# 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 $\alpha$.
• So, for a polarizability constant of $24\times 10^{-30}$ which I'm not sure is correct I got a total energy of 290kJ/mol which still seems too high. Is this correct? – Steven Stewart-Gallus Jul 25 '16 at 22:22
• Volume polarisability is correct. I had a look again and found values for benzene of 25 .10$^{-24}$ ml so that is consistent with your values in m$^3$. Carbon tetrachloride has a value 10.5 for example. ( I had only looked at nitrogen (1.73) and water (1.44), before, apologies). An interaction energy of approx 24000 cm$^{-1}$ is still huge, its as large as an electronic excitation, but I can't see what wrong :( The ionisation energy and bond length seem ok . – porphyrin Jul 27 '16 at 10:59