Question 482: Atoms & Nuclei

Question

When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the

Accepted Answer

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

Answer

second line of Balmer series

Answer

first line of Balmer series

Answer

first line of Pfund series

Answer

second line of Paschen series

Atoms & Nuclei

When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the

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