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4.4: Nuclear mass formula

  • Page ID
    15013
  • [ "article:topic", "binding energy (nuclear)", "Bulk energy (nuclear)", "Surface energy (nuclear)", "Pauli Energy (nuclear)", "authorname:nwalet" ]

    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 \(B_{\text{Coul}}=-\epsilon Z^2/A\).

    765px-Liquid_drop_model.svg.png

    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 \(\PageIndex{1}\).

    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


    mass_form1.png

    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

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

    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

     

    152477155170608508.png

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