1. Energy levels of a hydrogen-like atom

   The ऊर्जा of the $n$th orbit is $E_n = -\frac{K}{n^2}$ where $K$ is a constant for a given hydrogen-like atom.

2. Transition from $n$ to $(n-1)$

   The emitted photon has energy $h\nu = E_{n-1} - E_n$ Since energies are negative, $h\nu = -\frac{K}{(n-1)^2} - \left(-\frac{K}{n^2}\right)$ $h\nu = K\left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right)$

3. Simplify the expression

   = \frac{n^2-(n-1)^2}{n^2(n-1)^2}$$ Now, $$n^2-(n-1)^2 = n^2-(n^2-2n+1)=2n-1$$ Hence, $$h\nu = K\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, $h\nu \propto \frac{2n}{n^2\cdot n^2} = \frac{2}{n^3}$

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

5. Check options

   • A: $\frac{1}{n}$ ❌
   • B: $\frac{1}{n^2}$ ❌
   • C: $\frac{1}{n^{3/2}}$ ❌
   • D: $\frac{1}{n^3}$ ✅

Therefore, the correct option is D.