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

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(D) \(10\)

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As we know

\({E}_{k_{a}}=E_{k}-E_{L}\)

\(\frac{h c}{\lambda_{k_{a}}}=E_{k}-E_{L}\)

\(E_{L}= E_{k}-\frac{h c}{\lambda_{k_{a}}}\)

\(=27.5 {KeV}-\frac{12.42 \times 10^{-7}\, {eV\,m}}{0.071 \times 10^{-9}\, {m}}\)

\(E_{L}=(27.5-17.5)\, {keV}\)

\(=10\,{keV}\)

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