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(A) The wavelength of a spectral line in the Lyman series is given by $\frac{1}{\lambda_{L}} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}\right)$,where

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$5/27$

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(A) The wavelength of a spectral line in the Lyman series is given by $\frac{1}{\lambda_{L}} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}ight)$,where $n = 2, 3, 4, \dots$.For the longest wavelength in the Lyman series,we take $n = 2$:$\frac{1}{\lambda_{L}} = R\left(1 - \frac{1}{4}ight) = \frac{3R}{4} \implies \lambda_{L} = \frac{4}{3R}$.The wavelength of a spectral line in the Balmer series is given by $\frac{1}{\lambda_{B}} = R\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}ight)$,where $n = 3, 4, 5, \dots$.For the longest wavelength in the Balmer series,we take $n = 3$:$\frac{1}{\lambda_{B}} = R\left(\frac{1}{4} - \frac{1}{9}ight) = R\left(\frac{9-4}{36}ight) = \frac{5R}{36} \implies \lambda_{B} = \frac{36}{5R}$.The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:$\frac{\lambda_{L}}{\lambda_{B}} = \frac{4/3R}{36/5R} = \frac{4}{3R} 	imes \frac{5R}{36} = \frac{5}{27}$.

In the spectrum of hydrogen,the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

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