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Question
The Kα X-ray emission line of tungsten occurs at λ = 0.021 nm. The energy difference between K and L levels in this atoms is about
Options
A.0.51 MeV
B.1.2 MeV
C.59 keV
D.13.6 eV
Solution
λKα = 0.021 amu = 0.21
Since, λKα corresponding to the transition of an electron from L-shell to K-shell,
therefore
EL EK = (in eV) =
or ᐃE = 59 keV.
Since, λKα corresponding to the transition of an electron from L-shell to K-shell,
therefore
EL EK = (in eV) =
or ᐃE = 59 keV.
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