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(C) The radioactive decay follows the first-order kinetics equation: $t = \frac{2.303}{\lambda} \log \frac{N_0}{N_t}$.Here,$\lambda = \frac{0.693}{t_{

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$7.92 	imes 10^3 \ years$

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(C) The radioactive decay follows the first-order kinetics equation: $t = \frac{2.303}{\lambda} \log \frac{N_0}{N_t}$.Here,$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5000} \ min^{-1}$.Given $N_0 = 15 \ counts/min/gm$ and $N_t = 5.0 \ counts/min/gm$.Substituting the values: $t = \frac{2.303 	imes 5000}{0.693} 	imes \log \frac{15}{5}$.$t = \frac{2.303 	imes 5000}{0.693} 	imes \log 3$.$t = \frac{2.303 	imes 5000}{0.693} 	imes 0.4771 \approx 7927 \ years$.Thus,the age of the specimen is approximately $7.92 	imes 10^3 \ years$.

$A$ wood specimen from an archeological site shows a $_6^{14}C$ activity of $5.0 \ counts/min/gm$ of carbon. If a freshly cut wood gives $15 \ counts/min/gm$ of carbon,what is the age of the specimen? (Given: $t_{1/2}$ for $_6^{14}C$ is $5000 \ years$)

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