The quantum number corresponding to the excited state of ion if on transition to the ground state th — Atomic Structure Chemistry Question

💡 Solution & Explanation

The total energy emitted is the sum of the energies of the two photons: $\Delta E = \frac{1240}{108.5} + \frac{1240}{30.4} \approx 11.43 \text{ eV} + 40.79 \text{ eV} \approx 52.22$ eV. The energy of the $n^{th}$ state for $He^+$ ($Z=2$) is $E_n = \frac{-13.6 \times 2^2}{n^2} = \frac{-54.4}{n^2}$ eV. The ground state energy is -54.4 eV. The initial energy is $E_n = -54.4 + 52.22 = -2.18$ eV. Equating this gives $\frac{-54.4}{n^2} = -2.18 \implies n^2 \approx 25 \implies n = 5$. Therefore, correct answer is 5.