For hydrogen atom, the number of revolutions of the electron per second in the orbit of quantum numb — Atomic Structure Chemistry Question

💡 Solution & Explanation

Frequency of revolution $f = v/(2\pi r)$. Since $v \propto 1/n$ and $r \propto n^2$, we have $f \propto (1/n) / n^2 = 1/n^3 = n^{-3}$. Therefore, correct answer is C.