or
k Sn ¼
25
ð Þ
2
49
ð Þ
2
 k Fe
¼
25
ð Þ
2
49
ð Þ
2
 1.93 = 0:5 ˚
A
Making a similar calculation by combining platinum and barium, we can obtain
k Ba ¼ 0.37 ˚
A :
Example 10 Wavelength of X-ray K a line of an unknown element is 0.7185 Å.
Find the element.
Solution: Given: k K a
ð Þ¼ 0.7185 ˚
A ¼ 0.7185 Â 10
À10 m, element = ?
Let us use Moseley’s law given by
m¼
c
k
¼
m e
4 Z À b
ð
Þ
2
8e 2
0 h
3
1
n 2
1
À
1
n 2
2
Since, K a transition takes place from n = 2 to n = 1 and b = 1. Therefore,
substituting n 1 ¼ 1; n 2 ¼ 2 in the above equation, we obtain
k K a
ð Þ ¼
4
3
8e
2
0 ch
3
me 4 Z À 1
ð
Þ
2
!
or
Z - 1
ð
Þ
2 ¼
32 Â e
2
0  c  h
3
3m  e 4  k K a
ð Þ
¼
32 Â 8:85 Â 10
À12
ð
Þ
2 Â3 Â 10
8
 6:626  10
À34
ð
Þ
3
3 Â 9:1 Â 10 À31 Â 1:6 Â 10 À19
ð
Þ
4 Â0:7185 Â 10 À10
¼ 1701:54
or
Z ¼ 1 þ 41:25 ffi 42
⟹ The required element is molybdenum.
Example 11 Determine the wavelengths of K a , K b and L a lines for copper
(Z = 29).
Solution: Given: Z (copper) = 29, K a ¼ ?; K b ¼ ?; L a ¼ ?
7.1 Production of X-Rays
271
k Sn ¼
25
ð Þ
2
49
ð Þ
2
 k Fe
¼
25
ð Þ
2
49
ð Þ
2
 1.93 = 0:5 ˚
A
Making a similar calculation by combining platinum and barium, we can obtain
k Ba ¼ 0.37 ˚
A :
Example 10 Wavelength of X-ray K a line of an unknown element is 0.7185 Å.
Find the element.
Solution: Given: k K a
ð Þ¼ 0.7185 ˚
A ¼ 0.7185 Â 10
À10 m, element = ?
Let us use Moseley’s law given by
m¼
c
k
¼
m e
4 Z À b
ð
Þ
2
8e 2
0 h
3
1
n 2
1
À
1
n 2
2
Since, K a transition takes place from n = 2 to n = 1 and b = 1. Therefore,
substituting n 1 ¼ 1; n 2 ¼ 2 in the above equation, we obtain
k K a
ð Þ ¼
4
3
8e
2
0 ch
3
me 4 Z À 1
ð
Þ
2
!
or
Z - 1
ð
Þ
2 ¼
32 Â e
2
0  c  h
3
3m  e 4  k K a
ð Þ
¼
32 Â 8:85 Â 10
À12
ð
Þ
2 Â3 Â 10
8
 6:626  10
À34
ð
Þ
3
3 Â 9:1 Â 10 À31 Â 1:6 Â 10 À19
ð
Þ
4 Â0:7185 Â 10 À10
¼ 1701:54
or
Z ¼ 1 þ 41:25 ffi 42
⟹ The required element is molybdenum.
Example 11 Determine the wavelengths of K a , K b and L a lines for copper
(Z = 29).
Solution: Given: Z (copper) = 29, K a ¼ ?; K b ¼ ?; L a ¼ ?
7.1 Production of X-Rays
271
