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

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$5:27$

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(A) The wavelength $\lambda$ for a transition in a hydrogen-like 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 first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$:$\frac{1}{\lambda_{L_1}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies \lambda_{L_1} = \frac{4}{3R}$.For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$:$\frac{1}{\lambda_{B_1}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36} \implies \lambda_{B_1} = \frac{36}{5R}$.The ratio of the wavelengths is $\frac{\lambda_{L_1}}{\lambda_{B_1}} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.

Find the ratio of the first wavelengths of the Lyman series and the Balmer series.

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