Question

Co-60 is a beta emitter with a half-life of 5.3 years. Approximately what fraction of Co–60 atoms will remain in a particular sample after 26.5 years?

Answer 1

The equation for radioactive decay is given by

`A=A_oe^((-0.693t)/T_(1/2))`

where `T_(1/2)` is the half-life of the material and `t` is the period of time that elapses in a given case.
In this case, it is actually the ratio `A/(A_o)` that we are looking for, so this will be found by evaluating the exponential function

`A/(A_o)= e^(((-0.693(26.5))/5.3) = e^(-3.465)=0.03125`

So, the portion remaining is 0.03125 or 3.125 % which is `1/32` as a fraction.

Note that I used the exponential form of the equation. You could also use the (simpler) form

`A=A_o(1/2)^(t/T_(1/2))`

if you rather.