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Binding energy, Nuclear fusion

  1. Jun 22, 2008 #1
    I've got a question on nuclear physics, specifically binding energy.

    I understand that binding energy is the energy needed to separate a nucleus into its constituents. But something has been bugging me.


    For example,

    Two deuterium nuclei fuse together to form a Helium-3 nucleus, with the release of a neutron.
    The binding energies per nucleon for deutrium is 1.09 MeV and for Helium-3, 2.54 MeV.

    If I'm not mistaken, binding energy is also defined as the energy that would be released should a nucleus be formed from its separate nucleons.

    If my calculations are correct, the Helium-3 nucleus has a higher binding energy than the 2 deuterium nuclei. Why is this so?

    Also, the energy released would be 3.26 MeV. That would mean that the Helium-3 nucleus has a higher mass than the 2 deuterium nuclei put together. But why? Shouldn't the mass of the 2 deuterium nuclei be more than the Helium-3 nucleus? Therefore, the mass defect would result in the energy released.
     
  2. jcsd
  3. Jun 22, 2008 #2
    My textbook (notes rather) indicates that the energy released in a nuclear fission and fusion can be found by subtracting the binding energies of the reactants from the binding energies of the products.

    How did they get this? I'm totally clueless.
     
  4. Jun 22, 2008 #3

    dynamicsolo

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    Unfortunately, there are two conventions for discussing binding energy. One looks simply at the absolute value of it, because it is understood what the definition actually is.

    The definition for binding energy is that it is negative. The total energy of a system is usually defined as a sum of kinetic plus potential energy. In a system whose constituents are acted on by an attractive force, the potential energy is negative. When the total negative potential energy exceeds the total positive kinetic energy of the constituents, the total energy is negative and the system is said to be 'bound' (the members don't have enough energy collectively to all escape...)

    In a nuclear system, the total energy would be the mass-energy of the constituents (ultimately the quarks) plus the mass-energy of the 'chromodynamic' force attracting them. When the negative mass-energy of this attractive force exceeds the positive mass-energy of the quarks, the nucleus is bound and (at least somewhat) stable. But the total energy is negative; this is the "binding energy". If a rearrangement of the nucleons (or at least the quarks, if a nucleon is transformed) releases some particle with a positive kinetic energy, the rest of the system now has a more negative energy (a "greater binding energy") and we say it has become "more tightly bound".

    You should find on checking the masses of protons and neutrons, and comparing the appropriate sums to the masses of deuterons and helium-3 nuclei (try the NIST website), you will find that the nuclei have lower masses than the sums for their individual constituent nucleons. The negative difference ("mass deficit") is the binding energy. You should find that the negative difference for the He-3 is greater than the negative difference for the two deuterons. The difference between those differences would be the positive kinetic energy carried off by the neutron in the fusion reaction.
     
  5. Jun 23, 2008 #4
    With reference to the first part, the mass-energy of the constituents would be the positive kinetic energy and the mass-energy of the attractive force would be the negative potential energy. So, when the mass-energy of the attractive force exceeds the mass-energy of the constituents, the difference would be the binding energy; there is a net negative attractive force keeping them together. Have I got it right?

    But these are the parts that I'm uncertain about:

    1. The sum of the masses of the individual constituent nucleons are larger than the mass of the nucleus.
    2. The mass deficit is the binding energy.
    3. The difference between the mass deficits would be the kinetic energy carried off by the neutron.
     
    Last edited: Jun 23, 2008
  6. Jun 23, 2008 #5
    I believe I have got #3 of my last post.

    If the binding energy of the "products" is more than the binding energy of the "reactants" it basically means that the BE (binding energy), of the "products" is more negative than the BE of the "reactants". Therefore, the difference is actually the positive kinetic energy that has been carried off by the neutrons.

    That about right?
     
  7. Jun 23, 2008 #6

    dynamicsolo

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    (I had to read that twice to be sure I understood you: the minus sign is easy to get confused over...) Yes, that's essentially how it goes: we have conservation of "mass-energy" which is now counted among particles and fields, so it is really just an extension of the classical energy conservation principles.

    If the "reactants" had "masses" of, say, 1800 MeV each, and fused to form a single nucleus of 2650 MeV and released a neutron with a mass of 940 MeV (rounded off just to keep the arithmetic simple), we'd find that it left with a kinetic energy of

    1800 + 1800 - 2650 - 940 MeV = 10 MeV .

    Again, to keep the math simple, let's call the proton mass 940 MeV also. Then the individual reactant nuclei have binding energies of

    1800 - 940 (p) - 940 (n) = -80 MeV each,

    for a total binding energy of -160 MeV (some people drop the signs in discussion, since binding energies are negative by definition). The product nucleus has a binding energy of

    2650 - 2·940 (2p) - 940 (n) = -170 MeV .

    So we have

    2 · (-80 MeV) BE = -170 MeV BE + 10 MeV KE,

    meaning that the free neutron makes off with 10 MeV of kinetic energy which has been extracted from the chromodynamic field binding the nuclear quarks together.

    In real nuclear reactions, these calculations have to be done much more precisely, since the changes are often rather small. But this, I hope, will convey the idea...
     
    Last edited: Jun 23, 2008
  8. Jun 23, 2008 #7
    Thanks, I understood that (most of the parts anyway), except this.

    Why is the sum of the masses of the individual constituent nucleons more than the mass of the nucleus itself? Following that, why is the "mass deficit" = binding energy of the nucleus? How is it related to what you posted - binding energy is when the negative mass-energy of the attractive force exceeds the positive mass energy of the constituents?
     
    Last edited: Jun 23, 2008
  9. Jun 23, 2008 #8
    I was thinking about it and thought if the separate mass of protons and neutrons included the "chromodynamic" force and is left out in the mass of the nucleus. Thereby obtaining the binding energy by finding the difference in mass. Which also explains why separate protons and neutrons have a higher mass collectively, than the nucleus as a whole.

    I'm not sure if I've got it but that was the most logical explanation I could come up with.
     
    Last edited: Jun 23, 2008
  10. Jun 23, 2008 #9

    dynamicsolo

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    The attractive force between quarks has what would classically be described as a negative potential energy. In modern terms, such a negative potential energy corresponds to a negative "mass-energy". The effect is that the observed mass of the nucleus is less than the mass of the "particle" constituents, since the observed mass would be the sum of the positive mass-energies of the nucleons plus the negative mass-energy of the field drawing them together.

    An example where it is reasonably clear to see this is the carbon-12 nucleus, which is to define the atomic mass unit. The mass of a nucleus of this isotope is defined (since 1961) to be exactly 12.000 amu, which has an energy-equivalence to

    12 ·(931.494 MeV) = 11,177.9 MeV .

    This nucleus is made up of 6 protons and 6 neutrons, so the mass of its constituent particles is

    (6 · 938.272 MeV) + (6 · 939.565 MeV) = 11,267.0 MeV (energy-equivalent).

    The bound system has a mass measured at an energy-equivalence of 89 MeV lower than the mass of its twelve nucleons. This is the result of the binding field among the nucleons (ultimately, their constituent quarks). [Thus, further, the amu itself is based on a measure taking such binding fields into account.]

    As with atomic and molecular bonds, the rearrangement of which leads to the release or absorption of energy in chemical reactions, it is the rearrangement of the nuclear force bonds (ultimately the chromodynamic bonds among quarks -- apart from transformation of quarks in weak interactions) that can lead to release of energy in nuclear reactions.
     
  11. Jun 23, 2008 #10
    "In a nuclear system, the total energy would be the mass-energy of the constituents (ultimately the quarks) plus the mass-energy of the 'chromodynamic' force attracting them. When the negative mass-energy of this attractive force exceeds the positive mass-energy of the quarks, the nucleus is bound and (at least somewhat) stable. But the total energy is negative; this is the "binding energy"."


    Thanks for your reply.

    Though what you wrote made sense to me, the first post you made doesn't seem to be in sync with it.

    If the negative mass-energy exceeds the positive mass-energy, then the system would be bound. But now, you're saying that the system when bound will have a positive mass-energy.
     
  12. Jun 23, 2008 #11

    dynamicsolo

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    OK, I think I see where I confused the issue. I probably should not have made the analogy that I did. For the nuclear system, the observed mass will be the sum of the positive mass-energy of the "particles" plus the negative mass-energy of the "binding field", giving a total mass-energy lower than the total positive mass-energy of the individual "particles". The system is bound if there is a "mass deficit", which is that negative mass-energy.

    Sorry for not re-reading some of my earlier posts before going on; there is an inconsistency between how I started talking about this and where I went with it... (The point I'd intended to make was that attractive forces has negative potential energy, speaking classically, and this transfers over to a negative mass-equivalence for the binding field.)
     
  13. Jun 23, 2008 #12
    It's clear to me now.

    Thank you so much dynamicsolo, your help is much appreciated!
     
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