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Question

(A) For the same radioactive substance,the decay constant $\lambda$ is identical for all lumps.The activity $A$ of a radioactive sample is given by $A

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$1 : 2 : 3$

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(A) For the same radioactive substance,the decay constant $\lambda$ is identical for all lumps.The activity $A$ of a radioactive sample is given by $A = \lambda N$,where $N$ is the number of radioactive nuclei present at time $t$.Since $\lambda$ is constant,the activity is directly proportional to the number of nuclei: $A \propto N$.At any time $t$,the number of nuclei in each lump is $N(t) = N_0 e^{-\lambda t}$,where $N_0$ is the initial number of nuclei.Therefore,the ratio of activities at any time $t$ is:$A_1 : A_2 : A_3 = (\lambda N_{0,1} e^{-\lambda t}) : (\lambda N_{0,2} e^{-\lambda t}) : (\lambda N_{0,3} e^{-\lambda t})$$A_1 : A_2 : A_3 = N_{0,1} : N_{0,2} : N_{0,3}$Since the initial ratio of activities was $1 : 2 : 3$,the ratio of the initial number of nuclei $N_{0,1} : N_{0,2} : N_{0,3}$ must also be $1 : 2 : 3$.Thus,the ratio of their activities at any future date remains $1 : 2 : 3$.

There are three lumps of a given radioactive substance. Their activity is in the ratio of $1 : 2 : 3$ now. What will be the ratio of their activities at any further date?

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