The activity of a radioactive element decreas | vedclass

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Question

(C) The activity of a radioactive substance is given by $R = R_0 e^{-\lambda t}$.Given that after $t = 9$ years,the activity becomes $R = \frac{R_0}{3

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$R_0/9$

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(C) The activity of a radioactive substance is given by $R = R_0 e^{-\lambda t}$.Given that after $t = 9$ years,the activity becomes $R = \frac{R_0}{3}$.Substituting these values: $\frac{R_0}{3} = R_0 e^{-\lambda 	imes 9}$,which implies $e^{-9\lambda} = \frac{1}{3}$ ... $(i)$.After a further $9$ years,the total time elapsed is $t = 9 + 9 = 18$ years.The new activity $R'$ is given by $R' = R_0 e^{-\lambda 	imes 18}$.This can be written as $R' = R_0 (e^{-9\lambda})^2$.Substituting the value from equation $(i)$: $R' = R_0 \left(\frac{1}{3}ight)^2 = \frac{R_0}{9}$.

The activity of a radioactive element decreases to one-third of its original activity $R_0$ in $9$ years. After a further $9$ years,its activity will be

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