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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

$A$ doubly ionised $Li$ atom is excited from its ground state $(n = 1)$ to $n = 3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratios $\lambda_{32}/\lambda_{31}$ and $\lambda_{21}/\lambda_{31}$ are,respectively:

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The ratio of the frequencies of the long wavelength limits of Lyman and Balmer series of hydrogen spectrum is

When an electron in a hydrogen atom jumps to the innermost orbit,the radiation emitted belongs to which one of the following series?

Let $F_1$ be the frequency of the second line of the Lyman series and $F_2$ be the frequency of the first line of the Balmer series. Then,the frequency of the first line of the Lyman series is given by:

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