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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.After $280$ days,the number of half-lives elapsed is $n = 280 / 140 = 2$.Let $A_{280}$ be the activity after $280$ days,which is $6000 \, dps$.Using the formula $A_{280} = A_{initial} 	imes (1/2)^n$,we get $6000 = A_{initial} 	imes (1/2)^2$.$6000 = A_{initial} 	imes (1/4)$.$A_{initial} = 6000 	imes 4 = 24000 \, dps$.

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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