Question

Approximately ____ days have passed when 20% of Cobalt (Co) remains?

Answer 1

The half-life of Cobalt 60 is 5.27 years. This means that after this time, only 50 % of the initial amount of Cobalt is left. The relationship between the number of nuclei left at time t and the time, for a radioactive decay, is given by
$N(t) = N_0 ( \frac{1}{2})^{ \frac{t}{t_{1/2} }$
where $N(t)$ is the number of nuclei left at time t, $N_0$ is the number of nuclei at t=0, and $t_{1/2}$ is the half-life. For cobalt, $t_{1/2}=5.27~y$.

We can re-arrange the formula as
$\frac{N(t)}{N_0} = ( \frac{1}{2} )^{ \frac{t}{t_{1/2}} }$
and then
$t=t_{1/2} log_{1/2}( \frac{N(t)}{N_0} )$
the problem asks for the time t at which only 20% of cobalt is left, that means the time t at which
$\frac{N(t)}{N_0} = 0.20$
therefore, using this into the previous equation, we get
$t=5.27~y \cdot log_{1/2} (0.20) = 12.24~y$
so, after a time of 12.24 years, only 20% of cobalt is left.