Q 12-11-106JEE MainJEE Main 2022 (26 Jun, Shift 1)Medium
An electron with speed $v$ and a photon with speed $c$ have the same de Broglie wavelength. If the kinetic energy and momentum of the electron are $E_e$ and $P_e$ and that of the photon are $E_{ph}$ and $P_{ph}$ respectively, which of the following is correct?
Answer: (B) $\dfrac{E_e}{E_{ph}} = \dfrac{v}{2c}$
Same wavelength means the same momentum $p = \dfrac h\lambda$, so $\dfrac{P_e}{P_{ph}} = 1$.
$E_e = \dfrac12mv^2 = \dfrac{pv}{2}$ and $E_{ph} = pc$, so $\dfrac{E_e}{E_{ph}} = \dfrac{v}{2c}$.
Solution by Sreeraj P, M.Sc Physics