Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>

3.8 /

=

((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
equal to its atomic number