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Activity of a radioactive sample decreases to $\left(\frac{1}{3}\right)^{r d}$ of its original value in 3 days. Then in 9 days its activity reduces to
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Verified Answer
The correct answer is:
$\left(\frac{1}{27}\right)$ of the original value
Activity after time $t$ given that half-life is $T$ :
$A=\frac{A_0}{2^{t / T}}$
$\frac{A_0}{3}=\frac{A_0}{2^{3 / T}}$
Activity in 9 days,
$A=\frac{A_0}{2^{t / T}}=\frac{A_0}{2^{9 / T}}=\frac{A_0}{\left(2^{3 / T}\right)^3}=\frac{A_0}{27}$
$A=\frac{A_0}{2^{t / T}}$
$\frac{A_0}{3}=\frac{A_0}{2^{3 / T}}$
Activity in 9 days,
$A=\frac{A_0}{2^{t / T}}=\frac{A_0}{2^{9 / T}}=\frac{A_0}{\left(2^{3 / T}\right)^3}=\frac{A_0}{27}$
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