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Binding energy per nucleon / fission / fusion

  1. Oct 30, 2009 #1
    What I'm looking to understand is why fission and or fusion result in the release of energy.

    I understand that:

    By looking at the Binding Energy per Nucleon Curve, due to the strong force acting at very small length scales and falling off as 1/x^3 and the electro-magnetic repulsion of the protons falling off as inverse square, we end up with the curve that we shape. Lighter elements are building up their nuclei to even build up a force, but than after 4 nuclei diameter, the EMF starts to have substantial input.

    I also have sort of understood that by looking at the scenario:

    1 AMU is measured as 1/12 of the mass of the carbon-12 atom, but the mass of the H-1 atom is 1.007825 AMU, therefore, what we say is that the nucleons in the carbon atom have lost some of their mass in the form of binding energy.

    So this leads me to understand that...just like the analogy of a star colliding into the planet and either a. bouncing off with kinetic energy, or b. the gravitational pull helping to store that kinetic energy as potential energy by overcoming that kinetic energy with gravitational force and holding that planet tight to it (extreme case of dampening)...
    the nucleons ultimately have some of their mass "tied up" in binding energy per nucleon in the middle area of hte curve.

    Where I've not made a connection yet:

    Why would Fusion or Fission Yield Net Energy. If the binding energies per nucleon are ultimately higher in the middle of the curve, then wouldn't they have more of their mass tied up into the binding energies of the nucleus, and therefore the mass defect that would exist could not be released as energy, but would be tied up into binding energy?
  2. jcsd
  3. Oct 30, 2009 #2


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    The mass isn't tied up as you describe it. It is released as radiation.
  4. Nov 1, 2009 #3
    I would imagine it releases kinetic energy of a neutron plus some other radiation based on E=hf. Still strange concept to me though, but that would explain why things like Iron and Nickel are on the top of the curve, and are the most stable elements are exist in the core of the earth, and really are in fact stable. Are there any good books that would provide a good coverage of this topic of Nuclear Science?

  5. Nov 8, 2009 #4
    Remember mass-energy equivalence. As Mathman said, that mass is carried away as energy. Think in the extreme case as with Annihilation: you take an electron and a positron and kaboom... you get all of that mass released as energy, half of which is in the form of neutrinos, and half of which is (mostly) in the form of photons. That is the whole point of E=+-MC^2
  6. Nov 8, 2009 #5


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    From Wikipedia - electron positron Annihilation: Note neutrino production is rare at low energies. It is mostly 2 photons.

    Low energy case
    There are only a very limited set of possibilities for the final state. The most possible is the creation of two or more gamma ray photons. Conservation of energy and linear momentum forbid the creation of only one photon. In the most common case, two photons are created, each with energy equal to the rest energy of the electron or positron (511 keV).[1] A convenient frame of reference is that in which the system has no net linear momentum before the annihilation; thus, after collision, the gamma rays are emitted in opposite directions. It is also common for three to be created, since in some angular momentum states, this is necessary to conserve C parity.[2] It is also possible to create any larger number of photons, but the probability becomes lower with each additional photon because these more complex processes have lower quantum mechanical amplitudes.

    Since neutrinos also have a smaller mass than electrons, it is also possible — but exceedingly unlikely — for the annihilation to produce one or more neutrino/antineutrino pairs. The same would be true for any other particles, which are as light, as long as they share at least one fundamental interaction with electrons and no conservation laws forbid it. However, no other such particles are known.

    [edit] High energy case
    If the electron and/or positron have appreciable kinetic energies, other heavier particles can also be produced (e.g. D mesons), since there is enough kinetic energy in the relative velocities to provide the rest energies of those particles. It is still possible to produce photons and other light particles, but they will emerge with higher energies.

    At energies near and beyond the mass of the carriers of the weak force, the W and Z bosons, the strength of the weak force becomes comparable with electromagnetism.[2] This means that it becomes much easier to produce particles such as neutrinos that interact only weakly.

    The heaviest particle pairs yet produced by electron-positron annihilation in particle accelerators are W+/W− pairs. The heaviest single particle is the Z boson. The driving motivation for constructing the International Linear Collider is to produce Higgs bosons in this way.
  7. Nov 9, 2009 #6


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    Here is an explanation at a fairly elementary level: http://www.lightandmatter.com/html_books/4em/ch02/ch02.html#Section2.5 [Broken] See sections 2.5-2.7.

    This is incorrect. There is no simple closed-form expression for the distance-dependence of the nuclear force. At large distances, it's exponential, not a power law.

    I think the basic answer to your question is that you have the sign flipped. A more bound system has a *lower* binding energy, not a higher one.
    Last edited by a moderator: May 4, 2017
  8. Nov 9, 2009 #7
    I based my statement on this article from NASA http://gltrs.grc.nasa.gov/reports/1996/TM-107030.pdf" [Broken], which I must admit does deal with annilihation in the context of propulsion, and I should not have extrapolated that 50% loss of propulsion = 50% of the product of annihilation being neutrinos. I stand corrected. Thanks very much for the cogent explanation!
    Last edited by a moderator: May 4, 2017
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