If an electron in a hydrogen atom jumps from | vedclass

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

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

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(B) The Rydberg formula for the wavelength of emitted radiation is given by:$\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For a hydrogen atom,$Z = 1$.For the transition from the $3^{	ext{rd}}$ orbit to the $2^{	ext{nd}}$ orbit $(n_2=3, n_1=2)$:$\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{9} ight) = R_H \left( \frac{9-4}{36} ight) = \frac{5 R_H}{36} \quad \dots(1)$For the transition from the $4^{	ext{th}}$ orbit to the $3^{	ext{rd}}$ orbit $(n_2=4, n_1=3)$,let the wavelength be $\lambda'$:$\frac{1}{\lambda'} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R_H \left( \frac{1}{9} - \frac{1}{16} ight) = R_H \left( \frac{16-9}{144} ight) = \frac{7 R_H}{144} \quad \dots(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{\lambda'}{\lambda} = \frac{5 R_H / 36}{7 R_H / 144} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$Therefore,$\lambda' = \frac{20}{7} \lambda$.

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

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