Atoms and Nuclei

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:$E_n=-13.6 \; \frac{Z^2}{n^2}$For $\mathrm{Li} ^{2+}$, 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:$\;E_2=-13.6 \times \;\frac{3^2}{2^2}=-13.6 \times \;\frac{9}{4}$$\;E_2=-30.6 \text{eV}$Since $E_2+E=0$, we have:$-30.6+E=0$Thus, solving for $E$ :$E=+30.6 \text{eV}$Therefore, the energy required to remove the electron from the first excited state of $\mathrm{Li} ^{2+}$ is 30.6 eV .