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(C) The frequency of a spectral line in the hydrogen atom is given by $f = R c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydb

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$\frac{7 f}{20}$

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(C) The frequency of a spectral line in the hydrogen atom is given by $f = R c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R$ is the Rydberg constant and $c$ is the speed of light.For the Lyman series,$n_1 = 1$. The first line is $n_2 = 2$ and the second line is $n_2 = 3$.$f_1 = R c \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R c \left( 1 - \frac{1}{4} ight) = \frac{3}{4} R c$$f_2 = R c \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = R c \left( 1 - \frac{1}{9} ight) = \frac{8}{9} R c$The difference is $f = f_2 - f_1 = R c \left( \frac{8}{9} - \frac{3}{4} ight) = R c \left( \frac{32 - 27}{36} ight) = \frac{5}{36} R c$. Thus,$R c = \frac{36 f}{5}$.For the Balmer series,$n_1 = 2$. The first line is $n_2 = 3$ and the second line is $n_2 = 4$.$f'_1 = R c \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R c \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5}{36} R c$$f'_2 = R c \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R c \left( \frac{1}{4} - \frac{1}{16} ight) = \frac{3}{16} R c$The difference is $f' = f'_2 - f'_1 = R c \left( \frac{3}{16} - \frac{5}{36} ight) = R c \left( \frac{27 - 20}{144} ight) = \frac{7}{144} R c$.Substituting $R c = \frac{36 f}{5}$,we get $f' = \frac{7}{144} 	imes \frac{36 f}{5} = \frac{7 f}{4 	imes 5} = \frac{7 f}{20}$.

If the difference in the frequencies of the first and second lines of Lyman series of hydrogen atom is $f$,then the difference in frequencies of the first and second lines of Balmer series of hydrogen atom is

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