Q. Let $\lambda_{\alpha},\lambda_{\beta}$ and $\lambda_{\alpha}^{'}$ denote the wavelengths of the X-rays of the $K_{\alpha},K_{\beta}$ and $L_{\alpha}$ lines in the characteristic $X$-rays for a metal. Then
Explanation
(c) According to the energy diagram of $X$-ray spectra
$\because\mathrm{\Delta}E = \frac{hc}{\lambda} \Rightarrow \lambda \propto \frac{1}{\mathrm{\Delta}E}$
($\mathrm{\Delta}E =$ Energy radiated when $e^{-}$ jumps from, higher energy orbit to lower energy orbit)
$\because(\mathrm{\Delta}E)_{k_{\beta}} > (\mathrm{\Delta}E)_{k_{\alpha}} > (\mathrm{\Delta}E)_{L_{\alpha}}\ \therefore\lambda_{\alpha}^{'} > \lambda_{\alpha} > \lambda_{\beta}$
Also $(\Delta E)_{k_{\beta}} = (\Delta E)_{k_{\alpha}} + (\Delta E)_{L_{\alpha}}$
$\Rightarrow \frac{hc}{\lambda_{\beta}} = \frac{hc}{\lambda_{\alpha}} + \frac{hc}{\lambda_{\alpha}^{'}} \Rightarrow \frac{1}{\lambda_{\beta}} = \frac{1}{\lambda_{\alpha}} + \frac{1}{\lambda_{\alpha}^{'}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.