Determine the first excitation potential for a He^+ ion.

The energy of an electron in the $n$-th orbit of a hydrogen-like ion is given by:

$E_n = -13.6 \dfrac{Z^2}{n^2} \text{ eV}$

For a $\text{He}^+$ ion, the atomic number $Z = 2$.

The energy in the ground state ($n = 1$) is:

$E_1 = -13.6 \times \dfrac{2^2}{1^2} = -54.4 \text{ eV}$

The energy in the first excited state ($n = 2$) is:

$E_2 = -13.6 \times \dfrac{2^2}{2^2} = -13.6 \text{ eV}$

The first excitation energy is the energy required to excite the electron from $n = 1$ to $n = 2$:

$\Delta E = E_2 - E_1 = -13.6 - (-54.4) = 40.8 \text{ eV}$

The first excitation potential is the potential difference required to provide this energy, which is $40.8 \text{ V}$.