In this question, it is given that
Curium-243 has a half-life of 28.5 days. In a sample of 5.6 grams of curium-243.
And we have to use the following formula, which is
[tex]A(t) = A_{0} (1/2)^{t/h}[/tex]
Here t=12 days, half life, h =28.5 days . And the initial value ,A(0)=5.6 gms
So we will get
[tex]A(t) = 5.6(1/2)^{12/28.5}[/tex]
[tex]A(t) = 5.6* 0.747 =4.2 gm[/tex]
Therefore after 12 days, 4.2 gmsof curium-243 remains .
