Image charge derivation for Schottky effect

Thermionic emission in metals

In a metal, conduction electrons fill states up to $E_F$ at 0 K. To leave the metal, an electron needs extra energy equal to the work function $\Phi$. When the metal is heated, the Fermi-Dirac tail spreads above $E_F$, and the few electrons whose kinetic energy normal to the surface exceeds $E_F + \Phi$ escape. This is thermionic emission. The emitted current density is given by the Richardson-Dushman equation

$$J = B_0 T^2 \exp\left(-\frac{\Phi}{kT}\right), \qquad B_0 = \frac{4\pi m e k^2}{h^3} \approx 1.2 \times 10^{6}\ \text{A m}^{-2}\text{K}^{-2}$$

$J$ rises very fast with $T$ and is larger for a small $\Phi$ (hence oxide- or thorium-coated cathodes).

Image charge method

When an electron is at distance $x$ outside a flat conducting surface, the charges it induces on the metal act like a positive image charge $+e$ at distance $x$ inside the surface, i.e. $2x$ away from the electron.

   metal  |  vacuum
          |
   (+e) - | - (-e)
    <--x-->|<--x-->
   image  |  electron

Attractive image force:

$$F_{im} = -\frac{e^2}{4\pi\varepsilon_0 (2x)^2} = -\frac{e^2}{16\pi\varepsilon_0 x^2}$$

Taking the potential energy as zero at infinity, the PE at $x$ is minus the work needed to pull the electron from $x$ out to infinity against this force:

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

Measured from $E_F$, the electron's PE without a field is $\Phi - \dfrac{e^2}{16\pi\varepsilon_0 x}$, which rises smoothly to $\Phi$ far from the surface.

Effect of an applied field

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

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

Maximum of the barrier:

$$\frac{d(PE)}{dx} = \frac{e^2}{16\pi\varepsilon_0 x^2} - e\mathcal{E} = 0 \quad\Rightarrow\quad x_m = \sqrt{\frac{e}{16\pi\varepsilon_0\mathcal{E}}}$$

Substituting $x_m$:

$$\begin{aligned} PE_{max} &= \Phi - \frac{e^2}{16\pi\varepsilon_0}\sqrt{\frac{16\pi\varepsilon_0\mathcal{E}}{e}} - e\mathcal{E}\sqrt{\frac{e}{16\pi\varepsilon_0\mathcal{E}}} \\ &= \Phi - 2\sqrt{\frac{e^3\mathcal{E}}{16\pi\varepsilon_0}} = \Phi - \sqrt{\frac{e^3\mathcal{E}}{4\pi\varepsilon_0}} \end{aligned}$$

So the effective work function is $\Phi_{eff} = \Phi - \beta_S\sqrt{\mathcal{E}}$, where $\beta_S = \sqrt{e^3/(4\pi\varepsilon_0)}$.

Emission current density (Schottky effect)

Replacing $\Phi$ by $\Phi_{eff}$ in the Richardson-Dushman equation:

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

where $J_0$ is the zero-field current density. Thus $\ln J$ increases linearly with $\sqrt{\mathcal{E}}$. Example: at $\mathcal{E} = 10^8$ V/m, $\Delta\Phi \approx 0.38$ eV and $x_m \approx 1.9$ nm.