$A$ piece of wood from the ruins of an ancient building was found to have a $^{14}C$ activity of $12$ disintegrations per minute per gram of its carbon content. The $^{14}C$ activity of the living wood is $16$ disintegrations per minute per gram. How long ago did the tree,from which the wooden sample came,die? Given half-life of $^{14}C$ is $5760$ years.

(C) The radioactive decay law is given by $I = I_0 e^{-\lambda t}$,where $I$ is the activity at time $t$,$I_0$ is the initial activity,and $\lambda$ is the decay constant.
Given $I = 12$ disintegrations/min/g,$I_0 = 16$ disintegrations/min/g,and $T_{1/2} = 5760$ years.
The decay constant $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5760} \text{ year}^{-1}$.
Substituting the values into the decay equation:
$12 = 16 e^{-\lambda t}$
$\frac{12}{16} = e^{-\lambda t} \implies 0.75 = e^{-\lambda t}$
Taking the natural logarithm on both sides:
$\ln(0.75) = -\lambda t$
$t = -\frac{\ln(0.75)}{\lambda} = -\frac{\ln(0.75) \times 5760}{0.693}$
Since $\ln(0.75) \approx -0.2877$:
$t = \frac{0.2877 \times 5760}{0.693} \approx 2391 \text{ years}$.
Thus,the tree died approximately $2391$ years ago.