# How do you calculate the ionization energy of lithium?

Apr 21, 2018

By using a computer... we obtained a value of $\text{5.39271223 eV}$, compared to the true value of $5.39171495 \left(\pm 0.00000004\right) \text{eV}$. Not bad.

Lithium clearly has more than one electron; that makes it so the ground-state energy is not readily able to be calculated by hand, since the electronic coordinates are mutually dependent due to the inherent electron-electron correlation.

Instead, we would have to supply input files to a computer software and calculate the ground-state energies that way, of $\text{Li}$ and ${\text{Li}}^{+}$.

Using the so-called Feller-Peterson-Dixon method to get practically perfect accuracy, one would have to calculate (or consider):

$\Delta {E}_{\text{IE" = DeltaE_("IE",0) + DeltaE_"corr" + DeltaE_"CBS" + DeltaE_"CV" + DeltaE_"QED" + DeltaE_"SR" + DeltaE_"SO" + DeltaE_"Gaunt}}$

where:

• $\Delta {E}_{\text{IE} , 0}$ is the initial ionization energy calculated from Multi-Configurational Self-Consistent Field (MCSCF) theory.
• $\Delta {E}_{\text{corr}}$ is the dynamic correlation energy contribution not accounted for in Multi-Configurational Self-Consistent Field theory, but recovered in Multi-Reference Configuration Interaction (MRCI).
• $\Delta {E}_{\text{CBS}}$ is the energy contribution from extrapolating to the limit of an infinite set of basis functions that represent atomic orbitals.
• $\Delta {E}_{\text{CV}}$ is the energy contribution from correlating the core electrons with the valence electron(s).
• $\Delta {E}_{\text{QED}}$ is the energy contribution from the so-called Lamb Shift, a quantum electrodynamics interaction present primarily among $s$ orbitals.
• $\Delta {E}_{\text{SR}}$ is the energy contribution from relativistic effects. This is negligible in $\text{Li}$ but is automatically accounted for using the 2nd order Douglas-Kroll-Hess (DKH) Hamiltonian for light atoms (3rd order DKH for heavy atoms).
• $\Delta {E}_{\text{SO}}$ is the energy contribution from spin-orbit coupling.
• $\Delta {E}_{\text{Gaunt}}$ is the energy contribution from high-order two-electron correlation in the relativistic scheme.

That WOULD be extremely involved for a heavier atom... Here are some things that save time:

• No electron correlation is present here since only one state is possible (spin up in a $2 s$ orbital!).

So we can get by from a simple Hartree-Fock calculation. There will be a tiny bit of core-valence ($1 s \text{-} 2 s$) correlation, so $\Delta {E}_{\text{CV}} \ne 0$ and that can be taken care of with an MRCI using a weighted-core basis set.

• The model potentials for QED only are made for $Z \ge 23$ (quote: "Fails completely for $Z < 23$"), so there is no point in including the Lamb Shift at all.

• $\Delta {E}_{\text{SR}}$ is included by default by the 2nd order DKH Hamiltonian.

• The $\Delta {E}_{\text{SO}}$ contribution can be included but it has been done before... It is $\text{0.000041 eV}$, or $\text{0.000945 kcal/mol}$. Gaunt is unimportant here, based on how small the spin-orbit value is.

Here are the (not so interesting) results:

From this, we had gotten that:

color(blue)(DeltaE_"IE") = "123.195465 kcal/mol" + "0.000000 kcal/mol" + "0.000000 kcal/mol" + "1.162450 kcal/mol" + "0.000000 kcal/mol" + "accounted for" + "0.000945 kcal/mol" + "0.000000 kcal/mol"

$=$ $\textcolor{b l u e}{\underline{\text{124.358860 kcal/mol}}}$

$=$ $\textcolor{b l u e}{\underline{\text{5.39271223 eV}}}$

whereas the value on NIST is practically the same, at $5.39171495 \left(\pm 0.00000004\right) \text{eV}$, a mere 0.01849% error...