Q 12-11-068JEE MainJEE Main 2024 (29 Jan, Shift 1)Medium
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25\%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
Answer: (B) $\dfrac18$
Equal wavelengths mean equal momenta $p$.
Electron: $K_e = \dfrac{p^2}{2m} = \dfrac{pv}{2}$. Photon: $E = pc$.
$$\frac{K_e}{E} = \frac{v}{2c} = \frac{0.25}{2} = \frac18$$
Solution by Sreeraj P, M.Sc Physics