The ratio of minimum wavelengths of Lyman and | vedclass

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(C) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$.For the minimum wav

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$0.25$

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(C) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$.For the minimum wavelength (shortest wavelength),the transition occurs from $m = \infty$ to the ground state $n$ of the series.For the Lyman series,$n = 1$ and $m = \infty$:$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \implies \lambda_{L} = \frac{1}{R}$.For the Balmer series,$n = 2$ and $m = \infty$:$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4} \implies \lambda_{B} = \frac{4}{R}$.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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