The half-life for radioactive decay of ^{14}C | vedclass

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(B) Radioactive decay follows first-order kinetics.The decay constant $(\lambda)$ is given by $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \

Vedclass · Accepted Answer

$1845$

Vedclass · Answer

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

The half-life for radioactive decay of $^{14}C$ is $5730 \ years$. An archaeological artifact containing wood had only $80\%$ of the $^{14}C$ found in a living tree. The age of the sample is ....... $years$.

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