Is it possible to make a metal plate emit protons (not electrons) when it is illuminated with light? If possible, how?
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$\begingroup$ You could improve this question by stating what you believe and your reasoning/rationale. But I do like the question! $\endgroup$– jerk_dadtCommented Mar 4, 2014 at 8:09
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$\begingroup$ Probably not with normal visible light. Protons are either tightly bound in a nucleus or tightly bound with an electron. $\endgroup$– Brandon EnrightCommented Mar 4, 2014 at 8:37
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3 Answers
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No. Electrons are very loosely bound while protons are are held together by the strong force. It would take an enormous amount of energy to eject a proton and does not actually surmount the potential barrier of the strong force it merely tunnels through it (this is called proton decay). For electrons, which are bound to the nucleus by electromagnetic force, have a potential barrier that is much, much easier to surmount. This is why electrons get ejected and not protons.
The strong force is roughly 137 times stronger than the electromagnetic force. This approximately means the barrier that the proton needs to over come is 137 times larger than the electrons barrier.
Please note the comments for additional insight. Thanks @JohnRennie, @yatima2975, and @BrandonEnright!
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1$\begingroup$ I suppose if you could make your photon arbitrarily energetic there would come a point where protons were ejected. However it wouldn't be a clean process like photoemission - you'd just be destroying the atoms on the surface. $\endgroup$ Commented Mar 4, 2014 at 8:18
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$\begingroup$ The nuclear binding energy per nucleon is on the order of a couple of MeVs, while the electron binding energy for the outermost electron is in the tens of eVs. I don't know why you're mentioning the inverse of the fine structure constant here! $\endgroup$ Commented Mar 4, 2014 at 12:57
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$\begingroup$ @yatima2975 : you can say that the electromagnetic force is "proportional" to the fine-structure constant (or that the fine-structure constant is an adimensional measurement of the strength of a force). In the case of the strong force, this value is close to 1. This said, the fundamental difference here is the distance: nucleons are tightly packed, whereas electrons are far far away, and partially screened. Therefore, the energies have much higher ratios, as you said. $\endgroup$– DavidmhCommented Mar 4, 2014 at 14:07
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As jerk_dadt said, protons are much more tightly bounded in the nucleus by the strong nuclear force. Furthermore, photoelectric effect happens typically on metals, where the most external electrons are "floating" in no-mans land between the atoms, not tightly bound to any of them. So, the bonding for the protons is way stronger than 137 times. But we can assume you can generate a photon sufficiently energetic.
Let's revisit what happens when you shoot a UV photon to your metal. It first encounters the electrons, and if it gets absorbed, it will increase its energy, move at a high speed and be scattered until it gets outside of the metal. The shift in momentum (from going towards the surface to going away from it) gets absorbed by the rest of the structure. It is really tiny, and dispersed across several scattering processes, so not a big deal.
But now you have your super-energetic photon. The cross-section for the photon-electron scattering decreases with the energy of the photon, so you could assume it passes through the electronic shell without a problem. It now hits the nucleus, that offers a much broader target and gets (at least partially) absorbed. Then, suddenly, the protons inside will have a high energy and start moving. The strong interactions are quite complicated and a detailed calculation of what would be exactly going on is difficult, but for sufficiently high energy photons, you will excite the atom so much it would split. If the energy is high enough, you will also create pairs of particles than will take away part of the energy.
The fragments will follow what is energetically favorable, so my guess is you will get two fragments roughly similar in size, as you do in most radioactive decays.
Let's assume the original metal was iron, the most efficiently packed nucleus. The new fragments have a much higher energy than the original, so the fragments will not have much kinetic energy and move slowly. They will attract the electrons around them and try to bond chemically with the material around.
But if you had say, uranium, the two fragments will have much less energy than the original, so they can actually move fast (minus the energy taken by neutrinos, photon emission and so on), and they could get to escape the solid.
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I would think it is possible that a photoelectric effect might occur in proton conductors where the protons are not tightly bound, but free to move around.
Another situation where they are relatively free to move around is in certain hydrated metals, particularly palladium. If the loading factor of hydrogen is high enough in a palladium lattice, perhaps a photoelectric-like effect for protons can be observed at the surface. I suppose that the electrons are still the primary current carriers in this case, and so hydrated metals are not considered "proton conductors", despite the high mobility of the protons. If I were to venture a guess, it would be that the work function for the mobile protons would be about the same as for the electrons, and so the ratio of the number of protons emitted to the number of electrons emitted photo-electrically would just be proportional to the ratio of the density of states for the two particles at the surface.
