JEE Main 2022 · 27 June · morning · Single correct · +4 / −1

A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :
(Given, Planck's constant = 6.6 × 10⁻³⁴ Js).

  1. (a) 2.10 × 10⁻³⁴ Js
  2. (b) 1.05 × 10⁻³⁴ Js
  3. (c) 3.15 × 10⁻³⁴ Js
  4. (d) 4.2 × 10⁻³⁴ Js
Answer and solution

Answer (b) 1.05 × 10⁻³⁴ Js

Solution

Physics sir · 3 steps

First let the 10.2 eV tell you which orbit the electron reaches, then use L = nh/2π for the jump.

  1. Find the new orbit. In hydrogen, Eₙ = -13.6/n² eV. Starting from -13.6 eV and adding 10.2 eV gives -3.4 eV = -13.6/4 eV, so the electron goes from n = 1 to n = 2.
  2. Angular momentum rule. Bohr's postulate says L = nh/2π, so L grows in equal steps of h/2π. Going up one orbit adds exactly one step.

    Δ L = (2-1)h/2π = (6.6 × 10⁻³⁴)/(2 × 3.14) ≈ 1.05 × 10⁻³⁴ J s

  3. So. The angular momentum increases by 1.05 × 10⁻³⁴ J s, which is option (b).

The idea to keep. In Bohr's model angular momentum depends on n, not n². Each jump of one orbit adds exactly h/2π, so Δ L = Δ n · h/2π.

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 2022.