The ratio of minimum wavelengths of Lyman and | vedclass

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(B) The wavelength of 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_

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$0.25$

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(B) The wavelength of 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 minimum wavelength,the transition must be from $n_2 = \infty$ to $n_1$.For the Lyman series,$n_1 = 1$,so $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,which gives $\lambda_L = \frac{1}{R}$.For the Balmer series,$n_1 = 2$,so $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$,which gives $\lambda_B = \frac{4}{R}$.The ratio of the minimum wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

The ratio of minimum wavelengths of Lyman and Balmer series will be

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