The thermal velocity of the free electron in a metallic conductor varies from $10^5\ \mathrm{m/s}$ to $10^6\ \mathrm{m/s}$. In spite of high velocity, free electrons fail to escape from the metallic surface. Why is that?
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$\begingroup$ Is that velocity from $\sqrt{k_bT/2m}$? That is assuming a classical gas which occurs at a density of electrons much less then that in a solid. $\endgroup$– Shane P KellyCommented Jul 15, 2017 at 16:45
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12$\begingroup$ They do escape if the temperature is high enough; it's called thermionic emission and it has quite a few industrial uses. $\endgroup$– zwolCommented Jul 15, 2017 at 18:27
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$\begingroup$ Do they not also escape if you happen to provide them with an "easier" path to ground than what's offered by the conductor? $\endgroup$– arothCommented Jul 17, 2017 at 4:43
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1$\begingroup$ It is not just the magnitude of the velocity (which you have emphasized), but also its direction. If the electron has a high velocity in the direction of the conductor, it will never "escape" from the conductor. To escape, the velocity must have a component perpendicular to the conductor! $\endgroup$– GuillCommented Jul 19, 2017 at 4:53
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5 Answers
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Electrons are bound to the metal by the attraction of the nuclei. After screening of the nuclei by other electrons in the metal, there is a net electric field creating a potential barrier for the electrons to escape. This potential barrier is called the work function and is defined with respect to the Fermi energy of the electrons. The work function is usually around a couple electron volts, while the fermi energy is usually around $10\,\mathrm{eV}$. This means there is around a $12\,\mathrm{eV}$ potential barrier for the electrons to over come before they can escape.
On a temperature scale, the fermi energy corresponds to $T=E/k_b\approx10\,000\,\mathrm{K}$ This means at room temperature, and up to a couple $1000\,\mathrm{K}$, the "free" electron gas is a degenerate fermi gas, and the electrons rarely have energy greater then $10\,\mathrm{eV}$.
The relevant velocity in degenerate electron systems is the fermi velocity or the velocity of the fastest electron in the degenerate electron gas:
$$v_f=\sqrt{E_f/2m}=\sqrt{10\,\mathrm{eV} ~/~ (2\times 0.5\,\mathrm{MeV}/c^2)}\approx 10^{-3}c,$$
which is even faster than the velocity you quoted.
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$\begingroup$ Okay fine, but in this context, why free electron is not emitted in case of photo-electric emission ?? $\endgroup$ Commented Jan 23, 2019 at 15:15
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The electrons in the conductor which are not free are also travelling at high speed but they are bound to particular atoms. It requires energy to remove them. The 'free' electrons in the conductor are not really free. They are not bound to individual atoms but they are shared by and bound to a large number of atoms which form a microscopic crystal called a 'grain'. It still requires energy to remove them from a grain, but not as much as to remove electrons attached to individual atoms.
Grains are very close together so it requires very little energy for an electron to overcome the energy barrier and 'hop' from one grain to another along the conductor. There are usually no adjacent grains outside of the conductor, so electrons do not leap out of it unless, for example, another conductor is placed in close contact with it, or the electric field at the surface of the conductor is strong enough to overcome the force keeping electrons inside the grains, or a collision with a photon gives the electron more energy. The minimum potential difference for this to happen is called the Work Function.
answered Jul 15, 2017 at 16:24
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If you remove $n$ electrons off the metal, the metal becomes $+ne$ positively charged. An electron which tries to leave the surface will be pulled back by the attractive forces. In other words, an electron does not have sufficient energy to over come the potential barrier.
You can supply energy to the electrons by heating or by shining light on the metal. This will give enough energy for the electrons to overcome the potential barrier. The phenomenons are known as thermionic emission and photoelectric effect respectively.
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2$\begingroup$ Well, this doesn't really explain it. If the metal is hot enough, it may start emitting electrons. $\endgroup$– RuslanCommented Jul 15, 2017 at 20:03
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1$\begingroup$ I think that is included in the last sentence. If thermionic emission has to be mentioned, then why not photoelectric effect? I think the last statement clearly establishes the fact there is a potential barrier which an electron has to overcome and it generally does not have sufficient energy to do so. $\endgroup$– YashasCommented Jul 16, 2017 at 2:48
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I can't provide the exact numerical data just yet, but I can give a basic explanation. Simply put, the electrons in a metal lattice form a bounded cloud,free to move inside the lattice. This cloud of free electrons forms metallic bonds with the positively charged nuclei.These bonds hold a metal together. Now, based on spectroscopic analysis, the energy of the electrons is simply not enough to overcome the potential barrier-leaving the metal surface.
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If a electron were able to just escape the conductor, then it would be pulled back by the positive charge left behind (An image charge potential, one could say). Due to this it would be pulled back and will not be able to escape until it is able to obtain enough energy to become completely free.
Electrons can attain energy by the photoelectric effect, or by heating the conductor (Thermionic emissions).Two example of Thermionic emissions is the Child-Langmuir Law and The Richardson Law. (This PDF gives you a short intro to both ).
Also, note that if a photoelectric effect were continuously illuminated, and were not connected to a circuit, then after some time the photoelectric emission would stop since the positive charges left behind would accumulate and increase the potential barrier required to escape.
