hc
k K a
ð Þ
= E L - E K
or
k K a
ð Þ¼
hc
E L À E K
¼
6:626 Â 10
À34
 3  10
8
1:6 Â 10 À19 Â 66 Â 10 3 ¼ 0.19 ˚
A
Similarly, from Eq. (ii), we can obtain k K b
ð Þ ¼ 0:17 ˚
A :
Example 7 An unknown element is found to emit L a X-rays of wavelength
4.174 Å, while the wavelength of L a (X-ray line) of platinum (atomic number 78) is
1.321 Å. Determine the atomic number of unknown element where the value of b
for L a is 7.4.
Solution: Given: Wavelength of the unknown element (say x) k a = 4.174 Å,
k L a
ð Þ of Pt = 1:321 ˚
A; At. No. (Pt) = 78, b = 7.4, At. No. of element (x) = ?
We know that the characteristic frequency emitted by an element and its atomic
number is related through
m ¼ [a(Z À b)]
2
where a, b are constants.
Thus, for Pt, this equation can be written as
m Pt ¼
c
k Pt
¼ a 78 À 7.4
ð
Þ
½
Š
2
ðiÞ
Similarly, for unknown element
m x ¼
c
k x
¼ a x À 7:4
ð
Þ
½
Š
2
ðiiÞ
Dividing Eqs. (ii) from (i) and solving for x, we obtain
x ¼ 7.4 þ
k Pt
k x
1=2
 70.6
¼ 7.4 þ
1321
4174
1=2
 70.6 = 47:1
Thus the unknown element is silver.
Example 8 The wavelength of K a (X-ray line) of tungsten is 210 Å. Calculate its
value for copper if the atomic number of tungsten is 74 and that of copper is 29.
7.1 Production of X-Rays
269