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Question-binding energy per atom

  1. Aug 19, 2007 #1
    Thanx to all who helped me with my first posts. Now on to bigger and better things! The Binding Energy Curve shows Helium at about 7. As there are 4 nucleons in Helium, would the atom have a total binding energy of 28? Iron has a binding energy of about 8.75, so would it's total binding energy per atom be 56 x 8.75=490??
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  3. Aug 19, 2007 #2


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    See - http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html

    The binding energy for an alpha particle (nucleus of He atom) is about 7 MeV per nucleon or 28.3 MeV for the entire nucleus. In theory, if one were to impart 28.3 MeV to a He nucleus, it would dissociate into 2 protons and 2 neutrons.

    For Fe, the binding energy is about 8.75 or 8.8 MeV per nucleon, and with a mass of ~56 amu, it would have a binding energy of 490 MeV. But that's just a number. In reality, one would find spallation reactions more probable than complete dissociation of the Fe nucleus into protons and neutrons.

    It doesn't make much practical sense to talk about binding energy per atom, since heavy atoms do not form from direct combination of individual protons and neutrons, as far as we can tell. Do as others suggested in the other posts and look up "stellar nucleosynthesis".

    Or look here - http://csep10.phys.utk.edu/astr162/lect/energy/bindingE.html and subsequent pages.
  4. Aug 19, 2007 #3
    Perhaps I am confused, but the reason I wanted to know about the energy of the total atom is to find out the amount of energy left over for us after fusion of 2 hydrogen isotopes. I am thinking that after the total energy of the "missing mass" is calculated (say using deutrium and deutrium), then 28.3 MeV must be subtracted from this total (this is the energy needed to bind the four nucleons into a helium atom). The energy left over is for us: (????MeV - 28.3MeV=the energy left over for us MeV). This concept seems to easy---I am almost sure I must misunderstand something or another!! lol
    Last edited: Aug 19, 2007
  5. Aug 19, 2007 #4


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    There is perhaps some misconception about the binding energy and what's available from fusion.

    The binding energy is the nuclear energy released in a fusion or fission reaction, and that energy is released in the form of gamma-rays and/or as kinetic energy of the products, which could also include beta/positron particles and neutrinos, depeding on the type of reaction.

    The fusion reaction involving protons - the pp chain - starts with the reaction p + p -> d + e+ + neutrino, then d + p -> He3, then He-3 + He-3 -> He-4 + 2 p.


    the CNO cycle is part of the story

    but that requires some formation of C, N, O or an initial amount.

    Now all that occurs in stars.

    On earth it is not so simple, because we cannot reproduce the particle densities and concommitant pressures that occur in the core of stars.

    So we use d + t -> He-4 + n + 17.6 MeV, of which 14.1 MeV goes to the neutron and 3.5 MeV goes to the He-4 nucleus. So not all of the 28.3 MeV of the binding energy of He-4 is available. Some was already released in the formation of the d and t.

    Other reactions included d + d -> t + p or He3 + n, each reaction having about 50% probability, and releasing a few MeV.

    If we get a higher temperature, we could use d + He3 -> He4 + p, which is aneutronic. The only problem is the expense of He3, which is not readily available on earth. The He-3 reaction can occur as a result of about half of the d + d reactions, and would happen as soon as the He3 is produced in reactor - assuming we ever perfect such as device.
  6. Aug 19, 2007 #5
    This is frustrating because the more I THINK I understand, the more I find out I misunderstood something! I thought that the "missing mass" energy would amount to MORE than 28.3. Then, when the 28.3 was used to bind the nucleons into helium, whatever was left over was for us, or the "yield" I guess would be the proper term But after trying the math, I find that the "missing mass", which is .025604u, times the mass/energy conversion number, which is 931.494MeV/u, produces an answer that is 28.3! Now if binding the nucleons requires 28.3, I don't understand why anything is left over for us. Is the Helium never produced, therefore the binding energy doesn't need to bind anything? I also have seen the number 17.6 MeV a lot. What is this number all about? Keep in mind I have NO experience in nuclear physics. I took a bunch of Math in college, so I can do that part. But obviously there is some basic principle involved with fusion, binding energy, yield, etc. that I do NOT understand. Do all of the pieces (protons and neutrons) disappear, leaving energy behind instead of Helium? If anybody has the time, showing the math, with explanations at each step of where mass and/or energy is going or came from, would perhaps clear this up for me. I have seen some formulas that do this, but they have all been pretty much just numbers, with no explanations of why this number is what it is.
  7. Aug 20, 2007 #6


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    The binding energy, 28.3 MeV in the case of He-4, would be released upon formation of He-4 from the combination of 2p and 2n. The binding energy is related to the mass difference between the products and reactants, and is the energy given off when products are combined in the nuclear reaction.

    However, it is essentially impossible to take 2 p and 2 n and combine them simultaneously. Rather, He-4 is formed in a sequence of intermediate steps, each one releasing some nucler energy.

    See this page - http://hyperphysics.phy-astr.gsu.edu/hbase/astro/procyc.html

    Another way of looking at the binding energy is that it is the energy that would have to be put into a nucleus in order to remove a nucleon (binding energy/nucleon) or dissociate the atoms into protons and neutrons (binding energy/ atom), i.e. unbind the nucleons. Taking the simplest case, when a neutron and proton combine to form a deuteron, some energy is given off in the form of a gamma ray, and that is the binding energy. To separate the proton and neutron, one would have to add energy equivalent to the binding energy.

    In nuclear reactions, one normally uses small particles, e.g. neutrons, protons, deuterons, alpha particles, etc, on larger targets. In the case of d+t fusion, taking the mass of d and t, and subtracting the mass of He-4 and n yields 17.6 MeV, which is related to the mass defect, which is a manifestation of the 'binding energy'.

    See - http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/fusion.html#c2


    The He is produced when the d and t merge and reform very quickly into He + n. The individual protons and neutrons are still there, they are simply reorganized and in doing so release energy.

    Well the problem is that it's mostly numbers.

    We use something we call the 'Q-value', which is based upon the difference in masses of nuclear reactants and products.

    Q = (mass (reactants) - mass (products) ) * c2, which makes use of E = mc2 equivalent of mass and energy. This is a mathematical statement, but does explain the nuclear processes involved inside the nucleus. We do know that if one calculates the masses of reactants and products in a nuclear reaction, the mass of the products is usually less, and that mass difference is the binding energy.

    Take d + t. The rest mass of the deuteron = 2.013553 u, and the rest mass of the t is 3.015501 u for a total mass of 5.029054 u. The rest mass of an alpha particle is 4.001503 u and that of the neutron is 1.008665 u for a total mass of 5.010168 u. The difference in mass of products 5.029054 u and reactants 5.010168 u is 0.018886 u, which is equivalent to 17.6 MeV (based on 931.481 MeV/u).
  8. Aug 20, 2007 #7
    I have to go to work, so just a quick note--I was laboring under the thought that the binding of the nucleons would CONSUME energy, not RELEASE energy. I don't see why it is the other way around, but I'll deal with that later. It seems if you join two pieces of your model airplane together, it CONSUMES "glue" (binding energy), rather than RELEASE "glue". Hmmmmm, counting cards in Blackjack is nothing compared to this stuff! :)
  9. Aug 21, 2007 #8
    Obviously the Binding Energy thing is one of the roots of my problem in getting this. Refering to my example above with the model airplane parts and glue, why do the nucleons RELEASE energy when they are bound together, INSTEAD of USING energy in the process of binding the nucleons together??? The airplane parts don't give glue back to us, they take it from us, in order to bind. Yes, I know glue is a chemical reaction, not a nuclear one :)
  10. Aug 21, 2007 #9


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    Perhaps glue is a bad analogy for nuclear exchange forces.

    It's more like the nucleons fall into a potential well. To get a nucleon out of the well, one has to add energy.

    If one drop's something into a well, it gives up some gravitational potential energy. To retrieve the something from the well, one has to apply work or effort.

    Neutrons enter the nucleus quite easily since they are neutral. Protons or other + charged particles/nuclei require sufficient kinetic energy to overcome the coulombic repulsion.
  11. Aug 22, 2007 #10
    Astronuc, I appreciate your patience with me--most would have given up by now!! This would be so much easier in a face-to-face. But, it is not, so I continue to attempt to understand. Let me try a new track--one simple (??) question at a time.

    The two protrons must be FORCED (by us) together, because of the magnetic repellsion (sp?) between two positively charged entities. But, once they are close enough, a "nuclear force" emits from the both of them, and is strong enough to overpower the magnetic repellsion. The two protons snap together. I can visualize this in my head--it probably LOOKS like two identical magnets--positive end of one facing the negative end of the other--moving closer and closer together until they suddenly snap together. Remember I said LOOKS---I understand the two processes are not the same thing (probably closer to being the opposite things!). Anyways, back to the protrons. The strength of the magnetic repellsion--(the opposite of the attraction in the visual magnets used in example above)--this magnetic repellsion has been overpowered by this "nuclear force". The two protrons are stuck together.

  12. Aug 22, 2007 #11


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    Hey Potaire, de nada. :smile: Now we'll have to consider quark models of nucleons.

    This is a nice start - http://hyperphysics.phy-astr.gsu.edu/hbase/particles/quark.html

    and http://hyperphysics.phy-astr.gsu.edu/hbase/particles/qrkdec.html

    Binding energy is similar an energy difference, but it is a manifestation or consequence of the nuclear forces.

    The electrostatic repulsion between protons is the Coulomb force - like charges repel - not magnetic force.

    This is the other part - but this is about beta decay.

    There is a similar reaction for positron emission, which is what happens in the pp-cycle in stars.
    p + p -> (pn) + e+ + [itex]\nu_e[/itex], where (pn) = d, deuteron.

    The reaction happens almost instantaneously.

    Although the illustration show particles as spherical, that's arbitrary because we don't really know what nucleons and other subatomic particles look like. We can't because they are two small. But we do know that their effects, like scattering behavior, are very localized.
    Last edited: Aug 23, 2007
  13. Aug 22, 2007 #12
    Perhaps my problem lies in understanding "nuclear potential energy". I understand about the stretched rubberband--- elastic(?) potential energy. The bowling ball held up off of the floor is gravitational potential energy. A can of gasoline is chemical potential energy. Now, the energy from the rubberband is released when you let go of one end of the rubberband. The energy of the bowling ball is realeased when it cracks the floor after dropping it. The energy from the gasoline is released in the explosion after you throw a match into the can. But the release of energy from putting things together stumps me. I easily understand the release of energy in fission--the energy that binds the atom together is released when the atom is broken into pieces. Piece of cake!!! But it appears that energy is not only released when breaking atoms apart, but even MORE energy is released if you put atoms together??????? Huh???? WHY???? This is my hangup in a nutshell--I don't understand why energy is released when putting an atom together. I would think that it would consume energy (the "missing mass" energy) in order to hold the new atom together
  14. Aug 22, 2007 #13


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    Well, combustion or detonation involve breaking atomic bonds and reforming them. The energy in an explosion comes from the 'release' of stored (potential) chemical energy, and that energy is released as the kinetic energy of the molecules which is manifested as heat and pressure (momentum).

    Similarly, when a nuclear reaction releases energy, the reacting particles are 'falling' into a lower energy state.

    When a free electron is captured by an ion, a photon is emitted, i.e. energy is released as the electron and ion form a neutral atom. One has to put energy into an atom to ionize it.

    There are also endothermic fusion reactions, which do require adding more energy. Let me dig up some examples.
  15. Aug 22, 2007 #14
    I still do not understand WHY energy is released when the nucleons are fused together. WHY, WHY, WHY????
  16. Aug 22, 2007 #15


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    Because they fall into a lower energy state. It's that simple.

    If a mass lifted up the graviational force field, one has to put energy in. If something falls in a gravitational field, it transforms energy from gravitational energy to kinetic energy.
    Last edited: Aug 23, 2007
  17. Aug 23, 2007 #16
    "Because they fall into a lower energy state. It's that simple."

    Hmmmmmmm, certainly you realize that is not much of an answer. It certainly does NOT explain the WHY-----WHY does the fusing of 2 atoms release huge energy, and where does this energy come from?? Perhaps the "missing mass" energy accounts for the energy required to hold the new atom together on a permanent basis (Binding Energy? 28.3 MeV??). But, WHY is there MORE energy than that, which is released to us, and WHERE did the MORE energy come from???
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