Thermionic emission and Richardson equation

Thermionic emission and work function

Work function $\Phi$ is the minimum energy needed to remove an electron from the Fermi level of a metal to the vacuum just outside it ($\Phi = E_{vac} - E_F$; e.g. W 4.5 eV, thoriated W 2.6 eV).

Thermionic emission is the emission of electrons from a heated metal. At high $T$ the Fermi-Dirac tail extends above $E_F + \Phi$, so some electrons have enough energy, in the direction normal to the surface, to escape. It is used in cathodes of vacuum tubes, CRTs and X-ray tubes.

 Energy
  ^  vacuum level -------------
  |                 ^
  |                 | Phi
  |  EF   ----------v----  <- top filled states
  |       /////////////////
  |  metal  (filled band)

Derivation of Richardson-Dushman equation

Take the surface normal to $x$. Electrons escape if $\tfrac{1}{2}m v_x^2 \ge E_F + \Phi$. For these high-energy electrons $E - E_F \gg kT$, so $f(E) \approx \exp[-(E - E_F)/kT]$. The number of electrons per unit volume with velocity in $dv_x\,dv_y\,dv_z$ (including spin) is

$$dn = 2\left(\frac{m}{h}\right)^3 \exp\left[-\frac{\tfrac12 m(v_x^2 + v_y^2 + v_z^2) - E_F}{kT}\right] dv_x\,dv_y\,dv_z$$

The current density is $J = e\int v_x\, dn$ over all $v_y, v_z$ and over $v_x \ge v_{x0}$, where $\tfrac12 m v_{x0}^2 = E_F + \Phi$. Using $\int_{-\infty}^{\infty} e^{-mv^2/2kT}dv = \sqrt{2\pi kT/m}$ for $v_y$ and $v_z$:

$$J = 2e\left(\frac{m}{h}\right)^3 e^{E_F/kT} \cdot \frac{2\pi kT}{m} \int_{v_{x0}}^{\infty} v_x e^{-mv_x^2/2kT}dv_x$$

The last integral equals $\frac{kT}{m}e^{-(E_F+\Phi)/kT}$. Therefore

$$J = \frac{4\pi m e k^2}{h^3} T^2 \exp\left(-\frac{\Phi}{kT}\right) = B_0 T^2 e^{-\Phi/kT}$$

with $B_0 = \dfrac{4\pi m e k^2}{h^3} = 1.20 \times 10^{6}\ \text{A m}^{-2}\text{K}^{-2}$. This is the Richardson-Dushman equation. Real metals have smaller $B_0$ (surface reflection).

Schottky effect (field-assisted emission)

With an accelerating field $\mathcal{E}$ at the cathode, the electron's potential energy outside the surface is (image force + applied field):

$$PE(x) = \Phi - \frac{e^2}{16\pi\varepsilon_0 x} - e\mathcal{E}x$$

(measured from $E_F$). Setting $d(PE)/dx = 0$ gives the maximum at $x_m = \sqrt{e/(16\pi\varepsilon_0\mathcal{E})}$, and the barrier is lowered by

$$\Delta\Phi = \sqrt{\frac{e^3\mathcal{E}}{4\pi\varepsilon_0}} = \beta_S\sqrt{\mathcal{E}}, \qquad \beta_S = \sqrt{\frac{e^3}{4\pi\varepsilon_0}}$$

So the effective work function is $\Phi_{eff} = \Phi - \beta_S\sqrt{\mathcal{E}}$ and

$$J = B_0 T^2 \exp\left[-\frac{\Phi - \beta_S\sqrt{\mathcal{E}}}{kT}\right]$$

 PE
  ^       ___ max (lowered)
  |  Phi /   \___
  |     /        \___  with field
  |    /             \___
  | EF|____________________> x
     surface   xm

Thus the emission current rises with field even at fixed temperature. Example: $\mathcal{E} = 10^8$ V/m lowers $\Phi$ by about 0.38 eV.