In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n - 1). If n >> 1, the frequency of radiation emitted is proportional to :

For a hydrogen-like atom, the stationary‐state energies are given by the well-known Bohr formula

$$E_n \;=\; -\,\frac{R_H\,Z^{2}}{n^{2}}\;,$$

where $$R_H$$ is Rydberg’s constant expressed in energy units, $$Z$$ is the atomic number (for hydrogen itself $$Z=1$$) and $$n$$ is the principal quantum number.

The electron is said to jump from the level having quantum number $$n$$ to the neighbouring lower level $$n-1$$. The energy released in this downward transition equals the difference of the two level energies, so

$$\Delta E \;=\; E_{n-1} \;-\; E_{n}.$$

First write each energy explicitly from the formula:

$$E_{n-1} \;=\; -\,\frac{R_H Z^{2}}{(n-1)^{2}}, \quad E_{n} \;=\; -\,\frac{R_H Z^{2}}{n^{2}}.$$

Substituting these expressions into the definition of $$\Delta E$$ gives

$$\Delta E \;=\; -\,\frac{R_H Z^{2}}{(n-1)^{2}} \;-\;\Bigl(\,-\,\frac{R_H Z^{2}}{n^{2}}\Bigr) \;=\; -\,\frac{R_H Z^{2}}{(n-1)^{2}} \;+\;\frac{R_H Z^{2}}{n^{2}}.$$

Factorising the constant $$R_H Z^{2}$$:

$$\Delta E \;=\; R_H Z^{2}\left(\frac{1}{n^{2}} \;-\;\frac{1}{(n-1)^{2}}\right).$$

Simplify the bracket by taking a common denominator:

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

Expand $$\;(n-1)^{2}$$ to get $$n^{2}-2n+1$$, and then subtract $$n^{2}$$:

$$(n^{2}-2n+1) \;-\; n^{2} \;=\; -\,2n \;+\; 1.$$ Hence

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

The negative sign merely indicates that energy is released; the frequency will depend on the magnitude, so we take the absolute value:

$$\lvert\Delta E\rvert \;=\; R_H Z^{2}\,\frac{2n-1}{n^{2}(n-1)^{2}}.$$

The question states that $$n \gg 1$$, i.e. $$n$$ is very large. Under this condition we may approximate

$$2n-1 \;\approx\; 2n, \quad (n-1)^{2} \;\approx\; n^{2}.$$

Substituting these large-$$n$$ equivalents gives

$$\lvert\Delta E\rvert \;\approx\; R_H Z^{2}\,\frac{2n}{n^{2}\,n^{2}} \;=\; 2\,R_H Z^{2}\,\frac{1}{n^{3}}.$$

Planck’s relation links the emitted energy and the frequency:

$$\Delta E \;=\; h\nu.$$

Therefore

$$\nu \;=\;\frac{\Delta E}{h} \;\propto\; \Delta E \;\propto\;\frac{1}{n^{3}}.$$

Thus, when $$n$$ is very large, the frequency of the photon emitted in the transition $$n \rightarrow n-1$$ is inversely proportional to the cube of the quantum number $$n$$.

Hence, the correct answer is Option B.