The liquid drop model and its analysis
Mathematical analysis of the theory delivers an equation which attempts to predict the binding energy of a nucleus in terms of the numbers of protons and neutrons it contains. This equation has five terms on its right hand side. These correspond to the cohesive binding of all the nucleons by the strong nuclear force, the electrostatic mutual repulsion of the protons, a surface energy term, an asymmetry term (derivable from the protons and neutrons occupying independent quantummomentum states) and a pairing term (partly derivable from the protons and neutrons occupying independent quantumspin states). If we consider the sum of the following five types of energies, then the picture of a nucleus as a drop of incompressible liquid roughly accounts for the observed variation of binding energy of the nucleus:
Volume energy. When an assembly of nucleons of the same size is packed together into the smallest volume, each interior nucleon has a certain number of other nucleons in contact with it. So, this nuclear energy is proportional to the volume.
Surface energy. A nucleon at the surface of a nucleus interacts with fewer other nucleons than one in the interior of the nucleus and hence its binding energy is less. This surface energy term takes that into account and is therefore negative and is proportional to the surface area.
Coulomb Energy. The electric repulsion between each pair of protons in a nucleus contributes toward decreasing its binding energy. Asymmetry energy (also called Pauli Energy). An energy associated with the Pauli exclusion principle. If it wasn’t for the Coulomb energy, the most stable form of nuclear matter would have N=Z, since unequal values of N and Z imply filling higher energy levels for one type of particle, while leaving lower energy levels vacant for the other type. Pairing energy. An energy which is a correction term that arises from the tendency of proton pairs and neutron pairs to occur. An even number of particles is more stable than an odd number.
In the following formulae, let A be the total number of nucleons, Z the number of protons, and N the number of neutrons. The mass of an atomic nucleus is given by
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image001.gif[/IMG]
where mp and mn are the rest mass of a proton and a neutron, respectively, and EB is the binding energy of the nucleus. The semi-empirical mass formula states that the binding energy will take the following form: [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
Each of the terms in this formula has a theoretical basis, as will be explained below.
The term aVA is known as the volume term. The volume of the nucleus is proportional to A, so this term is proportional to the volume, hence the name.
The basis for this term is the strong nuclear force. The strong force affects both protons and neutrons, and as expected, this term is independent of Z. Because the number of pairs that can be taken from A particles is [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image003.gif[/IMG], one might expect a term proportional to A2. However, the strong force has a very limited range, and a given nucleon may only interact strongly with its nearest neighbors and next nearest neighbors. Therefore, the number of pairs of particles that actually interact is roughly proportional to A, giving the volume term its form. The coefficient aV is smaller than the binding energy of the nucleons to their neighbours Eb, which is of order of 40 MeV. This is because the larger the number of nucleons in the nucleus, the larger their kinetic energy is, due to Pauli’s exclusion principle. If one treats the nucleus as a Fermi ball of A nucleons, with equal numbers of protons and neutrons, then the total kinetic energy is [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image004.gif[/IMG], with εF the Fermi energy which is estimated as 38 MeV. Thus the expected value of aV in this model is [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image005.gif[/IMG], not far from the measured value. The term aSA2 / 3 is known as the surface term. This term, also based on the strong force, is a correction to the volume term.
The volume term suggests that each nucleon interacts with a constant number of nucleons, independent of A. While this is very nearly true for nucleons deep within the nucleus, those nucleons on the surface of the nucleus have fewer nearest neighbors, justifying this correction. This can also be thought of as a surface tension term, and indeed a similar mechanism creates surface tension in liquids. If the volume of the nucleus is proportional to A, then the radius should be proportional to A1 / 3 and the surface area to A2 / 3. This explains why the surface term is proportional to A2 / 3. It can also be deduced that aS should have a similar order of magnitude as aV.
The term [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image006.gif[/IMG]is known as the Coulomb or electrostatic term.
The basis for this term is the electrostatic repulsion between protons. To a very rough approximation, the nucleus can be considered a sphere of uniform charge density. The potential energy of such a charge distribution can be shown to be [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image007.gif[/IMG]
where Q is the total charge and R is the radius of the sphere. Identifying Q with Ze, and noting as above that the radius is proportional to A1 / 3, we get close to the form of the Coulomb term. However, because electrostatic repulsion will only exist for more than one proton, Z2 becomes Z(Z − 1). The value of aC can be approximately calculated using the equation above:
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image008.gif[/IMG]
Quantum charge integers:
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image009.gif[/IMG]
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image010.gif[/IMG]
Integration by substitution:
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image011.gif[/IMG]
Potential energy of charge distribution:
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image012.gif[/IMG]
Electrostatic Coulomb constant:
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image013.gif[/IMG]
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image014.gif[/IMG]
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image015.gif[/IMG]
The term [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image018.gif[/IMG]is known as the asymmetry term. The theoretical justification for this term is more complex. Note that as A = N + Z, the parenthesized expression can be rewritten as (N − Z). The form (A − 2Z) is used to keep the dependence on A explicit, as will be important for a number of uses of the formula.
The Pauli exclusion principle states that no two fermions can occupy exactly the same quantum state in an atom. At a given energy level, there are only finitely many quantum states available for particles. What this means in the nucleus is that as more particles are "added", these particles must occupy higher energy levels, increasing the total energy of the nucleus (and decreasing the binding energy). Note that this effect is not based on any of the fundamental forces (gravitational, electromagnetic, etc.), only the Pauli exclusion principle. Protons and neutrons, being distinct types of particles, occupy different quantum states. One can think of two different "pools" of states, one for protons and one for neutrons. Now, for example, if there are significantly more neutrons than protons in a nucleus, some of the neutrons will be higher in energy than the available states in the proton pool. If we could move some particles from the neutron pool to the proton pool, in other words change some neutrons into protons, we would significantly decrease the energy. The imbalance between the number of protons and neutrons causes the energy to be higher than it needs to be, for a given number of nucleons. This is the basis for the asymmetry term. The actual form of the asymmetry term can again be derived by modelling the nucleus as a Fermi ball of protons and neutrons. Its total kinetic energy is
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image019.gif[/IMG]
where Np, Nn are the numbers of protons and neutrons and [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image020.gif[/IMG], [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image021.gif[/IMG]are their Fermi energies. Since the latter are proportional to [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image022.gif[/IMG]and [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image023.gif[/IMG], respectively, one gets [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image024.gif[/IMG]for some constant C.
The leading expansion in the difference Nn − Np is then
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image025.gif[/IMG]
At the zeroth order expansion the kinetic energy is just the Fermi energy [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image026.gif[/IMG]multiplied by [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image027.gif[/IMG]. Thus we get [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image029.gif[/IMG]
The first term contributes to the volume term in the semi-empirical mass formula, and the second term is minus the asymmetry term (remember the kinetic energy contributes to the total binding energy with a negative sign).
εF is 38 MeV, so calculating aA from the equation above, we get only half the measured value. The discrepancy is explained by our model not being accurate: nucleons in fact interact with each other, and are not spread evenly across the nucleus. For example, in the ****l model, a proton and a neutron with overlapping wavefunctions will have a greater strong interaction between them and stronger binding energy. This makes it energetically favourable (i.e. having lower energy) for protons and neutrons to have the same quantum numbers (other than isospin), and thus increase the energy cost of asymmetry between them. One can also understand the asymmetry term intuitively, as follows. It should be dependent on the absolute difference | N − Z | , and the form (A − 2Z)2 is simple and differentiable, which is important for certain applications of the formula. In addition, small differences between Z and N do not have a high energy cost. The A in the denominator reflects the fact that a given difference | N − Z | is less significant for larger values of A. The term δ(A,Z) is known as the pairing term (possibly also known as the pairwise interaction). This term captures the effect of spin-coupling. It is given by: [IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image030.gif[/IMG]
where
[IMG]file:///C:/Users/MASTER%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image031.gif[/IMG]
Due to Pauli exclusion principle the nucleus would have a lower energy if the number of protons with spin up will be equal to the number of protons with spin down. This is also true for neutrons. Only if both Z and N are even, both protons and neutrons can have equal numbers of spin up and spin down particles. This is a similar effect to the asymmetry term. The factor A − 1 / 2 is not easily explained theoretically. The Fermi ball calculation we have used above, based on the liquid drop model but neglecting interactions, will give an A − 1 dependence, as in the asymmetry term. This means that the actual effect for large nuclei will be larger than expected by that model. This should be explained by the interactions between nucleons; For example, in the ****l model, two protons with the same quantum numbers (other than spin) will have completely overlapping wavefunctions and will thus have greater strong interaction between them and stronger binding energy. This makes it energetically favourable (i.e. having lower energy) for protons to pair in pairs of opposite spin. The same is true for neutrons.
