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Question

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

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$\frac{20}{27} \lambda_0$

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(B) 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 $n_f = 2$,the wavelength is $\lambda_0$:$\frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$.For the transition from $n_i = 4$ to $n_f = 2$,let the wavelength be $\lambda_1$:$\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{4-1}{16} ight) = \frac{3R}{16}$.Now,taking the ratio $\frac{\lambda_1}{\lambda_0}$:$\frac{\lambda_1}{\lambda_0} = \frac{1/\lambda_0}{1/\lambda_1} = \frac{5R/36}{3R/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$.Therefore,$\lambda_1 = \frac{20}{27} \lambda_0$.

The wavelength of radiation emitted is $\lambda_0$ when an electron jumps from the third to the second orbit of a hydrogen atom. For the electron to jump from the fourth to the second orbit of the hydrogen atom,the wavelength of radiation emitted will be

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