JEE Main 2022 · 30 June · morning · Single correct · +4 / −1

A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :

  1. (a) 1.05 × 10⁻³⁴ Js
  2. (b) 2.11 × 10⁻³⁴ Js
  3. (c) 3.16 × 10⁻³⁴ Js
  4. (d) 4.22 × 10⁻³⁴ Js
Answer and solution

Answer (b) 2.11 × 10⁻³⁴ Js

Solution

Physics sir · 3 steps

First find which level 12.09 eV lifts the electron to, then use Bohr's rule L = nħ.

  1. Find the new level. The ground-state energy is -13.6 eV. After absorbing 12.09 eV, the energy becomes -13.6 + 12.09 = -1.51 eV.
  2. Match to n. Since Eₙ = -13.6/n² eV, setting -1.51 = -13.6/n² gives n² = 9. So the electron jumps from n = 1 to n = 3.

    Δ L = (n₂ - n₁)h/2π = 2 × (6.63 × 10⁻³⁴)/2π

  3. So. With h/2π = 1.055 × 10⁻³⁴ J s, the increase is Δ L = 2.11 × 10⁻³⁴ J s, which is option (b).

The idea to keep. In Bohr's model the angular momentum is nħ, so its change is just Δ n × ħ. Get Δ n from the energy levels first.

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.