antimatterfusionmass-energynuclear-physics
Is fission/fusion of any element to iron-56 (or nickel-62?) the best way to convert mass to energy, that doesn't involve black holes?
In other words, will we be always limited to convert only about 1% of the mass available to energy? Are there other ways (using strangelets? antimatter?) to go beyond that limit?
I exclude black holes as, as I understand, you can only extract a finite amount of energy by reducing their spin, so they are not viable for energy production on a cosmological scale.
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:
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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.
If you look at our observable universe, there is no bulk antimatter. Physicists discovered antimatter experimentally, and the present standard model of physics posits the existence of antimatter.
BUT to generate antimatter , one has to spend at least the energy of generating twice the mass, particle and antiparticle, and this can only happen in an accelerator which will have to be fed with a lot more extra energy so that the beams, colliding or hitting targets, can generate the particle antiparticle pair. They come in pairs because of conservations laws, lepton number for electrons and positrons, baryon number for protons and antiprotons.
Fusion utilizes the existing nuclear state taking advantage of the binding energy curve, rearanging existing particles not creating new ones.
With this in mind, :
can antimatter ever actually be superior to fusion in terms of being used to generate electricity?
No.
is trying to develop the most efficient form of fusion (possibly combining a fission cycle) the best way to go about it
In terms of energy in energy out yes, it is much more efficient.
if we can bypass engineering issues, in principle, antimatter electricity generation could one day be superior (more useful energy out for what we put in
Not for cost, creating antiparticles is very costly.
For usefulness in space travel, if antimatter could be stored, which I find doubtful, it might be efficient. At the moment though magnetic fields are needed to store antimatter plasma, and those are also energy eaters. All in all fusion wins in the energy balance sheet.
Best Answer
Matter-antimatter annihilation, such as an electron annihilating with a positron to form two high-energy photons, can convert 100% of the mass into radiation. So fission and fusion are far from the most efficient ways to convert mass into other forms of energy. Unfortunately, the universe appears to contain almost no antimatter.