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:

`(dm)/(dt) = -0.01m \\\\ (dm)/(m) = -0.01dt \\\\ \int _{m_(0)}^(m) (dm)/(m) = -0.01 \int_(0)^(t) dt \\\\ ln (m)/(m_(0)) =-0.01t \\\\ m(t)=m_(0) e^(-0.01t)`

At half-life, the time is given by

`e^(-0.01t) = (m_(0)/2)/(m_(0)) = (1)/(2) \\\\ -0.01t = ln(0.5) \\\\ t = - (ln(0.5)x)/(-0.01) = 69.315`

Answer: The half-life is 69.3 days.