The ground state energy of a hydrogen atom is -13.6 eV . The energy required to ionize a Li ²⁺ ion from its first excited state is

Step-by-step solution

For a hydrogen-like species, the energy of an electron in the n^ th orbit is given by E_n = -13.6 Z^2 n^2 eV . For a Li ²⁺ ion, the atomic number Z = 3 . The first excited state corresponds to the principal quantum number n = 2 . Substituting these values, the energy of the electron in the first excited state is: E₂ = -13.6 3^2 2^2 = -13.6 9 4 = -30.6 eV . The ionization energy is the energy required to remove the electron from this state to infinity ( E_ = 0 ). Ionization energy = E_ - E₂ = 0 - (-30.6 eV ) = 30.6