Can an atom absorb a photon, yet its total kinetic energy is decreased?

[sophiecentaur]

I only said it is a matter of semantics whether we associate the amount of energy to the electron or to tha atom(the system). The amount is the same!

The fact that a proton has 2000 times the mass of the electron means that you can treat the proton as 'stationary' in a mechanical collision - but that's all. Your arguing would need to change direction rapidly in the case of a diatomic molecule absorbing a photon which has energy corresponding to a change in vibrational energy. So why not talk in a consistent way, rather than hanging onto a wording that's around 100 years old to back up your preferences?

 

[Vanadium 50]

Please give a specific reference, with context

The old classic, "My Physics Textbook". Page 984.

You seem to be getting upset. The problem isn't that the OP hasn't read the answer, it's that he doesn't believe the answer. That's not something we can fix.

Sure, there will likely be negative consequences down the road, but maybe that will provide the necessary impetus for a change of his perspective.

 

[Dale]

We need keqe2/r2 = mv2/r.

Details of the calculation:
mv2 = keqe2/r, so the kinetic energy of the electron is
KE(r) = ½mv2 = ½keqe2/r.
The potential energy of the electron in the field of the positive proton point charge is U(r) = -qeV(r) = - keqe2/r.

As you mentioned “in the real world” to me earlier, I will do the same here. In the real world people, including textbook authors, often say one thing and do another. Although this author said “energy of the electron” note what they actually did in the math.

First, they used the variable $r$. In context, $r$ describes the atom, it is not the radius of the electron nor the radius of the nucleus, it is the radius of the atom.

Second, $m$ in the KE formula is constant. If the PE belonged to the electron then as it gained PE it would also gain mass. Since an electron is so small, this would be noticeable, small but noticeable.

So although in English they say “potential energy of the electron” in math they actually do “the potential energy of the atom”.

 

[avicenna]

As you mentioned “in the real world” to me earlier, I will do the same here. In the real world people, including textbook authors, often say one thing and do another. Although this author said “energy of the electron” note what they actually did in the math.

First, they used the variable $r$. In context, $r$ describes the atom, it is not the radius of the electron nor the radius of the nucleus, it is the radius of the atom.

Second, $m$ in the KE formula is constant. If the PE belonged to the electron then as it gained PE it would also gain mass. Since an electron is so small, this would be noticeable, small but noticeable.

So although in English they say “potential energy of the electron” in math they actually do “the potential energy of the atom”.

I surrender!

My level of physics has not reached the level where the mass of the electron may change. Give me time to do some catch-up!!!

 

[PeterDonis]

I have never ever said that an electron on its own can have potential energy. I only said it is a matter of semantics whether we associate the amount of energy to the electron or to tha atom(the system).

You're quibbling. These two claims are the same, because associating the energy with just the electron is saying that the electron on its own can have potential energy. Which it can't.

From my physics textbook

Which textbook?

From my physics textbook, pg 984:

From: University of Tennessee, Knoxville

As I said, you have to read these sources in context. You're not doing that.

Nobody is disputing the math; that's not the issue. The issue is that it's not "a matter of semantics" what the energy is associated with. It has to be associated with the system--in this case the atom--not just the electron. The quotes you give do not contradict that; they are just being sloppy: because they are using a frame in which the proton is at rest, they are describing the PE as though it belonged to the electron, even though it doesn't. But switch frames and this sloppiness no longer works. The actual physics is that the PE is a property of the system, not the electron, regardless of what sloppiness you can get away with in a particular frame.

As for why the actual physics is important, that should be obvious from the discussion in this thread: it explains why the answer to your original question has to be "no".

 

[PeterDonis]

My level of physics has not reached the level where the mass of the electron may change.

It can't--that is precisely @Dale's point. The electron is not a bound system that can even have potential energy on its own. Its mass is a constant and can't change. Advancing your level of understanding won't change that.

 

[Structure seeker]

@avicenna Bohr's theory was meant for electron orbit states around an atom. Imagine a proton without electrons (a H+ ion) absorbing a photon. The proton is moving to the right, the photon to the left. It's simply the law of momentum conservation that the atom will be moving slower after the encounter.

In the case of electrons orbiting an atom moving contrary to an incoming photon: the electron may enter a more excited state, but due to its electronic attraction to the atom core and the gained momentum in contrary direction at first, the atom core will move slightly slower afterwards. The gained energy for the electron is less than the lost kinetic energy of the atom core. The atom's "rest energy" (with the electron still excited) might have increased but its total energy reduced.

 

[Drakkith]

My level of physics has not reached the level where the mass of the electron may change. Give me time to do some catch-up!!!

The mass of the electron doesn't change, and neither does the mass of the nucleus. This is similar to boosting the Moon into a higher orbit around the Earth. Energy has been added, and thus mass, but the mass of the Moon and the Earth don't change. How could they? An observer on the surface of either body would see no physical change in the body, no temperature increase, nowhere that mass or energy has been added. Therefore it must be the mass of the Earth-Moon system as a whole that changes, not the mass of either body.

 

[vanhees71]

I have never ever said that an electron on its own can have potential energy. I only said it is a matter of semantics whether we associate the amount of energy to the electron or to tha atom(the system). The amount is the same!

It is not semantics but a basic misunderstanding. In an atom in an energy eigenstate the electrons and the atomic nucleus are in an entangled state and thus inseparable. You have an energy of the atom as a whole. To talk about distribution of this energy to various parts of the atom doesn't make sense.

Amazingly this is not discussed in the textbooks, although it's an (if not the) important feature of quantum theory. See the following AJP paper

https://arxiv.org/abs/quant-ph/9709052
https://doi.org/10.1119/1.18977

 

[bob012345]

It is not semantics but a basic misunderstanding. In an atom in an energy eigenstate the electrons and the atomic nucleus are in an entangled state and thus inseparable. You have an energy of the atom as a whole. To talk about distribution of this energy to various parts of the atom doesn't make sense.

Amazingly this is not discussed in the textbooks, although it's an (if not the) important feature of quantum theory. See the following AJP paper

https://arxiv.org/abs/quant-ph/9709052
https://doi.org/10.1119/1.18977

Why do you think it's the most important feature of quantum theory?

 

[vanhees71]

If there's one feature which is really "quantum" in contradistinction to "classical physics", it's entanglement, I'd say, but that's of course a pretty subjective opinion.

 

[bob012345]

If there's one feature which is really "quantum" in contradistinction to "classical physics", it's entanglement, I'd say, but that's of course a pretty subjective opinion.

OK, but would there be any difference from using the usual Schrödinger approach by considering the proton and electron as an entangled state vs. the electron in a coulomb potential?

 

[vanhees71]

The usual Schrödinger approach directly leads to this result. The point is to write the Hamiltonian in terms of center-of-mass and relative coordinates. The center-of-mass motion is free, and the relative-coordinate piece leads to the usual hydrogen atom for a quasiparticle around the center (which is the center of mass of electron and proton) with the reduced mass, $\mu=m_{\text{e}} m_{\text{p}}/(m_{\text{e}}+m_{\text{p}})$. It's as in classical mechanics (Kepler problem). While a product state in center-of-mass and relative coordinates, rewritten again in electron and proton observables, it's showing that electron and proton are in an entangled state.

 

[hilbert2]

Maybe it's also possible that some exotic particle in high-energy physics can absorb a photon to become a higher-mass particle without the KE or PE changing?

 

[Dale]

Maybe it's also possible that some exotic particle in high-energy physics can absorb a photon to become a higher-mass particle without the KE or PE changing?

You would need to be clear what you mean by PE not changing in that case. But yes, in principle that could happen.

 

[PeterDonis]

some exotic particle in high-energy physics can absorb a photon to become a higher-mass particle without the KE or PE changing?

If the particle is free, it is impossible for it to absorb a single photon without its momentum (and hence its KE) changing.

 

[Dale]

If the particle is free, it is impossible for it to absorb a single photon without its momentum (and hence its KE) changing.

No. You can change momentum without changing KE. It would be pretty fine tuned though. The momentum would have to start out along e.g. the negative x axis and change to the positive x axis at whatever velocity corresponds to the same KE. There should be one frame where this is true for any absorption by a free particle.

 

[PeterDonis]

You can change momentum without changing KE.

Hm, yes, for a very fine-tuned case this is possible.

 

[Dale]

Hm, yes, for a very fine-tuned case this is possible.

Yes, not only is it fine tuned, but also I cannot think of any reason why you would use that frame naturally. You would only use it specifically to make this point.

 

[Vanadium 50]

Since the decay $\Sigma^0 \rightarrow \Lambda + \gamma$ occurs, the inverse process $\gamma + \Lambda \rightarrow \Sigma^0$ must also occur.

However, both baryons cannot be at rest. There exist a frame where the baryon kinetic energies are the same and a (different) frame where the momentum magnitudes are the same. This example also relies on the baryons being composite particles.