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(D) According to the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$In the first case,the electron jumps from

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$\frac{20}{7}$

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(D) According to the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} ight]$In the first case,the electron jumps from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:$\frac{1}{\lambda_1} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{9} - \frac{1}{16} ight] = R \left[ \frac{16 - 9}{144} ight] = \frac{7}{144} R$    .... $(i)$In the second case,the electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{9 - 4}{36} ight] = \frac{5}{36} R$    .... $(ii)$To find the ratio $\frac{\lambda_1}{\lambda_2}$,we divide the expression for $\frac{1}{\lambda_1}$ by $\frac{1}{\lambda_2}$:$\frac{\lambda_2}{\lambda_1} = \frac{\frac{5}{36} R}{\frac{7}{144} R} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$Therefore,the ratio $\frac{\lambda_1}{\lambda_2} = \frac{20}{7}$.

An electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases is

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