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Where does energy from fussion/fission come from?

  1. Apr 29, 2012 #1
    When a nucleus fissions into two smaller ones, the total binding energy pr. nucleus is increased. Somewhat similarily when two nucleus fusion into one larger nucleus the binding energy is increased. Thus both fusion and fission are exothermic processes. This is the explaination I get in my school book for where the energy comes from in fusion and fission.
    The book seems to forget that all though the processes are exothermic, the energy created is in the shape of binding energy in the nucleus! How can we profit from this energy? How do they do it at nuclear power plants? Is binding energy the same as the kinetic energy of the atoms?
     
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  3. Apr 29, 2012 #2

    jtbell

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    Binding energy is negative. When the magnitude of the binding energy increases, it makes energy available for release as kinetic energy of the products. Making up some numbers blindly, here's an example:

    Initial binding energy = final binding energy + kinetic energy of products
    -990 MeV = -1000 MeV + 10 MeV
     
  4. Apr 29, 2012 #3
    How is the binding energy released as kinetic energy?

    Or are you saying that the transformed mass in fusion and fission, is converted into both binding energy AND kinetic energy?
     
  5. Apr 29, 2012 #4

    jtbell

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    The difference in binding energy (before versus after) is simply the energy associated with the mass difference, via ##E = mc^2##.
     
  6. Apr 29, 2012 #5
    I am aware of that. But how is binding energy used to boil water at power plants? Dont we need kinetic energy for that?
     
  7. Apr 29, 2012 #6

    D H

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    Binding energy is always positive (or zero in the special case of 1H), but that's just a quibble over how "binding energy" is defined. Nuclear binding energy is the amount of energy that must be added to a nucleus to separate the constituent particles to infinite distance. The amount of energy bound up in the nucleus is negative.

    Another way to look at it: The mass of the nucleus is always less than the sum of the masses of the neutrons and protons that form the nucleus.


    Start with fission. A 235U that captures a neutron will usually (about 82% of the time) undergo fission. It splits into, for example, 89Kr, 144Ba, 3 neutrons, and photons (gammas). Most of the energy that results from this fission is in the form of kinetic energy in the product nuclides. Those product nuclides will quickly collide with other nuclides, dispersing this kinetic energy. In other words, the kinetic energy quickly becomes thermal energy. Those product nuclides are highly unstable. They decay into something else, and this too becomes thermal energy. The same goes for the gammas. Most of them are also absorbed, their energy becoming thermal energy. Now it's just a matter of converting that thermal energy into useful energy. This is now just a problem of good old thermodynamics. Typically that thermal energy is used to boil water. The steam drives a steam turbine, and the turbine creates electricity.

    Conceptually, fusion works in a similar fashion. A 4He nucleus has less mass than do (for example) a tritium nucleus and a proton. Conservation of energy (mass is energy) dictates that that excess mass of 3H+1H versus 4He be converted to some other form of energy, kinetic energy + photons + neutrinos. Most of this quickly becomes thermal energy. In practice, it is a lot harder to harness fusion in a sustained, controlled manner than it is to harness fission.
     
  8. Apr 29, 2012 #7

    Astronuc

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    Thermal energy is the kinetic energy of the fission products in the fuel.

    Binding energy is the energy required to separate a nucleon from an atom.

    When a fissile atom absorbs a neutron, the configuration becomes unstable and then nucleus may fission, in which case, the energy is release as kinetic energy of fission products and 2 or 3 neutrons, OR the nucleus may emit a gamma ray.

    Very few isotopes are fissile. U-233, U-235, Pu-239, and Pu-241 are fissile, and only U-235 is found naturally on earth. Other isotopes (U-238, Th-232) are fissionable, i.e., they can fission but require fast neutrons. Still other isotopes can undergo spontaneous fission.
     
  9. Apr 29, 2012 #8

    jtbell

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    Yeah, binding energy is always given as a positive number, but when you use it in energy-conservation calculations, you usually end up having to subtract it. A minus sign has to come in, one way or another. This always confuses students. I tell students that binding energy isn't energy that a nucleus has, it's energy that a nucleus is missing, compared to the energy of its separated components.
     
  10. Apr 29, 2012 #9
    Well that would explain alot. But what about the increase in the product-nuclides binding energi? According to the "binding energy pr. nucleon curve", the binding energy pr. nucleon is greater in the product nuclides of fission and fusion processes. Thus some of the transformed mass must have been made into binding energy, and not only kinetic energy. Do I sound completely confused or what?

    So saying that binding energy, generated by the strong nuclear force, is what keeps the nucleus together is nonsence? Why on earth would physists call it binding energy, if its the energy needed to SPLIT the nucleus? Its like calling a dog a cat! Your absolutely right, this is really confusing for students like myself.

    But then what about the "binding energy pr. nucleon curve"? It seems that in this case the increase in binding energy pr. nucleon is used to explain that fission and fusion are both exothermic. If binding energy is the energy needed to split a nucleus then how can the curve be used to explain this? An increase in binding energy, would then really just be an increase in the "amount of energy needed to split the nucleus", in which case it would be a measure of increase in the strong nuclear force - not in energy.
     
  11. Apr 29, 2012 #10

    D H

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    You sound completely confused. You are thinking of binding energy as something nuclides have. With binding energy defined the way it is, it is perhaps better to think of it as something nuclides don't have.

    It's just a sign convention. There are lots of other places where arbitrary sign conventions can come into play with regard to energy. One example: The first law of thermodynamics. Sometimes you'll see it written as [itex]dU = \delta Q - \delta W[/itex], the change in internal energy of a system is equal to heat transferred to the system less the work done by the system. Other times you see it as [itex]dU = \delta Q + \delta W[/itex], the change in internal energy of a system is equal to heat transferred to the system plus the work done on the system. Another example is gravitational potential energy. Physicists in general tend to represent gravitational potential energy as a negative quantity that drops to zero as distance goes to infinity, but geophysicists like to represent gravitational potential energy as a positive quantity that drops to zero as distance goes to infinity.

    Don't get confused by sign conventions. Does it really matter if you add a negative quantity or subtract a positive one? The end result is the same. It's just a convention. As defined, binding energy is the difference between the total energy of the dissociated protons and neutrons and the energy of the nuclide itself.
     
  12. Apr 29, 2012 #11
    Okay I'll keep that definition in mind then.

    I understood your fusion example ealier, and find it very intuitive that the decrease in mass must mean an increase in energy, thus prooving that the fusion proces was exothermic. My textbook explains it in some other way though..

    In my text book the author argues that fusion is exothermic by looking at the binding enegy before and after a fusion proces. He concludes that because the binding energy of helium is greater than that of deuterium and tritium combined, the reaction is exothermic. Is that correct? If binding energy rather than being energy that the nuclides have is energy that they lack, how can it be an argument for fusion being exothermic? Wikipedia explains it similarily..
     
  13. Apr 29, 2012 #12

    Astronuc

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    Firstly, not all fusion reactions are exothermic.

    Secondly d+t => 2He5* => He4 + n. He-4 is very stable, and it takes a lot of energy to remove a neutron or dissociate into 2 protons and 2 neutrons. In fusion, the difference in masses of the reactants and products gives the energy which is 'released' or manifest in the kinetic energy of the products.

    The binding energies of the d and t are much less than that of He-4. It would require 28.3 MeV to dissociate an alpha particle into its constituents.

    This gives a good explanation.
    http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html#c1
     
  14. Apr 29, 2012 #13

    D H

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    So let's look at 2H+3H→4He+n in terms of binding energy. (Somewhat) arbitrarily define the energy of the five nucleons (2 protons and 3 neutrons) each separated by an infinite distance as zero. The binding energy of deuterium and tritium provides the information needed for calculating the energy of the left hand side configuration while that of helium 4 provides the information needed for calculating the energy of the right hand (the lone neutron tautologically has a binding energy of zero).

    On the left hand side, the binding energy of deuterium is 1.112283 MeV / nucleon; for tritium it is 2.827266 MeV / nucleon. The energy of the deuterium+tritium is -(2*1.112283 MeV + 3*2.827266 MeV) = -10.706364 MeV. On the right hand side, helium 4 has a binding energy of 7.073915 MeV / nucleon, so the energy here is -(4*7.073915 MeV) = -28.29566 MeV. The right hand side has less energy than does the left hand side. To conserve energy, the reaction must release 17.5893 MeV. The reaction is exothermic.
     
  15. Apr 29, 2012 #14
    but I thought we defined the binding energy as the energy that the nucleus did NOT have?

    On the link posted by Astronuc it says in the very top that "the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together.

    How come there seems to be different definitions of what the binding energy is?! Some say its the energy that holds the nucleons together. Others say that its the energy that the nucleus does not have, in comparison to its seperate nucleons. Surely there must be one firm definition, or are they basically saying the same thing?
     
  16. Apr 29, 2012 #15

    Dale

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    But the point is that the product nucleus does NOT have even more energy than the reactant nuclei do NOT have :smile:.

    Think of it this way. Two single-parent nuclear families each have loans of $1000. Those loans represent money that they do not have. The parents get married resulting in a much larger family. They go to the bank and wind up with a new consolidated loan of $2500. This also represents money that the new family does not have. However, since the new loan is larger than the sum of the previous two loans they now have $500 of available cash. The cash is enough to boost their KE to move to a new neighborhood.
     
  17. Apr 29, 2012 #16
    We are trying to say the same thing! Binding energy is the energy (or mass, E=MC^2) a nucleon will give up to become part of an atom.

    If we take Carbon 12 {which weighs 12.0000} and add a neutron {which weighs 1.00866} we get Carbon 13 {which weighs 13.00335} So if Energy = Mass at 931MeV/amu

    Adding the Carbon and the neutron:
    12+1.00866=13.00866
    But Carbon 13 only weighs 13.00335
    13.00866 - 13.00335 = .00531
    So the neutron&carbon-12 seems to have given up .00531 amu's or [.00531*931=4.8] ~4.8Mev to form Carbon-13. That energy given up is the Binding Energy. We would have to add that energy to the atom to get our neutron back (so that is how we can look at it as energy the atom doesn't have) When two light particle fuse together that is where the energy comes from.

    For fission it is the same energy but with a slight twist. Notice your book talks about the binding energy per nucleon. That is because when you get above ~60 the short range nuclear force {that creates binding energy} from adding another neutron does not have as big of an effect. So when you have really big atoms {i.e. Uranium} and we split those, the binding energy (per nucleon) is allowed to have a bigger effect, thus from the earlier example when Uranium 235 + 1 neutron gets together and "blows apart" into Krypton 89, Barium 144 and 3 neutrons, the Kr, Ba, 3N have less mass than the U, N. That mass was turned into Kinetic Energy of the fission fragments.

    And that is how nuc plants profit from fission. The splitting uranium turns into lighter elements and kinetic energy. The kinetic energy heats up the reactor, which heats up the water/gas, which is tunred into mechanical and eventually electrical energy.

    Don't give up - if you need more help let us know!
     
  18. Apr 29, 2012 #17

    Astronuc

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    Some people aren't too careful with the terminology/definitions or the grammar/semantics.

    The nuclear exchange force between nucleons holds them together against the repulsion of the Coulomb force.

    Effectively when nucleons bind they lose some 'mass', usually in the form of a gamma ray, which means the configuration is more stable at a lower potential energy (the reference being a free nucleon).

    The binding energy is the energy deficit, or the energy required for a nucleon to become free. There are some measures of binding energy per nucleon, which is the average binding energy, or total binding energy average over the number of nucleons. There is also the binding energy of the last nucleon, i.e., it's the energy given up by the last nucleon (usually a neutron) to establish the particular nuclide, which is the same as the energy that one would have to add to the nucleus to liberate a nucleon, usually a neutron.

    If one can produce a strong enough gamma ray, then one can remove a neutron from a nucleus, as in photo-emission. In the case of a deuteron, one can cause photo-dissociation of the deuteron into a proton and neutron.

    One can also break up nuclei with the collision of high energy particles in process we call spallation.

    In nuclear reactions, one has to look at the rest masses of the reactants and products, and whatever energy is required to obtain a certain result.
     
  19. Apr 29, 2012 #18
    Bravo! I think I got it lol. This is how I explain it to myself on basis of having read all your posts..

    The binding energy is the energy recuired to split a nucleus into its components. That is the same as saying that the binding energy is the energy that the components (nucleons) radiated upon becomming a part of the nucleus. An increase in binding energy pr. nucleon, thus means that more energy is needed to split the nucleus. Which again following the above two way definition, means that more energy has been radiated by the components upon shaping the new nucleus. This energy comes from the transformation of the mass defect into kinetic energy, via the formula E = Q = - mc^2.

    I hope i'm right? Thank you VERY much regardless.
     
    Last edited: Apr 29, 2012
  20. Apr 30, 2012 #19
    Did you mean "... to zero" for the geophysicists?
     
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