We know that the transitions corresponding to K a , K b and L a are:
K a : n ¼ 2
ð
Þ! n ¼ 1
ð
Þ; K b : n ¼ 3
ð
Þ! n ¼ 1
ð
Þand L a : n ¼ 3
ð
Þ!ðn ¼ 2Þ
Let us find the required wavelengths, one by one by using the equation
m ¼
1
k
¼
m
c
¼
me
4
Z À b
ð
Þ
2
8e 2
0 h 3
1
n 2
1
À
1
n 2
2
or
k¼
8e
2
0 ch
3
me 4 Z
2
Â
1
n
2
1
À
1
n
2
2
Substituting different values of constants, we obtain
k K a
ð Þ ¼
8 Â 8:85 Â 10
À12
ð
Þ
2 Â3 Â 10
8
 6:626  10
À34
ð
Þ
3
9:1 Â 10 À31 Â 1:6 Â 10 À19
ð
Þ
4 Â 29
ð Þ
2 Â
1
1 2 À
1
2 2
À
Á
¼ 1.22 Â 10
À10 m ¼ 1:22 ˚
A
Similarly,
k K b
ð Þ ¼
8 Â 8:85 Â 10
À12
ð
Þ
2 Â3 Â 10
8
 6:626  10
À34
ð
Þ
3
9:1 Â 10 À31 Â 1:6 Â 10 À19
ð
Þ
4 Â 29
ð Þ
2 Â
1
1 2 À
1
3 2
À
Á
¼ 1:45 Â 10
À10 m ¼ 1:45 ˚
A
and
k L a
ð Þ ¼
8 Â 8:85 Â 10
À12
ð
Þ
2 Â3 Â 10
8
 6:626  10
À34
ð
Þ
3
9:1 Â 10 À31 Â 1:6 Â 10 À19
ð
Þ
4 Â 29
ð Þ
2 Â
1
2 2 À
1
3 2
À
Á
¼ 7:84 Â 10
À10 m ¼ 7:84 ˚
A
7.2 X-Ray Diffraction by Crystals
(a) The well-known Bragg’s equation is derived on the basis of the following:
(i) A crystal is made up of various sets of equidistant parallel planes without
taking into consideration the actual distribution of atoms in them.
272
7 Diffraction of Waves and Particles by Crystal