Compton Scattering (Back-Scattering)

In summary, if high-energy gamma rays are perfectly back-scattered at 180 degrees, the detected scattered photons will all have the same energy of 0.26 MeV, regardless of the incident gamma ray energy as long as it is much larger than the rest-mass energy of an electron. This can be proven using the Compton scattering equation and converting it to an energy equation.
  • #1
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Homework Statement



Consider the situation where high-energy gamma rays are striking a detector after
scattering off of the electrons in the material surrounding a detector. Show that if
the gamma rays are perfectly back-scattered by the material, such that the scattering
angle is θ = 180 degrees, then the detected scattered photons will all have essentially
the same energy of 0.26 MeV, independent of the precise energy of the incident gamma
rays as long as the incident gamma ray energy is much larger than the rest-mass energy
of an electron.

Homework Equations



[tex]\lambda[/tex]final - [tex]\lambda[/tex]initial = [tex]\frac{h}{(me)c}[/tex] (1 - cos[tex]\theta[/tex])

The Attempt at a Solution



Well for [tex]\theta[/tex] = 180 degrees, the compton scattering eqn becomes
[tex]\lambda[/tex]final - [tex]\lambda[/tex]initial = [tex]\frac{2h}{(me)c}[/tex].
I'm confused about where to go from here since [tex]\lambda[/tex]final - [tex]\lambda[/tex]initial needs to be proven it is 0, and also that [tex]\lambda[/tex]final = [tex]\lambda[/tex]initial = 0.26MeV.
I have a feeling I need to to use the binomial expansion somewhere since it is saying that the incident gamma ray energy has to be much larger than the rest-mass energy
of an electron. Can someone push me in the right direction?
 
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  • #2
Hi person in the same PHYS242 class at Queen's I'm in.

You're on the right track. Just convert the equation you have right now to an energy equation using ∆E and you will get ∆E = 1/2mec2. Solve this and you will get 0.26 MeV with the assumptions that your initial lambda is much much smaller than your final lambda (and therefore your ∆E can be considered a final E rather than delta).
 

Related to Compton Scattering (Back-Scattering)

1. What is Compton Scattering (Back-Scattering)?

Compton Scattering (Back-Scattering) is a physical phenomenon that occurs when a photon (particle of light) collides with an electron, causing the photon to lose energy and change direction. This process is also known as inelastic scattering, as the energy of the photon is not conserved.

2. How is Compton Scattering (Back-Scattering) different from regular scattering?

In regular scattering, the photon interacts with the entire atom and changes direction without losing energy. In Compton Scattering (Back-Scattering), the photon interacts with only one electron and loses energy in the process, resulting in a change in wavelength.

3. What is the significance of Compton Scattering (Back-Scattering) in science?

Compton Scattering (Back-Scattering) is significant in science because it provides evidence for the particle-like behavior of light, as well as the wave-particle duality of matter. It also allows scientists to study the structure of atoms and the behavior of electrons.

4. What are some real-world applications of Compton Scattering (Back-Scattering)?

Compton Scattering (Back-Scattering) is used in medical imaging techniques such as X-rays and CT scans, where the scattering of high-energy photons can provide valuable information about the internal structure of the body. It is also used in materials science to study the properties of materials at the atomic level.

5. How does the angle of scattering in Compton Scattering (Back-Scattering) relate to the energy of the photon?

The angle of scattering in Compton Scattering (Back-Scattering) is directly related to the energy of the photon. As the energy of the photon increases, the angle of scattering also increases. This relationship is described by the Compton Scattering formula, which can be used to calculate the change in wavelength of the scattered photon.

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