JEE Main 2025 Apr 3 Shift 2, Physics Q24: Bohr Model of the Hydrogen Atom

The fourth excited state corresponds to $n_i=5$ (ground state is $n=1$, so first excited $=2$, ..., fourth excited $=5$).

Photon energy from Bohr's relation:

$\Delta E=13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ eV.

$2.86=13.6\left(\frac{1}{n^2}-\frac{1}{25}\right)$.

$\frac{1}{n^2}=\frac{2.86}{13.6}+\frac{1}{25}=0.2103+0.04=0.2503\approx \frac{1}{4}$.

$\therefore n^2=4\Rightarrow n=2$.