Why fissions are mostly asymmetric rather than symmetric??
I mean of course at low energies...
Even at high energies ~1 MeV, the fissions are still asymmetric.
Probably because of the QM effects binding energy and sub-nuclear energy levels.
First of all thx for the reply...
I meant energies a lot higher than 1 Mev, with increasing energies the probability for symmetric fission gets higher...
But can we get deeper, i mean what are these effects that leads to that fact??
You mean something like 14 MeV? Well, certainly as the energy increases, the proportion of more symmetric fissions increases.
The most symmetric fission reaction would by 92U236* -> 2 46Pd117 + 2 n, but the probability of that particular outcome, as well as any particular outcome, is relatively small (given that there are many possible outcomes or combinations of products).
Well, looking at the initiating event - one neutron 'hits' the U-235 nucleus - it is asymmetric! Even the U-235 is asymmetric because it has an odd number of neutrons.
The most symmetric the reaction could be is if the neutron 'hits' the nucleus dead-on-center, but still the U-235 maintains some asymmetry. However, the probability of a precisely dead-on-center collision is quite small. Even then, there is the spin of the neutron which would necessarily make an interaction asymmetric.
That's what i meant, exactly.
This makes sense, but i still don't get it, i mean it's not as simple as hitting a board, where you can hit your target or get some deviation...
Where's that center?? Does the shell model explain this?? OR why would the neutron spin make it necessarily asymmetric??
Maybe i'm a little thick headed, but can't help it i must ask...
In the center. :uhh: The center of mass, which may shift according to whatever resonances the nucleus is experiencing. Remember, there is a finite probability of 'spontaneous' fission.
Possibly. I'd have to dig around to find some expermental results which would verify that the 'shell model' is consistent with the distribution of fission products.
Because it has a preferred direction.
Thinking on the classic scale, if one has a spinning top and it collides as a result of translational motion, it scatters with a preferred direction (which is asymmetric) by virtue of the spin (angular momentum).
:rofl: yeah i know the center is the center, but i meant i can't pin point that center, and hence maybe i can't aim for a symmetric fission. Forget it if it sounds ridiculous.
No - it is unlikely that a neutron would hit on-center. Also, think of Rutherford scattering of alpha particles or Compton scattering of gamma or X-rays, it is rare to get a perfect recoil in the incident direction and that indicates the relatively low probability of a particle colliding on a perfect on-center strike.
I was under the impression that the collision dynamics between the neutron and the nucleus had an insignificant effect on the way in which the nucleus fissions. This is because the kinetic energy of a thermal neutron is much less than the binding energy of that neutron upon capture by the nucleus. Am I wrong on that?
Well, certainly yes for thermal neutrons, and there are three outcomes, absorption with fission, absorption activation (gamma-emission) and scattering.
The aborption of a thermal neutron, or any neutron, is asymmetric since it 'enters' (interacts) at a particular location at the nucleus.
I was thinking mainly of fast neutrons, which result in more symmetric fissions as is evidenced by the proportion of the f.p. spectrum near Z=46, i.e. midway between Z=36 and 56, which are common f.p.s
However, even spontaneous fissions are asymmetric which would indicated that internal resonances are asymmetric. Really what I was trying to explain is that there is essentially no way to achieve symmetry.
This is a very good question, and an interesting one.
It has been 33 years since I did any work in nuclear physics and I am not sure I understood it that well even then, but here is my understanding of the greater asymmetry of fission produced by thermal neutrons. I will leave it up to the real experts here, such as Morbius or Astronuc, to correct me if anything I am about to say is wrong.
The neutron capture cross-section of the U235 nucleus is large for thermal neutrons. This means that a thermal neutron can be captured when it collides with the nucleus at a wide range of angles (not just head-on 90 degrees).
When a neutron is captured, there is a release of about 6 MeV of energy within the nucleus. The release of 6 MeV of neutron binding energy from the capture of the neutron excites the nucleus and the excited nucleus undergoes conformational changes.
Given that the nucleus of U235 is quite large compared to a neutron, the location in the nucleus where that neutron is captured should affect the resulting distribution of that energy, at least initially, within the nucleus.
It seems to me that the distribution of energy would determine whether and where the nucleus breaks apart. The point at which the attractive nuclear force is exceeded by the repulsive coulomb force determines where the break occurs. If the energy distribution is asymmetric (ie. closer to one side because the neutron is absorbed at an angle less than 90 degrees) the nucleus will break apart in unequal pieces.
The capture cross-section for fast neutrons is much smaller. This means that fast neutrons striking at an oblique angle will bounce off rather than be captured. Consequently, the fast neutron has to strike at close to 90 degrees ie. at or very near the centre of the the nucleus in order to be captured. This is why fast neutrons will produce fission products with similar masses ie. closer to 1/2 the mass of the U235 nucleus.
Still learning but sounds good to me...
Edit: Okay i've been doing some reading today, and well in fission after the compound nucleus is formed according to the model, it is not about the center, since it becomes "memoryless".
IT doesn't remember where the neutron hit it, and if that's true, then it has nothing to do with the formation of asymmetric/symmetric fission products.
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