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As indicated in the answer to this physics.stackexchange question the total amount of nucleons is preserved during fission. As a result the atomic number of a daughter product can be predicted if another was already known. In the case of $U_{92}^{235}$ fissioning to $Cs_{55}^{137}$, we know the atomic number of the second fission product is given by: ...


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Fission reaction produce a distribution of several different nuclei. Wikipedia (User:JWB) has a nice graph that shows the relative probabilities. However, in the case of a single reaction, where it is given that a Cs-137 nucleus is produced, you can probably be more specific because there are correlations between the fission products, but it will require ...


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Actually fission processes are stochastic, so we can't predict exact products of it. The most probable fission products are Cs-137 and Sr-90. The total amount of neutrons emitted is also quite unpredictable, but as I recall it usually lies in the range between 2 and 4.


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Is there a method for producing neutrons on demand? As dmckee said in his comment google "neutron generator". I was surprised at the miniaturization. How do quarks convert protons into neutrons? They do not. Energy has to be supplied to reorder the groupings of quarks during scattering, and they regroup according to the quantum mechanical ...


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When an isotope of an element $A$ is fissioned, it breaks up in a number of pairs of daughter products, say $B$ and $C$, where $B$ and $C$ are isotopes of comparable atomic mass. Often 2 or 3 neutrons are also released (opening up the possibility of chain reactions). However, the total number of nucleons, that is protons plus neutrons, is preserved during ...


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Fission reaction tend to produce a range of potential daughter nuclei, so the decay path is not just one single pair. There is a broad literature on fission yields, much of it from the 1950's and 60's (not surprisingly). For example, the 1965 IAEA symposium on Physics and Chemistry of Fission is available on-line at the IAEA. A search on Cs137 finds papers ...


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With regard to your first question, the transmitted plane wave doesn't undergo any scattering from the potential. This is made explicit by the representation of the scattered wavefunction as a sum of an incident planewave and an outgoing spherical wave. As such, the transmitted wave doesn't have anything to tell us about the scattering event. All of that ...


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Note that the combined rest masses of the quarks (~10 MeV/$c^2$) account for about 1% of the proton and neutron mass (~938 MeV/$c^2$), the main contribution to the mass are the gluons from the Strong Force. Since the composition of the proton and neutron are different, so is the force that binds them.


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There are, of course, limitations to the independent particle model (IPM) version of the shell model. First of all, the configuration you get from the shell model depends on the assumed ordering of the single-particle orbits. If one uses single-particle orbits calculated for 208Pb, then the ordering for neutron orbits above N=50 is 1g7/2, 2d5/2, 2d3/2, ...


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The range of attraction between two protons is short. So if you have a large number of protons only the long range repulsive Coulomb force will dominate and the nucleus will not be stable. So you need neutrons which are free from this repulsive force.


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I'm sure this has been answered elsewhere on the site, so I won't provide too much detail. The nucleus is held together by the strong nuclear force between nucleons (protons and neutrons). The force is short range and effectively only acts between adjacent nucleons. At the same time, the Coulomb repulsion between protons acts over a long range - so that all ...


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The spin of the neutron was measured by the Stern-Gerlach experiment by Sherwood, Stephenson and Bernstein (1954) (sadly paywalled, free links welcome), Abstract: A neutron beam was polarized by total reflection from a magnetized iron mirror. The beam was then analyzed by passing it through an inhomogeneous magnetic field. From the deflection pattern ...



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