Q 12-12-095JEE MainJEE Main 2021 (31 Aug, Shift 2)Easy
A free electron of $2.6$ eV energy collides with a $\text{H}^+$ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. ($h = 6.6\times10^{-34}$ J s)
Answer: (B) $1.45\times10^9$ MHz
First excited state: $n = 2$, $E_2 = -3.4$ eV.
Photon energy $= 2.6 - (-3.4) = 6.0$ eV.
$$f = \frac{6.0\times1.6\times10^{-19}}{6.6\times10^{-34}} \approx 1.45\times10^{15}\ \text{Hz} = 1.45\times10^9\ \text{MHz}$$
Solution by Sreeraj P, M.Sc Physics