Why is iron-56 the most stable nuclei?

This probably seems naiive, but why? My main question being that why isn't hydrogen the most stable. It is the simplest and should be, right? If everything wants to arrive at the lowest possible energy state, why do nuclei want to arrive at iron? I can understand splitting atoms (fission) perfectly well, it makes sense. Fusion of atoms also makes some sense, except for hydrogen, again. Can someone tell me why this is true? I am sure i am missing something crucial here...

This probably seems naiive, but why? My main question being that why isn't hydrogen the most stable. It is the simplest and should be, right? If everything wants to arrive at the lowest possible energy state, why do nuclei want to arrive at iron? I can understand splitting atoms (fission) perfectly well, it makes sense. Fusion of atoms also makes some sense, except for hydrogen, again. Can someone tell me why this is true? I am sure i am missing something crucial here...

The single-proton hydrogen nucleus is certainly the most stable with respect to fission, and it can't emit an alpha or beta particle either so it is perfectly stable with respect to those. But it is unstable with respect to fusion. Fusion with another hydrogen to make helium gives off energy, so fusion is energetically favored. Under intense pressure this is what happens.

For an iron nucleus to undergo fission or fusion it has to absorb energy, so the odds are against it doing either of these things.

This probably seems naiive, but why? My main question being that why isn't hydrogen the most stable. It is the simplest and should be, right? If everything wants to arrive at the lowest possible energy state, why do nuclei want to arrive at iron? I can understand splitting atoms (fission) perfectly well, it makes sense. Fusion of atoms also makes some sense, except for hydrogen, again. Can someone tell me why this is true? I am sure i am missing something crucial here...

You are mixing up two different concepts here. Fe-56 is just as stable as every other non-radioactive isotope, including Hydrogen. What I assume you are asking is why Fe-56 has the lowest binding energy per nucleon. Actually it is Ni-62 which is the most tightly bound nucleus, but Fe-56 is close.

The reason why the binding energy per nucleon curve has a peak is because there are two competing forces at work - the protons trying to push each other apart, and the strong nuclear force trying to hold things together. These two forces have a crossover point because the strong nuclear force has a very short range but the electric force has a very long range. As the nucleus gets bigger, the electric force starts to win out.

The reason why the binding energy per nucleon curve has a peak is because there are two competing forces at work - the protons trying to push each other apart, and the strong nuclear force trying to hold things together. These two forces have a crossover point because the strong nuclear force has a very short range but the electric force has a very long range. As the nucleus gets bigger, the electric force starts to win out.

This argument doesn't explain why there is a most-stable N for a given Z. It implies that stability would always increase when you increased N. See the two links I gave in #2. The following may also be helpful.

FAQ: Why does the line of stability have the average over-all shape it does?

For light nuclei, the line of stability hugs the N=Z line, and this is because of the Pauli exclusion principle. If you have N=8 and Z=8 (16O), you can put the 8 neutrons in the 8 lowest energy states, and the 8 protons in the 8 lowest energy states. With N=10 and Z=6 (16C), the exclusion principle forces you to put those last few neutrons in high-energy states that weren't occupied in 16O.

For heavy nuclei, the mutual electrical repulsion of the protons breaks the symmetry in the way the strong nuclear force treats neutrons and protons. This effect favors higher N/Z ratios, so the line of stability bends away from N=Z.

The line of stability also has little wiggles superimposed on top of its broad over-all curve. These are caused by quantum mechanical shell effects, the nuclear analogs of the ones in atomic physics that make the noble gases so chemically stable. These shell effects have nothing to do with the over-all shape of the line of stability. For example, the nucleus 100Sn (N=50, Z=50) has two closed shells, but it is very far from the line of stability.

This argument doesn't explain why there is a most-stable N for a given Z. It implies that stability would always increase when you increased N. See the two links I gave in #2. The following may also be helpful.

FAQ: Why does the line of stability have the average over-all shape it does?

For light nuclei, the line of stability hugs the N=Z line, and this is because of the Pauli exclusion principle. If you have N=8 and Z=8 (16O), you can put the 8 neutrons in the 8 lowest energy states, and the 8 protons in the 8 lowest energy states. With N=10 and Z=6 (16C), the exclusion principle forces you to put those last few neutrons in high-energy states that weren't occupied in 16O.

For heavy nuclei, the mutual electrical repulsion of the protons breaks the symmetry in the way the strong nuclear force treats neutrons and protons. This effect favors higher N/Z ratios, so the line of stability bends away from N=Z.

The line of stability also has little wiggles superimposed on top of its broad over-all curve. These are caused by quantum mechanical shell effects, the nuclear analogs of the ones in atomic physics that make the noble gases so chemically stable. These shell effects have nothing to do with the over-all shape of the line of stability. For example, the nucleus 100Sn (N=50, Z=50) has two closed shells, but it is very far from the line of stability.

Is my explanation is fundamentally incorrect? Or just an over-simplification which is incomplete?

As Ben pointed out, your explanation is Ok as far as it goes, but you also need to mention what is called the asymmetry energy. With the nucleon-nucleon attraction term and the proton-proton repulsion term alone, it would be energetically favorable to replace all protons by neutrons. A term is needed to keep the number of neutrons and protons approximately equal.

The asymmetry energy takes into account the Fermi nature of the particles. Adding neutrons becomes more and more costly as it drives up their Fermi level. This effect will be minimized if N ≈ Z, equalizing the Fermi levels for neutrons and protons. The asymmetry energy term is usually written (A - 2Z)2/A.