Modern Physics Ques 161
- 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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Answer:
Correct Answer: 161.(c)
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 \\ \text { or } \quad \Delta E \approx 59 keV \end{aligned} $
BETA
