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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 smallest energy transition,which is from $n_2 = 2$. Thus,$\frac{1}{\lambda_{L, \max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$,so $\lambda_{L, \max} = \frac{4}{3R}$.For the Balmer series,$n_1 = 2$. The longest wavelength corresponds to the transition from $n_2 = 3$. Thus,$\frac{1}{\lambda_{B, \max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$,so $\lambda_{B, \max} = \frac{36}{5R}$.The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is $\frac{\lambda_{L, \max}}{\lambda_{B, \max}} = \frac{4/3R}{36/5R} = \frac{4}{3R} 	imes \frac{5R}{36} = \frac{4 	imes 5}{3 	imes 36} = \frac{20}{108} = \frac{5}{27}$.

In the spectrum of the 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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