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

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Escape peaks from gamma rays in annihilation processes occur when one of the annihilation photons, typically at 511 keV, escapes detection. This results in a detectable energy value of E_peak, which is equal to the gamma ray energy minus 511 keV. The discussion raises questions about the momentum of the electron-positron system before annihilation, particularly why it may have a vanishing total momentum leading to two emitted photons. Additionally, the interaction of the incident photon with a nucleus is questioned regarding its role in producing a particle-antiparticle pair with vanishing momentum. Understanding these dynamics is crucial, especially in applications like PET scans, where most detected photons are 511 keV and emitted back-to-back.
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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