Answer
$8.6\times10^{-7}$.
Work Step by Step
Find the ratio of the decay rates.
$$\frac{\left(\frac{\Delta N}{\Delta t}\right)_{218}}{\left(\frac{\Delta N}{\Delta t}\right)_{214}}$$
Use equation 30–3b for the decay rate, $\frac{\Delta N}{\Delta t}=-\lambda N$.
$$\frac{\left(\frac{\Delta N}{\Delta t}\right)_{218}}{\left(\frac{\Delta N}{\Delta t}\right)_{214}}=\frac{-\lambda_{218}N_{218}}{-\lambda_{214}N_{214}}$$
Assume there are equal numbers of each isotope.
$$\frac{\left(\frac{\Delta N}{\Delta t}\right)_{218}}{\left(\frac{\Delta N}{\Delta t}\right)_{214}}=\frac{\lambda_{218} }{\lambda_{214} }$$
Use equation 30-6 to bring in the half-lives.
$$\frac{\left(\frac{\Delta N}{\Delta t}\right)_{218}}{\left(\frac{\Delta N}{\Delta t}\right)_{214}}=\frac{T_{1/2,\;214} }{ T_{1/2,\;218 }}$$
$$\frac{\left(\frac{\Delta N}{\Delta t}\right)_{218}}{\left(\frac{\Delta N}{\Delta t}\right)_{214}}=\frac{1.6\times10^{-4}s }{(3.1m)(60s/m)}= 8.6\times10^{-7}$$
