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(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \r

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$\frac{27}{32} \lambda$

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(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R$ is the Rydberg constant.For the Lyman series,$n_1 = 1$.The first member corresponds to the transition from $n_2 = 2$ to $n_1 = 1$:$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies R = \frac{4}{3\lambda}$.The second member corresponds to the transition from $n_2 = 3$ to $n_1 = 1$:$\frac{1}{\lambda'} = R \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = R \left( 1 - \frac{1}{9} ight) = \frac{8R}{9}$.Substituting $R = \frac{4}{3\lambda}$ into the equation for $\lambda'$:$\frac{1}{\lambda'} = \frac{8}{9} 	imes \left( \frac{4}{3\lambda} ight) = \frac{32}{27\lambda}$.Therefore,$\lambda' = \frac{27}{32} \lambda$.

If the wavelength of the first member of the Lyman series of hydrogen is $\lambda$,what is the wavelength of the second member?

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