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(C) The activity of a radioactive sample follows the law $A = A_0 (1/2)^n$, where $n$ is the number of half-lives.Given that the activity becomes $(1/

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$1/256$

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(C) The activity of a radioactive sample follows the law $A = A_0 (1/2)^n$, where $n$ is the number of half-lives.Given that the activity becomes $(1/4)$ of the original value in $4 	ext{ days}$, we have $(1/4) = (1/2)^n$, which implies $n = 2$.Since $2$ half-lives correspond to $4 	ext{ days}$, the half-life $T_{1/2} = 4 / 2 = 2 	ext{ days}$.Now, for a total time $t = 16 	ext{ days}$, the number of half-lives $n'$ is $n' = t / T_{1/2} = 16 / 2 = 8$.The activity after $16 	ext{ days}$ will be $A = A_0 (1/2)^8$.$A = A_0 (1/256)$.Therefore, $x = 1/256$.

Activity of a radioactive sample decreases to $(1/4)^{th}$ of its original value in $4 \text{ days}$. Then in $16 \text{ days}$ its activity will become $x$ times the original value. The value of $x$ is

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