# Question #d1d52

Jul 19, 2017

$\text{54.4 eV}$

#### Explanation:

For starters, the ionization energy of a hydrogen atom is equal to $\text{13.6 eV}$ because the ground state has an energy of $\text{-13.6 eV}$ in a hydrogen atom.

So for the hydrogen atom, the energy of a given excited state, which is characterized by $n > 1$, is given by

${E}_{n} = - \frac{\text{13.6 eV}}{n} ^ 2$

For the ground state, you have $n = 1$ and

${E}_{\text{ground state" = -"13.6 eV}}$

Now, for a hydrogen-like atom like the helium cation, ${\text{He}}^{+}$, the energy of a given excited state is given by

${E}_{n} = - \frac{\text{13.6 eV}}{n} ^ 2 \cdot {Z}^{2}$

Here $Z$ represents the atomic number of the element. Consequently, the energy of the ground state will be given by

${E}_{\text{ground state" = -"13.6 eV}} \cdot {Z}^{2}$

Since a helium atom has

$Z = 2$

you can say that its ground state will be at

${E}_{\text{ground state He" = -"13.6 eV}} \cdot {2}^{2}$

${E}_{\text{ground state He" = -"54.4 eV}}$

This implies that in order to remove the electron from a ${\text{He}}^{+}$ ion to form a ${\text{He}}^{2 +}$ ion, you have to supply $\text{54.4 eV}$ of energy.

${\text{He"^(+) + "54.4 eV" -> "He"^(2+) + "e}}^{-}$