# How do you calculate the second ionization energy of Helium?

##### 1 Answer
Aug 9, 2017

You can do it by calculating the first ionization energy of ${\text{He}}^{+}$, which is a hydrogen-like atom.

${\text{He"(g) -> "He}}^{+} \left(g\right) + {e}^{-}$, $\text{ "" } \Delta {H}_{I {E}_{1}}$

${\text{He"^(+)(g) -> "He}}^{2 +} \left(g\right) + {e}^{-}$, $\text{ } \Delta {H}_{I {E}_{2}}$

As ${\text{He}}^{+}$ is hydrogen-like, i.e. it has one electron, and is thus isoelectronic with $\text{H}$... we can invoke Koopman's approximation theorem, which in simplified terms, states...

For a one-electron orbital, the potential energy of that orbital is opposite in sign to the ionization energy from that orbital.

Thus, we just need to get the magnitude of ${E}_{1}$:

${E}_{n} = - {Z}^{2} \cdot \frac{\text{13.61 eV}}{n} ^ 2$

where $Z$ is the atomic number and $n$ is the principal quantum number. And naturally, $- \text{13.61 eV}$ is the ground-state energy of hydrogen atom.

color(blue)(DeltaH_(IE_2)) = |E_1("He"^(+))| = |-2^2cdot"13.61 eV"/1^2|

$=$ $\underline{\textcolor{b l u e}{\text{54.44 eV}}}$