Q 12-12-099JEE MainJEE Main 2021 (26 Feb, Shift 2)Medium
The recoil speed of a hydrogen atom after it emits a photon in going from $n = 5$ state to $n = 1$ state will be
Answer: (A) $4.17$ m s$^{-1}$
Photon energy: $E = 13.6\left(1 - \dfrac{1}{25}\right) = 13.06$ eV $= 2.09\times10^{-18}$ J.
Momentum conservation: $mv = \dfrac{E}{c}$.
$$v = \frac{E}{mc} = \frac{2.09\times10^{-18}}{1.67\times10^{-27}\times3\times10^8} \approx 4.17\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics