Question

The $K_a$ X-ray emission line of tungsten occurs at $\lambda =0.021nm$. The energy difference between $K$ and $L$ levels in this atom is about

• A. $0.51MeV$
• B. $1.32MeV$
• C. $59KeV$
• D. $13.6eV$

Answer 1

The correct option is C $59KeV$

Line $K_{\alpha }$ corresponds to transfer of electron from L-shell to K-shell
Here, $\lambda =0.021nm=2.1\times {10}^{-11}m$
By the relation, $E=\frac{hc}{\lambda }$ or $E\left(eV\right)=\frac{hc}{e\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 2.1\times {10}^{-11}}=5.9\times {10}^{4}eV=59KeV$