Let us first find the frequency of revolution of electron in the orbit classically. In Bohr’s model velocity of electron in nth orbit, υ = $\frac{nh}{2\pi mr}$ where radius r = $\frac{n^{2}h^{2}\varepsilon_{0}}{\pi me^{2}}$ Thus orbital frequency of electron in nth orbit is υ = $\frac{v}{2\pi r}$ = $\frac{\frac{nh}{2\pi mr}}{2\pi r}$ ⇒ υ = $\frac{nh}{4\pi^{2}mr^{2}}$ = $\frac{nh}{4\pi^{2}m}\left[\frac{\pi me^{2}}{\pi^{2}h^{2}\varepsilon_{0}}\right]^{2}$ or υ = $\frac{me^{4}}{4n^{3}h^{3}\varepsilon_{0}^{2}}$ ... (i) Now, let us find the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). υ = $\frac{mr^{4}}{8\varepsilon_{0}^{2}h^{3}}\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]$ ⇒ υ = $\frac{me^{4}}{8\varepsilon_{0}^{2}h^{3}}\left[\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}\right]$ or υ = $\frac{me^{4}}{8\varepsilon_{0}^{2}h^{3}}\left[\frac{2n-1}{n^{2}(n-1)^{2}}\right]$ For large n, 2n – 1 ≈ 2n and n – 1 ≈ n, frequency, υ = $\frac{me^{4}}{8\varepsilon_{0}^{2}h^{3}}\left[\frac{2n}{n^{4}}\right]$ or υ = $\frac{me^{4}}{4n^{3}h^{3}\varepsilon_{0}^{2}}$ ... (ii) Equation (i) and (ii) are equal, hence for large value of n, the classical frequency of revolution of electron in nth orbit is same as frequency of radiation when electron de-excite from level n to (n – 1).