# Moseley's Law

## Problems from IIT JEE

Problem (IIT JEE 2003): Characteristic X-rays of frequency ${4.2\times{10}^{18}}\;{Hz}$ are produced when transitions from $L$-shell to $K$-shell take place in a certain target material. Use Moseley's law to determine the atomic number of the target material. (Rydberg's constant $={1.1\times{10}^{7}}\;\mathrm{m^{-1}}$.)

Solution: The characteristic X-ray is emitted when an electron in $L$ shell makes a transition to the vacant state in $K$ shell. In Moseley's law, \begin{align} \sqrt{\nu}=a(Z-b), \end{align} the parameter $b\approx 1$ for this transition because electron from $L$ shell finds nuclear charge $Ze$ shielded by remaining one electron in $K$ shell i.e., effective nuclear charge is $(Z-1)e$. Thus, by substituting values, \begin{align} \frac{1}{\lambda}&=\frac{\nu}{c}=\frac{4.2\times{10}^{18}}{3\times{10}^{8}}=R(Z-1)^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \nonumber\\ &=1.1\times{10}^{7} (Z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right],\nonumber \end{align} which gives, $Z=42$.