Answer
The activity of that sodium-24 sample after one half-life is equal to 6.0 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:
$12$ $mCi$ $\times \frac{1}{2} = 6.0$ $mCi$
