JEE Main 2026 — Atomic Physics Question with Solution

The angular momentum of an electron in the $n$-th orbit of a hydrogen atom is given by Bohr's quantization rule:

$L=\frac{nh}{2\pi }$

Given that the angular momentum is $\frac{3h}{\pi }$, we can equate the two expressions:

$\frac{nh}{2\pi }=\frac{3h}{\pi }$

Solving for $n$, we get:

$n=6$

The energy of an electron in the $n$-th orbit of a hydrogen atom is given by the formula:

$E_n=-\frac{13.6}{n^2}\text{ eV}$

Substituting $n=6$ into the energy formula:

$E_6=-\frac{13.6}{6^2}\text{ eV}$

$E_6=-\frac{13.6}{36}\text{ eV}$

$E_6\approx -0.377\text{ eV}\approx -0.38\text{ eV}$

Answer: $-0.38$