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I recently learned that gold gets its yellow color because electrons in the 5d orbital absorb 'blue' photons and transition to the 6s orbital. The drop in the blue end of the visual spectrum makes the reflected light appear yellow.

This is fine, but raises a number of questions I haven't been able to find answers to. I will continue to use gold as the running example but I imagine there are more general implications:

  1. The 6s orbital can hold 2 electrons, but one spot is already taken in the ground state for gold. So each gold atom can effectively absorb one 'blue' photon, yes? So what happens when the next 'blue' photon comes along?

  2. When the electron in 6s decays to its ground state in 5d, it must emit a photon. What wavelength does that photon have? Shouldn't it be pretty close to the 'blue' photon that excited it in the first place? Shouldn't this negate the effect on the overall color of gold?

  3. What's the relationship between (1) and (2) above? How quickly does an electron return to ground state, and how many 'blue' photons can a gold atom absorb per second?

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The outer-shell 6S orbital is significantly larger than the others, and the outer orbitals of adjacent atoms hybridize to form a band with a continuous distribution of energies, centered around the energy of the 6S orbital of an isolated Au atom. This half-filled band (half filled because the ground state of an isolated atom has one 6S electron) is what makes Au a conductor. The conduction band describes the conductive sea of electrons that are spread out between the Au atomic cores and which are not, in general, tightly bound to any one Au nucleus.

This means that in a macroscopically sized sample of Au there is a continuum of available states that the 5D electrons can be excited into, and so there is range of blue light that can be absorbed, not just a single isolated wavelength. Not all the blue light that impinges on a piece of Au is absorbed, but enough of it does get absorbed to deplete that part of the spectrum, leaving behind the golden color of the reflected light.

In this process, a given photon doesn't have to interact with a single atom that it is aimed at. The wavelength of the light is much longer than the interatomic distance, so even a very narrow pulse of blue light will still fall on a region containing many atoms. If a 5D electron from every single near-surface atoms were promoted to the 6S conduction band, then the absorption would indeed cease, but unless you expose the Au to extremely intense light, this essentially never occurs, because the electrons in the 6S band fall quickly back into the 5D orbitals.

Sometimes, this transition back to the 5D state is accompanied by the emission of a blue photon, which essentially cancels out the absorption that occurred before. If every transition back to the 5D state radiated this way, then you would not get a net golden color, since the absorbed and reemitted photons would quickly balance out. However, it is also possible for the 6S-band electrons to be de-excited in other ways. In particular, an electrons can collide with an atomic core and excite a oscillation in the nuclear lattice (a phonon). (The fact that the 6S orbitals are spread out in a band actually also helps make it easier to have this kind of collisional de-excitation.) With this kind of process, the net transfer of energy is from the original blue photon into lattice vibrations, or heat.

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