d hkl ¼
a
h
2
þ k
2
þ l
2
À
Á 1=2
ðiiÞ
Combining these two equations, we have
sin h ¼
k
2d hkl
¼
h
2
þ k
2
þ l
2
À
Á 1=2
2a
From Table 9.1, we know that the first four lines and their corresponding planes
for simple cubic structure are: 1(100), 2(110), 3(111) and 4(200).
Therefore, the 2h angles corresponding to these (hkl) planes are
sin h 100 ¼
k
2d 100
¼
1
2
þ 0
2
þ 0
2
À
Á 1=2
2a
¼
1 Â 1:54
2 Â 4
or 2h 100 ¼ 2 sin
À1
k
2d 100
¼ 2 sin
À1 1 Â 1:54
2 Â 4
¼ 22:2
Similarly,
sin h 110 ¼
k
2d 110
¼
1
2
þ 1
2
þ 0
2
ð
Þ
1=2
2a
¼
ffiffi ffi
2
p  1:54
2 Â 4
or 2h 110 ¼ 2 sin
À1
k
2d 110
¼ 2 sin
À1
ffiffi ffi
2
p  1:54
2 Â 4
¼ 31:6
sin h 111 ¼
k
2d 111
¼
1
2
þ 1
2
þ 1
2
ð
Þ
1=2
2a
¼
ffiffi ffi
3
p  1:54
2 Â 4
or 2h 111 ¼ 2sin
À1
k
2d 111
¼ 2 sin
À1
ffiffi ffi
3
p  1:54
2 Â 4
¼ 39:0
and
sin h 200 ¼
k
2d 200
¼
2
2
þ 0
2
þ 0
2
ð
Þ
1=2
2a
¼
2 Â 1:54
2 Â 4
or 2h 200 ¼ 2 sin
À1
k
2d 200
¼ 2 sin
À1 2 Â 1:54
2 Â 4
¼ 45:3
Example 3 Determine the (hkl) and 2h values of the first four lines on the powder
patterns obtained by using CuKa radiation of wavelength 1.54 Å corresponding to a
body-centered cubic structure with a = 4 Å.
Solution Given: k = 1.54 Å, body-centered cubic structure with a = 4 Å, (hkl) and
2h values of the first four lines = ?
342
9 Determination of Crystal Structure Parameters