Conservation of electrical energy on a conductor

[G Cooke]

I'm having trouble seeing how electric potential energy production on a conductor follows conservation of energy.

Let's use the photoelectric effect as an example. A photon with energy E = hν strikes a conductor, ejecting a photoelectron with a maximum kinetic energy of hν - φ. Assuming the maximum kinetic energy is achieved, this leaves only φ to be transferred from the photon into the conductor.

However, we know that electrical energy follows the equation E = QV = Q²/C, which implies that the electric potential energy produced on the conductor by the photoelectric effect is equal to E_prod = E_fin - E_init = (Q_init + Q_elec)²/C - Q²_init/C, which clearly increases with initial charge. Since φ is a constant, this implies that it is possible for E_prod to be greater than φ, which appears to violate conservation of energy since φ was the total amount of energy that entered the conductor.

Clearly I'm missing something. Any help would be much appreciated.

 

[mfb]

If the target for the photoeffect has a positive charge, φ increases. Typically the electrons have some path to flow back to it or the total number of electrons knocked out is negligible so this is rarely an issue.

 

[G Cooke]

If the target for the photoeffect has a positive charge, φ increases.

That is true; however, even if we assume zero initial charge, there appears to be a problem.

With Q_init = 0, we have φ at its base value (i.e., no stopping potential) and E_prod = E_fin - E_init = Q²_elec/C - 0. So φ is a constant, and E_prod depends on C, which is arbitrary. Thus, it is possible that E_prod is greater than φ, again appearing to violate conservation of energy.

 

[mfb]

You can't make C small enough to make E_prod relevant compared to φ. In fact, φ will simply include this tiny amount of energy already.

 

[G Cooke]

You can't make C small enough to make E_prod relevant compared to φ.

I'm not so sure about that. To make E_prod > φ, algebra says all we need is C < Q²_elec/φ, which is about 1e-38F...pretty small, but impossible?

In fact, φ will simply include this tiny amount of energy already.

Could you elaborate on this?

 

[mfb]

I'm not so sure about that. To make E_prod > φ, algebra says all we need is C < Q²_elec/φ, which is about 1e-38F...pretty small, but impossible?

It is impossible. Even a spherical capacitor with "infinite" separation from the rest of the universe will have a capacitance of $C=4 \pi \epsilon_0 r$. Solving for r, we get 10^-28 meters. That is 13 orders of magnitude smaller than a proton. You can't make a metallic sphere of that size.

Could you elaborate on this?

φ is the energy needed to remove an electron from the material. This includes the electrostatic attraction from the metal that now has a single positive charge - in principle, in practice it is negligible.

 

[G Cooke]

It is impossible. Even a spherical capacitor with "infinite" separation from the rest of the universe will have a capacitance of $C=4 \pi \epsilon_0 r$. Solving for r, we get 10^-28 meters. That is 13 orders of magnitude smaller than a proton. You can't make a metallic sphere of that size.

I see. But actually, my math is 19 orders of magnitude off because I forgot that φ will be in units of Joules rather than eV here. So I guess this makes the sphere 6 orders of magnitude larger than a proton. Still impossible, I presume?

φ is the energy needed to remove an electron from the material. This includes the electrostatic attraction from the metal that now has a single positive charge - in principle, in practice it is negligible.

Interesting. I was actually just about to type that this would all be cleared up if (1) the maximum kinetic energy of the photoelectron is hν - φ and (2) φ is defined as the material-specific work function W plus the FUTURE stopping potential E_fin, i.e., the electric potential that will be on the conductor after the photoelectron is ejected. That would make intuitive sense anyway since it would seem to be the post-ejection potential that actually pulls on the photoelectron, not the pre-ejection potential.

So it looks like my whole problem was that I originally had φ defined as just the material-specific work function W rather than W + E_fin.

For what it's worth, if it were the pre-ejection potential, then conservation of energy could be proven violated assuming divergence of the sequence E_{prod, n} = (2n+1)Q²_elec/(W + (W/C)E_{prod, n-1}), E_{prod, 0} = Q²_elec/W.

 

[mfb]

I see. But actually, my math is 19 orders of magnitude off because I forgot that φ will be in units of Joules rather than eV here. So I guess this makes the sphere 6 orders of magnitude larger than a proton. Still impossible, I presume?

0.7 nm (assuming φ=2 eV) Still impossible to have a metal with the usual electron bands with that size. Instead of the photoeffect you can get ionization of individual atoms. There the attraction is important - the whole ionization energy comes from this attraction.

 

[G Cooke]

So I realize now that energy is conserved because φ includes the post-ejection electric potential energy E_fin of the the conductor, which it must because the newly ejected photoelectron is caught inside the electric field from the now positively charged conductor and must overcome its force in order to escape.

But what if the geometry of the conductor is such that the electric field is zero in the region where the photoelectron ejects from it? For example, if the photon strikes the inner surface of a spherical shell, the charge left by the ejected photoelectron will distribute evenly around the shell, creating a zero net electric field inside the shell (Gauss's law) so that the photoelectron feels no attractive force. That implies φ here does not include E_fin. Yet, it would seem that the conductor still acquires electric potential energy E_fin due to its charge (assume the photoelectron gets decelerated by collisions with air molecules inside the shell so that it doesn't cancel out the conductor's charge). It appears that we can now have E_fin > φ and, by extension, E_prod > φ, yet again appearing to violate conservation of energy.

 

[mfb]

the charge left by the ejected photoelectron will distribute evenly around the shell

That process takes some time.

φ is a description of a macroscopic quantity. It doesn't help to search for effects of individual charges on it. That's like asking what the temperature of a macroscopic object is if one particular atom in it has a slightly higher energy at this point in time.

 

[ZapperZ]

So I realize now that energy is conserved because φ includes the post-ejection electric potential energy E_fin of the the conductor, which it must because the newly ejected photoelectron is caught inside the electric field from the now positively charged conductor and must overcome its force in order to escape.

But what if the geometry of the conductor is such that the electric field is zero in the region where the photoelectron ejects from it? For example, if the photon strikes the inner surface of a spherical shell, the charge left by the ejected photoelectron will distribute evenly around the shell, creating a zero net electric field inside the shell (Gauss's law) so that the photoelectron feels no attractive force. That implies φ here does not include E_fin. Yet, it would seem that the conductor still acquires electric potential energy E_fin due to its charge (assume the photoelectron gets decelerated by collisions with air molecules inside the shell so that it doesn't cancel out the conductor's charge). It appears that we can now have E_fin > φ and, by extension, E_prod > φ, yet again appearing to violate conservation of energy.

As someone who studies photoemission and does experimental work in it, I find this whole thread to be extremely confusing.

First of all, the work function of a metal that we know and love includes a number of physical effects, INCLUDING the image-charge potential of the emitted photoelectrons when it is very close to the metal surface. This is part of the potential that it has to overcome to escape the bulk material. It is why electrons emitted with little to no KE will never escape, because the image potential will suck it back in. So to ask about a region where there will be no "electric field" on the surface upon emission is puzzling.

Secondly, the photoelectric effect is done on metals that are grounded. It means that (i) there is no charging effects and (ii) the band structure is not modified due to photoelectron emission. If it isn't grounded and charging effects occur, then in the simplest cases, the potential barrier will continue to increase with each electron emission until it is high enough that photons with a specific energy will no longer cause any photoelectron emission.

I still do not understand this energy non-conservation issue here. What comes in is what goes out, minus the "toll" that one has to pay to get out. Where is the non-conservation part?

Zz.

 

[G Cooke]

That process takes some time.

I see. So somehow it is necessarily true that the electric field from the conductor's new positive charge will drain the kinetic energy of the photoelectron before that positive charge can distribute around the inner surface of the shell. Interesting.

Actually, now that I think about it, I guess the the reason why the charge wouldn't distribute first is because it itself is attracted to the photoelectron that was just ejected, so it will want to stay at the point closest to the photoelectron, and if it does that, then the photoelectron doesn't get to escape as it would if there were an even charge distribution.

 

[G Cooke]

As someone who studies photoemission and does experimental work in it, I find this whole thread to be extremely confusing.

Sorry! Especially disregard that sequence divergence statement. I realized afterwards it's incorrect in many different ways. Probably should have it deleted so it doesn't confuse everybody who comes across this thread...

First of all, the work function of a metal that we know and love includes a number of physical effects, INCLUDING the image-charge potential of the emitted photoelectrons when it is very close to the metal surface. This is part of the potential that it has to overcome to escape the bulk material. It is why electrons emitted with little to no KE will never escape, because the image potential will suck it back in. So to ask about a region where there will be no "electric field" on the surface upon emission is puzzling.

Right. I just blindly applied the electrostatics concept of charge distribution as if it happens instantaneously. I see now that that can't apply if there is a floating charge (the photoelectron) inside the shell because the charge on the shell will stay at the point closest to the photoelectron due to mutual attraction, pulling the photoelectron back unless the photoelectron does actually have enough kinetic energy to overcome that pull, i.e., φ > E_fin, implying that energy is conserved.

Secondly, the photoelectric effect is done on metals that are grounded. It means that (i) there is no charging effects and (ii) the band structure is not modified due to photoelectron emission. If it isn't grounded and charging effects occur, then in the simplest cases, the potential barrier will continue to increase with each electron emission until it is high enough that photons with a specific energy will no longer cause any photoelectron emission.

True. I see this now that my definition of φ has been corrected.

I still do not understand this energy non-conservation issue here. What comes in is what goes out, minus the "toll" that one has to pay to get out. Where is the non-conservation part?

Agreed. I'm aware that energy non-conservation is equivalent to saying you can get more than 2 out of 1+1. I just couldn't see where the electric energy was coming from at first since I was thinking of φ as W only or sometimes W plus the initial potential rather than W plus the final potential.