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13.Nuclei
medium
Activity of a radioactive sample decreases to $(1/3)^{rd}$ of its original value in $3\, days$. Then, in $9\, days$ its activity will become
A
$(1/27)$ of the original value
B
$(1/9)$ of the original value
C
$(1/18)$ of the original value
D
$(1/3)$ of the original value
(AIIMS-2009)
Solution
$R=R_{0} e^{-\lambda t}$
$\Rightarrow \frac{1}{3}=e^{-\lambda \times 3}=e^{-3 \lambda}$ ………$(1)$
Let activity in $9$ days be $R'$. Then
$\frac{R^{\prime}}{R_{0}}=e^{-\lambda \times 9}=e^{-9 \lambda} e^{-\lambda \times 3}=\left(e^{-3 \lambda}\right)^{3}$
$=\left(\frac{1}{3}\right)^{3}, \quad$ from $(1)$
$=\frac{1}{27} \Rightarrow R^{\prime}=\frac{R_{0}}{27}.$
Standard 12
Physics
