# Symmetric or antisymmetric spontaneous fission

1. Jan 6, 2016

### Noddi

Is it possible for a nucleus to undergo antisymmetric spontaneous fission? And if so, what is the process responsible?

Thanks

2. Jan 6, 2016

### Staff: Mentor

What is "antisymmetric spontaneous fission"?
Usually fission produces one larger and one smaller fragment, so it is asymmetric. That is not the same as antisymmetric.
The responsible process is the combination of strong and electromagnetic interaction.

3. Jan 6, 2016

### Noddi

Sorry! Yes I meant to write asymmetric. As in why would spontaneous fission produce two nuclei of different masses instead of producing two nuclei of the same mass?

4. Jan 6, 2016

### Staff: Mentor

Why do you expect two nuclei of the same mass?

5. Jan 6, 2016

### Noddi

Because if a nucleus undergoes fission into two nuclei of the same mass, the energy released would be higher than if the same parent nuclei undergoes fission to produce two nuclei of different masses. So wouldn't that mean it is more energetically favorable for a parent nuclei to undergo fission into two nuclei of the same mass?

6. Jan 6, 2016

### Staff: Mentor

The final energy is not the only relevant quantity, the intermediate steps are important as well. I don't know the details, but a symmetric splitting should have a higher energy barrier.

7. Jan 6, 2016

Staff Emeritus
Alpha decay is a kind of spontaneous fission with very asymmetric fragments.

8. Jan 7, 2016

### Staff: Mentor

There is also the rare cluster decay - like alpha, but with carbon, oxygen or similar fragments instead of helium nuclei.

I didn't check the numbers, but Wikipedia suggests that uneven fission product sizes actually release a bit more energy than symmetric products. Higher-energetic nuclei seem to produce more symmetric fission products.
Nuclear fission product

9. Jan 7, 2016

### Noddi

Good points about the alpha decay and rare cluster decays. I had never thought of it that way before.

I'm a bit muddled about the idea that uneven fission product sizes could release more energy than symmetric products. I.e surely the fission of a parent nuclei into two evenly sized nuclei would release more energy than an alpha decay of the same parent nuclei? Unless you are talking about the energy released over multiple uneven decays?

Would I be correct in saying that both symmetric and asymmetric fission products are present in nature, but the higher energy barrier of the symmetric splitting means, on average, it is less likely to happen? However, when it does, a single decay into symmetric nuclei will release more energy than a single decay into asymmetric nuclei?

Thanks you both for your help, I am enjoying thinking about it further.

10. Jan 7, 2016

### Staff: Mentor

I checked the numbers. Let's take ${}^{235}_{~92}U$ with an additional neutron. The symmetric fission gives $2\cdot {}^{117}_{~46}Pd + 2n$, with a combined mass of 2*116.9178 u = 233.8356 u plus neutrons.

A possible asymmetric fission reaction is ${}^{96}_{39}Y + {}^{138}_{~53}I + 2 n$, with a combined mass of 95.9159 u + 137.9223 u = 233.8382 u plus the neutrons. 2.8 MeV more, and the other combinations I tested gave similar results.

11. Jan 7, 2016

### Noddi

Exactly, the total mass of the asymmetric fission products are higher than the total mass of the symmetric products. But that means the binding energy released in the fission is higher for the symmetric products, since difference in mass of the parent nuclei and its products is larger.

For the case of symmetric products, the mass difference is about 0.2003u so the energy released is about 187MeV.
For the case of asymmetric products, the mass difference is about 0.1977u so the energy release is about 184MeV.

12. Jan 7, 2016

### bcrowell

Staff Emeritus
Both symmetric and asymmetric fission exist in nature. I don't think it's true that the Q value always favors one over the other, nor is it true that the Q value is always the decisive factor. And even so, note that the Q value is going to be influenced by shell effects, so you can't just reason based on expectations from the liquid drop model.

A better way to think about all of this is the following. We can parametrize the shape of a nucleus using a couple of parameters, call them $a_2$ and $a_3$. The $a_2$ parameter measures elliptical stretching, while $a_3$ measures asymmetric deformation into a pear shape. A sphere has $a_2=0$ and $a_3=0$. A uranium nucleus in its ground state might have $a_3=0$ (it's symmetric) and $a_2=0.2$ (a slight prolate deformation like an American football).

Using techniques such as Strutinsky smearing, you can calculate a potential energy $U$ that is a function $U(a_2,a_3)$ of the deformation. You can visualize this as a surface. In its ground state, the nucleus sits in a metastable minimum of this function. On the average, the behavior of this function is that of the liquid drop model, but for a particular nucleus it incorporates shell corrections. (That's what the Strutinsky smearing does for you.) Because of the Heisenberg uncertainty principle, the nucleus in its ground state can't just sit at a single well-defined point in this landscape, with a well-defined shape. There are zero-point fluctuations in the shape. Large, low-probability fluctuations will carry the nucleus into classically forbidden areas. By tunneling, it can break out into the region where the deformation is so large that it will break up. The route that it takes will essentially be the one that minimizes the height and width of the potential barrier, so that the WKB tunneling probability is maximized. If this route heads out to nonzero values of $a_3$, then there is spontaneous symmetry breaking, and you get asymmetric fission. Otherwise, you get symmetric fission.

The Q value will certainly be *correlated* with the WKB tunneling probability, but it's not a perfect correlation.

This paper gives a good overview: http://www.nature.com/nature/journal/v409/n6822/full/409785a0.html

To see some real calculated potential energy surfaces, you can go here: https://t2.lanl.gov/nis/data/astro/molnix96/peseps2gamma-fis.html . The parameter $\epsilon_2$ they use is basically the same as the $a_2$ I was referring to in the generic description above. Their $\gamma$ isn't equivalent to $a_3$ but describes a different type of asymmetry. I believe the surface actually shows the PE at each point that is minimized with respect to a parameter like $a_3$, but that parameter's actual value is hidden on these plots. The Nature paper goes into a lot more detailed description of asymmetric and symmetric fission, but it doesn't actually show potential energy surfaces.

13. Jan 7, 2016

### Noddi

That's a really cool idea to think of the potential energy as a function of deformation as a surface. I think I understand the process more clearly now so thank you very much for that.