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(A) The Rydberg formula for the wavelength of emitted photons is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For

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$20/7$

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(A) The Rydberg formula for the wavelength of emitted photons is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the first jump from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:$\frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16 - 9}{144} ight) = R \left( \frac{7}{144} ight)$.For the second jump from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9 - 4}{36} ight) = R \left( \frac{5}{36} ight)$.Now,calculating the ratio $\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{R(5/36)}{R(7/144)} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$.

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths,$\lambda_1/\lambda_2$,of the photons emitted in this process is

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