1. Use the Bohr energy formula

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

2. Energy emitted in transition $(n+1) \to n$

   The emitted photon energy is $\Delta E = E_n - E_{n+1}.$

   Using the formula, $\Delta E = -\frac{13.6}{n^2} - \left(-\frac{13.6}{(n+1)^2}\right)$ $\Delta E = 13.6\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right).$

3. Simplify the expression

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

4. Use the condition $n \gg 1$

   For very large $n$, $2n+1 \approx 2n, \qquad (n+1)^2 \approx n^2.$

   Therefore, $\Delta E \propto \frac{2n}{n^2\cdot n^2} = \frac{2}{n^3}.$

   Hence, $\Delta E \propto \frac{1}{n^3}.$

5. Relate energy to frequency

   Since photon energy is $\Delta E = h\nu,$ frequency $\nu$ is proportional to $\Delta E$.

   Therefore, $\nu \propto \frac{1}{n^3}.$

6. Option check

   • A: $\dfrac{1}{n}$ — incorrect
   • B: $\dfrac{1}{n^2}$ — incorrect
   • C: $\dfrac{1}{n^3}$ — correct
   • D: $\dfrac{1}{n^4}$ — incorrect

Therefore, the correct option is C.