4 ms·

So if I'm a gold atom, and I absorb some photons, a bunch pass through me, and some get reflected - and the photons that are reflected from me over a period of

by jwecker 17y ago

So if I'm a gold atom, and I absorb some photons, a bunch pass through me, and some get reflected - and the photons that are reflected from me over a period of time happen to be average out to something a cone cell in the retina would interpret as gold- couldn't I say that I'm gold-colored? In other words, replacing the billion-atom aggregation with a simple time series? It seems to me it's an argument about semantics, not physics.

olavk 17y ago
I think there is some terminology confusion. The image is not a "photo" - it is not a recording of emitted light. Hence the colors used to display the image for us have nothing to do with the color (reflected wavelengths) of the molecules.
asdlfj2sd33 17y ago
Ah but protons aren't particles... except some times they are, other times they are a waves. The bottom line is individual atoms don't reflect protons.
- antaeos 17y ago
  Did you mean photons?
  - asdlfj2sd33 17y ago
    Embarrassingly, yes. Obviously I can neither type nor spell well.
graphene 17y ago
that's not the case, the physics in fact is very different for a single atom than for a lump of metal. A single atom has very specific set of electron energies as a consequence of the electrons being confined to the area around the nucleus. This means that it can only absorb or emit photons with energies corresponding to the spacings between any two energy levels (and the absorption/emission must be accompanied by a jump/fall of an electron between corresponding energy levels. Any type of atom will thus have a secific pattern of absorption/emission peaks, this is incidentally how we can tell the composition of stars and dust clouds in distant galaxies. For a lump of metal, a good approximation is to consider the electrons as being able to freely roam throughout the entire crystal. As a consequence, when they are excited by an incoming photon (of any energy), they immediately relax back to their original state, emitting the photon again at the same energy. This is why metal films act as mirrors and why metals in general are shiny. As for the goldish color of Gold, that has to do with some relativistic effects which I don't understand (yet ;-) ). I believe the same is true for Copper.

by jwecker 17y ago

olavk 17y ago

I think there is some terminology confusion. The image is not a "photo" - it is not a recording of emitted light. Hence the colors used to display the image for us have nothing to do with the color (reflected wavelengths) of the molecules.

asdlfj2sd33 17y ago

Ah but protons aren't particles... except some times they are, other times they are a waves. The bottom line is individual atoms don't reflect protons.

antaeos 17y ago
Did you mean photons?
- asdlfj2sd33 17y ago
  Embarrassingly, yes. Obviously I can neither type nor spell well.

graphene 17y ago

that's not the case, the physics in fact is very different for a single atom than for a lump of metal. A single atom has very specific set of electron energies as a consequence of the electrons being confined to the area around the nucleus. This means that it can only absorb or emit photons with energies corresponding to the spacings between any two energy levels (and the absorption/emission must be accompanied by a jump/fall of an electron between corresponding energy levels. Any type of atom will thus have a secific pattern of absorption/emission peaks, this is incidentally how we can tell the composition of stars and dust clouds in distant galaxies. For a lump of metal, a good approximation is to consider the electrons as being able to freely roam throughout the entire crystal. As a consequence, when they are excited by an incoming photon (of any energy), they immediately relax back to their original state, emitting the photon again at the same energy. This is why metal films act as mirrors and why metals in general are shiny. As for the goldish color of Gold, that has to do with some relativistic effects which I don't understand (yet ;-) ). I believe the same is true for Copper.