Question

Fluorine-18 undergoes positron emission with a half-life of 1.10 x 102 minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.) 

Answer 1

Decay constant \(\mathrm{k}=\ln (2) /\) half life

\(=\ln (2) / 1.10 \times 10^{2}=0.0063013 \mathrm{~min}^{-1}\)

\(\ln (M / M o)=-k t\)

where \(M\) is mass at time \(t\), Mo is initial mass

\(\ln (83 / 248)=-0.0063013 \times t\)

Time \(\mathrm{t}=174 \mathrm{~min}=1.74 \times 10^{2} \mathrm{~min}\)