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I can't understand why a free electron can't interact with a photon (because of a violation in the law of conservation of momentum) when the reflectivity of metals depends upon photons interacting with free electrons. This seems like a contradiction. Can someone one please explain this to me.

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    $\begingroup$ "free electron can't interact with a photon due to [...]" You have misunderstood what you've read. A free electron can't simply absorb (or simple emit) a photon, with no other consequences, but that doesn't mean that it can't interact in more complicated ways. Not that the photon picture is the clearest way to understand reflectivity in metals which is better analyzed in the classical wave model. $\endgroup$ – dmckee --- ex-moderator kitten Jul 1 '18 at 17:22
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    $\begingroup$ Also, electrons that are responsible for reflection are not really free as in single electron in vacuum. They are bound to the metal, free to move within it, but still constrained to not go out of it, unless something rips them out. $\endgroup$ – Ján Lalinský Jul 1 '18 at 17:24
  • $\begingroup$ Is there any condition to absorb a photon? $\endgroup$ – Anns Jul 1 '18 at 17:42
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    $\begingroup$ Typically (I think), electrons in a conductor (or semi-conductor) that can participate in an electric current are called mobile electrons rather than free electrons. $\endgroup$ – Alfred Centauri Jul 1 '18 at 18:29
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First you have to learn that you are asking about two types of electron states.

You are talking about a free electron.

A free electron is unbound in vacuum, away from any other particle or atom.

That free electron cannot absorb or emit photons.

Now the electron inside the metal, that electron is not free. It is still bound to the metal, it is free to move inside the metal, and that is why it is a conductor.

Now the high reflection of metals, it is the reason for metals (most of them) have silver color.

In metals this is due to electrons on d levels (as per QM) to move to s orbitals. This needs lot of energy (ultraviolet) and so visible light does not have enough energy to match this energy level and so photons with the wavelengths of visible light will not be absorbed but elastically scattered.

That is how reflection works, elastic, Rayleigh scattering, and that is how a mirror image is built, because that is the only way a photon keeps its energy and phase.

That is why it is silver, because it reflects all visible lights, it does not emit any of its own light, it does not have its own color.

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  • $\begingroup$ rayleigh scattering of infrared waves is very low,then can it pass through metal,at the least some of the initial intensity. $\endgroup$ – Anns Jul 1 '18 at 17:53
  • $\begingroup$ are you asking if infrared can pass through metal? $\endgroup$ – Árpád Szendrei Jul 1 '18 at 18:10
  • $\begingroup$ Yes,i am asking if infrared can pass through metal. $\endgroup$ – Anns Jul 1 '18 at 18:13
  • $\begingroup$ The answer is no, it cannot. 90-95 percent gets reflected, the rest goes into the metal, and some of it will create heat the metal. Please see here: quora.com/… $\endgroup$ – Árpád Szendrei Jul 1 '18 at 18:15
  • $\begingroup$ Why metals are highly reflective,I know it is due to presence of free electrons but what is the mechanism,can you give me a link?. $\endgroup$ – Anns Jul 1 '18 at 18:20

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