JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Modern Physics

Energy of incident photon:

$E=\frac{hc}{\lambda }=\frac{1.237\times 10^{-6}}{970\times 10^{-10}}=12.75$ eV

The electron in ground state ($n=1$, $E_1=-13.6$ eV) absorbs this energy:

$E_{\text{final}}=-13.6+12.75=-0.85$ eV

This corresponds to the $n=4$ level ($E_4=-13.6/16=-0.85$ eV).

The electron is excited to $n=4$. In the emission spectrum, transitions can occur from $n=4$ to all lower levels:

Number of spectral lines = ${}^4{C}_2=\frac{4\times 3}{2}=6$

Note: The source indicates the unit of $hc/e$ in the original paper is incorrect, but the calculation proceeds as shown.