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(C) Given: Initial activity $A_0 = 20$ decays/min.Final activity $A = 2$ decays/min.Half-life $T_{1/2} = 5730$ years.We know that the activity $A$ at

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(C) Given: Initial activity $A_0 = 20$ decays/min.Final activity $A = 2$ decays/min.Half-life $T_{1/2} = 5730$ years.We know that the activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$,where $\lambda = \frac{0.693}{T_{1/2}}$.Rearranging for $t$,we get $t = \frac{1}{\lambda} \ln\left(\frac{A_0}{A}ight) = \frac{T_{1/2}}{0.693} 	imes 2.303 	imes \log_{10}\left(\frac{A_0}{A}ight)$.Substituting the values: $t = \frac{5730}{0.693} 	imes 2.303 	imes \log_{10}\left(\frac{20}{2}ight)$.Since $\log_{10}(10) = 1$,we have $t = \frac{5730 	imes 2.303}{0.693} \approx 5730 	imes 3.322$.$t \approx 19039$ years.

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