During the testing process for some pharmaceuticals, the drug is \"tagged\" with
ID: 2155920 • Letter: D
Question
During the testing process for some pharmaceuticals, the drug is "tagged" with a radioactive material. This way researchers can determine if the pharmaceutical is going to other parts of the body than the intended target and what effect it has on the non-target areas. By adding this radioactive tag to the pharmaceutical, researchers can pinpoint all parts of the body and the concentration that accumulates in non-target areas.One possible such tag is technetium-99, 99Tc, which has a half-life of 6.05 hours.
a) How many 99Tc nuclei are required to give an activity of 1.7 ?Ci?
N =
b) What is the decay constant for this isotope?
?= hour-1
c) Suppose the drug containing the amount of 99Tc calculated in part (a), were injected into the patient 1.5 hours after it was prepared, what would its activity be at that time?
R = ?Ci
Explanation / Answer
You should not use µCi which is a forbidden unit , use becquerels (Bq) which is 1 disintegration/s 1µCi = 3.7*10^4 Bq so 1.7µCi corresponds to 1.7*3.7*10^4=62900Bq So 62900=6.29*10^4 atoms shoud disintegrated by second. lAs 1 second is small against 6.05 hours use the derivative formula N=N0*decay constant *t (t=1s, N number of atoms disintegrated, N0 apounts of atoms present) I calculate first b decay constant formula decay constant = ln2/half life = 0.693/(6.05*3600)=3.182*10^-5 d/s and 6.29*10^4= N0/3.182*10^-5 so N0 = 1.976*10^9 atoms (this is answer a) c) formula N= N0*2^(-t/half life) A= A0*2^(-1.5/6.05)=1.7*2^-0.25=1.7*.84=1.4… µCi
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