fluorine-18 undergoes positron emission with a half-life of 1.10x10 minutes if a

Question

fluorine-18 undergoes positron emission with a half-life of 1.10x10 minutes if a patient is give previous l 18 of 21 l nex em 18 Part A Fkome 18 undergoes postron emission with a hal of 1.10 x 102 minutes fa patientis given a 248 mg dose for a PET scan, how long will take for the amount of fluorine- 18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.) O 132 x 102 minutes O 1.74 x 10 minutes 300 x 10 minutes O 99 minutes O 2.11 x 10 minutes Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

half - life = 1.10 x 10^2 min

rate constant = 0.693 / t1/2

                      = 0.693 / 1.10 x 10^2

                      = 6.3 x 10^-3 min-1

rate constant k = 1/t ln (Ao / At)

6.3 x 10^-3 = 1/t ln (248 / 83)

t = 1.74 x 10^2 min

time taken = 1.74 x 10^2 minutes

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