AIIMS 2018 Physics Spectral Series MCQ Question

The first spectral line in the Balmer series for hydrogen corresponds to the transition from $n_2 = 3$ to $n_1 = 2$, yielding a wavelength of 6534 Å. For singly ionized helium (He$^+$), the second spectral line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$, using the formula $\frac{1}{λ} = R(2)² \left[ \frac{1}{2²} - \frac{1}{4²} \right]$. This results in a wavelength of 2430 Å, which is not listed as an option. However, the first spectral line for He$^+$ is actually calculated from the transition $n_2 = 3$ to $n_1 = 2$, yielding 1215 Å, making option A the only valid choice. Other options are incorrect as they do not correspond to the calculated wavelengths for the specified transitions in the Balmer series.