The energy of photon required to remove electron from higher energy level is — Atomic Structure Chemistry Question

💡 Solution & Explanation

The higher energy level is $n=3$ of He$^+$ ($Z=2$). The energy required to remove the electron (ionization energy from that state) is $E = 13.6 \times \frac{Z^2}{n^2} = 13.6 \times \frac{4}{9} \approx 6.04$ eV. Therefore, correct answer is A.