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

`(\mathrm dU)/(\mathrm dt)=-kU`

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

`\frac{\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=-(\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)=-\frac{\ln(0.5)}k=(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.