Solved Examples
Example 1 In a cubic crystal system, the distance between the consecutive
(111) plane is 2Å. Determine its lattice parameter.
Solution: Given: Crystal plane (111), d = 2Å, a = ?
For a cubic crystal (Table 4.3), we have
d ¼
a
h
2
þ k
2 + l
2
À
Á 1=2
Substituting different values, we get
d ¼
a
1
2
þ 1
2 + 1
2
À
Á 1=2 ¼
a
ffiffi ffi
3
p
or a ¼ d
ffiffi ffi
3
p ¼ 2
ffiffi ffi
3
p ˚
A ¼ 3:46 ˚
A
Example 2 In a tetragonal crystal, the lattice parameters a = b = 2.42Å, and
c = 1.74Å. Deduce the interplanar spacing between the consecutive (101) planes.
Table 4.4 Interplanar spacing in various crystal systems
Crystal system d hkl
Cubic
a h
2 þ k
2 þ l
2
À
Á À1=2
Tetragonal
h
2 þ k
2
a
2
þ l
2
c
2
h
i À1=2
Orthorhombic
h
2
a
2 þ k
2
b
2 þ l
2
c
2
h
i À1=2
Hexagonal
4=3ðh
2 þ hk þ k
2 Þ
a
2
þ l
2
c
2
! À1=2
Rhombohedral
a 1À3 cos
2 a þ 2 cos
3 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 aÀcos a
ð
Þ
Â
à 1=2
Monoclinic
h
2
k
2 þ l
2
c
2 À
2hl cos b
ð
Þ
ac
sin
2 b
þ k
2
b
2
2
4
3
5
À1=2
Triclinic
h
a
h=a cos c cos b
k=b
1
cos a
l=c cos a
1
þ k
b
1
h=a cos b
cos c k=b cos a
cos b l=c
1
þ l
c
1
cos c h=a
cos c
1
k=b
cos b cos a l=c
2
4
3
5
À1=2
1
cos c cos b
cos c
1
cos a
cos b cos a
1
À1=2
4.4 Interplanar Spacing
157