[Review Topics] [References Use the References to access important values if nee

Question

[Review Topics] [References Use the References to access important values if needed for this question. Some Radioactive Isotopes Useful in Med Mode of Decay Half-ife 20.3 m 109 m 14.3 d 27.7 d 44.5 d 78.3 h 118 d 13.3 s ?? Carbon-11 Fluorine 18 Phosphorus-32 E.C., ? Cr Chromium-51 2Pe Iron-59 TGa Gallium-67 Se Selenium-75 81 Kr Krypton-81m Sr game Technetium-99m ?.?., ? E.C., ? Strontium-81 6.01 h 8.04 d ? ?Hg 2:ITI Iodine-131 Mercury-197 Thallium-201 E.?., ? ?.?., 64.1 h 3.05 d The radioactive isotope thallium-201 is used in medical imaging as indicated on the table above. If 58.1 milligrams of thallium-201 is administered to a patient, how many milligrams are left in the body after 12.2 days? mg

Explanation / Answer

The radiactive decay of Thallium-201(TI) is a 1st order reaction. For 1st order reaction

rate of decay (-d(TI)/dt is proportional to mass of TI, the proportionality being the rate constant

-d(TI)/dt= KTI, K is also known as the rate constant. When the equation is integrated noting that t=0, initial mass of TI in the body is TIo mg and at t=t, mass of Ti is Ti,

lnTI= lnTIo-Kt (1)

half life is represented as (t1/2) which is defined as the time required for the mass to be reduced to 50% of the initial mass, at t= t1/2, TI=TIo2/

when these values are sustituted in Eq.1,

ln(TIo/2)= ln(Tio)-K*t1/2

or K=0.693/t1/2, from the data on Ti, half life t1/2= 3.05 days

hecne K= 0.693/3.05=0.2272/day

when this value od K is substituted in Eq.1, Eq.1 becomes

ln(TI) = lnTIo- 0.2272*t (2)

given TIO= 58.1 mg and t=12.2 day, Substituting these values in Eq.2 enables one to determine, Ti at 12.2 days

ln(TI) = ln(58.1)-0.2272*12.2

Ti= 3.63 mg

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