Question

Use Moseley's law with b = 1 to find the frequency of the $K_{\alpha }$ X -ray of La(Z=57) if the frequency of the $K_{\alpha }$ X -ray of Cu(Z-29) is known to be 1.88 $\times {10}^{18}$ Hz.

Answer 1

Given b= 1

For La, a = (57)

$\sqrt{f}$ = a (Z-b)

$\Rightarrow \sqrt{f}$ = a (57-1)

= b $\times$ 56 ...(1)

for Cu(29)

$\sqrt{1.88}\times {10}^{18}$ = a (29-1) = 28 a ...(2)

Divinding (1) and (2)

$\sqrt{\frac{f}{1.88\times {10}^{18}}}=\frac{a\times 56}{a\times 28}$ = 2

$\Rightarrow f=1.88\times {10}^{18}(2{)}^2$

= $4\times 1.88\times {10}^{18}$

= 7.52 $\times {10}^{18}$ Hz.