지수적 증가와 감소

The amount of Radium-221 in a sample will be

$$\begin{aligned} Q(t)&=Ce^{-kt} \\ \text{where $C$ is the amount in the sample at time $t=0$, and $k$ is some positive constant. We know the half-life of the isotope, so we can find $e^{-k}$:} \\ \frac{C}{2}=Q(30)&=Ce^{-k\cdot 30}\\ \frac{1}{2}&=\left(e^{-k}\right)^{30}\\ 2^{-\tfrac{1}{30}}&=e^{-k} \\ \text{So,} \\ Q(t)&=C\left(e^{-k}\right)^t=C\cdot2^{-\tfrac{t}{30}} \\ \text{When only 0.01\% of the original sample is left, $Q(t)=0.0001C$:} \\ 0.0001C=Q(t)&=C\cdot 2^{-\tfrac{t}{30}}\\ 0.0001&=2^{-\tfrac{t}{30}}\\ \log(0.0001)&=\log\left(2^{-\tfrac{t}{30}}\right)\\ \log\left(10^{-4}\right)&=-\frac{t}{30}\log2\\ -4\log 10&=-\frac{t}{30}\log2\\ t&=120\cdot\frac{\log10}{\log2}\approx 398.6 \end{aligned}$$

It takes about 398.6 seconds (that is, roughly 6 and a half minutes) for all but $0.01$% of the sample to decay.

Remark: we can do another reality check here. The half-life is 30 seconds. 6 and a half minutes represents 13 half-lives. So, the sample is halved 13 times: $\left(\tfrac{1}{2}\right)^{13}\approx 0.00012=0.012\%$. So these 13 half-lives should reduce the sample to about 0.01% of its original amount, as desired.