Question

a) Two photons per atomic decay are emitted, striking the detectors mounted on linearly opposite sides of a ring that surrounds the patient in the PET scanner. How does the conservation of linear momentum explain the motion of these two photons?

b) Discuss what you understand from errors in determining the location of emitting nucleus due to positron range and non-collinearity in PET scans. Should E_max be big or small, to reduce the positron range error?

thank youu

Answer 1

a)

The total momentum of the isotope before decay is equal to the total momentum of the two released photons after decay. The two photons which are released travel in opposite directions to strike the plates, as velocity is opposite, the linear momentum will be zero after they are releases ( obviously the mass will be same). Before decay , the isotope was at rest, so momentum was zero,

therefore, linear momentum before decay and after decay was zero.

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b)

Non collinearity is independent of radio nuclide used. However, the error in range of position is dependent on the energy of emitted positrons. The error is generally determined from separation of detectors.