
#1
September 16th, 2012,
05:23
Jimson asks over on
Quote:
I've got some A2 physics homework and unfortunately i can't seem to do the first question :
Quote:
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:
\[t= \; \frac{\log(0.8)}{\log(2)} \times 100\]

September 16th, 2012 05:23
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