Charcoal prepared from a living tree exhibits a disintegration rate...

The radioactive decay law is given by:

t = \dfrac{2.303}{\lambda} \log \left(\dfrac{A_0}{A} \right)

The decay constant $\lambda$ is:

$\lambda =$\dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{5730} \text{ y}^{-1}

Given:

$A_0 = 15.3 \text{ dis min}^{-1} \text{ g}^{-1}$

$A = 12.3 \text{ dis min}^{-1} \text{ g}^{-1}$

Substituting the values into the decay equation:

$t =$\dfrac{2.303 \times 5730}{0.693} \log \left( \dfrac{15.3}{12.3} \right)$<br><br>$t = \dfrac{13196.19}{0.693} \log (1.2439)$<br><br>$t = 19042.1 \times 0.0948$<br><br>$t \approx 1805 \text{ y}