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(C) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the Lyman s

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$\frac{5}{27}$

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(C) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the Lyman series,$n_1 = 1$. The longest wavelength corresponds to the smallest energy transition,which occurs at $n_2 = 2$.$\frac{1}{\lambda_{\max(L)}} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3R}{4} \implies \lambda_{\max(L)} = \frac{4}{3R}$.For the Balmer series,$n_1 = 2$. The longest wavelength corresponds to the smallest energy transition,which occurs at $n_2 = 3$.$\frac{1}{\lambda_{\max(B)}} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{9-4}{36} ight] = \frac{5R}{36} \implies \lambda_{\max(B)} = \frac{36}{5R}$.The ratio of the longest wavelengths is $\frac{\lambda_{\max(L)}}{\lambda_{\max(B)}} = \frac{4}{3R} 	imes \frac{5R}{36} = \frac{4 	imes 5}{3 	imes 36} = \frac{20}{108} = \frac{5}{27}$.

Ratio of longest wavelength corresponding to Lyman and Balmer series in hydrogen spectrum is

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