Photo-electric effect: why did they expect this?

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The discussion centers on the photoelectric effect and the relationship between light intensity and electron ejection velocity, questioning how wave theory can explain this phenomenon. Participants explore the differences between classical electromagnetic theory and quantum mechanics, noting that while Maxwell's equations describe light as a wave, the photoelectric effect suggests a particle-like behavior of light. There is debate over whether Lorentz's force applies to photon-electron interactions and how the classical wave model can coexist with the quantum photon model. Some argue that the standard photoelectric effect model is often misapplied to scenarios that don't fit its parameters, emphasizing the need to adhere to its original scope. The conversation highlights the complexities in reconciling classical and quantum theories in understanding light and its interactions with matter.
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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?
 
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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.
 
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?
 
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.
 
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.
 
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.
 
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.
 
mr. vodka said:
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?

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.
 
mr. vodka said:
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.
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.

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.
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
mr. vodka said:
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.

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
Thank you all.

y33t said:
Frequency oscillation clock (acceleration) of electron movement is the frequency.

Can you enlighten me on what you mean with this?

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

atyy said:
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
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
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
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
ZapperZ said:
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!
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
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
kith said:
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?

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.

Philip Wood said:
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.]

Correct.

Zz.
 
  • #17
ZapperZ said:
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.

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
mr. vodka said:
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

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

Zapperz said:
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!

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
Oh okay then, my apologies.

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.
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
ZapperZ said:
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.
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.
 
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  • #21
mr. vodka said:
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?

I will have to do an extensive search on the sources from several years ago.

The only one that I have on hand easily is the J.J. Thorn paper that has a couple of references, including the Lamb's paper.

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

Edit: Try Tim Boyer's paper Phys. Rev. D 11, 790–808 (1975)

kith said:
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.

Classical solids?

The don't give better agreement in the standard photoelectric effect. It is only when you stop considering, say, standard metals, or when you start using single-crystal metals, then in experiments such as angle-resolved photoemission (ARPES), I have never seen the electromagnetic wave model (and this is on a quantum solid) being used for that phenomenon. If someone has a reference where such a thing is used, I'd love to see it.

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?

But in the classical system, increasing the frequency while keeping the amplitude constant does NOT increase the energy per cycle. It only increases the energy per unit time. If you argue that that's the same as increasing the amount of energy given to the solid, then you are arguing that one can accumulate enough energy over a period of time to cause emission. This then will not be consistent to the experimental observation of a cut-off frequency below which, no emission can be seen no matter how long one waits for 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.

I don't see this as being a problem because historically, that WAS the case (there are many other issues we can pick on if we want to be picky about such accuracy - relativistic mass being another). If it is well-known to have a classical explanation, it wouldn't have been such a puzzle back then. Ironically, the photon picture that is partly responsible for the birth of QM is required for the solid involved in the classical wave explanation of the photoelectric effect!

Zz.
 
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  • #22
ZapperZ said:
But in the classical system, increasing the frequency while keeping the amplitude constant does NOT increase the energy per cycle. It only increases the energy per unit time. If you argue that that's the same as increasing the amount of energy given to the solid, then you are arguing that one can accumulate enough energy over a period of time to cause emission.
I don't claim that. Increasing the energy per unit time should lead to electrons with higher mean kinetic energies beeing emitted. If we keep the amplitude constant, this simply means that there will be fewer electrons. This is equivalent to the standard description, where the photon number has to be lowerd if we want to keep the beam energy constant while increasing the frequency.

ZapperZ said:
I don't see this as being a problem because historically, that WAS the case (there are many other issues we can pick on if we want to be picky about such accuracy - relativistic mass being another). If it is well-known to have a classical explanation, it wouldn't have been such a puzzle back then. Ironically, the photon picture that is partly responsible for the birth of QM is required for the solid involved in the classical wave explanation of the photoelectric effect!
I agree, that this is kind of pedantic and I acknowledge that the photon explanation from Einstein has played a major role in the development of the formalism, we can use today to describe the photo-electric effect via classical waves. I still find it valuable to point out this alternate explanation.
 
  • #23
  • #24
kith said:
I don't claim that. Increasing the energy per unit time should lead to electrons with higher mean kinetic energies beeing emitted. If we keep the amplitude constant, this simply means that there will be fewer electrons. This is equivalent to the standard description, where the photon number has to be lowerd if we want to keep the beam energy constant while increasing the frequency.

But this isn't what happened. Increasing the energy per unit time by simply increasing the amplitude doesn't do anything to the mean kinetic energy.

Zz.
 
  • #25
ZapperZ said:
But this isn't what happened. Increasing the energy per unit time by simply increasing the amplitude doesn't do anything to the mean kinetic energy.
I didn't say anything about increasing the amplitude in the post you cited. I talked about increasing the frequency, which was identified with increasing the energy per unit time by you. I don't really get what you want so say. If you think my explanations are wrong, please provide the correct semiclassical description (electromagnetic wave + quantum states).
 
  • #26
kith said:
I didn't say anything about increasing the amplitude in the post you cited. I talked about increasing the frequency, which was identified with increasing the energy per unit time by you. I don't really get what you want so say. If you think my explanations are wrong, please provide the correct semiclassical description (electromagnetic wave + quantum states).

The two ways to increase "energy per unit time" of a classical wave is either by increasing frequency, or increasing amplitude.

1. Increasing frequency.

I already argued this in a previous post. You increase energy per unit time, but you cannot increase the energy PER CYCLE (keeping amplitude constant). But by saying that ("increase energy"), it means that you count and accumulate the energy in each cycle. This means that eventually, as you accumulate enough energy, you should liberate a photoelectron, regardless of the work function.

This is not true and not what happens.

2. Increasing amplitude

Increasing amplitude without increasing energy clearly has been shown to not change the KE distribution of the photoelectrons.

Both scenarios using classical wave do not fit observations.

Zz.
 
  • #27
I don't have time to answer for a few days, but I don't consider this discussion to be over. ;-)
 
  • #28
ZapperZ, my apologies for the long delay but now I'm back.

To finish this more quickly, I'll start again with your first objection:
ZapperZ said:
But in the classical system, increasing the frequency while keeping the amplitude constant does NOT increase the energy per cycle.
I don't see how this relates to my explanation. Increasing the frequency of the wave leads to a higher final state in the free continuum beeing in resonance with the wave, and therefore to a higher kinetic energy of the electrons. Why do you think the energy per cycle is relevant for this?
 
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