Question

The activity of a radioactive sample is $1.6$ curie, and its half-life is $2.5$days. Its activity after $10$ days will be

• A. $0.8$ curie
• B. $0.4$ curie
• C. $0.1$ curie
• D. $0.16$ curie

Answer 1

The correct option is C $0.1$ curie

Activity $=\frac{dN}{dt}=\lambda N_o{e}^{-\lambda t}$
Also, $\lambda =\frac{ln2}{2.5}$
At $t=0$, activity $=\lambda N_o=1.6$ curie
Thus at $t=10$
Activity $=\lambda N_o{e}^{-\lambda 10}=1.6e^{-4ln2}=0.1$ curie