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  4. The Kα 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) [h=4.14 × 10-15 eV s , c=3 × 108 ms -1]

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Q. 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)
$\left[h=4.14 \times 10^{-15} \,eV\,s , c=3 \times 10^{8}\, ms ^{-1}\right]$

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Solution:

$E_{k_{a}}=E_{k}-E_{L}$
$\frac{h c}{\lambda_{k_{a}}}=E_{k}-E_{L}$
$E_{L}=E_{k}-\frac{h c}{\lambda_{k_{a}}}$
$=27.5 \,KeV -\frac{12.42 \times 10^{-7}\, eV\, m }{0.071 \times 10^{-9} m }$
$E_{L}=(27.5-17.5)\, keV$
$=10\, keV$