Answer
(a) $2.83\times 10^9y$
(b) $1.16\times 10^9y$
Work Step by Step
We can find how old the rock is as follows:
$\lambda=\frac{\ln 2}{T_{\frac{1}{2}}}=\frac{\ln 2}{1.2\times 10^9}=5.78\times 10^{-10}y^{-1}$
We know that
$N=N_{\circ}e^{-\pi t}$
and $t=\frac{\ln \frac{N}{N_{\circ}}}{\lambda}=\frac{1}{5.78\times 10^{-10}y^{-1}}(\ln (0.195))=2.83\times 10^9y$
(b) We know that
$t=\frac{1}{\lambda}\frac{N}{N_{\circ}}=\frac{1}{5.776\times 10^{-10}y^{-1}}(\ln \frac{0.100}{0.195})=1.16\times 10^9y$
