From first table, make use of h. The values of sin
2
h , common factor (c.f) are
found and added into other columns of the next table. The values of N for allowed
reflections give us that the unknown cubic form is the body-centered cubic
(bcc) structure. Finally, we get the value of lattice parameter “a.”
Line
h (°)
sin
2 h
Common factor (c.f)
N ¼
sin
2 h
c.f
ffiffiffi ffi
N
p
a ¼ d
ffiffiffi ffi
N
p
1
19.30
0.1092
2
ffiffi ffi
2
p
3.30
2
27.26
0.2183
4
ffiffi ffi
4
p
3.36
3
34.85
0.3265
0.0546
6
ffiffi ffi
6
p
3.31
4
41.28
0.4352
8
ffiffi ffi
8
p
3.31
5
47.50
0.5436
10
ffiffiffiffiffi
10
p
3.31
6
53.84
0.6518
12
ffiffiffiffiffi
12
p
3.31
The average value of “a” from the Table is found to be 3.316 Å. The above data
indicates that the element is Niobium.
(ii) Laue Method
If the crystal to film distance is taken to be D and the distance of the diffraction spot
(Laue spot) from the center of the photographic film (Fig. 9.2) is r, then
tan 2h ¼
r
D
where h is the Bragg’s angle
Solved Examples
Example 1 Bragg’s reflection is observed at an angle of 5° in a transmission Laue
pattern for KCl crystal. The crystal to film distance is 4 cm; determine the location
of the spot with reference to center of the film.
Solution: Given: Bragg’s angle h = 5°, crystal to film distance D = 4 cm, distance
of reflection from the center of the pattern, r = ?
Fig. 9.2 Relationship among
reflecting plane, it is normal
and the direction of diffraction
beam
9.1 Steps in Crystal Structure Determinations
355