Why are electrons emitted almost instantaneously during the photoelectric effect?

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From: https://opentextbc.ca/universityphysicsv3openstax/chapter/photoelectric-effect/

"The photoelectric effect has three important characteristics that cannot be explained by classical physics: (1) the absence of a lag time, (2) the independence of the kinetic energy of photoelectrons on the intensity of incident radiation, and (3) the presence of a cut-off frequency."

How to interpret the absence of a lag time, ie instantaneous momentum.
Mechanically speaking, if we place an object on the edge of a table, any slightest action on the table will immediately cause the object to fall. However someone had to put the object on the edge of the table.
I think the example is equivalent to an electron in a photo effect. A certain amount of energy is needed to allow the electron to be very weakly bound to the surrounding atoms.
 

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Greetings,
A certain amount of energy is needed to allow the electron to be very weakly bound to the surrounding atoms.
That condition is satisfied by the electronic structure of the atom itself. No "preparation" by the incident radiation field is required.

Best regards,
ES
 
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PeterDonis
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A certain amount of energy is needed to allow the electron to be very weakly bound to the surrounding atoms.
No. The electrons emitted in the photoelectric effect are already weakly bound to the surrounding atoms. At least, that's the case in the most common experiments to demonstrate the effect, which use metals. In experiments with non-metals, the cutoff frequency is higher because the electrons that are emitted are more tightly bound to the atoms, so it takes more energy to free them.
 
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Thanks for the answers.
To summarize - a weak electron bond at the surrounding atoms is the cause of the absence of delay time.
Best regards,
BS
 
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PeterDonis
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To summarize - a weak electron bond at the surrounding atoms is the cause of the absence of delay time.
No. The absence of delay time is there even for photoemission from non-metals, where the electrons are more strongly bound to the atoms (and the cutoff frequency is higher).

The absence of delay time is because in QM, the energy that is taken up by the electron arrives all at once, in a single "packet" (the photon), instead of being gradually given to it by incoming EM waves (which is what the classical model predicts). That is true regardless of whether the photon carries a smaller amount of energy (for photoemission from metals) or a larger amount of energy (for photoemission from nonmetals).
 
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vanhees71
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It is one of the most persistent legends that the photoeffect is necessarily explained only by photons, i.e., quantization of the electromagnetic field. That's not true, because on the level we discuss it here, it follows from the semiclassical approach in 1st-order time-dependent perturbation theory. "Semiclassical" means that we describe the bound electron quantum-mechanically and the electromagnetic field ("light") as a classical wave.

That there is no (or rather negligible) "delay time" but a sharp threshold in frequency for the emission of the electron is due to the fact that you can emit the electron only when the frequency of the light is in resonance with some energy difference between the electron's bound state and one of its scattering state. It is not possible to "accumulate" this ionization energy from the em. wave as the fully classical model suggests.
 
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vanhees71 said:
That there is no (or rather negligible) "delay time" but a sharp threshold in frequency for the emission of the electron is due to the fact that you can emit the electron only when the frequency of the light is in resonance with some energy difference between the electron's bound state and one of its scattering state. It is not possible to "accumulate" this ionization energy from the em. wave as the fully classical model suggests.
This is not clear to me. According to this, it seems that the photoelectric effect occurs only at one (resonant) frequency or at specific frequencies, and not at a frequency that provides sufficient energy for the electron to jump out.
 
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hutchphd
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According to this, it seems that the photoelectric effect occurs only at one (resonant) frequency
No. The scattered states belong to the continuum of free states so the possible energy difference is essentially continuous above threshold. Only if both states are bound do you get discreet differences.
 
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Unfortunately I’m not that far off with quantum mechanics yet, especially when it comes to the photo effect. It’s great that you directed me to take the first steps in that direction. If I'm not mistaken, today's quantum mechanics interprets the photo effect as follows (quote):

"In the photoelectric effect, or photoeffect, a photon is absorbed by a target atom and, as a result, an atomic electron is emitted or promoted to a bound open orbital thus leaving the residual ion or atom in an excited state."
I don't know if that's today's official interpretation?

Such an interpretation seems absolutely logical to me. A photon is more likely to hit an atom (the nucleus of an atom) than the electron itself.

This is a completely different interpretation from the classical interpretation that can be seen in almost all basics of physics, where a photon hits an electron.

Whether talking about resonance or excitation of atoms, both require some time. However, as you said, we can ignore that "delay time".
 
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vanhees71
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This is not clear to me. According to this, it seems that the photoelectric effect occurs only at one (resonant) frequency or at specific frequencies, and not at a frequency that provides sufficient energy for the electron to jump out.
No, according to this the photoeffect occurs at any frequency above the threshold frequency. The unbound electron states have a continuous energy spectrum!
 
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