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(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \righ

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$4: 1$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the shortest wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.For the Lyman series,$n_1 = 1$. Thus,$\frac{1}{\lambda_L} = R (1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.So,$\lambda_L = \frac{1}{R}$.For the Balmer series,$n_1 = 2$. Thus,$\frac{1}{\lambda_B} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$.So,$\lambda_B = \frac{4}{R}$.The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is $\frac{\lambda_B}{\lambda_L} = \frac{4/R}{1/R} = 4: 1$.

The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for a hydrogen atom is:

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