The ratio of minimum wavelengths of Lyman and | vedclass

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Question

(B) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the minimum wavele

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the minimum wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.For the Lyman series,$n_1 = 1$,so $\frac{1}{\lambda_L} = R_H (\frac{1}{1^2} - 0) = R_H$,which means $\lambda_L = \frac{1}{R_H}$.For the Balmer series,$n_1 = 2$,so $\frac{1}{\lambda_B} = R_H (\frac{1}{2^2} - 0) = \frac{R_H}{4}$,which means $\lambda_B = \frac{4}{R_H}$.The ratio of minimum wavelengths $\frac{\lambda_L}{\lambda_B} = \frac{1/R_H}{4/R_H} = \frac{1}{4} = 0.25$.

The ratio of minimum wavelengths of Lyman and Balmer series will be $:-$

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