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As far as I know, scattering occurs when light excites the atoms or molecules to their higher energy state(virtual state for scattering) followed by emitting photons corresponding to energy differences. I think this process is not quite different with usual absorption and following emission process. What can be the difference btw them?

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When a photon interacts with an atom, three things can happen:

  1. elastic scattering, the photon keeps its energy and changes angle (mirror reflection)

https://en.wikipedia.org/wiki/Elastic_scattering

  1. inelastic scattering, the photon keeps part of its energy and changes angle (photon transferring vibrational and rotational energies to the molecules, heat up the material)

https://en.wikipedia.org/wiki/Inelastic_scattering

  1. absorption, the photon gives all its energy to the absorbing atom/electron, and the photon ceases to exist

https://en.wikipedia.org/wiki/Absorption_(electromagnetic_radiation)

When there is spontaneous emission (like in your case), the electron relaxes, emitting a photon. Before that, the electron needs to be excited, by another photon, that is absorption.

So the answer to your question is that scattering is when the photon interacts with the atom, and it can lead to elastic scattering, inelastic scattering or absorption.

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  • $\begingroup$ What about pair production for photons of enough energy? $\endgroup$ – jmh Oct 14 at 2:02
  • $\begingroup$ @jmh you are correct, during pair production, the photon converts its energy into a electron positron pair, near an atomic nucleus. The photon and the atom interact only in terms of momentum conservation, the atom receives a recoil. In these terms it should be among elastic scattering. $\endgroup$ – Árpád Szendrei Oct 14 at 2:27

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