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Models of Nuclear Structure

There are nine general models of nuclear structure:

These models are not mutually exclusive. Each may capture an element of nuclear structure. The Liquid Drop Model, the Shell Model and the Model of Independent Particles in a Potential Well are dealt with elsewhere.

The Substructure Model of Nuclear Structure

The Substructure Model is based upon the notion that within the nucleus there are combinations of nucleons which are maintained and interact with each other. 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  ZmN 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  ZmN. The difference is called the mass deficit Δ


 ZΔN =  Z(1m0) +  N(0m1) −  ZmN
 

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


ZmN + γ → Z'mN' + Z"mN"
 

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


ZΔN − γ = Z'ΔN' + Z"ΔN"
 

Therefore the energy required to remove a subunit is


γ = ZΔNZ'ΔN'Z"ΔN"
 

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 = ZΔNZ-1ΔN'
and
γ = ZΔNZΔN-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 zmn subunit is not


Z+zΔN+nZΔN
but instead
γ = Z+zΔN+nZΔNzΔn
 

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.

Energy Required to Dislodge a Neutron
ZNElementChange in
Binding Energy
(MeV)
11Hydrogen2.225
22Helium20.578
33Lithium5.663
44Beryllium18.900
55Boron8.436
66Carbon18.722
77Nitrogen10.553
88Oxygen15.664
99Fluorine9.149
1010Neon16.865
1111Sodium11.070
1212Magnesium16.531
1313Aluminum11.365
1414Silicon17.180
1515Phosphorus11.319
1616Sulfur15.042
1717Chlorine11.508
1818Argon15.255

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.

Energy Required to Dislodge a Proton
ZNElementChange in
Binding Energy
(MeV)
11Hydrogen2.225
22Helium19.814
33Lithium4.588
44Beryllium17.255
55Boron6.586
66Carbon15.957
77Nitrogen7.551
88Oxygen12.127
99Fluorine5.607
1010Neon12.844
1111Sodium6.740
1212Magnesium11.693
1313Aluminum6.306
1414Silicon11.585
1515Phosphorus5.595
1616Sulfur8.864
1717Chlorine5.143
1818Argon8.507

In contrast to the case for the dislodgment of a neutron the values for the dislodgment of a proton are not reaching 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.,

Binding Energies of
the Last Alpha Particle
in Z=N Nuclei
ElementZ=NBinding
Energy
(MeV)
He20.0
Be4− 0.09
C67.4
O87.2
Ne104.7
Mg129.3
Si149.9
S166.9
K186.7
Ca207.2
Ti225.1
Cr247.7
Fe267.9
Ni289.0
Zn302.7
Ge320.8
Se342.2
Kr361.2
Sr382.7
Zr403.9

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;

Binding Energies of
the Last Deuteron Particle
in Z=N Nuclei
ElementZ=NBinding
Energy
(MeV)
H10.0
Li31.5
B56.0
N710.3
F97.6
Na1111.3
Al1311.4
P1511.9
Cl1711.9
K1911.7
Sc2110.4
V2212.7
Mn2512.9
Co2712.8
Cu2912.9
Ga3110.8
As3310.3


The Binding Energies of Nuclei Which Could Contain an Integral Number of Alpha Particles

ElementNeutronsProtonsBinding
Energy
Number of
Alpha Particles
Binding
Energy
Difference
He2228.295674128.2956740
Be4456.49951256.5913480.091838
C6692.161728384.8870227.274706
O88127.6193364113.18269614.43664
Ne1010160.6448595141.4783719.166489
Mg1212198.256896169.77404428.482846
Si1414236.536897198.06971838.467172
S1616271.780668226.36539245.415268
Ar1818306.71579254.66106652.054634
Ca2020342.05210282.9567459.09526
Ti2222375.474711311.252464.22229
Cr2424 411.46212 339.548088 71.913912
Fe2626 447.69713 367.843762 79.853238
Ni2828 483.98814396.139436 87.848564
Zn3030 514.99215 424.43511 90.55689
Ge3232 545.9516 452.730784 93.219216
Se3434 576.417 481.026458 95.373542
Kr3636 607.118 509.322132 97.777868
Sr3838 638.119 537.617806 100.482194
Zr4040 669.820 565.91348 103.88652
Mo4242 700.921 594.209154 106.690846
Ru4444 731.422 622.504828 108.895172
Pd4646 762.123 650.800502 111.299498
Cd4848 793.424 679.096176 114.303824
Sn5050 824.925 707.39185 117.50815

The data in the above table suggests that there are structures of the alpha particles. There is no significant increase in binding energy for two alpha particles but for three there is. The additional binding energy for the number of alpha particles above two is roughly constant at about 7 MeV per additional alpha particle until a level of 14 alpha particles is reached. Thereafter the increase is about 3 MeV per additional alpha particle, as shown below.

The stability of the increments in excess binding energy is shown below.

Alpha Particle Substructures in Nuclei

Another way of illustrating the existence of alpha particle substructures in nuclei is to plot the effect of adding additional nucleons (alternating between protons and neutrons) to a nucleus that could contain an integral number of alpha particles. The graph is as shown.

There are actually six lines in this graph but pairs of lines so nearly coincide that they appear to be one.

The data upon which the graph is based are shown below.

The Binding Energies (MeV) of Nuclei
As Additional Nucleons Are Added
to Nuclei Containing an Integral
Number Alpha Particles
Number of
Additional Nucleons
P N P N P N
0 0 0 28.30 28.30 56.50 56.50
1 0 0 26.33 27.41 56.31 58.16
2 2.22 2.22 31.99 31.99 64.75 64.75
3 7.72 8.48 37.60 39.24 73.44 76.20
4 28.30 28.30 56.50 56.50 92.16 92.16

What the first column represents is the sequence of adding a proton, then a neutron, then a proton and finally a neutron. The second column is the effect of the sequence of adding a neutron, then a proton, then a neutron and finally a proton. The other pairs of columns represent the same pairs of sequences. The plots of the data are nearly parallel in the graph.

It is notable that the creation of a singleton proton, one that is not part of an alpha particle or deuteron, creates a smaller binding energy than adding a singleton neutron. However, the magnitude of the difference is barely noticeable in the graph.

The second and third tier of plots show a similar pattern to the first given above.

The relevant quantity is the increase in binding energy over that of the nucleus with an integral number of alpha particles. When this increment in binding energy is plotted the results are as shown below.

If the value of the binding energy derived solely from the formation of alpha particles in nuclei the above lines would all be the same. While they are not exactly the same they are quite similar. The lines for the creation of the first two alpha particles are very close. The third deviates marginally and this could be from the formation of some configuration of alpha particles within the nucleus. Likewise the plots for the third and fourth cases (which are shown in the two separate graphs) are quite close.

Energy Efficient Arrangements
of Alpha Particles in Nuclei

A final construction confirms the nature of the source of the binding energy on nuclei. In the graph below is shown the binding energy resulting from three different augmentations of a Helium 4 nuclide (an alpha particle). The first is the successive addition of protons. This is the lower line. It shows no increase in binding energy, in fact there are substantial decreases and this augmentation cannot go very far. The middle line is for the successive addition of neutrons. This shows some small increases in binding energy but the amount is weak compared with the third form of augmentation. The third form involves alternating additions of protons and neutrons; first a proton then a neutron. This shows a big jump when the additions amount to an alpha particle.

Binding Energies for Nuclei Resulting
from the Addition of Nucleons
Number of
Additional
Nucleons
Protons
only
Neutrons
only
Alternating
protons and
neutrons
0 28.2957 28.2957 28.2957
1 27.4100 26.3300 27.4100
2 29.2691 26.9240 31.9946
3 28.8200 24.7200 39.2445
4 31.4080 24.7820 56.49951
5 30.2600not stable 58.1649
6 30.3400not stable 64.7507

Conclusions Concerning the Alpha Particle Model

The conclusion from the data shown so far is that there is some sort of structure within nuclei involving alpha particles with possible subsidiary nucleon pairs (proton/neutron, neutron-neutron or proton-proton). The binding energy of a nuclide is the binding energy of the nuclide that contains an integral number of alpha particles plus the binding energy of any additional nucleon pairs plus the binding energy due to any singleton nucleon. The binding energy of a nuclide is composed of two parts; the intrinsic binding energies of the substructures and the structural binding energy due to the interaction of the substructures. For material on the statistical analysis of the binding energies (mass deficits) of the 2931 nuclides see Mass Deficits.

The Lattice Models of Nuclear Phenomena

The lattice models involve simulation of nuclear phenomena using massive computation. They start with simplistic models of nuclear structure and compute all of the interactions. This is in the nature of Monte Carlo simulations. By varying the parameters of the model and the simulation the investigators try to reproduce experimental results. For example, lattice models have been used to investigate the fragmentation of nuclei due to nuclear projectiles. A simple-cubic packed lattice is randomly filled or fully-filled with randomly formed nearest-neighbor bonds. A projectile particle bores a hole through the arrangement and the fragmentation of the target is computed. The results are averaged and compared with experimental results. The parameters are varied until the simulation results give a sufficiently close approximation of the experimental results.

The Fluctuating Combinations Model

This model conceives of nuclei having no set structure but instead fluctuating between all feasible substructures. One of the most convincing bits of evidence for the substructure model is that alpha particles are sometimes ejected from nuclei. Scattering experiments do not support the notion of alpha particles existing within nuclei. In the fluctuating combinations model alpha particles are not permanent elements of a nucleus but appear from time to time and thus are there when a nucleus breaks up.

The Collective Model

This model was developed by Aage Bohr and Benjamin Mottelson based upon prior work by James Rainwater. It holds that the surface of a neucleus is like that derived from the liquid drop model but that the internal motions of the nucleons give rise to the general shape. That general shape is in some cases non-spherical, such as ellipsoids. Bohr and Mottelson found that the non-spherical shapes can be explained in terms of rotations of the nuclei. For more on this model see Collective Model.

The Interacting Boson Model

This model is the creation of F. Iachello and A. Arima. It was first introduced in 1974 and has connections with the Collective Model of Aage Bohr and Ben R. Mottelson. It is also a descendant of the Shell Model. The group theoretic methods of particle physics used to explain the physics of nuclei.

(To be continued.)

For a new model of nuclear structure see New Model.


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