4
$\begingroup$

I hope someone could correct the mistakes I'm making in my speech. Thanks.

Some say that when matter heats up, its atoms start accelerating faster in random directions exchanging energy by bumping into eachother.

Accelerating atoms inside a material are supposed to emit EM waves. So when matter heats up, it starts emiting more energy as EM radiation. If we accelerate a free charge it starts emiting light. The same is true for accelerating ions which are just atoms with an unbalanced charges. Does the same happen for atoms if they are not bound in a material?

That seems to me like both electrons and protons emit light as they accelerate in the same directions inside the atom which is bound to a material. But its EM waves destructively interfere so they should emit nothing. But wouldn't that mean they lose energy anyways because they've all emited it?

Other people say that heated atoms don't really accelerate but the electrons jump to a higher energy state, and when they fall down, they emit light. But that would mean that accelerating ions then shouldn't emit EM waves.

What is correct interpretation then?

Both can't be correct.

$\endgroup$
6
$\begingroup$

Accelerated charged particles emit electromagnetic radiation. The emitted power is proportional to the square of the acceleration. Given a fixed charge, then an accelerating Lorentz force produces an acceleration inversely proportional to mass and so radiated power is inversely proportional to the square of the mass.

As ions are much more massive than free electrons, then their radiative output (by this mechanism) can usually be neglected. There is no question of constructive or destructive interference here because there is no reason that the light emitted by separate particles should have any particular phase relationship.

Heated atoms and ions do move faster and heat may result in electrons occupying higher energy states in both (if the ions have any remaining electrons). You appear to be confused between two mechanisms of producing radiation. One consists of acceleration of free charged particles - free-free emission that could be thermal in nature or caused by external fields; the other is transitions within atoms/ions - bound-bound transitions between bound energy states.

In general both these things (and bound-free and free-bound radiation) are occurring.

$\endgroup$
  • $\begingroup$ Ok, so electron and an ion which has 1 extra electron both emit the same radiation if they have exactly the same acceleration (they move side by side the whole time)? But how does the extra electron in the ion know to start emiting em waves, while the rest don't? I'm just puzzled what stops the accelerating free atoms from radiating EM waves? $\endgroup$ – MaDrung May 31 '17 at 8:28
  • $\begingroup$ Atoms cannot be treated as multiple individual charges each accelerating in a particular way. A classical treatment does not work for bound atoms or ions. @MaDrung $\endgroup$ – Rob Jeffries May 31 '17 at 9:48
  • $\begingroup$ Ok, but ions do radiate because of accelerating, right? Which is wierd then if it's not because of all individual charges radiating. hmmm $\endgroup$ – MaDrung May 31 '17 at 9:51
  • $\begingroup$ >> There is no question of constructive or destructive interference here because there is no reason that the light emitted by separate particles should have any particular phase relationship. -- Why is this? Even if extremely small, there ought to be at least some interference pattern, no? $\endgroup$ – MPath May 31 '17 at 10:35
  • $\begingroup$ @mikey Why do you think that there should be any fixed phase relationship between the light emitted by any particular ion and electron (even ignoring the fact that that the amplitude of light emitted from the ion is millions of times smaller)? $\endgroup$ – Rob Jeffries May 31 '17 at 10:44
1
$\begingroup$

The optical vibrational modes of salts are in the infrared. Those absorb and reflect strongly (Reststrahlen), so they will also radiate. In the same region of the spectrum, covalent materials can be transparent, so the black-body radiation will be low.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.