Photoelectric and Compton effects

In summary, the photoelectric and Compton effects cannot be explained with the wave model of radiation due to the fact that the energy of the emitted electrons only depends on the frequency of the EM radiation, not its intensity. This contradicts the expected behavior of a wave, where increasing the intensity would increase the energy. This can be explained by the quantum effects present in the microscopic system of an electron interacting with an EM wave, where the threshold frequency of the electron's energy must be surpassed in order for it to be released from its bound state. This cannot be explained by the classical wave theory and a modern description is needed.
  • #1
TrickyDicky
3,507
27
What exactly in the photoelectric and Compton effects can't be explained with the wave model of radiation?
For instance in the photoelectric effect, why the fact that the energy of the electrons emited from a material after being subjected to certain radiation depends only on the frequency of the EM radiation and not on its intensity is not expected of a wave?
I mean if a EM wave carries momentum, wouldn't conservation of momentum and energy considerations make the outcome of the interaction between the EM wave and the bound electron depend on the energies of the electron and the EM wave and their respective momenta?, being the momentum of the EM wave: p=hf/c and therefore proportional to its frequency just as its energy=nhf is also proportional to the radiation frequency.
 
Physics news on Phys.org
  • #2
Doesn't the energy of a wave depend on its intensity? In that case the kinetic energy of the photoelectrons should depend on the intensity of the light, which it doesn't. Therefore it is impossible to explain the photolectric effect with a wave theory.

In Compton scattering, the wavelength changes before and after scattering - is this really in accordance with a wave explanation? I'm not so sure.
 
  • #3
kloptok said:
Doesn't the energy of a wave depend on its intensity? In that case the kinetic energy of the photoelectrons should depend on the intensity of the light, which it doesn't. Therefore it is impossible to explain the photolectric effect with a wave theory.
Good point, I'm aware that in the classic Maxwell theory of EM radiation, the energy depends only on its intensity, but at the time it was formulated most of the processes of emision,absortion and interaction of light with matte were unknown. I wonder if the old wave theory ha been expanded toaccount for the dependency of waves energy on frequency.
 
  • #4
For instance the energy (Poynting vector) of a EM wave according to classic theory is calculated thru the average intensity and since involves the average of the square of a sine function (with its frequency and wavelength) it can be switched to a 1/2.
But if a wave interacts with matter I don't think its energy can be defined by its average intensity any longer. Does this make any sense?
 
  • #5
TrickyDicky said:
What exactly in the photoelectric and Compton effects can't be explained with the wave model of radiation?
For instance in the photoelectric effect, why the fact that the energy of the electrons emited from a material after being subjected to certain radiation depends only on the frequency of the EM radiation and not on its intensity is not expected of a wave?

Look at a mass oscillating at the end of the spring. How do you increase the energy in that system? By increasing its amplitude. In fact, if you look at the wave equation, its energy and intensity can be increased by doing nothing more than increasing the amplitude of oscillation.

In the standard photoelectric effect, doing such a thing doesn't affect the energy of the outgoing photoelectrons. In fact, if the frequency is below some threshold, increasing the intensity does nothing to cause any kind of electron emission.

Zz.
 
  • #6
ZapperZ said:
Look at a mass oscillating at the end of the spring. How do you increase the energy in that system? By increasing its amplitude. In fact, if you look at the wave equation, its energy and intensity can be increased by doing nothing more than increasing the amplitude of oscillation.

In the standard photoelectric effect, doing such a thing doesn't affect the energy of the outgoing photoelectrons. In fact, if the frequency is below some threshold, increasing the intensity does nothing to cause any kind of electron emission.

Zz.

And wouldn't that different behaviour be explained by the fact that the spring system is macroscopic, and behaves classically while the electron-EM wave is a microscopic system where quantum effects makes the frequency of the inciding wave play a threshold role, related to the fact that the electron's wavefunction(with energy eigenvalues dependent on frequency) doensn't allow the electron to be released from its bound state until some frequency threshold is surpassed so that it can go free?

I mean that in the macroscopic situation the quantum effects are averaged out, so it behaves classically and in this setting the energy can be increased just by increasing the amplitude, but in the microscopic setting, involving just one electron, quantum effects are noticeable and only after the threshold of frequency of the electron's energy is overcome, does the increase in intensity of the EM wave manifests as increasing numbers of electrons displaced.
 
  • #7
TrickyDicky said:
And wouldn't that different behaviour be explained by the fact that the spring system is macroscopic, and behaves classically while the electron-EM wave is a microscopic system where quantum effects makes the frequency of the inciding wave play a threshold role, related to the fact that the electron's wavefunction(with energy eigenvalues dependent on frequency) doensn't allow the electron to be released from its bound state until some frequency threshold is surpassed so that it can go free?

This makes no sense. Both systems are being described by the same form of the wave equation!

Aren't you asking why the old classical description doesn't work for the photoelectric effect? It appears that you are revising history and trying to "modify" the classical description to include what it doesn't have!

I don't think I want to play this game.

Zz.
 
  • #8
ZapperZ said:
Aren't you asking why the old classical description doesn't work for the photoelectric effect? It appears that you are revising history and trying to "modify" the classical description to include what it doesn't have!
You surely read strange things between lines. Revising history never entered my mind. But actually I was kind of asking why are we clinging to the "old classical description", when we have a modern description, if that is a sinful question and has no place here, I'll take it back, but I hope you don't want to turn this into some kind of Inquisition.
ZapperZ said:
I don't think I want to play this game.
What game? but hey, serve yourself.
 
  • #9
TrickyDicky said:
why are we clinging to the "old classical description", when we have a modern description

For most practical applications of electromagnetism, it's much much much [...] much much easier to use classical electrodynamics than quantum electrodynamics. :smile:

Similarly, we don't use general relativity to calculate the trajectory of a baseball, or even, usually, things like the orbits of planets in the solar system.
 
  • #10
jtbell said:
For most practical applications of electromagnetism, it's much much much [...] much much easier to use classical electrodynamics than quantum electrodynamics. :smile:

Similarly, we don't use general relativity to calculate the trajectory of a baseball, or even, usually, things like the orbits of planets in the solar system.

Thanks jtbell, that's exactly my point. For most practical applications (i.e. classical behaviour) no doubt it's easier, but for the specific cases mentioned in the OP (photoelectric and Compton effects), they can only be treated with QM, and my question was not if the classic treatment of waves should be "revised", but whether with the current quantum mechanical model of waves interacting with matter we can fully account for these experimental results.
 
  • #11
TrickyDicky said:
Thanks jtbell, that's exactly my point. For most practical applications (i.e. classical behaviour) no doubt it's easier, but for the specific cases mentioned in the OP (photoelectric and Compton effects), they can only be treated with QM, and my question was not if the classic treatment of waves should be "revised", but whether with the current quantum mechanical model of waves interacting with matter we can fully account for these experimental results.

What is the "quantum mechanical model of waves interacting with matter"?

The EM wave in classical model is a real, physical wave. The "wavefunction" (if this is what you are referring to as the "quantum mechanical model of waves") isn't! Furthermore, if you look at the interaction via Feynman diagrams, what 'waves' are we talking about here?

Zz.
 
  • #12
ZapperZ said:
What is the "quantum mechanical model of waves interacting with matter"?

The EM wave in classical model is a real, physical wave. The "wavefunction" (if this is what you are referring to as the "quantum mechanical model of waves") isn't! Furthermore, if you look at the interaction via Feynman diagrams, what 'waves' are we talking about here?

Zz.
I'd be delighted to discuss the philosophy of QM with you and nitpick about semantics, but it won't happen, I'm actually more interested in the practical side of physiscs.
But since you mention Feynman diagrams, they are just a device to make calculations easier, a way to simplify the mathematical apparatus of QFT, you don't actually believe there's something real about the diagrams, do you? They are a graphic representation, one of various ways to represent a reality. Do you also consider virtual particles to be real because they appear in the diagrams?
 
  • #13
TrickyDicky said:
I'd be delighted to discuss the philosophy of QM with you and nitpick about semantics, but it won't happen, I'm actually more interested in the practical side of physiscs.
But since you mention Feynman diagrams, they are just a device to make calculations easier, a way to simplify the mathematical apparatus of QFT, you don't actually believe there's something real about the diagrams, do you? They are a graphic representation, one of various ways to represent a reality. Do you also consider virtual particles to be real because they appear in the diagrams?

But it is as "real" as the wavefunction. You don't actually think that we "measure" the wavefunction, do you? QM can be represented equally using all those different formulations. I can easily do without those wavefunction and only deal with 2nd quantization, matrix formulation, etc.. etc!

You never addressed the point I raised that the physical wave of EM radiation is DIFFERENT than the wavefunction in QM! The QM wavefunction is not physical - it "lives" in configuration space! But somehow, you think that just because the word "wave" is in there, they are of the same beast!

Zz.
 
  • #14
ZapperZ said:
You never addressed the point I raised that the physical wave of EM radiation is DIFFERENT than the wavefunction in QM! The QM wavefunction is not physical - it "lives" in configuration space! But somehow, you think that just because the word "wave" is in there, they are of the same beast!

Zz.
And where did I say(or even implied) that they are the same? I basically asked questions. I never did say that an EM wave and a wave funtion are the same thing! , perhaps I didn't express myself correctly, the wave function is a mathematical abstraction and is basically used with matter particles, is not even well defined for photons. But obviously it must represent something physical, otherwise QM wouldn't work as a physical theory.
 
  • #15
TrickyDicky said:
And where did I say(or even implied) that they are the same? I basically asked questions. I never did say that an EM wave and a wave funtion are the same thing! , perhaps I didn't express myself correctly, the wave function is a mathematical abstraction and is basically used with matter particles, is not even well defined for photons. But obviously it must represent something physical, otherwise QM wouldn't work as a physical theory.

Why is it not "well defined" for photons? When Marcella[1] used QM to arrive all the interference patterns using slits, what do you think he was using? Ping pong balls? You should write a rebuttal to his paper by telling him that he's using something that isn't well-defined.

Zz.

[1] T.V. Marcella Eur. J. Phys. v.23, p.615 (2002).
 
  • #16
TrickyDicky said:
What exactly in the photoelectric and Compton effects can't be explained with the wave model of radiation?
For instance in the photoelectric effect, why the fact that the energy of the electrons emited from a material after being subjected to certain radiation depends only on the frequency of the EM radiation and not on its intensity is not expected of a wave?
I mean if a EM wave carries momentum, wouldn't conservation of momentum and energy considerations make the outcome of the interaction between the EM wave and the bound electron depend on the energies of the electron and the EM wave and their respective momenta?, being the momentum of the EM wave: p=hf/c and therefore proportional to its frequency just as its energy=nhf is also proportional to the radiation frequency.



http://books.google.com.kw/books?id...esnum=9&ved=0CDMQ6AEwCDgK#v=onepage&q&f=false


The classical EM vector potential is quantized(FFT), written in the hamiltonian, and solved perturbatively. P.A, and transition probabilities calculated. The field is called interaction of radiation with matter.
 
  • #17
ZapperZ said:
Why is it not "well defined" for photons? When Marcella[1] used QM to arrive all the interference patterns using slits, what do you think he was using? Ping pong balls? You should write a rebuttal to his paper by telling him that he's using something that isn't well-defined.

Zz.

[1] T.V. Marcella Eur. J. Phys. v.23, p.615 (2002).

Once again you are interpreting my words in a twisted way that fits your preconceptions, that something isn't well defined doesn't mean that a wave function can not be built for photons, photon wave function is a very useful concept, for instance in optics, but you won't find it in most books about QM or QFT, it has its limitations maybe explained for historical reasons,quote from THE PHOTON WAVE FUNCTION by Iwo Bialynicki-Birula "I believe that the reasons why a first-quantized theory of photons has never been fully developed are mainly historical. Had Dirac discovered his relativistic wave equation [1] prior to his quantization of the electromagnetic field [2], he would have noticed and most probably further explored a great similarity between the wave equation for the electron (or even better for the neutrino) and the Maxwell equations. As it happened, this similarity was noticed later (for the first time apparently by Majorana)".
When I said not well defined I should have been more specific, what is not well defined for some authors (others don't see this as a problem) and for some QM texts is the photon's (or any massless particle) position as an observable, but certainly what is well defined is their interaction with other particles so your ping-pong game should work just fine.
BTW, I love your nitpicking, as long as it clears my points, sometimes I'm lazy.

But let me remind you that my comment about wave functions and photons was triggered by your claim that "the physical wave of EM radiation is DIFFERENT than the wavefunction in QM!" , next you are equally interested in shouting that EM radiation(and please don't nitpick that photons have nothing to do with EM radiation) can be formalized with the QM wave function.
What game are YOU playing?
 
  • #18
TrickyDicky said:
Once again you are interpreting my words in a twisted way that fits your preconceptions, that something isn't well defined doesn't mean that a wave function can not be built for photons, photon wave function is a very useful concept, for instance in optics, but you won't find it in most books about QM or QFT, it has its limitations maybe explained for historical reasons,quote from THE PHOTON WAVE FUNCTION by Iwo Bialynicki-Birula "I believe that the reasons why a first-quantized theory of photons has never been fully developed are mainly historical. Had Dirac discovered his relativistic wave equation [1] prior to his quantization of the electromagnetic field [2], he would have noticed and most probably further explored a great similarity between the wave equation for the electron (or even better for the neutrino) and the Maxwell equations. As it happened, this similarity was noticed later (for the first time apparently by Majorana)".
When I said not well defined I should have been more specific, what is not well defined for some authors (others don't see this as a problem) and for some QM texts is the photon's (or any massless particle) position as an observable, but certainly what is well defined is their interaction with other particles so your ping-pong game should work just fine.
BTW, I love your nitpicking, as long as it clears my points, sometimes I'm lazy.

This makes it even less clear. What EXACTLY about the physics here that makes not well-defined? All you can tell me is some quote, as if you don't quite understand the physics itself!

What exactly is it about classical maxwell wave equation for EM radiation that is "similar" to the Dirac equation? What exactly is the evidence that you can get the identical set of experimental results from the photoelectric effect using only the classical EM wave equation?

All I see here are not "evidence", but rather "quotations". That isn't physics. That's mimicking! A parrot can do that without understanding the content of what it is saying.

So do this:

1. Show what you mean by your "wave" equation

2. Show what you mean by your QM "wavefunction"

3. And finally, show evidence that #1 can reproduce the standard photoelectric effect*.

Failure to do #3 would be sufficient to answer your original question, does it not?

Zz.

*One needs only to look at, say, the Pittman et al. paper for references that shows that under some stochastic conditions, one can use the classical wave to reproduce the standard photoelectric effect results. This has been discussed ad nauseum on this forum already and why this is only a special case so far. However, the physics used in such a description is nowhere similar to what you've been trying to argue here. Even people who have doubts about the photo picture would do a double take on what you've been arguing.
 
  • #19
qsa said:
http://books.google.com.kw/books?id...esnum=9&ved=0CDMQ6AEwCDgK#v=onepage&q&f=false


The classical EM vector potential is quantized(FFT), written in the hamiltonian, and solved perturbatively. P.A, and transition probabilities calculated. The field is called interaction of radiation with matter.

Thanks for your contribution, certainly in a way my question is solved (or perhaps dissolved would be more correct) by QED and interaction fields since second quantization of the classical EM wave is what my OP was pointing at it as it could be viewed as the modern "quantum mechanical model of EM waves interacting with matter".
 
  • #20
ZapperZ said:
This makes it even less clear. What EXACTLY about the physics here that makes not well-defined? All you can tell me is some quote, as if you don't quite understand the physics itself!

What exactly is it about classical maxwell wave equation for EM radiation that is "similar" to the Dirac equation? What exactly is the evidence that you can get the identical set of experimental results from the photoelectric effect using only the classical EM wave equation?

All I see here are not "evidence", but rather "quotations". That isn't physics. That's mimicking! A parrot can do that without understanding the content of what it is saying.

So do this:

1. Show what you mean by your "wave" equation

2. Show what you mean by your QM "wavefunction"

3. And finally, show evidence that #1 can reproduce the standard photoelectric effect*.

Failure to do #3 would be sufficient to answer your original question, does it not?

Zz.

*One needs only to look at, say, the Pittman et al. paper for references that shows that under some stochastic conditions, one can use the classical wave to reproduce the standard photoelectric effect results. This has been discussed ad nauseum on this forum already and why this is only a special case so far. However, the physics used in such a description is nowhere similar to what you've been trying to argue here. Even people who have doubts about the photo picture would do a double take on what you've been arguing.

Pitiful. And this is not physics either.


Feel free to lock it, I made my point sufficiently clear. Thanks for your "useful" contri anyway.
 
  • #21
TrickyDicky said:
Thanks for your contribution, certainly in a way my question is solved (or perhaps dissolved would be more correct) by QED and interaction fields since second quantization of the classical EM wave is what my OP was pointing at it as it could be viewed as the modern "quantum mechanical model of EM waves interacting with matter".

The quantization of classical EM is called first quantization. The word second quantization is reserved for quantizing the QM wave equations. There is a subtlty in the use of the vector potential. The same Vector potential is elevated to a symmetry in QFT, but not looked upon as a classical notion, it is understood(interpreted) as a guage.
 
  • #22
qsa said:
The quantization of classical EM is called first quantization. The word second quantization is reserved for quantizing the QM wave equations.

You are completely right, I meant first and wrote second, perhaps because the concept second quantization is generally related to QFT.
qsa said:
There is a subtlty in the use of the vector potential. The same Vector potential is elevated to a symmetry in QFT, but not looked upon as a classical notion, it is understood(interpreted) as a guage
Would you please explain what that change from the classical notion to the gauge means physically?
 
  • #23
TrickyDicky said:
Would you please explain what that change from the classical notion to the gauge means physically?


Well, it might sound strange but in modern physics the word physical takes on a different notion(i.e. if particles move faster than light then they are considered unphysical) and sometimes become very obscure. take for instance the vector potential that we are discussing, while it is a classical description yet it is considered not "physical"( some do citing Ahranov phenomenon). in QM world the situation becomes even more abstract, in QFT even the position probability of the particle becomes undefined.

So, the new physics is about modeling the system so that we can predict outcomes of experiments and not concerned with entities making sense according with our intuitive classical world.So, usually a lagrangian is setup with an ansatz(guess) symmetry, and if that leads to mathematically consistent and agrees with experiment we are happy with that.

the reason why we left the classical world is that they broke down at the edge of experiments. Now, trying to devise a quantum gravity theory it is suspected that current models are also suffering shortcomings but reverting to classical looks extremely unlikely.
 
  • #24
I'll take a stab at the OP.

The Compton effect is purely classical. I will illustrate with a simple thought experiment.

A blackened sheet of copper has been charged with 1 Coulomb and is floating motionless in space.

An intense pulse of green light lasting 1 second strikes the sheet and accelerates it a minute amount.

While the sheet is accelerating, it is radiating energy at a very low frequency corresponding to it's minute acceleration.

The green pulse has transferred some of it's momentum to the plate and the plate in turn has reradiated some of it away at much longer wavelengths.

The photoelectric effect is tricker but is also completely classical. Thought experiment number two.

I have a very large wall perhaps ten miles on a side. The wall consists of a closely packed array of square waveguides each 1 meter across. The waveguides are a mile deep and lossless. On the backside there is an array of diodes and lamps which light up when there is incident RF energy.

Now the funny thing is, no matter how much power I direct at the wall with my transmitter, nothing on the other side lights up unless I turn up the frequency of the transmitter to just over 300 MHz. As long as the frequency is greater than that, I can heat a filament and eject electrons out the backside.
 
Last edited:
  • #25
Antiphon said:
The Compton effect is purely classical. I will illustrate with a simple thought experiment.

...

The photoelectric effect is tricker but is also completely classical. Thought experiment number two.

That the photoelectric is completely classical is not correct, as explained earlier in this thread. Classically, the kinetic energy of the photoelectrons should depend on the intensity of the incoming light, which it doesn't - thus contradicting the classical approach. I can't say much about your thought experiment, but you should explain where those waveguides come from? I don't see how this thought experiment is an example of the photoelectric effect.

Concerning the Compton effect I've thought intensely about whether it is possible or not to explain it without quantum physics. It seems that the shift in wavelength of the photons cannot be explained by classical wave theory. This is what Wikipedia's article on Compton scattering says:

"The effect is important because it demonstrates that light cannot be explained purely as a wave phenomenon. Thomson scattering, the classical theory of an electromagnetic wave scattered by charged particles, cannot explain low intensity shifts in wavelength (Classically, light of sufficient intensity for the electric field to accelerate a charged particle to a relativistic speed will cause radiation-pressure recoil and an associated Doppler shift of the scattered light[2], but the effect would become arbitrarily small at sufficiently low light intensities regardless of wavelength.) Light must behave as if it consists of particles to explain the low-intensity Compton scattering. Compton's experiment convinced physicists that light can behave as a stream of particle-like objects (quanta) whose energy is proportional to the frequency."
 
  • #26
Antiphon said:
I'll take a stab at the OP.

The Compton effect is purely classical. I will illustrate with a simple thought experiment.

A blackened sheet of copper has been charged with 1 Coulomb and is floating motionless in space.

An intense pulse of green light lasting 1 second strikes the sheet and accelerates it a minute amount.

While the sheet is accelerating, it is radiating energy at a very low frequency corresponding to it's minute acceleration.

The green pulse has transferred some of it's momentum to the plate and the plate in turn has reradiated some of it away at much longer wavelengths.

The photoelectric effect is tricker but is also completely classical. Thought experiment number two.

I have a very large wall perhaps ten miles on a side. The wall consists of a closely packed array of square waveguides each 1 meter across. The waveguides are a mile deep and lossless. On the backside there is an array of diodes and lamps which light up when there is incident RF energy.

Now the funny thing is, no matter how much power I direct at the wall with my transmitter, nothing on the other side lights up unless I turn up the frequency of the transmitter to just over 300 MHz. As long as the frequency is greater than that, I can heat a filament and eject electrons out the backside.
No, both effect are not classical (in case, the second could be defined "semi-classical", in its simpler form, but certainly not "classical").
In the first case you should show how to derive classically the formula of Compton Scattering with your model.
In the second you should show how, classically, the energy of the electrons is proportional to the RF frequency.

(Hint: you can't :smile:)
 
  • #27
lightarrow said:
No, both effect are not classical (in case, the second could be defined "semi-classical", in its simpler form, but certainly not "classical").
In the first case you should show how to derive classically the formula of Compton Scattering with your model.
In the second you should show how, classically, the energy of the electrons is proportional to the RF frequency.

(Hint: you can't :smile:)

Of course you're right. What I was trying to do was bring attention to the fact that the answers being posted in the thread were good but not conclusive as qualitative explanations. In the photoelectric case, it was being said that a wavelength threshold was needed to be exceeded and intensity alone is not enough to cause emission. That's true but you can (and I did) construct classical devices that also do this.
 
  • #28
Antiphon said:
Of course you're right. What I was trying to do was bring attention to the fact that the answers being posted in the thread were good but not conclusive as qualitative explanations. In the photoelectric case, it was being said that a wavelength threshold was needed to be exceeded and intensity alone is not enough to cause emission. That's true but you can (and I did) construct classical devices that also do this.

But your "classical device" is nowhere similar to "metallic surface" in which the photoelectric effect is done. Unless you can show that your "thought experiment" is identical to what is going on on a metallic surface, your comparison isn't valid. In fact, I don't see any kind of similarity whatsoever, considering that one only has certain discrete frequency of propagation for such a waveguides. This is in contrast to the fact that for the photoelectric effect, ANY frequency above the threshold will allow for photoemission.

Zz.
 

Related to Photoelectric and Compton effects

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to electromagnetic radiation, such as light. This occurs when the energy of the radiation is high enough to overcome the binding energy of the electrons in the material.

2. How does the photoelectric effect support the particle nature of light?

The photoelectric effect provides evidence for the particle nature of light, as it shows that light behaves as individual packets of energy called photons. These photons transfer their energy to the electrons in the material, causing them to be emitted.

3. What is the Compton effect?

The Compton effect is a phenomenon in which photons lose energy when they collide with electrons. This results in a change in the wavelength of the photon, known as the Compton shift.

4. How does the Compton effect support the wave-particle duality of light?

The Compton effect supports the wave-particle duality of light, as it shows that light can behave as both a wave and a particle. The change in wavelength of the photon is evidence of its wave-like nature, while the transfer of energy to the electron is evidence of its particle-like behavior.

5. What are the practical applications of the photoelectric and Compton effects?

The photoelectric and Compton effects have many practical applications, including in solar panels, photomultiplier tubes, and X-ray imaging. The photoelectric effect is also used in sensors and detectors, while the Compton effect is utilized in radiation therapy for cancer treatment.

Similar threads

Replies
5
Views
693
Replies
10
Views
2K
Replies
5
Views
859
Replies
13
Views
1K
  • Quantum Physics
Replies
12
Views
786
Replies
8
Views
949
Replies
0
Views
149
Replies
6
Views
1K
Replies
3
Views
822
Replies
30
Views
2K
Back
Top