Do photons lose energy due to interaction with the particles in a medium? If yes, then does the frequency of the component electromagnetic waves change? ( since energy of light is directly proportional to it's frequency)


It is possible for light to be absorbed in a medium (most real world materials are not perfectly transparent). When this happens, most individual photons are fully absorbed - the ones that are not absorbed continue with the same energy.

The one exception to this: there are certain scattering mechanisms in solids (most notably, Compton scatter and Raman scatter) in which a photon scatters off an electron, and only part of its energy is passed on to the electron. Such a photon then continues in a new direction, and with less energy (longer wavelength). For visible light, the probability of Compton scatter is extremely small - most interactions of photons with matter in the 400 nm - 800 nm range result in absorption.

Also, because the energy of the scattered photon is related to the energy of the incident photon by the Compton equation:

$$E' = \frac{E}{1+\frac{E}{m_0 c^2}(1-\cos\theta))}$$

When $E\ll m_0 c^2$, very little energy will be lost according to this equation: in the limit, this results in Thomson scatter, where no energy is exchanged. In situations where electrons are sufficiently bound to their atoms that the energy of the photon is insufficient to knock them free, this approximation becomes exact: the electron can't absorb "just a little bit" of energy, as that is quantum-mechanically forbidden.

Raman scattering on the other hand provides a mechanism for inelastic scattering in the visible and near-UV regime - so yes, some photons will lose part of their energy and keep going. Raman spectroscopy is used for material characterization especially because it's so specific in its mechanism (and there are no other mechanisms that would cause scatter, and that would drown the result).

In reality, for visible light you are firmly in that regime. At higher energies, the above comes into play. So it depends a bit on your definition of "light".

  • $\begingroup$ Can electromagnetic wave of higher frequency undergo scattering and lose energy to turn into visible light? $\endgroup$ – D.K. Jun 6 '17 at 15:15
  • $\begingroup$ One mechanism that does something like that is fluorescence - strictly speaking the photon is completely absorbed, and another (lower energy) is emitted; but given the time scales of the event it looks like a photon is scattered and loses energy. $\endgroup$ – Floris Jun 6 '17 at 15:29
  • $\begingroup$ But the process you are really asking about is Raman scattering - an actual inelastic scattering that produces a photon of lower energy, and that operates at this range of wavelengths / energies. $\endgroup$ – Floris Jun 6 '17 at 15:31
  • $\begingroup$ Is a high frequency wave travelling in a denser medium more likely to get scattered? $\endgroup$ – D.K. Jun 6 '17 at 15:34
  • $\begingroup$ Certainly density helps; the scattering cross section is a function of energy, but typically higher energy radiation is more penetrating (so less likely to interact). However, when you look at X rays traveling through matter: at low energy they are more likely to be absorbed, while at high energy, if they interact at all, the interaction is likely to be scatter. Sorry that's not a "yes/no" answer... $\endgroup$ – Floris Jun 6 '17 at 15:36

Yes. They do.

But this depends on $2$ things:

  1. the frequency of the incident radiation, ($\nu$) and
  2. the strength of the bond between the particles of the medium, (E)

Generally, light quantum or the concerned photon is absorbed when $E>> h\nu$. However, if $E\approx h\nu$ or, $E<< h\nu$, we have scattering of light, an example of the particle nature of light.

At low energy, this scattering phenomenon is called Photoelectric Effect, followed by Rayleigh Scattering; at comparatively higher energy, there is Compton Scattering and at extremely high energies, Pair Production.

The scattered photon loses energy in the process, has its wavelength increased and as usual, its frequency decreased.

  • $\begingroup$ you mean wavelength decreased? $\endgroup$ – D.K. Jun 6 '17 at 15:17
  • $\begingroup$ @D.K. No, wavelength INCREASED, since energy is inversely proportional to wavelength. $\endgroup$ – SchrodingersCat Jun 6 '17 at 15:22
  • $\begingroup$ of sorry, I meant frequency decreased? @SchrodingersCat $\endgroup$ – D.K. Jun 6 '17 at 15:24

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