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(C) The wavelength $\lambda$ for a hydrogen atom transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{

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$4: 1$

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(C) The wavelength $\lambda$ for a hydrogen atom transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the shortest wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.Thus,$\frac{1}{\lambda} = \frac{R}{n_1^2}$,or $\lambda = \frac{n_1^2}{R}$.For the Balmer series,$n_1 = 2$,so $\lambda_{Balmer} = \frac{2^2}{R} = \frac{4}{R}$.For the Bracket series,$n_1 = 4$,so $\lambda_{Bracket} = \frac{4^2}{R} = \frac{16}{R}$.The ratio of the shortest wavelength of the Bracket series to the Balmer series is $\frac{\lambda_{Bracket}}{\lambda_{Balmer}} = \frac{16/R}{4/R} = \frac{16}{4} = 4:1$.

The ratio of the shortest wavelengths of Bracket and Balmer series of hydrogen atom is

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