What is the energy of the London dispersion force holding together two atoms of the same element? I read somewhere that the energy is around $U=-\frac{3\alpha I}{4r^6}$ where $\alpha$ is the polarizability and $I$ is the ionization energy but I'm obtaining way too large energies for a lithium crystal.

The shortest distance between two lithium atoms in the crystal is $304 pm$

The first ionization energy of lithium is $520.2 kJ/mol$

I've found various polarizability constants online in papers but I can't really figure them out or the units they are using.

Also, I suspect that I want the dispersion force between $\ce{Li^+}$ ions and not neutral $\ce{Li}$ atoms.
Polarisability as used here has dimensions of volume. The London formula is $$U \approx -\frac{3}2\frac{I_aI_b}{I_a+I_b}\frac{\alpha _a \alpha _b}{r^6}$$  for two species a and b so you just need to square $\alpha$.