The activity of a sample of radioactive mater | vedclass

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(C) Let $A_0$ be the initial activity of the radioactive sample.The activity at any time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\l

Vedclass · Accepted Answer

$A_2 = A_1 e^{(t_1 - t_2)/T}$

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(C) Let $A_0$ be the initial activity of the radioactive sample.The activity at any time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$,where $\lambda$ is the decay constant.At time $t_1$,the activity is $A_1 = A_0 e^{-\lambda t_1}$.At time $t_2$,the activity is $A_2 = A_0 e^{-\lambda t_2}$.Dividing the two equations: $\frac{A_2}{A_1} = \frac{A_0 e^{-\lambda t_2}}{A_0 e^{-\lambda t_1}} = e^{-\lambda(t_2 - t_1)}$.Since the mean life $T = \frac{1}{\lambda}$,we can substitute $\lambda = \frac{1}{T}$.Thus,$\frac{A_2}{A_1} = e^{-(t_2 - t_1)/T} = e^{(t_1 - t_2)/T}$.Therefore,$A_2 = A_1 e^{(t_1 - t_2)/T}$.

The activity of a sample of radioactive material is $A_1$ at time $t_1$ and $A_2$ at time $t_2$ $(t_2 > t_1)$. If its mean life is $T$,then which of the following is true?

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