Question

In preparation for a PET scan, a 75 kg patient is injected with fludeoxyglucose, a special type of glucose containing radioactive Fluorine-18 (18F). The injected solution contains 2.6 mCi of 18F, which has a half life of 109.8 minutes. Each radioactive particle has a mean energy of 249.8 keV, 90% of which is absorbed by the patient’s body. (1 mCi= 3.7*107 Bq.)

a) Draw a graph showing the activity of the solution as a function of time. Label the half life and tau = λ^-1 .

b) Calculate the number of radioactive 18F nuclei in the initial solution.

Answer 1

Nl Xt 0.693 1452 xi0

Please rate it up thanks :)