A piece of wood from an archaeological sample | vedclass

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(B) Given: Initial activity $(A_0)$ = $15.0 	ext{ counts min}^{-1} 	ext{ g}^{-1}$ Final activity $(A)$ = $5.0 	ext{ counts min}^{-1} 	ext{ g}^{-1}

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$9,200 	ext{ yr}$

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(B) Given: Initial activity $(A_0)$ = $15.0 	ext{ counts min}^{-1} 	ext{ g}^{-1}$ Final activity $(A)$ = $5.0 	ext{ counts min}^{-1} 	ext{ g}^{-1}$ Half-life $(t_{1/2})$ = $5770 	ext{ yr}$ The age of the sample $(t)$ is given by the radioactive decay formula: $t = \frac{2.303}{\lambda} \log\left(\frac{A_0}{A}ight)$ Since $\lambda = \frac{0.693}{t_{1/2}}$,$t = \frac{2.303 	imes t_{1/2}}{0.693} \log\left(\frac{A_0}{A}ight)$ $t = \frac{2.303 	imes 5770}{0.693} \log\left(\frac{15.0}{5.0}ight)$ $t = \frac{2.303 	imes 5770}{0.693} \log(3)$ Using $\log(3) \approx 0.4771$: $t = \frac{2.303 	imes 5770 	imes 0.4771}{0.693} \approx 9148 	ext{ yr}$ The closest value is $9,200 	ext{ yr}$.

$A$ piece of wood from an archaeological sample has $5.0 \text{ counts min}^{-1} \text{ g}^{-1}$ of $^{14}C$, while a fresh sample of wood has a count of $15.0 \text{ counts min}^{-1} \text{ g}^{-1}$. If the half-life of $^{14}C$ is $5770 \text{ yr}$, the age of the archaeological sample is:

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