Example 5 Calculate the atomic density in (100), (110) and (111) planes of bcc
iron whose lattice parameter is 2.87 Å.
Solution: Given: Crystal planes are: (100), (110) and (111). Crystal structure is bcc,
so that n = 2, a = 2.87 Å = 2.87 Â10
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 bcc structure
d 100 ¼
a
2
; d 110 ¼
a
ffiffi ffi
2
p and d 111 ¼
a
2
ffiffi ffi
3
p ; V ¼ a
3
Therefore,
q ð100Þ ¼
2 Â d 100
a 3
¼
2a
2a 3 ¼
1
a 2 ¼
1
2:87 Â 10 À10
ð
Þ
2
¼ 1:21 Â 10
19 atoms=m
2
Similarly,
q ð110Þ ¼
2 Â d 110
a 3
¼
2a
ffiffi ffi
2
p
a 3
¼
ffiffi ffi
2
p
a 2 ¼
ffiffi ffi
2
p
2:87 Â 10 À10
ð
Þ
2
¼ 1:72 Â 10
19 atoms=m
2
q ð111Þ ¼
2 Â d 111
a 3
¼
2a
2
ffiffi ffi
3
p
a 3
¼
1
ffiffi ffi
3
p
a 2
¼
1
ffiffi ffi
3
p
2:87 Â 10 À10
ð
Þ
2
¼ 7:0 Â 10
18 atoms=m
2
Fig. 3.11 Atomic linear density along a [110] direction in fcc b [111] direction in bcc
3.5 Atomic Density in Crystals
117