Electrons Absorbing Photons - Extra Energy

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Legaldose
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Okay, so when a photon travels near an electron orbiting some atom, the electron then absorbs the photon (given that it has enough energy) and causes the electron to jump up an energy level. My question is what happens when a photon has a little more than enough energy to bump up the electron, where does the remaining energy go? Is it correct to say that another photon gets released that carries the remaining energy? And if so, then is that partly why objects are certain colors? If I have this totally wrong please feel free to correct me. Thanks.
 
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objects are certain colours due to the fact that they REFLECT that colour of light. such that there phonons are opaque to that frequency.
a photon is absorbed by an atom causing it to be excited and an electron moves to a higher energy state. photons will only be absorbed by atoms if they have the exact energy required to go from one energy state to another. if not they simply won't be absorbed. and you are correct the energy is released in the form of another photon of the EXACT same energy as the original due to conservation of energy.
 
So you are saying that if it takes 1J to move an electron up an energy level, and a photon comes along that has 1.1J of energy, then the photon will not be absorbed?
 
For monatomic gases, it is fair to say that for practical purposes electronic transitions do not occur with appreciable intensity unless the photon frequency matches the gap between energy states - although this does not have to be "exact", due to the uncertainty principle.

Photons can,however, undergo electronic Raman scattering even in this situation (in other words, the transition takes place, and a red-shifted photon carries away the excess energy). Such transitions are rather weak, and not easily observed (laser or synchrotron sources are needed).

In molecules and condensed phases a variety of mechanisms exists to take up any excess photon energy with appreciable probability, leading to broad absorption lines (as seen in the UV-visible spectra taken in the chemistry lab).
 
Thanks gadong, this clears things up quite a bit for me. And even better, it makes good sense as well.