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(B) The wavelength of the spectral lines in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1

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Balmer

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(B) The wavelength of the spectral lines in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Balmer series,$n_1 = 2$. The visible region of the hydrogen spectrum corresponds to the Balmer series.The wavelength $4860 \mathring A$ (or $486 \ nm$) corresponds to the $H_{\beta}$ line of the Balmer series,which occurs when an electron transitions from $n_2 = 4$ to $n_1 = 2$.Since this wavelength falls within the visible spectrum,it belongs to the Balmer series.

Which of the following spectral series in a hydrogen atom gives a spectral line of $4860 \mathring A$?

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