Question

A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 present at any time. After 10 years, 0.03% of the radioactivity has dissipated (that is, 0.9997 of the initial amount remains). Suppose that initially there is 180 pounds of this substance. Find the half-life. (Round your answer to the nearest whole number.)

Answer 1

Let $U(t)$ denote the amount of uranium-238 in the reactor at time $t$. As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

$\dfrac{\mathrm dU}{\mathrm dt}=-kU$

where $k>0$ is constant. Separating variables and integrating both sides gives

$\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}$

Suppose we start some amount $u$. This means that at time $t=0$ we have $U(0)=u$, so that

$u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}$

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if $t$ is taken to be in years) for some starting amount $u$,

$0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}$

The half-life is the time $t_{1/2}$ it takes for the starting amount $u$ to decay to half, $0.5u$:

$0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}$

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.