JEE Main 2023 · 1 February · evening · Single correct · +4 / −1

An electron of a hydrogen like atom, having Z=4, jumps from 4ᵗʰ energy state to 2ⁿᵈ energy state. The energy released in this process, will be :
(Given Rch = 13.6 eV)
Where R = Rydberg constant
c = Speed of light in vacuum
h = Planck's constant

  1. (a) 10.5 eV
  2. (b) 40.8 eV
  3. (c) 13.6 eV
  4. (d) 3.4 eV
Answer and solution

Answer (b) 40.8 eV

Solution

Physics sir · 4 steps

Hydrogen formula, but scale it by Z² = 16: then it is just the difference between two levels.

  1. Energy of a level. For a hydrogen-like atom, Eₙ = -(Rch Z²)/n². With Z = 4, every level is Z² = 16 times deeper than in hydrogen.

    Δ E = Rch Z²(1/n_f²-1/nᵢ²)

  2. Put in numbers. Rch Z² = 13.6 × 16 = 217.6 eV. The bracket is 1/2²-1/4² = 1/4-1/16 = 3/16.
  3. Multiply. Δ E = 217.6 × 3/16 = 40.8 eV. This energy is released, because the electron moves to a lower level.
  4. So. The energy released is 40.8 eV, option (b).

The idea to keep. For any hydrogen-like ion, use the hydrogen result multiplied by Z². A transition always means the difference of two levels, never the energy of one level.

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.