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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 Lyman-alpha $(Ly-\alpha)$ line,the transition is from $n_2 = 2$ to $n_1 = 1$:$\frac{1}{\lambda_{Ly-\alpha}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.For the Balmer-alpha $(Ba-\alpha)$ line,the transition is from $n_2 = 3$ to $n_1 = 2$:$\frac{1}{\lambda_{Ba-\alpha}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$.Now,calculating the ratio $\frac{\lambda_{Ly-\alpha}}{\lambda_{Ba-\alpha}}$:$\frac{\lambda_{Ly-\alpha}}{\lambda_{Ba-\alpha}} = \frac{\frac{1}{\lambda_{Ba-\alpha}}}{\frac{1}{\lambda_{Ly-\alpha}}} = \frac{\frac{5R}{36}}{\frac{3R}{4}} = \frac{5}{36} 	imes \frac{4}{3} = \frac{5}{27}$.Wait,re-evaluating the ratio: $\frac{\lambda_{Ly-\alpha}}{\lambda_{Ba-\alpha}} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{36}{5} = \frac{48}{5}$.Re-checking the question requirement: Ratio of Lyman-alpha to Balmer-alpha wavelength is $\frac{\lambda_{Ly-\alpha}}{\lambda_{Ba-\alpha}} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{36}{5} = \frac{48}{5}$.Given the options,there is a discrepancy. Let's re-calculate: $\lambda_{Ly-\alpha} = \frac{4}{3R}$,$\lambda_{Ba-\alpha} = \frac{36}{5R}$. Ratio = $\frac{4/3}{36/5} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.

In the hydrogen spectrum,the ratio of the wavelengths for Lyman-alpha radiation to Balmer-alpha radiation is

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