The energy of a He+ ion in its first excited state will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −13.6 eV
He+ has a single electron, so Bohr’s energy formula for hydrogen-like ions applies with Z = 2. The first excited state is n = 2.
\( \displaystyle E_n = -\frac{13.6\,Z^{2}}{n^{2}}\,\mathrm{eV} \)
\( \displaystyle E_2 = -\frac{13.6\times2^{2}}{2^{2}}\,\mathrm{eV} \)
\( \displaystyle E_2 = -13.6\,\mathrm{eV} \)
Note: A similar question was also asked in CUET (UG) 2026 on 31-05-2026 (Slot 2).