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
[tex]N(t) = N_0 ( \frac{1}{2})^{ \frac{t}{t_{1/2} } [/tex]
where [tex]N(t)[/tex] is the number of nuclei left at time t, [tex]N_0[/tex] is the number of nuclei at t=0, and [tex]t_{1/2}[/tex] is the half-life. For cobalt, [tex]t_{1/2}=5.27~y[/tex].
We can re-arrange the formula as
[tex] \frac{N(t)}{N_0} = ( \frac{1}{2} )^{ \frac{t}{t_{1/2}} }[/tex]
and then
[tex]t=t_{1/2} log_{1/2}( \frac{N(t)}{N_0} )[/tex]
the problem asks for the time t at which only 20% of cobalt is left, that means the time t at which
[tex] \frac{N(t)}{N_0} = 0.20 [/tex]
therefore, using this into the previous equation, we get
[tex]t=5.27~y \cdot log_{1/2} (0.20) = 12.24~y[/tex]
so, after a time of 12.24 years, only 20% of cobalt is left.
