2
$\begingroup$

"In the case of neutron-induced fission reactions, an incident neutron provides additional energy to a target nucleus in the form of kinetic energy and nuclear binding energy. Neutrons have the principal advantage, and they do not need to overcome the coulomb forces as in the case of charged particles."

Source: https://www.nuclear-power.com/nuclear-power/fission/critical-energy-threshold-energy-for-fission/

Does anyone have an idea as to how exactly a neutron causes a nucleus to split? I know from above that the neutron increases binding energy, which can be used to overcome the excitation energy required for fission, but wouldn't an increase in binding energy mean that the nucleus is more tightly held together, i.e. more stable? Where exactly does the free energy come from to pull the nucleons apart against the nuclear force, assuming that the kinetic energy of the neutron is negligible?

$\endgroup$

1 Answer 1

2
$\begingroup$

In the case of fission, a neutron is a convenient way of delivering a punch to a marginally stable nucleus, perturbing it from a more-or-less spherical shape in the process of either scattering off the nucleus or getting captured.

That nucleus can hold itself together only if it remains spherical. If it begins to wobble into an elliptical shape, then the minor axis tends to neck down and pinch off into two nuclei plus a few loose neutrons. Those two daughter nuclei then repel each other electrostatically and fly off with great speed.

This is a somewhat simplified model for fission. A more detailed but still very readable account is furnished in Serber's book, The Los Alamos Primer.

$\endgroup$
6
  • $\begingroup$ Thank you for your response. I assume that the wobbling is due to nucleus entering some kind of excited state. Does this have anything to do with the nuclear shell? For instance, if U-235 absorbs a neutron and becomes U-236, is the extra neutron somehow incompatible with the other neutrons in the nucleus, causing it to convert to a proton and therefore increasing the repulsive forces? I am looking for something a little bit more beyond the liquid drop model, which is what you seemed describe. $\endgroup$ Commented Feb 14, 2022 at 5:25
  • $\begingroup$ @BrunoNowak Serber's book goes into this in detail- neutron energies, activation energies, cross-sections, etc for 235, 238, 239.... $\endgroup$ Commented Feb 14, 2022 at 6:47
  • $\begingroup$ We can also notice that the fission of 236U is not systematic: 236U is produced in thermal neutron reactors . There are many books about " theory of fission " . $\endgroup$ Commented Feb 14, 2022 at 7:38
  • $\begingroup$ @nielsnielsen I read through the book your recommended and was not able to find an explanation for how the binding energy of the last neutron is used to induce fission. From my understanding, the binding energy is the energy that holds nucleons together. This energy comes from the loss of mass of the nucleons. How could the binding energy of the last neutron both bound it to the nucleus and excite the nucleus? Wouldn't that create, essentially, 2x the binding energy? I feel like I am missing something major here. $\endgroup$ Commented Feb 15, 2022 at 23:29
  • $\begingroup$ @BrunoNowak, you are, but I do not know how to clear this up for you. $\endgroup$ Commented Feb 16, 2022 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.