Answer to Question #9118 Submitted to "Ask the Experts"
Category: Radiation Basics — Photons
The following question was answered by an expert in the appropriate field:
As a medical physicist and a PET (positron emission tomography) user, I wonder what happens to a positron when it goes through matter.
The positron, created by the transformation of a proton to a neutron, a positron, and a neutrino in the nucleus of a radionuclide that decays by the beta-plus mode, results in the emission of the positron from the nucleus, with some kinetic energy, the amount depending on the energy available from the decay process and how that energy is distributed between the positron and the neutrino. In the case of the commonly used PET radionuclide, 18F, the positron may have a maximum energy of 633 keV; the average positron energy is about 250 keV.
The typical positron will expend most of its kinetic energy undergoing collision interactions, producing ionization and excitation in atoms of the medium through which it passes. When the positron has dissipated all of its kinetic energy, it will annihilate with a conventional electron. Such an event produces two 511 keV annihilation photons, which carry energy away from the interaction site in opposite directions. The 511 keV is the exact energy of each of the annihilation photons if the positron was at rest (having energy determined only by the temperature of its environment). While it is true that elevations in temperature produce increased kinetic energies of atoms and constituents, at the temperatures of typical concern, the typical increases are so slight as to be insignificant in terms of any effect on the observed energies of annihilation photons.
In a low-atomic-number medium such as soft tissue there will be a small amount of positron energy loss by bremsstrahlung x-ray production and by the process of in-flight annihilation. In the latter process, the positron annihilates with a conventional electron and whatever excess kinetic energy the positron possessed at the time of annihilation gets transferred to one or both of the two annihilation photons that result.
When such an event happens, it is likely that each of the annihilation photons will be left with different energies. As an example, if a 400 keV positron annihilates in flight, the 400 keV of kinetic energy may be split between the resulting annihilation photons. If 150 keV goes to one photon and 250 keV to the other, the respective annihilation photons would have energies of 661 keV and 761 keV. In addition, since the photons must behave so as to conserve the forward momentum of the positron at the time of the annihilation event, these photons will typically not be moving in opposite directions as is the case when the positron annihilation occurs after the positron has lost all of its excess kinetic energy.
It is approximately true that the average fraction of positron kinetic energy that is lost via in-flight annihilation is about the same as the fraction of the kinetic energy that is lost by bremsstrahlung (when the positron slows to rest). Considering our 400 keV positron, if we look up the bremsstrahlung radiation yield in water (near soft tissue equivalent) we find a value of 1.66 x 10-3, which represents the fraction of the positron energy that is lost, on average, as bremsstrahlung when the positron loses all its energy as it slows to rest. We may use this as an estimate of the average fraction of the 400 keV positron’s energy that is lost through in-flight annihilation. This would imply that about one out of every 600 positrons, each of 400 keV, would interact via in-flight annihilation. Thus, this is not a very important process for many practical situations.
The annihilation photons produced by in-flight annihilation would clearly not be useful in PET scanning since they are not generally emitted in directional opposition to each other, and their energies are different. Such photons would have a small negative effect in increasing unwanted detection events in the radiation detectors.
Thanks for an interesting question. I hope this answer is sufficient for your purposes.
George Chabot, PhD, CHP