JEE Main 2025 · 28 January · evening · Single correct · +4 / −1
The frequency of revolution of the electron in Bohr's orbit varies with n , the principal quantum number as:
- (a) 1/n⁴
- (b) 1/n²
- (c) 1/n³
- (d) 1/n
Answer and solution
Answer (c) 1/n³
Solution
Physics sir · 3 steps
Frequency is speed over circumference, so combine how v and r each scale with n.
- Bohr scalings. In the nth Bohr orbit the radius grows as rₙ ∝ n² and the speed falls as vₙ ∝ 1/n (for a fixed atom, so Z is constant).
- Define frequency. Frequency is the number of revolutions per second. One revolution covers the circumference 2π rₙ, so the frequency is speed divided by circumference.
fₙ = vₙ/(2π rₙ) ∝ (1/n)/n² = 1/n³
- So. The frequency of revolution varies as 1/n³, which is option (c).
The idea to keep. Remember the three Bohr scalings: r ∝ n², v ∝ 1/n, E ∝ 1/n². Anything else, like frequency or time period, you build from these.
Answer from the NTA final answer key. Solution by PrepWiser.