Question

For Molybdenum, the wavelength of $K_{\alpha }$ line is 0.72 nm and of $K_{\beta }$ line is 0.60 nm. From this information wavelength of $L_{\alpha }$ line in nm is

Answer 1

$E_M-E_K=\frac{hc}{0.6}\text{and}{E}_L-E_K=\frac{hc}{0.72}$
$E_M-E_L=hc(\frac{1}{0.6}-\frac{1}{0.72})$
$E_M-E_L=\frac{hc}{\lambda }$
$\frac{1}{\lambda }=\frac{1}{0.6}-\frac{1}{0.72}\Rightarrow \lambda =\frac{0.6\times 0.72}{0.12}$
$\lambda =3.6nm$