# 4.4: Nuclear mass formula

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- 15013

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:

*Bulk energy*: This is the term studied above, and saturation implies that the energy is proportional to \(B_{\text{bulk}}=\alpha A\).*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 \(B_{\text{surface}}=-\beta A\).*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 \(B_{\text{symm}}=-\gamma (N/2-Z/2)^2/A^2\).*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 \(B_{\text{Coul}}=-\epsilon Z^2/A\).

*Illustration of the terms of the semi-empirical mass formula in the liquid drop model of the atomic nucleus. Image used with permission (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 \(\PageIndex{1}\).

parameter | value |
---|---|

\(\alpha\) | 15.36 MeV |

\(\beta\) | 16.32 MeV |

\(\gamma\) | 90.45 MeV |

\(\epsilon\) | 0.6928 MeV |

**Figure \(\PageIndex{1}\): **Difference between fitted binding energies and experimental values (color), as a function of \(N\) and \(Z\).

In Table \(\PageIndex{2}\) 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,

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

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

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

parameter |
value |
---|---|

\(\alpha\) | 15.36 MeV |

\(\beta\) | 16.32 MeV |

\(\gamma\) | 90.46 MeV |

\(\delta\) | 11.32 MeV |

\(\epsilon\) | 0.6929 MeV |

* Table \(\PageIndex{3}\): *\(B/A\) versus \(A\), mass formula subtracted