College Algebra, Chapter 5, 5.5, Section 5.5, Problem 20

A sample of radon-222 has decayed to $58 \%$ of its original amount after 3 days.

(a) Find the half-life of radon-222.

(b) How long will it take the sample to decay to $20 \%$ of its original amount?




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

a.) To solve for the half-life, we must find $r$, so


$
\begin{equation}
\begin{aligned}

0.58 m_0 =& m_0 e^{-r(3)}
&& \text{Divide each side by } m_0
\\
\\
0.58 =& e^{-3r}
&& \text{Take $\ln$ of each side}
\\
\\
\ln (0.58) =& -3r
&& \text{Recall } \ln e = 1
\\
\\
r =& \frac{- \ln (0.58)}{3}
&& \text{Divide each side by } -3
\\
\\
r =& 0.1816
&&

\end{aligned}
\end{equation}
$



Now the half-lfe is..



$
\begin{equation}
\begin{aligned}

r =& \frac{\ln 2}{h}
\\
\\
h =& \frac{\ln 2}{r}
\\
\\
=& \frac{\ln 2}{0.1816}
\\
\\
=& 3.82 \text{ days}

\end{aligned}
\end{equation}
$


b.) Then, if $m(t) = 0.20 m_0$, then



$
\begin{equation}
\begin{aligned}

0.20 m_0 =& m_0 e^{-0.1816 t}
&& \text{Divide each side by } m_0
\\
\\
0.20 =& e^{-0.1816 t}
&& \text{Take $\ln$ of each side}
\\
\\
\ln (0.20) =& -0.1816 t
&& \text{Recall } \ln e = 1
\\
\\
t =& \frac{\ln(0.20)}{-0.1816}
&& \text{Solve for } t
\\
\\
t =& 8.86 \text{ days}
&&

\end{aligned}
\end{equation}
$