# What are the ionization potentials of hydrogen-like helium and lithium?

Jul 20, 2017

${\text{He"^(+)(g) -> "He}}^{2 +} \left(g\right) + {e}^{-}$

DeltaH_(IP)("He"^(+)(g)) ~~ "54.42 eV"

${\text{Li"^(2+)(g) -> "Li}}^{3 +} \left(g\right) + {e}^{-}$

DeltaH_(IP)("Li"^(2+)(g)) ~~ "122.45 eV"

Well, since these are hydrogen-like atoms with one electron each, they ought to have similar ionization potentials that differ only by the atomic number.

In atomic units, the energy of $\text{H}$-like atoms is given by

$\boldsymbol{{E}_{n} = - \frac{{Z}^{2} \cdot \text{13.6057 eV}}{{n}^{2}}}$,

where $- \text{13.6057 eV}$ IS the ground-state energy of $\text{H}$ atom, $Z$ is the atomic number, and $n$ is the principal quantum number.

In this case, only one electron is in ${\text{He}}^{+}$ or ${\text{Li}}^{2 +}$, so we really only need to look at $n = 1$.

The ionization potential is then the energy put into the atom to remove that single electron. We can check each of these on NIST.

$\frac{\textcolor{b l u e}{\Delta {H}_{I P} \left(\text{He"^(+)(g))) = +((2^2)("13.6057 eV}\right)}}{{1}^{2}}$

$\approx$ $\textcolor{b l u e}{\underline{\text{54.42 eV}}}$

for

${\text{He"^(+)(g) -> "He}}^{2 +} \left(g\right) + {e}^{-}$

[ (row 2)

$\frac{\textcolor{b l u e}{\Delta {H}_{I P} \left(\text{Li"^(2+)(g))) = +((3^2)("13.6057 eV}\right)}}{{1}^{2}}$

$\approx$ $\textcolor{b l u e}{\underline{\text{122.45 eV}}}$

for

${\text{Li"^(2+)(g) -> "Li}}^{3 +} \left(g\right) + {e}^{-}$

[ (row 3)