In the treatment of cancer of the thyroid, the radioactive liquid Iodine-131 is often used. Suppose that after one day in storage, analysis shows that an initial amount A0 of iodine-131 in a sample has decreased by 8.3%. (a) Find the amount of iodine-131 remaining in the sample after 8 days.

First, I need to find the equation to use. \f[\frac{dx}{dt}=-kx=>\frac{1}{x}dx=-kdt=>\int\frac{1}{x}dx=-\intkdt=>\f]

\f[ln|x|=-kt+C=>e^{ln|x|}=e^{-(CKt)}=>x=-Ce^{kt} \f]

The half life of Iodine-131 is 8 days so even though it decreased by 8.3% the percentage at the end of the 8 days will still be 50% decrease. This is approximately 8.3% per day.  So you can use the equation \f[x=Ce^{(-kt)}; \f] thus the equation can be used for longer periods of time like years and be rewritten as \f[k=1-day; x=The-percentage-Output,t=time,C=Constant=percentage;find=>x \f]

\f[ln|(x)|=-Ckt=>ln||(x)|=e^{(-(0.083)(1)(8))}=>e^{ln|x|}=e^{((-0.083*1*8))}=> \f]

\f[x=e^{(-0.664)}=>x=0.514788*100 =>51.4788-percent=or=51.5=percent\f]

That is fairly close to half life just above half.


(b) Explain the significance of the result in part (a).


I decided to do a table to compare the results.  The significance is that the same equation can be used for the half life of Iodine every time.  Keep in mind that Iodine decays into different forms of radioactive material each time making it a different radioactive material. Part (a) just proves that the decay of Iodine 131 is approximately half of the previous amount stored.

Radioactive Decay Table
Day Percent Start Percentage
1 100.00000 91.7
2 91.70000 84.0889
3 84.08890 77.1095
4 77.10952 70.7094
5 70.70943 64.8405
6 64.84055 59.4588
7 59.45878 54.5237
8 54.52370 49.9982