Nucleus

The author has discovered that isobars and isotops highlight a particularity. Moving from one nucleus to its adjacent nucleus points out a binary structure. This confirms the "Wave Model" described in the second page of this website. This webpage covers only this discorery.

Isobars

Usually, nucleus graphs are plotted from the atomic number "A", the neutron number "N" or the proton number "Z". The following figure was drawn by Excel© on a u quark basis. The u quarks inside the d quarks were taken into account. That is to say, each proton is made up of three u quarks and one electron, and each neutron is made up of three u quarks and two electrons, as previously explained.

This graph covers the first nuclei, those for which the mass number goes from 3 to 16. The X-coordinates thus go from 9 to 48 since each nucleon, proton or neutron, has three up quarks. The mass of each nucleus was initially divided by the mass number A. An offset of 930,9 MeV was subtracted from each element in order to make the graph more readable (see the note).

This figure shows a simplified graph. A more precise graph emphasizes that the lowest point of each isobar's group is always reached when the number of electrons is equal to half the number of u quarks, including the d quark electrons.

Note: There have been many studies of atoms. However, it is the interpretation that is particularly interesting because this study has a new basis. In fact, the binary structure of the nucleus leads to consider that the d quark is made up of a u quark surrounded by an electron. The following figures leave no doubt about this binary structure.

Isotopes

The lowest point noted with the isobars is repeated with the isotopes. However, examination of the curves shows that the mass of each isotope oscillates with a period of two elements.

In order to better emphasize this oscillation, the difference between two adjacent isotopes, the derivative, has been calculated and plotted with Excel©. Thus, every other time, the following figures show a negative derivative. The object of these graphs is to know what the electron of the d quarks becomes inside the neutron. For that, it is necessary that the number of protons does not vary.

On these graphs, the mass increases and decreases alternately in steps of two neutrons. On the other hand, these graphs show an anomaly at 126, which is a "magic number".

This binary structure shows that the transition from a neutron to a proton and conversely leads to a modification of mass, i.e. of the closed volume, of the nucleus (see Part 1). The immediate deduction, and also the only possible explanation, is to consider that the nucleus has a deuteron structure, as explained in the following section.

Binary Structure
of the Nucleus

When a neutron meets a proton inside the nucleus, the outer-shell electron of the neutron "phagocytes" the proton to make a deuteron. Deuterons would not, therefore, be composed entirely of a proton and a neutron, but of two protons and an outer-shell electron, which act as a strong nuclear force, keeping the two protons locked inside the deuteron.

This scheme could explain:

1. The binary structure observed,
2. The strong nuclear force of the nucleus.

Moreover, the structure in two protons and one electron of the deuteron is more homogeneous and logical than the structure of one proton and one neutron. It should be noted that other structures, like e-(e-(e-(u u u u u u))) or e-(e-(u u u (e-(u u u)))) (see the last page "Interactions") are also possible but improbable.

The He Nucleus

Since alpha particles are helium nuclei, we suspect the He structure to be one of the basic structures of the nucleus.

Within the "Wave Model", many configurations are possible for the He structure. However, taking into account the great stability of this nucleus, it is judicious to think that the following scheme is the most probable. This configuration is very close to the previous deuteron scheme. It has 6 electrons and 8 positrons since 2 positrons makes 3 u quarks. So, the overall charge is -6 +2 = +2, i.e. the charge of the 4He nucleus.

This scheme seems confirmed by the experimentation because the two outer-shell electrons of the 4He nucleus make it particularly strong. Alpha particles are also very strong. This explains alpha radioactivity.

H Isotopes

The following figure covers the possible configurations of H isotopes where electrons act as the strong nuclear force. Electrons go in the periphery of the nucleus to act as the strong nuclear force. The binary structure of the nucleus suggest that some possibilities are most probable than others.

Please note that the following diagrams are only for teaching purposes. These figures have been intentionally simplified.

4H Isotopes

The 4H isotope may lead to different schemes using protons with one, two and three outer-shell electrons.

Intuitively, the binary structure of the nucleus suggests that, inside the nucleus, protons and deuterons toggle. So, the most probable configurations are when electrons surround protons. The correct configuration would require some investment of time, and must be in accordance with many parameters: decay modes, binding energy, volume differences from one isotope to another, the mass derivative, quadripolar moment etc...

It should be pointed out, once more, that if a decay or radioactivity produces protons and neutrons such as alpha particles, it does not mean that these particles were parts of the nucleus before the interaction per se. Maybe the nucleus contains a "soup" of u quarks with all electrons in its periphery. Protons, neutrons, d quarks are produced during the interaction. On the other hand, since waves and particles are both created from spacetime, it is necessary to bear in mind that what we see is not necessarily what really exists. The only thing we can be sure of is that all these particles and waves come from spacetime (see Part 2).

Quarks or Positrons?

In the most probable configuration of the previous figure and in some other configurations, it is possible that the 12 u quarks are separated inside the nucleus and move freely.

It is also possible that a re-combination occurs so that the u quarks come back to their original state, i.e. positrons. In such a case, we would have 8 positrons instead of 12 u quarks.

In reality, no one knows this configuration. Some experimentations in that way are necessary to exactly know the internal structure of the nucleus.

Possible Explanation
of the Binary Structure

The binary structure suggests that, when a group of isotopes is examined, the nucleus is created in two phases.

• Phase 1
  The first neutron takes its place in the nucleus as a proton. It is stripped from its electron. The latter joins the other electrons on the nucleus periphery.
• Phase 2
  The second neutron takes its place in the nucleus as a proton. Its electron also surrounds the preceding proton, making a deuteron, instead of going on the periphery of the nucleus.

In both cases, the volume of the nucleus increases since it contains one more proton. This volume difference comes from the location of electrons.

When the electron goes on the nucleus's periphery, it produces an increase in volume. When it is used to make a deuteron, the increase in volume is different. This could explain the binary steps .

These two phases are repeated in a loop. Thus, we have a succession of increasing and decreasing volumes in a same isotope group.

It is also possible that the electrons go on the periphery, two per outer layer, such as in the orbitals of the atom, as the Pauli Principle states. This process could also explain the periodicity of two.

We could suppose that the electron decreases the Coulomb Field inside the nucleus and the repulsion force between protons is decreased too. However, this is only an assumption.

Bethe - Weizsäcker Formula

This formula determines the binding energy of a nucleus of mass m (A, Z):

B  =  a_vA  -  a_sA^2/3  -  a_cZ²/A^1/3  -  a_a (N - Z)²/A  +  C

• The first term is the volume energy (a_v = 15.56 MeV),
• The second term is the surface energy (a_s = 17.23 MeV),
• The third term comes from the Coulomb Force (a_c = 0.7 MeV),
• The fourth term is an asymmetry energy (a_a = 23.6 MeV),
• C is an adjustment constant.

The traditional strong nuclear force is not in accordance with this formula. The problem lies in the two following terms:

1. Surface energy (see the note)
   The strong nuclear force supposes linking protons and neutrons inside the nucleus. Under no circumstances is this force a "surface force". In this way, the Bethe-Weizsäcker Formula should not have a surface term.
   Within the Spacetime Model, the surface component term is perfectly logical because it matches exactly our model of outer-shell electrons acting like a rubber band to explain the strong nuclear force.
2. The Coulomb Force A similar problem is met with the Coulomb term. Since the Coulomb Force is far less important than the strong nuclear force, this term is unexplainable in the Bethe-Weizsäcker Formula.
   Within the Spacetime Model, the nucleus volume comes from the repulsion force between protons. The presence of a Coulomb term in this formula is, therefore, perfectly logical. It is even a necessity.
   Another point must also be considered. The nuclear volume, i.e. the mass, and the binding energy increase both as A, the atomic number. Currently, physicists think that the nuclear forces are saturated since each nucleon interacts only with its neighbours. This "theory of saturation"does not have logical and rational basis.

If we consider that the neutrons are transformed into protons inside the nucleus, the atomic number A relates the overall number of protons, i.e. addition of the original protons and the protons coming from neutrons.

In other words, it is highly probable that

For the explanation of M, please see the previous pages of this website.

All these protons make a repulsive Coulomb Force between them, which creates the closed volume (see Part 1). It is, therefore, normal that the volume increases as the atomic number does. It is only a simple Coulomb problem, and not a complex and unexplained phenomenon of saturated forces.

Note: The explanation of the surface energy usually uses the Van Der Walls Model. This model is not convincing. It is a good comparison but not a reliable explanation of the phenomenon.