Light as an EM wave to explain photoelectric effect?

[physJUNKIE89]

OK, so recently I have learned that in the early 20th century, while Max Planck was attempting to explain the quantum nature of light, two men named Philip Lenard and Heinrich Hertz discovered the photoelectric effect.
They found that an electron must absorb a specific amount of energy (frequency) in order to be energized highly enough to be ejected from their previous locations. However, the electrons would not absorb radiation of any other frequency, such as mechanical waves that accumulate energy into the reciever. Instead, the amount of energy imparted over time is not as important as the one specific energy level needed to eject an electron.
-----This proved to be a strike against the wave model and in favor of the quantum model.

Now, I had also recently learned about resonant frequencies and sympathetic vibration for sound waves. This is the property of any object, dependent on its mass, density, and shape, to cast its own unique frequency when energy is imparted to it.
In sympathetic vibration, the natural frequency of the object is met by a sound wave of the same frequency, thus causing the object to vibrate even more wildly and out of control until the object finally becomes broken or disfigured.
-----This can only happen if the outside waves match the object's natural frequency.

After learning these two separate, but fascinating topics, I couldn't help but to see the relationship, and I was forced to draw an analogy between an object breaking and an electron ejecting.

Contrary to the quantum theory, I was forced to wonder if the photoelectric effect could be explained by this wave property of resonance.

Of course sound waves are mechanical and light is electromagnetic, but this is an attempt to make for myself better sense of this topic and maybe draw a relationship between the two.

I humbly thank all who respond.

 

[A. Neumaier]

Contrary to the quantum theory, I was forced to wonder if the photoelectric effect could be explained by this wave property of resonance.

Of course sound waves are mechanical and light is electromagnetic, but this is an attempt to make for myself better sense of this topic and maybe draw a relationship between the two.

It is indeed a resonance effect. The loosely bound electrons in the receptive surface of the detector have a discrete spectrum (ground state and excited state) and a continuous spectrum (electron flying away from the detector).

In the absence of an external field, the electrons are in the ground state, but the interaction with the electromagnetic fields causes them to pick up energy if the incident frequency matches the difference of frequencies to another state of the spectrum. In particular, if that frequency is large enough to reach the continuum states, an electron may be excited so much that it leaves the receptor surface, and then can be magnified and detected.

See also the entry ''The photoelectric effect'' in Chapter A4 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#photodetection

 

[ZapperZ]

OK, so recently I have learned that in the early 20th century, while Max Planck was attempting to explain the quantum nature of light, two men named Philip Lenard and Heinrich Hertz discovered the photoelectric effect.
They found that an electron must absorb a specific amount of energy (frequency) in order to be energized highly enough to be ejected from their previous locations. However, the electrons would not absorb radiation of any other frequency, such as mechanical waves that accumulate energy into the reciever. Instead, the amount of energy imparted over time is not as important as the one specific energy level needed to eject an electron.
-----This proved to be a strike against the wave model and in favor of the quantum model.

Now, I had also recently learned about resonant frequencies and sympathetic vibration for sound waves. This is the property of any object, dependent on its mass, density, and shape, to cast its own unique frequency when energy is imparted to it.
In sympathetic vibration, the natural frequency of the object is met by a sound wave of the same frequency, thus causing the object to vibrate even more wildly and out of control until the object finally becomes broken or disfigured.
-----This can only happen if the outside waves match the object's natural frequency.

After learning these two separate, but fascinating topics, I couldn't help but to see the relationship, and I was forced to draw an analogy between an object breaking and an electron ejecting.

Contrary to the quantum theory, I was forced to wonder if the photoelectric effect could be explained by this wave property of resonance.

Of course sound waves are mechanical and light is electromagnetic, but this is an attempt to make for myself better sense of this topic and maybe draw a relationship between the two.

I humbly thank all who respond.

The analogy only goes so far. For example, the amount of energy that is absorbed from the wave, and how that translates into what the photoelectron has in the end, is not clear from your model. Can you account for the kinetic energy spectrum of the outgoing photoelectrons? Such a spectrum is an essential part of techniques such as photoemission spectroscopy.

It is indeed a resonance effect. The loosely bound electrons in the receptive surface of the detector have a discrete spectrum (ground state and excited state) and a continuous spectrum (electron flying away from the detector).

In the absence of an external field, the electrons are in the ground state, but the interaction with the electromagnetic fields causes them to pick up energy if the incident frequency matches the difference of frequencies to another state of the spectrum. In particular, if that frequency is large enough to reach the continuum states, an electron may be excited so much that it leaves the receptor surface, and then can be magnified and detected.

Wait. Why is there a "discrete spectrum" for the detector? The conduction band of a metal is continuous, and it takes an infinitesimal amount of energy to excite it above the Fermi energy. The dispersion curve for a typical metal is a "parabola", not "discrete steps"!

Zz.

 

[BruceW]

electrons would not absorb radiation of any other frequency

Actually, the electron will be freed as long as it absorbs a photon which has any frequency greater than a certain "threshold" frequency. Therefore, its not really similar to classical resonance, since a wide range of frequencies can cause ejection of the electron.

 

[A. Neumaier]

Wait. Why is there a "discrete spectrum" for the detector? The conduction band of a metal is continuous, and it takes an infinitesimal amount of energy to excite it above the Fermi energy. The dispersion curve for a typical metal is a "parabola", not "discrete steps"!

The conduction band is not responsible for the photoeffect, but loosely bound electrons that must receive energy enough to escape from the location where they are bound. The surplus energy manifests itself as kinetic energy of the escaping electron.

 

[ZapperZ]

The conduction band is not responsible for the photoeffect, but loosely bound electrons that must receive energy enough to escape from the location where they are bound. The surplus energy manifests itself as kinetic energy of the escaping electron.

It is not? Since when?

Can you look at Valla et al.(Phys. Rev. Lett. 83, 2085–2088 (1999)) and tell me what it is that they're measuring on that Mo (110) surface? http://www.tcm.phy.cam.ac.uk/LocalOrbital/friday.html" also has, at the bottom of the page, the band structure for Cu. Would you like to show where the band structure is for your "loosely bound electrons" that is not part of the conduction band, and why those are the electrons that are participating in the photoemission process and NOT the conduction band?

Zz.

 

[A. Neumaier]

It is not? Since when?

Can you look at Valla et al.(Phys. Rev. Lett. 83, 2085–2088 (1999)) and tell me what it is that they're measuring on that Mo (110) surface? http://www.tcm.phy.cam.ac.uk/LocalOrbital/friday.html" also has, at the bottom of the page, the band structure for Cu. Would you like to show where the band structure is for your "loosely bound electrons" that is not part of the conduction band, and why those are the electrons that are participating in the photoemission process and NOT the conduction band?

A good description of what really happens is in the introductory section of the (over 1000 times cited) paper
Doniach, S. and Sunjic, M.,
Many-electron singularity in X-ray photoemission and X-ray line spectra from metals,
Journal of Physics C: Solid State Physics, 3 (1970), 285.

Of course talking of individual electrons is a semiclassical notion that needs interpretation, since in reality all there is is the electron field, in which electrons are indistinguishable.

From the quantum field perspective, the electron field is in equilibrium if there is no incindent e/m field. This equilibrium is described by the Fermi sea, which consists of the lower end of the conduction band. Excitations are described by electron-hole pairs, where both the electron and the hole are effective quasi-particle states. The electron, if it gets free, becomes of course an ordinary electron, and the hole state encoded the place where it left the metal. The conduction band (the Fermi sea) is unaffected, except for the additional hole potential created by the leaving electron.

If one inspects the physics underlying the effective particle-hole picture, one finds that the hole is indeed created by an electron leaving a usually occupied position in the mean-field solution of the field equations. This can be most easily seen by looking at a simpler, purely atomic system in the usual post-Hartree-Fock approximations, where the bare hole states are just bound states in easily accessible shells.

Doniach and Sunjic therefore use phrases such as ''knocking out an inner shell electron'' or ''captures an electron from an adjacent inner shell of the atom'', ''the hole moves ''up' to a higher atomic level''

All this justifies the semiclassical language that a loosely bound electron (on a lower atomic level, in an inner shell - not form the conduction electrons) gets liberated by the photoeffect.

 

[ZapperZ]

A good description of what really happens is in the introductory section of the (over 1000 times cited) paper
Doniach, S. and Sunjic, M.,
Many-electron singularity in X-ray photoemission and X-ray line spectra from metals,
Journal of Physics C: Solid State Physics, 3 (1970), 285.

NOT a good paper for this context. X-ray photoemission is NOT the same as standard photoelectric effect. Since when is standard photoelectric effect uses x-ray light source? Furthermore, x-ray photoemission probes the CORE LEVEL of the material. Standard photoelectric effect doesn't!

Of course talking of individual electrons is a semiclassical notion that needs interpretation, since in reality all there is is the electron field, in which electrons are indistinguishable.

From the quantum field perspective, the electron field is in equilibrium if there is no incindent e/m field. This equilibrium is described by the Fermi sea, which consists of the lower end of the conduction band. Excitations are described by electron-hole pairs, where both the electron and the hole are effective quasi-particle states. The electron, if it gets free, becomes of course an ordinary electron, and the hole state encoded the place where it left the metal. The conduction band (the Fermi sea) is unaffected, except for the additional hole potential created by the leaving electron.

If one inspects the physics underlying the effective particle-hole picture, one finds that the hole is indeed created by an electron leaving a usually occupied position in the mean-field solution of the field equations. This can be most easily seen by looking at a simpler, purely atomic system in the usual post-Hartree-Fock approximations, where the bare hole states are just bound states in easily accessible shells.

Doniach and Sunjic therefore use phrases such as ''knocking out an inner shell electron'' or ''captures an electron from an adjacent inner shell of the atom'', ''the hole moves ''up' to a higher atomic level''

All this justifies the semiclassical language that a loosely bound electron (on a lower atomic level, in an inner shell - not form the conduction electrons) gets liberated by the photoeffect.

That makes no sense based on all the ARPES data that we know of that probes within the first eV of the Fermi energy. You have avoided my question on (i) the Valla paper and (ii) the band structure of a typical metal. When you use visible light (which is what is normally used in a standard photoelectric effect), what energy exactly do you probe inside the band? You seem to think that all photoemission is core level photoemission, or worst still, always involves an Auger process! The paper you cite is NOT all there is to the story.

I've done extensive work in ARPES, and my Avatar is one example of an ARPES spectrum on Bi-2212 superconductor within the first 200 meV of the Fermi level. None of what you have described is consistent not only with the experiment, but also with the physics that is used in photoemission spectroscopy.

Zz.

 

[A. Neumaier]

NOT a good paper for this context.

I found the paper by following backwards a chain of references from the paper you cited, looking for an online source that had some explanations rather than only assertions about experiments. Perhaps you can point me to a better source that explains it from your point of view.

X-ray photoemission is NOT the same as standard photoelectric effect. Since when is standard photoelectric effect uses x-ray light source? Furthermore, x-ray photoemission probes the CORE LEVEL of the material. Standard photoelectric effect doesn't!

Both are electromagnetic waves. I don't see why the different energy range makes a big difference in the explanation. The explanation of the photoeffect in the quantum optics book by Mandel and Wolf (surely based on a simplified model, but still explaining the effect correctly) doesn't treat different energies differently.

That makes no sense based on all the ARPES data that we know of that probes within the first eV of the Fermi energy. You have avoided my question on (i) the Valla paper and (ii) the band structure of a typical metal. When you use visible light (which is what is normally used in a standard photoelectric effect), what energy exactly do you probe inside the band? You seem to think that all photoemission is core level photoemission, or worst still, always involves an Auger process!

I've done extensive work in ARPES, and my Avatar is one example of an ARPES spectrum on Bi-2212 superconductor within the first 200 meV of the Fermi level. None of what you have described is consistent not only with the experiment, but also with the physics that is used in photoemission spectroscopy.

Then please give me an online source (though not public, AIP sources are ok) where I can find a more thorough description on what happens on the level of a specific model for the electrons in the metal.

 

[ZapperZ]

I found the paper by following backwards a chain of references from the paper you cited, looking for an online source that had some explanations rather than only assertions about experiments. Perhaps you can point me to a better source that explains it from your point of view.

Would you prefer a photoemission text instead?

Both are electromagnetic waves. I don't see why the different energy range makes a big difference in the explanation. The explanation of the photoeffect in the quantum optics book by Mandel and Wolf (surely based on a simplified model, but still explaining the effect correctly) doesn't treat different energies differently.

And it should, because the MATERIAL makes a whole lot of difference on what one gets. Remember, YOU were using core level photoemission and you specifically DENIED the participation of conduction electrons! I pointed out the fallacy of your statement because you seem to have the impression that only CORE LEVELs are involved, which is wrong! ARPES experiments are mainly done using UV light sources. This is beyond the reach of core level energies! Your model of photoemission only accounts for ONE aspect of the photoemission process!

Then please give me an online source (though not public, AIP sources are ok) where I can find a more thorough description on what happens on the level of a specific model for the electrons in the metal.

http://iopscience.iop.org/1367-2630/7/1/E02

Maybe you need to familiarize yourself with the physics of the phenomenon first before offering a "FAQ" on the subject.

Edit: here's a simple presentation on photoemission process that also includes core-level photoemission. Note my point that core level photoemission is only ONE aspect of photoemission in general, and why it isn't relevant to a discussion on the standard photoelectric effect.

http://www.ccmr.cornell.edu/igert/modular/docs/IGERT_Lecture1.pdf Zz.

 

[Dickfore]

If I am not mistaken, ZapperZ's avatar is a Photoemission spectrum.

 

[ZapperZ]

If I am not mistaken, ZapperZ's avatar is a Photoemission spectrum.

It is, for a highly overdoped Bi-2212 high-Tc compound. I did that experiment myself, and it was published.

Zz.

 

[A. Neumaier]

Would you prefer a photoemission text instead?

An online reference is far easier for me to access. Textbooks are for the long-term use, and need more time to be acquired and read. (The references you already gave contain several references to textbooks.)

And it should, because the MATERIAL makes a whole lot of difference on what one gets.

I read more and I now understand the source of my mistake. The quantum optics literature (with which I am familiar) just _uses_ photodetectors_ to detect photons, which is the real interest. Unlike builders of photodetectors, they aren't really interested in the details of how these work. Thus the simplest models are being discussed. I also checked that Mandel & Wolf are actually silent about the electron spectrum; that it should be discrete was my unchecked addition to what they wrote - I had thought in terms of atoms rather than crystals.

On the other hand, those using the photoeffect to study materials are not really interested in the photons, but in the information these contain about the material, and in particular about the electronic state. I hadn't read anything about this before, and appreciate the references you gave.

Here the electron spectrum matters a lot, and as I know know there are two different regimes - the valence band electrons responsible for the response to light and UV, and the bound electrons for the Xray regime.

Remember, YOU were using core level photoemission and you specifically DENIED the participation of conduction electrons! I pointed out the fallacy of your statement because you seem to have the impression that only CORE LEVELs are involved, which is wrong! ARPES experiments are mainly done using UV light sources. This is beyond the reach of core level energies! Your model of photoemission only accounts for ONE aspect of the photoemission process!

You are right, I should have been more cautious, because I hadn't studied this aspect before. Thanks for correcting me.

http://iopscience.iop.org/1367-2630/7/1/E02

I found in particular the article http://iopscience.iop.org/1367-2630/7/1/097 from this issue very useful.

Maybe you need to familiarize yourself with the physics of the phenomenon first before offering a "FAQ" on the subject.

I hadn't written anything detailed on how electrons participate in the photoeffect before my partially mistaken contribution to this thread. In particular, my FAQ article on the photoeffect doesn't say anything particular about the electrons, and didn't contain my mistake. I'll probably add something to the FAQ once I am sure to understand everything relevant on this.

 

[ZapperZ]

I hadn't written anything detailed on how electrons participate in the photoeffect before my partially mistaken contribution to this thread. In particular, my FAQ article on the photoeffect doesn't say anything particular about the electrons, and didn't contain my mistake. I'll probably add something to the FAQ once I am sure to understand everything relevant on this.

I disagree. Your "FAQ" on photoelectric effect contains a LOT of misinformation. You seem to think that the ONLY way to do a photoelectric effect is to use a photodetector. this is puzzling. Any undergraduate student in physics would have seen a simple photoelectric effect done by using a simple anode to collect the photoelectrons - no "photomultiplier" involved. In fact, my setup right now that measures the quantum efficiencies of a photocathode only has an anode and a picoammeter!

And you still cite this "weakly bound electrons". As a physicist, I am shocked that you never bothered to check if such states existed in a typical metal's band structure. When I showed you the Cu band structure which clearly doesn't show such a thing (and which you couldn't identify), it should have rang many warning bells that you had things wrong. Independent of the photoelectric effect, claiming the presence of such a state is wrong as far as material science/solid state physics is concerned.

The Einstein's photoelectric model is MORE CORRECT THAN YOU THINK. The photoemission phenomenon is more involved and more complex than you think! You really should consult those who work in such a field before offering such a response. If no one here had an expertise in photoemission, your misinformation in this thread would have gone unchallenged.

Zz.

 

[A. Neumaier]

I disagree. Your "FAQ" on photoelectric effect contains a LOT of misinformation. You seem to think that the ONLY way to do a photoelectric effect is to use a photodetector. this is puzzling. Any undergraduate student in physics would have seen a simple photoelectric effect done by using a simple anode to collect the photoelectrons - no "photomultiplier" involved. In fact, my setup right now that measures the quantum efficiencies of a photocathode only has an anode and a picoammeter!

This is because in the context of experiments to check the foundations of quantum mechanics that I had in mind, one wants to detect single photons, which need a multiplier. Now I see that a number of my statements need more qualification so that they are not interpreted too broadly, or I must add more material about other uses.

And you still cite this "weakly bound electrons".

I just removed this phrase. (Changes may need up to an hour to propagate to the outside world.) It now reads:
''Sections 9.1-9.5 show that the electron field responds to a classical external electromagnetic radiation field by emitting electrons according to Poisson-law probabilities''
Is this good enough?

As a physicist, I am shocked that you never bothered to check if such states existed in a typical metal's band structure. When I showed you the Cu band structure

I hadn't found the band structure for Cu you had pointed to. I just returned to the post where you said
''Can you look at Valla et al.(Phys. Rev. Lett. 83, 2085–2088 (1999)) and tell me what it is that they're measuring on that Mo (110) surface? This link also has, at the bottom of the page, the band structure for Cu.''
I only now noticed that there was an embedded link (which is invisible with my konqueror browser unless one puts the mouse directly upon it). I thought you referred to the Valla paper where I didn't find anything about Cu, hence ignored the remark.

The Einstein's photoelectric model is MORE CORRECT THAN YOU THINK.

I didn't claim it was incorrect, only
''Einstein's explanation of the photoeffect is too simplistic, and is
not conclusive. Now, 100 years later, his picture is known to be
approximate only, and that currents in metals are in fact produced by
the continuous electron fields of QED. Discrete semiclassical particles
are just very rough approximations.''
What needs correction here?

The photoemission phenomenon is more involved and more complex than you think!
You really should consult those who work in such a field before offering such a response.

Well, you work in the field, so let me consult you.

If no one here had an expertise in photoemission, your misinformation in this thread would have gone unchallenged.

Then please help me to improve the FAQ. I can improve things only when I know that something is wrong and what needs correction. I strive for correct information and ask at the top of the FAQ to be informed about errors (which are almost unavoidable in a FAQ of this size - i can't work full-time on the FAQ).

Thus if you detail the sentences or phrases which are misleading in your view then I'll check them all against the literature (now where I know where to look this is not
difficult), and if you suggest improved formulations I'll consider them.

Then I'll upgrade the text accordingly over the next few days or weeks.