Thermionic Emission: Kinetic Energy Distribution of Emitted Electrons

I am having a conceptual problem understanding the kinetic energy of thermionically emitted electrons.

I know that in order to escape the surface the electrons must have energies of at least the Fermi Level plus the work function ($\varepsilon \ge \varepsilon_F + \phi$). However, is all that kinetic energy converted to potential energy once it leaves the material so that once it leaves the surface an electron has kinetic energy $T = \varepsilon - \varepsilon_F + \phi$?

The electron maintains the total energy $\epsilon$ that it had while inside the material. Inside the material it has some kinetic energy and some potential energy, but we don't really care exactly what those are since we know their sum. The point is, once it enters the vacuum, its total energy is still made up of kinetic and potential energy. Since the vacuum potential energy is $U = \epsilon_F + \phi$ just outside the surface, then indeed the kinetic energy is $T = \epsilon - U = \epsilon - \epsilon_F - \phi$. Of course if there are any electric fields in the vacuum, then the kinetic energy of the electron will change as it moves away from the surface.
• Yes, you're right Nanite. I did some more research and it's the same idea as ionization. The distribution of energies is the Fermi-Dirac Distribution, $f(\varepsilon) = \frac{1}{1 + \exp(\frac{\varepsilon - \epsilon_F}{k_B T})}$ Nov 16 '13 at 2:15