Question

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

• A. 0.51MeV
• B. 1.2 MeV
• C. 59 KeV
• D. 13.6 eV

Answer 1

The correct option is C 59 KeV

$E_K-E_L=\frac{hc}{\lambda }=\frac{(6.6\times {10}^{-34})(3\times {10}^8)}{(0.021\times {10}^{-9})(1.6\times {10}^{-19})}eV=59keV$