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(B) The wavelength $\lambda$ of radiation emitted during a transition from orbit $n_2$ to $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} =

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$n^3$

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(B) The wavelength $\lambda$ of radiation emitted during a transition from orbit $n_2$ to $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Here,$n_1 = n$ and $n_2 = n + 1$.Substituting these values: $\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) = R \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right)$.Simplifying the numerator: $(n^2 + 2n + 1 - n^2) = 2n + 1$.So,$\frac{1}{\lambda} = R \frac{2n + 1}{n^2(n+1)^2}$.For large $n$,$2n + 1 \approx 2n$ and $(n+1)^2 \approx n^2$.Thus,$\frac{1}{\lambda} \approx R \frac{2n}{n^2 \cdot n^2} = R \frac{2n}{n^4} = \frac{2R}{n^3}$.Therefore,$\lambda \propto n^3$.

The electron of a hydrogen atom makes a transition from the $(n + 1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

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