Classical Explanation for Photoelectric Effect

[musashi1029]

Hello.

I have a question regarding photoelectric effect. My textbook says that when photoelectric effect experiment was first performed, physicists could not explain the outcome with classical physics (e.g. maximum kinetic energy depends on frequency of light, not intensity). My question is, wasn't there even one thing that could be explained with classical physics about the results of this experiment? If anyone knows there was, please explain it to me.

Thank you in advance.

 

[mgb_phys]

Classically you can explain light knocking out an electron,what you can't explain is how you need just one photon of high enough energy to do the job when billions of slightly lower energy put together can't.

 

[atyy]

The original explanation was that light consisted of discrete particles called photons.

This turns out not to be necessary, and it is possible that it is just the energy levels in the atoms that are discrete.

So classical physics cannot explain the experiment, but you can choose to put the non-classicality in the atoms or the light.

However, the photon concept turns out to be necessary for many other phenomena, so it is still useful to regard photons as a correct explanation of the photoelectric effect.

 

[musashi1029]

Classically you can explain light knocking out an electron,what you can't explain is how you need just one photon of high enough energy to do the job when billions of slightly lower energy put together can't.

I see. So only the thing they could explained with classical physics was that light could knock out electrons out of matter under some circumstances. Thank you very much for the very concise answer!

The original explanation was that light consisted of discrete particles called photons.

This turns out not to be necessary, and it is possible that it is just the energy levels in the atoms that are discrete.

So classical physics cannot explain the experiment, but you can choose to put the non-classicality in the atoms or the light.

However, the photon concept turns out to be necessary for many other phenomena, so it is still useful to regard photons as a correct explanation of the photoelectric effect.

I see, so without photon theory explanation is very tough. Thank you!

 

[musashi1029]

Classically you can explain light knocking out an electron,what you can't explain is how you need just one photon of high enough energy to do the job when billions of slightly lower energy put together can't.

Actually, would you mind explaining to me why "billions of slightly lower enegergy put together" cannot cause electrons to leave matter?

 

[ZapperZ]

Hello.

I have a question regarding photoelectric effect. My textbook says that when photoelectric effect experiment was first performed, physicists could not explain the outcome with classical physics (e.g. maximum kinetic energy depends on frequency of light, not intensity). My question is, wasn't there even one thing that could be explained with classical physics about the results of this experiment? If anyone knows there was, please explain it to me.

Thank you in advance.

You may want to read the intro in J.J. Thorn et al. paper and references therein. You can get a copy of it here:

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

The standard photoelectric effect strongly points in favor of the photon picture, but cannot rule out light being a classical field. It is other phenomena (photon antibunching, multiphoton photoemission, which-way experiments, etc.) that would make the convincing evidence for the photon picture.

I'm a bit hesitant to point this out since it may complicate things beyond what you've asked for and what you wish to know, but what the hey... The Einstein photoelectric effect, at least the naive version of it, http://physicsandphysicists.blogspot.com/2007/11/violating-einsteins-photoelectric.html" , even when using the photon picture. I'm mentioning this because of the point being made that "... billions of slightly lower energy put together..." cannot produce a photoelectron. This is only true for the single-photon photoemission, which is what the standard photoelectric effect is. We have plenty of observations of multiphoton photoemission using photons with energy LESS than the work function.

Zz.

 

[atyy]

I'm a bit hesitant to point this out since they may complicate things beyond what you've asked for and what you wish to know, but what the hey... The Einstein photoelectric effect, at least the naive version of it, http://physicsandphysicists.blogspot.com/2007/11/violating-einsteins-photoelectric.html" , even when using the photon picture.

I remember asking my high school teacher about 20 years ago about two-photon effects in the photoelectric effect, and was told that if the intensity is high enough you see a second peak, exactly as expected (she didn't tell me about selection rules, but still a pretty good answer I think). More recently I read that Einstein himself mentioned two-photon effects in his paper. I haven't read the paper myself, and I can't quite remember what the source is ...

 

[musashi1029]

>ZapperZ, atyy
I see. So if the light intensity is high enough, an electron can obtain enough energy from multiple photons with energy lower than work function before the energy decays. It is very clear now, thank you so much!

 

[mgb_phys]

Not an expert but I thought you only have a 2photon photo electric effect in rather non-classical setups like nano-particles a surface plasmon or some metastable state.

I don't know that you can hit a normal electron with two photons and eject it.

 

[Redbelly98]

It can also happen with more than 2 photons, at least in the case of ionizing atoms.

http://www.google.com/search?hl=en&safe=off&q=multiphoton+ionization+nonresonant&btnG=Search

 

[granpa]

the photoelectric effect experiment has been done with extremely low light intensities. so low that it should have taken a single atom minutes to absorb enough energy to emit an electron. instead there is absolutely no noticable delay before electrons begin to be emitted by the metal. this does not rule out the possibility that light is a wave but of so would require that some of the atoms somehow start off with some random amount of energy already within them.

 

[malawi_glenn]

isn't this a pretty old thread?

 

[granpa]

you want me to start a new one?

 

[malawi_glenn]

Well the discussion about it seemed quite dead, OP's last activity was december 2008.

 

[cesiumfrog]

the photoelectric effect experiment has been done with extremely low light intensities. so low that it should have taken a single atom minutes to absorb enough energy to emit an electron. instead there is absolutely no noticable delay before electrons begin to be emitted by the metal. this does not rule out the possibility that light is a wave but of so would require that some of the atoms somehow start off with some random amount of energy already within them.

To clarify: Are you just arguing this proves the energy is concentrated at points that are randomly distributed over the beam's cross-section, rather than being uniformly spread over the beam cross-section (whilst making the assumption that the electron can only absorb power from its immediate vicinity and not the entire beam cross-section)? Or that it proves the energy is also longitudinally concentrated (in which case I take it the average delay matches a classical prediction but for some runs of the experiment the delay is longer or shorter)?

Regards your concession: wouldn't the frequency dependence (the total non-response to any duration of redder light, and the increasing excess kinetic energy with bluer light), etc, disprove the hypothesis of the atoms having random pre-accumulated quantities of energy?

Regards the competing model: is this behaviour truly not also predicted by the semi-classical approach of considering a quantum mechanical atom in a classical electromagnetic field?

 

[Bob S]

Classically you can explain light knocking out an electron,what you can't explain is how you need just one photon of high enough energy to do the job when billions of slightly lower energy put together can't.

The energy of oxidation of a simple sugar molecule C₆H₁₂O₆ is about 29 eV (electron volts). Visible light is about 2 to 3 eV. How does photosynthesis create a sugar molecule from CO₂ and H₂O when visible light photons have so little energy?

 

[mgb_phys]

The photoelectric effect applies to promoting electrons in atoms, not breaking bonds in molecules.
Photosynthesis involves a lot of intermediate states created by enzymes so the changes in energy at each step are fairly low. From what I remember of the Krebbs cycle the electrons are provided by the hydrogen, they aren't ejected directly from the atoms.

 

[alxm]

The energy of oxidation of a simple sugar molecule C₆H₁₂O₆ is about 29 eV (electron volts). Visible light is about 2 to 3 eV. How does photosynthesis create a sugar molecule from CO₂ and H₂O when visible light photons have so little energy?

Well, the reaction immediately catalyzed by Photosystem II is splitting water into molecular oxygen:
2 H₂O --> O₂ + 4 H⁺ + 4 e^-
But the question still stands, since that reaction requires more energy than a single photon provides as well; it takes four.

What happens in the first step, is that light is absorbed by the chlorophyll, which ejects an electron that goes off to reduce plastoquinone somewhere else*. So the photoelectric effect is in fact somewhat relevant here, in that the absorption of light leads to a charge separation:
Chlorophyll + photon --> Chlorophyll⁺ + e^-
Although unlike the classical photoelectric effect, the electron isn't freed entirely - there's not enough energy for that - it moves to another part of the enzyme, but it remains in a bound state.

This photo-oxidation of chlorophyll then leads to it being reduced back to its original state by a tyrosine amino-acid residue, which in turn is reduced by something else in a chain that ends over at another part of the enzyme, where there's a cluster of four manganese ions (and a calcium ion, although it probably doesn't take part). So one of the manganese ions gets reduced (either Mn(III) -> Mn(IV) or Mn(IV)->Mn(V), the exact details of the reaction are not entirely known). This is where the actual chemical reaction occurs.

This whole procedure occurs four times, until the manganese cluster is oxidized 'enough' to form the energetically-expensive O-O bond, producing hydrogen ions and re-reducing the manganese ions in the process. (A simplification of course - the reaction doesn't occur in a single step, but the O-O bond formation is the 'biggest' one)

The hydrogen ions (aka protons) form a concentration gradient across the chloroplast membrane, which is then used to power ATP-synthase, which powers most of everything else. Just the same as in our mitochondrial membrane.

(* The energy of the reduced plastoquinone (=plastoquinol) is extracted elsewhere, which also ends up producing protons for the gradient.)

 

[Vanadium 50]

the photoelectric effect experiment has been done with extremely low light intensities. so low that it should have taken a single atom minutes to absorb enough energy to emit an electron. instead there is absolutely no noticable delay before electrons begin to be emitted by the metal.

Reference, please?

 

[malawi_glenn]

The energy of oxidation of a simple sugar molecule C₆H₁₂O₆ is about 29 eV (electron volts). Visible light is about 2 to 3 eV. How does photosynthesis create a sugar molecule from CO₂ and H₂O when visible light photons have so little energy?

pick up your chemistry book?

 

[granpa]

the photoelectric effect experiment has been done with extremely low light intensities. so low that it should have taken a single atom minutes to absorb enough energy to emit an electron. instead there is absolutely no noticable delay before electrons begin to be emitted by the metal. this does not rule out the possibility that light is a wave but if so would require that some of the atoms somehow start off with some random amount of energy already within them.

To clarify: Are you just arguing this proves the energy is concentrated at points that are randomly distributed over the beam's cross-section, rather than being uniformly spread over the beam cross-section (whilst making the assumption that the electron can only absorb power from its immediate vicinity and not the entire beam cross-section)? Or that it proves the energy is also longitudinally concentrated (in which case I take it the average delay matches a classical prediction but for some runs of the experiment the delay is longer or shorter)?

Regards your concession: wouldn't the frequency dependence (the total non-response to any duration of redder light, and the increasing excess kinetic energy with bluer light), etc, disprove the hypothesis of the atoms having random pre-accumulated quantities of energy?

Regards the competing model: is this behavior truly not also predicted by the semi-classical approach of considering a quantum mechanical atom in a classical electromagnetic field?

A 'classical' explanation would require that the light be a wave. no concentration of light energy in any direction or at any points.

you are right about the frequency response.

one thing that never gets mentioned is that metals, being conductors, normally reflect light. so why does the metal in the photoelectric effect not do so?

also does the photoelectric effect occur in nonmetals? if not then what is it about metals that makes it happen. according to recent threads the difference between metals and nonmetals has to do with the existence of a band gap (within nonmetals). and someone else said that band gaps might be related to (proportional to) the effective mass of electrons within the material.

 

[ZapperZ]

one thing that never gets mentioned is that metals, being conductors, normally reflect light. so why does the metal in the photoelectric effect not do so?

That's because there's no such thing as a perfect reflector. But since metals are good reflector of light, at least over a large range around the visible spectrum, they are also not efficient photoemitter.

also does the photoelectric effect occur in nonmetals? if not then what is it about metals that makes it happen. according to recent threads the difference between metals and nonmetals has to do with the existence of a band gap (within nonmetals). and someone else said that band gaps might be related to (proportional to) the effective mass of electrons within the material.

Photoemission occurs in insulators and semiconductor as well. In fact, many of the photocathodes in photomultiplier/photodetectors are made of semiconductors.

Please note also that the photoemission process is not infinitely prompt. Despite the apparent instantaneous time scale, the "response time" for photoemission can range from the order of femtoseconds for metals, to picosecond/nanoseconds for semiconductors/insulators.

Zz.

 

[granpa]

dont know if this has anything to do with it or not:
http://en.wikipedia.org/wiki/Plasmon

 

[granpa]

whats the equivalent of a metals 'work function' in a semiconductor (as it relates to the photoelectric effect)?

its impossible not to imagine that the absorption of light by a macroscopic piece of material (whether metal, semiconductor, or insulator) and its frequency specific response (photoelectric effect) isn't related somehow to the absorption of light by, and the frequency specific response of, individual atoms (or rather individual electrons within atoms). its as if the outer electrons in a material somehow combine their individual wavefunctions into some single material-wide wavefunction. (valence/conduction bands?).

still wouldn't explain why individual electrons get ejected though.

 

[Bob S]

Well, the reaction immediately catalyzed by Photosystem II is splitting water into molecular oxygen:
2 H₂O --> O₂ + 4 H⁺ + 4 e^-
But the question still stands, since that reaction requires more energy than a single photon provides as well; it takes four... (more)

Thank You. It is apparent that the Calvin cycle (photosynthesis) is a lot more complex than the Krebs cycle.

 

[Vanadium 50]

the photoelectric effect experiment has been done with extremely low light intensities. so low that it should have taken a single atom minutes to absorb enough energy to emit an electron. instead there is absolutely no noticable delay before electrons begin to be emitted by the metal. this does not rule out the possibility that light is a wave but of so would require that some of the atoms somehow start off with some random amount of energy already within them.

Reference, please?

You sent the following via PM:

I'm afraid that I don't remember where I read that. but it was probably after a simple google search. you would stand a better chance of finding the reference than I would.

So, you post something truly remarkable - that you get the photoelectric effect without photons. When I ask for a reference, you tell me to look it up myself.

This is bogus. The onus of backing up a claim is on the person making it. Please provide your reference.

 

[granpa]

without photons?

my post said 'extremely low light intensities'. how do you get 'no photons' out of that?

yes, it is remarkable. that's why the photoelectric effect is so important and why it requires such an elaborate (nonclassical) explanation.

sounds to me like you just want to get me to do your homework for you.

also see post 22 by Zapper.

 

[Vanadium 50]

If it takes minutes for the first photons to come out - and make no mistake about it, that's exactly what it means if it takes minutes to accumulate enough energy to emit an electron, since the number of photons needed to eject an electron is one - before the minutes have elapsed, that means you have a photoelectric effect without photons.

If you can't post a reference to your claim, people may conclude that the reason you can't is because there isn't one.

 

[granpa]

the light energy falling on each individual atom would take several minutes to eject an electron. presumably the total light energy falling on the whole metal plate is sufficient though to eject electrons without delay.

 

[Matterwave]

I think what granpa is saying is that if you take the cross section of the atom, and use the classical forumla Power delivered = Flux*cross section, it would take minutes to accumulate enough energy eject an electron, but in reality it's nearly instantaneous.

 

[granpa]

even if it is absorbing energy from a square region equal in width to the wavelength of visible light (the maximum that it could classically absorb from) there should still be a noticeable delay. yet there isnt.

Zapper says that they have now succeeded in detecting a delay. I'll take has word for it.

 

[ZapperZ]

even if it is absorbing energy from a square region equal in width to the wavelength of visible light (the maximum that it could classically absorb from) there should still be a noticeable delay. yet there isnt.

Zapper says that they have now succeeded in detecting a delay. I'll take has word for it.

Note that I said "response time". I'm trying to use the proper terminology for it because it has a specific meaning, and I would hope that people can google it and find exactly what it means as far as photocathodes are concerned.

And since I've also mentioned elsewhere of the Spicer's 3-step model of photoemission, I would hope people would go look for that as well and figure out why there is such a thing as a "response time" in photoemission. It isn't as trivial as calling it "delay" time.

Zz.

 

[granpa]

I thought others might find this useful:

http://www.eurekalert.org/pub_releases/2009-05/ksu-kfl052109.php

"With the help of ultrashort laser pulses, the motion of electrons can now be followed in time. "

In agreement with a recent experiment, their calculation shows that electrons of a metal surface that are near atomic nuclei are photo-emitted with a delay of about 110 attoseconds relative to another type of electron. These conduction electrons are not attached to individual atoms and enable metals to conduct electricity.

While Einstein's model is not wrong, it is not a proof for the particle-character of light, Thumm said. Einstein published his model about two decades before modern quantum theory was developed. Modern quantum theory of matter predicts the emission of an electron even when light is regarded as a classical electro-magnetic wave.

heres another nice article on the basics of the photoemission process:
http://prola.aps.org/abstract/PRA/v9/i2/p976_1

this article seem to mention something about a time delay:
http://bjps.oxfordjournals.org/cgi/content/citation/III/10/109

http://adsabs.harvard.edu/abs/1968PhLA...27..343C

 

[Vanadium 50]

I think what granpa is saying is ...

Well, if he were to post a reference to what he is claiming, we could all read it for ourselves rather than guessing.

I think it's also important that we distinguish between a delay of nano-seconds and a pre-emission of minutes.

 

[ZapperZ]

http://www.eurekalert.org/pub_releases/2009-05/ksu-kfl052109.php

You have done this several times, providing a link that is tangential to the topic being discussed. So STOP IT. It is confusing the issue, and has caused this thread to go into a million different directions.

I do not see the point in this link, since you have made no attempt at extracting what is relevant in this PRESS RELEASE to the issue being discussed in this thread.

Zz.