How do escape peaks from gamma rays occur in annihilation processes?

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SUMMARY

The discussion centers on the phenomenon of escape peaks in gamma-ray detection, specifically in the context of annihilation processes involving positrons and electrons. It is established that escape peaks occur at an energy value of E_{peak} = E_{gamma} - 511 keV, where the undetected annihilation photon possesses an energy of 511 keV. The conversation highlights the importance of momentum conservation in the annihilation process, noting that the total momentum of the electron-positron system must vanish for two 511 keV photons to be emitted. Additionally, it emphasizes the role of positron slowing down in matter, which is relevant for applications like PET scans.

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Angelos K
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How do escape peaks occure?

I mean peaks originating from a gamma ray located at the Energy value of \ E_{peak} \equiv E_{gamma}-511KeV.

I read that an annihilation process takes place and one of the annihilation photons escapes detection.To arive at detecting an Energy of E_{peak} the undetected annihilation photon must have had the energy of 511KeV.

Is there any reason why the particle system before annihilation (electron/positron) had a vanishing total momentum, so that two photons of 511KeV were emitted? Why did the incident photon interact with a nucleus in a way such that the produced particle/antiparticle pair had vanishing momentum?

I'd appreciate help.
 
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The annihilation cross section is small at high energies, so most of the time positrons slow down in matter before annihilating.
PET scans use that, too - most of their photons are 511 keV and back-to-back, even though the original positron from beta decays can have a large momentum.
 

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