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(C) The activity $A$ of a radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$ or $A = A_0 \left(\frac{1}{2}\right)^{t/T}$,where $T$ is

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$\left(\frac{1}{27}\right)$ of the original value

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(C) The activity $A$ of a radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$ or $A = A_0 \left(\frac{1}{2}ight)^{t/T}$,where $T$ is the half-life. Given that in $t_1 = 3 \ days$,the activity becomes $A_1 = \frac{A_0}{3}$. Using the relation $A = A_0 \left(\frac{1}{k}ight)^{t/	au}$,where $	au$ is the time taken to become $\frac{1}{k}$ of the original value: $\frac{A_0}{3} = A_0 \left(\frac{1}{3}ight)^{3/3} \implies \frac{1}{3} = \left(\frac{1}{3}ight)^1$. This confirms that the activity reduces by a factor of $\frac{1}{3}$ every $3 \ days$. For $t_2 = 9 \ days$,the number of such intervals is $n = \frac{9}{3} = 3$. Therefore,the activity after $9 \ days$ will be $A_2 = A_0 \left(\frac{1}{3}ight)^n = A_0 \left(\frac{1}{3}ight)^3 = \frac{A_0}{27}$.

The activity of a radioactive sample decreases to $\left(\frac{1}{3}\right)$ of its original value in $3 \ days$. Then,in $9 \ days$,its activity reduces to:

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