For hydrogen atom, the number of revolutions of the electron per second in the orbit of quantum numb — Atomic Structure Chemistry Question
Question
For hydrogen atom, the number of revolutions of the electron per second in the orbit of quantum number, n, is ∝
Answer: C
💡 Solution & Explanation
The number of revolutions per second represents the frequency of revolution f = v / (2*π*r). Since velocity v is ∝ 1/n and radius r is ∝ n^2, the frequency f is ∝ (1/n) / n^2 = 1/n^3 = n^-3.
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