Question

The half-life of a radioisotope is $4$ days. A sample of the radioisotope was prepared $12$ days ago. The resent activity of the sample is $240$ counts $s^{-1}$. What was its activity $12$ days ago?

Answer 1

$\ln \frac{A_t}{A_o}=-ktk=\frac{0.693}{t_{1/2}}=\frac{0.693}{4}=\frac{\ln 2}{4}\ln \left(\frac{240}{A_o}\right)=-\frac{\ln 2}{4}\times 12\ln \left(\frac{240}{A_o}\right)=-3\ln 2\ln \left(\frac{A_o}{240}\right)=3\ln 2=\ln 2^3=\ln 8\frac{A_o}{240}=8A_o=8\times 240=1930count{s}^{-1}$