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?

Jun 15, 2017

$t = \ln \frac{4}{0.00011}$ years

Explanation:

The initial quantity at t = 0 is:

${Q}_{0} \left(0\right) = {Q}_{0} {e}^{- 0.00011 \cdot 0} = {Q}_{0}$

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

$Q \left(t\right) = {Q}_{0} / 4$

Substitute this expression into the function and the ${Q}_{0}$'s will cancel out:

$Q \left(t\right) = {Q}_{0} / 4 = {Q}_{0} {e}^{- 0.00011 t}$

$\Rightarrow \frac{\cancel{{Q}_{0}}}{4 \cancel{{Q}_{0}}} = {e}^{- 0.00011 t}$

$\frac{1}{4} = {e}^{- 0.00011 t}$

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

$\ln \left(\frac{1}{4}\right) = \cancel{\ln} {\cancel{e}}^{- 0.00011 t}$

$\ln \left(\frac{1}{4}\right) = - 0.00011 t$

$\ln \left(\frac{1}{4}\right) = \ln \left({4}^{-} 1\right) = - 1 \cdot \ln \left(4\right)$

$t = \ln \frac{4}{0.00011} \approx 12600$ years

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