Atoms & Nuclei

JEE Main level, numerical answer, medium

64 of 239

Question 64 of 239

The Kα{K}_{\alpha } X-ray of molybdenum has wavelength 0.071nm0.071nm. If the energy of a molybdenum atom with a KK electron knocked out is 27.5keV27.5keV, the energy of this atom when an LL electron is knocked out will be keVkeV. (Round off to the nearest integer ) [h=4.14×10−15eVs,c=3×108ms−1]\left[h=4.14\times {10}^{-15}eVs,c=3\times {10}^{8}{ms}^{-1}\right]

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Answer: 10

Ekα=Ek−EL{E}_{{k}_{\alpha }}={E}_{k}-{E}_{L}
hcλkα=Ek−EL\dfrac{hc}{{\lambda }_{{k}_{\alpha }}}={E}_{k}-{E}_{L}
EL=Ek−hcλkα{E}_{L}={E}_{k}-\dfrac{hc}{{\lambda }_{{k}_{\alpha }}}
=27.5KeV−12.42×10−7eVm0.071×10−9m=27.5KeV-\dfrac{12.42\times {10}^{-7}eVm}{0.071\times {10}^{-9}m}
EL=(27.5−17.5)keV{E}_{L}=(27.5-17.5)keV
=10keV=10keV

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