Answer
$T_{1/2} = 4.6\times 10^9~y$
Work Step by Step
The sample of uranium has $2.5\times 10^{18}~atoms$
The text says that 12 atoms would decay in 1 second.
We can find the disintegration constant:
$-\frac{dN}{dt} = \lambda~N$
$\lambda = \frac{-\frac{dN}{dt}}{N}$
$\lambda = \frac{-(-12~atoms/s)}{2.5\times 10^{18}~atoms}$
$\lambda = \frac{12~atoms/s}{2.5\times 10^{18}~atoms}$
$\lambda = 4.8\times 10^{-18}~s^{-1}$
We can find the half-life:
$T_{1/2} = \frac{ln(2)}{\lambda}$
$T_{1/2} = \frac{ln(2)}{4.8\times 10^{-18}~s^{-1}}$
$T_{1/2} = 1.444\times 10^{17}~s$
$T_{1/2} = (1.444\times 10^{17}~s)~[\frac{1~y}{(365)(24)(3600~s)}]$
$T_{1/2} = 4.6\times 10^9~y$
