College Algebra, Chapter 5, 5.5, Section 5.5, Problem 14

If we have a 22-mg sample, then the half-life of radium-226 is 1600 years.

(a) Determine a function that models the mass remaining after $t$ years.

(b) Find how much of the sample will remain after 4000 years.

(c) After how long will only 18 mg of the sample remain?



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

By substituting all the information, the model will be..

$m(t) = 22 e^{- \left( \frac{\ln 2}{1600} \right) t} $

b.)


$
\begin{equation}
\begin{aligned}

\text{if } t =& 4000 \text{ years, then}
\\
\\
m(4000) =& 22e^{- \left( \frac{\ln 2}{1600} \right) (4000)}
\\
\\
m(4000) =& 3.89 \text{ mg}

\end{aligned}
\end{equation}
$


c.)


$
\begin{equation}
\begin{aligned}

\text{if } n(t) =& 18 \text{ mg, then}
&&
\\
\\
18 =& 22 e^{- \left( \frac{\ln 2}{1600} \right) (t)}
&& \text{Divide each side by } 22
\\
\\
\frac{9}{11} =& e^{- \left( \frac{\ln 2}{1600} \right) t}
&& \text{Take $\ln$ of each side}
\\
\\
\ln \left( \frac{9}{11} \right) =& - \left( \frac{\ln 2}{1600} \right) (t)
&& \text{Divide each side by } \frac{-\ln 2}{1600}
\\
\\
t =& \frac{\displaystyle \ln \left( \frac{9}{11} \right)}{\displaystyle \frac{\ln 2}{1600}}
&& \text{Solve for } t
\\
\\
t =& 463.21 \text{ years or } 464 \text{ years}
&&

\end{aligned}
\end{equation}
$


It shows that after 464 years the mass will decrease to 18 mg.