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(D) 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 reduces from

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$24000$

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(D) 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 reduces from $6000\,dps$ to $3000\,dps$ in $140\,days$,the half-life $T_{1/2}$ is $140\,days$.At $t = 280\,days$,the number of half-lives elapsed is $n = 280 / 140 = 2$.Let $A_i$ be the initial activity. At $t = 280\,days$,the activity $A_{280} = 6000\,dps$.Using the formula $A_{280} = A_i (1/2)^n$,we have $6000 = A_i (1/2)^2$.$6000 = A_i / 4$.$A_i = 6000 	imes 4 = 24000\,dps$.

After $280\,days$,the activity of a radioactive sample is $6000\,dps$. The activity reduces to $3000\,dps$ after another $140\,days$. The initial activity of the sample in $dps$ is

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