Nuclear models occur under a set of hypotheses, each of which explains some aspect of nuclear behavior. The most basic is perhaps the “liquid drop model” which is used to account for nuclear fission, radioactivity. This is illustrated in Fig, 1 below:
Fig. 1: The Liquid Drop model showing U235 Fission
As the name implies, the nucleus is depicted in terms of a “liquid drop”, which analogous to a liquid macroscopic drop, exhibits surface tension and an excitation energy when disturbed.
This model is premised on the exceedingly short range of nuclear forces which requires that nearest neighbor attractions predominate. This is somewhat similar to the type of attraction between the molecules of a liquid which leads to the property of its surface tension. Thus, in a liquid drop nucleus each nucleon shares its total binding energy with every nucleon.
As the simplified model diagram (for U 235 fission) shows, we expect a spherical configuration in the minimum energy or unexcited state. The unstable (compound) nucleus then yields two “daughter nuclei” : La 139 and Mo 95, with two additional neutrons released
The total binding energy initially holding all the nucleons together in the liquid drop nucleus model is comprised of a sum of separate energies, including: surface or E(s), volume or E(v), and Coulombic repulsion or E(c) – the latter due to the electrostatic repulsion between protons. Thus, the binding energy:
E(B) = E(s) + (E(v) + E(c)
A reduction in any one or all of these constituent energies reduces the overall binding energy and hence the bonding energy between any pair of nucleons. To account for energy release in the fission of heavy nuclei we need only represent the nucleus as a liquid drop and its transition from the unexcited to the excited state as shown in Fig. 1.
(1) fo(Z, A) = 1.008142Z + 1.008982 (A – Z)
(5) f4 (Z, A) = + a4 (Z - A/2)2/ A
Fig. 2: The potential well for the Shell model
The Schrodinger Equation is then set up for this potential and solved. The solution yields stationary quantum states, somewhat like the states which are associated with electrons in the outer shells of the atom.
The form or expression for the energy of a nucleus is very closely approximated by the energy associated with a “square well” potential. The Fermi energy in more explicit form (which we will not elaborate upon too much beyond this) is:
It is found that the Schrodinger equation to solve becomes:
-ħ2/ 2m ( dy2/dx2 ) = Ey
dy2/dx2 + 2m/ ħ2 (Ey) = 0
Where all the Fermi shell energy levels are presumed occupied up to N/2. Use this to obtain a simplified estimate of the energy associated with the oxygen nucleus if we assume its shells filled up to N =4 and use an estimate for L as:
R = r o A 1/3
(Take m = 1.7 x 10 -27 kg)