Q 12-12-089JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
A hydrogen atom in its ground state absorbs $10.2$ eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of (Given, Planck's constant $= 6.6\times10^{-34}$ Js).
Answer: (B) $1.05\times10^{-34}$ Js
$E_2 - E_1 = -3.4 - (-13.6) = 10.2$ eV, so the electron goes from $n = 1$ to $n = 2$.
By Bohr's postulate $L = \dfrac{nh}{2\pi}$, so
$$\Delta L = \frac{h}{2\pi} = \frac{6.6\times10^{-34}}{2\times3.14} \approx 1.05\times10^{-34}\ \text{Js}$$
Solution by Sreeraj P, M.Sc Physics