When an electron in hydrogen atom jumps from the fourth Bohr orbit to second Bohr orbit we get the
When an electron in hydrogen atom jumps from the fourth Bohr orbit to second Bohr orbit we get the
Second line of Paschen series
First line of pfund series
Second line of Balmer
First line of Balmer series
Solution
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 $\mathrm{n}=3,4,5, \ldots$ 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.