# Why is the nucleus of an Iron atom so stable?

Lighter nuclei liberate energy when undergoing fusion, heavier nuclei when undergoing fission.

What is it about the nucleus of an Iron atom that makes it so stable?

Alternatively: Iron has the greatest nuclear binding energy - but why?

• It's greatest nuclear binding energy per nucleon. Crucial difference. Also it is Nickel that (just) wins that particular competition, not iron. – ProfRob Apr 15 '16 at 22:02

## 3 Answers

It all comes down to a balance between a number of different physical interactions.

The binding energy of a nucleus is commonly described with the semiempirical mass formula:

$$E(A, Z) = a_V A - a_S A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A-2Z)^2}{A} + \delta(A,Z)$$

where $A = Z + N$ is the total number of nucleons, $Z$ the number of protons, and $N$ the number of neutrons.

The different contributions have physical explanation as:

1. $a_V$ : volume term, the bigger the volume the more nucleons interact with each other through the strong interaction, the more they attract each other
2. $a_S$ : surface term, similar to the surface tension, some energy stored in there, reducing the binding interaction
3. $a_C$ : the Coulomb repulsion of the protons within the nucleus
4. $a_A$ : asymmetry term, rooted in the Pauli exclusion principle. Basically if there are more of one type of nucleon (generally of neutrons) then the overall energy is larger than needs to be thus decreasing the binding energy (note: $A-2Z = Z - N$)
5. $\delta$ : pairing term, depends on whether there are even or odd number of nucleons altogether and even or odd number of protons/neutrons. In empirical description usually modeled as a continuous variable $a_P/A^{1/2}$.

This is of the expression for the total binding energy, what is interesting is the binding energy per nucleon, as a measure of stability:

$$E(A, Z)/A \approx a_V - a_S \frac{1}{A^{1/3}} - a_C \frac{Z(Z-1)}{A^{4/3}} - a_A \frac{(A-2Z)^2}{A^2} + a_P \frac{1}{A^{3/2}}$$

To see which nucleus (what value of $A$) is the most stable one has to find for which $A$ is this function maximal. At this point $Z$ is arbitrary but we should chose a physically meaningful value. From theoretical point of view a good choice is the $Z$ that gives the highest binding energy for a given $A$ (the most stable isotope), for which we need to solve solve $\frac{\partial (E/A)}{\partial Z} = 0$. The results is $Z_{stable}(A) \approx \dfrac12\dfrac{A}{1+A^{2/3} \frac{a_C}{4 a_A}}$. After putting back the $Z_{stable}(A)$ into $E(A, Z)/A$ one can maximize the function value to get the "optimal number" of nucleons for the most stable element. Depending on the empirically determined values of $a_S, a_C, a_A, a_P$ the maximum will occur in the area $A \approx 58 \ldots 63$.

The interpretation of this result is something like this:

• for small atoms (small $A$) the biggest contribution is the surface term (they have a large surface-to-volume ratio), and they want increase the number of nucleons to reduce it - hence you have fusion
• for large atoms (large $A$) the Coulomb term increases because more protons mean more repulsion between them, and also, to keep everything together more neutrons are needed (thus $N \gg Z$ which makes the asymmetry term larger as well. By ejecting some nucleons (alpha decay), or converting between neutrons and protons (beta decay) the nucleus can reduce this terms.
• optimally bound $A$ (and $Z$) happens when these two groups of competing contributions balance each others out.
• Correct me if I'm wrong, but isn't it true that our knowledge of QCD isn't good enough to derive anything like the terms in this formula? I find this fascinating. – j.c. Nov 18 '10 at 3:01
• As the name suggests: it is "semi-empirical" formula, ie. some ideas are coming from theory, some parameters from experiment. The model is largely derived under the "liquid drop" description of the nucleus, and if you check the link above, some coefficients ($a_V$, $a_C$, $a_A$ in particular) have very good theoretically predicted values. In the end, of course, they are checked against the experiment and the coefficients are fitted based on the data. If there are discrepancies or more advanced theory, the model is/will be modified. – Gergely Nov 18 '10 at 6:33
• Forgive my impertinence, but isn't part of the way the terms are defined in the SEMF designed such that Fe-56 is a maximally stable nucleus? Isn't using it to answer this question rather circular? – Landak Nov 24 '15 at 17:41
• @Landak You always have circles - we don't know the basic rules of the universe. You observe some effect, build a theory to explain the observation, see if the advanced predictions of your theory check out with reality, and that's it. But there's no zero-point from which you could build, like in mathematics. Observation is always the starting point. Circular is fine as long as you don't count the same piece of evidence twice - something scientists are very careful about. Observations in the future make your model more or less confident, and in doing so, affect all the dependent models in turn. – Luaan Jan 31 '16 at 22:49

The bonding of nuclei is dominated by 2 main forces - the strong nuclear force, and the electromagnetic force. The strong nuclear force is much stronger than the electromagnetic force, but acts over much shorter distances.

For small nuclei (eg hydrogen and helium), if you're able to add more nucleons, they are likely to stick due to the attraction of the strong force. This is why smaller nuclei tend to fuse together. Sticking the particles together results in a lower-energy configuration so it is more stable.

For larger nuclei, the size of the nucleus means particles on one side don't feel much strong force attraction from particles on the other side, but they still feel electromagnetic repulsion (if they are charged, ie protons). This means that larger nuclei are less stable, and can form lower-energy configurations by splitting into smaller parts (fission).

Iron is at the middle point in terms of nucleus size, where either adding or removing particles would result in a higher-energy configuration, and so it is regarded as the most stable nucleus.

• I love easy to understand answers that make me feel smarter ;-). – Stein Åsmul Jun 23 '11 at 0:34

In some sense the nucleus of a Helium (He-4) is more stable, than the nucleus of an Iron. About 20 MeV is required to detach any particle from the nucleus of a Helium. But only about 10 MeV is enough to detach a nucleon from the nucleus of an Iron.

• Sure it takes such a large energy to detach one nucleus, since the energy curve strongly favors adding more nucleus instead of taking away. This does not say anything about stability. If instead of taking away you want to add more nucleus then there's your He-fusion (into Be-8) which is very much exothermic. On the other hand, there's no energy generating nuclear reaction from iron. – Gergely Nov 17 '10 at 6:31
• @Greg, He-fusion into Be-8 is not exothermic. Be-8 is not stable, it decays into He-4 with 0.1 MeV. – voix Nov 17 '10 at 16:36
• my bad, indeed the Be-8 is a small barrier, but one more He-4 and the resulting C-12 (and later down the chain the heavier elements) do produce energy. The original comment still stands: He-4 might be a local-maxima on the binding energy curve (due the high energy cost for taking nucleons away, and also the energy cost adding some because of the closed nucleon shells), but iron is still more stable. – Gergely Nov 18 '10 at 0:21
• @greg: adding free neutrons or protons to iron generates energy, only adding alpha particle doesn't. – Ron Maimon Apr 13 '12 at 5:52
• @RonMaimon Comming from Complex one-dimensional structures of space (see this please after the next link) I wrote hand waving About the distribution of electrons magnetic moments in atoms. It comes out that iron has a more symmetric distribution of the electrons than other elements except the nobel gases. – HolgerFiedler Feb 3 '16 at 10:36