If an electron in a hydrogen atom jumps from | vedclass

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \r

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$\frac{20}{7}\lambda$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} ight)$.For the transition from $n_i = 3$ to $n_f = 2$:$\frac{1}{\lambda} = R \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$.For the transition from $n_i = 4$ to $n_f = 3$:$\frac{1}{\lambda'} = R \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16-9}{144} ight) = \frac{7R}{144}$.Taking the ratio of the two equations:$\frac{\lambda'}{\lambda} = \frac{5R/36}{7R/144} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$.Therefore,$\lambda' = \frac{20}{7}\lambda$.

If an electron in a hydrogen atom jumps from the $3^{rd}$ orbit to the $2^{nd}$ orbit,it emits a photon of wavelength $\lambda$. When it jumps from the $4^{th}$ orbit to the $3^{rd}$ orbit,the corresponding wavelength of the photon will be

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