Electron Flux Example

A sharp tip with a radius of curvature $a=100nm$ is held at a potential of $V=1kV$. The work function of the tip material is $\varphi =2eV$. Assume a barrier thickness of 0.2nm for electron tunneling

---

Solution Part 1: Electric Field Calculation The electric field $E$ at the surface of the tip is approximated by:

$$E=\frac{V}{a}$$

Given values are:

• $V=1\mathrm{kV}=1000\mathrm{V}$
• $a=100\mathrm{nm}=100\times 10^{-9}\mathrm{m}=1\times 10^{-7}\mathrm{m}$

Substitute these values into the equation:

$$E=\frac{1000\mathrm{V}}{1\times 10^{-7}\mathrm{m}}=1\times 10^{10}\mathrm{V}/\mathrm{m}$$

Part 2: Current Density Calculation The current density $J$ from a field emitter is given by the Fowler-Nordheim equation:

$$J=A\frac{E^2}{\varphi }\exp \left(-B\frac{{\varphi }^{3/2}}{E}\right)$$

where:

• $J$ is the current density in $\mathrm{A}/{\mathrm{m}}^2$
• $E$ is the electric field at the emitter surface in V/m
• $\varphi$ is the work function of the emitter material in eV
• $A\approx 1.54\times 10^{-6}\mathrm{A}\mathrm{eV}{\mathrm{V}}^{-2}$ and $B\approx 6.83\times 10^9\mathrm{V}{\mathrm{m}}^{-1}{\mathrm{eV}}^{-3/2}$ are FowlerNordheim constants.

Given values are:

• $\varphi =2\mathrm{eV}$
• $E=1\times 10^{10}\mathrm{V}/\mathrm{m}$
• $A=1.54\times 10^{-6}\mathrm{A}\mathrm{eV}{\mathrm{V}}^{-2}$
• $B=6.83\times 10^9\mathrm{V}{\mathrm{m}}^{-1}{\mathrm{eV}}^{-3/2}$

Substitute these values into the Fowler-Nordheim equation:

$$\begin{array}{c}J=\left(1.54\times 10^{-6}\right)\frac{{\left(1\times 10^{10}\right)}^2}{2}\exp \left(-\left(6.83\times 10^9\right)\frac{(2{)}^{3/2}}{1\times 10^{10}}\right)\\ J=\left(1.54\times 10^{-6}\right)\frac{1\times 10^{20}}{2}\exp \left(-\left(6.83\times 10^9\right)\frac{(2\sqrt{2})}{1\times 10^{10}}\right)\\ J=\left(0.77\times 10^{14}\right)\exp (-(6.83\times 2\sqrt{2}\times 0.1))\\ J=\left(0.77\times 10^{14}\right)\exp (-(6.83\times 2.828\times 0.1))\\ J=\left(0.77\times 10^{14}\right)\exp (-1.93)\\ J=\left(0.77\times 10^{14}\right)\times (0.145)\\ J\approx 0.11\times 10^{14}\mathrm{A}/{\mathrm{m}}^2=1.1\times 10^{13}\mathrm{A}/{\mathrm{m}}^2\end{array}$$

Part 3: Energy Spread Estimation For field emission guns, typical energy spreads are in the range of $0.1-0.3\mathrm{eV}$.