Question

A certain isotope decays radioactively so that the mass satisfies the equation m′=−0.01m. Here the time is measured in days. What is the half life? [hint: solve the equation then find at what time your solution gives half the inital mass]

Answer 1

The given equation is
m' = -0.01m
where m =  the mass at time t, days.

Assume that m(0) = m₀, the initial mass.
The solution for the differential equation is as follows:
$\frac{dm}{dt} = -0.01m \\\\ \frac{dm}{m} = -0.01dt \\\\ \int _{m_{0}}^{m} \frac{dm}{m} = -0.01 \int_{0}^{t} dt \\\\ ln \frac{m}{m_{0}} =-0.01t \\\\ m(t)=m_{0} e^{-0.01t}$

At half-life, the time is given by
$e^{-0.01t} = \frac{m_{0}/2}{m_{0}} = \frac{1}{2} \\\\ -0.01t = ln(0.5) \\\\ t = - \frac{ln(0.5)x}{-0.01} = 69.315$

Answer: The half-life is 69.3 days.