Can electrons absorb a photon?

[Dale]

What conservation law would prohibit a free atom to absorb a photon, any photon, and change its speed accordingly?

Conservation of energy and conservation of momentum. Again, sit down and work it out, you will see that they cannot both be conserved.

the electrons are "receiving" the photons and, together with the rest of the atom, absorb them,

That seems to be a distinction without a difference. If it is the electron together with the rest of the atom then that is the atom.

I wrote electron in the atom (see below) and I explained above what I meant.

So do you understand the conservation issue?

If the energy of one photon could be used both for the transition of the electron (or more electrons) and for a translation of the whole atom (in addition to the one associated with the transition), the absorption of the light would be continuous and possibly complete.

This doesn’t follow. Again, the conservation of energy and momentum prohibit this. You need to actually work out some of these problems to get a better feel for what is going on. Start with the electron, work it out until you are convinced, then move on to the atom.

 

[DanMP]

... Again, the conservation of energy and momentum prohibit this. You need to actually work out some of these problems to get a better feel for what is going on. Start with the electron, work it out until you are convinced, then move on to the atom.

Ok, I think I got it. Thank you.

If I got it right, the same conservation laws prohibit multiple transitions (the energy of one photon to be used by the atom to raise more than one electron to a higher orbital). If I'm wrong, why it doesn't happen [frequently]? You said that "it must have a very low probability". Why?

...That seems to be a distinction without a difference. If it is the electron together with the rest of the atom then that is the atom...

There is a difference if the electron in the atom receives the photon: it may explain why only one electron get raised to a higher orbital ... Also, it is closer to a particle approach (we are talking about photons not light waves ...) than what DrClaude wrote:

... Basically, the point is that you need a dipole to absorb a photon. There has to be two opposite charges to couple to the electromagnetic field. ...

 

[Dale]

There is a difference if the electron in the atom receives the photon:

I disagree. “Electron together with the rest of the atom” = “the atom”. They are exactly the same thing. It is like you are saying “the letter A together with the rest of the alphabet” and I am saying “the alphabet”. It is the same thing.

We know that the electron without the rest of the atom cannot absorb a photon. We know that the electron with the rest of the atom can. The electron with the rest of the atom is the atom. So the atom absorbs the photon.

 

[sophiecentaur]

We know that the electron without the rest of the atom cannot absorb a photon.

Free Electrons can be caused to oscillate by a passing Radio Wave in the Ionosphere. So Energy can be transferred to an electron in a conventional sense. The Quantum steps in that case would be Zero or at least a lot less than the photon energy of the wave. The theory behind the phenomenon is 'correct' in that it makes good predictions so it's a bit difficult to resolve the apparent paradox in my mind. Perhaps the energy is reactive and leaves the electron unchanged when the wave has passed (or been refracted)?

 

[Dale]

Free Electrons can be caused to oscillate by a passing Radio Wave in the Ionosphere. So Energy can be transferred to an electron in a conventional sense.

Sure, energy can easily be transferred from a photon to an electron, but it is through scattering not absorption.

 

[DrClaude]

When thinking about the possibility of a single photon leading two two excited electrons in an atom, one must keep in mind that single-electron orbitals in multi-electron atoms correspond to a simplified model. Strictly speaking, in a n-electron atom, we only have n-electron electronic states. Assigning individual electrons to individual orbitals helps us understand some things, but it is not a perfect representation of reality.

That said, there are not that many stable electronic states with more than one excited electron. In helium, for instance, there are none: even the energy of the lowest possible doubly-excited electronic state, 2s², is above the single ionisation threshold. Then, to get single-photon transitions, there has to be allowed dipole transitons between two states, and there are probably not many of them.

That doesn't mean that these transitons do not exist, and indeed there is one that can be seen in copper: there is a line at 510.554 nm that corresponds to the 3d⁹4s² → 3d¹⁰4p transition. It is the basis for the copper vapour laser. A similar transition is also observed in gold.

 

[sophiecentaur]

Sure, energy can easily be transferred from a photon to an electron, but it is through scattering not absorption.

Ah yes. One is an electromagnetic interaction and the other is momentum and KE.

 

[Dale]

That doesn't mean that these transitons do not exist, and indeed there is one that can be seen in copper: there is a line at 510.554 nm that corresponds to the 3d94s2 → 3d104p transition. It is the basis for the copper vapour laser.

That is very interesting. So I was wrong in assuming that such transitions are unlikely, rather they are rare.

How about unstable states? Wouldn’t they have corresponding absorption lines even if they don’t have emission lines at that frequency?

Edit: or do they but because it is an unstable state it is so short lived that the line is very broad

 

[DanMP]

...That doesn't mean that these transitons do not exist, and indeed there is one that can be seen in copper: there is a line at 510.554 nm that corresponds to the 3d⁹4s² → 3d¹⁰4p transition. It is the basis for the copper vapour laser. A similar transition is also observed in gold.

In lasers it's about emission of photons. You are sure that there is also an absorption line for this double transition?

... Then, to get single-photon transitions, there has to be allowed dipole transitons between two states, and there are probably not many of them. ...

Transitions between states? That doesn't look like the atom as a whole is absorbing the photon and then use the energy to raise 2 electrons to a higher orbital ...

 

[Dale]

Transitions between states? That doesn't look like the atom as a whole is absorbing the photon

Why not? The states are states of the atom.

 

[DanMP]

I disagree. “Electron together with the rest of the atom” = “the atom”. They are exactly the same thing. ...

If my interpretation, that the electron in the atom receives the photon and absorbs its energy and momentum together with the rest of the atom, is the same thing with the mainstream interpretation that the atom absorbs the photon, it means that my interpretation is not necessarily wrong ...

 

[Dale]

If my interpretation, that the electron in the atom receives the photon and absorbs its energy and momentum together with the rest of the atom, is the same thing with the mainstream interpretation that the atom absorbs the photon, it means that my interpretation is not necessarily wrong ...

Do you understand that a free electron can not absorb a photon?

If so then if you wish to say “electron together with the rest of the atom” instead of just “atom” then you can do so. Nobody else is likely to think that it is worthwhile deliberately taking 8 words to say what 1 word says, but you are absolutely right that it is not necessarily wrong. As I said before, it is a distinction without a difference.

What would be necessarily wrong is to say that “the electron absorbs the photon”, or even that “the electron in the atom absorbs it”. It is “the electron together with the rest of the atom”, or just “the atom”.

I note that in your post you say “the electron in the atom”. This is wrong. The energy and momentum of the photon cannot be localized to the electron without violating conservation of energy or momentum. It requires the atom.

 

[DanMP]

... but you are absolutely right that it is not necessarily wrong. As I said before, it is a distinction without a difference. ...

It is good enough for me. Thank you.

... I note that in your post you say “the electron in the atom”. This is wrong. The energy and momentum of the photon cannot be localized to the electron without violating conservation of energy or momentum. It requires the atom.

I wrote "the electron in the atom receives the photon and absorbs its energy and momentum together with the rest of the atom". What I meant is that momentum (and some energy?) may be transferred from the receiving electron to the rest of the atom by virtual photons, as you can read in Wikipedia:

An isolated electron at a constant velocity cannot emit or absorb a real photon; doing so would violate conservation of energy and momentum. Instead, virtual photons can transfer momentum between two charged particles. This exchange of virtual photons, for example, generates the Coulomb force.[99] Energy emission can occur when a moving electron is deflected by a charged particle, such as a proton. The acceleration of the electron results in the emission of Bremsstrahlung radiation.[100]

By the way, when one electron descends to a lower orbital, the photon is emitted by the atom or by the electron?

 

[Dale]

By the way, when one electron descends to a lower orbital, the photon is emitted by the atom or by the electron?

By the atom.

What I meant is that momentum (and some energy?) may be transferred from the receiving electron to the rest of the atom by virtual photons

Those virtual photons are just as much part of the atom as the electrons and the nucleus. In fact, similar bosons for the nuclear forces in the nucleus make up most of the mass of an atom. So this is still “the electron together with the rest of the atom”.

 

[DrClaude]

That is very interesting. So I was wrong in assuming that such transitions are unlikely, rather they are rare.

How about unstable states? Wouldn’t they have corresponding absorption lines even if they don’t have emission lines at that frequency?

Edit: or do they but because it is an unstable state it is so short lived that the line is very broad

I would guess that individual transitions are not directly observed because they are too broad. See the discussion in the Wikipedia page on autoionization.

 

[DrClaude]

In lasers it's about emission of photons. You are sure that there is also an absorption line for this double transition?

Absorption and emission are symmetrical processes.

 

[DrClaude]

Transitions between states? That doesn't look like the atom as a whole is absorbing the photon and then use the energy to raise 2 electrons to a higher orbital ...

But the nucleus is still there, and its presence is required when writing the potential part of the Hamiltonian used to calculate those orbitals.

 

[DanMP]

Absorption and emission are symmetrical processes.

Well, they are, but still, it may be something similar to Sum-frequency generation, meaning that, in the copper vapour laser, 2 individual photons are in fact emitted by the (2 excited electrons in the) atom and then they "combine" into one.

Even if the above is/looks wrong/silly, I insist: you know for sure that there is an absorption line for this double transition? Maybe you can provide a link ...

 

[Dale]

I insist: you know for sure that there is an absorption line for this double transition?

By symmetrical processes he means that they both happen in a laser. So the laser he cited above demonstrates both absorption and emission.

 

[DanMP]

By symmetrical processes he means that they both happen in a laser. So the laser he cited above demonstrates both absorption and emission.

Sorry, but I'm not convinced. The laser in question may be "charged" both by one or by two photons, so I don't think that it proves without a doubt that one photon can cause a double transition. There must be an absorption experiment with that line present, in order to have an absolute proof. Do you know one? DrClaude? Anybody?

 

[Dale]

Sorry, but I'm not convinced.

I don’t think that is reasonable doubt. It seems to be a position based on ignorance of how a laser operates. If a material lases then it demonstrates both absorption and emission at the lasing wavelength.

Anyway, there are several promising sounding papers, but the ones I could find are all behind paywalls, so I couldn’t confirm their contents. You are welcome to do so.

https://doi.org/10.1063/1.435365

https://journals.aps.org/pr/abstract/10.1103/PhysRev.34.35

https://www.spiedigitallibrary.org/...es-in-active-media-of/10.1117/12.160510.short

http://iopscience.iop.org/article/10.1070/QE1980v010n03ABEH009986/pdf

I don't think that it proves without a doubt that one photon can cause a double transition.

Even so, you still have to contend with the clear proof that a double transition causes one photon. That clearly is not compatible with the electrons doing the emission. Since it is the atom doing the emission then why wouldn’t it be the atom doing the absorption? Particularly since the laws of physics governing the process are the same for both processes.

What is your reason for sticking with the “electron absorption” model? You seem to have a very strong attachment to it that I don’t understand. It doesn’t conserve momentum unless you introduce the rest of the atom. So why struggle to maintain it?

 

[DanMP]

... Anyway, there are several promising sounding papers, ...
...
What is your reason for sticking with the “electron absorption” model? You seem to have a very strong attachment to it that I don’t understand. It doesn’t conserve momentum unless you introduce the rest of the atom. So why struggle to maintain it?

Thank you very much for the links you provided. Unfortunately I can't see more than the abstracts, but maybe someone here can, and then help me/us to get a conclusive answer.

My main reason to insist on this (or any) matter is the fact that I need to understand what is really happening.

Yes, I preferred the idea that the electrons are absorbing/emitting the photons and that the momentum is transferred to the rest of the atom through virtual photons, but now I'm fine with your interpretation that the atom as a whole does the absorption/emission and transfers the energy to/from one or more electrons. Still, I want to know, if I can, which one is closer to reality.

LE: in both interpretations, overall, the atom does the absorption (/emission), but in mine one electron in the atom receives the photon, while in yours the photon is received by the atom as a whole. I wonder, if the atom as a whole receives the photons in absorption, why the Compton scattering is not on the atom as a whole as well?

 

[Mister T]

If my interpretation, that the electron in the atom receives the photon and absorbs its energy and momentum together with the rest of the atom, is the same thing with the mainstream interpretation that the atom absorbs the photon, it means that my interpretation is not necessarily wrong ...

Let's suppose an interaction between a photon of energy $hf$ and an atom results in the atom's energy increasing by $hf$. The photon disappears. We say it's absorbed, but in fact it ceases to exist, so that is different from the way a paper towel absorbs water. Without the atomic nucleus there would be no atomic energy levels. The potential energy is a property of the atom, not a property of anyone single component of the atom. It is the interaction between the components that gives rise to the potential energy, without that interaction there is no potential energy.

 

[DanMP]

... It is the interaction between the components that gives rise to the potential energy, without that interaction there is no potential energy.

My interpretation agrees with what you wrote. Even more, I think I mentioned this interaction (that takes place through virtual photons) when I cited from Wikipedia how the momentum is transferred to the rest of the atom.

 

[Dale]

My main reason to insist on this (or any) matter is the fact that I need to understand what is really happening.
...
Still, I want to know, if I can, which one is closer to reality.

Well, the usual model correctly predicts all of the observed phenomena, and (insofar as it differs at all) yours only predicts some of them, so I think that much is pretty clear.

I wonder, if the atom as a whole receives the photons in absorption, why the Compton scattering is not on the atom as a whole as well

Scattering and absorption are pretty different. Did you ever sit down and work out the conservation of momentum and energy as I recommended?

If so, then you should have seen that scattering can conserve both energy and momentum with just the electron, but absorption cannot. That is why a single electron can scatter a photon but cannot absorb it. It is all about conservation principles. The same conservation principles lead to both behaviors.

 

[Dale]

The potential energy is a property of the atom, not a property of anyone single component of the atom.

This is an important point that applies for potential energy in general. We sometimes speak as though it is localized to one part of a system, but generally such comments are misleading.

 

[DanMP]

... Did you ever sit down and work out the conservation of momentum and energy as I recommended?

If so, then you should have seen that scattering can conserve both energy and momentum with just the electron, but absorption cannot. That is why a single electron can scatter a photon but cannot absorb it. It is all about conservation principles. The same conservation principles lead to both behaviors.

Yes, I sat down and did what you recommended so many times. I even explained (cited from Wikipedia) how the momentum is transferred to the rest of the atom ...

Yes, I understand that (and why) a single electron can scatter a photon, but tell me: the whole atom can? If not, why not? If yes, what decides if the atom as a whole or one of its electrons would scatter an incident photon?

 

[Dale]

the whole atom can?

Any charged particle can participate in elastic scattering. So as long as the atom is ionized it can participate. Inelastic scattering can involve even an unionized atom.

If yes, what decides if the atom as a whole or one of its electrons would scatter an incident photon?

Whether or not the electron is bound or free. If the electron is free then the electron scatters, if the electron is bound then the atom scatters. A bound electron cannot be treated independently of the rest of the atom.

For loosely bound electrons and high energy photons the difference between the two is small. I.e. the scattering is elastic for free electrons and inelastic for bound photons, but the degree of inelasticity may be small. In such cases you may see some references blur the line.

 

[DanMP]

... Whether or not the electron is bound or free. If the electron is free then the electron scatters, if the electron is bound then the atom scatters. A bound electron cannot be treated independently of the rest of the atom. ...

So, in scattering it is always about the electron ... We say that the atom scatters, just because the electron is part of an atom. This is very close to how I interpreted the absorption: the electron "receives" the photon but, because it is a part of an atom and momentum & energy are transferred to the rest of the atom, we say that the atom scatters/absorbs ...

 

[Dale]

So, in scattering it is always about the electron .

No, it is not always about the electron. It is only about the electron if the electron is unbound. I was pretty clear about that.

the electron "receives" the photon

As far as I can tell the word “receives” has no meaning in this context.

At energies below pair production there are three basic interactions: a photon does not enter but does leave (emission), a photon enters but does not leave (absorbed), a photon enters and a photon leaves (scattering). “Receive” is not one of them.

 

[DanMP]

No, it is not always about the electron ...

Ok, thanks for your replies.