrm{vol} \\%\\) (i.e., 0.15). Using Equation the following results:\n\\[\n\\begin{array}{c}\nE=E_{0}\\left(1-1.9 P+0.9 P^{2}\\right) \\\\\n= \\left(265 \\mathrm{GPa}\\right)\\left[1-(1.9)(0.15)+(0.09)(0.15)^{2}\\right]=195 \\mathrm{GPa} \\quad\\left(28.4 \\times 10^{6} \\mathrm{psi}\\right)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 719,
|
||
"question": "The modulus of elasticity for titanium carbide (TiC) having 5 vol\\% porosity is \\(310 \\mathrm{GPa}\\) \\((45 \\times 10^{6} \\mathrm{psi})\\).\n(a) Compute the modulus of elasticity for the nonporous material.\n(b) At what volume percent porosity will the modulus of elasticity be \\(240 \\mathrm{GPa}(35 \\times 10^{6}\\) psi)?",
|
||
"answer": "(a) This portion of the problem requests that we compute the modulus of elasticity for nonporous TiC given that \\(E=310 \\mathrm{GPa}\\left(45 \\times 10^{6} \\mathrm{psi}\\right)\\) for a material having 5 vol\\% porosity. Thus, we solve Equation for \\(E_{0}\\) (the modulus of elasticity of the nonporous material) using \\(P=0.05\\), as follows:\n\\[\n\\begin{aligned}\nE_{0} & =\\frac{E}{1-1.9 P+0.9 P^{2}} \\\\\n& =\\frac{310 \\mathrm{GPa}}{1-(1.9)(0.05)+(0.9)(0.05)^{2}}=342 \\mathrm{GPa} \\quad\\left(49.6 \\times 10^{6} \\mathrm{psi}\\right)\n\\end{aligned}\n\\]\n(b) Now we are asked to compute the volume percent porosity at which the elastic modulus of TiC is \\(240 \\mathrm{MPa}(35 \\times 10^{6} \\mathrm{psi})\\). Since from part (a), \\(E_{0}=342 \\mathrm{GPa}\\), and using Equation, we may write the following:\n\\[\n\\frac{E}{E_{0}}=\\frac{240 \\mathrm{GPa}}{342 \\mathrm{GPa}}=0.702=1-1.9 P+0.9 P^{2}\n\\]\nOr, after rearrangement the following expression results:\n\\[\n0.9 P^{2}-1.9 P+0.298=0\n\\]Now, solving for the value of \\(P\\) using the quadratic equation solution yields\n\\[\nP=\\frac{1.9 \\pm \\sqrt{(-1.9)^{2}-(4)(0.9)(0.298)}}{(2)(0.9)}\n\\]\nThe positive and negative roots are as follows:\n\\[\n\\begin{array}{l}\nP^{+}=1.94 \\\\\nP^{-}=0.171\n\\end{array}\n\\]\nObviously, only the negative root is physically meaningful, and therefore the value of the porosity to give the desired modulus of elasticity is \\(17.1 \\mathrm{vol} \\%\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 720,
|
||
"question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows:\n\\begin{tabular}{ll}\n\\hline \\(\\sigma_{f_{S}}(\\mathbf{M P a})\\) & \\(\\boldsymbol{P}\\) \\\\\n\\hline 70 & 0.10 \\\\\n60 & 0.15 \\\\\n\\hline\n\\end{tabular}\n(a) Compute the flexural strength for a completely nonporous specimen of this material.\n(b) Compute the flexural strength for a 0.20 volume fraction porosity.",
|
||
"answer": "(a) Given the flexural strengths at two different volume fraction porosities, we are asked to determine the flexural strength for a nonporous material. Taking the natural logarithm of both sides of Equation 7.22 yields the following expression:\n\\[\n\\ln \\quad f_{S}=\\ln \\quad 0 \\quad n P\n\\]\nUsing the data provided in the problem statement, two simultaneous equations may be written in the above form as\\[\n\\begin{array}{l}\n\\ln \\left(70 \\mathrm{MPa}\\right)=\\ln \\quad 0 \\quad \\left(0.10\\right) n \\\\\n\\ln \\left(60 \\mathrm{MPa}\\right)=\\ln \\quad 0 \\quad \\left(\\begin{array}{ll}\n0 & 0\n\\end{array}\\right) n\n\\end{array}\n\\]\nSolving for \\(n\\) and \\(\\sigma_{0}\\) leads to the following values:\n\\[\n\\begin{array}{l}\nn=3.08 \\\\\n\\sigma_{0}=95.3 \\mathrm{MPa} .\n\\end{array}\n\\]\nFor the nonporous material, \\(P=0\\), and, from Equation 7.22, \\(\\sigma_{0}=\\sigma_{f s}\\). Thus, \\(\\sigma_{f s}\\) for \\(P=0\\) is 95.3 \\(\\mathrm{MPa}\\).\n(b) Now we are asked for \\(\\sigma_{f s}\\) at \\(P=0.20\\) for this same material. From Equation 7.22\n\\[\n\\begin{array}{l}\n\\sigma_{f s}=\\sigma_{0} \\exp (-n P) \\\\\n\\quad=\\left(95.3 \\mathrm{MPa}\\right) \\exp \\left[-(3.08)(0.20)\\right] \\\\\n\\quad=51.5 \\mathrm{MPa}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 721,
|
||
"question": "For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to\n\\[\n\\sigma(t)=\\sigma(0) \\exp \\left(-\\frac{t}{\\tau}\\right)\n\\]\nwhere \\(\\sigma(t)\\) and \\(\\sigma(0)\\) represent the time-dependent and initial (i.e., time \\(=0\\) ) stresses, respectively, and \\(t\\) and \\(\\tau\\) denote elapsed time and the relaxation time, respectively; \\(\\tau\\) is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation was suddenly pulled in tension to a measured strain of 0.5 ; the stress necessary to maintain this constant strain was measured as a function of time. Determine \\(E_{f}(10)\\) for this material if the initial stress level was \\(3.5 \\mathrm{MPa}(500 \\mathrm{psi})\\), which dropped to 0.5 \\(\\mathrm{MPa}(70 \\mathrm{psi})\\) after \\(30 \\mathrm{~s}\\).",
|
||
"answer": "In order to determine \\(\\sigma(10)\\), it is first necessary to compute \\(\\tau\\) from the data provided in the problem statement. In order to solve for \\(\\tau\\) from Equation it is first necessary to take natural logarithms of both sides of the equation as follows:\n\\[\n\\ln \\sigma(t)=\\ln \\sigma(0)-\\frac{t}{\\tau}\n\\]\nAnd solving this expression for \\(\\tau\\) leads to the following:\n\\[\n\\tau=\\frac{-t}{\\ln \\sigma(t)-\\ln \\sigma(0)}=\\frac{-t}{\\ln \\left[\\frac{\\sigma(t)}{\\sigma(0)}\\right]}\n\\]\nThe problem statement provides the following values:\n\\[\n\\begin{aligned}\n\\sigma(0) & =3.5 \\mathrm{MPa} \\\\\n\\sigma(t) & =\\sigma(30)=0.5 \\mathrm{MPa} \\\\\nt & =30 \\mathrm{~s}\n\\end{aligned}\n\\]Therefore, the value of \\(\\tau\\) is equal to\n\\[\n\\begin{aligned}\n\\tau & =\\frac{-t}{\\ln \\left[\\frac{\\sigma(t)}{\\sigma(0)}\\right]} \\\\\n& =\\frac{-30 \\mathrm{~s}}{\\ln \\left[\\frac{0.5 \\mathrm{MPa}}{3.5 \\mathrm{MPa}}\\right]}=15.4 \\mathrm{~s}\n\\end{aligned}\n\\]\nUsing Equation we compute the value of \\(\\sigma(10)\\) (that is, the value of \\(\\sigma\\) when \\(t=10 \\mathrm{~s}\\) ), as follows:\n\\[\n\\begin{aligned}\n\\sigma(10) & =\\sigma(0) \\exp \\left(-\\frac{10 \\mathrm{~s}}{\\tau}\\right) \\\\\n& =(3.5 \\mathrm{MPa}) \\exp \\left(-\\frac{10 \\mathrm{~s}}{15.4 \\mathrm{~s}}\\right)=1.83 \\mathrm{MPa}\n\\end{aligned}\n\\]\nWe now compute the value of \\(E_{f}(10)\\) using Equation as follows:\n\\[\nE_{f}(10)=\\frac{\\sigma(10)}{\\varepsilon_{0}}=\\frac{1.83 \\mathrm{MPa}}{0.5}=3.66 \\mathrm{MPa} \\quad \\text { (522 psi) }\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 722,
|
||
"question": "The following table gives a number of Rockwell G hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values.\n\\begin{tabular}{rrr}\n47.3 & 48.7 & 47.1 \\\\\n52.1 & 50.0 & 50.4 \\\\\n45.6 & 46.2 & 45.9 \\\\\n49.9 & 48.3 & 46.4 \\\\\n47.6 & 51.1 & 48.5 \\\\\n50.4 & 46.7 & 49.7\n\\end{tabular}",
|
||
"answer": "The average of the given hardness values is calculated using Equation as\n\\[\n\\begin{array}{c}\n\\stackrel{\\textstyle \\sum_{i=1}^{18} \\mathrm{HRG}_{i}}{\\stackrel{\\textstyle \\textstyle \\mathrm{HRG}}{=}} \\\\\n\\stackrel{\\textstyle \\textstyle =}{\\stackrel{\\textstyle \\textstyle =}{\\stackrel{\\textstyle \\textit{47.3}+52.1+45.6 \\ldots+49.7}{\\stackrel{\\textstyle \\textit{18}}{=}}}} \\\\\n\\stackrel{\\textstyle \\textit{47.3+52.1+45.6 \\ldots+49.7}{=}}{\\stackrel{\\textstyle \\textit{18}}{=}} \\\\\n\\text { And we compute the standard deviation using Equation as follows: } \\\\\n\\stackrel{\\textstyle \\textit{s}}{\\stackrel{\\textstyle \\textit{s}}{\\stackrel{\\textstyle \\textit{18}}{=}} \\stackrel{\\textstyle \\textit{18}}{\\stackrel{\\textstyle \\textit{18}}{=}} \\stackarrow{\\textstyle \\textit{18}} \\stackrel{\\textstyle \\textit{18}}{\\stackrel{\\textit{18}}{=}} \\stackrel{\\textit{18}}{\\stackrel{\\textit{18}}{=}} \\\\\n\\stackrel{\\textstyle \\textit{47.3-48.4})^{2}+\\left(52.1-48.4\\right)^{2}+\\ldots+\\left(49.7-48.4\\right)^{2} \\\\\n\\stackrel{\\textstyle \\textit{17}}{\\stackrel{\\textstyle \\textit{17}}{=}} \\\\\n\\stackrel{\\textstyle \\textit{64.95}}{=}\\stackrel{\\textstyle \\textit{17}}{=} 1.95",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 723,
|
||
"question": "The following table gives a number of yield strength values (in MPa) that were measured on the same aluminum alloy. Compute average and standard deviation yield strength values.\n\\begin{tabular}{rrr}\n274.3 & 277.1 & 263.8 \\\\\n267.5 & 258.6 & 271.2 \\\\\n255.4 & 266.9 & 257.6 \\\\\n270.8 & 260.1 & 264.3 \\\\\n261.7 & 279.4 & 260.5\n\\end{tabular}",
|
||
"answer": "The average of the given yield strength values is calculated using Equation as\n\\[\n\\begin{array}{l}\n\\frac{15}{\\sigma_{y}}=\\frac{i}{i=1}\\left(\\sigma_{y}\\right)_{i} \\\\\n\\frac{15}{\\sigma_{y}}=\\frac{274.3+267.5+255.4+\\ldots+260.5}{15}=265.9 \\mathrm{MPa} \\\\\n\\frac{274.3+267.5+255.4+0.5}{15}\n\\end{array}\n\\]\nAnd we compute the standard deviation using Equation as follows:\n\\[\n\\begin{array}{l}\ns=\\sqrt{\\left[\\sum_{i=1}^{15}\\left(\\sigma_{y}\\right)_{i}-\\bar{\\sigma}_{y}\\right]^{2}} \\\\\n=\\left[\\frac{(274.3-265.9)^{2}+(267.5-265.9)^{2}+\\ldots+(260.5-265.9)^{2}}{14}\\right]^{1 / 2} \\\\\n=\\sqrt{\\frac{753.75}{14}}=7.34 \\mathrm{MPa}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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^{\\circ}\\) and \\(35^{\\circ}\\), respectively, with the tensile axis. If the critical resolved shear stress is \\(6.2 \\mathrm{MPa}(900 \\mathrm{psi})\\), will an applied stress of \\(12 \\mathrm{MPa}\\) (1750 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=\\) \\(60^{\\circ}, \\lambda=35^{\\circ}\\), and that the values of the critical resolved shear stress and applied tensile stress are 6.2 MPa (900 psi) and \\(12 \\mathrm{MPa}\\) (1750 psi), respectively. From Equation\n\\[\n\\tau_{R}=\\sigma \\cos \\phi \\cos \\lambda=\\left(12 \\mathrm{MPa}\\right)\\left(\\cos 60^{\\circ}\\right)\\left(\\cos 35^{\\circ}\\right)=4.91 \\mathrm{MPa} \\quad \\text { (717 psi) }\n\\]\nSince the resolved shear stress (4.91 MPa) is less that the critical resolved shear stress (6.2 MPa), the single crystal will not yield.\nHowever, from Equation, the stress at which yielding occurs is\n\\[\n\\sigma_{y}=\\frac{\\tau_{\\text {cross }}}{\\cos \\phi \\cos \\lambda}=\\frac{6.2 \\mathrm{MPa}}{\\left(\\cos 60^{\\circ}\\right)\\left(\\cos 35^ \\circ\\right)}=15.1 \\mathrm{MPa} \\quad\\left(2200 \\mathrm{psi}\\right)\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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^{\\circ}\\) with the tensile axis. Three possible slip directions make angles of \\(30^{\\circ}, 48^{\\circ}\\), and \\(78^{\\circ}\\) with the same tensile axis.\n(a) Which of these three slip directions is most favored?\n(b) If plastic deformation begins at a tensile stress of \\(2.5 \\mathrm{MPa}(355 \\mathrm{psi})\\), determine the critical resolved shear stress for zinc.",
|
||
"answer": "We are asked to compute the critical resolved shear stress for \\(\\mathrm{Zn}\\). As stipulated in the problem, \\(\\phi=65^{\\circ}\\), while possible values for \\(\\lambda\\) are \\(30^{\\circ}, 48^{\\circ}\\), and \\(78^{0}\\).\n(a) Slip will occur along that direction for which (cos \\(\\phi \\cos \\lambda\\) ) is a maximum, or, in this case, for the largest \\(\\cos \\lambda\\). Cosines for the possible \\(\\lambda\\) values are given below.\n\\[\n\\begin{array}{l}\n\\cos \\left(30^{\\circ}\\right)=0.87 \\\\\n\\cos \\left(48^{\\circ}\\right)=0.67 \\\\\n\\cos \\left(78^{\\circ}\\right)=0.21\n\\end{array}\n\\]\nThus, the slip direction is at an angle of \\(30^{\\circ}\\) with the tensile axis.\n(b) From Equation, the critical resolved shear stress is just\n\\[\n\\begin{array}{c}\n\\text { crss }=\\quad y\\left(\\begin{array}{llll}\n\\cos & \\cos & )_{\\max }\n\\end{array}\\right. \\\\\n\\quad=(2.5 \\mathrm{MPa})\\left[\\cos \\left(65^{\\circ}\\right) \\cos \\left(30^{\\circ}\\right)\\right]=0.91 \\mathrm{MPa} \\quad(130 \\mathrm{psi})\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 728,
|
||
"question": "The lower yield point for an iron that has an average grain diameter of \\(1 \\times 10^{-2} \\mathrm{~mm}\\) is \\(230 \\mathrm{MPa}(33,000 \\mathrm{psi})\\). At a grain diameter of \\(6 \\times 10^{-3} \\mathrm{~mm}\\), the yield point increases to \\(275 \\mathrm{MPa}\\) \\((40,000 \\mathrm{psi})\\). At what grain diameter will the lower yield point be \\(310 \\mathrm{MPa}(45,000 \\mathrm{psi})\\) ?\n\\(\\underline{\\text {",
|
||
"answer": "The best way to solve this problem is to first establish two simultaneous expressions of Equation, solve for \\(\\sigma_{0}\\) and \\(k_{\\mathrm{y}}\\), and finally determine the value of \\(d\\) when \\(\\sigma_{\\mathrm{y}}=310 \\mathrm{MPa}\\). The data pertaining to this problem may be tabulated as follows:\n\\begin{tabular}{lcc}\n\\(\\sigma_{\\mathrm{y}}\\) & \\(d(\\mathrm{~mm})\\) & \\(d^{-1 / 2}(\\mathrm{~mm})^{-1 / 2}\\) \\\\\n\\(230 \\mathrm{MPa}\\) & \\(1 \\times 10^{-2}\\) & 10.0 \\\\\n\\(275 \\mathrm{MPa}\\) & \\(6 \\times 10^{-3}\\) & 12.91\n\\end{tabular}\nThe two equations thus become\n\\[\n\\begin{array}{l}\n230 \\mathrm{MPa}=\\sigma_{0}+(10.0) k_{\\mathrm{y}} \\\\\n275 \\mathrm{MPa}=\\sigma_{0}+(12.91) k_{\\mathrm{y}}\n\\end{array}\n\\]\nSolving these two simultaneous equations leads to \\(\\sigma_{0}=75.4 \\mathrm{MPa}\\) and \\(k_{y}=15.46 \\mathrm{MPa}(\\mathrm{mm})^{1 / 2}\\). At a yield strength of \\(310 \\mathrm{MPa}\\)\n\\[\n310 \\mathrm{MPa}=75.4 \\mathrm{MPa}+\\left[15.46 \\mathrm{MPa}(\\mathrm{mm})^{1 / / 2}\\right] d^{-1 / 2}\n\\]\nOr\n\\[\nd^{-1 / 2}=\\frac{310 \\mathrm{MPa}-75.4 \\mathrm{MPa}}{15.46 \\mathrm{MPa}(\\mathrm{mm})^{1 /}\n\\]\nThus,\\[\nd = \\left[\\frac{15.46\\ \\mathrm{MPa}(\\mathrm{mm})^{1/2}}{310\\ \\mathrm{MPa}\\ -75.4\\ \\mathrm{MPa}}\\right]^2 = \\, 4.34 \\, \\times \\, 10^{-3}\\ \\mathrm{mm}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 732,
|
||
"question": "(a) What is the driving force for recrystallization?\n(b) What is the driving force for grain growth?",
|
||
"answer": "(a) The driving force for recrystallization is the difference in internal energy between the strained and unstrained material.\n(b) The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 733,
|
||
"question": "The average grain diameter for a brass material was measured as a function of time at \\(650^{\\circ} \\mathrm{C}\\), which is shown in the following table at two different times:\n\\begin{tabular}{cc}\n\\hline Time (min) & Grain Diameter \\((\\mathrm{mm})\\) \\\\\n\\hline 40 & \\(5.6 \\times 10^{-2}\\) \\\\\n100 & \\(8.0 \\times 10^{-2}\\) \\\\\n\\hline\n\\end{tabular}\n(a) What was the original grain diameter?\n(b) What grain diameter would you predict after \\(200 \\mathrm{~min}\\) at \\(650^{\\circ} \\mathrm{C}\\) ?",
|
||
"answer": "(a) Using the data given and Equation (for \\(n=2\\) )—that is\n\\[\nd^{2} \\quad d_{0}^{2}=K t\n\\]\nwe may set up two simultaneous equations with \\(d_{0}\\) and \\(K\\) as unknowns, as follows:\n\\[\n\\begin{array}{lll}\n\\left(5.6 & 10^{2} \\mathrm{~mm}\\right)^{2} & d_{0}^{2}=(40 \\mathrm{~min}) K \\\\\n\\left(8.0 & 10^{2} \\mathrm{~mm}\\right)^{2} & \\begin{array}{l}\nd_{0}^{2}=(100 \\mathrm{~min}) K\n\\end{array}\n\\end{array}\n\\]\nSolution of these expressions yields a value for \\(d_{0}\\), the original grain diameter, of\n\\[\nd_{0}=0.031 \\mathrm{~mm}\n\\]\nand a value for \\(K\\) of\n\\[\nK=5.44 \\quad 10^{5} \\mathrm{~mm}^{2} / \\mathrm{min}\n\\](b) At \\(200 \\mathrm{~min}\\), the diameter \\(d\\) is computed using a rearranged form of Equation (incorporating values of \\(d_{0}\\) and \\(K\\) that were just determined) as follows:\n\\[\n\\begin{array}{c}\nd=\\sqrt{d_{0}^{2}+K t} \\\\\n\\\\\n=\\sqrt{\\left(0.031 \\mathrm{~mm}\\right)^{2}+\\left(5.44 \\times 10^{-5} \\mathrm{~mm}^{2} / \\mathrm{min}\\right)(200 \\mathrm{~min})} \\\\\n\\\\=0.109 \\mathrm{~mm}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 739,
|
||
"question": "For each of the following pairs of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.\n(a) Branched and atactic poly(vinyl chloride) with a weight-average molecular weight of \\(100,000 \\mathrm{~g} / \\mathrm{mol}\\); linear and isotactic poly(vinyl chloride) having a weight-average molecular weight of \\(75,000 \\mathrm{~g} / \\mathrm{mol}\\)\n(b) Random styrene-butadiene copolymer with \\(5 \\%\\) of possible sites crosslinked; block styrene-butadiene copolymer with \\(10 \\%\\) of possible sites crosslinked\n(c) Branched polyethylene with a number-average molecular weight of \\(100,000 \\mathrm{~g} / 1 \\mathrm{~mol}\\); atactic polypropylene with a number-average molecular weight of \\(150,000 \\mathrm{~g} / \\mathrm{mol}\\)",
|
||
"answer": "(a) 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.\n(b) 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.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 740,
|
||
"question": "For each of the following pairs of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.\n(a) Linear and isotactic poly(vinyl chloride) with a weight-average molecular weight of \\(100,000 \\mathrm{~g} / \\mathrm{mol}\\); branched and atactic poly(vinyl chloride) having a weight-average molecular weight of \\(75,000 \\mathrm{~g} / \\mathrm{mol}\\)\n(b) Graft acrylonitrile-butadiene copolymer with \\(10 \\%\\) of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with \\(5 \\%\\) of possible sites crosslinked\n(c) Network polyester; lightly branched polytetrafluoroethylene",
|
||
"answer": "(a) 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 \\mathrm{~g} / \\mathrm{mol}\\) versus 75,000 \\(\\mathrm{g} / \\mathrm{mol})\\), and tensile strength increases with increasing molecular weight.\n(b) 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.\n(c) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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 \\()\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 744,
|
||
"question": "An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of \\(40 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) (36.4 ksi \\(\\sqrt{\\mathrm{in}}\\).). It has been determined that fracture results at a stress of \\(300 \\mathrm{MPa}(43,500 \\mathrm{psi})\\) when the maximum (or critical) internal crack length is \\(4.0 \\mathrm{~mm}\\) (0.16 in.). For this same component and alloy, will fracture occur at a stress level of \\(260 \\mathrm{MPa}\\) (38,000 psi) when the maximum internal crack length is \\(6.0 \\mathrm{~mm}(0.24 \\mathrm{in.})\\) ? Why or why not?",
|
||
"answer": "We are asked to determine if an aircraft component will fracture for a given fracture toughness \\(\\left(40 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right)\\), stress level \\((260 \\mathrm{MPa})\\), and maximum internal crack length \\((6.0 \\mathrm{~mm})\\), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. (Note: Because the cracks are internal, their lengths are equal to \\(2 a\\).) It first becomes necessary to solve for the parameter \\(Y\\), using Equation, for the conditions under which fracture occurred (i.e., \\(\\sigma=300 \\mathrm{MPa}\\) and \\(2 a=4.0 \\mathrm{~mm}\\) ). Therefore,\n\\[\nY=\\frac{K_{I C}}{\\sigma \\sqrt{\\pi a}}=\\frac{40 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(300 \\mathrm{MPa})} \\sqrt{(2 \\pi)\\left(\\frac{4 \\times 10^{-3} \\mathrm{~m}}{2}\\right)}=1.68\n\\]\nNow we will solve for the product \\(Y \\sqrt{a}\\) for the other set of conditions, so as to ascertain whether or not this value is greater than the \\(K_{I C}\\) for the alloy. Thus,\n\\[\n\\begin{aligned}\nY \\sigma \\sqrt{\\pi a} & =(1.68)(260 \\mathrm{MPa}) \\sqrt{(2 \\pi)\\left(\\frac{6 \\times 10^{-3} \\mathrm{~m}}{2}\\right)} \\\\\n& =42.4 \\mathrm{MPa} \\sqrt{\\mathrm{m}} \\quad(39 \\mathrm{ksi} \\sqrt{\\mathrm{m}}) .\n\\end{aligned}\n\\]\nTherefore, fracture will occur since this value \\(\\left(42.4 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right)\\) is greater than the \\(K_{I C}\\) of the material, \\(40 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 746,
|
||
"question": "A structural component is fabricated from an alloy that has a plane-strain fracture toughness of \\(62 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\). It has been determined that this component fails at a stress of \\(250 \\mathrm{MPa}\\) when the maximum length of a surface crack is \\(1.6 \\mathrm{~mm}\\). What is the maximum allowable surface crack length (in \\(\\mathrm{mm}\\) ) without fracture for this same component exposed to a stress of \\(250 \\mathrm{MPa}\\) and made from another alloy that has a plane strain fracture toughness of \\(51 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\) ?",
|
||
"answer": "This problem asks us to determine the maximum allowable surface crack length without fracture for a structural component for a specified fracture toughness \\(\\left(51 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right)\\), given that fracture occurs for the same component using the same stress level \\((250 \\mathrm{MPa})\\) and another fracture toughness \\(\\left(62 \\mathrm{MPa} \\sqrt{\\mathrm{m}}\\right)\\). (Note: Because the cracks surface cracks, their lengths are equal to a.) The maximum crack length without fracture \\(a_{C}\\) may be calculated using Equation -i.e.,\n\\[\na_{C}=\\frac{1}{\\pi\\left(\\frac{K_{I_{C}}}{\\sigma Y}\\right)^{2}}\n\\]\nwhere \\(K_{I C}\\) is the plane strain fracture toughness, \\(\\sigma\\) is the applied stress, and \\(Y\\) is a design parameter. Using data for the first alloy \\(\\left[K_{I C}=62 \\mathrm{MPa} \\sqrt{\\mathrm{m}}, a_{C}=1.6 \\mathrm{~mm}(1.6 \\times 10^{-3} \\mathrm{~m}), \\sigma=250\\right.\\) \\(\\mathrm{MPa}\\) ], it is possible to calculate the value of \\(Y\\) using a rearranged form of the above equation as follows:\n\\[\n\\begin{aligned}\nY & =\\frac{K_{I C}}{\\sigma \\sqrt{\\pi a_{C}}} \\\\\n& =\\frac{62 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(250 \\mathrm{MPa}) \\sqrt{(\\sigma)(1.6 \\times 10^{-3} \\mathrm{~m})}}\n\\end{aligned}\n\\]\\[\n = 3.50\n\\]\n\nUsing this value of \\(Y\\), the maximum crack length for the second alloy is determined, again using Equation as\n\n\\[\n\\begin{aligned}\na_{c} & =\\frac{1}{\\pi\\left(\\frac{K_{I C}}{\\sigma Y}\\right)^{2}} \\\\\n& =\\frac{1}{\\pi}\\left[\\frac{51 \\mathrm{MPa} \\sqrt{\\mathrm{m}}}{(250 \\mathrm{MPa})(3.50)}\\right]^{2} \\\\\n& =0.00108 \\mathrm{~m}=1.08 \\mathrm{~mm}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 749,
|
||
"question": "Briefly answer the following: \\\\ (a) Why may there be significant scatter in the fracture strength for some given ceramic material? \\\\ (b) Why does the fracture strength increase with decreasing specimen size?",
|
||
"answer": "(a) 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.\n(b) The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 752,
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals: tin, molybdenum, iron, gold, zinc, and chromium.",
|
||
"answer": "Creep becomes important at about \\(0.4 T_{m}, T_{m}\\) being the absolute melting temperature of the metal.\nFor \\(\\mathrm{Sn}, 0.4 T_{m}=(0.4)(232+273)=202 \\mathrm{~K}\\) or \\(-71^{\\circ} \\mathrm{C}\\left(-96^{\\circ} \\mathrm{F}\\right)\\)\nFor Mo, \\(0.4 T_{m}=(0.4)(2617+273)=1049 \\mathrm{~K}\\) or \\(776^{\\circ} \\mathrm{C}\\left(1429^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Fe}, 0.4 T_{m}=(0.4)(1538+273)=724 \\mathrm{~K}\\) or \\(451^{\\circ} \\mathrm{C}\\left(845^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Au}, 0.4 T_{m}=(0.4)(1064+273)=535 \\mathrm{~K}\\) or \\(262^{\\circ} \\mathrm{C}\\left(504^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Zn}, 0.4 T_{m}=(0.4)(420+273)=277 \\mathrm{~K}\\) or \\(4^{\\circ} \\mathrm{C}\\left(39^{\\circ} \\mathrm{F}\\right)\\)\nFor \\(\\mathrm{Cr}, 0.4 T_{m}=(0.4)(1875+273)=859 \\mathrm{~K}\\) or \\(586^{\\circ} \\mathrm{C}\\left(1087^{\\circ} \\mathrm{F}\\right)\\)",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 753,
|
||
"question": "A cylindrical specimen \\(13.2 \\mathrm{~mm}\\) in diameter of an \\(S-590\\) alloy is to be exposed to a tensile load of \\(27,000 \\mathrm{~N}\\). At approximately what temperature will the steady-state creep be \\(10^{-3}\\) \\(\\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 \\(d_{0}=13.2 \\mathrm{~mm}(13.2 \\times 10^{-3} \\mathrm{~m})\\) and \\(F=27,000 \\mathrm{~N}\\), the stress is equal to\n\\[\n\\begin{aligned}\n\\sigma= & \\frac{27,000 \\mathrm{~N}}{\\pi\\left(\\frac{13.2 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}} \\\\\n& =197 \\times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}=197 \\mathrm{MPa}\n\\end{aligned}\n\\]\nor approximately \\(200 \\mathrm{MPa}\\). From Figure 9.40 the point corresponding \\(10^{-3} \\mathrm{~h}^{-1}\\) and \\(200 \\mathrm{MPa}\\) lies approximately midway between \\(730^{\\circ} \\mathrm{C}\\) and \\(815^{\\circ} \\mathrm{C}\\) lines; therefore, the temperature would be approximately \\(775^{\\circ} \\mathrm{C}\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 755,
|
||
"question": "Steady-state creep rate data are given in the following table for some alloy taken at \\(200^{\\circ} \\mathrm{C}(473 \\mathrm{~K})\\) :\n\\begin{tabular}{ll}\n\\hline\\(s\\left(h^{-1}\\right)\\) & \\(\\sigma[\\mathrm{MPa}(\\mathrm{psi})]\\) \\\\\n\\hline \\(2.5 \\times 10^{-3}\\) & \\(55(8000)\\) \\\\\n\\(2.4 \\times 10^{-2}\\) & \\(69(10,000)\\) \\\\\n\\hline\n\\end{tabular}\nIf it is known that the activation energy for creep is \\(140,000 \\mathrm{~J} / \\mathrm{mol}\\), compute the steady-state creep rate at a temperature of \\(250^{\\circ} \\mathrm{C}(523 \\mathrm{~K})\\) and a stress level of \\(48 \\mathrm{MPa}(7000 \\mathrm{psi})\\).",
|
||
"answer": "This problem gives \\(\\dot{\\varepsilon}_{s}\\) values at two different stress levels and \\(200^{\\circ} \\mathrm{C}\\), and the activation energy for creep, and asks that we determine the steady-state creep rate at \\(250^{\\circ} \\mathrm{C}\\) and \\(48 \\mathrm{MPa}\\) (7000 psi).\nTaking natural logarithms of both sides of Equation yields\n\\[\n\\ln \\dot{\\varepsilon}_{s}=\\ln K_{2}+n \\ln \\sigma-\\frac{Q_{c}}{R T}\n\\]\nWith the given data there are two unknowns in this equation--namely \\(K_{2}\\) and \\(n\\). Using the data provided in the problem statement we can set up two independent equations as follows:\n\\[\n\\begin{array}{l}\n\\ln \\left(2.5 \\times 10^{-3} \\mathrm{~h}^{-1}\\right)=\\ln K_{2}+n \\ln (55 \\mathrm{MPa})-\\frac{140,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(473 \\mathrm{~K})} \\\\\n\\ln \\left(2.4 \\times 10^{-2} \\mathrm{~h}^{-1}\\right)=\\ln K_{2}+r \\ln (69 \\mathrm{MPa})-\\frac{140,000 \\mathrm{J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} /\\left.\\mathrm{mol}-\\mathrm{K}\\right)(473 \\mathrm{~K})}\n\\end{array}\n\\]Now, solving simultaneously for \\(n\\) and \\(K_{2}\\) leads to \\(n=9.97\\) and \\(K_{2}=3.27 \\times 10^{-5} \\mathrm{~h}^{-1}\\). Thus it is now possible to solve for \\(\\varepsilon_{s}\\) at \\(48 \\mathrm{MPa}\\) and \\(523 \\mathrm{~K}\\) using Equation as\n\\[\n\\begin{aligned}\n\\dot{\\varepsilon}_{s} & =K_{2} \\sigma^{n} \\exp \\left(-\\frac{Q_{c}}{R T}\\right) \\\\\n& =(3.27 \\times 10^{-5} \\mathrm{~h}^{-2})(48 \\mathrm{MPa})^{9.97} \\exp \\left[-\\frac{140,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K})(523 \\mathrm{~K})}\\right] \\\\\n& =1.94 \\times 10^{-2} \\mathrm{~h}^{-1}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 756,
|
||
"question": "Steady-state creep data taken for an iron at a stress level of \\(140 \\mathrm{MPa}(20,000 \\mathrm{psi})\\) are given here:\n\\begin{tabular}{cc}\n\\hline\\(\\dot{\\epsilon}_{s}\\left(h^{-1}\\right)\\) & \\(T(K)\\) \\\\\n\\hline \\(6.6 \\times 10^{-4}\\) & 1090 \\\\\n\\(8.8 \\times 10^{-2}\\) & 1200 \\\\\n\\hline\n\\end{tabular}\nIf it is known that the value of the stress exponent \\(n\\) for this alloy is 8.5 , compute the steady-state creep rate at \\(1300 \\mathrm{~K}\\) and a stress level of \\(83 \\mathrm{MPa}(12,000 \\mathrm{psi})\\).",
|
||
"answer": "This problem gives \\(\\dot{\\varepsilon}_{s}\\) values at two different temperatures and \\(140 \\mathrm{MPa}(20,000 \\mathrm{spi})\\), and the value of the stress exponent \\(n=8.5\\), and asks that we determine the steady-state creep rate at a stress of \\(83 \\mathrm{MPa}(12,000 \\mathrm{spi})\\) and \\(1300 \\mathrm{~K}\\).\nTaking natural logarithms of both sides of Equation yields\n\\[\n\\ln \\dot{\\varepsilon}_{s}=\\ln K_{2}+n \\ln \\sigma-\\frac{Q_{c}}{R T}\n\\]\nWith the given data there are two unknowns in this equation--namely \\(K_{2}\\) and \\(Q_{c}\\). Using the data provided in the problem statement we can set up two independent equations as follows:\n\\[\n\\begin{array}{l}\n\\ln \\left(6.6 \\times 10^{-4} \\mathrm{~h}^{-1}\\right)=\\ln K_{2}+(8.5) \\ln (140 \\mathrm{MPa})-\\frac{Q_{c}}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{K})(1090 \\mathrm{~K})} \\\\\n\\ln \\left(8.8 \\times 10^{-2} \\mathrm{~h}^{-1}\\right)=\\ln K_{2}+(0.5) \\ln (140 \\mathrm{MPa})-\\frac{Q}{(8.31 \\mathrm{~J} / \\mathrm{mol}-\\mathrm{L})(1200 \\mathrm{~K})}\n\\end{array}\n\\]Now, solving simultaneously for \\(K_{2}\\) and \\(Q_{c}\\) leads to \\(K_{2}=57.5 \\mathrm{~h}^{-1}\\) and \\(Q_{c}=483,500 \\mathrm{~J} / \\mathrm{mol}\\). Thus, it is now possible to solve for \\(\\dot{\\varepsilon}_{s}\\) at \\(83 \\mathrm{MPa}\\) and \\(1300 \\mathrm{~K}\\) using Equation as\n\\[\n\\begin{aligned}\n\\dot{\\varepsilon}_{s} & =K_{2} \\sigma^{n} \\exp \\left(-\\frac{Q_{c}}{R T}\\right) \\\\\n& =(57.5 \\mathrm{~h}^{-1})(83 \\mathrm{MPa})^{8.5} \\exp \\left[-\\frac{483,500 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K})(1300 \\mathrm{~K})}\\right] \\\\\n& =4.31 \\times 10^{-2} \\mathrm{~h}^{-1}\n\\end{aligned}\n\\]",
|
||
"is_select": 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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 758,
|
||
"question": "It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, and that the value of \\(n\\) in the exponential is 5.0. If, at some temperature, the fraction recrystallized is 0.30 after 100 min, determine the rate of recrystallization at this temperature.",
|
||
"answer": "This problem gives us the value of \\(y(0.30)\\) at some time \\(t(100 \\mathrm{~min})\\), and also the value of \\(n\\) (5.0) for the recrystallization of an alloy at some temperature, and then asks that we determine the rate of recrystallization at this same temperature. It is first necessary to calculate the value of \\(k\\). We begin by rearranging Equation as\n\\[\n\\exp \\left(-k t^{n}\\right)=1-y\n\\]\nAnd then take natural logarithms of both sides:\n\\[\n-k t^{n}=\\ln (1-y)\n\\]\nNow solving for \\(k\\) gives\n\\[\nk=-\\frac{\\ln (1-y)}{t^{n}}\n\\]\nWhich, using the values cited above for \\(y, n\\), and \\(t\\) yields\n\\[\nk=-\\frac{\\ln (1-0.30)}{(100 \\mathrm{~min})^{5}}=3.57 \\times 10^{-11}\n\\]At this point we want to compute \\(t_{0.5}\\), the value of \\(t\\) for \\(y=0.5\\), which means that it is necessary to establish a form of Equation in which \\(t\\) is the dependent variable. Rearrangement of Equation above leads to\n\\[\nt^{n}=\\frac{\\ln (1 \\quad y)}{k}\n\\]\nAnd solving this expression for \\(t\\) leads to\n\\[\nt=\\left[-\\frac{\\ln (1-y)}{k}\\right]^{1 / n}\n\\]\nFor \\(t_{0.5}\\), this equation takes the form\n\\[\nt_{0.5}=\\left[-\\frac{\\ln (1-0.5)}{k}\\right]^{1 / n}\n\\]\nAnd incorporation of the value of \\(k\\) determined above, as well as the value of \\(n\\) cited in the problem statement (5.0), we compute the value of \\(t_{0.5}\\) as follows:\n\\[\nt_{0.5}=\\left[-\\frac{\\ln (1-0 .5)}{3.57 \\times 10^{-11}}\\right]^{1 / 5}=114.2 \\text { min }\n\\]\nTherefore, from Equation, the rate is just\n\\[\n\\begin{array}{l}\n\\text { rate }=\\frac{1}{t_{0.5}}=\\frac{1}{114.2 \\min }=8.76 \\times 10^{-3}(\\min )^{-1}\n\\end{array}\n\\]",
|
||
"is_select": 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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 761,
|
||
"question": "Briefly cite the differences among pearlite, bainite, and spheroidite relative to microstructure and mechanical properties.",
|
||
"answer": "The microstructures of pearlite, bainite, and spheroidite all consist of \\(\\alpha\\)-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an \\(\\alpha\\)-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.\nBainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 764,
|
||
"question": "For each of the following pairs of polymers do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.\n(a) Branched polyethylene having a number-average molecular weight of \\(850,000 \\mathrm{~g} / \\mathrm{mol}\\); linear polyethylene having a number-average molecular weight of \\(850,000 \\mathrm {~g} / \\mathrm{mol}\\)\n(b) Polytetrafluoroethylene having a density of \\(2.14 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and a weight-average molecular weight of \\(600,000 \\mathrm{~g} / \\mathrm{mol}\\); PTFE having a density of \\(2.20 \\mathrm{~g} / \\mathrm{cm}^{3}\\) and a weight-average molecular weight of \\(600,000 \\mathrm{~g} / \\mathbf{m o l}\\)\n(c) Linear and syndiotactic poly(vinyl chloride) having a number-average molecular weight of \\(500,000 \\mathrm{~g} / \\mathrm{mol}\\); linear polylethylene having a number-average molecular weight of \\(225,000 \\mathrm{~g} / \\mathrm{mol}\\)\n(d) Linear and syndiotactic polypropylene having a weight-average molecular weight of \\(500,000 \\mathrm{~g} / 0 \\mathrm{~mol}\\); linear and atactic polypropylene having a weight-average molecular weight of \\(750,000 \\mathrm{~g} / \\mathrm{mol}\\)",
|
||
"answer": "(a) 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.\n(b) Yes, it is possible to determine which polymer has the higher melting temperature. Of these two polytetrafluoroethylene polymers, the PTFE with the higher density \\(\\left(2.20 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\) 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.\n(c) 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 aseasily 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.\n(d) 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 \\(T_{m}\\), since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene \\((500,000 \\mathrm{~g} / \\mathrm{mol})\\) is less than for the atactic material \\((750,000\\) \\(\\mathrm{g} / \\mathrm{mol}\\) ); a lowering of molecular weight generally results in a reduction of melting temperature.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 765,
|
||
"question": "(a) Compute the electrical conductivity of a cylindrical silicon specimen \\(7.0 \\mathrm{~mm}(0.28 \\mathrm{in}\\).) diameter and \\(57 \\mathrm{~mm}(2.25 \\mathrm{in}\\).) in length in which a current of 0.25 A passes in an axial direction. A voltage of \\(24 \\mathrm{~V}\\) is measured across two probes that are separated by \\(45 \\mathrm{~mm}\\) (1.75 in.).\n(b) Compute the resistance over the entire \\(57 \\mathrm{~mm}\\) (2.25 in.) of the specimen.",
|
||
"answer": "This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen.\n(a) We use Equations for the conductivity, as follows:\n\\[\n\\begin{aligned}\n\\sigma & =\\frac{1}{\\rho} \\\\\n& =\\frac{1}{V A}=\\frac{I l}{V A} \\\\\n& =\\frac{I l}{V \\pi\\left(\\frac{d}{2}\\right)^{2}}\n\\end{aligned}\n\\]\nSince the cross-sectional area of a cylinder having a diameter \\(d\\) is\\[\nA=\\pi\\left(\\frac{d}{2}\\right)^{2}\n\\]\nNow, incorporation of values for the several parameters provided in the problem statement, leads to the following:\n\\[\n\\begin{array}{l}\n\\sigma=\\frac{I l}{V \\pi\\left(\\frac{d}{2}\\right)^{2}} \\\\\n\\quad=\\frac{(0.25 \\mathrm{~A})(45 \\times 10^{-3} \\mathrm{~m})}{(24 \\mathrm{~V})(\\pi)\\left(\\frac{7.0 \\times 10^{-3} \\mathrm{~m}}{2}\\right)^{2}}=12.2 \\mathrm{~(Q\\cdot \\mathrm{m})}^{-1}\n\\end{array}\n\\]\n(b) The resistance, \\(R\\), over the entire specimen length may be computed by combining and rearranging Equations as follows:\n\\[\n\\begin{array}{l}\nR=\\frac{l \\rho}{A} \\\\\n\\quad=\\frac{l\\left(\\frac{1}{\\sigma}\\right)}{A}=\\frac{l}{\\sigma A} \\\\\n\\quad=\\frac{l}{\\sigma \\pi\\left(\\frac{d}{2}\\right)^{2}}\n\\end{array}\n\\]\nThe value of \\(R\\) may be computed upon entering values given in the problem statement into the preceding expression:\\[\n\\begin{array}{l}\nR=\\displaystyle\\frac{l}{\\sigma x}\\left(\\displaystyle\\frac{d}{2}\\right)^2\\\\\n\\\\=\\frac{57\\times10^{-3}\\,\\,\\mathrm{m}}{\\left[12.2\\,\\,(\\Omega\\cdot\\mathrm{m})^{-1}\\right](\\pi)\\left(\\frac{7.0\\times10^{-3}\\,\\,\\mathrm{m}}{2}\\right)^2}=121.4\\,\\,\\Omega\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 767,
|
||
"question": "In terms of electron energy band structure, discuss reasons for the difference in electrical \\\\ conductivity among metals, semiconductors, and insulators.",
|
||
"answer": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 769,
|
||
"question": "Briefly state what is meant by the drift velocity and mobility 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.\nThe 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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 770,
|
||
"question": "(a) Calculate the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is \\(500 \\mathrm{~V} / \\mathrm{m}\\).\n(b) Under these circumstances, how long does it take an electron to traverse a \\(25-\\mathrm{mm}\\) (1in.) length of crystal?",
|
||
"answer": "(a) The drift velocity of electrons in Si may be determined using Equation. Since the room temperature mobility of electrons is \\(0.145 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\), and the electric field is \\(500 \\mathrm{~V} / \\mathrm{m}(\\) as stipulated in the problem statement), the drift velocity is equal to\n\\[\n\\begin{array}{c}\nv_{d}=\\mu_{e} E \\\\\n=(0.145 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s})(500 \\mathrm{~V} / \\mathrm{m})=72.5 \\mathrm{~m} / \\mathrm{s}\n\\end{array}\n\\]\n(b) The time, \\(t\\), required to traverse a given length, \\(l(=25 \\mathrm{~mm})\\), is just the length divided by the drift velocity, as follows:\n\\[\nt=\\frac{l}{v_{d}}=\\frac{25 \\times 10^{-3} \\mathrm{~m}}{72.5 \\mathrm{~m} / \\mathrm{s}}=3.45 \\times 10^{-4} \\mathrm{~s}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 771,
|
||
"question": "At room temperature the electrical conductivity and the electron mobility for aluminum are \\(3.8 \\times 10^{7}(\\Omega \\cdot \\mathrm{m})^{-1}\\) and \\(0.0012 \\mathrm{~m}^{2} / \\mathrm{V} \\cdot \\mathrm{s}\\), respectively.\n(a) Compute the number of free electrons per cubic meter for aluminum at room temperature.\n(b) What is the number of free electrons per aluminum atom? Assume a density of \\(2.7\\) \\(\\mathrm{g} / \\mathrm{cm}^{3}\\).",
|
||
"answer": "(a) The number of free electrons per cubic meter for aluminum at room temperature may be computed using rearranged form of Equation as follows:\n\\[\n\\begin{aligned}\nn & =\\frac{\\sigma}{|e| \\mu_{e}} \\\\\n& =\\frac{3.8 \\times 10^{7}(\\Omega \\cdot \\mathrm{m} )^{-1}}{\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)\\left(0.0012 \\mathrm{~m}^{2} / \\mathrm{N} \\cdot \\mathrm{s}\\right)} \\\\\n& =1.98 \\times 10^{29} \\mathrm{~m}^{-3}\n\\end{aligned}\n\\]\n(b) In order to calculate the number of free electrons per aluminum atom, we must first determine the number of copper atoms per cubic meter, \\(N_{\\mathrm{Al}}\\). From Equation (and using the atomic weight for \\(\\mathrm{Al}\\) found inside the front cover-viz. \\(26.98 \\mathrm{~g} / \\mathrm{mol}\\) ) then the values of \\(N_{\\mathrm{Al}}\\) is determined as follows:\n\\[\nN_{\\mathrm{Al}}=\\frac{N_{\\mathrm{A}} \\rho^{\\prime}}{A_{\\mathrm{Al}}}\n\\]\\[\n\\begin{array}{c}\n\\frac{\\left(6.022 \\times 10^{23} \\mathrm{atoms} / \\mathrm{mol}\\right)\\left(2.7 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(10^{6} \\mathrm{~cm}^{3} / \\mathrm{m}^{3}\\right)}{26.98 \\mathrm{~g} / \\mathrm{mol}} \\\\\n\\\\=6.03 \\times 10^{28} \\mathrm{~m}^{-3}\n\\end{array}\n\\]\n(Note: in the above expression, density is represented by \\(\\rho^{\\prime}\\) in order to avoid confusion with resistivity which is designated by \\(\\rho\\).) And, finally, the number of free electrons per aluminum atom \\(\\left(n / N_{\\mathrm{Al}}\\right)\\) is equal to the following:\n\\[\n\\frac{n}{N_{\\mathrm{Al}}}=\\frac{1.98 \\times 10^{29} \\mathrm{~m}^{-3}}{6.03 \\times 10^{28} \\mathrm{~m} \\mathrm{~m}^{-3}}=3.28 \\mathrm{electrons} / \\mathrm{Al} \\mathrm{atom}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 774,
|
||
"question": "Predict whether each of the following elements will act as a donor or an acceptor when added to the indicated semiconducting material. Assume that the impurity elements are substitutional.\n\\begin{tabular}{cc}\n\\hline Impurity & Semiconductor \\\\\n\\hline\\(N\\) & \\(S i\\) \\\\\n\\(B\\) & \\(G e\\) \\\\\n\\(S\\) & InSb \\\\\nIn & CdS \\\\\n\\(A s\\) & ZnTe \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "Nitrogen will act as a donor in \\(\\mathrm{Si}\\). Since it \\((\\mathrm{N})\\) is from group VA of the periodic table, and an \\(\\mathrm{N}\\) atom has one more valence electron than an \\(\\mathrm{Si}\\) atom.\nBoron will act as an acceptor in \\(\\mathrm{Ge}\\). Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom.\nSulfur will act as a donor in InSb. Since \\(S\\) is from group VIA of the periodic table, it will substitute for \\(\\mathrm{Sb}\\); an \\(\\mathrm{S}\\) atom has one more valence electron than an \\(\\mathrm{Sb}\\) atom.\nIndium will act as a donor in \\(\\mathrm{CdS}\\). Since In is from group IIIA of the periodic table, it will substitute for \\(\\mathrm{Cd}\\); an In atom has one more valence electron than a \\(\\mathrm{Cd}\\) atom.\nArsenic will act as an acceptor in \\(\\mathrm{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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 775,
|
||
"question": "Germanium to which \\(10^{24} \\mathrm{~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).\n(a) Is this material n-type or p-type?\n(b) Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and \\(0.05 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\), respectively.",
|
||
"answer": "(a) This germanium material to which has been added \\(10^{24} \\mathrm{~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.)\n(b) 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} \\mathrm{~m}^{-3}\\), and, as given in the problem statement, \\(\\mu_{e}=0.1 \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\\). Thus, the conductivity is equal to\n\\[\n\\begin{array}{c}\n\\sigma=n|e| \\mu_{e} \\\\\n=\\left(10^{24} \\mathrm{~m}^{-3}\\right)\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)\\left(0.1 \\mathrm{~m}^{2} / \\mathrm{V}-\\right. \\\\\n\\left.=1.60 \\times 10^{4}(\\Omega \\cdot \\mathrm{m})^{-1}\n\\end{array}\n\\]",
|
||
"is_select": 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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 777,
|
||
"question": "At temperatures between \\(540^{\\circ} \\mathrm{C}(813 \\mathrm{~K})\\) and \\(727^{\\circ} \\mathrm{C}(1000 \\mathrm{~K})\\), the activation energy and preexponential for the diffusion coefficient of \\(\\mathrm{Na}^{+}\\)in \\(\\mathrm{NaCl}\\) are \\(173,000 \\mathrm{~J} / \\mathrm{mol}\\) and \\(4.0 \\times 10^{-4}\\) \\(\\mathrm{m}^{2} / \\mathrm{s}\\), respectively. Compute the mobility for an \\(\\mathrm{Na}^{+}\\)ion at \\(600^{\\circ} \\mathrm{C}(873 \\mathrm{~K})\\).",
|
||
"answer": "For this problem, we are given, for \\(\\mathrm{NaCl}\\), the activation energy \\((173,000 \\mathrm{~J} / \\mathrm{mol})\\) and preexponential \\(\\left(4.0 \\times 10^{-4} \\mathrm{~m}^{2} / \\mathrm{s}\\right)\\) for the diffusion coefficient of \\(\\mathrm{Na}^{+}\\)and are asked to compute the mobility for a \\(\\mathrm{Na}^{+}\\)ion at \\(873 \\mathrm{~K}\\). The mobility, \\(\\mathrm{Na}^{+}\\), may be computed using Equation; however, this expression also includes the diffusion coefficient \\(D_{\\mathrm{Na}}{ }^{+}\\), which is determined using Equation as follows:\n\\[\n\\begin{aligned}\nD_{\\mathrm{Na}}{ }^{+} & =D_{0} \\exp \\left(-\\frac{Q_{d}}{R T}\\right) \\\\\n& =\\left(4.0 \\times 10^{-4} \\mathrm{~m} \\mathrm{~m}^{2} / \\mathrm{s}\\right) \\exp \\left[-\\frac{173,000 \\mathrm{~J} / \\mathrm{mol}}{(8.31 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K})(873 \\mathrm{~K})}\\right] \\\\\n& =1.76 \\times 10^{-14} \\mathrm{~m}^{2} / \\mathrm{s}\n\\end{aligned}\n\\]\nNow solving for \\(\\mathrm{Na}^{+}\\)using Equation (assuming that \\(n_{\\mathrm{Na}}{ }^{+}=1\\), since the valence of the \\(\\mathrm{Na}^{+}\\) ion is +1 ) yields the following:\n\\[\n\\begin{aligned}\n\\mu_{\\mathrm{Na}}{ }^{+} & =\\frac{n_{\\mathrm{Na}} \\cdot e D_{\\mathrm{Na}}{ }^{+}}{k T} \\\\\n& =\\frac{(1)\\left(1.602 \\times 10^{-19} \\mathrm{C} / \\mathrm{atom}\\right)\\left(1.76 \\times 10^{-14} \\mathrm{~m} \\mathrm{~m}^{2} / \\mathbf{s}\\right)}{\\left(1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{atom}-\\mathrm{K}\\right)(873 \\mathrm{~K})} \\\\\n& =2.34 \\times 10^{-13} \\mathrm{~m}^{2} / \\mathrm{V}-\\mathrm{s}\n\\end{aligned}\n\\]\\[\n\\left\\{\n\\begin{array}{l}",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 779,
|
||
"question": "Consider a parallel-plate capacitor having an area of \\(3225 \\mathrm{~mm}^{2}(5 \\mathrm{in} .2)\\), a plate separation of \\(1 \\mathrm{~mm}(0.04 \\mathrm{in}\\).), and a material having a dielectric constant of 3.5 positioned between the plates.\n(a) What is the capacitance of this capacitor?\n(b) Compute the electric field that must be applied for \\(2 \\times 10^{-8} \\mathrm{C}\\) to be stored on each plate.",
|
||
"answer": "(a) We are first asked to compute the capacitance. Combining Equations leads to the following expression:\n\\[\n\\begin{aligned}\nC & =\\varepsilon \\frac{A}{l} \\\\\n& =\\frac{\\varepsilon_{r} \\varepsilon_{0} A}{l}\n\\end{aligned}\n\\]\nIncorporation of values given in the problem statement for parameters on the right-hand side of this expression, and solving for \\(C\\) yields\n\\[\n\\begin{aligned}\nC & =\\frac{\\varepsilon_{r} \\varepsilon_{0} A}{\\left.l\\right.} \\\\\n& =\\frac{(3.5)\\left(8.85 \\times 10^{-12} \\mathrm{~F} / \\mathrm{m}\\right)\\left(3225 \\mathrm{~mm}^{2}\\right)\\left(1 \\mathrm{~m}^{2} / 10^{6} \\mathrm{~mm}^{2}\\right)}{10^{-3} \\mathrm{~m}} \\\\\n& =10^{10} \\mathrm{~F}=100 \\mathrm{pF}\n\\end{aligned}\n\\]\n(b) Now we are asked to compute the electric field that must be applied in order for \\(2 \\times\\) \\(10^{-8} \\mathrm{C}\\) to be stored on each plate. First we need to solve for \\(V\\) in Equation as\\[\nV=\\frac{Q}{C}=\\frac{2 \\times 10^{-8}\\ \\mathrm{C}}{10^{-10}\\ \\mathrm{F}}=200\\ \\mathrm{V}\n\\]\nThe electric field \\(E\\) may now be determined using Equation, as follows:\n\\[\nE=\\frac{V}{l}=\\frac{200}{10^{-3} \\mathrm{\\ m}} \\mathrm{V}=2.0 \\times 10^{5} \\mathrm{~V} / \\mathrm{m}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 780,
|
||
"question": "The polarization \\(P\\) of a dielectric material positioned within a parallel-plate capacitor is to be \\(4.0 \\times 10^{-6} \\mathrm{C} / \\mathrm{m}^{2}\\).\n(a) What must be the dielectric constant if an electric field of \\(10^{5} \\mathrm{~V} / \\mathrm{m}\\) is applied?\n(b) What will be the dielectric displacement \\(D\\) ?",
|
||
"answer": "(a) We solve this portion of the problem by computing the value of \\(\\varepsilon_{r}\\) using Equation. Rearranging this equation such that \\(\\varepsilon_{r}\\) is the dependent parameter gives the following:\n\\[\n\\varepsilon_{r}=\\frac{P}{\\varepsilon_{0} E}+1\n\\]\nWhen values of the parameters on the right-hand side of this equation are inserted the value of \\(\\varepsilon_{r}\\) equal to\n\\[\n\\begin{aligned}\n\\varepsilon_{r}= & \\frac{4.0 \\times 10^{-6} \\mathrm{C} / \\mathrm {m}^{2}}{\\left(8.85 \\times 10^{-12} \\mathrm{~F} / \\mathrm {m}\\right)\\left(1 \\times 10^{5} \\mathrm{~V} / \\mathrm {m}\\right)}+1 \\\\\n& =5.52\n\\end{aligned}\n\\]\n(b) The dielectric displacement \\(D\\) may be determined using Equation, as follows:\n\\[\n\\begin{aligned}\nD & =\\varepsilon_{0} E+P \\\\\n& =\\left(8.85 \\times 10^{-12} \\mathrm{~V} / \\mathrm {m}\\right)\\left(1 \\times 10^{5}\\mathrm{~V} / \\mathrm {m}\\right)+4.0 \\times 10^{-6} \\mathrm{C} / \\mathrm m^{2} \\\\\n& =4.89 \\times 10^{-6} \\mathrm{C} / \\mathrm{m}^{\\mathbf{2}}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.\n(b) For gaseous argon, solid \\(\\mathrm{LiF}\\), liquid \\(\\mathrm{H}_{2} \\mathrm{O}\\), and solid \\(\\mathrm{Si}\\), what kind(s) of polarization is (are) possible? Why?",
|
||
"answer": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.\n(b) Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments.\nBoth 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.\nBoth electronic and orientation polarizations are found in liquid \\(\\mathrm{H}_{2} \\mathrm{O}\\). The \\(\\mathrm{H}_{2} \\mathrm{O}\\) molecules have permanent dipole moments that are easily oriented in the liquid state.\nOnly 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 782,
|
||
"question": "(a) List the four classifications of steels.\n(b) For each, briefly describe the properties and typical applications.",
|
||
"answer": "The four classifications of steels, their properties, and typical applications are as follows:\nLow Carbon Steels\nProperties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable.\nTypical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans.\nMedium Carbon Steels\nProperties: heat treatable, relatively large combinations of mechanical characteristics.\nTypical applications: railway wheels and tracks, gears, crankshafts, and machine parts.\nHigh Carbon Steels\nProperties: hard, strong, and relatively brittle.\nTypical applications: chisels, hammers, knives, and hacksaw blades.\nHigh Alloy Steels (Stainless and Tool)\nProperties: hard and wear resistant; resistant to corrosion in a large variety of environments.\nTypical applications: cutting tools, drills, cutlery, food processing, and surgical tools.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 783,
|
||
"question": "Compare gray and malleable cast irons with respect to\n(a) composition and heat treatment\n(b) microstructure\n(c) mechanical characteristics.",
|
||
"answer": "This question asks us to compare various aspects of gray and malleable cast irons.\n(a) With respect to composition and heat treatment:\nGray iron--2.5 to \\(4.0 \\mathrm{wt} \\% \\mathrm{C}\\) and 1.0 to \\(3.0 \\mathrm{wt} \\%\\) Si. For most gray irons there is no heat treatment after solidification.\nMalleable iron--2.5 to \\(4.0 \\mathrm{wt} \\% \\mathrm{~C}\\) and less than \\(1.0 \\mathrm{wt} \\% \\mathrm{Si}\\). White iron is heated in a nonoxidizing atmosphere and at a temperature between 800 and \\(900^{\\circ} \\mathrm{C}\\) for an extended time period.\n(b) With respect to microstructure:\nGray iron--Graphite flakes are embedded in a ferrite or pearlite matrix.\nMalleable iron--Graphite clusters are embedded in a ferrite or pearlite matrix.\n(c) With respect to mechanical characteristics:\nGray iron--Relatively weak and brittle in tension; good capacity for damping vibrations.\nMalleable iron--Moderate strength and ductility.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 784,
|
||
"question": "Give the distinctive features, limitations, and applications of the following alloy groups: titanium alloys, refractory metals, superalloys, and noble metals.",
|
||
"answer": "This question asks us for the distinctive features, limitations, and applications of several alloy groups.Titanium Alloys}\nDistinctive features: relatively low density, high melting temperatures, and high strengths are possible.\nLimitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine.\nApplications: aircraft structures, space vehicles, and in chemical and petroleum industries.Refractory Metals}\nDistinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths.\nLimitation: some experience rapid oxidation at elevated temperatures.\nApplications: extrusion dies, structural parts in space vehicles, incandescent light filaments, \\(\\mathrm{x}\\)-ray tubes, and welding electrodes.Superalloys}\nDistinctive features: able to withstand high temperatures and oxidizing atmospheres for long time periods.\nApplications: aircraft turbines, nuclear reactors, and petrochemical equipment.Noble Metals}\nDistinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile.\nLimitation: expensive.\nApplications: jewelry, dental restoration materials, coins, catalysts, and thermocouples.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 786,
|
||
"question": "Compare sand, die, investment, lost-foam, and continuous casting techniques.",
|
||
"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.\nFor 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.\nFor 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.\nFor 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.\nFor 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 789,
|
||
"question": "Contrast compression, injection, and transfer molding techniques that are 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. For transfer molding, the solid materials (normally thermosetting in nature) are first melted in the transfer chamber prior to being forced into the die. And, 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 790,
|
||
"question": "Why must fiber materials that are melt-spun and then drawn be thermoplastic? Cite two reasons.",
|
||
"answer": "Fiber materials that are melt spun must be thermoplastic because: (1) In order to be melt spun, they must be capable of forming a viscous liquid when heated, which is not possible for thermosets. (2) During drawing, mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively brittle, they are not easily elongated.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 791,
|
||
"question": "Estimate the maximum and minimum thermal conductivity values for a cermet that contains 90 vol\\% titanium carbide (TiC) particles in a nickel matrix. Assume thermal conductivities of 27 and \\(67 \\mathrm{~W} / \\mathrm{m} \\cdot \\mathrm{K}\\) for TiC and \\(\\mathrm{Ni}\\), respectively.",
|
||
"answer": "This problem asks for the maximum and minimum thermal conductivity values for a TiC\\(\\mathrm{Ni}\\) cermet. Using a modified form of Equation the maximum thermal conductivity \\(k_{\\max }\\) is calculated as follows:\n\\[\n\\begin{aligned}\nk_{\\max } & =k_{m} V_{m}+k_{p} V_{p}=k_{\\mathrm{Ni}} V_{\\mathrm{Ni}}+k_{\\mathrm{TiC}} V_{\\mathrm{TiC}} \\\\\n& =(67 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})(0.10)+(27 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})(0 .90)=31.0 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K}\n\\end{aligned}\n\\]\nUsing a modified form of Equation, the minimum thermal conductivity \\(k_{\\min }\\) is\n\\[\n\\begin{aligned}\nk_{\\min } & =\\frac{k_{\\mathrm{Ni}} k_{\\mathrm{TiC}}}{V_{\\mathrm{Ni}} k_{\\mathrm{TiC}}+V_{\\mathrm{TiC}} k_{\\mathrm{Ni}}} \\\\\n& =\\frac{(67 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})(27 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})}{(0.10)(27 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})+(0.90)(67 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K})} \\\\\n& =28.7 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K}\n\\end {aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 792,
|
||
"question": "A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.70 and 0.30 , respectively, estimate the upper limit for the specific stiffness of this composite, given the data that follow.\n\\begin{tabular}{lcc}\n\\hline & Specific Gravity & Modulus of Elasticity (GPa) \\\\\n\\hline Copper & 8.9 & 110 \\\\\nTungsten & 19.3 & 407 \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "There are two approaches that may be applied to solve this problem. The first is to estimate the upper limits for both the elastic modulus \\(\\left[E_{\\mathrm{c}}(u)\\right]\\) and specific gravity \\(\\left(\\rho_{\\mathrm{c}}\\right)\\) for the composite, using expressions of the form of Equation, and then take their ratio. Using this approach\n\\[\n\\begin{array}{c}\nE_{\\mathrm{c}}(u)=E_{\\mathrm{Cu}} V_{\\mathrm{Cu}}+E_{\\mathrm{W}} V_{\\mathrm{W}} \\\\\n=(110 \\mathrm{GPa})(0.30)+(407 \\mathrm{GPa})(0.70) \\\\\n=318 \\mathrm{GPa} \\\\\n\\text { and for the specific gravity } \\\\\nc=\\underset{c}{\\mathrm{Cu}} V_{\\mathrm{Cu}}+\\underset{W}{\\mathrm{w}} V_{\\mathrm{W}} \\\\\n=(8.9)(0.30)+(19.3)(0.70)=16.18\n\\end{array}\n\\]\nTherefore, we compute the specific stiffness as follows:\n\\[\n\\text { Specific Stiffness }=\\frac{E_{\\mathrm{c}}(u)}{\\rho_{\\mathrm{c}}}=\\frac{318}{16.18} \\frac{\\mathrm{GPa}}{16.18}=19.65 \\mathrm{GPa}\n\\]With the alternate approach, the specific stiffness is calculated, again employing a modification of Equation, but using the specific stiffness-volume fraction product for both metals, as follows:\n\\[\n\\begin{array}{l}\n\\text { Specific Stiffness }=\\frac{E_{\\mathrm{Cu}}}{\\rho_{\\mathrm{Cu}}} V_{\\mathrm{Cu}}+\\frac{E_{\\mathrm{W}}}{\\rho_{\\mathrm{W}}} V_{\\mathrm{W}} \\\\\n\\quad=\\frac{110 \\mathrm{\\ GPa}}{8.9}(0.30)+\\frac{407 \\mathrm{\\ GPa}}{19.3}(0.70)=18.47 \\mathrm{\\ GPa}\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 793,
|
||
"question": "For a continuous and oriented fiber-reinforced composite, the moduli of elasticity in the longitudinal and transverse directions are 33.1 and \\(3.66 \\mathrm{GPa}\\left(4.8 \\times 10^{6}\\right.\\) and \\(\\left.5.3 \\times 10^{5} \\mathrm{psi}\\right)\\), respectively. If the volume fraction of fibers is 0.30 , determine the moduli of elasticity of fiber and matrix phases.",
|
||
"answer": "This problem asks for us to compute the elastic moduli of fiber and matrix phases for a continuous and oriented fiber-reinforced composite. We can write expressions for the longitudinal and transverse elastic moduli using Equations, as follows:\n\\[\n\\begin{array}{c}\n\\text { For the longitudinal modulus: } \\\\\nE_{c l}=E_{m}\\left(1-V_{f}\\right)+E_{f} V_{f} \\\\\n33.1 \\mathrm{GPa}=E_{m}\\left(1-0.30\\right)+E_{f}\\left(0.30\\right)\n\\end{array}\n\\]\nWhile transverse modulus expression is as follows:\n\\[\nE_{c t}=\\frac{E_{m} E_{f}}{\\left(1-V_{f}\\right) E_{f}+V_{f} E_{m}}\n\\]\n\\[\n3.66 \\mathrm{GPa}=\\frac{E_{m} E_{f}}{(1-0.30) E_{f}+0.30 E_{m}}\n\\]\nSolving these two expressions simultaneously for \\(E_{m}\\) and \\(E_{f}\\) leads to the following values:\n\\[\n\\begin{array}{l}\nE_{m}=2.6 \\mathrm{GPa} \\quad \\left(3.77 \\quad 10^{5} \\mathrm{psi}\\right) \\\\\nE_{f}=104 \\mathrm{GPa} \\quad \\left(15 \\quad 10^{6} \\mathrm{psi}\\right)\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.\n(a) Using the data provided, determine the volume fraction of fibers required.\n(b) What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is \\(50 \\mathrm{MPa}(7250 \\mathrm{psi})\\).\n\\begin{tabular}{ccc}\n\\hline & Modulus of Elasticity [GPa (psi)] & Tensile Strength [MPa (psi)] \\\\\n\\hline Carbon fiber & \\(260(37 \\times 10^{6})\\) & \\(4000(580,000)\\) \\\\\nNylon 6,6 & \\(2.8(4.0 \\times 10^{5})\\) & \\(76(11,000)\\) \\\\\n\\hline\n\\end{tabular}",
|
||
"answer": "(a) Given some data for an aligned and continuous carbon-fiber-reinforced nylon 6,6 composite, we are asked to compute the volume fraction of fibers that are required such that the fibers carry \\(97 \\%\\) of a load applied in the longitudinal direction. From Equation\n\\[\n\\frac{F_{f}}{F_{m}}=\\frac{E_{f} V_{f}}{E_{m} V_{m}}=\\frac{E_{f} V_{f}}{E_{m}\\left(1-V_{f}\\right)}\n\\]\nNow, using values for \\(F_{f}\\) and \\(F_{m}\\) from the problem statement\n\\[\n\\frac{F_{f}}{F_{m}}=\\frac{0.97}{1-0.97}=32.3\n\\]\nAnd substitution of the values for \\(E_{f}\\) and \\(E_{m}\\) given in the problem statement into the first equation above leads to\n\\[\n\\frac{F_{f}}{F_{m}}=32.3=\\frac{(260 \\mathrm{GPa}) V_{f}}{(2.8 \\mathrm{GPa})\\left(1-V_{f}\\right)}\n\\]And, solving for \\(V_{f}\\) yields, \\(V_{f}=0.258\\).\n(b) We are now asked for the tensile strength of this composite. The computation requires the use of Equation—viz.,\n\\[\n\\sigma_{c l}^{*}=\\sigma_{m}^{*}\\left(1-V_{f}\\right)+\\sigma_{f}^{*} V_{f}\n\\]\nWe now substitute into this expression the following values:\n\\[\n\\begin{array}{l}\n\\sigma_{m}^{*}=50 \\mathrm{MPa} \\text { (given in the problem statement) } \\\\\nV_{f}=0.258 \\text { (computed above) } \\\\\n\\sigma_{f}^{*}=4000 \\mathrm{MPa} \\text { (given in the problem statement) }\n\\end{array}\n\\]\nFinally we determine the composite tensile strength as follows:\n\\[\n\\begin{array}{l}\n\\sigma_{c l}^{*}=\\sigma_{m}^{*}\\left( 1-V_{f}\\right) +\\sigma_{f}^{*} V_{f} \\\\\n\\quad=(50 \\mathrm{MPa})(1 \\quad 0.258)+\\left(4000 \\mathrm{MPa}\\right)(0.258) \\\\\n\\quad=1070 \\mathrm{MPa} \\quad(155,000 \\mathrm{psi})\n\\end{array}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 795,
|
||
"question": "(a) Cite several reasons why fiberglass-reinforced composites are used extensively.(b) Cite several limitations of this type of composite.",
|
||
"answer": "(a) 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.(b) 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 796,
|
||
"question": "(a) What is a hybrid composite? (b) List two important advantages of hybrid composites over normal fiber composites.",
|
||
"answer": "(a) A hybrid composite is a composite that is reinforced with two or more different fiber materials in a single matrix.(b) Two advantages of hybrid composites are: (1) better overall property combinations, and (2) failure is not as catastrophic as with single-fiber composites.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.\n(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 \\(\\left[E_{m}=4.0 \\mathrm{GPa}\\left(6 \\times 10^{5} \\mathrm{psi}\\right)\\right]\\).",
|
||
"answer": "(a) 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:\n\\[\nE_{c l}=E_{m} V_{m}+E_{f 1} V_{f 1}+E_{f 2} V_{f 2}\n\\]\nin which the subscripts \\(f 1\\) and \\(f 2\\) refer to the two types of fibers.\n(b) Now we are asked to compute the longitudinal elastic modulus for a glass- and aramid-fiber hybrid composite. From Table 15.4, the elastic moduli of aramid and glass fibers are, respectively, \\(131 \\mathrm{GPa}(19 \\times 10^{6} \\mathrm{psi})\\) and \\(72.5 \\mathrm{GPa}\\left(10.5 \\times 10^{6} \\mathrm{psi}\\right)\\). Thus, from the previous expression\n\\[\n\\begin{aligned}\nE_{c l} & =(4 \\mathrm{GPa})(1.0 \\quad 0.25 \\quad 0.35)+(131 \\mathrm{GPa})(0.25)+(72.5 \\mathrm{GPa})(0.35) \\\\\n& =59.7 \\mathrm{GPa} \\quad(8.67 \\quad 10^{6} \\mathrm{psi})\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 798,
|
||
"question": "Briefly describe laminar composites. What is the prime reason for fabricating these \\\\ materials?",
|
||
"answer": "Laminar composites are a series of sheets or panels, each of which has a preferred high- \\(^{2}\\) strength direction. These sheets are stacked and then cemented together such that the orientation of the high-strength direction varies from layer to layer.\nThese composites are constructed in order to have a relatively high strength in virtually all directions within the plane of the laminate.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 799,
|
||
"question": "An \\(\\mathrm{Fe} / \\mathrm{Fe}^{2+}\\) concentration cell is constructed in which both electrodes are pure iron. The} \\(\\mathrm{Fe}^{2+}\\) concentration for one cell half is \\(0.5 \\mathrm{M}\\) and for the other cell half is \\(2 \\times 10^{-2} \\mathrm{M}\\). Is a voltage generated between the two cell halves? If so, what is its magnitude and which electrode will be oxidized? If no voltage is produced, explain this result.",
|
||
"answer": "This problem calls for us to determine whether or not a voltage is generated in a \\(\\mathrm{Fe} / \\mathrm{Fe}^{2+}\\) concentration cell, and, if so, its magnitude. Let us label the \\(\\mathrm{Fe}\\) cell having a \\(0.5 \\mathrm{M} \\mathrm{Fe}^{2+}\\) solution as cell 1 , and the other as cell 2 . Furthermore, assume that oxidation occurs within cell 2 , wherein \\(\\left[\\mathrm{Fe}_{2}^{2+}\\right]=2 \\times 10^{-2} M\\). Hence, we write the electrochemical reaction as follows:\n\\[\n\\mathrm{Fe}_{2}+\\mathrm{Fe}_{1}^{2+} \\rightarrow \\mathrm{Fe}_{2}^{2+}+\\mathrm{Fe}_{1}\n\\]\nand, employing Equation leads to\n\\[\n\\begin{aligned}\n\\Delta V & =-\\frac{0.0592}{2} \\log \\left[\\frac{\\mathrm{Fe}_{2}^{2+}}{\\left[\\mathrm{Fe}_{1}^{2+}\\right]}\\right. \\\\\n& =-\\frac{0.0592}{2} \\log \\left[\\frac{2 \\times 10^{-2} M}{0.5 M}\\right]=+0.0414 \\mathrm{~V}\n\\end{aligned}\n\\]Therefore, a voltage of \\(0.0414 \\mathrm{~V}\\) is generated when oxidation occurs in cell 2, the one having the lower \\(\\mathrm{Fe}^{2+}\\) concentration - i.e., \\(2 \\times 10^{-2} M\\).",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 800,
|
||
"question": "For the following pairs of alloys that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.\n(a) Aluminum and cast iron\n(b) Inconel and nickel\n(c) Cadmium and zinc\n(d) Brass and titanium\n(e) Low-carbon steel and copper",
|
||
"answer": "This problem asks, for several pairs of alloys that are immersed in seawater, to predict whether or not corrosion is possible, and if it is possible, to note which alloy will corrode. If both of the alloys in the pair reside within the same set of brackets in this table, then galvanic corrosion is unlikely. However, if the two alloys do not lie within the same set of brackets, then that alloy appearing lower in the table will experience corrosion.\n(a) For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode.\n(b) 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).\n(c) For the cadmium-zinc couple, corrosion is possible, and zinc will corrode.\n(d) For the brass-titanium pair, corrosion is possible, and brass will corrode.\n(e) For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 802,
|
||
"question": "The corrosion rate is to be determined for some divalent metal \\(M\\) in a solution containing hydrogen ions. The following corrosion data are known about the metal and solution:\n\\begin{tabular}{cc}\n\\hline For Metal M & For Hydrogen \\\\\n\\hline\\(V_{(\\mathrm{MM}^{2 *})}=\\) & \\(0.90 \\mathrm{~V}\\) \\\\\n\\(i_{0}=10^{-12} \\mathrm{~A} / \\mathrm{cm}^{2}\\) & \\(V_{\\left(\\mathrm{H}^{-} / \\mathrm{H}_{2}\\right)}=0 \\mathrm{~V}\\) \\\\\n\\(\\beta=+0.10\\) & \\(i_{0}=10^{-10} \\mathrm{~A} / \\mathrm{cm}^{2}\\) \\\\\n\\hline\n\\end{tabular}\n(a) Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal \\(M\\left(\\right.\\) in \\(\\left.\\mathrm{mol} / \\mathrm{cm}^{2} \\cdot \\mathrm{s}\\right)\\).\n(b) Compute the corrosion potential for this reaction.",
|
||
"answer": "(a) This portion of the problem asks that we compute the rate of oxidation for a divalent metal \\(\\mathrm{M}\\) given that both the oxidation and reduction reactions are controlled by activation polarization, and also given the polarization data for both \\(\\mathrm{M}\\) oxidation and hydrogen reduction. The first thing necessary is to establish relationships of the form of Equation for the potentials of both oxidation and reduction reactions. Next we will set these expressions equal to one another, and then solve for the value of \\(i\\) which is the corrosion current density, \\(i_{\\mathrm{C}}\\). Finally, the corrosion rate may be calculated using Equation. The two potential expressions are as follows:\nFor hydrogen reduction\n\\[\nV_{\\mathrm{H}}=V_{\\left(\\mathrm{H}^{+} / \\mathrm{H}_{2}\\right)}+\\beta_{\\mathrm{H}} \\log \\left(\\frac{i}{i_{\\mathrm{OH}}}\\right)\n\\]\nAnd for M oxidation\n\\[\nV_{\\mathrm{M}}=V_{\\left(\\mathrm{M} / \\mathrm{M}^{2 *}\\right)}+\\beta_{\\mathrm{M}} \\log \\left(\\frac{i}{i_{\\mathrm{OM}}}\\right)\n\\]Setting \\(V_{\\mathrm{H}}=V_{\\mathrm{M}}\\) and solving for \\(\\log i\\) (that is, \\(\\log i_{\\mathrm{C}}\\) ) leads to the following expression:\n\\[\n\\log i_{\\mathrm{C}}=\\left(\\frac{1}{\\beta_{\\mathrm{M}}-\\beta_{\\mathrm{H}}}\\right)\\left[V_{\\left(\\mathrm{H}^{+} / \\mathrm{H}_{2}\\right)}-V_{\\left(\\mathrm{M} / \\mathrm{M}^{2+}\\right)}-\\beta_{\\mathrm{H}} \\log i_{0_{\\mathrm{H}}}+\\beta_{\\mathrm{M}} \\log i_{0_{\\mathrm{M}}}\\right]\n\\]\nAnd, incorporating values for the various parameters provided in the problem statement results in the following:\n\\[\n\\begin{aligned}\n\\log i_{\\mathrm{C}}= & {\\left[\\frac{1}{0.10-(-0.15)}\\right]\\left[0-(-0.90)-(-0.15)\\left\\{\\log \\left(10^{-10}\\right)\\right\\}+(0.10)\\left\\{\\log \\left(10^{-12}\\right)\\right\\}\\right] } \\\\\n& =-7.20\n\\end{aligned}\n\\]\nFrom which \\(i_{\\mathrm{C}}\\) may be determined as follows:\n\\[\n\\begin{aligned}\ni_{\\mathrm{C}} & =10^{-7.20}=6.31 \\times 10^{-8} \\mathrm{~A} / \\mathrm{cm}^{2} \\\\\n& =6.31 \\times 10^{-8} \\mathrm{C} / \\mathrm{s} \\cdot \\mathrm{cm}^{2}\n\\end{aligned}\n\\]\nNow, from Equation\n\\[\nr=\\frac{i_{c}}{n F}\n\\]\nwe compute the oxidation rate as follows:\n\\[\nr=\\frac{6.31 \\times 10^{-8} \\mathrm{C} / \\text { s-cm }^{2}}{(2)(96,500 \\mathrm{C} / \\mathrm{mol})}=3.27 \\times 10^{-13} \\mathrm{~mol} / \\mathrm{cm}^{2} \\text {-s }\n\\](Note: in the above discussion, we denote the Faraday constant by an \" \\(F\\) \" rather than with the symbol used in the textbook.)\n(b) Now it becomes necessary to compute the value of the corrosion potential, \\(V_{c}\\). This is possible by using either of the above equations for \\(V_{\\mathrm{H}}\\) or \\(V_{\\mathrm{M}}\\) and substituting for \\(i\\) the value determined above for \\(i_{c}\\). This is done as follows (using the expression for hydrogen reduction):\n\\[\n\\begin{aligned}\nV_{c} & =\\mathrm{V}_{\\left(\\mathrm{H}^{+} / \\mathrm{H}_{2}\\right)}+\\beta_{\\mathrm{H}} \\log \\left(\\frac{i_{c}}{i_{0_{\\mathrm{H}}}}\\right) \\\\\n& =0+(-0.15 \\mathrm{~V}) \\log \\left(\\frac{6.31 \\times 10^{-8} \\mathrm{~A} / \\mathrm{cm}^{2}}{10^{-10} \\mathrm{~A} / \\mathrm{cm}^{2}}\\right)=-0.420 \\mathrm{~V}\n\\end{aligned}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 807,
|
||
"question": "For each of the following pairs of materials, decide which has the larger thermal conductivity. Justify your choices.\n(a) Pure silver; sterling silver (92.5 wt\\% Ag-7.5 wt\\% Cu)\n(b) Fused silica; polycrystalline silica\n(c) Linear and syndiotactic poly(vinyl chloride) (DP \\(=1000\\) ); linear and syndiotactic polystyrene \\((D P=1000)\\)\n(d) Atactic polypropylene ( \\(\\bar{M}_{\\mathrm{w}}=10^{6} \\mathrm{~g} / \\mathrm{mol}\\) ); isotactic polypropylene ( \\(\\bar{M}_{\\mathrm{w}}=10^{\\circ} 5 \\mathrm{~g} / \\mathrm{mol}\\) )",
|
||
"answer": "This question asks for us to decide, for each of several pairs of materials, which has the larger thermal conductivity and why.\n(a) 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.\n(b) 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.\n(c) 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.\n(d) 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).",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 808,
|
||
"question": "(a) Briefly explain why thermal stresses may be introduced into a structure by rapid heating or cooling.\n(b) For cooling, what is the nature of the surface stresses?\n(c) For heating, what is the nature of the surface stresses?",
|
||
"answer": "(a) 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.\n(b) For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser degree than the cooler surface.\n(c) For heating, the surface stresses will be compressive in nature since the interior expands to a lesser degree than the hotter surface.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 809,
|
||
"question": "The magnetic flux density within a bar of some material is 0.630 tesla at an \\(\\mathrm{H}\\) field of \\(5 \\times\\) \\(10^{5} \\mathrm{~A} / \\mathrm{m}\\). Compute the following for this material: (a) the magnetic permeability and (b) the magnetic susceptibility. (c) What type(s) of magnetism would you suggest is (are) being displayed by this material? Why?",
|
||
"answer": "(a) The magnetic permeability of this material may be determined according to a rearranged form of Equation as follows:\n\\[\n\\begin{aligned}\n\\mu & =\\frac{B}{H} \\\\\n& =\\frac{0.630 \\text { tesla }}{5 \\times 10^{5} \\mathrm{~A} / \\mathrm{m}}=1.260 \\times 10^{-6} \\mathrm{H} / \\mathrm{m}\n\\end{aligned}\n\\]\n(b) The magnetic susceptibility is calculated using a combined form of Equations as follows:\n\\[\n\\begin{aligned}\n\\chi_{m} & =\\mu_{r}-1=\\frac{\\mu}{\\mu_{0}}-1 \\\\\n& =\\frac{1.260 \\times 10^{-6} \\mathrm{H} / m}{1.257 \\times 10^{-6} \\mathrm{H} / \\mathrm{m}}-1=2.387 \\times 10^{-3}\n\\end{aligned}\n\\]\n(c) 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.",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 812,
|
||
"question": "The formula for samarium iron garnet \\(\\left(\\mathrm{Sm}_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right)\\) may be written in the form \\(\\mathrm{Sm}_{3}^{+} \\mathrm{Fe}_{2} \\mathrm{Fe}_{3}^{+} \\mathrm{O}_{12}\\), where the superscripts \\(a, c\\), and \\(d\\) represent different sites on which the \\(\\mathrm{Sm}^{3+}\\) and \\(\\mathrm{Fe}^{3+}\\) ions are located. The spin magnetic moments for the \\(\\mathrm{Sm}^{3+}\\) and \\(\\mathrm{Fe} 3^{+}\\)ions positioned in \\(a\\) and \\(c\\) sites are oriented parallel to one another and antiparallel to the \\(\\mathrm{Fe}^{3+}\\) ions in \\(d\\) sites. Compute the number of Bohr magnetons associated with each \\(\\mathrm{Sm}^{3+}\\) ion, given the following information: (1) each unit cell consists of eight formula \\(\\left(\\mathrm{Sm}_{3} \\mathrm{Fe}_{5} \\mathrm {O}_{12}\\right)\\) units; (2) the unit cell is cubic with an edge length of \\(1.2529 \\mathrm{~nm}\\); (3) the saturation magnetization for this material is 1.35 \\(\\times 10^{5} \\mathrm{~A} / \\mathrm{m}\\); and (4) there are 5 Bohr magnetons associated with each \\(\\mathrm{Fe}^{3+}\\) ion.",
|
||
"answer": "In this problem we are to determine the number of Bohr magnetons associated with each \\(\\mathrm{Sm}^{3+} \\mathrm{ion}\\) given that each unit cell consists of eight formula units, the unit cell is cubic with an edge length of \\(1.2529 \\mathrm{\\mathrm{nm}}\\), the saturation magnetization for the material is \\(1.35 \\times 10^{5} \\mathrm{~A} / \\mathrm{m}\\), and that there are 5 Bohr magnetons for each \\(\\mathrm{Fe}^{3+}\\) ion.\nThe first thing to do is to calculate the number of Bohr magnetons per unit cell, which we will denote \\(n_{\\mathrm{B}}\\). Solving for \\(n_{\\mathrm{B}}\\) in Equation, we get\n\\[\n\\begin{aligned}\nn_{\\mathrm{B}} & =\\frac{M_{\\mathrm{x}} a^{3}}{\\mu_{\\mathrm{B}}} \\\\\n& =\\frac{\\left(1.35 \\times 10^{5} \\mathrm{~A} / m\\right)\\left(1.2529 \\times 10^{-9} \\mathrm{~m}\\right)^{3}}{9.27 \\times 10^{-24} \\mathrm{~A} \\cdot \\mathrm{m}^{2} / \\mathrm{BM}} \\\\\n& =28.64 \\text { Bohr magnetons/unit cell }\n\\end{aligned}\n\\]\nNow, there are 8 formula units per unit cell or \\(\\frac{28.64}{8}=3.58\\) Bohr magnetons per formula unit.\nFurthermore, for each formula unit there are two \\(\\mathrm{Fe}^{3+}\\) ions on \\(a\\) sites and three \\(\\mathrm{Fe}^{3+}\\) on \\(d\\) sites which magnetic moments are aligned antiparallel. Since there are 5 Bohr magnetons associatedwith each \\(\\mathrm{Fe}^{3+}\\) ion, the net magnetic moment contribution per formula unit from the \\(\\mathrm{Fe}^{3+}\\) ions is 5 Bohr magnetons. This contribution is antiparallel to the contribution from the \\(\\mathrm{Sm}^{3+}\\) ions, and since there are three \\(\\mathrm{Sm}^{3+}\\) ions per formula unit, then\n\\[\n\\mbox{No. of Bohr magnetons} / \\mbox{Sm}^{3+}=\\frac{3.58 \\mbox{ BM }+5 \\mbox{ BM }}{3}=2.86 \\mbox{ BM }\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"idx": 814,
|
||
"question": "Cite the differences between hard and soft magnetic materials in terms of both hysteresis behavior and typical applications.",
|
||
"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.\nWith regard to applications:\nSoft 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.\nHard 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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"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.",
|
||
"is_select": 0
|
||
},
|
||
{
|
||
"idx": 817,
|
||
"question": "Visible light having a wavelength of \\(5 \\times 10^{-7} \\mathrm{~m}\\) appears green. Compute the frequency and energy of a photon of this light.",
|
||
"answer": "In order to compute the frequency \\(v\\) of a photon of green light, we use a rearranged form of Equation as follows:\n\\[\n\\begin{aligned}\nv & =\\frac{c}{\\lambda} \\\\\n& =\\frac{3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}}{5 \\times 10^{-7} \\mathrm{~m}} \\\\\n& =6 \\times 10^{14} \\mathrm{~s}^{-1}\n\\end{aligned}\n\\]\nNow, for the energy computation, we employ Equation as follows:\n\\[\n\\begin{aligned}\nE & =\\frac{h c}{\\lambda} \\\\\n& =\\frac{\\left(6.63 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}\\right)\\left(3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right)}{5 \\times 10^{-7} \\mathrm{~m}}\n\\end{aligned}\n\\]\\[\n = 3.98 \\times 10^{-19} \\mathrm{\\; J} \\; \\mathrm{(2.48 \\, eV)}\n\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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.",
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"is_select": 0
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},
|
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{
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"idx": 819,
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"question": "Can a material have a positive index of refraction less than unity? Why or why not?",
|
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"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.",
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"is_select": 0
|
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},
|
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{
|
||
"idx": 820,
|
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"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.",
|
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"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\\]",
|
||
"is_select": 1
|
||
},
|
||
{
|
||
"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}\\).",
|
||
"is_select": 1
|
||
}
|
||
]
|