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HW-02%20Solutions

# HW-02%20Solutions - VK 250 Summer 2008 Homework 2...

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3.13 In order to determine whether Nb has an FCC or a BCC crystal structure, we need to compute its density for each of the crystal structures. For FCC, n = 4, and a = 2 R 2 (Equation 3.1). Also, from Figure 2.6, its atomic weight is 92.91 g/mol. Thus, for FCC (employing Equation 3.5) ρ = nA Nb a 3 N A = nA Nb ( 2 R 2 ) 3 N A = (4 atoms/unit cell)(92.91 g/mol) (2) ( 1.43 × 10 -8 cm ) ( 2 ) [ ] 3 /(unitcell) ( 6.023 × 10 23 atoms /mol ) = 9.33 g/cm 3 For BCC, n = 2, and a = 4 R 3 (Equation 3.3), thus ρ = nA Nb 4 R 3 3 N A ρ = (2 atoms/unit cell)(92.91 g/mol) (4) ( 1.43 × 10 -8 cm ) 3 3 /(unitcell) ( 6.023 × 10 23 atoms/mol ) = 8.57 g/cm 3 which is the value provided in the problem statement. Therefore, Nb has a BCC crystal structure.

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3. 17 (a) We are asked to calculate the unit cell volume for Be. For HCP, from the solution to Problem 3.6 V C = 6 R 2 c 3 But, c = 1.568 a , and a = 2 R , or c = 3.14 R , and V C = (6)(3.14) R 3 3 = (6)(3.14) ( 3 ) 0.1143 × 10 -7 cm [ ] 3 = 4.87 × 10 23 cm 3 /unit cell (b) The theoretical density of Be is determined, using Equation 3.5, as follows: ρ = nA Be V C N A For HCP, n = 6 atoms/unit cell, and for Be, A Be = 9.01 g/mol (as noted inside the front cover). Thus, ρ = (6 atoms/unit cell)(9.01 g/mol) ( 4.87 × 10 -23 cm 3 /unit cell )( 6.023 × 10 23 atoms/mol ) = 1.84 g/cm 3 The value given in the literature is 1.85 g/cm 3 .