a.) Fluorine-18 undergoes positron emission with a half life of 1.10x10^2 minute
ID: 994242 • Letter: A
Question
a.) Fluorine-18 undergoes positron emission with a half life of 1.10x10^2 minutes. If a patient is given a 248 mg does for a PET scan, how long will it take for the amount of fluorine 18 to drop to 83mg? (Assume that none of the fluorine is excreted from the body). b.) Calculate the mass defect and the nucleir binding energy per nucleon in Mo-96 if mass of Mo-96 nucleus is 95.962 amu. The mass of a proton is 1.00728amu and the mass of a neutron is 1.008665 amu. Please explain the steps, thank you!Explanation / Answer
First let us find out rate constand by using following formula
Half life = 0.693 / k
k = 0.693 / 1.1 x102
k = 0.0063 Min-1
ln A = -kt + ln Ao
A0 = 248 mg (Initial concentration)
A = 83 mg ( concentration at the time of t)
ln 83 = - 0.0063 x t + ln 248
4.41 = -0.0063 x t + 5.51
0.0063 t = 1.09
t = 173.74 Min
173.74 Minutes will take to reduce the Fluorine-18 from 248 mg to 83 mg
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