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Q.
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
NTA AbhyasNTA Abhyas 2022
Solution:
$\lambda _{k_{\alpha }}=0.021nm=0.21\overset{^\circ }{A}$
The wavelength $\lambda _{k_{\alpha }}$ corresponds to the transition of an electron from $L$ shell to $K$ shell.
The energy is expressed as $E=\frac{h c}{\lambda }$
Therefore,
$E_{L}-E_{K}=\frac{12375}{\lambda \left(in \overset{^\circ }{A}\right)}eV=\frac{12375}{0 . 21}\approx58928eV$
or $\Delta E=59keV$ .