Question

What is the answer to:

The half-life of the nuclide 221At85 is 7.5 hours. If 0.100 mg of this nuclide is administrated for thyroid treatment, how long does it take to reduce the nuclide to 0.0125 mg?

Answer 1

Let \(C_0\) represent the initial amount of the nuclide. After 1 half life, there will be \(\frac{1}{2}C_0\) left. After 2 half lives, there will be \(\frac{1}{4}C_0\) left, and in general, after t hours, there will be \[C(t) = C_0(2)^{-t/7.5}\]We want to find \(t\) when \(\frac{C(t)}{C_0} = \frac{0.0125}{0.100} = 0.125\)
\[0.125 = (2)^{-t/7.5}\]Take the log of both sides, recalling that \(\log(a^n) = n \log a\) and solve for \(t\).

Here's a general graph of exponential decay. You can see that after about 3 1/3 half lives, only 10% remains, and after 6 2/3 half lives, only 1% remains. 10 half lives gets you 0.1%. Be glad you only have to wait 75 hours!