Imagine our environment is a vacuum. At which value of electric field or voltage, does field emission occur? I just want to know what is the maximum electric field we are allowed, to place between two electrodes, without field emission phenomenon. So, you can choose any metal that you wish in order to mention this maximum value.

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    $\begingroup$ I am afraid that there is no one sentence answer to your question. $\endgroup$
    – Farcher
    Oct 8, 2022 at 12:31
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    $\begingroup$ I know you have faced with totally general question, but, the winner is the person who define the environment so as to reach the maximum number for electric field. clear! $\endgroup$ Oct 8, 2022 at 12:35
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    $\begingroup$ Hello @Farcher, it would be great even if you mention a typical value for field emission phenomenon along with reference. $\endgroup$ Oct 8, 2022 at 12:46

2 Answers 2


Field emission is a quantum tunneling process, and as such can occur at any voltage, although for low enough fields it will be exponentially suppressed. Intuitively one should expect the emission rate to go as something like: $$dN/dt \propto \exp(-eEd/W)\equiv \exp(-E/E_c),\, E_c= W/ed$$ where $e$, $E$ and $W$ are the electron charge, applied electric field and the work function of the emitting material respectively. $d$ is the distance determining how much the bulk electrons orditals leak out into vacuum. You can estimate if as $d\approx \hbar v_F/W$, $v_F$ being fermi velocity. The work function $W$ is typically in the range of several electron-volts, i.e. comparable to typical Fermi energies, therefore $d$ is roughly equal lattice constant. So now that we have an idea of the order of magnitude of all involved quantities, we can estimate the characteristic field $E_c$ at which field emission becomes noticeable: $E_c$~$5eV/(e*1nm)=5$ V/nm$=5e7$ V/cm.

You can achieve such fields by either increasing your voltage or by reducing the distance between electrodes. If you work with an STM you can get filed emission already with several-to-ten volts of bias voltages. Also, note that the actual local field can be enhanced near surface inhomogeneities, that will facilitate field emisssion.


Well, I am a new master circuit student and I do know so much about physics.

The definition of work function is here:

In solid-state physics, the work function (sometimes spelled workfunction) is the minimum thermodynamic work (i.e., energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface.

And also, according to field electron emission:

Field electron emission, also known as field emission (FE) and electron field emission, is emission of electrons induced by an electrostatic field. The most common context is field emission from a solid surface into a vacuum.

So, we can say that field emission occurs when the electron can absorb enough energy to overcome the work function the metal.

So according to my research, I can say that based on Work functions of elements the maximum work functions of materials is 5.93eV (and I do not know a material with a higher work function is discovered or not). So, the voltage required to remove an electron from metal to a point in vacuum will be equal to:

5.93/(1.6e-19)=3.7e19 volts

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Oct 8, 2022 at 13:58
  • $\begingroup$ I've seen (unwanted) field emission at 10 kV in an experiment, so your estimate is way off. As @Farcher says, there isn't a simple answer. $\endgroup$
    – John Doty
    Oct 8, 2022 at 15:35

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