Physics - Modern Physics Question with Solution | TestHub

The wavelength of line in case of Balmer series is given by $\frac{1}{\lambda }=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{{\mathrm{n}}^2}\right)$, where $\mathrm{n}=3,4,5,\dots$ and $\mathrm{R}=$ Rydberg constant. So, for Balmer series, the transition takes from third orbit to second first line spectrum, fourth orbit to second for second line spectrum and so on. Hence, given transition represents second line of Balmer series.