How Long for 61Co to Reduce to 80% of Its Initial Value?

[CaptainBlack]

Jimson asks over on Yahoo Answers

I've got some A2 physics homework and unfortunately i can't seem to do the first question :

The question is: 61Co has a half life of 100 minutes. how long will it take for the radioactive isotope to reduce to 80% of the initial value?

The answer is 32 minutes, but of course i'll need the workings before it to actually prove i can do the question.

Thank you for any help given! (if any is anyway).

The key idea is that after 100 minutes 1/2 remains after 200 minutes 1/4 remains so the fraction remaining after \(t\) minutes is:

\[Q=2^{-t/100}\]

So if \(80\%\) remains we need to solve:

\[0.8=2^{-t/100}\]

which we do by taking logs (the base is unimportant as long as we use the sane base throughout):

\[\log(0.8)= -\; \frac{t}{100} \log(2)\]

So:

\[t=- \; \frac{\log(0.8)}{\log(2)} \times 100\]

 

[kuruman]

So since this thread is almost 10 years old, it is safe to provide the numerical answer, $$t=-\frac{100\log(0.8)}{\log(2)}=32~\text{min.}$$