# Photo-electric effect: why did they expect this?

1. Aug 14, 2011

### nonequilibrium

Hello,

How can one prove that (using the wave theory of light) a bigger intensity would give a greater exit velocity of the electron? I'm looking for something beyond the heuristic "more energy in the wave implies more energy in the electron".

Note: in case I'm giving of a crackpot vibe: I don't have the intention to go against the usual interpretation of this experiment (i.e. implying light are particles), I just want to understand its convincing power better.

EDIT: an aside Q: was it in the classical theory (i.e. Maxwell's equations describing light as wave) already clear that there was a relation between frequency and energy? Or is that only since quantum theory?

2. Aug 14, 2011

### xts

The intensity is proportional to square of the field.
Thus the electric field is proportional to square root of the intensity.
Thus the acceleration of the electron is proportional to the sqrt of intensity.
Thus (all other parameters of appropriate equations remains non changed) the final speed is proportional to sqrt of intensity.

3. Aug 14, 2011

### nonequilibrium

Hm, interesting in its obviousness. It seems I didn't think of how the light knocks the electrons out (was thinking intuitively of the sloshing of the waves). The thing I'm confused about now: Einstein proposed a particle model, but in the way he proposed it, it seems the electrons are simply being knocked out by the light particles, as if you're playing snooker? Shouldn't there instead be an electromagnetic force at work (like in the wave model), that has to do with the being charged of the particles? (cause I assume that if electrons were neutral, the photo-electric effect wouldn't happen) Do I need to learn QED to get this?

4. Aug 14, 2011

### xts

You don't need QED (at least now...) Pre WW-I physics is pretty sufficient.
Einstein's picture is that particles are knocked like by snooker balls. And intensity means more shots, so more electrons are knocked out. If the incoming ball has to low energy it is not able to knock the electron.

5. Aug 14, 2011

### nonequilibrium

I understand that, but doesn't that seem to throw whole Maxwell's theory out of the window? "Never mind the lorentz force, light just knocks out electrons like snooker balls"... This reasoning would even lead to suggest that it would also work on neutral particles.

6. Aug 14, 2011

### xts

Photons do not knock neutral particles, as their cross-section with them is zero. Welll - those are not quite snooker balls of fixed diameter. Here particles have different size depending what is the target.

And don't throw out Lorentz's force away. It is pretty useful when you think about yagi antenna receiving UHF broadcast. It would be infeasible to analyse this in terms of photon-electron interaction, as it is infeasible to "throw out Boyle's law" and analyse combustion engine in terms of single molecules bouncing of the piston.

7. Aug 14, 2011

### nonequilibrium

Hm, you seem to be missing my point. Of course I'm not actually advertising to throw out the Lorentz force, I'm just saying that the explanation "photons are particles that knock the electrons out like snooker balls" disregards anything having to do with the Lorentz force. I can't figure out if Lorentz' law simply isn't applicable to photon-electron interactions, or if it is but Einstein simply shut up about it.

And to get back to my original question: it was answered by using the laws of Maxwell. So Maxwell's laws were used to derive what would happen in this experiment. And the expectations turned out to be shattered. But how is the leap from "there's an error in the theory of Maxwell" to "there's an error in thinking light is a wave" justified? Maxwell's theory seems but one way to see light as a wave, it seems to me.

8. Aug 14, 2011

### y33t

If you consider one duty cycle, frequency is not directly related to energy in classical electromagnetics. As others stated, intensity is proportional to square of the electric field and is directly related to electron density flow. Frequency oscillation clock (acceleration) of electron movement is the frequency.

From quantum approach, E=h.v where v is the frequency of the individual photon and h is Planck constant. This expression gives the energy of a single photon. Classical electromagnetics govern electric and magnetic field vectors behaviors of a radiation where QED governs behaviors of individual photons.

Imagine a waveform belonging to an electric field. Divide the whole wave to almost infinitesimally small parts and analyze each part. More you increase the resolution more you will get close to extracting behavior of individual photons. QED is the quantized version of Maxwell equations.

9. Aug 14, 2011

### xts

It is not applicable to single-photon/single-electron interactions. But it is perfectly suitable statistical approximation for large-number-of-photons/many-(or-even-single)-electrons cases.

I owe my favourite view to Anton Zeilinger: "think about light as photons (snooker balls) when they interact, think about waves, as they propagate"

10. Aug 14, 2011

### atyy

Maxwell's equations are still correct - they just need to be quantized.

Incidentally, some aspects of the photoelectric effect don't require light to be quantized. They just need atoms to be quantized, and the classical Maxwell equations can still be used. http://www.physics.rutgers.edu/ugrad/313/Lamb Scully Photoelectric effect.pdf

A phenomenon that requires the quantum Maxwell field is the Lamb shift.

11. Aug 15, 2011

### nonequilibrium

Thank you all.

Can you enlighten me on what you mean with this?

Well that's a weird definition of "correct". Of course my point was not that they were totally wrong, but simply that they were wrong in some respect and that I did not get how upon establishing a (possibly small) error in the Maxwell equations, one goes to postulate light is actually a particle.

Ah the fact you went on to say this kinda implies you did get what I mean. Brilliant paper! I was struggling to see how the photo-electric effect logically implied the existence of photons. Good to see someone has proven it isn't. And also good to know there are experiments that absolutely require photons ;) So if I understand it correctly: the argument that you measure (low-intensity) light by clicks like a Geiger counter isn't conclusive to regard photons like particles, correct? (because the only "photon-counter" I know is the PMT which works on the principle of the photo-electric effect, unless I'm mistaken)

12. Aug 15, 2011

### Philip Wood

Thank you very much for the link to Lamb's paper. I reckon that, in the UK, the PE effect is cited as evidence for photons by almost all Physics teachers (pre-university, at least).

However, the PE effect as taught to students is (almost always) about emission from metal surfaces. Here we have a sea of electrons, whose pre-ejection energy levels are extremely closely spaced in the conduction band.

Do you think that Lamb's model of emission from individual atoms with discrete (bound state) energy levels is still a good enough one for metals, and that his argument still works for metals?

13. Aug 15, 2011

### ZapperZ

Staff Emeritus
We have had this discussion numerous times already, and I am a bit annoyed that people still use and try to "falsify" the standard photoelectric effect description into scenarios where it doesn't apply!

As I had said many times, and as Phillip Wood has mentioned, applying it to atoms or atomic gases is a bit silly because the phenomenon that is relevant is the photoemission of electrons from metallic photocathodes! So when you apply the rules of baseball to soccer, and then say that the rules don't quite work, who is to blame here?

I will flat out say that I violate the Einstein's photoelectric effect REGULARLY in my experiments! I get photoelectrons out of metallic photocathodes regularly using photons with energy below the work function, something that the photoelectric effect equation say cannot happen! Yet, do you see me jumping up and down and publishing papers saying that the photoelectric effect model is "wrong", or not needed? NOPE! Why? This is because I know the extent of the standard photoelectric effect model, and what I'm doing that actually goes beyond the scope of that model.

Let's stick to the standard photoelectric effect scope unless the whole topic is on the scenario that this model does not apply to.

Zz.

14. Aug 15, 2011

### kith

I don't think this is about falsifying the standard description. To my knowledge, the photoeffect can be explained either by using photons and a classical solid or by using classical waves and a quantum mechanical solid (and obviously by treating both parts quantum mechanically). Is this wrong?

15. Aug 15, 2011

### Philip Wood

So, ZapperZ, I take it that your answer to the question in the last para of my post (hash 12) would be 'No'. [No sarcasm intended, just wanted to be clear.]

16. Aug 15, 2011

### ZapperZ

Staff Emeritus
The standard, naive photoelectric effect is a strong indication of the photon model, but it cannot rule out the wave model completely. One can invoke some SED model and the like to arrive at the same description.

But as I've said already, that's like seeing a cow from very far. A model approximating it as a sphere would produce as good of an answer as the more accurate model that describes the actual cow. It is when one examines it closer do you detect the differences, or where the spherical cow model has no description. I've already mentioned previous various photoemission phenomena (multiphoton photoemission, resonant photoemission, etc.) that are devoid of any other description other than the photon model.

Correct.

Zz.

17. Aug 15, 2011

### nonequilibrium

ZapperZ - un-be-lievable! I took such care by stating more than once that I have no intention of saying Einstein was wrong and that light is actually a wave, all I'm interested in is in understanding why the photo-electric effect is (not?) a good indication for the particule-nature of light

Just as you are getting irritated by people making false claims to have overthrown Einstein, I am getting irritated that people are so quick to attack anybody that asks reasonable questions instead of just accepting what they're told

18. Aug 15, 2011

### ZapperZ

Staff Emeritus
Er.. hello? Let me emphasized what I said in the first sentence of my reply:

Unless I misread something, YOU DID NO SUCH THING, i.e. you were not the one who brought up the reference to a "photoelectric effect from atoms", did you? So how could my response be directed at your question?

Zz.

19. Aug 15, 2011

### nonequilibrium

Oh okay then, my apologies.

I'm not familiar with a SED model. Google tells me it's short for "spectral energy distribution"? I'm not seeing the link straight away?

What I'm mainly interested in is understanding/realizing how much the maxwell wave model must be changed in order to explain the photo-electric model.

Even more concrete: I'd like to convince my laymen friends of why the photo-electric effect is a strong (as you say) indication of the photon model. The normal treatment it is given in an introductory modern physics course seems unsatisfactory: it is simply assumed waves cannot exhibit such features. In the article you object to (on the previous page) it is argued that the photo-electric effect is not a strong indication. But you say it is, given you're looking at metal surfaces. It seems daft of them to have forgotten that; can there be more at play?

20. Aug 15, 2011

### kith

So the model photons + classical solid gives better agreement with the experiment than the model electromagnetic wave + quantum solid? If not, I don't see your point.

I think, both descriptions are possible. If I have an an incident classical electromagnetic wave and use quantum states for the solid, increasing the intensity should increase the dipole transition rates and thus lead to more elctrons being emitted. Increasing the frequency should lead to new possible final states in the free continuum, corresponding to electrons with higher kinetic energy. This is the correct behaviour, isn't it?

Nobody here wants prove the standard photo-electric effect description wrong and everybody acknowledges that there are experiments, which cannot be explained without QED. It would just be an interesting fact, that the photo-electric effect could be explained without photons. It is widely believed and teached, that it was the first phenomenon, where photons were actually needed. And this seems to be wrong.

Last edited: Aug 15, 2011