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

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(B) Radioactive decay follows first-order kinetics: $k = \frac{2.303}{t} \log \frac{[N]_0}{[N]_t}$.For radioactive decay, the rate constant $k$ is rel

Vedclass · Accepted Answer

$t = \frac{5730}{0.3} \log \frac{100}{80}$

Vedclass · Answer

(B) Radioactive decay follows first-order kinetics: $k = \frac{2.303}{t} \log \frac{[N]_0}{[N]_t}$.For radioactive decay, the rate constant $k$ is related to half-life $(t_{1/2})$ as $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 5730 	ext{ years}$, $[N]_0 = 100$, and $[N]_t = 80$.Substituting these values into the first-order equation: $t = \frac{2.303}{k} \log \frac{[N]_0}{[N]_t} = \frac{2.303 	imes 5730}{0.693} \log \frac{100}{80}$.Since $\frac{2.303}{0.693} \approx \frac{1}{0.3}$, the expression simplifies to $t = \frac{5730}{0.3} \log \frac{100}{80}$.

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

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