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The $\mathrm{K}_{\alpha} \mathrm{X}-ray$ of molybdenum has wavelength $0.071 \mathrm{~nm}$ . If the energy of a molybdenum atoms with a $\mathrm{K}$ electron knocked out is $27.5 \mathrm{keV}$ , the energy of this atom when an L electron is knocked out will be ________ $\mathrm{keV}.$ (Round off to the nearest integer)
$\left[\mathrm{h}=4.14 \times 10^{-15} \mathrm{eVs}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right]$

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best_answer

$E_{1}$ - Energy when an L shell-electron is knocked out
$E_{2}$ - Energy when a K shell-electron is knocked out
$E_{2}=27.5 \mathrm{keV}$
$K_{\alpha} \rightarrow$ wavelength emitted during knocking out of L-electron
$E_{K_{\alpha}} =\frac{1240}{\lambda_{k\alpha}}=\frac{1240\left ( ev \right )}{0.071}$
$E_{k \alpha }=\frac{1240}{71 \times 10^{-3}}=\frac{1240 \times 10^{3}}{71}(\mathrm{ev})$

$E_{2}=E_{1}+E_{k_{\alpha}}$

$27.5 \mathrm{eV}=E_{1}+17.5 \mathrm{keV}$

$E_{1}=10eV$

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vishal kumar

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