Spontaneous fission

Spontaneous fission (SF) is a mode of decay in which a heavy nucleus undergoes nuclear fission spontaneously, without energy input from external particles. This causes the nucleus to split in to daughter nuclei of variable size. Unlike other modes of decay, the decay products of spontaneous fission are quite random and are best described by a Probability distribution. It's a rather rare mode of decay, only observed in superheavy radionuclides like uranium and plutonium. Spontaneous fission is much less common than induced fission, as few nuclei are unstable enough to even permit it and even in those that do, its branching ratio is generally tiny compared to other decay modes. Induced fission, however, requires human intervention.

The nuclear reaction reads

where $\ce{X}$ is the original nuclide, $\ce{X'}$ and $\ce{X''}$ are the fission fragments and $\nu n$ are the $\nu$ additional neutrons that are emitted as part of the fission. The fragments are almost always extremely unstable and end up themselves decaying in a cascade of reactions.

Like alpha decay, spontaneous fission is governed by the opposition of the nuclear attraction due to the strong force and the repulsion due to electromagnetism. In fact, while it is not a cluster decay, it works in a similar manner.

An example of radionuclide that undergoes spontaneous fission is $^{238}\text{U}$, which has a binding energy of approximately $\sim7.6\text{ MeV/nucleon}$. Normally it alpha decays, but it may also split. When it does, it splits into two smaller nuclei sampled from two distributions, both of which have higher binding energy.

Mechanism#

Consider some superheavy nucleus ($A\gtrsim 230$). It's binding energy per nucleon is very low compared to much smaller nuclei, generally in the range of $7.5-7.6\text{ MeV/nucleon}$. As such, breaking into smaller nuclei, which generally have binding energies in the range of $8.4-8.5\text{ MeV/nucleon}$, is potentially very energy efficient. In practice, however, the strong attraction creates a deep well of potential energy in which then nucleus sits comfortably in and electrostatic repulsion creates a major potential barrier that prevents the nuclei from escaping or entering.

Let $R_{0}$ be the radius of the nucleus. If it were to split in two, the fragments would have radii $R_{1}$ and $R_{2}$. At the moment of fission, they would be "touching", so the total "size" of the result would be $R=R_{1}+R_{2}$. This end state would need to have enough energy to overcome the energy barrier for fission to happen. The potential energy at this distance is

For fission to be energetically favored, the final state must have a more negative binding energy than the initial one: $\Delta E=E_{f}-E_{i}<0$. Actually, this is just the Q value, so we'll use $Q=-\Delta E$ instead. For conservation to hold, other fission products, be it gamma rays, individual protons or electrons being ejected or other must transfer away this energy through their kinetic energy: $E_{i}=E_{f}+E_\text{other}=E_{f}+Q$.

Fission products must overall hold $214\text{ MeV}$ of kinetic energy.

In practice, this fission is very unrealistic, as it does not contain other products besides the two palladium nuclei and it is extremely unlikely for a nucleus to split precisely in half. Still, it makes for an easy example.

You might wonder how a nucleus whose $Q$ is lower than the potential barrier can cross the barrier. If you know alpha decay, you already know the answer: quantum tunneling. Despite the available energy being insufficient to fully cross the barrier, since the barrier is not that wide for nuclear scales, there's always a probability that the fragments can tunnel through and escape anyway. The higher $Q$ is, the higher the tunneling chance gets. This is why so few nuclides get to the fission point: $Q$ just isn't remotely high enough in some cases. Even when it is, it's often just barely enough to be noticeable: the uranium-238 from before has a fission half-life somewhere in the $\sim 10^{16}$ years.

Deformation and energy#

Fission is a more involved process than something like alpha decay. It involves a heavy deformation of the nucleus until it cracks in two pieces. The process is somewhat reminiscent of cellular mitosis: from its original shape (sometimes a sphere, sometimes more squished, depends on the nuclide), the nucleus gets deformed by the repulsive electrostatic forces into an ellipsoid. Eventually, the deformation is so heavy it splits into two pieces.

We can approximate this process by considering the deformed nucleus as being an ellipsoid of volume $V=\frac{4}{3}\pi ab^{2}$, where $a$ and $b$ are the axes of the ellipsoid. These can be expressed in terms of the volume $R$ of a sphere according to

where $\epsilon\equiv\beta\sqrt{5/4\pi}$ is the eccentricity and $\beta$ is the deformation parameter. If we take the volume the constant all throughout (we're deforming, not adding or removing), the surface area of an ellipsoid can be expressed as a power series as

Electrostatic repulsion instead weakens by a factor

The reduction in binding energy can therefore be expressed by putting what we've found above in the semi-empirical mass formula:

(the surface and Coulomb terms are the only ones that don't cancel out; also $Z(Z-1)\simeq Z^{2}$). Thus, the $Q$ value of spontaneous fission is, to first order,

Fission becomes spontaneous when $Q>0$, which means

or equivalently

This is only true for superheavy atoms.

The formula for the $Q$ value above is theoretically interesting but isn't particularly useful for predictions. Instead, just use the definition:

This comes out to roughly always be in the range of $200\pm 20\text{ MeV}$, with not that much variation from splitting different nuclides. This is due to a combination of very few nuclides being able to undergo fission and the distribution of output fragments being pretty rigid (see Nuclear fission > Decay products fore more). The binding energy per nucleon of any nuclide that can spontaneously split is around 7.6 MeV. Meanwhile, the binding energy of resulting fragments is around 8.5 MeV. Thus, the $Q$ value is pretty much always 0.9 MeV per nucleon of the original nucleus, so it is roughly approximated by

For $\ce{^{235}U}$ this is 211.5 MeV and for $\ce{^{238}U}$ this is 214.2 MeV. Almost the same. Do know that these are overestimates (e.g. $\ce{^{235}U}$ actually has a mean $Q$ value of 202.8 MeV).

Most (80-85%) of the energy of the fission is expelled as kinetic energy of the fragments. The neutrons also carry a lot of kinetic energy for their size. The amount varies per nuclide, but it's an important metric, because these neutrons can collide with other nuclei and induce fission, thus initiating a chain reaction. This can only occur if the mean neutron energy is sufficiently high to overcome the activation energy of the nucleus (see Induced fission > Energy). In $\ce{^{235}U}$, the neutrons have a mean kinetic energy of about 2 MeV, enough to fission both $\ce{^{235}U}$ and $\ce{^{238}U}$. Finally, prompt gamma ray emission can also carry several MeV of energy. Delayed beta decays and gamma decays then emit even more, though this is not included in the $Q$ value of the fission.