Question

A common radiopharmaceutical in PET is fluorodeoxyglucose (FDG) which is an analog to glucose.

The chemical formula is C6H11FO5, where the fluoron is 18F. Apatient is injected with a saline solution containing FDG. 18F decays to

the stable isotope 18O via electron capture or positron emission. The latter process is 30 times more common than the former. The half-life is 110 min.

(a) What mass of FDG must the solution contain in order to administer an activity of 8.0 mCi? After 45 min a PET scan is initiated.

(b) What is the rate of annihilation photons exiting the patients body (you can safely assume that the photons do not interact with the patient)?

(c) The PET scan takes 35 minutes. How many annihilation photons escape the body during that time?

Answer 1

(a) After 45 min the activity of F18 will be

A=A0exp(-t) = A0exp(-0.693t/T1/2)

A= 8.0 * exp (- 0.693*45/110) = 6.025 mCi.

(b) the rate of annihilation photons will be,

8.0 mCi = 8.0 *3.7*107 dis/sec = 0.296 *109 dis/sec

each disintegration will produce 2 annihilation photons of 0.511 MeV

rate of annihilation photons = 2* 0.296 *109 /sec = 0.592 *109 /sec

(c) Activity before scan is 8.0 mCi, and activity after 35 min will be

A=A0exp(-t) = A0exp(-0.693t/T1/2)

A= 8.0 * exp (- 0.693*35/110) = 6.42 mCi.

the disintergration during 35 min is, 8.0- 6.42=1.58 mCi = 5.846* 107 /sec

so annihilation photons escape from body will be, 2* 5.846* 107 /sec= 1.1692* 107 /sec