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The nuclear binding energy, is the energy that is needed to seperate the nucleons in a nucleus. And binding energy is also defined as the energy given out when a nucleus forms from nucleons.

Also the larger the nucleus is, the more energy is required to break it apart, so why doesn't that mean that larger nuclei are more stable? I mean Uranium has a lower binding energy per nucleon than Iron, but there are many many more nucleons in Uranium that Iron so the total binding energy is going to be much greater.

Basically I don't understand why whether an element gives out energy by fusion or fission (why the lighter element provide energy by fusion not fission and vice versa for heavy elements) depends on binding energy per nucleon and not "total" binding energy

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  • $\begingroup$ I am in a hurry, but I can give a hint. Why do you assume that the mass of nucleons that go into fusion is equal to those that come out of fission? If you don't, it is obvious that fission can give energy too. $\endgroup$ – Cheeku Jun 15 '13 at 16:18
  • $\begingroup$ I don't? I tried to simplify my question. $\endgroup$ – Jonathan. Jun 15 '13 at 16:31
  • $\begingroup$ physics.stackexchange.com/questions/68138/fusion-vs-fission $\endgroup$ – Cheeku Jun 15 '13 at 16:35
  • $\begingroup$ @Cheeku, I saw that I'm specifically asking why Binding energy per nucleon rather just binding energy defines the fusion or fission "thing" $\endgroup$ – Jonathan. Jun 15 '13 at 17:05
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Your basic nuclear reaction conserves the number of nucleons present.1

That is important, because at a bit less than 1 GeV each the mass of the nucleons dominates the total energy of all these states.

So the only place available to get or lose energy in a reaction is by

  1. Changing the flavor of nucleons. Every neutron converted to a proton gets you a neutrino and some gammas (once the positron has captured and annihilated).

  2. By changing the total binding energy. Notice that getting one nucleon more bound doesn't help if another one gets less bound by a large amount. The decision to express this in terms of the average binding energy is completely arbitrary because for $N$ total nucleons (which doesn't change, remember?) $E_\text{total} = E_\text{AVG} * N$.

Another questions addresses what is so special about iron that it has the highest binding energy per nucleon.


1 This is more or less required by Baryon number conservation in the Standard Model, and we will ignore the need for Baryon non-conservation in most beyond the Standard Model candidate theories.

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  • $\begingroup$ I don't really understand still. Binding energy is defined as "the energy needed to completely split a nucleus up" (at least at my level/textbook) then how does fission produce energy as it is spliting which Byrne definition of binding energy requires energy? $\endgroup$ – Jonathan. Jun 15 '13 at 20:34
  • $\begingroup$ Ah...I see. The "completely" there is to render the nucleus down to loose protons and neutrons. So you can imagine ripping the progenitor up completely (which costs the binding energy of that nucleus) and then reassembling the fission product from them (during which you get their binding energies back). If the sum of the binding energies of the products exceeds the binding energy of the progenitor that you come out ahead, which is the for many nuclei with $A$ up above 200. $\endgroup$ – dmckee Jun 15 '13 at 20:44
  • $\begingroup$ ok thanks that helps a lot. however now, I understand that the binding energy per nucleon of say Iron is greater than say Uranium. But there is many more nucleons in Uranium, so surely the binding energy of Uranium is greater than that of iron? (and so the binding energy of the products does not exceed that of the progenitor) $\endgroup$ – Jonathan. Jun 15 '13 at 21:06
  • $\begingroup$ Yes The heavy nuclei are more bound in total, but they are less bound per nucleon and the total number of nucleons is constant, so it follows that two light nuclei taken together are more bound that the single heavy nucleus that gave rise to them (because the total binding for any given nucleus is the binding energy per nucleon times the number of nucleons). $\endgroup$ – dmckee Jun 15 '13 at 21:12
  • $\begingroup$ ahh, it sounds stupid but I'm forgotting there is more than one product :) thanks, what you say make perfect sense. (its rare to find someone knowledgable, able to communicate it well and take the time to do it for idiots like me, thanks :)) $\endgroup$ – Jonathan. Jun 15 '13 at 21:50
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Nuclear physics is in the realm of quantum mechanics. To first order one can think of "one nucleon" in the collective left over strong interaction field of all the rest, as a potential. As for strong interactions nuclei are indistinguishable to first order the binding energy per nucleon makes sense. The larger this energy the harder to extract a nucleon from the potential.

nuclear binding energy

The total binding energy will depend on second order effects coming in from the Pauli exclusion principle and the electromagnetic repulsion of protons, from clusterings within large nuclei etc. For example there are a large number of unstable isotopes when the electromagnetic repulsion of the protons is not balanced by the neutrality of neutrons or when there are too many neutrons. A simple curve cannot be defined the way it can with the binding energy per nucleon.

The binding energy per nucleon is a necessary condition to see whether fusion can happen, but it is not sufficient to make sure the reaction will happen as all quantum numbers have to be satisfied . In this article there is a discussion of conditions and possibilities for fusion.

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