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

For a hydrogen atom, the energy of the $$n^{\text{th}}$$ level is given by $$E_n = -\frac{13.6}{n^2}$$ eV.

When an electron transitions from the $$(n+1)^{\text{th}}$$ level to the $$n^{\text{th}}$$ level, the energy of the emitted photon is:

$$\Delta E = E_{n+1} - E_n = -\frac{13.6}{(n+1)^2} + \frac{13.6}{n^2} = 13.6\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$$

Simplifying the expression inside the parentheses:

$$\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{2n + 1}{n^2(n+1)^2}$$

So $$\Delta E = 13.6 \times \frac{2n + 1}{n^2(n+1)^2}$$.

Since the frequency of the emitted radiation is $$\nu = \frac{\Delta E}{h}$$, we have $$\nu \propto \frac{2n+1}{n^2(n+1)^2}$$.

Now we apply the approximation for $$n \gg 1$$:

When $$n$$ is very large, $$2n + 1 \approx 2n$$ and $$(n+1)^2 \approx n^2$$.

Therefore: $$\nu \propto \frac{2n}{n^2 \cdot n^2} = \frac{2n}{n^4} = \frac{2}{n^3}$$.

So the frequency of radiation emitted is proportional to $$\frac{1}{n^3}$$, which is Option C.