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Question

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

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

Vedclass · Answer

(D) 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 the $3^{rd}$ orbit to the $2^{nd}$ orbit: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$. For the transition from the $4^{th}$ orbit to the $3^{rd}$ orbit,let the wavelength be $\lambda'$: $\frac{1}{\lambda'} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16-9}{144} ight) = \frac{7R}{144}$. Now,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,$\lambda' = \frac{20}{7} \lambda$.

When an electron in a hydrogen atom jumps from the third orbit to the second orbit,it emits a photon of wavelength $\lambda$. When it jumps from the fourth orbit to the third orbit,the wavelength emitted by the photon will be

Similar Questions

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