These three models are not mutually exclusive. The Liquid Drop Model and Shell Model are dealt with elsewhere.

The Substructure Model of Nuclear Structure

The Substructure Model is based upon the notion that within the nucleus there are subnuclei. To get evidence of such substructures it is necessary to consider the concept of the binding energy of a nucleus.

Binding Energy

Let Z represent the number of protons in a nucleus and N the number of neutrons. The atomic number A is the sum of Z and N, the total number of nucleons in a nucleus. Let  _Zm_N denote the mass of a nucleus composed of Z protons and N neutrons. The mass of a proton in this notation is  (₁m₀) and that of a neutron  (₀m₁).

One might expect the mass of a (Z, N) nucleus to be Z ((₁m₀))+N ((₀m₁)), but generally that is not the value of  _Zm_N. The difference is called the mass deficit Δ

 _ZΔ_N =  Z(₁m₀) +  N(₀m₁) −  _Zm_N
 

When the mass deficit is expressed in energy units it is called the binding energy of the nucleus. Usually this binding energy is expressed in millions of electron volts (MeV). The binding energy for a subunit of a nucleus is the amount of energy that would be required to disintegrate that nucleus into the subunit and the rest of the nucleus. This energy represents the amount of energy gained by putting the component subunits together into that nucleus.

Information on how strongly a particular subunit is bound can be gained by looking at the generic reaction

_Zm_N + γ → _Z'm_N' + _Z"m_N"
 

where Z=Z'+Z" and N=N'+N".

Subtracting this equation from

Z(₁m₀)+N(₀m₁) = Z'(₁m₀)+N'(₀m₁) + Z"(₁m₀)+N"(₀m₁)
 

gives the result

_ZΔ_N − γ = _Z'Δ_{N' + Z"}Δ_N"
 

Therefore the energy required to remove a subunit is

γ = _ZΔ_N − _Z'Δ_N' − _Z"Δ_N"
 

For the case of the removal of a proton or a neutron the binding energies of a single proton or single neutron are zero so the formula is simpler; i.e.,

γ = _ZΔ_N − _Z-1Δ_N'
and
γ = _ZΔ_N − _ZΔ_N-1
 

Thus for the deuteron, ₁H₁, the energy required to remove a proton or a neutron is γ = 2.2245 MeV − 0 = 2.2245 MeV. For the Helium 4 nucleus the energy required to remove a neutron is

γ = 28.2957 − 7.7180 = 20.5777 MeV.

For Lithium 6, ₃Li₃,

γ = 31.9941− 26.3307 = 5.6634 MeV.

However, in general, the energy required to remove a _zm_n subunit is not

_Z+zΔ_N+n − _ZΔ_N
but instead
γ = _Z+zΔ_N+n − _ZΔ_N − _zΔ_n
 

For the removal of a helium 4, ₂He₂, nucleus from a carbon 6 nucleus,

γ = 92.1617 − 56.4995 − 28.2957 = 7.3665 MeV.

An interesting calculation is the energy to remove a neutron from a nuclei which have equal numbers of protons and neutron, Z=N.

The graph of the above data indicates a dependence on the odd-even-ness of the proton and neutron number. The pattern is a higher number for even values of Z/N and a lower value for odd values. The difference between the even Z=N nuclei and the odd Z=N is that in the case of the even Z=N the nucleus could be organized entirely into He 4 nuclei. For odd Z=N there has to be a proton/neutron pair that is not part of a He 4 nucleus.

As the graph shows the even Z=N values are generally declining toward a stable level, which is the same stable level the odd Z=N values are rising to.

A similar pattern prevails in the energies required to dislodge a proton from Z=N nuclei.

In contrast to the case for the dislodgment of a neutron the values for the dislodgment of a proton are not reaching a a stable level. Instead the values for both the even and odd cases are declining for large values of Z=N. This is explained by the increasing separation of protons in the nuclei which puts them in the range where the nuclear force is weaker and less able to counteract the electrostatic repulsion of protons for one another. Neutrons are not subject to this effect.

This is pervasive evidence that nuclei contain substructures of He 4 nuclei. This makes the information on the binding energies of the last He 4 particle of interest; i.e.,

The pattern here is low values for the first two, increased values for the next two to Z=N=8, a decreased value, then high but declining values of Z=N=18, then low but increasing values to Z=N=28, followed by relatively low values for the rest of the cases of Z=N. The break points of 2, 8, 18 and 28 correspond to the so-called magic numbers of nuclear structure. In the first range there are 2 cases, in the second range 3 cases, in the third range 4 cases, in the fourth range 5 cases and in the final range 6 cases. These ranges of Z=N seem to correspond to shell structures.

To round out the empirical analysis the binding energies for the last deuteron (proton/neutron pair) are shown below;