The activity of carbon-14 in a piece of an an | vedclass

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(B) The half-life $t_{1/2}$ of $^{14}C$ is $5760 \, years$. The decay constant $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} \, years^{-1}$.Us

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$172.81 	imes 10^2 \, years$

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(B) The half-life $t_{1/2}$ of $^{14}C$ is $5760 \, years$. The decay constant $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} \, years^{-1}$.Using the radioactive decay formula: $t = \frac{2.303}{\lambda} \log \left( \frac{N_0}{N_t} ight)$.Given that the remaining activity is $12.5\%$,we have $\frac{N_0}{N_t} = \frac{100}{12.5} = 8$.Substituting the values: $t = \frac{2.303 	imes 5760}{0.693} \log(8)$.Since $\log(8) = 3 \log(2) = 3 	imes 0.3010 = 0.9030$.$t = \frac{2.303 	imes 5760 	imes 0.9030}{0.693} \approx 17281 \, years$.Converting to scientific notation: $17281 = 172.81 	imes 10^2 \, years$.

The activity of carbon-$14$ in a piece of an ancient wood is only $12.5\%$. If the half-life period of carbon-$14$ is $5760 \, years$,the age of the piece of wood will be $(\log \, 2 = 0.3010)$

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