The liquid-drop model is based on the constant density of nuclei, independent of the number and quality of nucleons. The phenomenon is analogous to a liquid drop in which the molecules are subjected to the same van der Waals forces, indepen­dent of the size of the drop.

According to the liquid-drop model, the nucleus can be imagined as a rather compact, spherical structure (similar to a liquid drop), the constituents of which are subjected to strong interactions acting in a very small range (about 10—15m). Really, this is the nuclear force, the energy of which is approximately proportional to the mass number (A).

When the binding energy of a nucleus is calculated using this model, the nuclear energy has to be taken into account first. Let’s suppose that the nuclear energy between two nucleons is U0. In the closest geometric packing of spheres, one nucleon has 12 neighbors (the coordination number is 12). It should mean —12U0, the total energy for one nucleon; however, each nucleon is considered twice (nucleon pairs are investigated), so only —12U0/2 = — 6U0 is the nuclear energy for one nucleon. For a nucleus with a mass number of A, the total nuclear energy is —6U0A. This energy is shown as volume energy in Figure 2.4.

Of course, the peripheral nucleons have only six neighbors, decreasing the nuclear energy. Considering the thickness of peripheral layer a, the volume of this layer is 4R2na (surface X thickness). Since the volume of one nucleus is 4R3n/3 and the number of nucleons is A, the volume of one nucleon is 4R3n/3A. The num­ber of nucleons in the peripheral layer can be obtained by dividing the volume of the layer with the volume of one nucleon: 3aA/R. Therefore, the surface energy (Es) of the nucleus can be expressed as:

The nuclear energy corrected by the surface energy can also be seen in Figure 2.4.

Since the protons repulse each other, the energy of repulsion has to be taken into account. The energy of the electrostatic repulsion can be expressed as in Eq. (2.11). So, the binding energy per nucleon has to be corrected with the electro­static repulsion too (Figure 2.4):

The calculated binding energy per nucleon (Eq. (2.17)) is not accurately equal to the experimental value. The differences can be explained by two things. The first is that each value of A has a value of Z for which there is maximum stability (see Eqs. (2.10) and (2.10a)). When the ratio of the protons and neutrons is different from the maximum stability, a so-called asymmetry energy must also be taken into account because of the slightly different interaction energies of proton—proton, proton—neutron, and neutron—neutron pairs. The second is that the nuclei with even—even proton and neutron numbers are more stable than nuclei with odd—odd, even—odd, or odd—even nuclei. For even—odd and odd—even nuclei, this effect is taken to be zero, and for even—even nuclei, it is a negative number (increasing sta­bility), whereas for odd—odd nuclei, it is a positive number (decreasing stability), and it will be discussed in Chapter 3.

The total semiempirical formula by Weizsacker for the binding energy per nucleon is as follows:

where

and

— = ±a5A3/4 (2.20)

A

and y and a5 are constants. As seen in Figure 2.4, the binding energies calculated by Eq. (2.18) agree well with the experimental values.