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Page 1: Chap 3 Solns-5/E - che.uri.eduche.uri.edu/course/che332/hmwk_soln/HW1/callister7e_sm_ch03_20.pdf= 0.35 nm, c = 0.45 nm, and α = β = γ = 90°. (b) The crystal structure would be

3-23

Crystal Systems

3.20 (a) The unit cell shown in the problem statement belongs to the tetragonal crystal system since a = b

= 0.35 nm, c = 0.45 nm, and α = β = γ = 90°.

(b) The crystal structure would be called body-centered tetragonal.

(c) As with BCC, n = 2 atoms/unit cell. Also, for this unit cell

VC = (3.5 × 10−8 cm)2(4.5 × 10−8 cm)

= 5.51 × 10−23 cm3/unit cell

Thus, using Equation 3.5, the density is equal to

ρ = nA

VC N A

= (2 atoms/unit cell) (141 g/mol)

(5.51 × 10-23 cm3/unit cell)(6.023 × 1023 atoms/mol)

= 8.49 g/cm3

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