Why does iron consume more energy in the fusion process than it produces?

I understand that once a star starts fusing iron, it's doomed to collapse because iron fusion requires more energy than it releases in the process, allowing the opposing gravity of the star to cause it to collapse.

But why? What makes iron special in this way? It seems to occupy a pretty inconsequential spot in the middle of the period table among the transition metals. So, why is it that iron breaks the rule in play for all the elements before it when it comes to fusion?

I noticed that on another question someone said that it wasn't iron, but nickel that was the first element that required more energy to fuse than it released- but every documentary and book I've read claims that it's iron. So, if you're answer is "iron isn't the first element that requires more energy..." please explain why every other source I've ever heard is wrong!

It would be good if you referenced your sources, because you may be misunderstanding them. We'd be able to see what they actually say, and help you understand them.

Nucleosynthesis of iron does not use more energy than it produces.

It is, however often referred to as the heaviest element created in fusion that results in more energy produced than consumed.

However, that isn't quite true. Heavier elements can be produced by fusion that produce more energy than is used, except these fusion reactions don't occur in stars. (Eg 40Ca + 40Ca)

Also, it is possible for heavier nuclei to be fused in stars that result in more energy being produced than is used, but these are unstable isotopes and they decay quickly.

So, more accurately, iron is the heaviest element produced in stellar nucleosynthesis in any significant quantity that produces more energy in fusion than the fusion consumes.

This is called the alpha process ladder. Keep adding alpha particles to the newly generated nuclei, until you stop getting more energy out than you put in.

The last step in the alpha process that does produce energy is 52Fe + 4He => 56Ni(excuse the rubbish notation; if this answer is considered helpful at all I will try to tidy up the notation)

56Ni + 4He => 60Zn uses more energy than it produces.

56Ni has a very short half-life of just 6 days, decaying to 56Co which has a half life of 77 days, which decays to 56Fe, which is stable. So when a star is on the limit of collapsing, it will be producing a lot of iron in it's final stages - some of it from decaying heavier radioactive isotopes.

Why does the alpha process stop producing energy at this point? It is because that is where the peak binding energy is. More.

• No, it isn't the peak binding energy, it is the peak binding energy per nucleon. – ProfRob Jun 28 '20 at 20:02
• I defer to your superior understanding on this topic @RobJeffries and accept that my statement is too vague - can you make a suggestion as to how I should rephrase that last sentence to be more accurate? I'll edit it. – Jeremy Jun 30 '20 at 4:02
• The graph in the link you provide to "binding energy" seems clear enough. If you add an alpha particle to a 56Ni nucleus, you get 60Zn plus some energy. Because 60Zn has about 8 MeV more binding energy than 56Ni. And heavier elements have even more total binding energy. So your sentence that alpha capture onto Ni uses more energy than it produces is also incorrect in isolation. astronomy.stackexchange.com/questions/36719/… – ProfRob Jun 30 '20 at 6:37
• @RobJeffries You're the Professor of Astrophysics, so I accept your correction. I feel that I could fiddle with this answer a lot and still not get it as correct as you are able to. Perhaps you could re-write the offending sentences correctly, as a comment, and I will edit the reply to match your recommendation. Alternatively, you could add a better answer written from scratch to supercede this attempt of mine – Jeremy Jul 1 '20 at 8:28

As can be seen in this question: What effects besides "mass defect" cause the alpha ladder beyond iron-56/nickel-56 to be endothermic? It is not so straightforward to explain why fusion stops at iron (actually it stops at 56Ni and then radioactive decay produces 56Fe afterwards - i.e. 56Ni is the final fusion product, not 56Fe).

Adding alpha particles to 56Ni is still exothermic (in isolation)

The elements around 56Ni, 56Fe etc. occupy a special position, in that they are at the peak of curve of binding energy per nucleon.

What that means is if you have a bunch of nucleons and it is possible to rearrange them to form into nuclei of various kinds, then the natural tendency is to minimise the total energy density by maximising the binding energy per nucleon and forming nuclei that are at the peak of the BE per nucleon curve (i.e. the "iron-peak" elements).

Thus, as highlighted in my answer to the question referred to above, whilst it is energetically favourable (i.e. it releases energy) in a a core made of 52Fe, to break an alpha particle off one nucleus and fuse it with another 52Fe nuclei to make 56Ni, the same is not true if the core is made up of 56Ni. i.e. It consumes energy to break an alpha particle out of a 56Ni nucleus and fuse it with another 56Ni nuclei to make 60Zn.

i.e. There are two steps: (1) Remove an alpha perticle from an existing nucleus; (2) add it to another nucleus to create a heavier nucleus. The two steps taken together are endothermic if the original nucleus is as heavy or heavier than 56Ni. (NB Note that 62Ni is technically the nucleus with maximum BE per nucleon [just], but there is no easy route to get to it by fusion).

Now in the core of a massive star, there is plenty of energy available, so fusion beyond 56Ni could occur (endothermically). However the Coulomb barrier also increases as the nuclei get more protons, and the temperatures required to initiate fusion increase. At the temperatures required to fuse 56Ni to 60Zn then photodisintegration by energetic photons becomes a very important process and so 60Zn (although some is present) tends to be photodisintegrated as quickly as it is produced. Since photodisintegration is very endothermic (see step (1) above), then this usually spells the end of the road for the star and can trigger core collapse.

As for why iron-peak elements are at the peak of the BE per nucleon curve, you need to look at the basic nuclear physics. The strong nuclear force is highly attractive but only operates between nearest neighbours, but nuclei at the "surface" are bound less tightly. There are fewer nucleons at the surface (relative to the total) if you have bigger nuclei. The protons in a nucleus however repel all the other protons in a nucleus; the effect diminishes as the nucleus gets larger, but grows strongly (as $$Z^2$$) with the number of protons. So you have two competing effects; one favouring larger nuclei, the other disfavouring large nuclei with lots of protons. The binding energy per nucleon is maximised somewhere in the middle.