ts1201f06 - EMSE 201 — Introduction to Materials Science...

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Unformatted text preview: EMSE 201 — Introduction to Materials Science & Engineering 27 September 2006 Name: Test #1 — 75 minutes; 150 points; 5 questions; 6 pages; 15% of course grade Partial credit will be given for correct set-ups and reasoning. Give units on numerical answers where appropriate. Please write all answers on these pages; use the backs if needed. Boltzmann’s constant, kB = 1.381×10-23 J K-1 = 8.620×10-5 eV K-1 Avogadro’s number, NA = 6.023×1023 mol-1 gas constant, R: 8.314 J mol-1 K-1 1) The unit cell of a compound of molybdenum (Mo) and silicon (Si) has the following characteristics: 111 ,, 222 Mo: 0, 0, 0 a = b = 0.3202 nm Si: 0, 0, 1 3 0, 0, 2 3 111 ,, 226 115 ,, 226 c = 0.7851 nm α = β = γ = 90° Si a) (6 points) In the space at right, sketch the unit cell. Clearly identify the atoms. Mo b) (4 points) How many atoms of each type are in the unit cell? Mo: 2 4 Si: c) (4 points) What is the Bravais lattice of this structure? body-centered tetragonal d) (2 points) How many nearest-neighbor Si atoms does each Mo atom have? 10 1 111 111 (Hint: the Si atoms at 0, 0, and , , are almost exactly equidistant from the Mo atom at , , .) 3 226 222 5 e) (2 points) How many nearest-neighbor Mo atoms does each Si atom have? Sketch the arrangement of atoms on the specified planes in this structure. Label the atoms in your sketches. f) (5 points) (100) g) (7 points) (110) Si Si Mo Mo continued on next page Department of Materials Science and Engineering 1 of 6 Case Western Reserve University EMSE 201 — Introduction to Materials Science & Engineering 27 September 2006 1) h) (8 points) Compute the density of this material. The atomic weights of Mo and Si are 95.94 and 28.09 g mol–1, respectively. density = (mass of atoms per cell)/(volume of cell) (2 pts) = (2AWMo + 4AWSi)/(ca2×NA) (3 pts) = (2×95.94 + 4×28.09)[g mol–1]/{(0.7851×.32022)[10–21cm3]×6.023×1023[mol–1]} = 6.275 g cm–3 (1 pt for inserting values; 1 pt for answer; 1 pt for units) i) (10 points) A compound of tungsten (W) and silicon forms with the same stoichiometry and structure as the compound discussed above. Would you expect these two compounds to form a continuous solid solution with each other? Evaluate each of the Hume-Rothery rules to justify your answer. atomic radius, nm Mo W 0.136 0.137 electronegativit y 2.16 2.36 allowed valences 6, 5, 4, 3, 2 6, 5, 4, 3, 2 Same structure (given): √ (1+1 pts) ∆r/r < 15%: √ (1+1 pts) ∆EN < 0.4: √ (1+1 pts) similar valences: √ (1+1 pts) Yes, we expect MoSi2 and WSi2 to form a continuous solid solution (2 pts) (and they do). Department of Materials Science and Engineering 2 of 6 Case Western Reserve University EMSE 201 — Introduction to Materials Science & Engineering 27 September 2006 2) a) (3 points) For a cubic unit cell, circle planes that belong to the {100} family: (1 pt for each correctly circled or uncircled item) a) (100) b) (010) c) (001) b) (3 points) For a tetragonal unit cell, circle planes that belong to the {100} family: a) (100) b) (010) c) (001) c) (3 points) For an orthorhombic unit cell, circle planes that belong to the {100} family: a) (100) b) (010) c) (001) d) (3 points) For a hexagonal unit cell, circle planes that belong to the {0001} family: a) (10 1 0) b) (01 1 0) c) (0001) e) (3 points) For a hexagonal unit cell, circle directions that belong to the <11 2 0> family: a) [2 1 1 0] b) [ 1 2 1 0] c) [11 2 0] 3) a) (8 points) List four intrinsic defects that can occur in a compound such as NaCl. 2 pts each for any of the following: cation vacancy, anion vacancy, cation interstitial, anion interstitial, free electron, hole, OR anion Frenkel pair, cation Frenkel pair, Schottky pair, electron-hole pair b) (8 points) List four extrinsic defects that can occur in NaCl. 2 pts each for any of the following: cation substitutional impurity, anion substitutional impurity, cation interstitial impurity, anion interstitial impurity, screw dislocation, edge dislocation, grain boundary, twin Indicate whether the following characteristics apply to intrinsic defects, extrinsic defects, or all defects. c) (3 points) They disrupt the periodicity of the crystal structure. all defects d) (3 points) Their equilibrium concentrations can be computed using Boltzmann factors. intrinsic e) (3 points) Their concentrations depend primarily on how the material was synthesized or processed. extrinsic Department of Materials Science and Engineering 3 of 6 Case Western Reserve University EMSE 201 — Introduction to Materials Science & Engineering 27 September 2006 4) a) (7 points) Show that the atomic packing factor for the diamond cubic structure is 34%. For diamond cubic, atoms touch along <111> directions. With atoms at the corners, face centers, and half of the tetrahedral interstices, the relationship between the atomic radius r and the lattice parameter a is 8r = 3 a (3 pts). 8 atoms per unit cell (1 pt) volume of atoms per cell = 8×4πr3/3 (1 pt) 3 ⎛ 8r ⎞ volume of cell = a = ⎜ ⎟ (1 pt) ⎝ 3⎠ 3 APF = volume of atoms/volume of cell = 8 × 4š r 3 /3 = 34.0% (1 pt) (8r/ 3 )3 b) (5 points) Despite the low packing factor of this structure, the diamond form of carbon has the highest number of atoms per unit volume of any solid at normal pressure. Compute the number of atoms per unit volume in diamond, given that the density of diamond is 3.52 g cm–3 and the atomic mass of carbon is 12.011 g mol–1. # of atoms per unit volume = NA×(density)/(atomic weight) (2 pts) = 6.023×1023 mol-1×3.52 g cm–3 / 12.011 g mol–1 (1 pt) = 1.76×1023 cm–3 (1 pt answer + 1 pt units) c) (8 points) Compute the number of atoms per unit volume in beryllium (HCP, 1.85 g cm–3, 9.012 g mol–1) and aluminum (FCC, 2.71 g cm–3, 26.98 g mol–1), both of which have the highest possible atomic packing factor for elemental solids (74%). Using the same approach as for part b: For Be: 6.023×1023 mol-1×1.85 g cm–3 / 9.012 g mol–1 (2 pts) = 1.24×1023 cm–3 (1 pt answer + 1 pt units) For Al: 6.023×1023 mol-1×2.71 g cm–3 / 26.98 g mol–1 (2 pts) = 6.05×1022 cm–3 (1 pt answer + 1 pt units) (Note: both answers are smaller than that for diamond in part b.) Department of Materials Science and Engineering 4 of 6 Case Western Reserve University EMSE 201 — Introduction to Materials Science & Engineering + 27 September 2006 – d) (14 points) Compute the numbers of Li and F ions per unit volume of lithium fluoride (LiF, rock salt structure, 2.635 g cm–3, 25.939 g mol–1). Compute the atomic packing factor of lithium fluoride, given that the ionic radius of the fluorine ion is 0.133 nm. (Hint: compute the ionic radius for the lithium ion by first computing the lattice parameter.) Because LiF has the rock salt structure, there are 4 LiF molecules per unit cell (2 pts). Vcell = 4 molecules / cell × 25.94 g / mol × 10 21 nm 3 / cm 3 = 6.54 × 10 –2 nm 3 (2 pts) 3 23 2.635 g / cm × 6.023 × 10 molecules / mol The lattice parameter a is the cube root of the cell volume, i.e. 0.403 nm (1 pt). Assuming that the cations are touching the anions, a = 2(rLi + rF) (2 pts). Given that the ionic radius of F- = 0.133 nm, rLi = 0.068 nm (1 pt). 4 3 The volume of the atoms is VS = 4 × π (rLi + + rF3− )= 0.0447 nm 3 (1 pt) 3 3 V 0.0447 nm APF = S = = 68% (2 pts) VC 0.0654 nm 3 The number of Li ions per unit volume is the same as the number of F ions per unit volume (1 pt), which is 4 molecules / cell × 10 21 nm 3 / cm 3 = 6.12 × 10 22 molecules / cm 3 or 1.22×1023 atoms/cm3 (2 pts) 3 0.0654 nm / cell (comparable to Be but still lower than diamond). 5) A thin film of copper is placed between two pieces of nickel and allowed to diffuse into them. The resulting concentration CCu of copper in the nickel as a function of time t and distance x from the initial interface is given by ⎛ – x2 ⎞ B CCu = exp ⎜ ⎟ Dt ⎝ 4 Dt ⎠ where B is a constant and D is the diffusion coefficient for copper in nickel. The plot below shows the concentration profiles CCu(x) that result from such an experiment at three different times. a) (10 points) What is the flux of copper at x = 0 (that is, at the original interface between the two metals) at any time? Comment, in light of the fact that CCu at x = 0 is obviously changing with time. C t1 t 2 = 2t 1 t 3 = 4t 1 From Fick’s first law, flux J = –D∂CCu/∂t (3 pts) x At x = 0, ∂CCu/∂t = 0 at all times (2 pts). Therefore, the flux must be zero at x = 0 at all times (2 pts). In other words, copper is diffusing to the right and to the left at the same rates (3 pts). This explains how the flux can be zero even though copper is obviously being depleted from the interface. continued on next page Department of Materials Science and Engineering 5 of 6 Case Western Reserve University EMSE 201 — Introduction to Materials Science & Engineering 27 September 2006 Name: 5) b) (12 points) At any time t > 0, there are four values of x for which ∂2CCu/∂x2 = 0. Indicate these points on the curve on the previous page for t3. What is the significance of these points? The positions are at x = ±∞ and at the inflection points on the concentration profiles (i.e., where the curvature changes from concave upward to concave downward) (1.5 pt each). From Fick’s second law, whenever ∂2CCu/∂x2 = 0, ∂CCu/∂t must also be zero (3 pt). That is, there is no net accumulation or depletion of the diffusing species at these points. At x = ±∞, this is true because the copper has not penetrated that far into the nickel, i.e. J = 0 (1 pt). At the inflection points, there is flux, but the flux of arriving copper exactly equals the flux of leaving copper (1 pt), so that there is no net change in the concentration of copper at that point at that instant. This condition (flux in = flux out) is also called steady state (1 pt). c) (6 points) What is the boundary condition that must be met at x = ±∞ for the equation given on the previous page to be valid? In practice this condition can be met if the thickness of the nickel pieces in the x direction is greater than or equal to 10 Dt . Compute this thickness if the experiment is to run for 120 h at 650 °C. For copper diffusing in nickel, Do = 2.7×10–5 m2 s–1 and Qd = 256 kJ mol–1. The requested boundary condition is that the composition at the far ends of the bar remains unchanged; in mathematical terms, CCu(±∞, t) = CCu,o (or in this case, CCu(±∞, t) = 0) (either version: 2 pts). For the specified parameters, this condition will be met when the nickel thickness is at least ⎛ ⎞ –256,000 J • mol –1 –1 10 Dt = 10 2.7 × 10 –5 m 2 s –1 exp ⎜ ⎟ × 120 h × 3600 s • h –1 –1 ⎝ 8.314 J • mol • K × (650 + 273)K ⎠ = 10 8.77 × 10 –20 m 2 s –1 × 120 h × 3600 s • h –1 = 1.95 × 10 –6 m = 1.95 µm correct D: 2 pts; correct answer: 1 pt; correct units: 1 pt. (Please do not write in the space below.) 1 /48 2 /15 3 Department of Materials Science and Engineering /25 4 6 of 6 /34 5 TOTAL: /28 __ ___ _/150 Case Western Reserve University ...
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