The excititation energy of a hydrogen -like ion in its first excitedstate is $40.8eV$ Find the energy needed to remain the electron from the ion

Solution: The excitation energy in the first excited state is
$E=Rhc{Z}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
$=\left(13.6eV\right)\times Z^2\times \frac{3}{4}$
Equation this to $40.8eV$ we get $Z=2$ So, that in question is $H{e}^{+}$ The energy of the ion in the ground state is
$E=\frac{RhcZ6\left(2\right)}{1^2}=-4\times \left(13.6eV\right)$
$=-54.4eV$
Thus $54.4eV$ is required to remain the electron from the ion