Why aren't positive ions emitted at the same rate as electrons via thermionic emission?

Question

The Effective Work Functions of the Elements are very similar for removing an electron or a positive ion, e.g. about 4.5 eV for Tungsten. If their work functions are similar, why aren't positive ions emitted via thermionic emission at a similar rate to electrons when a metal wire is heated?

Adding to my confusion, according to Richardson's Law,

$$ J = A_{\mathrm{G}} T^2 \mathrm{e}^{-W \over k T}, \qquad A_{\mathrm{G}} = \lambda_R {4 \pi m_e k^2 q_\text{e} \over h^3},\qquad \lambda_R\sim0.5, $$

the thermionic emission rate for electrons is proportional to the electron mass. This naively suggests that very massive positive ions should be emitted more often, not less, which is clearly very, very wrong. Instead, thermal emission of ions seems to be an evaporative process depending on the thermionic work function, the ionizing potential, and the latent heat of evaporation of the metal, with evaporative rate $\propto 1/\sqrt{M_{ion}}$. (See, for example, Smith's 1930 article on "The Emission of Positive Ions From Tungsten and Molybdenum".)

I've always assumed that the difference is due to interatomic forces creating a much higher potential barrier for an ion (or atom) to escape, even though the ultimate net energy requirement (i.e. the work function) is similar to that needed to remove an electron. I can't, however, immediately find a clear discussion that either confirms this or provides the correct explanation.

My apologies that I couldn't find unpaywalled alternatives for the cited papers.

Answer 1

In the context of thermionic emission we are discussing only conduction electrons, which roughly correspond to the outer atomic shells. In other words, all the results mentioned in the Q. are obtained within the framework of band theory and accepting the approximations of the band theory. This notably includes Born-Oppenheimer approximation, which, in the context of solids, separates not only the nuclei from electrons, but rather nuclei and low-lying electrons from the outer electrons. In other words, we assume that the nuclei and the core electrons form together a lattice, whose motion is slow, and which creates a periodic potential. We then solve for the orbitals of the remaining electrons in this potential - obtaining the band structure. We can then study the thermo- and photo-emission of these band electrons, but we cannot really treat ions on the same footing, since there are no free-floating ions in the bands.

Answer 2

I might be totally misunderstanding your question, but I think when you see the value of the effective work function (in Tungsten for instance) to remove a positive ion vs an electron both being 4.5 eV, it is assuming the positive ion has already been broken free from the lattice potential. So when a metal wire is heated, the electron must only gain 4.5 eV to escape the wire while the positive ion must gain 4.5 eV Plus the energy to break out of the lattice.