The ratio of the longest wavelengths correspo | vedclass

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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 Lyman series,the longest wavelength corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\frac{1}{\lambda_{Lyman}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.For the Balmer series,the longest wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\frac{1}{\lambda_{Balmer}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$.Taking the ratio of the longest wavelengths: $\frac{\lambda_{Lyman}}{\lambda_{Balmer}} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.

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

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