Ekα=Ek−EL
λkαhc=Ek−EL
EL=Ek−λkαhc
=27.5KeV−0.071×10−9m12.42×10−7eVm
EL=(27.5−17.5)keV
=10keV
JEE Main 2021 — Physics Modern Physics
The Kα X-ray of molybdenum has wavelength 0.071nm. If the energy of a molybdenum atom with a K electron knocked out is 27.5keV, the energy of this atom when an L electron is knocked out will be keV. (Round off to the nearest integer ) [h=4.14×10−15eVs,c=3×108ms−1]
Held on 27 Jul 2021 · Verified 6 Jul 2026.
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