The ratio of frac{{}^{14}C}{{}^{12}C} in a pi | vedclass

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(D) The radioactive decay follows first-order kinetics: $\lambda t = \ln \frac{N_0}{N_t}$.Given that the ratio in the wood sample is $\frac{1}{8}$ of

Vedclass · Accepted Answer

$17190$

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(D) The radioactive decay follows first-order kinetics: $\lambda t = \ln \frac{N_0}{N_t}$.Given that the ratio in the wood sample is $\frac{1}{8}$ of the atmospheric ratio,we have $\frac{N_t}{N_0} = \frac{1}{8}$,which implies $\frac{N_0}{N_t} = 8$.Using the relation $t = \frac{2.303}{\lambda} \log \frac{N_0}{N_t}$ or $t = \frac{t_{1/2}}{0.693} \ln \frac{N_0}{N_t}$.Since $\ln 8 = \ln 2^3 = 3 \ln 2$,we get $\lambda t = 3 \ln 2$.Substituting $\lambda = \frac{\ln 2}{t_{1/2}}$,we have $\frac{\ln 2}{t_{1/2}} 	imes t = 3 \ln 2$.Therefore,$t = 3 	imes t_{1/2} = 3 	imes 5730 	ext{ years} = 17190 	ext{ years}$.

The ratio of $\frac{{}^{14}C}{{}^{12}C}$ in a piece of wood is $\frac{1}{8}$ part that of the atmosphere. If the half-life of ${}^{14}C$ is $5730 \text{ years}$,the age of the wood sample is $.....$ years.

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