HW2 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 02 Question 3.5: Show that the atomic packing factor for BCC is 0.68. Solution of 3.5: The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C Since there are two spheres associated with each unit cell for BCC V S = 2(sphere volume) = 2 4 π R 3 3 = 8 π R 3 3 Also, the unit cell has cubic symmetry, that is V C = a 3 . But a depends on R according to Equation 3.3, and V C = 4 R 3 3 = 64 R 3 3 3 Thus, APF = V S V C = 8 π R 3 /3 64 R 3 /3 3 = 0.68
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Question 3.6: Show that the atomic packing factor for HCP is 0.74. Solution 3.6: The APF is just the total sphere volume-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus V S = 6 4 π R 3 3 = 8 π 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. The area of ACDE is just the length of CD times the height BC . But CD is just a or 2 R , and BC = 2 R cos (30 ° ) = 2 R 3 2 Thus, the base area is just AREA = (3)( CD )( BC ) = (3)(2 R ) 2 R 3 2 = 6 R 2 3 and since c = 1.633 a = 2 R (1.633) V C = (AREA)( c ) = 6 R 2 c 3 (3.S1) = ( 6 R 2 3 ) (2)(1.633) R = 12 3 (1.633) R 3 Thus, APF = V S V C = 8 π R 3 12 3 (1.633) R 3 = 0.74
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Question 3.7: Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the experimental value found inside the front cover. Solution of 3.7: This problem calls for a computation of the density of iron. According to Equation 3.5 ρ = nA Fe V C N A For BCC, n = 2 atoms/unit cell, and V C = 4 R 3 3 Thus, ρ = nA Fe 4 R 3 3 N A = (2 atoms/unit cell)(55.85 g/mol) (4) ( 0.124 × 10 -7 cm ) / 3 [ ] 3 /(unit cell) ( 6.022 × 10 23 atoms/mol ) = 7.90 g/cm 3 The value given inside the front cover is 7.87 g/cm 3 .
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Question 3.11: Zirconium has an HCP crystal structure and a density of 6.51 g/cm 3 . (a) What is the volume of its unit cell in cubic meters? (b) If the c/a ratio is 1.593, compute the values of c and a. Solution of 3.11: a) The volume of the Zr unit cell may be computed using Equation 3.5 as V C = nA Zr ρ N A Now, for HCP, n = 6 atoms/unit cell, and for Zr, A Zr = 91.22 g/mol. Thus, V C = (6 atoms/unit cell)(91.22 g/mol) ( 6.51 g/cm 3 )( 6.022 × 10 23 atoms/mol ) = 1.396 × 10 -22 cm 3 /unit cell = 1.396 × 10 -28 m 3 /unit cell b) From Equation 3.S1 of the solution to Problem 3.6, for HCP V C = 6 R 2 c 3 but, since a = 2 R , (i.e., R = a /2) then V C = 6 a 2 2 c 3 = 3 3 a 2 c 2 but, since c = 1.593 a V C = 3 3 (1.593)
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This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.

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HW2 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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