Why doesn't the atom absorb heat energy when it is low?

[sophiecentaur]

I don't think quantum theory allows for a 'band' of energies for electron states. What is the range of this 'band'?

I am aware that the energy bands can change depending on electromagnetic effects (Zeeman/Stark), is that what you mean?

We all learn about 'The Hydrogen atom' and how the photon interaction only occurs for Exactly one frequency. However . . .
All the atoms in a region of gas have slightly different Energy levels due to the Pauli Exclusion Principle. They all have a slight effect on each others' fields. If the atoms are in motion then they will absorb photons with different frequencies in the same way that the emitted photons are affected. Line broadening is greater in high pressure gas because the 'bandwidth' is greater.
Edit: Plus what @Dale said.

 

[Dale]

Line broadening is greater in high pressure gas because the 'bandwidth' is greater.

Oh, yeah, I completely forgot about Doppler broadening too.

 

[sophiecentaur]

Oh, yeah, I completely forgot about Doppler broadening too.

There are so many additions to the elementary hydrogen atom model yet that is the model that nearly 'everyone' quotes when linking for explanations. The introductions to QM at school should really contain caveats very early on.

 

[cmb]

But distribution of atoms with different energies still doesn't affect this outcome, I think this is a 'red herring'. 'A' photon interacts with 'an' atom. 'That' atom has energy levels. How broad are those?

A photon may well have the right energy to interact with an atom in the middle of a volume of stuff, but the question is; will it?

What we started with in this thread was that if a photon wasn't the right amount of energy then it wouldn't interact with the atom. My query is that an 'analogue' level, say photon energy from BB, will never be exactly the level in 'an atom'. If there were a million atoms it'd simply mean it was different to a million of them. Either the statements earlier in the thread are wrong and BB radiation can interact with atoms, or the statements earlier were right and they can't. It can't be both.

If the excitation state for hydrogen is, let's say, exactly 2eV for the sake of the point, then will a photon of 2.1eV interact with it? If not, then 2.01eV. If not then 2.001eV. If not then ... stop me when it will interact ...

 

[Dale]

'A' photon interacts with 'an' atom. 'That' atom has energy levels. How broad are those?

That is what my response in post 72 talked about. In a solid, however, it is not usually true that a photon interacts with an individual atom. It interacts with the whole lattice.

If the excitation state for hydrogen is, let's say, exactly 2eV for the sake of the point, then will a photon of 2.1eV interact with it? If not, then 2.01eV. If not then 2.001eV. If not then ... stop me when it will interact ...

See here for details:
https://quantummechanics.ucsd.edu/ph130a/130_notes/node428.html
For the hydrogen 2p to 1s transition the linewidth is about 0.4 E-6 eV.

say photon energy from BB, will never be exactly the level in 'an atom'. ... Either the statements earlier in the thread are wrong and BB radiation can interact with atoms, or the statements earlier were right and they can't. It can't be both.

There is a third option, which is that your claim that BB radiation will never be the right level to be absorbed is wrong. Thus the statements are right and yet BB radiation can interact with atoms.

If I understand your reasoning it is simply that in a continuous distribution the probability of any single real number is 0. But the natural or intrinsic linewidth is finite, so this argument simply doesn’t apply.

 

[cmb]

There is a third option, which is that your claim that BB radiation will never be the right level to be absorbed is wrong. Thus the statements are right and yet BB radiation can interact with atoms.

If I understand your reasoning it is simply that in a continuous distribution the probability of any single real number is 0. But the natural or intrinsic linewidth is finite, so this argument simply doesn’t apply.

Thanks for the response and the additional information, which is good.

Yes, of course, if there were a 3rd option with a margin of uncertainty then that resolves the thing, but the statements above didn't say that. It's like I put But you've resolved that now, thanks.

I am still curious as to what happens to the 'excess/lacking' energy after an interaction? What I mean is that if you have a load of photons all bombarding a substance with a 'real' energy level right at the bottom of that margin and no photons above the median excitation level, where does the system get the extra energy from? Like a quantum version of Maxwell's demon?

 

[Dale]

Any extra energy can either go back into the EM field (inelastic scattering) or it can go into KE.

Edit: oops, on rereading I see you were actually asking about missing energy. There isn’t any missing energy either. Energy is uncertain. There is no “real” energy level until it is measured.