Transitions, Photon Emission & Excitation question from MHT CET 2026 (13 Apr, Shift 2)

Answer: (c)

1. Second excited state is $n=3$, first excited is $n=2$, third excited is $n=4$.
2. $3\to2$: $\frac{1}{\lambda_0} \propto \frac14-\frac19 = \frac{5}{36}$. $4\to2$: $\frac{1}{\lambda'} \propto \frac14-\frac1{16} = \frac{3}{16}$.
3. $\frac{\lambda'}{\lambda_0} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}$
4. So $x = 27$, answer (c) $27$.