In the spectrum of a hydrogen atom,the ratio | vedclass

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(A) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \fra

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$5/27$

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

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

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