The `K_(alpha)X`-ray of molybdenum has wavlength `71 "pm"`. If the energy of a molybdernum atom with a K electron knocked out is `23.32 keV`. What will be the energy of this atom when an `L`-electron is knocked out ?

Explanation

`K_a` X-Ray results from the transits of an electron from L shell to K shell. If the energy of the atom with a vacancy in the K shell is `E_K` and the energy with a vacancy in the L shell is`E_L` the energy of the photon emitted is `E_K-E_L`. The energy of the 71 pm photon is
`E = hc/lambda`
= (1242 eV nm)/(71 xx 10 ^ (-3) nm) = 17.5 keV.
Thus, E_k - E_L = 17.5 keV
or, E_L = E_K - 17.5 keV
= 23.32 keV - 17.5 keV = 5.82 keV.