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(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.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 $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the transition from $n_2 = 3$ to $n_1 = 2$,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 $n_2 = 4$ to $n_1 = 2$,let the wavelength be $\lambda$:$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{4} - \frac{1}{16} ight] = R \left[ \frac{4 - 1}{16} ight] = \frac{3R}{16}$.Dividing the two equations:$\frac{\lambda}{\lambda_0} = \frac{5R/36}{3R/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$.Therefore,$\lambda = \frac{20}{27}{\lambda _0}$.

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 jump from the fourth to the second orbit of the hydrogen atom,the wavelength of radiation emitted will be

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