B Do Electrons Get Excited and De-Excited Equally in Colored Compounds?

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Compounds exhibit color due to the excitation of electrons between atomic or molecular orbitals, resulting in the absorption of specific wavelengths and the perception of complementary colors. While it may seem that the number of excited electrons should equal those de-exciting, this is not the case because energy can be lost permanently through various interactions. In solid-state materials, energy transitions are more complex, involving energy bands rather than simple line transitions. Additionally, absorbed energy can be redistributed within the solid, leading to scattering that affects the wavelengths transmitted. This understanding clarifies why compounds are colored rather than colorless despite electron excitation and de-excitation processes.
PumpkinCougar95
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I am not able to understand why compounds have colour. I have read that it is due to excitation of electrons from one atomic/Molecular orbital to another because of which certain wavelengths are absorbed, And we can see the "complementary colour" of those wavelengths.

What I don't understand is that when an electron gets excited, it has to eventually de-excite and come back to its initial orbital. Now let's say that on average an electron remains excited for some time ##t##. After ##t## time, shouldn't there be an equal number of electrons getting excited per sec to those getting de-excited?

If this were to happen, the same wavelength that is absorbed by one atom should be emitted by another atom? Wouldn't this mean that the compound is colourless?

I am confused, and any clarification would be greatly appreciated.
 
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PumpkinCougar95 said:
it has to eventually de-excite and come back to its initial orbital.
But not necessarily via the same single transition. What is initially taught about EM interaction with atoms is not enough for this.
The energy at the incident frequency can be lost for ever (absorbed). In the solid state, there are not simple line transitions (you get energy bands) and energy passes around within the solid. Even in a low density gas, the energy that's absorbed from one direction will re radiate in all directions, causing energy at certain values to be scattered - less of the original wavelength passes through with the rest of the beam.
 
Thanks ,that makes sense.
 
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