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(D) For the Lyman series,the maximum wavelength occurs for the transition from $n = 2$ to $n = 1$.Using the Rydberg formula: $\frac{1}{\lambda} = R \l

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$\frac{5}{27}$

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(D) For the Lyman series,the maximum wavelength occurs for the transition from $n = 2$ to $n = 1$.Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.Thus,$(\lambda_L)_{\max} = \frac{4}{3R}$.For the Balmer series,the maximum wavelength occurs for the transition from $n = 3$ to $n = 2$.Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$.Thus,$(\lambda_B)_{\max} = \frac{36}{5R}$.The ratio is $\frac{(\lambda_L)_{\max}}{(\lambda_B)_{\max}} = \frac{4}{3R} 	imes \frac{5R}{36} = \frac{4 	imes 5}{3 	imes 36} = \frac{20}{108} = \frac{5}{27}$.

The ratio of the maximum wavelengths of the Lyman and Balmer series in the hydrogen spectrum is ........

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