3.4: Nuclear mass formula

There is more structure in Figure 4.3.1 than just a simple linear dependence on $A$. A naive analysis suggests that the following terms should play a role:

1. Bulk energy: This is the term studied above, and saturation implies that the energy is proportional to $B_{\text{bulk}}=\alpha A$.
2. Surface energy: Nucleons at the surface of the nuclear sphere have less neighbors, and should feel less attraction. Since the surface area goes with $R^2$, we find $B_{\text{surface}}=-\beta A$.
3. Pauli or symmetry energy: nucleons are fermions (will be discussed later). That means that they cannot occupy the same states, thus reducing the binding. This is found to be proportional to $B_{\text{symm}}=-\gamma (N/2-Z/2)^2/A^2$.
4. Coulomb energy: protons are charges and they repel. The average distance between is related to the radius of the nucleus, the number of interaction is roughly $Z^2$ (or $Z(Z-1)$). We have to include the term $B_{\text{Coul}}=-\epsilon Z^2/A$.

Figure $\PageIndex{1}$: Illustration of the terms of the semi-empirical mass formula in the liquid drop model of the atomic nucleus. (CC BY-SA; Daniel FR).

Taking all this together we fit the formula

$$B(A,Z) = \alpha A - \beta A^{2/3} - \gamma (A/2-Z)^2A^{-1} - \epsilon Z^2 A^{-1/3} \label{eq:mass1}$$

to all know nuclear binding energies with $A\geq 16$ (the formula is not so good for light nuclei). The fit results are given in Table $\PageIndex{1}$.

Table $\PageIndex{1}$: Fit of masses to Equation $\ref{eq:mass1}$.

parameter	value
$\alpha$	15.36 MeV
$\beta$	16.32 MeV
$\gamma$	90.45 MeV
$\epsilon$	0.6928 MeV

Figure $\PageIndex{2}$: Difference between fitted binding energies and experimental values (color), as a function of $N$ and $Z$.

In Table $\PageIndex{1}$ we show how well this fit works. There remains a certain amount of structure, see below, as well as a strong difference between neighbouring nuclei. This is due to the superfluid nature of nuclear material: nucleons of opposite momenta tend to anti-align their spins, thus gaining energy. The solution is to add a pairing term to the binding energy,

$$B_{\text{pair}} = \begin{cases} A^{-1/2} & \text{for $N$ odd, $Z$ odd}\\ - A^{-1/2} & \text{for $N$ even, $Z$ even}\end{cases} \nonumber$$

The results including this term are significantly better, even though all other parameters remain at the same position (Table $\PageIndex{2}$). Taking all this together we fit the formula

$$B(A,Z) = \alpha A - \beta A^{2/3} - \gamma (A/2-Z)^2A^{-1} - \delta B_{\text{pair}}(A,Z)-\epsilon Z^2 A^{-1/3} \label{eq:mass2}$$

Table $\PageIndex{2}$: Fit of masses to Equation $\ref{eq:mass2}$.

parameter	value
$\alpha$	15.36 MeV
$\beta$	16.32 MeV
$\gamma$	90.46 MeV
$\delta$	11.32 MeV
$\epsilon$	0.6929 MeV

Figure $\PageIndex{3}$: $B/A$ versus $A$, mass formula subtracted.