Question

The $k_{\alpha }$ x-ray emission line of tungsten occurs at $\lambda =0.2\stackrel{\circ }{A}$. If energy difference between K and L level in this atom is approximately (10n) keV, then n is [Take hc = 1200 eV-nm]

Answer 1

$10n\times {10}^{3}eV=\frac{1200eV-nm}{0.02nm}$
$n=\frac{1200}{0.02\times {10}^4}=6$