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Why is fusion more efficient than fission?

  1. Jan 29, 2012 #1
    If you need to put more energy into making a bigger nucleus, shouldn't bigger nuclei have more energy that can be released? How does putting energy into making hydrogen into helium even release that much more energy?
     
  2. jcsd
  3. Jan 29, 2012 #2

    Astronuc

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    Staff: Mentor

  4. Jan 30, 2012 #3
    I still don't completely understand. You have a binding energy, and that's the energy required to hold a nucleus together, and you put tons of energy into making a bigger nucleus...but after a certain point the bonds between nuclei are weaker which means not as much energy was put into making them? Even if you split them though, isn't it a total of more energy within the bonds of a bigger nucleus still bigger than whatever bonding energy is in hydrogen?
    Or perhaps because with fission, you can only break atoms down into specific pieces which take up most of the energy within the binding of the nuclei yielded, and with simply putting a lot of energy into fusing just two hydrogen atoms together, the extra neutron doesn't take up as much binding energy so there is more energy available to be turned into kinetic and electro-magnetic energy?
     
  5. Jan 30, 2012 #4
    Binding energy is the energy required to pull a nucleus APART, not hold it together.

    Yes, but not in general more than all other combinations of nucleons. There is more total binding energy in U-238 than in 238 separate protons and neutrons, sure, but not more than Th-234 + He4.

    I'm not quite sure what you are talking about, but perhaps it is the activation energy that is confusing you. You have to overcome the Coulomb repulsion between parent nuclei in order to get them close enough to fuse, but the energy you spend doing this does not go into the binding energy in the new nucleus.
     
  6. Jan 30, 2012 #5
    So your telling me that if I put energy into building a bigger nucleus and I don't get that amount energy back, the energy didn't go into holding a bigger nucleus together and instead it just... transforms into...photons? Or disappears?
    Also, how is the energy required to pull it apart not in some way related to the energy that's holding the nucleus together? Isn't that the force you have to overcome?
    Or are you trying to say that because it requires such a big force to blow it apart that the force of that single neutron and some photons emitted from two hydrogen atoms fusing is more energy?
    I guess what I was saying originally is that when you blow up a big nucleus like uranium, instead of a lot of energy being released in the form of photons (even though there are a lot o them), most of the energy is still contained in the left-over nuclei, where-as in hydrogen fusion it doesn't require a lot of energy to hold a helium nuclei together so there's more energy available outside of the system of the nuclei for w/e process.
     
  7. Jan 30, 2012 #6
    If the new nucleus has a lower binding energy than the combined binding energy of those you started with, this means you had to put in some energy to move it into this less stable configuration, yes. I was just stating that this energy is different to the activation energy. So in this situation, you put in energy to overcome the activation energy, some of which goes into lowering the binding energy of the nucleon configuration, and the rest of which remains as kinetic energy. The net result is that you have less total kinetic energy than before, and more stored in the nuclear field configuration.

    Force and energy are different. It takes force to hold you on the earth, but no energy. You spend or gain energy by moving up or down in the gravitational potential. Likewise no energy is expended holding a nucleus together, but you spend or gain energy by reconfiguring the nuclear potential. So binding energy is the energy required to go from a bound nucleus to an unbound set of particles, or the energy you gain by taking all those particles and putting them together into a nucleus. Is that more clear? All nuclei have more binding energy than their separate constituents, but iron has the highest binding energy per nucleon. A bigger nucleus than iron has more total binding energy than iron, and more than iron + the individual extra nucleons you need, but not more than iron + (a bound state of the extra nucleons).
     
  8. Jan 31, 2012 #7
    Assume the fusion of Deuterium (H2) and Tritium (H3) (isotopes of hydrogen having, respectively, one neutron and two neutrons). They can be fused together into a single Helium-4 atom and a free neutron.

    The binding energy of Deuterium is: 2224,52±0,20 keV (Wikipedia)
    The binding energy of Tritium is: 8481,821±0,004 keV (Wikipedia)
    The binding energy of Helium-4 is: 28300.7 keV (Wikipedia)

    The energy generated by fusion Deuterium and Tritium is therefor:
    28300,7 keV - 2224,52 keV - 8481,821 keV = 17,59 MeV

    http://upload.wikimedia.org/wikipedia/commons/5/53/Binding_energy_curve_-_common_isotopes.svg

    In the above graph, going upward will release energy (since Ferrum-56 binds the easiest). So for lighter atoms, fusion will release energy, while for larger atoms, fission will release energy.


    EDIT:
    Now to answer the question asked in your title: you should check the graph. Going from (any isotope of) Hydrogen to Helium (or even higher) will generally produce more energy then any kind of fission, for example that of Uranium.
     
  9. Jan 31, 2012 #8
    I think I'm getting a better picture with the fission efficiency itself, but about this statement, doesn't it take some kind of at least potential energy to exist in a state that isn't at the lowest possible state?
     
  10. Jan 31, 2012 #9
    Well the lowest possible state is the state with the lowest potential energy yes, and you need to put in energy to move to a state with higher potential energy, but no energy is doing anything when you just sit in the one state, whatever potential energy it has. It takes no energy to sit in a chair on top of a mountain, even though there is potential energy stored in the configuration of yourself and the Earth.
     
  11. Jan 31, 2012 #10
    Yes. A Uranium fission yields something like 200ish MeV, whereas an individual fusion yields 17ish MeV (as noted above). Here's what you're missing that will make it more clear: the important figure is the binding energy released per nucleon in the reaction. The energy you gain per mass unit of fuel consumed is much greater for fusion than fission. That 17 MeV involved 5 amu of fuel, so you get 3.4 MeV/amu (328 GJ/g) . With fission, that 200 MeV is liberated from a fuel nucleus that has a mass of 235 amu, so you get 0.85 MeV/amu (82 GJ/g).
     
  12. Feb 1, 2012 #11
    Good discussion; good explanations kurros....

    I can't see the graph in post #7, is it like the one here in Wikipedia:


    http://en.wikipedia.org/wiki/Nuclear_binding_energy#Nuclear_energy

    Also, there is a nice synopsis about binding energy in different size atoms here:
    http://en.wikipedia.org/wiki/Nuclear_binding_energy#Determining_nuclear_binding_energy

    "..For nuclei larger than about four nucleons in diameter, the additional repelling force of additional protons more than offsets any binding energy which results between further added nucleons as a result of additional strong force interactions; such nuclei become less and less tightly bound as their size increases, though most of them are still stable..."
     
    Last edited: Feb 1, 2012
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