JEE Main 2025 · 22 January · evening · Single correct · +4 / −1

An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

  1. (a) √(h/(π e B))
  2. (b) √((4 h)/(π e B))
  3. (c) √(h/(2 π e eB))
  4. (d) √((2 h)/(π e B))
Answer and solution

Answer (a) √(h/(π e B))

Solution

Physics sir · 4 steps

Magnetic force gives the momentum, Bohr's rule fixes angular momentum, and first excited means n = 2.

  1. Circular motion. The magnetic force evB provides the centripetal force mv²/r. Cancelling one v gives mv = eBr.
  2. Bohr's condition. Angular momentum is quantised: mvr = nh/2π. Put mv = eBr into it to get eBr² = nh/2π.

    rₙ = √(nh/(2π eB))

  3. First excited state. The ground state is n = 1, so the first excited state is n = 2. Then r₂ = √(2h/(2π eB)) = √(h/(π eB)).
  4. So. The radius is √(h/(π eB)), which is option (a).

The idea to keep. When a magnetic field holds the electron in orbit, mv = eBr, so Bohr's rule gives r² ∝ n, not the hydrogen result r ∝ n². First excited state means n = 2.

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.