Schottky effect and modified Richardson equation

Schottky effect is the lowering of the work function (potential barrier) of a metal surface when a strong accelerating electric field is applied to it. Because the barrier is lowered, the thermionic emission current increases with field even at constant temperature.

Barrier without field (image force)

An electron at distance $x$ outside a metal surface induces an image charge $+e$ at distance $x$ inside. The attractive image force and its potential energy are

$$F = -\frac{e^2}{4\pi\varepsilon_0(2x)^2} = -\frac{e^2}{16\pi\varepsilon_0 x^2}, \qquad PE_{im} = -\frac{e^2}{16\pi\varepsilon_0 x}$$

Taking energies from the Fermi level, the electron's PE outside is

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

which approaches $\Phi$ at large $x$. So the full barrier to be crossed is $\Phi$.

Barrier with applied field

An applied field $\mathcal{E}$ that pulls electrons out adds PE $= -e\mathcal{E}x$:

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

 PE
  ^   no field: -> Phi
  | Phi .......___________
  |  dPhi{  .-''-.
  |       /        '-._  with E
  |      /             '-._
  | EF  /___________________> x
     surface   xm

Finding the lowered barrier

The peak is where $d(PE)/dx = 0$:

$$\frac{e^2}{16\pi\varepsilon_0 x_m^2} = e\mathcal{E} \quad\Rightarrow\quad x_m = \left(\frac{e}{16\pi\varepsilon_0\mathcal{E}}\right)^{1/2}$$

Putting $x_m$ back:

$$\begin{aligned} PE_{max} &= \Phi - \frac{e^2}{16\pi\varepsilon_0 x_m} - e\mathcal{E}x_m \\ &= \Phi - \left(\frac{e^3\mathcal{E}}{16\pi\varepsilon_0}\right)^{1/2} - \left(\frac{e^3\mathcal{E}}{16\pi\varepsilon_0}\right)^{1/2} \\ &= \Phi - \left(\frac{e^3\mathcal{E}}{4\pi\varepsilon_0}\right)^{1/2} \end{aligned}$$

So the barrier is reduced by

$$\Delta\Phi = \beta_S\sqrt{\mathcal{E}}, \qquad \beta_S = \left(\frac{e^3}{4\pi\varepsilon_0}\right)^{1/2} = 6.08 \times 10^{-24}\ \text{J}\,(\text{V/m})^{-1/2}$$

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

Modified Richardson equation

The Richardson-Dushman equation $J = B_0T^2\exp(-\Phi/kT)$ now uses $\Phi_{eff}$:

$$J = B_0 T^2 \exp\left[-\frac{\Phi - \beta_S\sqrt{\mathcal{E}}}{kT}\right] = J_0 \exp\left(\frac{\beta_S\sqrt{\mathcal{E}}}{kT}\right)$$

where $J_0 = B_0T^2e^{-\Phi/kT}$ is the zero-field emission and $B_0 = 4\pi mek^2/h^3 \approx 1.2 \times 10^{6}\ \text{A m}^{-2}\text{K}^{-2}$.

Remarks: $\ln J$ is linear in $\sqrt{\mathcal{E}}$ (Schottky plot). For $\mathcal{E} = 10^8\ \text{V/m}$, $\Delta\Phi \approx 0.38\ \text{eV}$ and $x_m \approx 1.9\ \text{nm}$. At much higher fields ($>10^9$ V/m) the barrier becomes thin enough for tunnelling, i.e. field emission.