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?

1 Answer
Jan 20, 2017

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.