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:

  1. (a) 1/n⁴
  2. (b) 1/n²
  3. (c) 1/n³
  4. (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.

  1. 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).
  2. 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

  3. 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.

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Asked about 110 times in JEE Main since 2020. This question is from 2025.