nuclear physics - 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?

Answer

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.