The ratio of wavelengths of the second line i | vedclass

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(C) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \fra

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$4: 1$

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(C) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.For the second line of the Balmer series,$n_1 = 2$ and $n_2 = 4$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$. So,$\lambda_B = \frac{16}{3R}$.For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$. So,$\lambda_L = \frac{4}{3R}$.The ratio of the wavelengths is $\frac{\lambda_B}{\lambda_L} = \frac{16/3R}{4/3R} = \frac{16}{4} = 4:1$.

The ratio of wavelengths of the second line in the Balmer series and the first line in the Lyman series of a hydrogen atom is:

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