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(B) The wavelength $\lambda$ for a transition in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2}

Vedclass · Accepted Answer

$5/27$

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(B) The wavelength $\lambda$ for a transition 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 longest wavelength,the energy difference must be minimum,meaning the transition occurs from $n_2 = n_1 + 1$ to $n_1$.For the Lyman series,$n_1 = 1$,so the longest wavelength corresponds to $n_2 = 2$: $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.For the Balmer series,$n_1 = 2$,so the longest wavelength corresponds to $n_2 = 3$: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$.Taking the ratio $\frac{\lambda_L}{\lambda_B} = \frac{1/\lambda_B}{1/\lambda_L} = \frac{5R/36}{3R/4} = \frac{5}{36} 	imes \frac{4}{3} = \frac{5}{27}$.

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

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