#Q(t) = Q_(o)e^(-kt)# may be used to model radioactive decay. #Q# represents the quantity remaining after years; #k# is the decay constant, 0.00011. How long, in years, will it take for a quantity of plutonium-240 to decay to 25% of its original amount?

1 Answer
Jun 15, 2017

Answer:

#t=ln(4)/0.00011# years

Explanation:

The initial quantity at t = 0 is:

#Q_0(0)=Q_0e^(-0.00011*0)=Q_0#

The amount left after t years is one quarter of the initial:

#Q(t)=Q_0/4#

Substitute this expression into the function and the #Q_0#'s will cancel out:

#Q(t)=Q_0/4=Q_0e^(-0.00011t)#

#rArrcancel(Q_0)/(4cancel(Q_0))=e^(-0.00011t)#

#1/4=e^(-0.00011t)#

Take the natural log of both sides, this is the inverse operation of #e^x# and will undo it (they cancel out):

#ln(1/4)=cancel(ln)cancel(e)^(-0.00011t)#

#ln(1/4)=-0.00011t#

#ln(1/4)=ln(4^-1)=-1*ln(4)#

#t=ln(4)/0.00011~~12600# years

According to nuclear-power.net, the half-life of plutonium-240 is 6560 years.
The amount left after #n# half-lives is #1/n^2#. So after #n=2# half-lives we will have #1/4# of the initial amount, which would be 13120 years. Our answer is in the ball park.