In a hydrogen atom,the ratio of the shortest | vedclass

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(C) The shortest wavelength in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty} \ri

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$4: 9$

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(C) The shortest wavelength in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty} \right) = \frac{R}{4}$.Thus,$\lambda_B = \frac{4}{R}$.The shortest wavelength in the Paschen series is given by: $\frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - \frac{1}{\infty} \right) = \frac{R}{9}$.Thus,$\lambda_P = \frac{9}{R}$.The ratio of the shortest wavelength in the Balmer series to that in the Paschen series is:$\frac{\lambda_B}{\lambda_P} = \frac{4/R}{9/R} = \frac{4}{9}$.

In a hydrogen atom,the ratio of the shortest wavelength in the Balmer series to that in the Paschen series is:

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