A piece of wood was found to have the C^{14}/ | vedclass

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(B) Given the ratio of $C^{14}/C^{12}$ in the sample is $N_t/N_0 = 0.7$.The decay constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5

Vedclass · Accepted Answer

$2966$

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(B) Given the ratio of $C^{14}/C^{12}$ in the sample is $N_t/N_0 = 0.7$.The decay constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} \, yr^{-1}$.Using the first-order kinetics equation: $t = \frac{2.303}{k} \log \frac{N_0}{N_t}$.Substituting the values: $t = \frac{2.303 	imes 5760}{0.693} \log \frac{1}{0.7}$.Since $\log(1/0.7) = \log(1.428) \approx 0.1549$.$t = \frac{2.303 	imes 5760 	imes 0.1549}{0.693} \approx 2966 \, yrs$.

$A$ piece of wood was found to have the $C^{14}/C^{12}$ ratio $0.7$ times that in a living plant. The time period when the plant died is ....... $years$ (Half-life of $C^{14} = 5760 \, yrs$).

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