I understand that in a crystalline metallic structure, such as one making up a bar of gold, there are one, or more, valance electrons of each atom that have left their outer shell (became free electrons as part of a metallic bond) and they are the electrons absorbing and reflecting light. But what about the other electrons that remain within the orbitals of those positive ions? Can those electrons absorb light, and then reflect light as they drop to lower levels? Or are those particular electrons cancelled out by the positive charge of the ions, therefore no absorbtion/emmision?

  • $\begingroup$ They will defnately show such effects. Don't Noble gases reflect light. $\endgroup$ – Anubhav Goel Feb 9 '16 at 4:31

In addition to the quantum mechanical model of electrons being in bound energy states , in orbitals around an atom, the band theory for solids, a quantum mechanical model, is necessary to explain the interaction of metals with light.

The electrons around an atom occupy more and more bound energy states. A photon that does not have the appropriate energy to kick out an electron from an atom, interacts with the atom as a whole. Atoms have spill over electric and magnetic fields ( which form the chemistry of the elements) and the scattering of the photons that one sees as reflection happens with the spill over field, of the atom as a whole, and is consistent with the classical electromagnetic wave scattering.

In metals, the band of energy states are bands where the electrons are shared by the lattice as a whole, as if it were one huge "molecule". The energy differences between the levels in the band are very small, so an incoming photon will interact with the spill over field from the band of electrons and scatter from the field, not an individual electron, following the classical reflection of electromagnetic waves. If the photon happens to have the appropriate energy to interact with a single electron in the band, one gets the photoelectric effect, the electron is ejected. There is no elastic interaction with a single identifiable electron. The elastic interactions giving rise to reflection happen with the field of the whole band .

The electrons in shells that do not belong to the conduction band are more tightly bound, i.e. photons of higher energy are necessary to eject them. There is no elastic interaction to give reflection from lower bound states.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z May 5 '16 at 9:40

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.