Given: The binding energy for a hydrogen atom is 13.6 eV . We know: The energy of an electron in the n-th state is given by: En=−13.6‌
Z2
n2
For Li2+, the atomic number Z=3. To find the energy required to remove the electron from the first excited state (n=2) : Substitute into the equation: ‌E2=−13.6×‌
32
22
=−13.6×‌
9
4
‌E2=−30.6eV Since E2+E=0, we have: −30.6+E=0 Thus, solving for E : E=+30.6eV Therefore, the energy required to remove the electron from the first excited state of Li2+ is 30.6 eV .