Mse 500

Chapter 3 Homework

3.2

VC = 16R3√2 = (16)(0.175 x 10-9m)3 (√2) = 1.213 x 10-28 m3

3.5

BCC = 2 spheres APF = VS / VC

VS = (2) 4/3 π R3 = 8 π R3 / 3

VC = 4 R3 / √3 = 64 R3 / 3√3

APF = 8 π R3 / 3 = 0.68

64 R3 / 3√3

3.9

VC = 4 R3 / √3 = 64 R3 / 3√3

P = nATa / VCNA = nATa / (64 R3 / 3√3) NA

R = [3√3 nATa / 64p NA]1/3

= (3√3) (2 atoms/unit cell)(180.9 g/mol) 1/3 = 1.43 x 10 -8cm =.143nm

(64)(16.9 g/cm3)(6.023 x 1023 atoms/mol)

3.13

FCC: n = 4, a = 2 R√2, A = 92.91 g/mol

P = nANb / a3NA = nANb / (2 R √2)3 NA

= (4 atoms/unit cell)(92.91 g/mol) =

{[(2)(1.43 x 10-8 cm)(√2)]3 / (unit cell)}(6.023 x 1023 atoms/mol)

9.33 g/cm3

BCC: n = 2, a = 4 R / √3, A = 92.91 g/mol

P = nANb / a3 NA

= (2 atoms/unit cell)(92.91 g/mol) =

{[(4)(1.43 x 10-8 cm) / (√3)]3 / (unit cell)}(6.023 x 1023 atoms/mol)

8.57 g/cm3

* After review of the values of both FCC and BCC crystal structures, it is determined that Niobium has a BCC crystal structure.

3.23

List the point coordinates of both the sodium and chlorine ions for a unit cell of the sodium chloride crystal structure (Figure 12.2).

Chlorine Sodium Horizontal plane Top face

0 0 0 ½ 0 0 0 0 ½ ½ 0 1

100 1 ½ 0 1 0 ½ 1 ½ 1

1 1 0 ½ 1 0 ½ ½ ½ ½ 1 1

010 0 ½ 0 1 1 ½ 0 ½ 1

0 0 1 0 1 ½

1 0 1

1 1 1

0 1 1

½ ½ 0

½ ½ 1

1 ½ ½

0 ½ ½

½ 0 ½

½ 1 ½

3.30

3.31

A direct = [