The energy of a He+ ion in its first excited state would be

The correct answer is Option (4) → −13.6 eV

For a hydrogen-like ion (one electron, nuclear charge Ze) the Bohr energy of level n is

    \( \displaystyle E_n = -\frac{13.6\,Z^{2}}{n^{2}}\,\mathrm{eV} \)

For He⁺, Z = 2; the first excited state is n = 2 (n = 1 is the ground state).

    \( \displaystyle E_2 = -\frac{13.6 \times 2^{2}}{2^{2}}\,\mathrm{eV} \)

    \( \displaystyle E_2 = -13.6\,\mathrm{eV} \)

Note: A similar question was also asked in CUET (UG) 2026 on 18-05-2026 (Slot 2).