Hi, I need some help setting up this problem properly
Problem:
After 23 days a 10-milligram sample of radioactive material decays to 5 milligrams. After how many days will there be 1 milligram of the material?
thanks!
$\displaystyle m(t) = m_0 e^{- \lambda t}$ where m(t) is the mass of the material at a given time t, $\displaystyle m_0$ is the initial amount of mass m(0), $\displaystyle \lambda$ is the half-life, and t is the time elapsed.
So we know that
$\displaystyle 5 = 10 e^{- \lambda \cdot 23}$
We need to solve this for $\displaystyle \lambda$.
$\displaystyle \frac{1}{2} = e^{-23 \lambda}$
$\displaystyle ln (1/2) = -23 \lambda$
$\displaystyle \lambda = -\frac{ln(1/2)}{23} = \frac{ln(2)}{23} \approx 0.030136834$ /day
So
$\displaystyle m(t) = 10 e^{-0.030136834 t}$
We wish to find out when m(t) = 1 mg.
$\displaystyle 1 = 10 e^{-0.030136834 t}$
Solve this like the preceding problem. I get $\displaystyle t \approx 76.404346182$ days.
So 76.4 days or so.
-Dan