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

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.