Question

`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?

Answer 1

`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.