For second peak, we have
d 200 ¼
a
h
2
þ k
2
þ l
2
À
Á 1=2 ¼
a
2 2 þ 0 2 þ 0 2
ð
Þ
1=2
¼
a
2
or 2 d 200 ¼ a ¼ 1:53 ˚
A
However, from Bragg’s equation, we have
k max 2d
For the second peak, this condition is violated and hence cannot occur.
Example 12 In a diffraction experiment, the Bragg’s angle corresponding to a
reflection for which h
2
þ k
2
þ l
2
À
Á ¼ 8 is found to be at 14.35° when the X-ray of
wavelength 0.71 Å is used. Determine the lattice parameter of the crystal. If there
are two other reflections of smaller Bragg’s angles, determine the crystal structure.
Solution: Given: h
2
þ k
2
þ l
2
À
Á ¼ 8, h ¼ 14:35
, k = 0.71 Å, a = ?, structure = ?
For the given (hkl), the value of d from Bragg’s equation is
d ¼
k
2 sin h
¼
0:71
2 Â sin 14:35
¼ 1:43 ˚
A
Further, for a cubic system, the relationship between a and d is given by
a ¼ d h
2
þ k
2
þ l
2
À
Á 1=2 ¼
ffiffi ffi
8
p  0:71
2 Â sin 14:35
¼ 4:05 ˚
A
Since there are two reflections at smaller angles before 14:35
, this indicates that
given value 8 lies at third place. This is possible only if the structure is fcc.
Example 13 A monochromatic X-ray beam of wavelength 1.79 Å is used to study
the fcc aluminum powder sample with a camera of radius 57.3 mm, determine the
first four S-values.
Solution: Given: Crystal is fcc, a = 4.05 Å, k = 1.79 Å, R = 57.3 mm, first four
S-values = ?
We know that the ratios of h
2
þ k
2 + l
2
À
Á
for allowed reflections in fcc are:
3 : 4 : 8 : 11 : 12 : 16 : 19 : 20
They correspond to (111), (200), (220), (113), (222), (400), etc. planes,
respectively.
352
9 Determination of Crystal Structure Parameters