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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} \rightarrowwavelength 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

Posted by

vishal kumar

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