Answer
(a) $r=0.02828/day$
(b) $Q(t) = 100e^{-.02828t}$
(c) $t\approx 24.5$ days.
Work Step by Step
1.2.12. In general, a radioactive material decays at a rate proportional to the amount present: If $Q(t)$ denotes the amount of material at time $t$,
$dQ/dt = -rQ$,
where $r>0$ is the decay rate.
(a) If 100 mg. of $Th^{234}$ decays to 82.04 mg. in 1 wk. (= 7 days), find r for $Th^{234}$.
$\textit{Solution}$:
We have
(1) $\frac{dQ}{dt} = -rQ$,
(2) $Q_{0} = Q(0) = 100$ mg.,
(3) $Q_{1} = Q(7) = 82.04$ mg.,
time, $t$, measured in days. We assume $r \ne 0$ and $Q\ne 0$. Thus, rewriting (1),
(4) $\frac{dQ/dt}{Q} = -r$,
By integrating both sides of (4),
$\ln(Q) = -rt + C$, where $C$ is a constant of integration,
$Q=ce^{-rt}$, where $c = e^{C}$.
Now, when $t=0$, $Q=100$ (using (2)), $c = 100$.
Then, by (3),
$82.04 = 100e^{-r\cdot 7}$,
$\ln(82.04) = \ln 100 -7r$,
$7r = \ln(100/82.04)$,
$r = .02828/day$.
(b) Find an expression for the amount of $Th^{234}$ at any time, $t$.
$\textit{Solution}$:
We use the general solution found in (a):
$Q = ce^{-rt}$,
and use the particular values found for $c$ and $r$
$Q(t) = 100e^{-.02828t}$
(c) Find the time required for the $Th^{234}$ to decay to one half the original amount.
$\textit{Solution}$:
This requires us to solve the following equation for $t$.
$50 = 100e^{-.02828t}$,
Proceeding,
$\frac{1}{2} = e^{-.02828t}$
$t = -\ln(.5)/.02828$
$\approx24.5$ days.
