Modern Physics 4 Question 20
20. 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 atoms is about
(1997C, 1M)
(a) $0.51 MeV$
(b) $1.2 MeV$ (c) $59 keV$
(d) $13.6 eV$
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Solution:
- $\lambda _{k _{\alpha}}=0.021 nm=0.21 \AA$
Since, $\lambda _{k _{\alpha}}$ corresponds to the transition of an electron from $L$-shell to $K$-shell, therefore
$$ \begin{aligned} E _L-E _K=(\text { in eV }) & =\frac{12375}{\lambda(\text { in } \AA)} \\ & =\frac{12375}{0.21} \approx 58928 eV \end{aligned} $$
or
$$ \Delta E \approx 59 keV $$
