Question

The wavelengths of $K_a$ X-rays of two metals A and B are $\frac{4}{1875R}$ and $\frac{1}{675R}$ respectively, where R is Rydberg constant. The number of elements lying between A and B in the periodic table (according to their atomic numbers) is

Answer 1

Using $\frac{1}{\lambda }=R(z-1{)}^2[\frac{1}{n_2^2}-\frac{1}{n_1^2}]$
For $K_{\alpha }x-rays;n_1=2,n_2=1$
For metal $A;\frac{1875R}{4}=R(Z_1-1{)}^2(\frac{3}{4})\Rightarrow z_1=26$
For metal $B;675R$ $R(Z_2-1{)}^2(\frac{3}{4})$
$\Rightarrow z_2=31$
Therefore, 4 elements lie between A and B.