Question #d1d52

For starters, the ionization energy of a hydrogen atom is equal to `"13.6 eV"` because the ground state has an energy of `"-13.6 eV"` in a hydrogen atom.

![http://dev.physicslab.org/Document.aspx?doctype=3&filename=AtomicNuclear_BohrModelDerivation.xml]()

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

`E_n = -"13.6 eV"/n^2`

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

`E_ "ground state" = -"13.6 eV"`

Now, for a hydrogen-like atom like the helium cation, `"He"^(+)`, the energy of a given excited state is given by

`E_n = -"13.6 eV"/n^2 * Z^2`

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

`E_ "ground state" = -"13.6 eV" * Z^2`

Since a helium atom has

`Z = 2`

you can say that its ground state will be at

`E_ "ground state He" = -"13.6 eV" * 2^2`

`E_ "ground state He" = -"54.4 eV"`

This implies that in order to remove the electron from a `"He"^(+)` ion to form a `"He"^(2+)` ion, you have to supply `"54.4 eV"` of energy.

`"He"^(+) + "54.4 eV" -> "He"^(2+) + "e"^(-)`

![http://www.jyi.org/issue/bohr-revisited-model-and-spectral-lines-of-helium/]()