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Q.
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
Line $K_{\alpha}$ corresponds to transfer of electron from $L$ -shell to $K$ -shell.
Here, $\lambda =0.021 nm$
$=2.1 \times 10^{-11} m =\frac{h c}{E}=\frac{h c}{e E( eV )}$
$\Rightarrow E( eV ) =\frac{h c}{e \lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 2.1 \times 10^{-11}} eV$
$=5.9 \times 10^{4} eV =59\, keV$