College Algebra, Chapter 5, Review Exercise, Section Review Exercise, Problem 98

After 8 days, a sample of bismuth-210 decayed to $33 \%$ of its original mass.
(a) Determine the half-life of this element.
(b) Determine the mass remaining after 12 days.

a.) Recall the formula for radioactive decay,
$m(t) = m_0 e^{-rt}$ which $\displaystyle r = \frac{\ln 2}{h}$
Where,
$m(t) =$ the mass remaining at time $t$
$m_0 =$ initial mass
$r =$ rate of decay
$t =$ time
$h =$ half life
If the mass remaining is $33\%$ of the original mass, then

$\begin{equation} \begin{aligned} m(t) &= 0.33 m_0 \text{ ,so}\\ \\ 0.33 m_0 &= m_0 e^{-\left( \frac{\ln 2}{h} \right)(8)} && \text{Divide both sides by } m_0\\ \\ 0.33 &= e^{-\left( \frac{\ln 2}{h} \right)(8)} && \text{Take ln of each side}\\ \\ \ln (0.33) &= -\left( \frac{\ln (2)}{h} \right)(8) && \text{Recall } \ln e = 1 \\ \\ h &= \frac{-8 \ln 2}{\ln (0.33)} && \text{Multiply $h$ both sides}\\ \\ h &= 5.0022 \text{ days or 5 days} \end{aligned} \end{equation}$

The half life of bismuth $-210$ is approximately 5 days.

b.) If $t = 12$ days, then

$\begin{equation} \begin{aligned} m(12) &= m_0 e^{-r(12)} && \text{;where } r = \frac{\ln 2}{h} = \frac{\ln 2}{5}\\ \\ m(12) &= m_0 e^{-\left( \frac{\ln 2}{h} \right)(12)}\\ \\ m(12) &= 0.1895 m_0 \end{aligned} \end{equation}$

It shows that after 12 days, the mass remaining will be approximately $19\%$ of the original mass.