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(B) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \fra

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$5: 27$

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(B) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$,where $R$ is the Rydberg constant.For the Lyman series,the largest wavelength corresponds to the transition from $n_i = 2$ to $n_f = 1$:$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$.For the Balmer series,the largest wavelength corresponds to the transition from $n_i = 3$ to $n_f = 2$:$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R}$.The ratio of the largest wavelength of the Lyman series to that of the Balmer series is:$\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.Thus,the ratio is $5: 27$.

For a hydrogen atom,the ratio of the largest wavelength of the Lyman series to that of the Balmer series is:

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