Fundamentals of Materials Science and Engineering 5th ed - Solutions

Fundamentals of Materials Science and Engineering 5th ed - Solutions

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CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING 2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as #g/amu = µ 1 mol 6 . 023 × 10 23 atoms ¶µ 1 g/mol 1 amu/atom = 1 . 66 × 10 24 g/amu 2.14 (c) This portion of the problem asks that we determine for a K + -Cl ion pair the interatomic spacing ( r o ) and the bonding energy ( E o ). From Equation (2.11) for E N A = 1 . 436 B = 5 . 86 × 10 6 n = 9 Thus, using the solutions from Problem 2.13 r o = µ A nB 1 / (1 n) = · 1 . 436 (9)(5 . 86 × 10 6 ) ¸ 1 / (1 9) = 0 . 279 nm and E o =− 1 . 436 · 1 . 436 (9)(5 . 86 × 10 6 ) ¸ 1 / (1 9) + 5 . 86 × 10 6 · 1 . 436 (9)(5 . 86 × 10 6 ) ¸ 9 / (1 9) 4 . 57 eV 2.19 The percent ionic character is a function of the electronegativities of the ions X A and X B according to Equation (2.10). The electronegativities of the elements are found in Figure 2.7. For TiO 2 , X Ti = 1 . 5 and X O = 3 . 5, and therefore, %IC = £ 1 e ( 0 . 25)(3 . 5 1 . 5) 2 ¤ × 100 = 63 . 2% 1
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CHAPTER 3 STRUCTURES OF METALS AND CERAMICS 3.3 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 V C = 16R 3 2 = (16)(0 . 143 × 10 9 m) 3 2 = 6 . 62 × 10 29 m 3 3.7 This problem calls for a demonstration that the APF for HCP is 0.74. Again, the APF is just the total sphere-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus V S = 6 µ 4 p R 3 3 = 8 p R 3 Now, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACDE shown below. A B C D E a = 2R a = 2R a = 2R 60 30 The area of ACDE is just the length of CD times the height BC . But CD is just a or 2 R , and BC = 2R cos(30 ) = 2R 3 2 Thus, the base area is just AREA = (3)( CD)( BC) = (3)(2R) µ 2R 3 2 = 6R 2 3 and since c = 1.633 a = 2 R (1.633) V C = (AREA)(c) = 6R 2 c 3 = (6R 2 3)(2)(1 . 633)R = 12 3(1 . 633)R 3 2
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Thus, APF = V S V C = 8 p R 3 12 3(1 . 633)R 3 = 0 . 74 3.12. (a) This portion of the problem asks that we compute the volume of the unit cell for Zr. This volume may be computed using Equation (3.5) as V C = nA Zr r N A Now, for HCP, n = 6 atoms/unit cell, and for Zr, A Zr = 91.2 g/mol. Thus, V C = (6 atoms/unit cell)(91 . 2 g/mol) (6 . 51 g/cm 3 )(6 . 023 × 10 23 atoms/mol) = 1 . 396 × 10 22 cm 3 /unit cell = 1 . 396 × 10 28 m 3 /unit cell (b) We are now to compute the values of a and c , given that c / a = 1 . 593. From the solution to Problem 3.7, since a = 2 R , then, for HCP V C = 3 3a 2 c 2 but, since c = 1 . 593 a V C = 3 3(1 . 593)a 3 2 = 1 . 396 × 10 22 cm 3 /unit cell Now, solving for a a = · (2)(1 . 396 × 10 22 cm 3 ) (3)( 3)(1 . 593) ¸ 1 / 3 = 3 . 23 × 10 8 cm = 0 . 323 nm And finally c = 1 . 593a = (1 . 593)(0 . 323 nm) = 0 . 515 nm 3.17 In this problem we are given that iodine has an orthorhombic unit cell for which the a , b , and c lattice parameters are 0.479, 0.725, and 0.978 nm, respectively.
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Fundamentals of Materials Science and Engineering 5th ed - Solutions

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