Scintillator Compton Edge

In summary, according to these equations, the energy of the Compton edge should be the same as the energy of the photo-peak. However, this is not always the case. The angle in the Compton formula is the angle between the incident and scattered photons. At the "edge" that angle is 180 degress. However, when the gamma ray has the most energy, the electron is not scattered and the Compton edge occurs at a lower energy.
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I am trying to calculate the energy at which the Compton edge occurs for a sample of Cs137 using a NaI(Tl) scintillator. I know the original energy of the gamma ray was 661.6keV. and that for the Compton edge the angle between the electron and the line of incident must be 0degrees (for max energy of photon). but this just give the value of the photopeak (incident gamma ray). How do i calculate the energy of the emitted photons from Compton scattering at the Compton edge??

[i have calculated the beginning of the Compton continuum using an angle of 180 degrees between the electron and line of incident (minimum energy of the photon) and this gives a value which is about the same as my graph]
i have looked at a few sites but am fairly confused as to why the Compton edge is at a lower energy to the initial gamma ray photon.

Many thanks
 
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According to these equations the Compton edge should be the same energy as the photo-peak, as when theta=0 the equation is just 662keV/1.
i don't understand how it is possible to calculate the Compton edge, with these equations.
 
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rich86 said:
According to these equations the Compton edge should be the same energy as the photo-peak, as when theta=0 the equation is just 662keV/1.
i don't understand how it is possible to calculate the Compton edge, with these equations.
I think you are misinterpreting theta.

Thus, the maximum energy which may be deposited by a single scatter event is E - E' for = 180˚.

http://nucleus.wpi.edu/Reactor/Labs/R-scin2.html
 
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ok so plugging in theta=180 and E=2MeV to http://nucleus.wpi.edu/Reactor/pics/rscineq2.jpg gives E' as .23MeV as is shows in the diagram.
http://nucleus.wpi.edu/Reactor/pics/rscin3.jpg
But then theta=0 (ie gamma ray has most energy and electron is not scattered) E'=2MeV. this is not what is shown in the diagram.

how do you work out the 1.77MeV Compton edge?!
 
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  • #6
ok sorry i was being really stupid the 1.77MeV comes from 2MeV-0.23MeV.

many thanks for your help.
 

1. What is a scintillator Compton edge?

A scintillator Compton edge is a spectral feature that appears in the energy spectrum of a radioactive source. It is caused by Compton scattering, where a photon from the source interacts with an electron in the scintillator material, resulting in a decrease in energy of the scattered photon. The edge marks the maximum energy that a photon can lose in a Compton scattering event.

2. How is the scintillator Compton edge used in scientific research?

The scintillator Compton edge can be used in various applications, such as gamma-ray spectroscopy, medical imaging, and radiation detection. In gamma-ray spectroscopy, the edge can be used to identify the energy of the source and the type of radioactive decay. In medical imaging, the edge can help to distinguish between different types of tissues and to create images with higher contrast. In radiation detection, the edge can be used to determine the energy and direction of incoming gamma-rays.

3. How is the scintillator material chosen for a particular application?

The scintillator material is selected based on its properties such as density, light yield, decay time, and interaction cross-section with gamma-rays. In general, a material with a high density and high light yield is preferred for better energy resolution, while a short decay time is desirable for fast detection. The interaction cross-section determines the efficiency of the scintillator in detecting gamma-rays.

4. Can the scintillator Compton edge be affected by external factors?

Yes, the scintillator Compton edge can be influenced by external factors such as temperature, humidity, and radiation damage. Changes in temperature and humidity can affect the light output and decay time of the scintillator material, leading to a shift in the edge position. Radiation damage, on the other hand, can cause changes in the scintillator structure and affect its response to gamma-rays.

5. How can the scintillator Compton edge be used to improve radiation detection?

The scintillator Compton edge can be used to improve radiation detection by providing a more precise measurement of the energy of incoming gamma-rays. This allows for better identification and discrimination of different types of radiation, which is crucial for radiation safety and protection. Additionally, the edge can be used to calibrate the energy response of a detector, leading to more accurate measurements of radiation levels.

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