What is the ratio of the ionization energy of a hydrogen atom from its first excited state to the ionization energy of a He ⁺ ion from its second excited state?

Step-by-step solution

The ionization energy from the n^ th state of a hydrogen-like species is given by IE = 13.6 Z^2 n^2 eV . For a hydrogen atom ( Z=1 ) in its first excited state ( n=2 ): IE₁ = 13.6 1^2 2^2 = 13.6 4 eV . For a He ⁺ ion ( Z=2 ) in its second excited state ( n=3 ): IE₂ = 13.6 2^2 3^2 = 13.6 4 9 eV . The ratio of their ionization energies is: IE₁ IE₂ = 1/4 4/9 = 9 16 . Thus, the required ratio is 9:16 . Answer: 9:16