College Algebra, Chapter 5, 5.5, Section 5.5, Problem 16

The equation $m(t) = 40e^{-0.0277t}$ represents the mass $m(t)$ remaining after $t$ days from a 40-g sample of the thorium-234.

(a) Identify how much of the sample will remain after 60 days.

(b) After how long will only 10 g of the sample remain?

(c) Determine the half-life of thorium-234.



a.) Recall the formula for radioactive decay

$m(t) = m_0 e^{-rt}$ in which $\displaystyle r = \frac{\ln 2}{h}$

where

$m(t)$ = mass remaining at time $t$

$m_0$ = initial mass

$r$ = rate of decay

$t$ = time

$h$ = half-life



$
\begin{equation}
\begin{aligned}

\text{if } t =& 60 \text{ days, then}
\\
\\
m(60) =& 40e^{-0.0277(60)}
\\
\\
=& 7.59 \text{ g}


\end{aligned}
\end{equation}
$



b.)



$
\begin{equation}
\begin{aligned}

\text{if } m(t) =& 10 \text{ g, then}
&&
\\
\\
10 =& 40e^{-0.0277(t)}
&& \text{Divide each side by } 40
\\
\\
\frac{1}{4} =& e^{-0.0277 t}
&& \text{Take $\ln$ of each side}
\\
\\
\ln \left( \frac{1}{4} \right) =& -0.0277 t
&& \text{Divide both sides by } -0.0277
\\
\\
t =& - \frac{\displaystyle \ln \left( \frac{1}{4} \right) }{0.0277}
&& \text{Solve for } t
\\
\\
t =& 50.05 \text{ days or 50 days}

\end{aligned}
\end{equation}
$


c.) To solve for the half-life, we set $m(t) = 20$


$
\begin{equation}
\begin{aligned}

20 =& 40e^{-0.0277 t}
&& \text{Divide each side by } 40
\\
\\
\frac{1}{2} =& e^{-0.0277 t}
&& \text{Take $\ln$ of each side}
\\
\\
\ln \left( \frac{1}{2} \right) =& -0.0277 t
&& \text{Recall that } \ln e = 1
\\
\\
t =& \frac{\displaystyle \ln \left( \frac{1}{2} \right)}{-0.0277}
&& \text{Solve for } t
\\
\\
t =& 25.02 \text{ days}



\end{aligned}
\end{equation}
$