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How a nucleus can break into two
nuclei under the impact of a neutron

The nuclei of some nuclides are unstable and change spontaneously through the ejection of an electron or an alpha particle; or in very rare cases the ejection of a positron. Such changes involve a reduction of the energy level through an appropriate balancing of the number of numbers of neutrons and protons. (Balancing does not mean achieving an equal number of neutrons and protons, but instead more neutrons than protrons.) This phenomena is easily understandable. It is the fission of a nucleus under the impact of a neutron that is puzzling, the real enigma.

The fission of uranium 235 beautifully illustrates the problem.

92U235 + n → 56Ba144 + 36Kr89 + 3n + γ

where γ stands for a gamma photon.

There is actually a distribution of yields and the above one represents the high points for the distribution.

One puzzle is why the three neutrons are not incorporated into one of the created nuclei. It is this creation of extra neutrons which made the nuclear chain reaction possible.

It would be plausible to explain the fission on the basis that the addition of a neutron to a U235 nucleus creates an unsustainable nucleus because it contains too many neutrons. But is not the case; there are isotopes of Uranium which contain even more neutrons.

Not all isotopes of uranium fission. For example,

92U238 + n → 93Np239 + e + νe
and
93Np23994Pu239 + e + νe

where νe stands for an electron anti-neutrino.

One conventional model of nuclear physics, the liquid drop model, treats nuclear fission as being the result of a spherical drop changing into a dumbbell-shaped structure which then separates. This ignores the shell structure of nuclei.

Consider the shell structures of the nuclei involved in the above fission reaction.

The Shell Occupancies
NuclideNucleonShell 1-4Shell 5Shell 6Shell 7Shell 8
92U235Proton28223210
92U235Neutron2822324417
56Ba144Proton2822
56Ba144Neutron2822326
36Kr89Proton288
36Kr89Neutron28223

Note that half of the 92 protons of the 92U235 nucleus is 46 and the proton number for 36Kr89 is ten less and the proton number of 56Ba144 is ten more. The neutron number of 92U235 is 143 and half of that is 71.5. The neutron number of 56Ba144 is 88 and that is 71.5 plus 16.5. The neutron number of 36Kr89 is 53, which is 71.5 minus 18.5.

The 56 protons of the Barium nucleus could come from the filled 1st through 5th shells.The 36 protons of the Krypton nucleus could then have come from 6th and 7th proton shells of the 92U235 nucleus and its 53 neutrons from its 7th and 8th neutron shells..

Binding Energies

The binding energy of a 92U235 nucleus is 1783.8703 million electron volts (MeV). The combined binding energy of a 56Ba144 nucleus and a 36Kr89 nucleus is 1957.149 MeV so the transform of the uranium nucleus into the barium and krypton nuclei releases 173.2787 MeV of energy that goes to the three neutrons and the gamma photon. The question is whether the form of the reaction is the one that releases the most energy. The answer is no; if one neutron is transferred from the barium isotope to the krypton isotope there is an additional 0.4 MeV of energy released. However if two neutrons are transferred from barium to krypton 13.355 MeV less energy is released. Furthermore, if one neutron is transferred from the krypton isotope to the barium isotope there is 0.74 MeV less energy released.

If the two created neutrons were retained by the barium and krypton isotopes and at most the released energy should be 183.9467 MeV. However perhaps there is need for the two neutrons to preserve linear momentum and also to carry off some of the released energy. One neutron preserves the momentum of the neutron impinging upon the uranium nucleus. Such a constraint would explain the existence of the three neutrons created in the reaction. But there is no apparent reason that the barium and krypton nuclei can not fulfill this requirement.

If the products of the fission were 37Rb89 and 55Cs144 there would be 3.478 MeV less energy released than for the barium and krypton products. If the products were 35Br89 and 57La144 there would be a whopping 95.0298 MeV less energy released than for the barium and krypton products. Thus a maximum energy release criterion explains the proton numbers of the fission products.

Now consider the nonfission of a 92U238 under impact by a neutron. This reaction releases 5.2874 MeV of energy, whereas the creation of 93Np236 from the impact of a neutron on 92U235 would not release any energy, but would, in fact, require 0.906 MeV input of energy. The transition from 93Np239 to 94Pu239 involves essentially no change in binding energy. (There is a required input of energy of 0.06 MeV which could come from the decay of a neutron.)

However, a split of a 92U238 into 36Kr90 and 56Ba146 would release 171.232 MeV of energy. Likewise a split in 94Pu239 would release more than 170 MeV of energy. So the puzzle continues. It may be that 94Pu239 splits but slowly.

The proton numbers of the reaction products of the neutron-induced fission of a 92U238 nucleus can be attributed to some maximum energy yield criterion. .

(To be continued.)


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