In positron-emission tomography (PET) used in medical research and diagnosis, co

Question

In positron-emission tomography (PET) used in medical research and diagnosis, compounds containing unstable nuclei that emit positrons are introduced into the brain, destined for a site of interest in the brain. When a positron is emitted, it goes only a short distance before coming nearly to rest. It forms a bound state with an electron, called "positronium, which is rather similar to a hydrogen atom. The binding energy of positronium is very small compared to the rest energy of an electron. After a short time the positron and electron annihilate. In the annihilation, the positron and the electron disappear, and all of their rest energy goes into two photons (particles of light) which have zero mass; all their energy is kinetic energy. These high energy photons, called "gamma rays", are emitted at nearly 180° to each other. What energy of gamma ray (in MeV, million electron volts) should each of the detectors be made sensitive to? (The mass of an electron or positron is 9x10-31 kg. 1 ev - 1.6x10-19 joules.) Mev

Explanation / Answer

The energy of the system is mainly just the combined rest energies of the positron + electron.
So Eelectron + Epositron + Ephotons = 0, or more simply 2mc^2 + Ephotons = 0, since all of the rest energy of the positron and electron is converted to the combined energy of the two gamma rays.
Since the gamma rays are emitted at nearly 180 to each other, they must have equal momenta and therefore equal energies (since the initial momentum of the system was nearly zero), so each of the detectors should be made sensitive to the energy of one gamma ray, which is mc^2

=> E = mc^2/1eV
=> E = (9*10^-31*(3*10^8)^2)/1.6*10^-19
= 506250 eV
= 0.506 MeV

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