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(D) Let $t$ be the age of the Indus Valley civilization.The decay rate $R$ is given by $R = \lambda N$.Since $N = N_0 e^{-\lambda t}$, the activity at

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$4224$ years

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(D) Let $t$ be the age of the Indus Valley civilization.The decay rate $R$ is given by $R = \lambda N$.Since $N = N_0 e^{-\lambda t}$, the activity at time $t$ is $R = R_0 e^{-\lambda t}$, where $R_0$ is the initial decay rate.Given $R_0 = 15$ decays/minute and $R = 9$ decays/minute.Thus, $\frac{R}{R_0} = e^{-\lambda t} \implies \frac{9}{15} = e^{-\lambda t} \implies \frac{3}{5} = e^{-\lambda t}$.Taking the natural logarithm on both sides: $\ln(3/5) = -\lambda t \implies \lambda t = \ln(5/3)$.Since $\lambda = \frac{\ln 2}{T_{1/2}}$, we have $t = \frac{\ln(5/3)}{\ln 2} 	imes T_{1/2}$.Given $T_{1/2} = 5730$ years, $t = \frac{\ln(1.667)}{\ln 2} 	imes 5730 \approx \frac{0.5108}{0.6931} 	imes 5730 \approx 0.737 	imes 5730 \approx 4224$ years.

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