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(A) 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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(A) 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 defines the series.The first line of the Balmer series corresponds to the transition from $n = 3$ to $n = 2$.The second line of the Balmer series corresponds to the transition from $n = 4$ to $n = 2$.Therefore,the transition from the fourth Bohr orbit to the second Bohr orbit represents the second line of the Balmer series.

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

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