Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1153: 37

$1.44\times 10^4y$

We can find the age of the fire pit as follows: $\lambda=\frac{\ln 2}{T_{\frac{1}{2}}}=\frac{\ln 2}{5730y}=1.210\times 10^{-4}y^{-1}$ We know that: $0.175N_{\circ}=N_{\circ}e^{-\pi t}$ $\implies 0.175=e^{-\pi t}$ Taking the natural log and solving for t, we obtain: $t=\frac{\ln 0.175}{\pi}=-\frac{\ln 0.175}{1.210\times 10^{-4}y^{-1}}=1.44\times 10^4y$