Q 12-12-094JEE MainJEE Main 2021 (27 Jul, Shift 2)Medium
The $K_\alpha$ X-ray of molybdenum has wavelength $0.071$ nm. If the energy of a molybdenum atom 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) $[h = 4.14\times10^{-15}\ \text{eV s},\ c = 3\times10^8\ \text{m s}^{-1}]$
Numerical value type. Enter your answer.
Answer: 10
Energy of the $K_\alpha$ photon:
$$E = \frac{hc}{\lambda} = \frac{4.14\times10^{-15}\times3\times10^8}{0.071\times10^{-9}} \approx 17.5\ \text{keV}$$
$K_\alpha$ is emitted when the vacancy moves from $K$ to $L$, so $E_L = E_K - E_{K_\alpha} = 27.5 - 17.5 = 10$ keV.
Solution by Sreeraj P, M.Sc Physics