---
title: "Thermionic emission and Richardson equation"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:thermionic-richardson-schottky"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/thermionic-richardson-schottky
site: BCT Engineering
---

# Thermionic emission and Richardson equation

Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal

- Asked in: 2078 Bhadra, 8 marks; 2071 Chaitra, 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/thermionic-richardson-schottky
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

What is Thermionic emission and work function? Derive the Richardson's expression for the thermionic emission for Schottky effect.

## Short answer

Thermionic emission is electron ejection from a heated metal; work function Φ is the minimum energy from Fermi level to vacuum. The Richardson-Dushman current density is J = B0 T² exp(−Φ/kT) with B0 = 4πmek²/h³. The Schottky effect, due to an applied field ℰ, lowers the barrier by ΔΦ = √(e³ℰ/(4πε₀)), giving J = B0 T² exp[−(Φ−β_S√ℰ)/kT].

## Answer

#### 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.

```text
 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

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

```text
 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the work function?
   A. Minimum energy to remove an electron from the Fermi level to vacuum
   B. Energy of the bottom of the band
   C. Energy of highest speed
   D. Heat capacity
   Answer: A. Phi = E_vac - E_F. (Key point: Work function)

2. Why can f(E) be replaced by exp(-(E-E_F)/kT) for escaping electrons?
   A. They have E - E_F much greater than kT
   B. They are at 0 K
   C. They are below EF
   D. Spin is zero
   Answer: A. Only high-energy tail electrons escape, where E - E_F >> kT. (Key point: Why Boltzmann approximation)

3. What does the derivation give?
   A. J = (4 pi m e k^2 / h^3) T^2 exp(-Phi/kT)
   B. J = B T exp(-Phi/kT)
   C. J = B T^2 exp(+Phi/kT)
   D. J independent of T
   Answer: A. Integrating v_x dn over escaping electrons gives B0 T^2 e^(-Phi/kT). (Key point: Richardson result)

4. What does the applied field do to the barrier?
   A. Lowers it by sqrt(e^3 E/4 pi eps0)
   B. Raises it
   C. No effect
   D. Doubles it
   Answer: A. Image force plus field lowers Phi by beta_S sqrt(E), increasing emission. (Key point: Schottky effect)

## Related in Electrical Engineering Material

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