Q 12-12-080JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
A light of energy $12.75$ eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\dfrac x\pi\times10^{-17}$ eV s. The value of $x$ is ______ (use $h=4.14\times10^{-15}$ eV s, $c=3\times10^8\ \text{m s}^{-1}$).
Numerical value type. Enter your answer.
Answer: 828
$-13.6+12.75=-0.85$ eV, which is $n=4$.
$$L=\frac{4h}{2\pi}=\frac{2\times4.14\times10^{-15}}{\pi}=\frac{828}{\pi}\times10^{-17}\ \text{eV s}$$
Solution by Sreeraj P, M.Sc Physics