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(A) The activity $A$ of a radioactive substance is given by $A = \lambda N$, where $\lambda$ is the decay constant and $N$ is the number of radioactiv

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$1 : 2 : 3$

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(A) The activity $A$ of a radioactive substance is given by $A = \lambda N$, where $\lambda$ is the decay constant and $N$ is the number of radioactive nuclei present at time $t$.The number of nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.Therefore, the activity at time $t$ is $A(t) = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}$, where $A_0$ is the initial activity.Since all three lumps are of the same radioactive substance, they share the same decay constant $\lambda$.If the initial activities are $A_{01}, A_{02}, A_{03}$ in the ratio $1 : 2 : 3$, then at any later time $t$, the activities will be $A_1(t) = A_{01} e^{-\lambda t}$, $A_2(t) = A_{02} e^{-\lambda t}$, and $A_3(t) = A_{03} e^{-\lambda t}$.The ratio of their activities at time $t$ will be $A_1(t) : A_2(t) : A_3(t) = (A_{01} e^{-\lambda t}) : (A_{02} e^{-\lambda t}) : (A_{03} e^{-\lambda t}) = A_{01} : A_{02} : A_{03} = 1 : 2 : 3$.Thus, the ratio remains constant over time.

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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