Transitions, Photon Emission & Excitation question from JEE Main 2016 (9 Apr, Shift 1)

Answer: (c)

1. Photon energy (H, $2 \to 1$) $= 10.2\text{ eV}$.
2. Ionisation energy of Li$^{2+}$ from level $n$: $\frac{13.6 \times 9}{n^2} = \frac{122.4}{n^2}$ eV.
3. Need $\frac{122.4}{n^2} \le 10.2 \Rightarrow n^2 \ge 12 \Rightarrow n \ge 3.46$.
4. Least integer $n = 4$ (needs $7.65$ eV).
5. So the answer is (c) 4.