Question

The $K_{\alpha }$ $X-$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. $1.2MeV$
• C. $59keV$
• D. $13.6eV$

Answer 1

The correct option is C $59keV$

Energy emitted for $K_{\alpha }$ jump
$E=\frac{hc}{\lambda }=\frac{3\times {10}^8\times 6.64\times {10}^{-34}}{0.021\times {10}^{-9}}=9.48\times {10}^{-15}J=59.28KeV$

The $K_{\alpha }$ jumps is from $K$ to $L$.