Example 9 Calculate the atomic density in (100), (110) and (111) planes of fcc
aluminum whose lattice parameter is 4.05 Å.
Solution: Given: Crystal planes are: (100), (110) and (111). Crystal structure fcc, so
that n = 4, a = 4.05 Å = 4:05 Â 10
À10
m, q p ¼ ?
We know that the atomic density in a crystal plane is given by
q p ¼
nd
V
Further, for fcc structure
d 100 ¼
a
2
; d 110 ¼
a
2
ffiffi ffi
2
p and d 111 ¼
a
ffiffi ffi
3
p ; V ¼ a
3
Therefore,
q ð100Þ ¼
4 Â d 100
a 3
¼
4a
2a 3 ¼
2
a 2 ¼
2
4:05 Â 10 À10
ð
Þ
2
¼ 1:22 Â 10
19 atoms=m
2
Similarly,
q ð110Þ ¼
4 Â d 110
a 3
¼
4a
2
ffiffi ffi
2
p
a 3
¼
ffiffi ffi
2
p
a 2 ¼
ffiffi ffi
2
p
4:05 Â 10 À10
ð
Þ
2
¼ 8:62 Â 10
18 atoms=m
2
q ð111Þ ¼
4 Â d 111
a 3
¼
4a
ffiffi ffi
3
p
a 3
¼
4
ffiffi ffi
3
p
a 2
¼
4
ffiffi ffi
3
p
2:5 Â 10 À10
ð
Þ
2
¼ 1:41 Â 10
19 atoms=m
2
Example 10 Calculate the atomic density in (0001) plane of hcp zinc, whose
lattice parameter a = b = 2.66 Å and c = 4.95 Å.
Solution: Given: Crystal plane is: (0001). Crystal structure hcp,
a = b = 2.66 Å = 2.66 Â10
À10 m and c = 4.95 Å = 4:95 Â 10
À10 m, q p ¼ ?
We know that the atomic density in a crystal plane is given by
q p ¼
Effective number of atoms in a given plane
Area of the given plane
¼
n
0
A
¼
nd
V
3.5 Atomic Density in Crystals
119