Question

When an electron in an excited state of $Mo$ atom falls from $L$ to K-shell, and X-ray is emitted. These X-rays are diffracted at an angle of ${7.75}^{\circ }$ by planes with a separation of $2.64\stackrel{\circ }{A}$. What is the difference in energy between K-shell and L-shell in $Mo$, assuming the first-order diffraction? $(\sin {7.75}^{\circ }-0.1349)$.

Answer 1

According to Bragg's equation:
$n\lambda =2d\sin \theta$
$1\times \lambda =2\times 2.64\sin {7.75}^{\circ }=2\times 2.24\times 0.1349$
$\lambda =0.712\stackrel{\circ }{A}$
Energy difference between $K$ and $L-shell$ of $Mo$.
$=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{0.712\times {10}^{-10}}=2.791\times {10}^{-15}J$.