For a hexagonal crystal system, we know that a = b 6 ¼ c and the interplanar
spacing is given by
d ðhkilÞ =
4
3
.
h
2
þ hk + k
2
À
Á
a 2
þ
l
2
c 2
"
# À1=2
Therefore,
d ð1010Þ ¼
4
3
Â
1
16
! À1=2
¼
1
12
! À1=2
¼ 3:46 ˚
A
d ð0110Þ ¼
4
3
Â
1
16
! À1=2
¼
1
12
! À1=2
¼ 3:46 ˚
A
d ð1100Þ ¼
4
3
Â
1 À 1 þ 1
ð
Þ
16
! À1=2
¼
1
12
! À1=2
¼ 3:46 ˚
A
d ð1120Þ ¼
4
3
Â
1 þ 1 þ 1
ð
Þ
16
! À1=2
¼
1
4
! À1=2
¼ 2 ˚
A
Example 7 The lattice parameter of an ideal rhombohedrdron is 3Å. Determine
the interplanar spacing in (100), (110) and (111) planes.
Solution: Given: An ideal rhombohedron with a = b = c = 3Å. Also, we know that
for an ideal rhombohedron, a = 60°. d 100 ¼ ?0:1cmd 110 ¼ ?0:1cmd 111 ¼ ?
The interplanar spacing for rhombohedral crystal system is given by
d hkl
ð Þ ¼
a 1 þ cos
3
a À 3 cos
2
a
ð
Þ
1=2
h
2
þ k
2
þ l
2
À
Á
sin
2
a þ 2 hk þ kl þ lh
ð
Þcos 2 a À cos a
ð
Þ
Â
à 1=2
Therefore,
d 100
ð
Þ ¼
3 1 þ cos
3 60 À 3 cos
2 60
ð
Þ
1=2
sin 60
¼
3 1 þ 2 Â 0:125 À 3 Â 0:25
ð
Þ
1=2
0:886
¼
2:12
0:866
¼ 2:45 ˚
A
d 110
ð
Þ ¼
2:12
2 sin
2 60 þ 2 cos 2 60 À cos 60
ð
Þ
Â
à 1=2
¼
2:12
2 Â 0:75 þ 2 0:25 À 0:5
ð
Þ
½
Š
1=2
¼
2:12
1:5 À 0:5
½
Š
1=2
¼ 2:12 ˚
A
162
4 Unit Cell Representations of Miller Indices