The ratio of the longest wavelengths correspo | vedclass

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(A) The wavelength of spectral lines in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1

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$5/27$

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(A) The wavelength of spectral lines in the hydrogen spectrum 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 series,$n_1 = 1$. The longest wavelength corresponds to the smallest energy transition,which is $n_2 = 2$.$\frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$.For the Balmer series,$n_1 = 2$. The longest wavelength corresponds to the smallest energy transition,which is $n_2 = 3$.$\frac{1}{\lambda_B} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{9-4}{36} ight] = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R}$.The ratio of the longest wavelengths is:$\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3R} 	imes \frac{5R}{36} = \frac{4 	imes 5}{3 	imes 36} = \frac{20}{108} = \frac{5}{27}$.

The ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is:

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