The ratio of the maximum wavelengths of the L | vedclass

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(D) The wavelength $\lambda$ is related to the energy transition by the Rydberg formula: $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n

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$5 : 27$

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(D) The wavelength $\lambda$ is related to the energy transition by the Rydberg formula: $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For maximum wavelength,the energy difference must be minimum,which corresponds to the first line of the series.For the Lyman series $(n_1 = 1, n_2 = 2)$: $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.For the Balmer series $(n_1 = 2, n_2 = 3)$: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$.The ratio of wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{1/\lambda_B}{1/\lambda_L} = \frac{5R/36}{3R/4} = \frac{5}{36} 	imes \frac{4}{3} = \frac{5}{27}$.

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

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