In Positron Emission Tomography (or PET)scans,a very small amount of a radioactive drug is given to a patient.Fluorine-18 doped or labeled fluorodeoxyglucose,also called FDG,is a radioactive version of the sugar glucose with a half-life of 110 minutes.Once injected into the body,it will go to places where glucose is used for energy.For example,FDG will go to the brain,and,in particular,to those parts of the brain that are actively burning glucose for energy.A variety of different images are observed depending on the parts of the brain being used in certain tasks,such as reading or listening.If the patient came back at the same time the next day after their initial injection for a further scan,how much of the original Fluorine-18 activity would remain? A)zero B)1.15 $\times$ 10^-2 % C)3.53 $\times$ 10^-2 % D)5.41 $\times$ 10^-2 % E)5.41 $\times$ 10^-4 %

Question

Quizplus · Accepted Answer

The half-life of Fluorine-18 is 110 minutes. After 110 minutes, half of the original activity will remain. After another 110 minutes, half of the remaining activity will remain, and so on. After 24 hours (1440 minutes), the fraction of the original activity remaining will be (1/2)^(1440/110) = 1.15 x 10^-2.

In Positron Emission Tomography (Or PET)scans,a Very Small Amount of a Radioactive