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(C) The wavelength $\lambda$ of a spectral line emitted during a transition from orbit $n_i$ to $n_f$ is given by the Rydberg formula: $\frac{1}{\lamb

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$27 :5$

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(C) The wavelength $\lambda$ of a spectral line emitted during a transition from orbit $n_i$ to $n_f$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$,where $R$ is the Rydberg constant.For the transition $3 ightarrow 2$: $\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$. Thus,$\lambda_1 = \frac{36}{5R}$.For the transition $2 ightarrow 1$: $\frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$. Thus,$\lambda_2 = \frac{4}{3R}$.The ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{36/5R}{4/3R} = \frac{36}{5R} 	imes \frac{3R}{4} = \frac{9 	imes 3}{5} = \frac{27}{5}$.

The ratio of the wavelengths of the spectral lines emitted due to transitions $3 \rightarrow 2$ and $2 \rightarrow 1$ orbits in the hydrogen atom is

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