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(C) The wavelength of a spectral line in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2

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Second line of Balmer series

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(C) The wavelength of a spectral line in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$ and $R$ is the Rydberg constant.For the Balmer series,the transition to the $n=2$ orbit determines the line:- For $n=3$ to $n=2$,we get the first line of the Balmer series.- For $n=4$ to $n=2$,we get the second line of the Balmer series.Therefore,the transition from the fourth orbit to the second orbit corresponds to the second line of the Balmer series.

When an electron in a hydrogen atom jumps from the fourth Bohr orbit to the second Bohr orbit,we get the:

Similar Questions

If Rydberg's constant is $R$,the longest wavelength of radiation in Paschen series will be $\frac{\alpha}{7 R}$,where $\alpha=$ . . . . . .

What is the ratio of the wavelengths of radiations emitted when an electron in a hydrogen atom jumps from the fourth orbit to the second orbit and from the third orbit to the second orbit?

The ratio of maximum to minimum wavelength in the Balmer series of a hydrogen atom is

The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition of an electron from the:

The number of possible spectral lines in a hydrogen atom is:

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