# 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?

Jan 20, 2017

The equation for radioactive decay is given by

$A = {A}_{o} {e}^{\frac{- 0.693 t}{T} _ \left(\frac{1}{2}\right)}$

where ${T}_{\frac{1}{2}}$ 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 $\frac{A}{{A}_{o}}$ 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 $\frac{1}{32}$ as a fraction.

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

$A = {A}_{o} {\left(\frac{1}{2}\right)}^{\frac{t}{T} _ \left(\frac{1}{2}\right)}$

if you rather.