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JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Chemistry question on Atomic Structure

The ground state energy of hydrogen atom is −13.6-13.6 eV. The energy of second excited state of He+\text{He}^+ ion in eV is
  1. A.
    −54.4-54.4
  2. B.
    −3.4-3.4
  3. C.
    −6.04-6.04
  4. D.
    −27.2-27.2
Show answer and solution

Correct option: C

Solution

For hydrogen-like ions, energy = -13.6 × Z²/n² eV.

For He⁺, Z = 2. The second excited state means n = 3.

E = -13.6 × (4)/(9) = -54.4/9 ≈ -6.04 eV.

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