Q 12-12-117JEE MainJEE Main 2019 (12 Apr, Shift 1)Medium
An excited He$^+$ ion emits two photons in succession, with wavelengths $108.5$ nm and $30.4$ nm, in making a transition to the ground state. The quantum number $n$, corresponding to its initial excited state is (for a photon of wavelength $\lambda$, energy $E = \dfrac{1240\ \text{eV}}{\lambda\ (\text{in nm})}$)
Answer: (B) $n = 5$
Total energy emitted:
$$\frac{1240}{108.5} + \frac{1240}{30.4} = 11.43 + 40.79 = 52.2\ \text{eV}$$
For He$^+$ ($Z = 2$), $E_n = -\dfrac{54.4}{n^2}$ eV:
$$54.4\left(1 - \frac1{n^2}\right) = 52.2 \Rightarrow \frac1{n^2} \approx 0.04 \Rightarrow n = 5$$
Solution by Sreeraj P, M.Sc Physics