JEE Main 2023 · 12 April · morning · Single correct · +4 / −1

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

  1. (a) 2
  2. (b) 4
  3. (c) 3
  4. (d) 1
Answer and solution

Answer (c) 3

Solution

Physics sir · 4 steps

Find the highest level 12.5 eV can reach, then count every way the electron can fall back down.

  1. Room temperature atoms. At room temperature almost every hydrogen atom is in the ground state, n = 1, with energy -13.6 eV. The level energies are Eₙ = -13.6/n² eV.
    LevelEnergyEnergy needed from n = 1Reachable with 12.5 eV?
    n = 2-3.4 eV10.2 eVyes
    n = 3-1.51 eV12.09 eVyes
    n = 4-0.85 eV12.75 eVno, needs more than 12.5
  2. Electron, not photon. A colliding electron can hand over only part of its energy, so it does not need an exact match. It can lift the atom to n = 3 and keep the extra 0.41 eV itself.
  3. Count the lines. From n = 3 the atom can fall 3 → 1, 3 → 2 and 2 → 1. That is n(n-1)/2 = (3 × 2)/2 = 3 lines.
  4. So. The gas emits 3 spectral lines, option (c).

The idea to keep. For electron bombardment, the highest level reached is the one whose excitation energy is still below the beam energy. Then the number of lines is n(n-1)/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 2023.