The ratio of longest wavelengths of the spect | vedclass

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(A) The longest wavelength corresponds to the transition between the closest energy levels. For the Lyman series,the transition is from $n_2 = 2$ to $

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$5/27$

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(A) The longest wavelength corresponds to the transition between the closest energy levels. For the Lyman series,the transition is from $n_2 = 2$ to $n_1 = 1$. Using the Rydberg formula: $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$. Thus,$\lambda_L = \frac{4}{3R}$. For the Balmer series,the transition is from $n_2 = 3$ to $n_1 = 2$. Using the Rydberg formula: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$. Thus,$\lambda_B = \frac{36}{5R}$. The ratio of the longest wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.

The ratio of longest wavelengths of the spectral lines in the Lyman and Balmer series of hydrogen spectrum is

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