Excitation energy of a hydrogen like ion in its first excitation state is $40.8 \mathrm{eV}$. Energy needed to remove the electron from the ion in ground state is

Excitation energy of hydrogen like ion in its first excitation state, $\Delta E=40.8 \mathrm{eV}$ For the hydrogen like ion, energy in the ground state $(n=1)$, $$ E=\frac{-13.6 Z^{2}}{n^{2}} \mathrm{eV}=-13.6 Z^{2} \mathrm{eV} $$ Energy in the first excited state, $E_{2}=\frac{-13.6 Z^{2}}{n^{2}}$ $$ =\frac{-13.6 Z^{2}}{4} $$ Excitation energy, $\Delta E=E_{2}-E_{1}$ $$ \begin{aligned} &=\left(\frac{-13.6 Z^{2}}{4}+13.6 Z^{2}\right) \mathrm{eV} \\ &=(-3.4+13.6) Z^{2} \mathrm{eV} \\ \Rightarrow \quad 40.8 \mathrm{eV} &=10.2 Z^{2} \mathrm{eV} \\ \Rightarrow \quad \quad Z^{2} &=\frac{40.8}{10.2} \\ \Rightarrow \quad \quad Z^{2} &=4 \\ \Rightarrow \quad \quad Z &=2 \end{aligned} $$ Thus, energy of the electron in the ground state is given as $$ \begin{array}{ll} \Rightarrow & E_{1}=-13.6(2)^{2} \mathrm{eV} \\ \Rightarrow & E_{1}=-54.4 \mathrm{eV} \end{array} $$ Hence, the energy needed to remove the electron from the ion in ground state (ionisation energy) $$ =-(-54.4) \mathrm{eV}=54.4 \mathrm{eV} $$