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Considering that electrons are highly mobile inside of a metal, why do they have such a tough time getting out at the edge of it and continuing their trip ballistically?

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  • $\begingroup$ Look up something called the "work function". $\endgroup$ Aug 10 '13 at 12:01
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Matter is held together by the electrical attraction between the electrons and the nuclei. Within the bulk of a solid or liquid, an electron feels these attractions from all directions equally, and therefore the force it experiences equals zero on the average. But if an electron finds itself at the surface, this isotropy is broken. The electron feels attractive forces toward the interior, which are not canceled out by any forces from the outside. Normally this causes any electron that impinges on the surface to be reflected back in like a pool ball hitting a cushion. To extract the electron out beyond the surface, you have to do a certain amount of work, called the work function $W$. The work function for a metal is typically about 5 eV.

Why does a cathode have to be heated to emit electrons?

It's not actually true that it has to be heated -- cold-cathode devices do exist, and thermionic emission does occur at all temperatures. However, at room temperature, $kT\approx 0.03\ eV$, which is much smaller than $W$. That means that only a very tiny fraction of the electrons have more energy than $W$. The probability of having an energy $W$ at temperature $kT$ goes like $e^{-W/kT}$, and Richardson found in 1901 that the current from a cathode, in the absence of an externally applied electric field, was proportional to $T^2e^{-W/kT}$.

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  • $\begingroup$ In my opinion, very weak explanation. Given that electron feels attractive force once it "finds itself at the surface", it would take him some time to stop and reverse the speed. In this case, we could observe electrons "flying out of material", and these electrons could easily be extracted. Also, the net electric force which affects the electron once it is out of the lattice is very weak due to screening effects. $\endgroup$
    – Vasiliy
    Aug 10 '13 at 8:31
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    $\begingroup$ If the work function is around 5 eV, shouldn't cathodes that have a potential drop to the anode of more than 5V start emitting without heat? $\endgroup$
    – yippy_yay
    Aug 12 '13 at 20:37
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    $\begingroup$ @SebastianHenckel: No, what matters locally is the electric field. The electric field required in order to cancel the electric field holding the electrons in at the surface is $\sim(5\ \text{V})/(0.1\ \text{nm})$. This is basicallly the strength of the microscopic fields, and if you apply it externally to matter, the matter is destroyed. $\endgroup$
    – user4552
    Aug 12 '13 at 20:56
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Your main mistake is that you think about electrons being moving along the lattice ballistically.

There at least two answers to this question: the classical one, and the QM one.

The classical answer is that even when the electrons are "highly mobile inside of a metal", they just hop between adjacent atoms. When it is in the atom, it resides on one of the allowed potential energy levels - exactly the same situation as encountered with a single atom. In order to remove the electron from the atom it resides in, you must supply the required amount of energy (which is called Work Function; Work Function of a material differs from ionization energy of material's element).

The QM answer is that when you stick all these atoms together, the orbitals of the valence electrons change such that they span the entire lattice (in fact, these orbitals even go slightly beyond the volume defined by the lattice). The above statement means that the probability of finding an electron far from the lattice it belongs to is negligible.

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  • $\begingroup$ I don't think the band structure is particularly relevant, and quantum mechanics is not really needed to understand the behavior of Richardson's Law. $\endgroup$
    – user4552
    Aug 9 '13 at 23:53
  • $\begingroup$ @BenCrowell, as I said, there are at least two ways to explain this intuitively. To my best knowledge, quantitative explanation of thermionic emission requires extensive use of QM. I tried to provide some understanding as to "why this doesn't happen". $\endgroup$
    – Vasiliy
    Aug 10 '13 at 8:24
  • $\begingroup$ @BenCrowell, the question is not regarding Richardson law. OP wanted to know why electrons are not "shot" out of metal. $\endgroup$
    – Vasiliy
    Aug 10 '13 at 8:32

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