--- title: "Schottky effect and modified Richardson equation" subject: "Electrical Engineering Material" chapter: 2 key: "semester_3/electrical_engineering_material:schottky-modified-richardson" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/schottky-modified-richardson site: BCT Engineering --- # Schottky effect and modified Richardson equation Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal - Asked in: 2081 Bhadra, 2+6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/schottky-modified-richardson - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2 ## Question What is Schottky effect? Derive the modified Richardson's equation that shows the reduction in barrier potential due to the application of external field. ## Short answer The Schottky effect is the lowering of the metal's work function by an applied electric field. For an electron at distance x, the image potential is -e^2/(16 pi epsilon0 x) and field E adds -eEx. The barrier maximum gives an effective work function Phi_eff = Phi - sqrt(e^3 E/(4 pi epsilon0)). The modified Richardson equation is J = B0 T^2 exp[-(Phi - beta_s sqrt(E))/kT]. ## Answer **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 $= -e\mathcal{E}x$: $$ PE(x) = \Phi - \frac{e^2}{16\pi\varepsilon_0 x} - e\mathcal{E}x $$ ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Schottky effect? A. Increase of work function with temperature B. Lowering of the metal's work function by a strong accelerating field, raising emission at constant T C. Emission of electrons by tunnelling through a thin barrier D. Increase in the Fermi energy under an applied field Answer: B. A strong accelerating field lowers the barrier, so thermionic current rises even at fixed temperature. (Key point: Schottky effect) 2. What is the PE of an electron outside a metal with image force and applied field? A. Φ - e^2/(16πε0 x) - eEx B. Φ + e^2/(16πε0 x) - eEx C. Φ - e^2/(16πε0 x) + eEx D. Φ - eEx only Answer: A. The image force lowers the PE by e^2/(16πε0 x) and the accelerating field subtracts eEx. (Key point: Barrier with field) 3. The barrier reduction ΔΦ is A. β_S E B. β_S / sqrt(E) C. 2 e E x_m only D. β_S sqrt(E) with β_S = sqrt(e^3/(4πε0)) Answer: D. Substituting x_m gives two equal terms that add to sqrt(e^3E/(4πε0)). (Key point: Barrier reduction) 4. The Schottky-modified current density is A. J = J0 exp(-β_S sqrt(E)/kT) B. J = J0 exp(β_S E/kT) C. J = J0 exp(β_S sqrt(E)/kT) D. J = J0 (1 + β_S sqrt(E)/kT) Answer: C. Replacing Φ by Φ - β_S sqrt(E) in the exponent multiplies J0 by exp(β_S sqrt(E)/kT), so ln J is linear in sqrt(E). 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