5
$\begingroup$

Considering a light wave traveling from the vacuum to air, I am trying to find what will happen to its wavelength.

At first, using optics, we know that:

$$n=\frac{c}{v}$$

where $n$ represents the refractive index. As $n_{vac}=1$:

$$n_{air}= \frac{\lambda_{vac}}{\lambda_{air}} \iff \lambda_{air} = \frac{\lambda_{vac}}{n_{air}} $$

So the wavelength will decrease in air.

But when using Compton Scattering:

The wave will lose energy after colliding with an air particle and its wavelength will increase.

Why is there a difference? Which is right?

Improve this question
$\endgroup$
2

1 Answer 1

Reset to default
10
$\begingroup$

Both are right in their respective situations.

Optical refraction is a relatively low-energy (non-ionizing) effect which does not exchange energy with the medium. The atoms have electromagnetic fields which interact with that of the photon, increasing the electrical permittivity $\epsilon$ of the medium over that of free space, $\epsilon_0$ (I am not sure if the same happens for magnetic permeability $\mu$, I don't see why not). According to Maxwell's equations for electromagnetism, this reduces the speed of propagation. It is this speed change at different points along the wave front which is responsible for the shortened wavelength and for refraction.

But Compton scattering is different. The photon energy is much higher and it physically collides with an individual atom, imparting energy to it and causing it to recoil, typically knocking an electron away and ionizing it. The photon loses that energy so, according to $\lambda=hc/e$, its wavelength must increase.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Then, in an experimental situation, which would you choose? As an undergraduate, I always believed I should use optical refraction, but your explanation seems to say that the Compton-scattering will have a greater effect on the output wavelength. Or am I misunderstanding your answer? Sorry, I am merely curious, as I am not the original poster. $\endgroup$
    – user7077252
    Apr 2, 2020 at 12:14
  • 3
    $\begingroup$ It depends on whether the photons have enough energy to ionize the atoms. For example optical wavelengths are classed as non-ionizing radiation, gamma rays as ionizing radiation. $\endgroup$
    – Guy Inchbald
    Apr 2, 2020 at 12:17
  • $\begingroup$ To be precise, Compton scattering applies to a "free charged particle"; but when the photon's energy is significantly greater than an atomic electron's binding energy, then for the purposes of scattering, we can pretend that it's "free": en.wikipedia.org/wiki/Compton_scattering $\endgroup$
    – Lawnmower Man
    Apr 2, 2020 at 20:53
  • $\begingroup$ Why 9?......10! It is very good. $\endgroup$
    – Sebastiano
    Apr 7, 2020 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.