The half-life period of radioactive decay of | vedclass

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(A) Radioactive decay follows first-order kinetics. The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ year^{-1}$.Usi

Vedclass · Accepted Answer

$1845$

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(A) Radioactive decay follows first-order kinetics. The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ year^{-1}$.Using the first-order integrated rate equation: $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} ight)$.Here,$[A]_0 = 100$ and $[A]_t = 80$.Substituting the values: $t = \frac{2.303}{k} \log \left( \frac{100}{80} ight) = \frac{2.303 	imes 5730}{0.693} \log(1.25)$.$t = \frac{2.303 	imes 5730}{0.693} 	imes 0.0969 \approx 1845 \ years$.

The half-life period of radioactive decay of ${}^{14}C$ is $5730 \ years$. An archaeological artifact contains $80\%$ of ${}^{14}C$ as compared to a living tree. Calculate the age of the sample in $years$.

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