Question

How do you calculate the second ionization energy of Helium?

Answer 1

You can do it by calculating the first ionization energy of `"He"^(+)`, which is a hydrogen-like atom.

`"He"(g) -> "He"^(+)(g) + e^(-)`, `" "" "DeltaH_(IE_1)`

`"He"^(+)(g) -> "He"^(2+)(g) + e^(-)`, `" "DeltaH_(IE_2)`

As `"He"^(+)` is hydrogen-like, i.e. it has one electron, and is thus isoelectronic with `"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^2cdot"13.61 eV"/n^2`

where `Z` is the atomic number and `n` is the principal quantum number. And naturally, `-"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|`

`=` `ulcolor(blue)("54.44 eV")`