JEE Main 2025 · 7 April · morning · Single correct · +4 / −1

In a hydrogen like ion, the energy difference between the 2ⁿᵈ excitation energy state and ground is 108.8 eV . The atomic number of the ion is:

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

Answer (c) 3

Solution

Physics sir · 4 steps

Second excited state means n = 3, not n = 2. Then the gap is just 13.6 Z² eV times (1 − 1/9).

  1. Name the levels. The ground state is n = 1. The first excited state is n = 2, so the second excited state is n = 3.
  2. Energy of a level. For a hydrogen-like ion, Eₙ = -(13.6 Z²)/n² eV. The gap between n = 3 and n = 1 is therefore:

    E₃ - E₁ = 13.6 Z²(1 - 1/9) = 13.6 × 8/9 Z² eV

  3. Put in numbers. Set this equal to 108.8 eV. Since 108.8 / 13.6 = 8, you get 8/9 Z² = 8, so Z² = 9.
  4. So. Z = 3, which is the ion Li²⁺. The answer is (c).

The idea to keep. Count excited states from n = 2 upwards, so the k-th excited state is n = k + 1. Every hydrogen-like energy is the hydrogen value scaled by Z².

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.