D.29 The thermionic emission equation

This note derives the thermionic emission equation for a typical metal following [41, p. 364ff]. The derivation is semi-classical.

To simplify the analysis, it will be assumed that the relevant electrons in the interior of the metal can be modeled as a free-electron gas. In other words, it will be assumed that in the interior of the metal the forces from surrounding particles come from all directions and so tend to average out.

(The free-electron gas assumption is typically qualitatively reasonable for the valence electrons of interest if you define the zero of the kinetic energy of the gas to be at the bottom of the conduction band. You can also reduce errors by replacing the true mass of the electron by some suitable effective mass. But the zero of the energy drops out in the final expression, and the effective mass of typical simple metals is not greatly different from the true mass. See chapter 6.22.3 for more on these issues.)

Assume that the surface through which the electrons escape is normal to the $x$-​direction. Then the classical expression for the current of escaping electrons is

j = \rho e v_x

where $\rho$ is the number of electrons per unit volume that is capable of escaping and $v_x$ is their velocity in the $x$-​direction. Note that the current above is supposed to be the current inside the metal of the electrons that will escape.

An electron can only escape if its energy ${\vphantom' E}^{\rm p}$ exceeds

{\vphantom' E}^{\rm p}_{\rm {esc}}=\mu+e\varphi_{\rm {w}}

where $\mu$ is the Fermi level, because the work function $\varphi_{\rm {w}}$ is defined that way. The number of electrons per unit volume in an energy range ${\rm d}{\vphantom' E}^{\rm p}$ above ${\vphantom' E}^{\rm p}_{\rm {esc}}$ can be found as

e^{-(e\varphi_{\rm {w}}+{\vphantom' E}^{\rm p}-{\vphantom'...
...\sqrt{{\vphantom' E}^{\rm p}} {\,\rm d}{\vphantom' E}^{\rm p}

That is because the initial exponential is a rewritten Maxwell-Boltzmann distribution (6.21) that gives the number of electrons per state, while the remainder is the number of states in the energy range according to the density of states (6.6).

Normally, the typical thermal energy ${k_{\rm B}}T$ is very small compared to the minimum energy $e\varphi_{\rm {w}}$ above the Fermi level needed to escape. Then the exponential of the Maxwell-Boltzmann distribution is very small. That makes the amount of electrons with sufficient energy to escape very small. In addition, with increasing energy above ${\vphantom' E}^{\rm p}_{\rm {esc}}$ the amount of electrons very quickly becomes much smaller still. Therefore only a very small range of energies above the minimum energy ${\vphantom' E}^{\rm p}_{\rm {esc}}$ gives a contribution.

Further, even if an electron has in principle sufficient energy to escape, it can only do so if enough of its momentum is in the $x$-​direction. Only momentum that is in the $x$-​direction can be used to overcome the nuclei that pull it back towards the surface when it tries to escape. Momentum in the other two directions only produces motion parallel to the surface. So only a fraction, call it $f_{\rm {esc}}$, of the electrons that have in principle enough energy to escape can actually do so. A bit of geometry shows how much. All possible end points of the momentum vectors with a magnitude $p$ form a spherical surface with area $4{\pi}p^2$. But only a small circle on that surface around the $x$-​axis, with an approximate radius of $\sqrt{p^2-p_{\rm {esc}}^2}$, has enough $x$-​momentum for the electron to escape, so

f_{\rm {esc}} \approx \frac{\pi\sqrt{p^2-p_{\rm {esc}}^2}^...
... {\vphantom' E}^{\rm p}_{\rm {esc}}}{4{\vphantom' E}^{\rm p}}

where the final equality applies since the kinetic energy is proportional to the square momentum.

Since the velocity for the escaping electrons is mostly in the $x$-​direction, ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $\frac12{m_{\rm e}}v_x^2$, which can be used to express $v_x$ in terms of energy.

Putting it all together, the current density becomes

j =
\int_{{\vphantom' E}^{\rm p}={\vphantom' E}^{\rm p}_...
...\rm p}}{m_{\rm e}}\right)^{1/2} {\rm d}{\vphantom' E}^{\rm p}

Rewriting in terms of a new integration variable $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $({\vphantom' E}^{\rm p}-{\vphantom' E}^{\rm p}_{\rm {esc}})$$\raisebox{.5pt}{$/$}$${k_{\rm B}}T$ gives the thermionic emission equation.

If an external electric field ${\cal E}_{\rm {ext}}$ helps the electrons escape, it lowers the energy that the electrons need to do so. Consider the potential energy in the later stages of escape, at first still without the additional electric field. When the electron looks back at the metal surface that it is escaping from, it sees a positron mirror image of itself inside the metal. Of course, there is not really a positron inside the metal; rearrangement of the surface electrons of the metal create this illusion. The surface electrons rearrange themselves to make the total component of the electric field in the direction parallel to the surface zero. Indeed, they have to keep moving until they do so, since the metal has negligible electrical resistance in the direction parallel to the surface. Now it just so happens that a positron mirror image of the electron has exactly the same effect as this rearrangement. The escaping electron pushes the surface electrons away from itself; that force has a repulsive component along the surface. The positron mirror image however attracts the surface electrons towards itself, exactly cancelling the component of force along the surface exerted by the escaping electron.

The bottom line is that it seems to the escaping electron that it is pulled back not by surface charges, but by a positron mirror image of itself. Therefore, including now an additional external electrical field, the total potential in the later stages of escape is:

V = - \frac{e^2}{16\pi\epsilon_0 d} - e {\cal E}_{\rm ext} d + \mbox{constant}

where $d$ is the distance from the surface. The first term is the attracting force due to the positron image, while the second is due to the external electric field. The constant depends on where the zero of energy is defined. Note that only half the energy of attraction between the electron and the positron image should be assigned to the electron; the other half can be thought of as work on the image. If that is confusing, just write down the force on the electron and integrate it to find its potential energy.

If there is no external field, the maximum potential energy that the electron must achieve occurs at infinite distance $d$ from the metal surface. If there is an electric field, it lowers the maximum potential energy, and it now occurs somewhat closer to the surface. Setting the derivative of $V$ with respect to $d$ to zero to identify the maximum, and then evaluating $V$ at that location shows that the external field lowers the maximum potential energy that must be achieved to escape by $\sqrt{e^3{\cal E}/4\pi\epsilon_0}$.