An excited $\mathrm{He}^{+}$ion emits two photons in succession, with wavelengths 108.5 nm and 30.4

Question

An excited $\mathrm{He}^{+}$ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm , in making a transition to ground state. The quantum number $n$, corresponding to its initial excited state is (for photon of wavelength $\lambda$, energy $\mathrm{E}=\frac{124 \mathrm{eV}}{\lambda(\ln \mathrm{nm})}$ ).

Professor PC Thomas Classes · Answer

Total energy $E = E_1 + E_2 = 1240/108.5 + 1240/30.4$. Equating to Rydberg formula for He+ gives $n=5$.