Journal of Nuclear and Particle Physics

p-ISSN: 2167-6895    e-ISSN: 2167-6909

2016;  6(1): 10-23

doi:10.5923/j.jnpp.20160601.03

 

Hypotheses on Nuclear Physics and Quantum Mechanics: A New Perspective

Joshua Egbon

Department of Mechanical Engineering, Ambrose Alli University, Ekpoma, Nigeria

Correspondence to: Joshua Egbon, Department of Mechanical Engineering, Ambrose Alli University, Ekpoma, Nigeria.

Email:

Copyright © 2016 Scientific & Academic Publishing. All Rights Reserved.

This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/

Abstract

The purpose of this study is to bring out new approach for determining the distance between each nucleon in a nucleus, explain why nuclear force is a short range force, and why electrostatic force has greater range than magnetic force, predict the structure of Helium-4 nucleus through different tests, and combine four different energies equations—nuclear energy equation, magnetic potential energy equation, electrostatic energy equation, and gravitational energy equation to form one energy equation, which is the net energy of the nucleus.In this study mathematical models were used to arrive at various experimental results, and new equations were developed, which can be used to predict the net energy of the nucleus, the self energy of every nucleon, the energy released during nuclear fission, the size of the atom, and the contraction of the nucleons in any nucleus.

Keywords: Nuclear Physics

Cite this paper: Joshua Egbon, Hypotheses on Nuclear Physics and Quantum Mechanics: A New Perspective, Journal of Nuclear and Particle Physics, Vol. 6 No. 1, 2016, pp. 10-23. doi: 10.5923/j.jnpp.20160601.03.

1. Introduction

Various attempts have been made in the past to determine the size of the nucleus (Geoff, 2010), as well as unify all the fundamental forces and elementary particles to be written in terms of a single field (Hubert, 2004).
The great accomplishment by others in determining the radius of the proton, its magnetic moment and mass will be of tremendous benefit in this study (Mohr, 2011).
The nuclear binding energy equation may not be used to predict the centre-to-centre distance between each proton and neutron in a nucleus; neither can the electrostatic energy equation account for the energy released in the fission of Uranium-235 (Gopal, 2010). Magnetic potential energy equation cannot be used to determine the energy of the revolving electron; and gravitational energy is the weakest of these forces in the nucleus.
A proper understanding of these forces/energies acting in the nucleus in terms of their range and strength is needed for precise prediction to be made on the effect of these forces/energies within and outside the nucleus (Varadarajan, 2004). These forces which all act in the nucleus will be used in determining the centre-to-centre distance apart between each neutron and proton in the nucleus, the atomic radius of revolving electron, the energy released in the fission of Uranium-235, the net energy of every nucleus system, and the contraction of the nucleons which leads to a loss in mass in the nucleus.
This study help bring together four different energies equations—nuclear energy equation, magnetic potential energy equation, electrostatic energy equation, and gravitational energy equation to form one energy equation, which is the net energy of the nucleus.
In this study mathematical models were used to arrive at various experimental results (Billings, 2013), and new equations were developed, which can be used to predict the net energy of the nucleus, the self energy of every nucleon, the energy released during nuclear fission, the size of the atom, and the contraction of the nucleons in any nucleus

2. Test for the Structure of Helium-4 Nucleus

Fundamental Knowledge/Mathematical Derivations
Some basic knowledge of orbiting bodies in gravitational field is needed here. From gravitational law the velocity of a satellite around an orbit is given by:
(1.1)
(1.2)
(1.3)
Where gi = acceleration due to gravity at that orbit
g = gravity of the orbited body at its surface
R = centre radius apart between orbited body and satellite
= Multiple of force producer radius to distance apart
= force producer radius
Escape velocity at any orbit,
(1.4)
From
(1.5)
Ve-escape velocity at surface of force producer
Also,
Which is times less than Veo
V0 – orbiting velocity round the earth
(1.6)
(1.7)
Kinetic energy of an orbiting body
(1.8)
Where V = Vo and M- mass of orbiting body
(1.9)
The least energy it will use to escape that orbit is
here V = Veo
(1.9.1)
Table 1.1. Relationship between Vo, Veo, rm, K.Eo, and K.Eeo
     
When two bodies are almost of similar weight, it is impossible for one to orbit the other.
When like poles of magnets are placed near each other, both repel themselves and escape at the least escape velocity around that orbit.
If two magnets M1 and M2 of different poles are attracted to each other at a distance apart, M2 of different poles are attracted to spring balances at a distance apart, M2 will record more reading on the spring balance if its mass is less than M1. This means that M2 is more likely to move towards M1 at the velocity around M1 orbit.
Figure 1.1. Extension of spring under magnetic field
With this basic knowledge the nuclear binding energy can be studied.
Nuclear force is a short range force (like magnetic force is short range) that exists between proton-proton, proton-neutron, and neutron-neutron.
Nuclear energy is the energy required to split the nucleus of an atom into its component part. It is the binding energy.
The protons of hydrogen combine to helium only if they have enough velocity to overcome each other’s mutual repulsion sufficiently to get within range of the strong nuclear attraction.
The Potential Energy (P.E) of an object at any point from the centre of the earth is given by
(1.9.2)
But
(1.9.3)
is the least energy it will use to escape that orbit.
Therefore the P.E of any system is equal to the least energy needed to separate the system.
For two magnets of like poles, the energy needed to separate them is equal to their P.E and the magnets will move at a velocity away from each other, which is equal to the escape velocity at that orbit.
A body may be considered a system if no external force is acting on it, or if external force does not have effect on it.
Total Energy in a system: the Binding Energy
Consider a system to be formed of N magnets having equal masses, and the distance between any two closest magnets is the same. The binding energy or P.E for all the bodies is the total energy used to hold them together.
Total Binding energy
(1.9.4)
Where EB = total binding energy
Eb = binding energy between any two magnet
N = total number of magnets
Figure 1.2. Multiple magnets in a force field
(1.9.5)
rm – multiple of radius of one magnet with the distance between the closest magnet
Ve – escape velocity at r1
Veo – escape velocity at r1 x rm = R
This basic knowledge will be used in dealing with the binding energy of various nucleuses of elements.
A nucleus will be considered a close system since no external force is acting on it. The electrostatic force between revolving electrons do not affect the position of the nucleus because of the greater weight of the nucleus.
In the nucleus system, the proton which has a mass slightly lesser than the neutron is used as the object because the force acting on it is the least force needed to separate the system.
Fundamental knowledge of electrostatic potential energy is also required. The electrostatic potential energy between the electron and the proton if the electron is on the surface of the proton is:
(1.9.6)
Where N- total number of charge
where Me – mass of electron
The escape velocity of electron at proton surface is
(1.9.7)
From Bohr’s theory on atomic model, the electron at the lowest energy level of a hydrogen atom has energy of 13.6eV (21.8 x 10-19J). This energy is the energy of the orbiting electron (E0)
(1.9.8)
From
This is the centre radius between the orbiting electron and the proton in a hydrogen atom – or the radius of the hydrogen atom.
Mathematical Models for Helium-4 Structure
Helium-4 nucleus is extremely stable. To know the structure of this nucleus, some possible outcome of its structure will be subjected to two different tests:
a. Atomic Radius Test
b. Nuclear Fission (electromagnetic potential ) Energy Test
P – Proton
N – Neutron
Model i)
Figure 2.1. Model I
Model ii)
Figure 2.2. Model II
Model iii)
Figure 2.3. Model III
Model iv)
Figure 2.4. Model IV
Test A: Atomic Radius Test
Using Mercury as a case study:
Mercury – N = 201; np = 80; nn = 121
Where N - number of protons and neutrons in the nucleus
np – number of protons in the nucleus
nn – number of neutrons in the nucleus
Atomic mass M2 = 200.9703023u; Lowest energy level or ground state of the electron = 10.4eV; empirical atomic radius Ra= 1.51 x 10-10m
Atomic nucleus radius
(2.1)
This formula is used for heavier nuclei (N>20); where
ro = 1.2 – 1.5fm; and N- total number of protons and neutrons in the nucleus.
(2.2)
Test B: Nuclear Fission Energy Test
In nuclear fission, the absorption of a neutron by a heavy nuclide (e.g U-235) causes the nuclide to become unstable and break into light nuclides and additional neutrons. This positively charged light nuclide then repel, releasing electromagnetic potential energy.
The fission of one atom of U-235 generates 202.5 MeV = 3.24 x10-11J
The Helium-4 model must meet these criterion or have values that are near rm = 19829 and Er = 3.24 x 10-11J
(Where Er = energy released).
Model I
Test A: At rpp = 3.33r1 for the helium nucleus where rpp = p-p distance.
Then from
Helium-4 has atomic mass of 4.00260325415U = 6.64648 x10-27kg
Actual mass of 2 protons + 2 electrons + 2 neutrons = 6.697 x 10-27kg
Total Binding Energy
Number of nucleons in Helium-4 is 4
(2.3)
Using this formula, the distance between any two closest protons in a mercury nucleus can be determined.
Actual mass of 80 protons + 80 electrons + 121 neutrons in a Mercury–201 atom = 1.338097424 x 10-25 + 7.28 x10-29+ 2.026662094 x 10-25
M1 = 3.365487518 x 10-25kg
Actual mass of mercury–201 =200.9703023U = 3.3372 x 10-25kg
Radius between any two closest proton in a mercury-201nucleus is given by
If an electron is brought to the mercury nucleus, and placed 3.45754 x 10-15m away from any proton, thereby increasing the total number of charge of the nucleus, then the electrostatic energy of the nucleus will be:
Where Npe – total number of protons + 1 electron added to the nucleus
Energy between a proton and this electron:
Velocity the electron will use to escape from this distance of 3.45754 x10-15m from the closest proton:
Since this velocity is the velocity at the nucleus, it can be regarded as the escape velocity Ve when compared to that used by the electron to orbit the nucleus 1.51 x 10-10m away.
Recall that the velocity of the orbiting electron in Mercury–201 atom is 1912366m/s
This compares well with the known 1.51 x 10-10m (empirical radius)
Test B: Nuclear fission (Electromagnetic potential) Energy Test
Energy released = Electromagnetic potential energy in nucleus
The nucleus in question here is Uranium–235, whose isotope mass number is 235.0439299U = 3.903 x 10-25kg, atomic number = 92; number of neutron = 143
Actual mass of 92 protons + 92 electrons + 143 neutrons
M1 = 3.9348 x 10-25kg
Mass loss = M1-M2 = (3.9348 x 10-25) – (3.903 x10-25)
= 3.179535 x 10-27Kg
Radius between any two closest protons in the nucleus
Magnetic potential Energy
(2.4)
Where = magnetic moment of a proton, which is equal to 1.410606743 x 10-26J/T
(2.5)
From electrostatic energy, Vat distance R is
Magnetic fields are produced by electric currents. The protons in a nucleus has magnetic properties because of the charge they carry. The electrostatic energy between the protons will cause them to move away from each other at a velocity (V which is equal to 8747195m/s for the uranium-235); thereby generating magnetic energy.
The magnetic field strength
Since each proton repels the other 91 protons in the nucleus, the magnetic potential energy for the nucleus is given by
This compares well with the 3.24 x 10-11J energy released during fission of Uranium-235
Results
Table of Result for all the Models:
Table 2.1. Results for all models
     
Expected result: rm = 19829; Ra = 1.51 x 10-10m, and Er = 3.24 x 10-11J
Conclusions
From the various models of Helium-4 Nucleus Model I compares best with the known results of rm = 19829, Ra = 1.51 x 10-10m, and Er = 3.24 x 10-11 J
From Model I Test II, it can be said that the energy released is the magnetic potential energy and not electrostatic potential energy or an addition of both.
Description of Model I: (Helium-4 Nucleus)
v The centre distance between the two proton in a Helium-4 nucleus is 3.33 x the radius of a proton
v The distance between any proton and the closest neutron is 2 x the radius of the proton
v The distance between the two neutrons is 2.43 x the radius of a proton
v The binding energy of Helium-4 nucleuses is 4.547 x 10-12J
v The electrostatic potential energy is 7.885 x 10-14J
v The magnetic potential energy is 4.9 x 10-13J
v The centre distance between the farthest proton and neutron is 2 x the radius of a proton
From this model an equation for the N-N centre distance as well as the P-N centre distance can be written
N-N:
Where Mn- mass of neutron
N- number of nucleons =4
rnn – 2.43
(2.6)
P-N:
Where Mp- mass of Proton
N- number of nucleus
rpn – 2
(2.7)
Table 2.2. Bond type and the binding energy of any nucleus
     
Centre–to–Centre Radii of the Nucleons of Some Elements
Using the formula
Note the distance R = r1 x rmwhere rm = rpp or rpn or rnn
From the above table, it can be seen that P-N, P-P and N-N are closest at Iron–56. This means Iron-56 from the table above has the greatest bond between any two nucleons in the nucleus.

3. Hypothesis on Nuclear Forces

The binding energy between any two nucleons is given by
Where m = mass of the smaller nucleon and
From Tab 2.3 one can look for the binding energy for each nucleus and find the relationship between them. It can be seen from the table that is increasing.
A good relationship between the binding energy between any two nucleons in a nucleus and the distance apart will be used to formulate a new equation.
Mathematical Models:
i) Magnesium–24 has binding energy between any two closest proton and neutron
ii) Molybdenum–96 has binding energy between any two closest proton and neutron
iii) Uranium–238 has binding energy between any two closest proton and neutron
Table 2.3. Centre-to-centre radii of the nucleons of some elements
     
Relationship:
Mathematical Model:
What is the centre-centre multiple distance for Iron-56 if the binding energy between any proton and neutron in an Iron-56 nucleus is 1.432 x 10-12J. (Take binding energy of Magnesium-24 and centre–to–centre multiple of its closest proton and neutron to be 1.383 x 10-12J and 2.177 respectively.)
This can be compared with 2.102 which is in the table.
Mathematical Model:
An unknown element has binding energy between two of its closest nucleons as , and the centre distance between the two nucleons is 2.469. What will be the binding energy between any two closest nucleons of Molybdenum-96, whose centre–to–centre distance multiple is 2.146
This compares well with the 1.403 x of the Molybdenum–96
Conclusion:
From the above, it can be concluded that the binding energy between any two nucleons is inversely proportional to their distance apart
Also the total binding energy in any nucleus is given by
EB = Eb (N-1)
Energy = force x Distance = FR
Hypothesis on Nuclear Forces states that
The force of attraction between any two nucleons held by nuclear forces, separated a distance R is proportional to the product of their masses and inversely proportional to the square root of their distance apart.
(3.1)
Where NB – nuclear constant, and varies for P-P, P-N and N-N
Energy is needed to bring a mass from infinity to the point in question, R.
(3.2)
Recall that it was earlier stated that
This present itself in a new form as:
where
Self Energy of Every Nucleon
This is the self potential of every nucleon, and it is given by
(3.3)
Where r1= radius of the nucleon
Using Helium-4 which is a very stable nucleus as a model, NB can be gotten.
For P-P: Helium-4 Rpp = r1 x rpp = (0.8775 x 10-15) x (3.33)
= 2.922075 x 10-15m
The nuclear force hypothesis when N> 2 for P-P, P-N and N-N
(3.4)
(3.5)
(3.6)
To know the energy between any two P-P or P-N or N-N, one multiply by R, so that becomes
The total binding energy EB of any nucleus is given by:
Mathematical Model: What is the centre-to-centre distance between any two closest neutrons in Tungsten–184 nucleus? Binding energy of Tungsten nucleus is 2.3631 x 10-10J.
From the hypothesis of nuclear forces, when N>2,
This compares well with the 2.847 that is in Tab 2.3.

4. Hypothesis of Superior Forces / Energy

A charged particle at rest experiences a force in an electric field, but none in a magnetic field. A magnetic field doesn’t speed up or slow down a particle: because it doesn’t act on the parallel component of velocity, it only acts on the perpendicular component of velocity, so it can change the particle’s direction.
Two instances can be sited:
i. A magnet placed near a beam of high velocity electrons moving in a straight line, causes the electrons to move in a circular path without altering the velocity of the electrons
ii. Planets are caused to orbit the sun because of the sun’s magnetic field. The forces of the revolving planets are not altered by the magnetic field, but only the gravitational force is considered.
Added to these, electrons revolve round the nucleus under electrostatic force. The magnetic field that exist at the atomic level only cause the electron to move in circular path without slowing or increasing the velocity of the electron.
It was earlier proved that the energy released during nuclear fission of Uranium–235 is magnetic potential energy and not the small electrostatic potential energy which is
times smaller than the magnetic potential energy in the nucleus.
Neither is the energy released an addition of the magnetic and electrostatic potential energy.
In fact, the above three examples shows that the force of attraction on a revolving body is not a sum of the electrical and magnetic force or magnetic and gravitational force.
Mathematical Models to Prove that the Magnetic Potential Energy is Dominant than the Electrical Energy at the Nucleus
To prove why the magnetic field is dominant at the nucleus level, a simple nucleus (Hydrogen) shall be tested mathematically:
Hydrogen: N = 1, proton radius = r1= 0.8775 x 10-15m, radius of atom Ra = 5.27 x 10-11m
If a proton is placed near the proton (nucleus) of a hydrogen atom, at 2r1 away, then the electrostatic force of attraction will be
The escape velocity of this proton is
Also from
Note: From equation (2.2)
At 2r1, Veo = 12528971m/s is taken as the escape velocity Ve, so that
The Magnetic Potential Energy at both points
Where
Results:
EE at 1.755 x 10-15m = 1.3128 x 10-13 J
and at 5.27 x 10-11m = 4.372 x 10-18J
Em at 1.755 x 10-15 = 1.053 x 10-12J
and at 5.27 x 10-11m = 2 x 10-19J
Conclusion: It is seen that the magnetic potential energy is greater than the electrostatic potential energy at the nucleus level, but at the atomic (radius) level the electrical energy is greater than the magnetic potential energy.
Hypothesis of Superior Forces/Energystates: The potential or kinetic energy possessed by a body under a non–contact (electric, magnetic, gravitational or nuclear) field is a measure of the superior force field.
Explanation: Beyond the atomic level magnetic force brings about bending effect, whereas gravitational and electrostatic force causes continuous movement.
Net Energy in the Nucleus
To know the net forces acting on any nucleus, it is assumed that the hypothesis of superior forces does not hold.
Next, the forces (or total force) acting on the nucleus is divided into two—attracting and repulsive forces.
The net force on the nucleus system is therefore
(4.1)
Table 4.1. Division of Forces in a Nucleus
     
But when compared to the other forces/energy gravity is very weak at the atomic level.
Example: The gravitational energy between the protons in a Plutonium-94 nucleus is:
This is negligible when compared with the electrostatic potential energy, which is:
The new equation for the net energy then becomes:
(4.2)
(4.3)
But and Np = number of protons in nucleus
(4.4)
But if the hypothesis of superior energy is put into consideration, the superior repulsive energy is Em. Then the net Energy becomes:
(4.5)
Summary of Net energy acting on nucleus with or without considering the Hypothesis of Superior Energy
Table 4.2. Net energy equation, with and without superior energy
     

5. Hypothesis on Nucleons Contraction, Mass Loss and Binding Energy

The protons of hydrogen combine to helium only if they have enough velocity to overcome each other’s mutual repulsion sufficiently to get within range of the strong nuclear attraction.
If two or more nucleons are to be brought together to be at a distance apart from each other, from Lorentz transformation, if they are brought together at a great speed, which can be compared with the speed of light c, then the length of the particles will appear to be less than its original length.
Mathematical Model
Using Lithium-7 as a case study:
EB = 6.2961 x 10-12J, Rpn = 2.5175475 x 10-15m, rpn = 2.869,
r1 = 0.8775x10-15m
The potential (nuclear) energy holding the protons and neutrons in Lithium-7 nucleus is equal to the Kinetic energy at which they will separate.
From Lorentz transformation, new length due to V is given by the equation:
For the proton and neutron which has escape velocity compared to C, there will be a change in their radius, so that the new radius will be:
Where V= escape velocity of the nucleon
r1 = original radius of the nucleon (proton or neutron)
From
The new radius of the proton will be
But the total self energy of every nuclear in the nucleus is constant provided no external energy disturb or is acting on the system.
The self energy for one nucleon (proton):
Before change in radius = r1
After change in radius = r2
Before change in radius
From equation (3.5) when N>2, = 9.495 x 1026Jm/Kg2
From Lorentz transformation there will be change in radius of the nucleus. This change in radius will lead to a change in mass for the self energy to remain constant since no external force is acting on the system.
So that
If (Mp1- Mp2) is the mass loss between the proton and neutron, total mass loss is given by
This is in agreement with the accepted value 6.2960508 x 10-12J
It is represented thus:
R = uncharged, but r1 and r2 changes

6. Reasons for Nuclear force Being Short Range

Mathematical Models
Using Mercury-201 as a case study, change in mass is 2.83x10-27kg.
From equation (2.7) centre-centre radius apart between a proton and neutron in the nucleus is given by:
Mercury-201 has its electron lowest energy level at 1.51 x 10-10m which is 19829 x the radius of the nucleus
If a neutron is brought to the Mercury nucleus, and placed 2.074523833 m away from any proton, thereby increasing the mass of the nucleus, then the nuclear energy of the nucleus will be:
Where NNn--is the sum of the nucleons in the nucleus and the neutron added to it
Energy between a proton and this neutron added to the nucleus:
Ve is the velocity the added neutron will use to escape from the nucleus to away from the centre of the nucleus, which is the region with the lowest energy level of the electron.
But
Recall that the radius multiple rm, between Mercury-201 nucleus and its empirical atomic radius is 19829
Therefore,
New radius of the neutron at a distance of will be:
To know the mass loss, we first find the self potential energy:
Also
Solving for and when the added neutron is away from a proton in the nucleus:
Recall that the escape velocity at this point is 39129306m/s.
From Lorentz Transformation,
To know the mass loss, we first find the self potential energy:
Also
This binding energy is in agreement with the earlier gotten
Results
Table 5.1. Change in mass, radius and velocity
     
Conclusions
From Lorentz transformation:
i) When V is very small as compared to c, V2/c2 will be negligible in comparison to unity
Therefore, M1=M2 and r1=r2
ii) When V is comparable to c, then will bring about a considerable change in mass and radius
Therefore M1> M2 and r1> r2
For two or more nucleons to be held by nuclear energy, they must be brought under high velocity that is, they will have enough velocity to overcome each other’s repulsion (when there is more than one proton)
- There must be a considerable change in mass
- There must be a considerable change in radius of the nucleons
- The least velocity at which they will be separated after they are joined together, must be comparable to the speed of light
From the table of result above, it can be seen that the loss in mass, change in velocity and radius of the neutron when it is away is very small when compared to when it is from the nearest proton.

7. Discussion: A New Perspective of the Nucleus

It was earlier said that nucleons contract under nuclear force and this lead to a change in mass in order to balance the self energy (potential) of every nucleon in a nucleus. The greater the nuclear binding energy the greater will be the contraction, as well as the loss in mass. The further apart the nucleons are from each other the less loss in mass, as well as the contraction of the nucleons.
If the contracted nucleons are separated outside the nuclear field, they may never regain their original radius. Thus:
i) The centre distance between any two closest nucleons may be unchanged, whereas their radii may change
ii) The nuclear self potential of any nucleon in a nucleus system is constant.

Physical Constant Used

To a good approximate, below are some constant used (all units are in S.I)
Proton Mass = 1.67262178 X 10-27kg
Proton Charge radius = 0.8775X10-15m
Proton Magnetic Moment = 1.410606743 X 10-26J/T
Proton Charge = 1.6X10-19C
Neutron radius = 0.8775 X 10-15m
Electron charge = 1.6 X 10-19C
Planck Constant = 6.626 X 10-34JS
Neutron mass = 1.67492735 X 10-27kg
Electron Mass = 9.1 X 10-31Kg
Atomic Mass Unit = 1.66053892 X 10-27Kg
Vacuum Permittivity = 8.854 X 10-12F/m
Gravitational Constant = 6.7 X 10-11Nm2/kg2

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[43]  en.m.wikipedia.org/wiki/Relativistic_quantum_mechanics
[44]  www.schoolphysics.co.uk/age16-19/Atomic%20physics/Electron%20physics/text/Electron_motion_in_electric_and_magnetic_fields/index.html
[45]  www.physicsoftheuniverse.com/topics_relativity_special.html-ag-physics.org/electron/
[46]  staff.orecity.k12.or.us/les.sitton/Nuclear/313.htm
[47]  applet-magic.com/He4.htm
[48]  en.m.wikipedia.org/wiki/Unifeid_field_theory
[49]  en.m.wikipedia.org/wiki/Uranium-235