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(B) In a hydrogen atom,the energy of the $n^{	ext{th}}$ level is given by $E_n = -\frac{E_0}{n^2}$,where $E_0$ is the ionization energy of hydrogen.F

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$\frac{1}{n^{3}}$

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(B) In a hydrogen atom,the energy of the $n^{	ext{th}}$ level is given by $E_n = -\frac{E_0}{n^2}$,where $E_0$ is the ionization energy of hydrogen.For a transition from $(n+1)$ to $n$,the energy of the emitted radiation $\Delta E$ is the difference between the energy levels:$\Delta E = E_{n+1} - E_n = E_0 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} ight)$Using the relation $\Delta E = h
u$,we have:$h
u = E_0 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} ight) = E_0 \left( \frac{2n+1}{n^2(n+1)^2} ight)$For $n >> 1$,we can approximate $(n+1) \approx n$ and $(2n+1) \approx 2n$:$h
u \approx E_0 \left( \frac{2n}{n^2 \cdot n^2} ight) = E_0 \left( \frac{2n}{n^4} ight) = \frac{2E_0}{n^3}$Thus,the frequency $
u$ is proportional to $\frac{1}{n^3}$.

In a hydrogen atom,the electron makes a transition from the $(n+1)^{\text{th}}$ level to the $n^{\text{th}}$ level. If $n >> 1$,the frequency of the radiation emitted is proportional to:

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