Are there articles on Classical computation of the Photoelectric Effect?

  • Thread starter thaiqi
  • Start date
  • #1
thaiqi
161
8
photoelectric effect is now explained in quantum idea. Why the classical model fails? Are there articles computing it in detail using classical model?
 

Answers and Replies

  • #2
sophiecentaur
Science Advisor
Gold Member
27,824
6,330
Summary:: photoelectric effect is now explained in quantum idea, how about the classical?

Why the classical model fails?
It's hardly surprising that some phenomena just cannot be explained without QM - that's why we had to invent it!
Einstein got his first recognition (iirc) for explaining the photoelectric effect. If you want electrons to be emitted from a metal surface then it needs to be hot enough for a significant number of them to have enough energy. Thermionic emission needs a HOT cathode. Photo emission would just not be expected to happen unless there are discrete packets of energy (i.e. photons) to target individual electrons and give them enough energy to escape from a cold surface.
 
  • #3
thaiqi
161
8
It's hardly surprising that some phenomena just cannot be explained without QM - that's why we had to invent it!
Einstein got his first recognition (iirc) for explaining the photoelectric effect. If you want electrons to be emitted from a metal surface then it needs to be hot enough for a significant number of them to have enough energy. Thermionic emission needs a HOT cathode. Photo emission would just not be expected to happen unless there are discrete packets of energy (i.e. photons) to target individual electrons and give them enough energy to escape from a cold surface.

Are there articles computing it in detail using classical model?
 
  • #4
sophiecentaur
Science Advisor
Gold Member
27,824
6,330
How could there be a classical model of a quantum effect?
 
  • #5
Lord Jestocost
Gold Member
919
764
photoelectric effect is now explained in quantum idea. Why the classical model fails? Are there articles computing it in detail using classical model?

Is it the following you might be thinking of?

Willis E. Lamb, Jr. and Marlan O. Scully in “The Photoelectric Effect Without Photons”:

In conclusion, we understand the photoeffect as being the result of a classical field falling on a quantized atomic electron. The introduction of the photon concept is neither logically implied by nor necessary for the explanation of the photoelectric effect.
 
Last edited:
  • Informative
  • Like
Likes etotheipi and sophiecentaur
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,025
12,932
In the classical approach you assume that the electron must accumulate at least the binding energy of a few eV to get out of the metal from the classical em. wave. You can estimate the time this needs by assuming that the electron is bound around its nucleus within the metal within a sphere of about the Bohr radius. Then you have the area over which the radiation energy must be accumulated and you can estimate how much time it takes until the photo current starts after switching on the light. That time for usual light sources is far longer than the about ##10^{-9} \text{s}## observed in experiment, while it's easily explained by the naive photon model of the old quantum theory a la Einstein. Ironically that work, for which they gave him his Nobel prize, is the only piece of his oevre which doesn't stand the progress of science ;-).

Of course Scully and Lamb are right, the photoelectric effect is perfectly understandable within the semiclassical theory (classical light shining on a quantized electron). Also see my corresponding Insights article: It's really understandable from 1st-order time-dependentent perturbation theory as taught in the QM 1 lecture:

https://www.physicsforums.com/insights/sins-physics-didactics/
 
  • #7
sophiecentaur
Science Advisor
Gold Member
27,824
6,330
the photoelectric effect is perfectly understandable within the semiclassical theory (classical light shining on a quantized electron).
The insights article makes sense, I guess. I haven't read it in detail but is it not just replacing one aspect of probability with another, in order to account for one particular electron achieving the work function. Isn't it just a longer winded alternative description - in the same way that there is always an alternative way to deal with an 'obviously wave' phenomenon in terms of photons and vice versa. Diffraction is hard work to explain in terms of probability but you can get there.

Perhaps a "semiclassical' theory doesn't grab me. It's rather like being a bit pregnant.
 
  • #8
Nugatory
Mentor
14,115
7,882
Perhaps a "semiclassical' theory doesn't grab me.
It doesn't really excite anyone, it's mostly interesting because of how it interacts with the history. The photoelectric effect is widely cited as proof of the quantization of light and was accepted as such at the time. BUt is it really that, or was it another example of the common experience of arriving at the right conclusion by the wrong path?
 
  • #9
sophiecentaur
Science Advisor
Gold Member
27,824
6,330
arriving at the right conclusion by the wrong path?
Only the winners get to write the history. We won't know who was actually right for some while. I think.
 
  • #10
thaiqi
161
8
...That time for usual light sources is far longer than the about 10−9s observed in experiment,...
How to compute the time? It is not an possible task for me. Are there any articles on it?(I can't download Scully's article fully.)
 
  • #11
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,025
12,932
The insights article makes sense, I guess. I haven't read it in detail but is it not just replacing one aspect of probability with another, in order to account for one particular electron achieving the work function. Isn't it just a longer winded alternative description - in the same way that there is always an alternative way to deal with an 'obviously wave' phenomenon in terms of photons and vice versa. Diffraction is hard work to explain in terms of probability but you can get there.

Perhaps a "semiclassical' theory doesn't grab me. It's rather like being a bit pregnant.
Alternative to what? The photoeffect can be properly understood only with quantum mechanics. The most simple explanation is the one given in the Insight article, using the semiclassical theory. On this level you don't gain very much using the full quantum field theory. Also one should keep in mind photons are always a wave phenomenon.

For me it's a sin to tell students about the "old quantum theory" a la Einstein 1905 implying somehow a photon were a kind of localized particle. Photons cannot be localized at all. You cannot even define a position observable for them. In teaching them "old quantum theory" does not only lead to wrong quantitative predictions but, and that's even worse, provides a wrong intuition and wrong qualitative pictures about quantum theory.
 
  • Like
Likes sophiecentaur
  • #12
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,025
12,932
How to compute the time? It is not an possible task for me. Are there any articles on it?(I can't download Scully's article fully.)
The idea is to assume classical electrodynamics. Then take a light source which radiates light with a power ##P##. The intensity is this power per unit area. Assuming a isotropic radiation you get ##I=P/(4 \pi r^2)## for the intensity at a distance ##r## from the light source. The time it takes to accumulate the binding energy ##W## is ##\Delta t=W/(I A)##, where ##A## is the area over which the electron is distributed when bound at an atom, which can be estimated by using the Bohr radius ##r_{\text{B}}##, i.e., ##A=\pi r_{\text{B}}^2##. So finally you get
$$\Delta t=4 \pi r^2 W/(\pi r_{\text{B}}^2 P).$$
From Halliday-Resnick I get some numbers, where this is given as an example: Consider a light source with ##P=1.5 \; \text{W}##, ##r=3.5 \; \text{m}## and ##W=2.3 \; \text{eV}## (for potassium) and ##r_{\text{B}} \simeq 5 \cdot 10^{-11} \; \text{m}##. Then you get ##\Delta t \simeq 4580 \; \text{s} \simeq 1.3 \; \text{h}##. So it would take more than an hour until you get an electron out of the metal.

One observes however that the photocurrent starts immediately, which is another hint that the photoelectric effect cannot be explained with classical electrodynamics.
 
  • Like
Likes sophiecentaur, etotheipi and A.T.
  • #13
thaiqi
161
8
The idea is to assume classical electrodynamics. Then take a light source which radiates light with a power ##P##. The intensity is this power per unit area. Assuming a isotropic radiation you get ##I=P/(4 \pi r^2)## for the intensity at a distance ##r## from the light source. The time it takes to accumulate the binding energy ##W## is ##\Delta t=W/(I A)##, where ##A## is the area over which the electron is distributed when bound at an atom, which can be estimated by using the Bohr radius ##r_{\text{B}}##, i.e., ##A=\pi r_{\text{B}}^2##. So finally you get
$$\Delta t=4 \pi r^2 W/(\pi r_{\text{B}}^2 P).$$
From Halliday-Resnick I get some numbers, where this is given as an example: Consider a light source with ##P=1.5 \; \text{W}##, ##r=3.5 \; \text{m}## and ##W=2.3 \; \text{eV}## (for potassium) and ##r_{\text{B}} \simeq 5 \cdot 10^{-11} \; \text{m}##. Then you get ##\Delta t \simeq 4580 \; \text{s} \simeq 1.3 \; \text{h}##. So it would take more than an hour until you get an electron out of the metal.

One observes however that the photocurrent starts immediately, which is another hint that the photoelectric effect cannot be explained with classical electrodynamics.
Thank you very much for such a detailed reply.
So this model treats light as a spherical wave(?) How is it if we imagine the light is full of wave trains(that emitted one by one by the atoms of the light source)?
 
  • #14
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,025
12,932
There's no difference. The point is that you have to treat at least the electrons relativistically to get the photoeffect described right, i.e., in accordance with observations.
 
  • #15
thaiqi
161
8
There's no difference. The point is that you have to treat at least the electrons relativistically to get the photoeffect described right, i.e., in accordance with observations.
Some textbooks say that in the classical model the light absorption by electron should only have something to do with the light density and have nothing to do with the light frequency. I think this is incorrect. The electron should oscillate within the light wave and will have something to do with the frequency, am I right?
 
  • #16
sophiecentaur
Science Advisor
Gold Member
27,824
6,330
The electron should oscillate within the light wave and will have something to do with the frequency, am I right?
I think that mechanical view is not particularly relevant. I don't think the idea of a wave 'shaking an electron loose' helps. Apart from the threshold photon energy, there is no 'resonance' involved (the dreaded Hydrogen Atom model comes to mind here and it's not appropriate at all in a solid metal). Whatever the frequency of the incoming photon, the KE of the electron will be (up to and including) the surplus energy.
 
  • #17
thaiqi
161
8
I think that mechanical view is not particularly relevant. I don't think the idea of a wave 'shaking an electron loose' helps. Apart from the threshold photon energy, there is no 'resonance' involved (the dreaded Hydrogen Atom model comes to mind here and it's not appropriate at all in a solid metal). Whatever the frequency of the incoming photon, the KE of the electron will be (up to and including) the surplus energy.

I found in Esposito's book: Advanced Concepts in Quantum Mechanics section 2.2.1 (page 21) it discussed about the classical model of photoelectric effect handled by Sommerfeld and Debye(1913). It is rather difficult for me, though.
 
  • #18
thaiqi
161
8
There's no difference. The point is that you have to treat at least the electrons relativistically to get the photoeffect described right, i.e., in accordance with observations.

One wave train is emitted by one electron transition, so conversely its energy should be able to kick one electron in the metal surface if it can be absorbed( serveral eVs). Thus isn't photoelectric effect very easy to understand?
 
  • #19
A.T.
Science Advisor
11,657
2,951
One wave train is emitted by one electron transition, ...
We already have the word "photon". Why introduce the confusing term "wave train"?
 
  • #20
thaiqi
161
8
"wave train" is the word used in physical optics, that is, classical theory.
 
  • #21
sophiecentaur
Science Advisor
Gold Member
27,824
6,330
We already have the word "photon". Why introduce the confusing term "wave train"?
I can see why it annoys you but the word photon is a bit entrenched and needs qualifying to remind people it's not a little bullet. The term 'wave train' is fine for me and it extends into concepts of coherence.
 
  • #22
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,025
12,932
The word photon must not be used in the semiclassical theory, where the em. field is treated as classical. A photon is a Fock state. In classical electrodynamics there's no Fock space!
 
  • Like
Likes weirdoguy, etotheipi and sophiecentaur

Suggested for: Are there articles on Classical computation of the Photoelectric Effect?

Replies
3
Views
940
  • Last Post
Replies
1
Views
618
  • Last Post
Replies
14
Views
116
Replies
2
Views
869
Replies
4
Views
1K
  • Last Post
Replies
14
Views
548
Replies
18
Views
13K
Replies
1
Views
587
Replies
2
Views
1K
Top