Answer
After three half-lives, the activity of that sodium-24 sample is equal to 1.5 mCi.
Work Step by Step
1. After a half-life, one-half of the sodium-24 decays. Therefore, the activity after that time is equal to $\frac{1}{2}$ of the initial activity.
Initial $^{24}_{11}Na$ activity: 12 mCi.
After one half-life:
$12$ $mCi$ $\times \frac{1}{2} = 6.0$ $mCi$
2. The same will occur on the following half-lives, so, repeat the process:
After the second half-life:
$6.0$ $mCi$ $\times \frac{1}{2} = 3.0$ $mCi$
After the third half-life:
$3.0$ $mCi$ $\times \frac{1}{2} = 1.5$ $mCi$
