--- title: "Image charge derivation for Schottky effect" subject: "Electrical Engineering Material" chapter: 2 key: "semester_3/electrical_engineering_material:schottky-image-charge" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/schottky-image-charge site: BCT Engineering --- # Image charge derivation for Schottky effect Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal - Asked in: 2081 Chaitra (new course), 6 marks; 2076 Chaitra, 2+6 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/schottky-image-charge - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2 ## Question Explain the thermionic emission in metal. Using image charge method, derive an expression of emission current density for Schottky effect. ## Short answer In thermionic emission, heated electrons with energy normal to the surface exceeding the work function Phi escape. Using the image charge method, an electron outside the metal sees an image potential -e^2/(16 pi epsilon0 x); an applied field E adds -eEx, lowering the barrier by Delta Phi = sqrt(e^3 E/(4 pi epsilon0)). The emission current density is J = B0 T^2 exp[-(Phi - beta_s sqrt(E))/kT] with beta_s = sqrt(e^3/(4 pi epsilon0)). ## Answer #### 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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In thermionic emission, which electrons escape the metal? A. All electrons at E_F, as soon as the metal is heated B. Those with kinetic energy normal to the surface greater than E_F + Φ, from the heated Fermi-Dirac tail C. Only electrons moving parallel to the surface D. Electrons with energy greater than Φ measured from the band bottom Answer: B. Heating spreads the Fermi-Dirac tail above E_F, and only electrons whose normal kinetic energy exceeds E_F + Φ can leave. (Key point: Thermionic emission) 2. What is the image force on an electron at distance x outside a metal surface? A. Attraction e^2/(16πε0 x^2), as if a +e charge were 2x away B. Repulsion e^2/(4πε0 x^2) from a charge -e at x C. Attraction e^2/(4πε0 x^2), from a +e charge at distance x D. Attraction e^2/(16πε0 x), independent of x^2 Answer: A. The image charge +e lies x inside, so the separation is 2x, giving e^2/(4πε0 (2x)^2) = e^2/(16πε0 x^2), attractive. (Key point: Image force) 3. Setting d(PE)/dx = 0 for PE = Φ - e^2/(16πε0 x) - eEx gives x_m equal to A. sqrt(e/(4πε0 E)) B. e/(16πε0 E) C. sqrt(16πε0 E/e) D. sqrt(e/(16πε0 E)) Answer: D. The image force term equals eE at the peak: e^2/(16πε0 x^2) = eE, so x_m = sqrt(e/(16πε0 E)). (Key point: Position of barrier maximum) 4. What is the effective work function with applied field E? A. Φ - eE x_m only B. Φ - β_S E (linear in E) C. Φ - sqrt(e^3 E/(4πε0)), i.e. Φ - β_S sqrt(E) D. Φ + β_S sqrt(E) Answer: C. The image term and field term at x_m are equal, so the total lowering is twice one of them, sqrt(e^3E/(4πε0)), proportional to sqrt(E). 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