What is the nature of nuclear energy? This is closely related to the correct explanation of mass defect.

I did some research of that topic and cannot come to a single comprehensive and consistent description.

Below are related statements I gathered or can think of, describing the problem area. To me, each of them seems to make sense, but some of them are contradictory, so obviously - wrong. Please kindly point out and explain these errors.

  1. The more nucleons in a nucleus, the bigger the nucleus is, so the average distance of a nucleon to each another is higher, hence the long-distance electromagnetic repulsion tends to overcome short-distance strong nuclear force, up to the point of occasional alpha decay in elements of transuranic end of the spectrum.

  2. The less nucleons in a nucleus, the closer in average they are, so the strong force per each is higher and easily overcomes electromagnetic repulsion.

  3. Given (1) and (2), the smaller the nucleus, the stronger it is bound.

  4. The stronger the nucleus is bound, the higher its binding energy.

  5. The higher the binding energy of nucleus, the more energy is stored per nucleon in the system.

  6. The higher the binding energy per nucleon, the more difficult it is to split the atom.

  7. The more difficult it is to split the atom, the more stable the atom is.

  8. The more energetic a binding, the more difficult it is to break the binding, the more stable the atom made out of such bindings.

  9. Natural systems tend to evolve to lower, not higher energy states.

  10. The literature presents the binding energy chart per element, with its peak at iron (~56 nucleons). Both the lighter and heavier elements tend to have smaller binding energy.

  11. Iron is the most stable element. It is abundant in the universe, as natural atomic evolution tends to get close to it from both ends of atomic number spectrum.

  12. The surplus energy in the nuclear reaction is achieved when heavy elements are split (fission), or light element are fused (fusion).

  13. The surplus energy is the energy taken out of the system, i.e. average energy per nucleus is higher before the reaction and lower after the reaction.

  14. After the reaction - the resulting elements are closer to the iron atomic number

  15. Mass defect is directly proportional to the binding energy. The stronger the binding energy per nucleon, the less mass per nucleon.

Example for neutron, proton and them bound together in deuterium:
$$ \begin{array}{c} \begin{alignat}{7} m_\text{n}&=1.008665 \, \mathrm{u} & \hspace{50px} & m_\text{p}=1.007276 \, \mathrm{u} \\ m_{\text{n}+\text{p}}&=2.015941 \, \mathrm{u} & & m_\text{d}=2.013553 \, \mathrm{u} \end{alignat} \\[5px] \Delta_m =0.002388 \, \mathrm{u} = 2.224\ \frac{\mathrm{MeV}}{c^2} \end{array} $$

The explanation of below (seemingly) contradictions somehow eludes me. Hope it is apparent to you:

  • why the lightest elements are not the ones which are most strongly bound? Possibly the peak at iron might be caused the geometry factor - i.e. when accounting for nucleons 3D positions, the average forces between them are no longer proportional to just the number of nucleons.

  • why does iron seem to need in its properties: the highest energy per nucleon as the most strongly bound, most stable element; and on the other hand: the lowest energy per nucleon, for the surplus energy be given off when approaching iron in fission/fusion reactions from either side of atomic number spectrum?

Also - what is the definition and explanation of the mass defect:

  • mass defect is the surplus energy given off from fusion/fission and hence is the difference between total mass-energy of the system before and after reaction

  • mass defect comes from different proportion of mass vs energy in an atom depending on its binding energy. When the binding energy is higher in an atom, the more of the total mass-energy of the system is stored in the binding of the nucleons and less in their mass - and the total stays the same. If so then why does the total mass-energy change after the split/fuse reaction?

  • $\begingroup$ Regarding J/K: Ni-62 is more tightly bound per nucleon than Fe-56. This is a common misconception, and it goes to show you shouldn't use charts that don't distinguish between different isotopes when it comes to nuclear processes. The prevalence of iron comes in part from the decay Ni-56 $\to$ Co-56 $\to$ Fe-56. Ni-56 is a preferred byproduct of supernovae in part due to having equal numbers of protons and neutrons (nuclear reactions have kinetics to consider as well as thermodynamics). $\endgroup$
    – user10851
    Nov 2, 2012 at 0:41
  • $\begingroup$ @Chris: thanks for the update! I did not know that.. $\endgroup$ Nov 3, 2012 at 10:02

2 Answers 2


To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from.

The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for protons or neutrons or for nuclei. For example, a proton has a mass of approximately $938 \frac{\mathrm{MeV}}{c^2}$, of which the rest mass of its three valence quarks only contributes about $11\frac{\mathrm{MeV}}{c^2}$; much of the remainder can be attributed to the gluons' quantum chromodynamics binding energy. (The gluons themselves have zero rest mass.) So most of the "energy" from the rest mass energy of the universe is actually binding energy of the quarks inside nucleons.

When nucleons bind together to create nuclei it is the "leakage" of this quark/gluon binding energy between the nucleons that determines the overall binding energy of the nucleus. As you state, the electrical repulsion between the protons will tend to decrease this binding energy.

So, I don't think that it is possible to come up with a simple geometrical model to explain the binding energy of nuclei the way you are attempting with your $\left(1\right)$ through $\left(15\right)$ rules. For example, your rules do not account for the varying ratios of neutrons to protons in atomic nuclei. It is possible to have the same total number of nucleons as $\sideset{^{56}}{}{\text{Fe}}$ and the binding energies will be quite different the further you move away from $\sideset{^{56}}{}{\text{Fe}}$ and the more unstable the isotope will be.

To really understand the binding energy of nuclei it would be necessary to fully solve the many body quantum mechanical nucleus problem. This cannot be done exactly but it can be approached through many approximate and numerical calculations. In the 1930's, Bohr did come up with the Liquid Drop model that can give approximations to the binding energy of nuclei, but it does fail to account for the binding energies at the magic numbers where quantum mechanical filled shells make a significant difference. However, the simple model you are talking about will be incapable of making meaningful predictions.

EDIT: The original poster clarified that the sign of the binding energy seems to be confusing. Hopefully this picture will help:

This graph shows how the potential energy of the neutron and proton that makes up a deuterium nucleus varies as the distance between the neutron and proton changes. The zero value on the vertical axis represents the potential energy when the neutron and proton are far from each other. So when the neutron and proton are bound in a deuteron, the average potential energy will be negative which is why the binding energy per nucleon is a negative number - that is we can get fusion energy by taking the separate neutron and proton and combining them into a deuteron. Note that the binding energy per nucleon of deuterium is $-1.1 \, \mathrm{MeV}$ and how that fits comfortably in the dip of this potential energy curve.

The statement that $\sideset{^{56}}{}{\text{Fe}}$ has the highest binding energy per nucleon means that lighter nuclei fusing towards $\text{Fe}$ will generate energy and heavier elements fissioning towards $\text{Fe}$ will generate energy because the $\text{Fe}$ ground state has the most negative binding energy per nucleon. Hope that makes it clear(er).

By the way, this image is from a very helpful article which should also be helpful for understanding this issue.

  • 2
    $\begingroup$ While you are right that this is the source of the binding, there is an insightful and correct first approximation to get the binding energy curve from first principles. This was worked out by Neils Bohr in the 1930s, and is called the "liquid drop model". It models the nucleus as a liquid drop with charge, and it explained the binding curve and predicted fission. The only major deviations from liquid drop are the magic numbers, which are the shell-filling effects, but you can ignore this to first order. $\endgroup$
    – Ron Maimon
    Jul 24, 2012 at 2:01
  • $\begingroup$ Hi, thanks for your valuable comments. Yes, these rules are too simple to account for all intricacies behind nuclear energy. They are also far too little to predict e.g. the shape of the binding energy curve per element. I do not even attempt to achieve such. On the contrary - I know these even produce contradictions, but which are wrong when they seem right? Because some must be. $\endgroup$ Jul 25, 2012 at 16:24
  • $\begingroup$ I just want to straighten out at least the high-level crude framework of this area. The main point is why is binding energy lowering element mass if it is supposed to store energy? Shouldn't adding energy to a system increase the total? If Iron has highest binding energy then why is approaching it through nuclear reactions actually releasing energy per nucleon not gathering it? Wouldn't it mean that Iron stores least energy per nucleon? But then why is it most stable and difficult to split? Regards $\endgroup$ Jul 25, 2012 at 16:25
  • $\begingroup$ @Pawel - I edited the answer to hopefully help with your confusion. $\endgroup$
    – FrankH
    Jul 25, 2012 at 19:49
  • 1
    $\begingroup$ @Pawel: The binding energy is the negative energy, the amount of energy less than the energy of an equal number of free protons and neutrons. The sign is what you are confused about. $\endgroup$
    – Ron Maimon
    Jul 26, 2012 at 1:55

Before I begin, though, I wanted to point out that the original poster talks about some of his points being "wrong" because of conflicting info. If you want to have a good discussion about the fundamentals of nature (which is almost purely philosophical, mind you), it is better to abandon absolutes. There is no right or wrong, good or bad. Forget about law and think only theory. Scientists, so far, have mostly agreed that the simplest (i.e., easiest to understand) explanation is best and that nothing is resistant to change or evolution. Though, some scientists in the main stream have been behaving poorly in my opinion. Big money and politics is a poor environment for good science.

Ok, with that out of the way, I'd like to add in some points about $\sideset{^{56}}{}{\text{Fe}}$. One of the comments on the question states:

The prevalence of iron comes in part from the decay Ni-56 → Co-56 → Fe-56. Ni-56 is a preferred byproduct of supernovae in part due to having equal numbers of protons and neutrons (nuclear reactions have kinetics to consider as well as thermodynamics).

I'm unaware of $\sideset{^{56}}{}{\text{Ni}}$ as a major byproduct of supernovae, BUT I am familiar with stellar evolution theory which states that when a star begins to burn (i.e., fusion) $\sideset{^{56}}{}{\text{Fe}}$ in its core, the sudden reduction in energy output related to the lower per nucleon binding energy of the product from fusing $\sideset{^{56}}{}{\text{Fe}}$ results in structural collapse of the star and hence the supernovae. Here's a source for more info: "Q: Why does iron kill stars?".

Now, why does $\sideset{^{56}}{}{\text{Fe}}$ have the "strongest" binding energy per nucleon? Well, I asked that question to a very knowledgeable professor in my undergraduate days and the answer was simple and elegant. The physical diameter of the $\sideset{^{56}}{}{\text{Fe}}$ nucleus is approximately equal to the distance over which the strong force can act before EM begins to take over. This means that every nucleon (proton, neutron) in the $\sideset{^{56}}{}{\text{Fe}}$ "feels" maximum pull via strong force due to no one nucleon being "out of range" of another nucleon's pull. This idea of a "stuffed" spherical arrangement of nucleons in combination with the idea that the strong force is dependent on neutron/proton proportions (which $\sideset{^{56}}{}{\text{Fe}}$ has the "super magic number" or whatever it is called) creates a situation where every nucleon in a $\sideset{^{56}}{}{\text{Fe}}$ nucleus reaches "maximum" binding energy (i.e., pull from all other nucleons) according to theory.


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