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(A) The activity $R$ of a radioactive sample is given by the radioactive decay law: $R = R_{0} e^{-\lambda t}$,where $R_{0}$ is the initial activity a

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$R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}$

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(A) The activity $R$ of a radioactive sample is given by the radioactive decay law: $R = R_{0} e^{-\lambda t}$,where $R_{0}$ is the initial activity at $t=0$.At time $t_{1}$,the activity is $R_{1} = R_{0} e^{-\lambda t_{1}}$.At time $t_{2}$,the activity is $R_{2} = R_{0} e^{-\lambda t_{2}}$.Dividing the expression for $R_{1}$ by $R_{2}$,we get:$\frac{R_{1}}{R_{2}} = \frac{R_{0} e^{-\lambda t_{1}}}{R_{0} e^{-\lambda t_{2}}} = e^{-\lambda t_{1} - (-\lambda t_{2})} = e^{-\lambda(t_{1}-t_{2})}$.Therefore,$R_{1} = R_{2} e^{-\lambda(t_{1}-t_{2})}$.

In a radioactive material,the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$,it is $R_{2}$. If the decay constant of the material is $\lambda$,then:

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