The ratio of the minimum wavelengths of the L | vedclass

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(B) For the Lyman series,the minimum wavelength occurs when $n_1 = 1$ and $n_2 = \infty$. The formula is $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} -

Vedclass · Accepted Answer

$0.25$

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(B) For the Lyman series,the minimum wavelength occurs when $n_1 = 1$ and $n_2 = \infty$. The formula is $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For Lyman series: $\frac{1}{\lambda_{L}} = R (\frac{1}{1^2} - 0) = R$,so $\lambda_{L} = \frac{1}{R}$.For Balmer series: $n_1 = 2$ and $n_2 = \infty$. So,$\frac{1}{\lambda_{B}} = R (\frac{1}{2^2} - 0) = \frac{R}{4}$,so $\lambda_{B} = \frac{4}{R}$.The ratio $\frac{\lambda_{L}}{\lambda_{B}} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

The ratio of the minimum wavelengths of the Lyman and Balmer series is ...... .

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