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(B) Given,for $^{14}C$:Initial activity $A_{0} = 16$ disintegrations $	ext{min}^{-1} 	ext{g}^{-1}$.Final activity $A = 12$ disintegrations $	ext{mi

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$2391$

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(B) Given,for $^{14}C$:Initial activity $A_{0} = 16$ disintegrations $	ext{min}^{-1} 	ext{g}^{-1}$.Final activity $A = 12$ disintegrations $	ext{min}^{-1} 	ext{g}^{-1}$.Half-life $t_{1/2} = 5760$ years.The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} 	ext{ year}^{-1}$.Using the radioactive decay formula $A = A_{0} e^{-\lambda t}$,we have $\frac{A_{0}}{A} = e^{\lambda t}$.Taking the natural logarithm on both sides: $\ln\left(\frac{A_{0}}{A}ight) = \lambda t$.Therefore,$t = \frac{1}{\lambda} \ln\left(\frac{A_{0}}{A}ight) = \frac{t_{1/2}}{0.693} 	imes 2.303 \log_{10}\left(\frac{A_{0}}{A}ight)$.Substituting the values: $t = \frac{5760}{0.693} 	imes 2.303 	imes \log_{10}\left(\frac{16}{12}ight)$.$t = \frac{5760 	imes 2.303}{0.693} 	imes \log_{10}(1.333)$.$t \approx 19142.8 	imes 0.1249 \approx 2391$ years.

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