Answer
At the end of 280 days, the decay rate falls to one-fourth of its initial value.
Work Step by Step
We can find the disintegration constant:
$\lambda = \frac{ln(2)}{T_{1/2}}$
$\lambda = \frac{ln(2)}{140~d}$
$\lambda = 0.004951~d^{-1}$
We can find $t$ when the decay rate is $\frac{1}{4}$ of its initial value:
$R = R_0~e^{-\lambda t} = \frac{R_0}{4}$
$e^{\lambda t} = 4$
$\lambda t = ln(4)$
$t = \frac{ln(4)}{\lambda}$
$t = \frac{ln(4)}{0.004951~d^{-1}}$
$t = 280~days$
At the end of 280 days, the decay rate falls to one-fourth of its initial value.
