[worlov (OP)]

Hello!

It is explicitly stated in the literature that classical electrodynamics can not explain the photoelectric effect. In my view, there is at least one possibility.

How do quasi-free electrons move in the metals? I imagine a following model. At room temperature, the conducting electrons revolve around the metal ions, but at the points of contact between the metal atoms they pass from one atom to another without or with relatively little resistance.

Statistically nothing changes in principle if we disregard the short-term movement of the electrons between the atoms. In this way, we get a simplified model of the metal, which consists of hydrogen-like atoms. For the alkali metals, ie the metals belonging to the same group as hydrogen, this may even be quite true.

Further, a hydrogen atom is a resonator (the resonance frequency is 1.42 GHz). Therefore, the system of metal ion and quasi-free electron should also behave as a resonator. This raises the question of whether there is a relationship between the natural frequencies of such resonators and the corresponding cutoff frequencies of the photo effect.

Has this research been done?

[Bored chemist]

Statistically nothing changes in principle if we disregard the short-term movement of the electrons between the atoms.

What changes is that you get a model which is wrong.

The sort of experiment you are talking about is done regularly.
https://en.wikipedia.org/wiki/Photoemission_spectroscopy

You also rather miss the point that those frequencies, such as the 1.42 GHz hydrogen line are the results of quantum effects.

[worlov (OP)]

What changes is that you get a model which is wrong.

The models are very common in physics. They allow to make the estimates... And I think my model is plausible. The conducting electrons will not fly through the atoms, they will certainly move at the edges of atoms (figure below).

The sort of experiment you are talking about is done regularly.

I mean purely classic approach. So I understand correctly that no one has thought about it yet.

[Bored chemist]

they will certainly move at the edges of atoms

No
The uncertainty principle prevents that.

The models are very common in physics.

Models that work are very common.

[alancalverd]

Not much photoelectricity occurs with photon frequencies below 10¹³ Hz, a lot more than 1.42 GHz.

Why do people persist with classical physics? Chemists stopped talking about caloric and phlogiston before BC was born. Quantum mechanics and relativity are older than the aeroplane. Nobody goes on holiday in a stagecoach, so why do they mess about with mathematical models that don't work?

[worlov (OP)]

The uncertainty principle prevents that.

The conducting electrons will not fly through the atoms, otherwise it would then come to the generation of X-rays.

[worlov (OP)]

Why do people persist with classical physics? ... Nobody goes on holiday in a stagecoach, so why do they mess about with mathematical models that don't work?

And why should any alternative idea be rejected immediately? Are not you curious? Maybe something was overlooked 100 years ago.

[Bored chemist]

The uncertainty principle prevents that.

The conducting electrons will not fly through the atoms, otherwise it would then come to the generation of X-rays.

According to classical physics, they should be doing that.
That's why we know classical physics is wrong.

[Bored chemist]

Not much photoelectricity occurs with photon frequencies below 10¹³ Hz, a lot more than 1.42 GHz.

Why do people persist with classical physics? Chemists stopped talking about caloric and phlogiston before BC was born. Quantum mechanics and relativity are older than the aeroplane. Nobody goes on holiday in a stagecoach, so why do they mess about with mathematical models that don't work?

Almost all the engineering/ design/ physics I have actually done (as opposed to reading about) was classical physics.
Similarly, sending a man to the Moon didn't need to take account of relativity, but adding the relativistic terms would have cluttered up their very limited computational power.

So, people often use classical physics because, at low speeds, for big things, it works well enough.

[worlov (OP)]

According to classical physics, they should be doing that.

Why? The conducting electron is the outermost electron of the atom. The other electrons within the atom will repel the outer electron. Yes, the positively charged metal ion generally attracts the outer electron, but the other electrons form the barriers on the way to the nucleus.

[Bored chemist]

Why? The conducting electron is the outermost electron of the atom.

And it should "fall" into the nucleus, emitting X Rays as it goes.

[jeffreyH]

@worlov You have been given answers by people who really do know what they are talking about. It's their profession. What exactly do you think you are going to teach them?

[alancalverd]

If classical electrodynamics applied to atoms, they would shrink to the size of their nuclei and the electron binding force would prevent the formation of molecules. Physics would be trivial, chemistry could not happen, and we would not be having this conversation.

[evan_au]

a hydrogen atom is a resonator (the resonance frequency is 1.42 GHz)

You are describing the 21cm "Hydrogen line", used by radio astronomers to map the concentration of neutral hydrogen gas in the galaxy and across the universe.
https://en.wikipedia.org/wiki/Hydrogen_line

Rather than being a natural, strong resonator, this is an incredibly weak "forbidden" resonance, which is what makes it so useful to astronomers.
- Hydrogen is so abundant in the universe that if it were a strong resonance, radio astronomers could not see out of the Solar System, because there is so much hydrogen in the Solar wind.
- In fact, this transition is extremely rare, and it takes a massive cloud of gas (thousands of times more massive than the Sun) to produce a measurable signal in radio telescopes.
- Natural emission of this hydrogen line is considered impossible to detect in the lab, but it can be artificially triggered
- On the other hand, the photoelectric effect can be readily detected in the lab - but with energies billions of times greater (UV photons instead of microwave photons) 

I imagine a following model. At room temperature, the conducting electrons revolve around the metal ions, but at the points of contact between the metal atoms they pass from one atom to another without or with relatively little resistance

What you are describing is similar to the Fermi surface, which describes the permissible momentum of electrons in the conduction band of a metal.
- If you join a few of the units together, It starts to look like a Swiss cheese
- The inner electrons are strongly attached to the nucleus, and don't participate in electrical conductivity

See: https://www.google.com/search?tbm=isch&q=fermi+surface+aluminum
https://en.wikipedia.org/wiki/Fermi_surface

the system of metal ion and quasi-free electron should also behave as a resonator. Has this research been done?

Yes, this is how the Fermi surface was initially mapped. It is typically done at very low temperatures, in a very strong superconducting magnet.

A more recent method is to bombard the metal with positrons.

For the alkali metals, ie the metals belonging to the same group as hydrogen, this may even be quite true.

Hydrogen as we have it on earth is a non-conductor.
- However, at the enormous pressures at the core of Jupiter, physicists believe that Hydrogen enters a metallic state, producing Jupiter's considerable magnetic field.
- Recently, scientists made a (somewhat controversial) claim to have produced metallic hydrogen in the lab (under enormous pressures in a diamond anvil). The most visible change was that it was no longer transparent.
- So the behavior of non-conductive hydrogen is totally different from metallic hydrogen.

Statistically nothing changes in principle if we disregard the short-term movement of the electrons between the atoms. In this way, we get a simplified model of the metal, which consists of hydrogen-like atoms.

I am afraid that conduction between atoms totally changes the behavior of a substance when irradiated by electromagnetic waves.
- For non-conductive materials, electromagnetic waves tend to pass right through (with a bit of refraction, as the speed of light is lower in matter than it is in a vacuum)
- For highly conductive materials like metals, electromagnetic waves cause movement of the surface electrons, inducing a current in the surface of the metal which is opposite and equal to the external field (by Lenz's Law). This cancels the incoming wave, and generates an outgoing wave where the angles follow the law of reflection that you learnt in High School.

the corresponding cutoff frequencies of the photo effect

The cutoff frequency of the photoelectric effect is determined by the Work Function of the metal.

It is closely related to the Fermi energy of the electrons.
It is not closely related to the Hydrogen Line.
See: https://en.wikipedia.org/wiki/Work_function

[worlov (OP)]

Thank you for detailed answer. But I continued to work on my model. To be able to treat it simply mathematically, I consider the electron as a uniformly charged ball. In this case, the formula for the resonance frequency is



R₀ is the atomic radius. For different elements there is the table (last column)

http://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page [nofollow])

When I set the values, I get the resonance frequencies close to the cutoff frequencies of the photo effect. There is supposed to be a connection.

[worlov (OP)]

Furthermore, the frequencies correlate within the groups and the periods. Below are the diagrams presented. f ^c  is the cutoff frequency of the photoeffect and f ^r  is the resonant frequency according to my model.

[Bored chemist]

Furthermore, the frequencies correlate within the groups and the periods.

So do the weights, but that  doesn't mean the effects are dominated by gravity.

the formula for the resonance frequency is

What resonance is that?

[evan_au]

I notice that the Group 2 graph (Be, Ca, Mg, Sr, etc) consists entirely of elements with an even number of electrons, and an even number of protons. Several other graphs also have this property.

However, the 21cm/1.4GHz atomic Hydrogen line occurs because there is:
- Just 1 unpaired electron, which acts like a tiny magnet
- Just 1 unpaired proton, which acts like a tiny magnet
- It is the tiny interaction between these two tiny magnets (parallel/anti-parallel) which produce the tiny energy transition of 5.8 μeV.

However, when there are:
- 2 electrons in the S orbital, their spins cancel each other out, eliminating Hydrogen's tiny magnetic field from electrons
- An even number of protons in the nucleus, their spins cancel each other out, eliminating Hydrogen's tiny magnetic field from the nucleus
- This also eliminates the tiny energy transition which generates the 21cm Hydrogen line
- And the quoted resonance formula becomes irrelevant

Also, when isolated atoms form up into molecules, the electrons tend to pair up so that their spins cancel, eliminating this supposed resonance. And the photoelectric effect can be seen with molecules as well as metals.

Please identify the source of "the formula for the resonance frequency" - I am sure there are a lot of caveats there that you have ignored!

I would be especially interested to see the correlation in Group 4 (Carbon to Lead), which makes the transition from non-metal to metal, and involves significant changes in the Fermi Surface and Fermi Energy.

The correlation in the graphs may have more to do with the dependence of the Work function on the atomic radius.
- ie correlation does not show causation

[worlov (OP)]

I derived the formula due to the following simplification. The electron is a negatively uniformly charged ball and the metal ion is a positively uniformly charged ball. The two are about the same size - the size of the atom in the metal bond. An external electric field causes the shift of the charge: the electron ball shifts relative to the metal ion ball. Between them arises an electric field, which is equal to the outer field but opposite, so that the outer field is completely compensated. The electric field strength in the region of overlap between the two balls



d - the shift of the electron ball relative to the metal ion ball. This field strength acts on the charge of the electron, resulting in the force that tries to bring the electron back to its rest position



As we can see, retroactive force is directly proportional to the shift. So we're getting to Hooke's Law. Therefore, we have for the spring constant



Then the resonance frequency



Logically, when the frequency of the external electric field coincides with the natural frequency of the conducting electrons, the resonential phenomenon occurs and the conducting electron is thrown out of the atom and finally out of the Metall.

[evan_au]

Logically, when the frequency of the external electric field coincides with the natural frequency of the conducting electrons, the resonential phenomenon occurs and the conducting electron is thrown out of the atom and finally out of the Metal.

The energy of the Hydrogen 1.4GHz resonance is 5.8 μeV.
The energy of the photoelectric work function is typically around 3-5 eV.

It is not logical to think that bombarding an electron with nx5.8 μeV could trigger an event requiring an energy a million times greater.

One of the principals of the photoelectric effect is that it is an "all-or-nothing" effect.
- One photon kicks out an electron, or it does not
- Atoms don't accumulate energy from radiation until they reach a threshold and then spit out an electron
- The whole idea of a resonance is that it works for some frequencies, and not others
- The "right" energies are in the Ultraviolet part of the spectrum
- Energies in the microwave band are irrelevant - by 6 orders of magnitude!

See: ]List of Work functions