The ratio of the wavelength of the last line | vedclass

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

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$\frac{9}{4}$

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(A) The wavelength $\lambda$ of a spectral line in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.For the last line of the Paschen series,$n_1 = 3$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = \frac{R}{9}$,which gives $\lambda_P = \frac{9}{R}$.For the last line of the Balmer series,$n_1 = 2$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$,which gives $\lambda_B = \frac{4}{R}$.The ratio of the wavelength of the last line of the Paschen series to that of the Balmer series is $\frac{\lambda_P}{\lambda_B} = \frac{9/R}{4/R} = \frac{9}{4}$.

The ratio of the wavelength of the last line of the Paschen series to that of the Balmer series is:

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