Burnout time

The half-life of a radioactive nuclide, t, is defined as the time it takes for half of the atoms of a radioactive source to undergo transformation. If one takes the standard decay equation:

$dN(t)/dt = -\lambda N$

where $N(t)$ is the number of atoms at the time t, and $\lambda$ is the decay constant, the solution is $N(t) = N(0) exp(-\lambda t)$. It follows that $1/2 = exp(-\lambda \tau)$ and hence $\tau =ln2/\lambda$. In a reactor, the rate of disappearance of a nuclide is given by

$dN(t)/dt = -\sigma_a N \phi$

where $N(t)$ is the number of atoms at the time t, $\sigma_a$ the absorption cross section and $\phi$ the neutron flux of the reactor. The absorption cross section is the sum of the capture and fission cross-sections and in addition the cross sections for n,2n, n,3n, n,p and n,alpha reactions i.e.

$\sigma_a = \sigma_{n,g} +\sigma_{n,2n}+ \sigma_{n,3n}+\sigma_{n,p}+\sigma_{n,\alpha}+ \sigma_{n,fission}$.

Similarly to the definition of the half-life, a "burnout time" $\tau_{bo}$ can defined as the time it takes for half the atoms of a fuel to transmute i.e. $\tau_{bo} = ln2/\sigma_a\phi$. The above relations can be combined to give a relation for the overall rate of change due to decay and reaction of a nuclide in a neutron flux i.e.:

$dN(t)/dt = -\lambda N -\sigma_a N \phi == -\left ( \frac{ln2}{\tau} + \frac{ln2}{\tau_{bo}}\right )N$

In this equation, when $t >> \tau_{bo}$ (long half-life, short burnout time), one can neglect the decay process as the absorption of a neutron occurs much faster than the decay, and when $\tau_{bo} >> t$ (long burnout time and short half-life), one can neglect the nuclear reaction as the nuclides has decayed long before a neutron is absorbed.

In the case of a 238U nucleus in a reactor, the 238U will transform mainly by neutron absorption to give 239U, since the burnout time for 238U is much shorter than its half-life of 4.47x109 years. This 239U then decays to 239Np since the half-life of 239U (23.4 minutes) is shorter than its burnout time. The 239Np than decays to 239Pu as its half-life of 2.35 days is much shorter than its burnout time (for a standard neutron flux of 5x1013 neutrons cm-2 s-1) of 1.18 years. The 239Pu then absorbs neutrons and goes on to 240Pu and 241Pu which all have a half-life of several years, so that the burnout time is much shorter and the neutron reaction takes place.

One has to be careful with the neglection of a particular process. If the half-life and the burnout time for the standard neutron flux of 5x1013 neutrons cm-2 s-1 aren't very different, only a little change in the neutron flux will change the behaviour of the nuclide. Changing the neutron flux can change the reaction path. This can be illustrated with the nuclide 233Pa. This nuclide has a half-life of 26.9 days and a burnout time of 1.09 years. In this case, the 233Pa decays to 233U. But if the neutron flux is increased to 8x1014 neutrons cm-2 s-1, the burnout time decreases to 25 days, so that the probability for the neutron reaction is higher than for the decay. This means that 233Pa absorbs a neutron and gives 234Pa more often than decaying to 233U.