# The Photoelectric Effect

1. Dec 15, 2007

### jumbogala

My physics class is currently studying the photoelectric effect, and I was wondering if anyone could clarify this for me:

"Only photons of a high-enough frequency, (above a certain threshold value) could knock an electron free. For example, photons of blue light had sufficient energy to free an electron from the metal, but photons of red light did not. More intense light above the threshold frequency could release more electrons, but no amount of light below the threshold frequency could release an electron."

What is it about the more intense light that gives it the capability to release more electrons?

2. Dec 15, 2007

### cesiumfrog

There's nothing terribly deep here. If you halved the amount of light (eg. with a half silvered mirror: the light getting through is dimmer) you'd expect to halve the rate of electron emission, right?

Last edited: Dec 15, 2007
3. Dec 15, 2007

### mjsd

an electron can only be knocked off by a particular photon of high enough energy. More intense light, means there are more photons (each with high enough energy) per unit time, and so more electrons may be knocked off per unit time.
when the energy is below the threshold, it doesn't matter how many photons you have per unit time, none of them will be able to knock any electrons out.

the fact that there is a threshold frequency demonstrates the particle-like nature of light. it you think about it, the "wave-model of light" does not predict this outcome.

4. Dec 15, 2007

### monish

I'm going to object to this statement. How does the wave model fail to account for the frequency threshold? It is well known that all atomic transitions have a characteristic coupling frequency (from the Schroedinger equation) including those from the bound state to the free states. Doesn't this suggest why light might need to have a certain frequency in order to be effective?

I think that I especially object to the idea that you can debunk the wave theory of light by just "thinking about it". Wouldn't you need to know how to do some pretty sophisticated calculations in order to detemine what the wave theory of light would predict in this experiment?

Marty

5. Dec 15, 2007

### mjsd

I was talking about it in terms of why increase in light intensity cannot help liberate any electrons if the frequency of light is not high enough. Using the wave-model, and consider how energy is transferred, one would immediately expect that if you shine light on the plate for long enough, you would be able to liberate the electrons eventually as energy coming from the light waves build up. And certainly, that also implies higher intensity light would make this happen faster.

The fact that this doesn't happen means that photoelectric effect cannot be explained by the wave-model. And one must turn to the particle-model where electrons and photons interact like billard balls. In this picture, only billard balls with high enough energy (determined by the threshold frequency) can kick the electron out. If none of them are with high enough energy, no electrons can be kicked out...no matter how many of them there are or how long you shine the metal with the source.

sure, but "thinking about it" gives you a good intuitive start.

such calculations are not usually shown to students when they first learn about photoelectric (which I suspect that the OP is in that category). I have already given you a reason above as to why the wave-model fails in this occasion. One can go on and do a calculation on it but the conclusion should be fairly clear.

6. Dec 15, 2007

### ZapperZ

Staff Emeritus
WE need to make sure we answer and address this question within the CONTEXT and the level of expertise of the person asking this question. Otherwise, we can get into a whole bunch of confusing and contradictory responses that do nothing to help the OP. I can, for example, go on and on on how to "violate" the Einstein's photoelectric effect description (I've done it myself experimentally). But such "intrusion" into what the OP had asked will only add a level of confusion that is not helping in answering the question.

While there CAN be a quasi-classical description that is consistent with the photoelectric effect (which means that the photoelectric effect is no longer the definitive "evidence" for the photon picture), we also know for certain that currently, there are no classical means of describing the observations derived from angle-resolved photoemission, resonant photoemission, core-level photoemission, multiphoton photoemission, etc... etc. All of the physics, and experimental results using those techniques, are exclusively described via the photon description. Thus, until such classical descriptions can account for everything that the photon picture can, we tend to accept that this is a more valid description of such phenomena. And within that context, it isn't unrealistic to accept the photoelectric effect as a strong evidence for the Einstein's photon picture.

Zz.

7. Dec 16, 2007

### monish

What doesn't happen? Don't you liberate more electrons the longer you shine the light? and doesn't higher intensity make this happen faster?

I think these things do happen. According to you, isn't that consistent ("one would immediately expect") with the wave model?

I don't follow your reasoning. Why don't you tell us what calculation you're thinking of? You don't need to do the calculation with numbers, just tell us what inputs you're thinking of and what is the general drift of it?

8. Dec 16, 2007

### mjsd

Did you read ZapperZ's post? Firstly, be aware that the situation can be a lot more complicated. Needless to say, to be "absolutely correct" (no such thing by the way, you can only be "more correct"), you would have to think a lot harder than what I shall present below. But at the level that I have been discussing this issue, a few intuitive ideas could suffice in constructing a fairly consistent view. Let me try again.

What doesn't happen:
when the incoming light is of frequency lower than the threshold, then it doesn't matter how long you shine the light nor what is the intensity, no photoelectrons are emitted (this is an experimental observation)

what would a wave-model predict:
firstly, the argument I used last time was based on the idea of "energy transfer". I thought you would understand what that means. Anyway, a (classical) wave-model would predict that even when frequency is below the threshold, if you shine light on the metal plate for long enough, electrons should be liberated, and the higher the intensity of the source, the less time this process would take. Also, this implies that the kinetic energy of the photoelectrons should be a function of the light intensity (which is not what we see in experiement, in fact kinetic energy of photoelectrons are function of incoming light's frequency).

ok, given these observations, let's go back and talk about how does a classical wave transfer energy. Clearly if you stand under the sun for too long, you will "heat up" . Why? the energy travelling from the sun in the form of EM radiation continuously strikes your body (the parameter here is in energy flow per unit area which can be described by the Poynting's vector) and you gradually absorbed more and more of it. If it is on a cloudy day (ie. less intensity), it will take longer for you to "heat up" to the same level. If you are an electron residing inside some metal, you will expect that the longer you are being shone on, the more energy you will absorb and eventually gain enough energy to get out of the metal.

A classic example of such "build up of energy" can be illustrated by a simple experiment like producing a sound wave of the correct frequency near a glass bowl/cup until it shatters. The correct frequency of course is the resonant frequency (fundamental or its harmonics) of the bowl/cup. The loudness of the sound is what we called "intensity" of the wave (or amplitude). So, if you turn the sound on and eventually the glass bowl/cup will break (well.... we did an experiment on this to show the students this idea, but we did choose a type of glass that wouldn't break as easy... so that we won't have to keep buying new glass for every class , but one could clearly see the oscillation of the bowl slowly becoming more violent).

Anyway, the idea here is clear: energy from the sound wave is transfered to the bowl very effectively (because it is at resonant frequency), and the vibration builds up and ultimately breaking the bowl. Should the amplitude of the sound wave is bigger, this will happen much faster. Mind you, you may turn the amplitude down to very tiny value, but eventually it will gain enough energy to break it. This is an example of energy transfer via a classical wave.

The kind of calculations you may be thinking of here is firstly work out how much energy per unit area is being transfered per unit time, and then see how many electrons are there per unit area... plus few other stuffs like how does the electrons absorb and redistribute the energy within the lattice etc. All done assuming energy is transferred as a continuous wave medium. Not going to be an easy calculation, but since all we are after is the question: would any of the electron receive enough energy to be liberated? Therefore, from the glass bowl analogy we would expect the wave-model to predict that as long as we wait long enough, it will absorb enough energy.

Now, the fact that a wave-model seems to predict a different story for the photoelectric effect, it is an indication that another model is needed,.. it turns out that the particle model works fine. Note that the resonant frequency is not like the threshold frequency we were talking about. The former comes in discrete values, while the latter is like a cutoff value above which all frequencies will work.

what would the particle model predict:
now, here each photon is like a billard ball carry energy defined by its frequency. When such billard ball of enough energy strikes an electron, energy is transferred to from the photon to the e- and so the e- is liberated. The idea here is that if the photon does not carry enough energy, the e- will not gain enough to be freed. And it doesn't matter how many of such photons are around, you still can't do it. But why? couldn't you have two or more of such low energy photons hitting the same electron in a short space of time (so that the e- hasn't lost the energy from the previous photon strikes through vibration in the lattice) and hence eventually building up enough energy just like in the wave-model? The difference here lies on the fact that it is now a discrete source rather than a continuous source like in the wave-model. So, the calculation can now become a bit more technical where you will have to take into account of the probability of hitting the same e- within a certain time frame. Needless to say some ppl have tried doing this calculation before (with various assumptions, otherwise it is probably intractable), and as far as I was told by the lecturer I was working with, the probability that this mechanism can successfully build up enough energy for a particular e- is extremely low. In fact on average, you would have to wait longer than the age of the universe to see it.

The moral is that in the photoelectric effect, light behaves more like a particle than a wave. And the existence of a threshold frequency (and other observations) is indicative of that.

Last edited: Dec 16, 2007
9. Dec 16, 2007

### malawi_glenn

"What doesn't happen? Don't you liberate more electrons the longer you shine the light? and doesn't higher intensity make this happen faster?"

yes if the energy of the photons are above the energy threashold, the number of electrons emitted per time is prortional to the intensity of the light (the number of photons per area per time).

Just as bullets hitting a target, making a mark on it. But if the bullets lacks energy to make a mark (penetrate/distort the targets surface), then no marks will be created, no matter how big your intensity of bullets is.

The wave-model of light, the electron can sit and gather energy from the wave, til it has enough to leave the atom.

10. Dec 16, 2007

### monish

If you are concerned about not misleading the OP, why do you object to my pointing out the truth?

You are going farther than me so far. I only said that a simple analysis of the photoelectric effect fails to disprove the wave theory. You seem to be saying that
the wave theory can actually provide a valid explanation of the experiment. If this is true, then why shouldn't the OP or anybody else know about it?

Speak for yourself. You may know these things for certain, but I don't.

Why don't you take just one of the many examples you cite above and explain why it succeded in debunking the wave theory of light where the traditional photo-electric effect fails? I would be interested in seeing a valid physical argument for the existence of light particles.

Wouldn't it be more realistic, and more correct, to simply say that while Einstein's explanation of the photo-electric effect was historically important in the evolution of our ideas, the fact is that there is no clear explanation at the undergraduate level which decisivley rules out the wave theory of light?

11. Dec 16, 2007

### malawi_glenn

monish: Can you demonstrate that "a simple analysis" of the photoelectric effect fails to disprove the wave theory? I cant get it, and have studied much quantum physics.

12. Dec 16, 2007

### ZapperZ

Staff Emeritus
What "truth"?

All I said is that within the simple setup of the photoelectric effect, it is OK to indicate that this is a strong argument for the photon picture. It is now up to you to show that there is validity in what you are claiming. Show, for example, the evidence that below threshold, a low intensity light source can, in fact, produce a photoelectron with any detectable probability as you increase the time of exposure.

Because of exactly what I have cited as a series of experiments that have no explanations using the wave picture! Think about it. You have one experiment that points strongly towards the photon picture, but cannot completely rule out the wave explanation. But then, as you make a more complex type of experiment, the wave picture completely has no such explanation. Furthermore, all the which-way experiments that are independent of the photoemission family of experiments rules out the wave picture. Such knowledge is the justification for what I had claimed. It is now up to you to justify why you think your view should be considered on equal grounds.

And that is why I made an effort to inform you. When you make such claim, but you are not in full awareness of the full body of knowledge, then you should consider that your "conclusion" or claims could be based on some degree of ignorance of what we do know now. I have not only spent time designing undergraduate laboratory exercise in the standard photoelectric effect experiment, I have also done research work using angle-resolved photoemission spectroscopy, and also on schottky-enabled photoemission. So I am not just talking about these things simply based on some superficial knowledge that I read somewhere without actually doing these things. It is based on that knowledge that I claim that the standard photoelectric effect experiment (technically considered as a single-photon photoemission that ignores band-structure effects) loses no generalities if it is exclusively treated as a strong evidence for the photon picture based everything we currently know.

You mean other than the fact that none of the experiments that I've listed have any wave description being used? (For most people, that is a glaring enough evidence). Look at the photocurrent produced as a function of intensity in a multiphoton photoemission measurement[1,2]. The change in the slope of the photocurrent occurs in discrete amount depending on whether 2, 3, 4, etc... photons are involved in the photoemission process. Wave picture has no means of describing this, and people who advocate such picture are strangely silent in attempting to do such a thing, as is the case for all the experiments that I've mentioned.

The which-way experiment has become almost an easy experiment done at the undergraduate level. In fact, if someone wants to do a search on here, I've highlighted one such experiment published in AJP that was done in an undergraduate lab[3].

And I haven't even started talking about single-photon sources yet.

Zz.

[1] K. Giesen et al., Phys. Rev. Lett. v.55, p.300 (1985).
[2] W.S. Fann et al., Phys. Rev. B v.44, p.10980 (1991).
[3] J.J. Thorn et al. Am. J. Phys. v.72 p.1210 (2004).

Last edited: Dec 16, 2007
13. Dec 16, 2007

### monish

OK. You've said a lot of things in your last post which I'd like to respond to but first I think we should clear up one misunderstanding.

mjsd said: "the wave theory of light does not account for this outcome".

I said: "I'm going to object to this statement".

YOU said: "there CAN be a quasi-classical description that is consistent with the photoelectric effect (which means that the photoelectric effect is no longer the definitive "evidence" for the photon picture)..."

At this point I thought you were agreeing with me. But you thought it was wrong for me to raise this point because it would "confuse" the OP. So I said:

"If you don't want to confuse OP, why do you object to me pointing out the truth?"

By "truth", I was only referring to what I thought you already agreed with: that the photo-electric effect does not in itself rule out the wave theory of light. Isn't that what you said?

14. Dec 16, 2007

### Peter Morgan

To attempt a different way to discuss the relevance of the photoelectric effect and other experiments that involve light: They identify the experimental regimes in which we can use quantum optics descriptions of experiments effectively.
There are numerous experiments that involve light that can be described easily using classical electromagnetism, there are numerous others that require rather more ingenuity for them to be described using classical electromagnetism, and there are some that have so far resisted any description using classical electromagnetism. Greenstein&Zajonc's book, "The Quantum Challenge: Modern Research on the Foundations of Quantum", is a very good undergraduate level resource.

The lack of a nice description using only classical electromagnetism, so far, does not mean there will never be such a description, but classical electromagnetism has to do the job. Here, I take "nice" to mean that the classical electromagnetism model must be as easy to use as quantum optics formalisms are, for engineering and other technological purposes. It's not sufficient to say that classical electromagnetism can produce a model, if that model is horrible -- that's the lesson that the existence and de facto rejection of de Broglie-Bohm models teaches us. That criterion also goes a long way to ruling out Stochastic Electrodynamics (SED) as a usable model for light, even without its other failings, but note that the notion of Lorentz invariant vacuum fluctuations that enters into SED is almost certainly essential as a feature of classical electrodynamic models for experiments in the quantum regime.

The killer for the discussion here (which, as ZapperZ points out, sadly ignores the OP) is that the effectiveness of quantum optics in Physics and in Engineering does not warrant talk about particles or about waves. Insofar as quantum optics is anything, it is an algebra of (noninteracting field) observables. The Hilbert space of states (in the vacuum sector) over the noninteracting field algebra has a discrete structure determined by the particle number operator, but that does not properly extend to an interacting field, nor to a noninteracting field that interacts with optically active physical objects, and in any case particle number does not warrant talk about photons without at least mentioning that they are bosons, so that they are not at all like classical particles. No.

There is a level, appropriate to the OP, at which post #2 gets it right: if there's less incident energy, there will be less photoelectrically emitted energy. Given the phrasing, and highlighting, of the OP, that seems a reasonable answer. Perhaps not much more to say unless Jumbogala would like to ask a supplementary question.

If we take the unhighlighted part of the OP to task, thereby revisiting all of the foundations of quantum theory, then I guess there's more to say. Restricting myself, however, there is one source of possible confusion that seems to leap out, that higher energy light is supposed to have higher frequency, $$E=h\nu$$ and all that. And yet a light source can also deliver more energy per second by being more "intense". There are two measurements of the light source that affect the intensity here, taking intensity to be energy per second, both "the frequency of the light" and the "number of photons per second". So Jumbogala, my suggestion is to keep those two separate. I hope that helps. Remember, however, that "the number of photons per second" and the "frequency of the light" are quantum mechanical observables of the beam of light that is being used in the experiment, so they aren't anything like properties of anything like a stream of classical particles.

15. Dec 16, 2007

### ZapperZ

Staff Emeritus
And it doesn't, at least as far as our STANDARD wave theory, as practiced by the beginning of the 20th century. If this is such a common knowledge and so well-known, then why the heck couldn't they explain both the BB radiation and the Photoelectric effect phenomenon back then?

The ability of the "classical wave picture" to weasel out of this is by invoking, as Peter Morgan has mentioned, some 'stochastic' processes, and by also invoking the band structure of materials, which is obtained using quantum mechanics in the first place! This is not completely kosher to me. Thus, for a student learning the straight-forward classical wave theory and not knowing any QM, I do not see anything wrong with what has been said about wave theory not being able to account for such an outcome. It was what physicists understood back then, and it is still the similar wave theory that the student has been taught.

Now if you want to argue that we should teach a student just learning about such-and-such physics EVERYTHING that we know of today, then you need to figure out if this has more to do with a pedagogical need, or simply something to satisfy some other reason that has nothing to do with the process of learning. Do you also object the use of the simple conservation of energy principle in intro physics? Or what about the "validity" of Thermo's Second Law, especially under non-equilibrium situation?

There are many instances in physics where what we present in basic intro physics that aren't quite right as one pushes the envelope further. Most of us learned physics this way, and it still didn't 'brainwashed' us into sticking to them as we learn more. So I do not see the purpose of all this.

I still want to know where you got the idea that below threshold and using low intensity light, you can get detectable photoelectrons in a standard photoelectric effect experiment. Can you please give me a reference for when this was shown to occur?

Zz.

Last edited: Dec 16, 2007
16. Dec 16, 2007

### monish

I don't think I'm going way out on a limb here. I was referring to msjd's comment that the threshold frequency demonstrates the particle nature of light. I simply pointed out that the origin of frequency dependencies in general is well understood based on the Schroedinger equation for the electron. To prove that you can't get this result with classical e-m, wouldn't you have to solve the Schroedinger wave function of an electron in a metal lattice, and then analyze its interaction with a time-varying electric field?

If you did all this work, isn't it possible that you might get the same answer as given by experiment? And even if you got the wrong result, thereby supposedly disproving the wave theory, I wouldn't call it a "simple analysis".

So to repeat my statement, I don't think you can disprove the wave theory of light with any kind of simple analysis of the photoelectric effect. I therefore strongly objected to mjsd's advice to the OP that he could arrive at this conclusion by just "thinking about it". (And I don't think you can get useful insights from his wine-glass analogy either.)

17. Dec 16, 2007

### DaTario

Let me gently put my spoon inside this pudim.

In some sense it is reasonable to think of photoelectrons being detected with low intensity under threshold light.

One has only to agree that:
1) there is no perfect monochromatic radiation
2) there is possibly no beam of radiation with spectrum completeley depleted of thermal effects which allow for a broad band content and the corresponding phenomena.
3) there is experimental evidence for high order processes in which low frequency photons join to give birth to processes which can only be obtained through high frequecy photon.
4) experimentalists should be more careful in analysis when the physical statement under consideration is that "something doesn't exists because it hasn't been detected". Of course, "Existence because it was detected" is another class of statement which also deserves carefully analysis, but it is another subject.

best wishes

DaTario

18. Dec 16, 2007

### monish

Did I say that?

19. Dec 16, 2007

### ZapperZ

Staff Emeritus
In post #7:

Zz.

20. Dec 16, 2007

### ZapperZ

Staff Emeritus
If this is true, then the photoelectric effect would not have been such a mystery, and there would be zero impetus for stating the below threshold condition in the first place since such absence of photoelectrons would not have occurred. Considering back then the quality of light sources is worse than what we have now, the would tend to NOT have a cleaner signal than what we have now.

Yet, history has shown that this is exactly what had happened, and the photoelectric effect was a mystery. You may also want to read Milikan's paper that clearly verified exactly the Einstein model, despite the fact that Milikan himself was skeptical of the model and set out to disprove it.

Zz.