3.2 If the atomic radius of aluminum is 0.143 nm, calculate the volume of its unit cell in cubic meters.

Solution

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

VC = 16R3 2 = (16) (0.143 10-9 m)3( 2) = 6.62 10-29 m3

3.5 Show that the atomic packing factor for BCC is 0.68.

Solution

The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or

APF =

VSVC

Since there are two spheres associated with each unit cell for BCC

VS = 2(sphere volume) = 2

4R33

=

8R33

Also, the unit cell has cubic symmetry, that is VC = a3. But a depends on R according to Equation 3.3,

and

VC =

4R3

3

= 64 R3

3 3

Thus,

APF =

VSVC

= 8 R3 /3

64 R3 /3 3= 0.68

(001)

4.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327C

(600 K). Assume an energy for vacancy formation of 0.55 eV/atom.

Solution

In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation 4.1. As stated in the problem, Qv = 0.55 eV/atom. Thus,

N vN

= exp QvkT

= exp

0.55 eV / atom(8.62 105 eV / atom - K) (600 K)

= 2.41 10-5