Question

The$K_{\alpha }$ X -rays of aluminium (Z= 13) and zinc (Z=30) have a wavelengths 887 pm and 146 pm respectively. Use Moseley's law $\sqrt{v}$ = a(Z-b) to find the wavelength of the $K_{\alpha }$ X-ray of iron (Z= 26).

Answer 1

Given $\lambda$ = 887 pm

f =$\frac{c}{\lambda }=\frac{3\times {10}^8}{887\times {10}^{-12}}$

= $3.382\times {10}^{-17}$

= 33.82 $\times {10}^{16}$

${\lambda }_2$ = 146 pm

f = $\frac{3\times {10}^8}{146\times {10}^{-12}}$

= 0.02054 $\times {10}^{20}$

= 2.054 $\times {10}^{18}$

We know f= a(Z-b)

5.815 $\times {10}^8$ = a(13-b) ...(1)

21.4331 $\times {10}^9$ = a(30-b) ...(2)

Divinding (1) by (2)

$\frac{13-b}{30-b}=\frac{5.815\times {10}^{-1}}{21.4331}$

= 0.4057

$\Rightarrow 30\times$0.4057 - 0.4057 b+b = 13

b = $\frac{0.829}{0.5943}$ = 1.39491

a = $\frac{5.815\times {10}^8}{11.33}$

= 0.51323 $\times {10}^8=5\times {10}^7$

For 'Fe'

f= 5 $\times {10}^7$ (26-1.39)

= 5 $\times 24.61\times {10}^7$

= 123.05 $\times {10}^7$

and $\frac{c}{\lambda }=5141.3\times {10}^{14}$

$\lambda =\frac{3\times {10}^8}{5141.3\times {10}^{14}}$

= 0.000198 $\times {10}^{-5}$ m

= 198 $\times {10}^{-12}$ = 198 pm