For h
2
þ k
2
þ l
2
À
Á ¼ 3, we have
d 111 ¼
a
ffiffi ffi
3
p ¼
4:05
ffiffi ffi
3
p
Further, from Bragg’s law, we have
h 111 ¼ sin
À1
k
2d 111
¼ sin
À1
ffiffi ffi
3
p  1:79
2 Â 4:05
¼ 22:5
Therefore, for 57.3 mm camera radius
S 1 ¼ 4h 111 ¼ 4 Â 22:5 ¼ 90:01 mm
Similarly, for h
2
þ k
2
þ l
2
ð
Þ¼4; 8; 11 we can obtain
d 200 ¼
a
2
¼ 2:025 ˚
A; h 200 ¼ 26:229982 and S 2 ¼ 105:0 mm
d 220 ¼
a
ffiffi ffi
8
p ¼ 1:4318912 ˚
A; h 220 ¼ 38:685670 and S 3 ¼ 154:7 mm
d 113 ¼
a
ffiffi ffi
1
p
1
¼ 1:2211209 ˚
A; h 113 ¼ 47:132856 and S 4 ¼ 188:5 mm
Example 14 First reflection obtained from a sample of copper powder (fcc) has the
S-value of 86.7 mm, by using CuKa radiation. Determine the camera radius.
Solution: Given: k K a
ð Þ ¼ 1:54 ˚
A, S = 86.7 mm, crystal is fcc, so for first reflection, the value of h
2
þ k
2
þ l
2
À
Á ¼ 3, for copper a = 3.61 Å, camera radius = ?
For fcc crystal, we have
d 111 ¼
a
h
2
þ k
2
þ l
2
À
Á 1=2 ¼
a
1 2 þ 1
2
þ 1
2
À
Á 1=2 ¼
3:61
ffiffi ffi
3
p
Further, from Bragg’s law, we have
h 111 ¼ sin
À1
k
2d 111
¼ sin
À1
ffiffi ffi
3
p  1:54
2 Â 3:61
¼ 21:68
Therefore, the value of S is
S 1 ¼ 4h 111 ¼ 4 Â 2168 ¼ 86:72 mm
This is equal to the given S-value. Since S ¼ 4h corresponds to the camera
radius 57.3 mm, therefore, 57.3 mm is the required camera radius.
9.1 Steps in Crystal Structure Determinations
353