A sample of fluorine-20 has an activity of 5.94 mCi. If its half-life is 11.0 s, what is its activity after 55.0 s?

Question

Quizplus · Accepted Answer

The activity of a radioactive sample decreases exponentially with time, following the formula $A = A_0 \times (1/2)^{t/T}$, where $A$ is the final activity, $A_0$ is the initial activity, $t$ is the elapsed time, and $T$ is the half-life of the substance. After 55.0 s, which is 5 half-lives (since 55.0 s / 11.0 s/half-life = 5), the activity of the sample would be $5.94 \, \text{mCi} \times (1/2)^5 = 5.94 \, \text{mCi} \times 1/32 = 0.186 \, \text{mCi}$.

A Sample of Fluorine-20 Has an Activity of 5