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(B) The Rydberg formula for the wavelength $\lambda$ of a transition is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \righ

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$\frac{5}{27}$

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(B) The Rydberg formula for the wavelength $\lambda$ of a transition is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,the longest wavelength corresponds to the transition from $n_2 = 2$ to $n_1 = 1$. Thus,$\frac{1}{\lambda_{	ext{Lyman}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.For the Balmer series,the longest wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$. Thus,$\frac{1}{\lambda_{	ext{Balmer}}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$.Taking the ratio: $\frac{\lambda_{	ext{Lyman}}}{\lambda_{	ext{Balmer}}} = \frac{5R/36}{3R/4} = \frac{5}{36} 	imes \frac{4}{3} = \frac{5}{27}$.

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is :-

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