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thaiqi

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- Thread starter thaiqi
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thaiqi

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- #2

sophiecentaur

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It's hardly surprising that some phenomena just cannot be explained without QM - that's why we had to invent it!Summary::photoelectric effect is now explained in quantum idea, how about the classical?

Why the classical model fails?

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

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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

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How could there be a classical model of a quantum effect?

- #5

Lord Jestocost

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Is it the following you might be thinking of?

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

„

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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

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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.the photoelectric effect is perfectly understandable within the semiclassical theory (classical light shining on a quantized electron).

Perhaps a "semiclassical' theory doesn't grab me. It's rather like being a bit pregnant.

- #8

Nugatory

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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?Perhaps a "semiclassical' theory doesn't grab me.

- #9

sophiecentaur

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Only the winners get to write the history. We won't know who was actually right for some while. I think.arriving at the right conclusion by the wrong path?

- #10

thaiqi

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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.)...That time for usual light sources is far longer than the about 10−9s observed in experiment,...

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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.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.

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.

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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 getHow 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.)

$$\Delta t=4 \pi r^2 W/(\pi r_{\text{B}}^2 P).$$

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.

- #13

thaiqi

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Thank you very much for such a detailed reply.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.

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)?

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- #15

thaiqi

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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

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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.The electron should oscillate within the light wave and will have something to do with the frequency, am I right?

- #17

thaiqi

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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

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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.

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We already have the word "photon". Why introduce the confusing term "wave train"?One wave train is emitted by one electron transition, ...

- #20

thaiqi

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"wave train" is the word used in physical optics, that is, classical theory.

- #21

sophiecentaur

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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.We already have the word "photon". Why introduce the confusing term "wave train"?

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