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How we see things, relating to Crystal field theory.

  1. Jul 4, 2011 #1
    Sorry this is largely chemistry based, but delves into electron theory.

    I'm a little confused about how the colour of transition metal compounds arises.

    I understand that say for [Cu(H2O)6]2+, you have an octahedral arrangement around the central atom, and the specific positioning of these relative to the 3d orbitals causes the formation of non-degenerate orbitals, 3 at a 'lower' energy level, say xy,xz yz and 2 at a higher level, x2-y2, z2.
    With a bit of research and looking around everyone seems to agree that when visible light hits one of these compounds, the photons whose energy corresponds to the energy difference between the two orbitals will be absorbed, so if the difference was equal to a 'photon of red light' red light would be absorbed, and all others transmitted, resulting in a blue colour perceived.

    My question is what happens to the energy that the 'excited' electron receives is it lost as heat?
    From my (limited) understanding of electrons and stuff, when you excite an electron, and it falls back down, it releases a photon of specific energy corresponding to the electron's initial energy. So why wouldn't an electron, excited by a 'photon of red light' not transmit red light as it 'falls' back down to its ground state?
  2. jcsd
  3. Jul 4, 2011 #2
    You reason as if you were in vapor, when this excited electron could disexcite before the next collision with another molecule. But you are in aqueous solution, where the mechanical means for disexcitation are many and quick. It is the general phenomenon of thermalisation. The recipient of the red photon does not remain isolated.
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