Ionisation energy of He+ is 19.6 × 10⁻⁸ J/atom. The energy of the first stationary state (n=1) of Li²⁺ is:

IEHe+ = 19.6 × 10⁻⁸ J/atom
The ionization energy of a hydrogen-like species is given by:
IE = 13.6 × Z²/n²
where Z is the atomic number and n is the principal quantum number.
For He⁺ (Z = 2, n = 1):
IEHe+ = 13.6 × 2²/1² = 54.4 eV = 54.4 × 1.602 × 10⁻¹⁹ J = 8.71 × 10⁻¹⁸ J
Given IEHe+ = 19.6 × 10⁻⁸ J/atom = 19.6 × 10⁻⁸ × 6.022 × 10²³ eV/atom ≈ 1.18 × 10¹⁶ eV/atom
For Li²⁺ (Z = 3, n = 1):
IELi²+ = 13.6 × 3²/1² = 122.4 eV = 122.4 × 1.602 × 10⁻¹⁹ J = 1.96 × 10⁻¹⁷ J
Energy of first stationary state (n = 1) of Li²⁺ is given by:
E₁ = -13.6 × Z²/n² = -13.6 × 3²/1² = -122.4 eV
E₁ = -122.4 × 1.602 × 10⁻¹⁹ J = -1.96 × 10⁻¹⁷ J
The magnitude of this energy is 1.96 × 10⁻¹⁷ J.
However, the given ionization energy for He+ seems to be incorrect. Let's use the given value and calculate the energy for Li2+ using the ratio of ionization energies:
IEHe+ / IELi²+ = (ZHe+²/ZLi²+²) = (2²/3²) = 4/9
Therefore, IELi²+ = (9/4) × IEHe+ = (9/4) × 19.6 × 10⁻⁸ J/atom = 44.1 × 10⁻⁸ J/atom = 4.41 × 10⁻⁶ J/atom
Therefore, the energy of the first stationary state (n=1) of Li²⁺ is 4.41 × 10⁻⁶ J/atom