# Momentum conservation in the photoelectric effect

1. Aug 21, 2009

### SergioPL

Electromagnetic radiated fields have both E and H fields perpendicular to the wave vector k. Therefore in photons the electric and magnetic fields are also perpendicular to k.

This means that when photons are absorbed by some electron, the Lorenz Force will be mostly perpendicular to the wave vector k (the speed of electrons uses to be much lesser than the speed of light so the electric field causes much more force than magnetic field).

However, the photon carries an inertial momentum P = (h•f)/c in the direction k and in photoelectric effect, the incident photon is absorbed by an electron and no other photon is generated.

So if the force caused by the photon on the electron is mainly perpendicular to k, how is possible to keep the inertial momentum of the system?

2. Aug 21, 2009

### ZapperZ

Staff Emeritus
Er.. this is not correct.

A photoelectric effect doesn't occur in pure free electron gas. It occurs in a solid. The bullk crystal structure absorbs a lot of the momentum, especially the recoil momentum of the emitted photoelectrons, and the momentum of the photons. Photon momentum, actually, is quite negligible when compared to the momentum of the photoelectrons.

Zz.

Last edited: Aug 21, 2009
3. Aug 22, 2009

### SergioPL

Thanks a lot ZapperZ, now I understand. The total electron momentum is much bigger than the photon momentum but it is compensated with the nucleus recoil.

4. Aug 22, 2009

### SergioPL

For me this means that photons can interact with more than a particle at time, since a single photon apply their electromagnetic field both on the electron and the nuclei, before disappearing.

Both particles have absorbed the photon although the electron gets almost all the energy.

5. Aug 22, 2009

### vin300

How could you explain energy conservation if the nucleus recoils?

6. Aug 22, 2009

### ZapperZ

Staff Emeritus
The particles do not "absorb" the photons, the same way a free electron cannot absorb a photon. It is the WHOLE SYSTEM that does this. Without the conduction electrons being inside the bulk material, you do not have a photoemission phenomenon.

Secondly, there's no evidence of 1 photon exciting more than 1 electron in a photoemission spectrum. There are "multiphoton" photoemission, but not multi-electron photoemission from a single photon.

The "nuclei" are well-shielded. You seem to not know about not only the conduction band, but also the inner shell electrons that are still localized at each of the lattice sites.

When you make "guesses" like what you are attempting here, you need to step back and ask yourself "do I have experimental evidence to back what I'm claiming?" This is, after all, still PHYSICS, which still requires empirical evidence. If you do not have that, then you're making a speculative conjecture that has no support. We do not allow that here in this forum.

Zz.

7. Aug 22, 2009

### ZapperZ

Staff Emeritus
Hit a ball at a wall. The KE of incoming ball is practically identical to the KE of the bounced ball. The wall DID recoil to preserve the conservation of momentum, didn't it? But compare the masses of the two. Do you think the recoil of the wall is noticeable enough to cause the KE of the ball to change?

Same thing with "electron" and "the whole lattice".

Zz.

8. Aug 22, 2009

### vin300

Blow a bomb. The momentum of all the paricles that blow away certainly sum to zero, but energy doesn't,it's scalar. The photon cannot produce more energy than it already had to increase the magnitude of momentum both ways.

9. Aug 22, 2009

### ZapperZ

Staff Emeritus
And you think that the energy of the outgoing photoelectrons are identical to the incoming photon? Since when?

Many of these things, including the image charge potential, are LUMPED into the work function!

I strongly suggest you read the Spicer papers on these things, that addressed both the momentum and energy conservation. If you disagree, you're welcome to write your own rebuttal and we can then revamped the whole photoemission theory.

Zz.

10. Aug 22, 2009

### vin300

Oh yes,right. It already has energy.

11. Aug 22, 2009

### ZapperZ

Staff Emeritus
Huh? What energy?

Zz.

12. Aug 22, 2009

### lightarrow

Be careful when you say "in photons" because we don't have any (recognized) model of the photon's internal structure, yet. We only know how to relate some properties of the photons to some properties of the em radiation, nothing else.

Alt, you are mixing a quantum tratment (photons) with a classical one (em radiation and classical electron). That's wrong.

I haven't understood this. If we want to reason classically, an em radiation makes a Lorentz force on a point-like electron exactly in the direction of k. Why do you say it should be perpendicular?

As ZapperZ wrote, it's not possible that a free particle could *completely* absorb a photon, because energy and momentum couldn't conservate simultaneously. If you want to have a classical analogy, think to a big, massive ball being hit by a very light ball and consider only kinetic energy (= elastic collision): the second ball will always recoil, it couldn't stop and loose all its kinetic energy after the collision.

13. Aug 22, 2009

### vin300

All the energies that the electron must escape from.I'm not in agreement with "the nucleus recoils", there's nothing that causes the nucleus to recoil. The nucleus and electron don't start repelling.

14. Aug 22, 2009

### vin300

Sure, but the analogy relates to photon-electron, not electron-nucleus

15. Aug 22, 2009

### Born2bwire

*shrug*
At least in this case, the first-order force wll be perpendicular to the k vector. The first-order effect is the Lorentz force due to the electric field of the electromagnetic wave, which is normal to the k vector. Though once the electric field's force imparts a velocity on the electron, then the electron will experience a force from the magnetic fields in the direction of the k vector.

16. Aug 23, 2009

### ZapperZ

Staff Emeritus
And you were worried about energy conservation but not momentum conservation? What gives? And what's with the "nucleus and electron don't start repelling"? We are talking about bulk material. The conduction electrons are not attached to any nucleus!

You are welcome to do your own literature search and write a rebuttal to all these papers[1,2,3,4].

And for your information, and to get the FACTS straight, notice that I kept saying that it is the LATTICE (not the "nucleus") that does the "recoil". YOU inserted the "nucleus".

Zz.

[1] M. Voss et al. Phys. Rev. B 78, 024301 (2008).
[2] T. Fujikawa et al. J. El. Spec. Rel. Ph. 162, 146 (2008).
[3] Y. Takata et al. Phys. Rev. Lett. 101, 137601 (2008).
[4] Christian Skou Søndergaard, Ph.D Thesis, University of Aarhus, Denmark (2001). See discussion on Pg. 7.

Last edited: Aug 23, 2009
17. Aug 23, 2009

### lightarrow

Infact I was exactly talking about this, in answer to the OP's question about photons and electrons. For this reason, as ZapperZ wrote, it is necessary to consider also the entire crystal's momentum.

Last edited: Aug 23, 2009
18. Aug 24, 2009

### SergioPL

I Know there is no evidence of multi-electron photoemission but what I wanted to remark is that, in order to keep the conservation laws, two particles –an electron and a proton- should be altered by the photon action.

The reason to say that these particles should be an electron and a proton is because I suppose photons interact with particles only via the Lorenz force and therefore it only interacts with charged particles.

Besides, since the photon momentum is in the wave vector direction and the electromagnetic force is mainly perpendicular, the particles should be with opposite charges, so an electron and a proton.

This supposition fits quite well with the fact that the nuclei recoils .

I have tried to use the classical electromagnetism formulae for photons which is quite speculative but since the radiated field is composed by photons it may have sense, though it’s still speculative.

It’s important to remark that the classical Coulomb force (or electromagnetic induction field caused by the existence of charges) must not be considered to be composed by photons since it has quite different properties than the radiated field, originated by the acceleration of charges. I only suggest that the radiated field is composed by photons and therefore these photons are the agents of the Lorentz force (i.e. electromagnetic interaction).

19. Aug 24, 2009

### ZapperZ

Staff Emeritus
Again, if you care to look at the references I've given, the accounting for the photon momentum is NEGLIGIBLE. You throw a ball at a wall. You see no "alteration" on the wall-earth system because the ball bounces back - or do you?

The reason for the "recoil" of the lattice has more to do with the ELECTRON EMISSION, not the photon momentum!

Zz.

20. Aug 24, 2009

### lightarrow

It's because with "lorentz force" I intended only the term qvXB (at least this is what they taught me at uni and what is written in the book I have). But I see in wiki and others sites that now they mean the entire term qE + qvXB!