Area of the plane ¼ 4
ffiffi ffi
2
p  10
À10 m
2 ¼ 32 Â 10
À20 m
2
Therefore, the planar atomic density is
q p ¼
2atoms
32 Â 10 À20 m 2 ¼ 6:25 Â 10
18 atoms=m
2
Example 3 Calculate the linear atomic density along [100], [110] and [111]
directions in fcc copper whose lattice parameter a = 3.61 Å.
Solution: Given: Crystal structure is fcc, directions are: [100], [110] and [111],
a = 3.61 Å = 3.61 Â10
À10 m, q l ¼ ?
The number of atoms intersected along [100] or [111] is one, therefore
q ½100Š ¼
1
a
¼
1
3.61 Â 10
À10
¼ 2:77 Â 10
9 atoms=m
The number of atoms intersected along [110] is 2 (Fig. 3.11a), therefore
q ½110Š ¼
2
ffiffi ffi
2
p
a
¼
ffiffi ffi
2
p
a
¼
ffiffi ffi
2
p
3:61 Â 10 À10 ¼ 3:92 Â 10
9 atoms=m
q ½111Š ¼
1
ffiffi ffi
3
p
a
¼
1
ffiffi ffi
3
p  3:61  10 À10
¼ 1:60 Â 10
9 atoms=m
Example 4 Calculate the linear atomic density along [100], [110] and [111]
directions in bcc iron whose lattice parameter a = 2.87 Å.
Solution: Given: Crystal structure is bcc, directions are: [100], [110] and [111],
a = 2.87 Å = 2.87 Â10
À10 m, q l ¼ ?
The number of atoms intersected along [100] or [110] is one, therefore
q ½100Š ¼
1
a
¼
1
2:87 Â 10 À10 ¼ 3:48 Â 10
9 atoms=m
q ½110Š ¼
1
ffiffi ffi
2
p
a
¼
1
ffiffi ffi
2
p ð2:87 Â 10 À10 Þ
¼ 2:46 Â 10
9 atoms=m
The number of atoms intersected along [111] is 2 (Fig. 3.11b), therefore
q ½111Š ¼
2
ffiffi ffi
3
p
a
¼
2
ffiffi ffi
3
p ð2:87 Â 10 À10 Þ
¼ 4:02 Â 10
9 atoms=m
116
3 Unit Cell Calculations