In the case of elastic scattering or absorption between photons and electrons, the incident light is either fully reflected or fully absorbed and hence it either retains its wavelength or ceases to exist. But how does inelastic scattering between them work? Specifically, if light has a wavelength, how exactly does that wavelength change in such a way that the light transfers only some of its energy and hence its wavelength increases?

  • $\begingroup$ Are you asking how it's possible for a photon to lose some amount of energy and hence have its wavelength increase? Or are you asking what the specific mechanism is that causes a photon to lose a specific amount of energy when interacting with an electron? Also, I think perhaps you're confused about what inelastic scattering means -- inelastic, in this context, means some of the energy involved in the "collision" between the particles escapes to the outside environment. AFAIK, it has nothing to do with whether all the photon's energy is transferred to the electron or only some of it. $\endgroup$ Commented May 27, 2023 at 4:06
  • $\begingroup$ Im asking about both of the questions above how is it possible and if yes then how does it do so? Also inelastic scattering is used in Raman spectroscopy so the wavelength recieved from that is higher or lower than original so by conservation of energy the photon must have transferred partial energy and must exist since it's scattering rather than absorption $\endgroup$
    – Razz
    Commented May 27, 2023 at 4:10
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    $\begingroup$ @MikaylaEckelCifrese but during Raman spectroscopy experiment we shine a monochromatic laser such that rarely inelastic scattering occurs so I'm not sure if its > energy probably it's the same energy as elastic scattering and I will check out Compton scattering thank you $\endgroup$
    – Razz
    Commented May 27, 2023 at 6:23
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    $\begingroup$ You might find this exposition on Raman scattering helpful. gravityandlevity.wordpress.com/2013/08/12/… $\endgroup$
    – KF Gauss
    Commented Jun 1, 2023 at 5:17
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    $\begingroup$ I found that so so much helpful but it would be great if you posted that as an answer @KFGauss $\endgroup$
    – Razz
    Commented Jun 4, 2023 at 14:05

1 Answer 1


One of the examples of inelastic light scattering is Raman scattering by an atom or molecule. During it, a photon with the energy $\hbar\omega$ is absorbed by an atom/molecule in the state $|i\rangle$, and a photon with some other energy $\hbar\omega'$ is emitted. The atom/molecule end up in the energy state $|f\rangle$ different from $|i\rangle$. The energy conservation relates the energies of the absorbed and emitted photon and the energies $E_i$ and $E_f$ of the atomic/molecular states as \begin{equation} E_i + \hbar\omega = E_f + \hbar\omega'. \end{equation} In the Wikipedia article on Raman spectroscopy such a process is illustrated for the case when the states $|i\rangle$, $|f\rangle$ are different vibrational states of a molecule.

The wavelength of the scattered light may by larger or smaller than the initial wavelength depending on whether $E_i < E_f$ (Stokes scattering) or $E_i > E_f$ (anti-Stokes scattering). This can be easily derived from the energy conservation law written above.

  • $\begingroup$ if absorption is taking place and emission , why do we call it scattering because scattering is such that the photon is reflected rather than absorbed and emitted? $\endgroup$
    – Naveen V
    Commented May 29, 2023 at 9:56
  • $\begingroup$ Yea I have the same question @NaveenV had , Also why is it so that the electron resides in a higher energy level than original ?? $\endgroup$
    – Razz
    Commented May 29, 2023 at 9:58
  • $\begingroup$ @Razz Why the electron resides in a higher energy level than original? Well, because it can. The electric field in general causes transitions between the energy levels which allowed by the dipole (quadrupole, etc.) matrix elements. In the case of Raman scattering, electric field in fact causes two transitions: first transition to a virtual intermediate energy level (absorption of a photon), and second one to a final energy level (emission of a photon). $\endgroup$
    – E. Anikin
    Commented May 29, 2023 at 10:13
  • $\begingroup$ @NaveenV I think it is a question of terminology. In my opinion, Raman scattering could be called something like "absorption - re-emission", but historically it is called "scattering". Also, note that there is no way to directly observe the absorption and re-emission processes that I've written above. These processes are virtual, like virtual particles in Feynman diagrams. In experiment, we have only incident light and emitted light. The emitted light has other direction and other wavelength than the incident light, so we call it "scattering". $\endgroup$
    – E. Anikin
    Commented May 29, 2023 at 10:19

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