Scintillator Compton Edge

rich86

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

OlderDan

The angle in the Compton formula is the angle between the incident and scattered photons. At the "edge" that angle is 180 degress. These two sources are consistent, and the second one might be more generally useful for your problem. Either one will give you the energy of the scattered gamma photons.

http://hyperphysics.phy-astr.gsu.edu.../comptint.html

http://nucleus.wpi.edu/Reactor/Labs/R-scin2.html

rich86

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.

OlderDan

I think you are misinterpreting theta.

rich86

ok so plugging in theta=180 and E=2MeV to gives E' as .23MeV as is shows in the diagram.

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?!

rich86

ok sorry i was being really stupid the 1.77MeV comes from 2MeV-0.23MeV.

many thanks for your help.