Why does helium have such a low van der Waals coefficient but not as low a Lennard-Jones coefficient?

Question

Helium has the lowest polarizability of any atom, and therefore ought to have the smallest London dispersion force. Indeed, if you look at the van der Waals constant of helium, you find that it has the lowest value of $a$ by a lot. $a$ roughly corresponds to the amount attraction between particles.

The long-distance force between neutral particles can be roughly modelled by the Lennard-Jones potential. Even though helium should have the weakest $r^{-6}$ term of any atom, if you look up the experimental values of the Lennard-Jones coefficients, you find that they're the same for Helium as they are for the rest of the noble gases.

What gives? Why is its Lennard-Jones potential roughly the same when its London dispersion force is so weak? (Perhaps my source is just wrong?)

Answer 1

Lennard-Jones Parameters

To quote the original source of those Lennard-Jones parameter values (Gordon and Kim),

For all the systems involving atoms larger than helium, the predictions appear quite reliable... Our approach thus provides the first successful prediction of the intermolecular potentials for the rare gases (except helium)

So I would not put much faith in the He-He LJ parameters in the table you linked. The analysis was good in 1972, but even still wasn't great for helium! The authors go on to state that

Polarizing (induction) forces are not included

which, arguably, is the strongest evidence not to trust the LJ parameters too much. The Lennard-Jones (6-12) potential's strongest theoretical justification is probably that dispersive forces can be shown to approximately follow $r^{-6}$ dependence. If dispersion's not even included, then that's a big red flag since the $r^{-12}$ part has no theoretical justification. It seems they just tried to fit the LJ potential to their calculations, probably for reasons of comparison with previous results. The experimental value that they compare their calculations to is $16.5 \times 10^{-16}$ ergs, which would be 1.0 in the table instead of 3.9, and thus would continue the expected trend.

van der Waals parameters

I will just be brief and say that the quantum gases, including helium, are not amenable to the use of the classical van der Waals equation. Helium's critical temperature is 5.2 kelvin, and there is a bunch of quantum stuff going on in that regime. You just can't compare the quantum gas parameters with those of fluids that behave classically; it's quantum apples and classical oranges.