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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Radiation Power
Equilibrium as key to
Nuclear Forces
H. Vic Dannon
[email protected]
May 2016
ABSTRACT: Assuming that the Hydrogen Proton is at rest, the
force on the orbiting hydrogen electron is
v e2
1 e2
.
me
=
re
4pe0 re2
However, the accelerating electron radiates energy towards the
proton, and would fall on the proton in a split second if that
energy would not be returned by the proton to the electron. Thus,
to prevent the annihilation of the Hydrogen Atom, the proton
must be accelerating and orbiting the electron, the same as the
sun orbits the earth. Then, the force on the proton is
Vp2
1 e2
Mp
.
=
rp
4pe0 re2
The total ignorance of the proton’s equation, renders baseless any
Atomic analysis done since Bohr to these days. In particular, the
Schrödinger equation does not compensate for the missing
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
electrodynamics here, because that equation too assumes that the
proton is at rest.
We establish here, that the Proton’s Motion is the key to
Quantized Angular Momentum in the nucleus and in the Atom, to
Nuclear Energy, to the Nuclear Forces, and to the Structure of the
Nucleus, and the Atom.
In particular, we emphasize Nuclear Forces because the enigma of
Nuclear Forces have never been resolved. To date there is no
explanation why the closely packed positive protons that repulse
each other at extreme forces inversely proportional to the squared
distance, stay bonded together, and how the neutral neutrons
bond them.
The claims that nuclear forces are the “leaking” of the strong
forces between quarks that were never directly detected, are
speculative if not hallucinated.
And remind more of religious
arguments, than Physics.
Here, perceiving the neutron as a mini-Hydrogen Atom, we
establish the motion of the protons, and the radiation
power
equilibrium between nucleonic electrons and protons as the source
of the Nuclear Forces.
The nuclear force is not a special force that defies electrical
repulsion. It is the result of the motion of the Atomic protons, and
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
the Neutronic protons in the nucleus. The moving protons stay
packed together, bonded by the Neutronic electrons orbitals.
Using the Zinc Nucleus as an example, we compute (with
Numbers, Not Lagrangians) Nuclear Forces, and Energies for the
Zinc Nucleus.
Regarding Angular Momentum, the electron in the n th Hydrogen
orbit, has angular momentum n , and the Quantum of Angular
Momentum for the Hydrogen Atom is  .
This determines the n th orbit radius, the speed of the Hydrogen
electron in the n th orbit, and the frequency of the electron motion
in the n th orbit.
For instance, the force on the nth orbit Hydrogen electron is
vn2
1 e2
= me ,
rn
4pe0 rn2
e2
1 1
,
= (mevnrn )2
4pe0
me rn
And applying mevnrn = n
h
,
2p
e0 h 2
rn = n
.
mee 2 p
2
Thus, quantized angular momentum is essential to determining
the radius, speed, and frequency in electron orbits:
[Born, p.113] expresses the false belief that  is the quantum of
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
angular momentum for any atom:
“It seems now a natural suggestion, that we should
regard the quantization condition for the angular
momentum as an essential feature of the new mechanics.
We therefore postulate that it is universally valid.”
In fact, any Atom has its particular quantum of angular
momentum for its proton orbits, and its particular quantum of
Angular Momentum for its electron orbits, both different from  .
Consequently, the radius, speed, and frequency obtained based on
the assumption that the quantum of angular momentum is  , are
all wrong.
The existing theory requires the quantum of angular momentum
in order to obtain the radius, speed and frequency. And to obtain
that quantum of angular momentum we need to know the radius,
and the speed.
We resolve this paradox by utilizing radiation equilibrium which
is the reason for the quantized angular momentum.
As in [Dan5], we assume that the Neutron is a collapsed Hydrogen
Atom. Consequently, in the Zinc nucleus the 35 neutrons add 35
protons to the 30 atomic protons, and 35 electrons that orbit those
65 protons, within the nucleus boundary. Clearly, the 60 protons
are moving in their own orbits within the Nucleus.
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Gauge Institute Journal, Volume 12, No. 2, May 2016
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H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
The Zinc Nucleus has A=65 Nucleons: Z = 30 Protons, and
A - Z = 35 Neutrons. We assume that each neutron is a mini-
Hydrogen atom made of an electron and a proton.
The Nucleonic electrons orbitals bond the Protons, and the
Neutronic Protons of the Nucleus.
The 30 Protons, the 35 Neutronic Protons and the 35 Nucleonic
electrons constitute the Zinc Nucleus.
The 35 Nucleus electrons at orbit radius
rNuc , encircle the 65
Protons that orbit the center at radius rp .
We approximate the 35 Neutronic electrons by
a charge of 35e,
with mass 35me
orbiting the center at radius rNuc ,
and speed bNucc .
We approximate the 65 protons by
a charge of 65e,
with mass 65M p
orbiting the center at radius rp ,
and speed bpc .
The nucleus is stable because the power radiated by the
accelerating Neutronic electrons towards the Protons, equals the
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
power radiated by the accelerating protons towards the Neutronic
electrons
To
derive
the
orbits
speeds,
radii,
frequencies,
angular
momentums, and energies of the Neutronic electrons, and the
protons in the nucleus, we apply Einstein’s mass-energy equation,
and our Radiation Power Equilibrium between the 35 Neutronic
electrons, and the 65 protons.
ZINC-NUCLEUS RADIATION POWER EQUILIBRIUM
65 protons with mass
30 electrons with mass
65M p , charge 65e, and radius rp
35m e , charge 35e and radius rNuc
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Next, we assume that the Zinc Atom 30 spectral electrons interact
only with the Zinc’s 30 atomic protons, and those 30 protons orbits
define the Zinc’s nucleus boundary.
The 30 Atomic electrons orbitals orbit the 30 atomic protons, and
bond the Zinc Atom.
Since the 30 Atomic electrons do not interact with the interior of
the nucleus,
the 30 Atomic protons appear located at the
boundary of the nucleus, orbiting the center at radius rNuc .
We approximate the 30 atomic electrons by
a charge of 30e,
with mass 30m e
orbiting the center at radius re ,
and speed bec .
We approximate the 30 atomic protons by
a charge of 30e,
with mass 30M p
orbiting the center at radius rNuc ,
and speed bpNucc .
The Zinc atom is stable because the power radiated by the
accelerating atomic electrons towards the Atomic Protons, equals
the power radiated by the accelerating atomic protons towards the
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
atomic electrons
To
derive
the
orbits
speeds,
radii,
frequencies,
angular
momentums, and energies of the 30 Atomic electrons, and the 30
Atomic protons in the nucleus, we apply our Radiation Power
Equilibrium between the 30 Atomic electrons, and the 30 Atomic
protons.
ZINC-ATOM RADIAION POWER EQUILIBRIUM
30 Atomic protons with mass
30M p , and radius rNuc
30 Atomic electrons with mass
30m e , and radius re
In conclusion, we obtain numerical values for orbit speeds, radii,
frequencies, angular momentums, energies, and forces between
electrons and protons, in the nucleus, and in the atom. We sum
these up in the following Summary.
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
SUMMARY OF NUMERICAL RESULTS
Hydrogen Proton nth Orbit
Radius rH,n = n 2rH
Speed VH,n =
me
» n 2 (1.221173735)10-12 m
Mp
1
ö4
1 c æç me ÷÷
1
ç
÷÷ » 334, 528 m/sec
ç
n 137 çè M p ø
n
1
c
æ
ö
M
1 137 ç p ÷÷4
1
4.311097293 ´ 1016 cycles/sec
Frequency
=
çç
»
÷
3
3
÷
2p
n 2prH çè me ø
n
WH,n
1
æ M ö4
Angular Momentum M p rn2Wn = n ççç p ÷÷÷  = n(6.546018057)
èç m e ÷ø
Quantum of Angular Momentum (6.546018057)
Neutron’s Electron nth Orbit
Radius rne,n  n 2 (9.398741807)10-14 m
= n 2 (1.776104665)10-3 rH
Speed vne,n 
=
Frequency
wne,n
2p
1
51, 558,134 m/sec
n
vH
1
n 0.043132065
 8.73067052 ´ 1019 cycles/sec
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Gauge Institute Journal, Volume 12, No. 2, May 2016
=
1
n3
H. Vic Dannon
1.305362686 ´ 104
me
Angular Momentum
1-
2
bne,
n
wH
cycles/sec .
2p
vne,nrne,n = n(0.043132065)
Quantum of Angular Momentum (0.043132065)
Neutron’s Proton nth Orbit
Radius rnp,n  n 2 (2.209505336)10-15 ,
Speed Vnp,n 
Frequency
Wnp
2p
Angular Momentum
1
7, 905,145 m/sec ,
n
 5.69422884 ´ 1020 cycles/second .
Mp
2
1 - bnp,
n
(bnp,nc )rnp,n = n(0.277126027)
Quantum of Angular Momentum (0.277126027)
The Zinc Nucleus
1) is at Radiation Power Equilibrium 
me
2
1 - bNuc
2
rNuc
»
11
Mp
1 - bp2
rp2
Gauge Institute Journal, Volume 12, No. 2, May 2016
2)
Mp
me
1
1
e2
2
bNucrNuc =
=
bp2rp
65 1 - b 2
35 1 - b 2
107
Nuc
p
rNuc
3)
65bp2
Mp
rNuc

rp
me
» 42.5
2
bNuc
 42.5
5)
7)
rp
»
2
35bNuc
4)
6)
H. Vic Dannon
Wp
wNuc
65 2
b
35 p
1
35 æç M p ö÷÷4
ç
÷ » 4.803464029
65 ççè m e ÷ø

rp ÷ö
1 1
e 2 æç
÷÷
65 çç 1 +
Mass-Energy M n - M p - me »
7

2 10
rp çè
rNuc ÷ø
Dm
Zinc Nucleus Proton nth Orbit
Orbit Radius
Speed
Frequency
rp,n  n 2 (1.436178468)10-13
Vp,n 
Wp,n
2p
=
1
n
3
1
5, 801, 257 m/sec
n
6.42885789 ´ 1018 cycles/sec
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Mp
Angular Momentum
Vp,n rp,n = n(13.21701291)
2
1 - bp,
n
Quantum of Angular Momentum (13.21701291)
Nucleus Radiation Equilibrium  Quantized Angular Momentum
Zinc Nucleus Electron nth Orbit
Orbit Radius
rNuc,n » n 2
rp,n
35 2
bNuc,n
 n 2 (6.108688998)10-12
2
65
bp,n
vNuc,n 
Speed
1
51, 560,184 m/sec
n
 Neutron Electron Speed
Frequency
wNuc,n
2p
Angular Momentum
=
1
n
3
1.343341943 ´ 1018 cycles/sec
me
1-
2
bNuc,
n
vNuc,nrNuc,n = n(2.761794253)
Quantum of Angular Momentum (2.761794253)
The Zinc Atom
1) is at Radiation Power Equilibrium 
30me
1 - be2
re2 »
30M p + 35M n
2
1 - bpNu
c
13
rN2uc
Gauge Institute Journal, Volume 12, No. 2, May 2016
2)
30me
1-
be2re
be2
=
be2
re
4)
=
107
re
3)
rNuc
»
30M p + 35M n
2
1 - bpN
uc
2
bpNuc
rNuc
rNuc
2
bpN
uc
Mp
»
me
+
7 Mn
» 63
6 me
1 30 2
bNuc
63 65
be2 
5)
6)
(30e )2
H. Vic Dannon
1
æ M p 7 M ö÷4
n÷
= ççç
+
÷ »
çè me
6 me ø÷
WpNuc
we
63 » 7.937253933
Zinc Atomic Electron nth Orbit
Orbit Radius
1
4, 446,105 m/sec
n
v e,n 
Speed
Frequency
re,n  n 2 (3.903068097)10-10 m
we,n
2p
Angular Momentum
=
1
n
3
1.81298294 ´ 1015 cycles/sec
me
1 - be,2 n
14
v e,nre,n = n(14.99154238)
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Quantum of Angular Momentum (14.99154238)
Zinc Atomic Proton nth Orbit
rNuc,n  n 2 (6.108688998)10-12
Orbit Radius
Speed
Frequency
VpNuc,n 
WpNuc,n

2p
Angular Momentum
1
n
3
(1.439010597)1016 cycles/sec
Mp
1-
1
556, 225 m/sec
n
2
bpNuc,
n
VpNuc,nrNuc,n » n(53.89157159)
Quantum of Angular Momentum (53.89157159)
Atomic Radiation Equilibrium  Quantized Angular Momentum
Zinc Nucleus Zero Point Energy
1
a
- (35)(65)
w
 -268, 285 eV
2
bNuc Nuc
-
1
e2
(35)(65)
 -268, 498 eV
8pe0
rNuc
 19,743´(Hydrogen’s Zero Point Energy)
The Zinc Nucleus Zero Point Energy Frequency
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Gauge Institute Journal, Volume 12, No. 2, May 2016
2
H. Vic Dannon
268,498 eV
 1.298451396 ´ 1020 cycles/sec
h
(35)(65)
a wNuc
 (1.297990381)1020 cycles/sec
bNuc 2p
Zinc Nucleus Nuclear Binding Energy
1
a
- (35)(65) Wp = -11, 415, 889 eV
bp
2
1
e2
(35)(65)
 -11, 420, 788 eV
rp
8pe0
 42.5´(Nucleus Zero Point Energy Binding)
 19349´(Hydrogen’s Nuclear Energy Binding)
Zinc Nucleus Nuclear Force
 1836´(Zinc Nucleus Zero Point Energy Force)
 7,287,084´(Zinc Atomic Zero Point Energy Force)
 694,938,230´(Hydrogen’s Nuclear Force)
Zinc Atomic Nuclear Force
 3969´(Zinc Atomic Zero Point Energy Force)
Zinc Nuclear Binding Energy
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
 19,688  (Hydrogen Binding Energy)
 108 ´(The Zinc Atomic Binding Energy)
Force on a Zinc Nucleus Proton
 2,142  (Force on a Zinc Atomic Proton)
Force on a Zinc Nucleus Electron
 8845  (Force on a Zinc Atomic Electron)
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Keywords: Subatomic, Photon, Neutron Radius, electron Radius,
Composite Particles, Quark, electron, Proton, Proton Radius,
Neutron,
Graviton,
Radiation
Energy,
Kinetic
Energy,
Gravitational Energy, Rotation Energy, Electric Energy, Orbital
Magnetic Energy, Spin Magnetic Energy, Centripetal Force,
Lorentz Force, Electric Charge,
Mass-Energy, Wave-particle,
Inertia Moments, Nuclear Structure, Nucleus Stability, Nuclear
Force, Nuclear Bonding, Orbitals, Electromagnetic Spectrum, X
Ray, Bohr Model, Hydrogen Atom, Radiation Equilibrium,
Neutron Star,
PhysicsAstronomyClassificationScheme:
14.20-c,
14.20.Dh,
14.60Cd, 14.60.Ef, 14.60.Fg, 14.60. 14.65.-q, 14.65.Bt, 14.70.Bh,
47.32.-y, 47.32.C, 03.75.-b, 03.75.Be, 61.05.J-, 61.05.fm,
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Contents
0. Protons’ Motion as key to Quantized Angular Momentum &
Nuclear Energy
I. HYDROGEN
1.
The Hydrogen Electron Quantum of Angular Momentum
2.
Hydrogen Radiation Equilibrium, and Proton Quantized
Angular Momentum
3.
Hydrogen Proton nth Orbit Radius, Speed, and Frequency
II. NEUTRON
4.
The Neutron’s Electron Quantum of Angular Momentum
5.
The Neutron’s Proton Quantum of Angular Momentum
6.
The Neutron’s Proton nth Orbit Radius, Speed, and Frequency
III. ZINC NUCLEUS
7.
Nucleus Radiation Equilibrium
8.
rNuc
rp
9.
The Neutronic Electron Energy
10. The Protons’ Energy
11. The Nucleus Mass-Energy
12. The Protons’ Orbit Radius
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
13. The Protons’ Speed
14. The 1st Orbit Proton Frequency
15. The Nucleus Electron Speed from Radiation Equilibrium
16. The Nucleus Electron Speed from the Force Equation
17. Nucleus Electron 1st Orbit Radius
18. 1st Orbit Nucleus Electron Frequency
19. Nucleus Electron Quantum of Angular Momentum
20. The Nucleus Electron nth Orbit Radius, Speed, and Frequency
21. The Nucleus Proton Quantum of Angular Momentum
22. The Nucleus Proton nth Orbit Radius, Speed, and Frequency
23. Nucleus Radiation Equilibrium is Equivalent to the Proton
Quantized Angular Momentum
IV. ZINC ATOM
24. Atomic Radiation Equilibrium
25.
re
rNuc
26. The Nucleus Speed
27. The Atomic Electrons Speed from Radiation Equilibrium
28. The Atomic Electron Speed from the Force Equation
29. The Atomic Electron 1st Orbit Radius
30. 1st Orbit Atomic Electron Frequency
31. Atomic Electron Quantum of Angular Momentum
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
32. The Atomic Electron nth Orbit Radius, Speed, and Frequency
33. The Atomic Proton Quantum of Angular Momentum
34. Atomic Radiation Equilibrium is Equivalent to the Proton
Quantized Angular Momentum
V. NUCLEAR FORCES
35. Nucleus Zero Point Energy
36. Nucleus Nuclear Energy Binding
37. Nuclear Force and Zero Point Energy Force
38. Gamma Rays Origin is Neutrons’ Protons
39. The Nuclear Binding Energy
40. The Atomic Binding Energy
41. Nuclear Forces vs. Atomic Forces
VI. SUMMARY OF NUMERICAL RESULTS
References
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Protons’ Motion as key to
Quantized Angular Momentum &
Nuclear Forces
Assuming that the Hydrogen Proton is at rest, the force on the
orbiting hydrogen electron is
v e2
1 e2
.
me
=
re
4pe0 re2
However, the accelerating electron radiates energy towards the
proton, and would fall on the proton in a split second if that
energy would not be returned by the proton to the electron. Thus,
to prevent the annihilation of the Hydrogen Atom, the proton
must be accelerating and orbiting the electron, the same as the
sun orbits the earth. Then, the force on the proton is
Vp2
1 e2
.
Mp
=
4pe0 re2
rp
The ignorance of this equation, renders baseless any analysis done
since Bohr to these days. In particular, the Schrödinger equation
does not compensate for the missing electrodynamics here,
because that equation too assumes that the proton is at rest.
Only Proton’s Motion ensures Radiation Power Equilibrium which
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
is key to Quantized Angular Momentum in the nucleus and in the
Atom, to Nuclear Energy, to the Nuclear Forces, and to the
Structure of the Nucleus, and the Atom.
0.1 The Angular Momentum Quantum is h / 2 only for the
Hydrogen Electron.
It can be shown that The electron in a Hydrogen orbit m , has
angular momentum m , and that the
Quantum of Angular
Momentum for the Hydrogen Atom is  .
This enables the determination of the orbit radius, the speed, and
the frequency of the electron in the orbit:
For instance, the force on the nth orbit Hydrogen electron is
vn2
1 e2
= me ,
rn
4pe0 rn2
e2
1 1
,
= (mevnrn )2
4pe0
me rn
And applying mevnrn = n
h
,
2p
e0 h 2
.
rn = n
mee 2 p
2
As shown in [Dan4], already the Hydrogen Proton in its orbits has
a different Quantum of Angular Momentum.
[Born, p.113] expresses the false belief that  is the quantum of
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
angular momentum for any atom:
“It seems now a natural suggestion, that we should
regard the quantization condition for the angular
momentum as an essential feature of the new mechanics.
We therefore postulate that it is universally valid.”
In fact, any Atom has its own quantum of angular momentum for
its proton orbits, and its own quantum of Angular Momentum for
its electron orbits, which is different from  .
For instance,  is NOT the Angular Momentum Quantum of a
Zinc Electron.
The key to determining the Quantum of Angular Momentum is
the Radiation Power Equilibrium mandated by the Proton’s
Motion:
0.2 Nuclear Motion, and Radiation Power Equilibrium as
key to Quantized Angular Momentum
The Force on the Zinc electron is
v e2
1 30e 2
=
,
4pe0 re2
1 - be2 re
me
me
1 - be2
v e2re =
24
1
30e 2 ,
4pe0
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
Assuming
H. Vic Dannon
c2
10
7
30e 2 .
1 - be2 » 1 ,
(m ev e re )2
c2
30e 2 .
»
7
m ere
10
Assuming that mev e re = k  ,
c2
(k  )2
30e 2 .
»
7
m ere
10
The orbit radius is
re »
=
k 2 107  2
30 mec 2e 2
k2
5.29177249 ´ 10-11
30
= k 2 ⋅ 1.763924163 ´ 10-12
Radiation Power equilibrium mandated by the Nucleus motion
enables us to obtain in 25.5,
re  3.848474069 ´ 10-10 m
Therefore, we obtain
k =
3.848474069 ´ 10-10 m
1.763924163 ´ 10-12
,
= 14.77081054
Thus, the Zinc Quantum of Angular Momentum is
 (14.77081054)
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
0.3 Nuclear Orbital Motion as key to Nuclear Energy
The electron radiates its energy onto the proton and, if not
compensated for that lost energy, would fall on the proton in a
split second. To prevent the demise of Atom, the proton must
radiate that energy back onto the electron.
This mandates the acceleration of the proton in an orbit so that
power radiated in both directions will be in equilibrium.
In the Hydrogen Atom, radiation power equilibrium holds when
the Inertia Moments of the electron and the proton are equal:
M p rp2 = m ere2 .
The proton orbit radius rp , is smaller than the electron’s re : in the
Mp
re2
=
» 1836 .
me
rp2
Hydrogen Atom,
Thus, the Nuclear force,
1 e2
4pe0 rp2
is 1836 times greater than
the Atomic force
1 e2
.
4pe0 re2
1 e2
And the Nuclear energy of the proton’s ground orbit,
4pe0 rp
is
1836  42.5 times greater than
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
1 e2
the Zero Point Energy of the electron’s ground orbit,
.
4pe0 re
The protons’ orbitals pack the nuclear energy that bonds the
nucleus. Ignoring the protons’ motion, eliminates 97.7% of the
energy of the Hydrogen Atom. Consequently, any analysis that
attempts to follow Bohr’s method is not even wrong.
On the other hand, the Schrödinger equation is useless here,
because it too assumes that the proton is at rest.
0.4 Radiation Power Equilibrium at the Zinc Nucleus
The Zinc Nucleus has A=65 Nucleons: Z = 30 Protons, and
A - Z = 35 Neutrons. We assume that each neutron is a mini-
Hydrogen atom made of an electron and a proton.
Consequently, the 35 Zinc neutrons add 35 protons to the 30
atomic protons, and 35 electrons that orbit those 65 protons,
within the nucleus boundary. Clearly, the 65 protons are moving
in their own orbits within the Nucleus.
The Nucleonic electrons at orbit radius
rNuc encircle the 65
Protons that orbit the center at radius rp . The electronic orbitals
bond the Protons, and the Neutronic Protons of the Nucleus.
We approximate the 35 Neutronic electrons by
a charge of 35e,
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
with mass 35m e
orbiting the center at radius rNuc ,
and speed bNucc .
We approximate the 65 protons by
a charge of 65e,
with mass 65M p
orbiting the center at radius rp ,
and speed bpc .
ZINC-NUCLEUS RADIATION POWER EQUILIBRIUM
65 protons with mass
30 electrons with mass
65M p , charge 65e, and radius rp
35me , charge 35e and radius rNuc
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H. Vic Dannon
The nucleus is stable because the power radiated by the
accelerating Neutronic electrons towards the Protons, equals the
power radiated by the accelerating protons towards the Neutronic
electrons.
To
derive
the
orbits
speeds,
radii,
frequencies,
angular
momentums, and energies of the Neutronic electrons, and the
protons in the nucleus, we apply Einstein’s mass-energy equation,
and Radiation Power Equilibrium between the 35 Neutronic
electrons, and the 65 protons.
The enigma of Nuclear Forces have never before been resolved. To
date there is no explanation why the closely packed positive
protons, that repulse each other at extreme forces inversely
proportional to the squared distance, stay bonded together, and
how the neutral neutrons bond them.
The claims that nuclear forces are the “leaking” of the strong
forces between quarks that were never directly detected, are
speculative if not hallucinated.
And remind more of religious
arguments, than physics.
Here, perceiving the neutron as a mini-Hydrogen Atom, we
establish the motion of the protons, and the radiation
power
equilibrium between nucleonic electrons and protons as the source
of the Nuclear Forces. We compute (with numbers, Not
Lagrangians) Nuclear Forces, and Energies for the Zinc Nucleus.
29
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
0.5 Radiation Power Equilibrium at the Zinc Atom
Next, we assume that the Zinc Atom 30 spectral electrons interact
only with the Zinc’s 30 atomic protons, and those 30 protons orbits
define the Zinc’s nucleus boundary.
The 30 Atomic electrons orbit the 30 atomic Protons that orbit the
center. The electrons’ orbitals balance the 30 atomic protons, and
bond the Zinc Atom
Since the 30 Atomic electrons do not interact with the interior of
the nucleus,
the 30 Atomic protons appear located at the
boundary of the nucleus, orbiting the center at radius rNuc .
We approximate the 30 atomic electrons by
a charge of 30e,
with mass 30m e
orbiting the center at radius re ,
and speed bec .
We approximate the 30 atomic protons by
a charge of 30e,
with mass 30M p
orbiting the center at radius rNuc ,
and speed bpNucc .
The Zinc atom is stable because the power radiated by the
30
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
accelerating atomic electrons towards the Atomic Protons, equals
the power radiated by the accelerating atomic protons towards the
atomic electrons
ZINC-ATOM RADIAION POWER EQUILIBRIUM
30 Atomic protons with mass
30M p , and radius rNuc
30 Atomic electrons with mass
To
derive
the
orbits
speeds,
30m e , and radius re
radii,
frequencies,
angular
momentums, and energies of the 30 Atomic electrons, and the 30
Atomic protons in the nucleus, we apply our Radiation Power
Equilibrium between the 30 Atomic electrons, and the 30 Atomic
protons.
Our methods here apply to any other Atom.
***********************
31
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Our computations are not too accurate. Most were done by a hand
calculator,
and
sometimes
we
approximated
the
physical
quantities.
For instance, we use for the speed of light
300, 000, 000 m/sec ,
and for the Fine Structure Constant
1/137.
**********************
We account for relativistic speeds by using the relativistic mass
m
1 - b2
, where b = v / c .
32
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
I. HYDROGEN
1.
The Hydrogen Electron Quantum
of Angular Momentum
1.1 The Hydrogen Electron 1st Orbit Angular Momentum
mv1r1 = 
Proof: The Hydrogen electron 1st orbit Angular Momentum is
m w12r12
mv1r1 =
w1
=
1
mv12
w1
v12
1 e2
=
, we have
From m
r1
4pe0 r12
=
1 æç 1 e 2 ö÷
÷÷
ç
2pn1 ççè 4pe0 r1 ÷ø
=
1
( -E1 ) .
2pn1
Since -E1 = hRydberg = hn1 ,
33
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
H. Vic Dannon
1
hn =  . 
2pn1 1
It follows that
1.2 The Hydrogen Electron nth orbit Angular Momentum
mvnrn = n
and that,
1.3
The Hydrogen Electron Angular Momentum
Quantum is  .
1.4 The Hydrogen Electron 1st Orbit Radius, Speed, and
Frequency
rH =
e0 h 2
,
mee 2 p
= 5.29177249 ´ 10-11 m
vH =
e2
c
= ac »
,
2e0h
137
= 2,189, 781 m/sec
34
Gauge Institute Journal, Volume 12, No. 2, May 2016
nH =
H. Vic Dannon
h
4p 2merH2
= 6.58472424 ´ 1015 cycles/sec
Proof: The force on the Hydrogen 1st orbit electron is
vH2
1 e2
= me
,
rH
4pe0 rH2
e2
1 1
,
= (mevHrH )2
4pe0
me rH
Since mevHrH = h / 2p ,
e0 h 2
rH =
,
mee 2 p
vH =
h
,
2pmerH
=
h
,
e0 h 2
2pm e
mee 2 p
e2
c
=
= ac »
2e0h
137
= 2,189, 781 m/sec .
wH
v
= H ,
2p
2prH
35
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
h
2pmerH
=
2prH
h
=
4p 2merH2
Substituting
e0 = 8.854187817 ´ 10-12
Coulomb
,
(Volt)(meter)
h = 6.6260755 ´ 10-34 Joul ,
m e = 9.1093897 ´ 10-31 Kg ,
e = 1.60217733 ´ 10-19 ,
we obtain
rH = 5.29177249 ´ 10-11 m ,
n H = 6.58472424 ´ 1015 cycles/sec .
1.5 The Hydrogen Electron nth Orbit Radius, Speed, and
Frequency
rn = n 2rH ,
vn =
1
v ,
n H
wn
1 wH
.
=
2p
n 3 2p
Proof: The force on the electron in the nth Orbit in the Hydrogen
Atom is
36
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
vn2
1 e2
= me ,
rn
4pe0 rn2
e2
1 1
,
= (mevnrn )2
4pe0
me rn
Since mevnrn = n
h
,
2p
e0 h 2
rn = n
= n 2rH .
2 p
mee
2
vn = n
=n
h
,
2pm ern
h
2pm en 2rH
=
1
h
,
n 2pm erH
=
1
v ,
n H
wn
v
= n ,
2p
2prn
1
vH
n
=
2pn 2rH
=
1 vH
n 3 2prH
=
1 wH
.
n 3 2p
37
,
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
2.
Hydrogen Radiation Equilibrium
and Proton Quantized Angular
Momentum
The Hydrogen Electron orbits at angular velocity wn along a circle
of radius rn a proton charge located a the center of the orbit.
The Hydrogen Proton orbits along a circle of radius rn an electron
charge located at the center of the orbit.
The electron orbits with angular momentums
m wnrn2 = n ,
determine the proton orbit with angular momentums
M Wn rn2 .
At equilibrium, the power radiated by the electron onto the proton
equals the power radiated by the proton onto the electron.
In [Dan4] we showed that
 the hydrogen atom is electrodynamically stable if and only if
the inertia moments of the electron and the proton are equal:
mrn2 = M rn2
38
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
 The Angular velocities of the proton, and the electron are
related by
æM
Wn
= çç
çè m
wn
1
÷÷ö4 =
÷ø
4
1836.152701 = 6.546018057
2.1 Hydrogen’s Radiation Equilibrium is equivalent to the
Proton’s nth Orbit Quantized Angular Momentum
æM
mrn2 = M rn2  M rn2Wn = n çç
çè m
1
ö4
÷÷
÷ø
Proof: () The n th orbit of the Proton has Angular Momentum
W
M rn2 Wn = mrn2 wn n

 w
mrn2
n
n
1
æ M ö4
= n çç ÷÷÷ . 
èç m ø
() The force on the Hydrogen proton in the nth orbit is
Vn2
1 e2
= Mp
,
4pe0 rn2
rn
e2
1 1
= (MVn rn )2
,
M rn
4pe0
2
Since (MVn rn ) =
(M Wn rn2 )2
æ h ÷ö2
= n ç ÷÷
çè 2p ø
2ç
39
M
,
m
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
e0 h 2
Me 2 p
M
,
m
e0 h 2
=n
me 2 p
m
,
M
rn = n 2
2
= n 2rH
= rn
m
M
m
.
M
Squaring both sides, mrn2 = M rn2 . 
2.2 The Hydrogen Proton nth Orbit Angular Momentum
æM
M p rn2Wn = n çç
èç m
1
ö4
÷÷÷ = n(6.546018057)
ø
2.3 The Hydrogen Proton Quantum of Angular Momentum
is the Angular Momentum of the 1st Proton Orbit
M r12W1 = (6.546018057)
40
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
3.
Hydrogen Proton nth Orbit
Radius, Speed, and Frequency
3.1 Hydrogen Proton 1st Orbit Radius, Speed, & Frequency
rH = rH
m
,
M
1
ö4
c æç m
ç
137 çè M
÷÷÷ ,
ø
WH
w æM
= H çç
2p
2p çè m
÷÷ö4 .
÷ø
VH =
1
Proof: The force on the proton in the Hydrogen Atom 1st orbit is
V12
1 e2
= Mp
,
r1
4pe0 r12
e2
1 1
= (MV1r1 )2
,
4pe0
M r1
2
Since (MV1r1 ) =
(M W1r12 )2
æ h ÷ö2
= çç ÷÷
èç 2p ø
r1 =
e0 h 2
M pe 2 p
41
M
,
m
M
,
m
Gauge Institute Journal, Volume 12, No. 2, May 2016
= rH
h æç M
Since MV1r1 = M W1r12 =
ç
2p çè m
H. Vic Dannon
m
.
M
1
÷÷ö4
÷ø
h æç M
V1 =
ç
2p çè m
h
=
2pm
=
1
÷÷ö4 1 ,
÷ø M r
1
M
r
m H
h æç m
ç
2pmrH çè M
æm
= vH çç
çè M
1
ö4
÷÷÷
ø
c æç m
=
ç
137 çè M
1
÷ö÷4
÷ø
W1
V
= 1 ,
2p
2pr1
=
1
æm
vH çç
çè M
ö÷4
÷
ø÷
2prH
m
M
42
æM
çç
m çè m
M
1
ö4
÷÷
÷ø
1
÷÷ö4 ,
÷ø
Gauge Institute Journal, Volume 12, No. 2, May 2016
æM
= n H çç
çè m
H. Vic Dannon
1
÷÷ö4 ,
÷ø
where
rH = 5.29177249 ´ 10-11 m ,
n H = 6.58472424 ´ 1015 cycles/sec .
3.2 Hydrogen Proton nth-Orbit Radius, Speed, & Frequency
rn = n 2rH
m
,
M
1 æm
Vn = vH çç
n çè M
1
÷÷ö4 ,
÷ø
Wn
1 WH çæ M
=
ç
2p
n 3 2p çè m
1
ö4
÷÷÷ .
ø
Proof: The force on the Hydrogen nth orbit proton is
Vn2
1 e2
=M
,
rn
4pe0 rn2
e2
1 1
= (MVn rn )2
,
4pe0
M rn
2
Since (MVn rn ) =
(M Wn rn2 )2
æ h ö÷2
= n ç ÷÷
çè 2p ø
2ç
43
M
,
m
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
e0 h 2
Me 2 p
M
,
m
e0 h 2
=n
me 2 p
m
,
M
rn = n 2
2
= n 2rH
m
.
M
1
h æ M ö4
Since MVn rn = M Wn rn2 = n çç ÷÷÷
2p çè m ø
h æM
Vn = n çç
2p çè m
h
=n
2pm
=
1
ö÷4 1
,
÷
ø÷ M rn
M 2
n rH
m
1 h æç m
ç
n 2pmrH çè M
1 æm
= vH çç
n çè M
1 c æç m
=
ç
n 137 çè M
Wn
V
= n ,
2p
2prn
44
1
ö÷4
÷
ø÷
1
ö÷4
÷
ø÷
1
ö4
÷÷
ø÷
æ
çç M
m çè m
M
1
ö÷4
÷ ,
ø÷
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
=
45
H. Vic Dannon
1
1 æç m
v ç
n H çè M
ö÷4
÷
ø÷
2pn 2rH
m
M
1
ö4
1 WH æç M ÷
ç ÷÷
n 3 2p çè m ø
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
II. NEUTRON
4.
The Neutron’s Electron Quantum
of Angular Momentum
In [Dan5], we established that the Neutron is a CollapsedHydrogen Atom composed of an electron and a proton:
The Neutron’s Electron (to be denoted ne ) has
Orbit Radius rne  9.398741807 ´ 10-14 m ,
Speed vne  51, 558,134 m/sec ,
bne = 0.171860446
Angular Velocity wne  5.485642074 ´ 1020 radians/sec ,
Frequency n ne  8.73067052 ´ 1019 cycles/second .
The Nucleonic Electron Quantum of Angular Momentum is the
Angular Momentum of the electron’s 1st orbit
4.1 Quantum of Neutron’s Electron Angular Momentum
46
Gauge Institute Journal, Volume 12, No. 2, May 2016
me
1-
2
bne
H. Vic Dannon
vnerne = (0.043132065)
Proof:
1
2
1 - bne
=
1
1 - (0.171860446)2
= 1.015103413
me
2
1 - bne
=
vnerne =
(1.015103413)(9.1093897)10-31(51, 558,134)(9.398741807)10-14
(1.05457266)10-34

= (0.042490317) . 
4.2 The Neutron’s Electron 1st Orbit Radius, Speed, and
Frequency
rne  9.398741807 ´ 10-14 m
= 1.776104665 ´ 10-3 rH
vne =
=
c
m/sec
5.909092905
vH
0.043132065
wne
w
= 1.305362686 ´ 104 H
2p
2p
47
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Proof: The force on the Neutron’s 1st orbit electron is
me
2
1 - bne
2
vne
1 e2
,
=
2
rne
4pe0 rne
2
æ
ö
m
÷
e
ç
e
vnerne ÷÷÷
= çç
çè 1 - b 2
4pe0
ø÷
ne
2
Since
me
1-
1
me
2
1 - bne
(bnec )rne = (0.043132065) ,
2
bne
rne = (0.043132065)2
h 2 e0
2
, ( rH = 5.29177249 ´ 10-11 )
1 - bne
2
p mee


rH
= (0.043132065)2
(
)
1 - (0.171860446)2 rH
= 1.776104665 ´ 10-3 rH .
vne =
=
(0.043132065)
me
r
2 ne
1 - bne
(0.043132065)
me
1-
=
1
,
rne
2
bne
(0.043132065)2 rH 1 - bn2e

1
0.043132065 m erH

vH
48
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
1
v
0.043132065 H
=
1
1
c
0.043132065 137
=
1
c
5.909092905
H. Vic Dannon
wne
v
= ne
2p
2prne
1
vH
0.043132065
=
2p(1.776104665) ´ 10-3 rH
vH
105
=
(4.3132065)(1.776104665) 2prH
= 1.305362686 ´ 104
wH
2p
4.3 The Neutron’s Electron nth Orbit Radius, Speed, and
Frequency
rne,n  n 2 (9.398741807)10-14 m
= n 2 (1.776104665)10-3 rH
vne,n =
=
1
c
m/sec
n 5.909092905
vH
1
n 0.043132065
49
Gauge Institute Journal, Volume 12, No. 2, May 2016
wne,n
2p
=
1
n3
H. Vic Dannon
1.305362686 ´ 104
wH
cycles/sec .
2p
Proof: The force on the Neutron’s nth orbit electron is
2
vne,
n
me
2
r
1 - bne,
n ne,n
1 e2
=
,
2
4pe0 rne,
n
æ
ö÷2
ç
me
e
= çç
vne,nrne,n ÷÷÷
ç
2
÷
4pe0
èç 1 - bne,n
ø÷
2
Since
me
2
1 - bne,
n
1
me
2
1 - bne,
n
(bne,nc )rne,n = n(0.043132065) ,
rne,n = n 2 (0.043132065)2
h 2 e0
2
1 - bne,
n ,
2
p mee

rH
= n 2 (0.043132065)2
(
)
1 - (0.171860446)2 rH
= n 2 (1.776104665)10-3 rH .
vne,n =
=
n(0.043132065)
me
n 2rne,n
2
1 - bne,
n
1
n
(0.043132065)
me
2
1 - bne
(0.043132065)2 rH 1 - bn2e
50
1
rne,n
,
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
1
1

n 0.043132065 m erH

vH
=
1
1
v
n 0.043132065 H
=
1
1
1
c
n 0.043132065 137
=
1
1
c
n 5.909092905
wne
v
= ne
2p
2prne
1
1
vH
n
0.043132065
=
2pn 2 (1.776104665) ´ 10-3 rH
=
=
vH
105
n 3 (4.3132065)(1.776104665) 2prH
1
1
n
3
1.305362686 ´ 104 n H
51
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
5.
The Neutron’s Proton Quantum of
Angular Momentum
In [Dan5], we established that the Neutron is a CollapsedHydrogen Atom composed of an electron and a proton:
The Neutron’s Proton (to be denoted np ) has
Orbit Radius rnp  2.209505336 ´ 10-15 ,
Speed Vnp  7, 905,145 m/sec ,
bnp = 0.02635048394
Angular Velocity Wnp  3.577789504 ´ 1021 radians/sec ,
Frequency
Wnp
2p
 5.69422884 ´ 1020 cycles/second .
The Neutron’s Electron nth orbit has angular momentum
me
2
wne,nrne,
n.
2
1 - bne,
n
The Neutron’s Proton nth orbit has angular momentum
Mp
2
1 - bnp,
n
2
Wnp,n rnp,
n.
52
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
At equilibrium, the power radiated by the electron onto the proton
equals the power radiated by the proton onto the electron.
Similarly to [Dan5], we obtain that
 the Neutron is electrodynamically stable if and only if the
inertia moments of the nth orbit electron and proton are
equal:
me
1-
2
bne,
n
2
rne,
n =
Mp
1-
2
bnp,
n
2
rnp,
n
 The Angular velocities of the proton, and the electron are
related by
Wnp,n
wne,n
æM
= çç
èç m
1
÷÷ö4 =
÷ø
4
1836.152701 = 6.546018057
The Neutron’s Proton Quantum of Angular Momentum is the
Angular Momentum of its 1st orbit:
5.1 Quantum of Neutron’s Proton Angular Momentum
Mp
2
1 - bnp
(bnpc )rnp = (0.277126027)
53
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
6.
The Neutron’s Proton nth Orbit
Radius, Speed, and Frequency
6.1 The Neutron’s Proton 1st Orbit Radius, Speed, and
Frequency
rnp = 4.181142667 ´ 10-5 rH
 2.212565574 ´ 10-15 m
Vnp = (3.608466555)vH
= 7, 901, 752 m/sec
n np = (86, 303)n H
= (5.682814561)1020 cycles/sec
Proof: The force on the Neutron’s 1st orbit proton is
Mp
2
Vnp
2 r
1 - bnp
np
1 e2
,
=
2
4pe0 rnp
æ M
ö÷2
ç
e
p
= çç
Vnp rnp ÷÷÷
ç
2
÷
4pe0
çè 1 - bnp
ø÷
2
1
Mp
2
1 - bnp
54
1
,
rnp
Gauge Institute Journal, Volume 12, No. 2, May 2016
Since
Mp
2
1 - bnp
H. Vic Dannon
(bnpc )rnp = (0.277126027) ,
rnp = (0.277126027)2
h 2 e0
2 me
1 - bnp
,
Mp
p mee 2

rH
= (0.277126027)2
(
1 - (0.02635048394)2
1
r
) 1836.152701
= 4.181142667 ´ 10-5 rH .
Vnp =
=
(0.277126027)
Mp
rnp
2
1 - bnp
(0.277126027)
Mp
2
1 - bnp
=
2
(0.277126027)2 rH 1 - bnp
1

0.277126027 m erH

vH
= (3.608466555)vH
= (3.608466555)
1
c
137
= 7, 901, 752 m/sec
55
me
Mp
H
Gauge Institute Journal, Volume 12, No. 2, May 2016
Wnp
2p
=
=
H. Vic Dannon
Vnp
2prnp
(3.608466555)vH
2p(4.181142667) ´ 10-5 rH
= 105
3.608466555 vH
4.181142667 2prH
= (86, 303)
WH
2p
= (86, 303)(6.58472424)1015
= (5.682814561)1020
6.2 The Neutron’s Proton nth Orbit Radius, Speed, and
Frequency
rnp,n  n 2 (2.212565574)10-15 m
= n 2 (4.181142667)10-5 rH
Vnp,n =
=
Wnp,n
2p
=
1
n
3
1
7, 901, 752 m/sec
n
1
(3.608466555)vH
n
(5.682814561)1020 cycles/sec
56
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
1
n3
(86, 303)
H. Vic Dannon
WH
.
2p
Proof: The force on the Neutron’s nth orbit proton is
2
Vnp,
n
Mp
2
r
1 - bnp,
n np,n
1
e2
=
,
2
4pe0 rnp,
n
æ
ö2
M
÷
ç
e
p
Vnp,n rnp,n ÷÷÷
= çç
çç 1 - b 2
÷÷
4pe0
è
ø
np,n
2
1
Mp
1
rnp,n
,
2
1 - bnp,
n
Since
rnp,n
Mp
1-
2
bnp,
n
(bnp,nc )rnp,n = n(0.277126027) ,
me
h 2 e0
2
= n (0.277126027)
1 - bnp,
,
n
p mee 2
Mp

2
2
rH
= n 2 (0.277126027)2
(
1 - (0.02635048394)2
= n 2 4.181142667 ´ 10-5 rH .
Vnp,n =
n(0.277126027)
æ
ö÷
Mp
çç
rnp,n ÷÷÷
çç
2
÷
çè 1 - bnp,
ø÷
n
57
1
r
) 1836.152701
H
Gauge Institute Journal, Volume 12, No. 2, May 2016
n(0.277126027)
=
Mp
2
1 - bnp
=
2
n 2 (0.277126027)2 rH 1 - bnp
1
1

n 0.277126027 m erH

vH
=
1
(3.608466555)vH
n
=
1
1
(3.608466555)
c
n
137
=
1
7, 901, 752 m/sec
n
Wnp,n
2p
H. Vic Dannon
=
Vnp
2prnp
1
(3.608466555)vH
n
=
2pn 2 (4.181142667) ´ 10-5 rH
=
=
=
=
1
n3
1
n3
1
n
3
1
n
3
105
3.608466555 vH
4.181142667 2prH
(86, 303)
wH
2p
(86, 303)(6.58472424)1015
(5.682814561)1020
58
me
Mp
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
III. ZINC NUCLEUS
7.
Nucleus Radiation Equilibrium
The Zinc Nucleus has A=65 Nucleons: Z = 30 Protons, and
A - Z = 35 Neutrons. Each neutron is a mini-Hydrogen atom
made of an electron and a proton.
The Nucleonic electrons orbitals bond the Protons, and the
Neutronic Protons of the Nucleus.
The 30 Protons, the 35 Neutronic Protons and the 35 Nucleonic
electrons constitute the Zinc Nucleus.
The Nucleus electrons at orbit radius rNuc encircle the 65 Protons
that orbit the center at radius rp .
We approximate the 35 Neutronic electrons by
a charge of 35e,
with mass 35m e
orbiting the center at radius rNuc ,
and speed bNucc .
We approximate the 65 protons by
a charge of 65e,
59
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
with mass 65M p
orbiting the center at radius rp ,
and speed bpc .
The nucleus is stable because the power radiated by the
accelerating Neutronic electrons towards the Protons, equals the
power radiated by the accelerating protons towards the Neutronic
electrons
ZINC-NUCLEUS RADIATION POWER EQUILIBRIUM
65 protons with mass
30 electrons with mass
65M p , charge 65e, and radius rp
35m e , charge 35e and radius rNuc
7.1 The Nucleus Electrons Acceleration
60
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
The electrons are attracted to the protons by the force
35me
2
1 - bNuc
a Nuc »
1 (65e )(35e )
,
2
4pe0
rNuc
Thus, the electrons accelerate at
aNuc »
2
1 - bNuc
1 65e 2
,
2
me
4pe0 rNuc
and radiate photons into the Protons fields.
7.2 The Nucleus Electrons Radiation Power
(35e )2
6pe0c
3
2
aNuc
2
2
(35e )2 æçç 1 - bNuc
1 65e 2 ÷÷ö
»
÷÷
ç
2
me
4pe0 rNuc
÷ø
6pe0c 3 çè
7.3 The 65 Protons Acceleration
The 65 Protons of the Nucleus orbit with radius rp , at speed bp ,
the charge of the Nucleus Electron located at the center.
The Protons with mass 65M p , and charge 65e are attracted to the
35 Nucleus electron by the force
65M p
1 - bp2
Ap »
1 (65e )(35e )
4pe0
rp2
Thus, the protons accelerate by
61
Gauge Institute Journal, Volume 12, No. 2, May 2016
Ap »
1 - bp2
Mp
H. Vic Dannon
1 35e 2
,
4pe0 rp2
and radiate photons into the Nucleonic electrons fields.
7.4
The 65 Protons Radiation Power
æ 1 - b2
ö÷2
2
(65e ) 2
(65e ) çç
1 35e ÷
p
÷
Ap »
ç
3
3ç
2 ÷
÷
M
4
pe
6pe0c
6pe0c çè
p
0 rp ÷
ø
2
2
7.5 Nucleus radiation Equilibrium
At equilibrium, the Power radiated by the 35 Nucleus Electrons
onto the 65 Protons, equals the Power radiated by the protons
onto the Nucleus electrons.
2
æ 1 - b2
ö÷2
æ
ö
2
2
2
2
(35e ) çç 1 - bNuc-e 1 65e ÷÷
(65e ) çç
1 35e ÷
p
÷
»
÷
ç
÷
3ç
2
3ç
2 ÷
÷
ç
m
4
pe
M
4
pe
÷
6pe0c è
6pe0c çè
e
0 rNuc-e ø
p
0 rp ÷
ø
2
We use » because we assume one orbit for the protons, and one
orbit for the electrons, while there are 65 proton orbits, and 35
electron orbits.
This Electrodynamics equality leads to a surprisingly purely
mechanical relation between the Relativistic Inertia Moments of
the electron and the Nucleus:
62
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
7.6 The Zinc Nucleus’ Inertia Moments Balance
The Zinc Nucleus is Electrodynamically stable if and only the
Inertia Moments of its 1st orbit electron and 65 Protons are related
by
me
2
1 - bNuc
2
rNuc
»
63
Mp
1 - bp2
rp2
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
8.
rNuc
rp
8.1
The Nucleus Force Balance
Mp
me
e2
1
1
2
bNucrNuc =
=
bp2rp
65 1 - b 2
35 1 - b 2
107
Nuc
p
Proof:
The Force on the 1st orbit nucleus electron is
me
2
1 - bNuc
2
vNuc
1 (65e )(e )
.
=
2
4pe0 rNuc
rNuc
Dividing both sides by c 2 ,
me
1
e2
2
bNucrNuc =
,
2
65 1 - b 2
4
pe
c
0
Nuc
=
=
m0 2
e ,
4p
e2
10
7
The Force on the 1st Orbit Proton is
Mp
Vp2
1 - bp2 rp
=
64
1 (e )(35e )
4pe0 rp2
.
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Dividing both sides by c 2 ,
Mp
e2
1
bp2rp =
,
2
35 1 - b 2
pe
c
4
0
p
e2
=
107
.
Therefore,
Mp
me
1
e2
1
2
bNuc
rNuc =
=
bp2rp . 
7
65 1 - b 2
35 1 - b 2
10
Nuc
p
rNuc
8.2
2
35bNuc
»
rp
65bp2
Proof: Dividing the Nucleus Inertia Moments Balance 7.3,
me
2
1 - bNuc
2
rNuc
»
Mp
1 - bp2
rp2 ,
by the Nucleus Force balance 8.1,
35me
1-
8.3
2
bNuc
2
bNuc-e
rNuc »
rNuc

rp
Mp
me
65
65M p
1-
bp2
bp2rp . 
´ 0.99 » 42.5
Gauge Institute Journal, Volume 12, No. 2, May 2016
Proof:
me
By 2.4,
2
1 - bNuc
rNuc
»
rp
H. Vic Dannon
Mp
2
rNuc
»
rp2 ,
1 - bp2
2
M p 1 - bNuc
.
4
m e 1 - bp2
Even if bNuc = 0.5 , and bp = 0.1 , then,
4
2
1 - bNuc
1 - bp2
we’ll use 0.99 . Then,
rNuc

rp
Mp
me
´ 0.99 =
1836.152701 ´ 0.99
 42.5 . 
2
bNuc
 42.5
8.4
8.5
Proof:
Wp
wNuc
65 2
b
35 p
1
=
35 çæ M p ÷÷ö4
ç
÷ = 4.803464029
65 ççè m e ø÷
Divide the Force Balance,
35m e
2
1 - bNuc
2
3
wNuc
rNuc
=
66
65M p
1 - bp2
W2prp3
» 0.933  1 .
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
by the Inertia Moments Equilibrium,
me
2
1 - bNuc
2
rNuc
»
Mp
1 - bp2
rp2 .
Then,
2
35wNuc
rNuc = 65W2p rp
Wp
wNuc
=
35 rNuc
65 rp
=
35 æç M p ö÷÷4
ç
÷
65 ççè m e ø÷
1
= (0.733799385)(6.546018057)
= 4.803464029 . 
67
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
9.
The Neutronic Electron Energy
9.1 The Neutronic Electron’s Electric Binding Energy in
the 1st orbit
1 65e 2
=4pe0 rNuc
U electric
65e 2
c
=107 rNuc
1
2
9.2 The Neutronic Electron’s Magnetic Energy
U magnetic =
=
1
4
2
m0rNuce 2n Nuc
1
(4p)2
2
m0vNuc
e2
1
rNuc
Proof: The current due to the electron’s charge e that turns n Nuc
cycles/second is
I = en Nuc
The Magnetic Energy of this current is [Benson, p.486]
1
LI 2 .
2
By [Fischer, p.97]
L = m0
2
prNuc
= 12 m0rNuc .
2prNuc
68
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Thus, the magnetic energy due to the electron’s current is
11
m r (en Nuc )2
2 2 0 Nuc
1
4
=
2
m0rNuce 2 n Nuc

1
4 p2
=
1
(4p)2
2
wNuc
2
m0vNuc
e2
1
rNuc
.
9.3 The Neutronic Electron’s Magnetic Energy in its First
Orbit is negligible compared to its Electric Energy
Proof:
1
U magnetic
U electric
=
2
2 1
v
e
m
0 Nuc
rNuc
(4p)2
1 65e 2
4 pe0 rNuc
=
2
1 1 vNuc
65 4p c 2
=
1 1 2
b
65 4p Nuc
Even if bNuc = 0.5 ,
=
0.25
» 3 ´ 10-4 . 
65 ⋅ 4p
9.4 The Neutronic Electron Rotation Energy
69
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
me
1
1 c 2 65e 2
2
vNuc
=
2 1 - b2
2 107 rNuc
Nuc
Proof:
From the balance between the Centripetal and Electric
forces on the electron in its Neutron orbit,
me
2
1 - bNuc
2
vNuc
1 65e 2
,
=
2
rNuc
4pe0 rNuc
me
1
1 1 65e 2
2
vNuc =
.
2 1 - b2
2 4pe0 rNuc
Nuc
Substituting
1
4p
= c 2m0 , and m0 =
,
e0
107
1 c 2 65e 2
.
=
2 107 rNuc
9.5 The Neutronic Electron Total Energy
U electron » m ec 2 +
1 1 2 65e 2
c
2 107 rNuc
Proof:
U electron » mec 2 +
1
107
c2
65e 2 1 c 2 65e 2
rNuc
2 107 rNuc
1 1 2 65e 2
c
= mec +
.
2 107 rNuc
2
70
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
10.
The Protons’ Energy
10.1 The Neutronic Proton Electric Binding Energy
1 35e 2
=4pe0 rp
U electric
=-
1
107
c2
35e 2
rp
10.2 The Neutronic Proton Magnetic Energy
U magnetic =
=
1
4
m0 r6pe 2n p2
1
(4 p)2
m0Vp2e 2
1
rp
Proof: The current due to the Neutronic Proton charge -e that
turns n p cycles/second is
I = -en p
The Magnetic Energy of this current is [Benson, p.486]
1
LI 2 .
2
By [Fischer, p.97]
L = m0
prp2
2prp
71
= 12 m0rp .
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Thus, the magnetic energy due to the Nucleus current is
11
m r (en p )2
22 0 p
=
1
4
m0 rpe 2 n p2

1
4 p2
=
1
(4p)2
W2p
m0Vp2e 2
1
.
rp
10.3 The Neutronic Proton Magnetic Energy in its Orbit
is negligible compared to its Electric Energy
Proof:
1
U magnetic
U electric
=
2 2 1
m
V
e
0
p
rp
(4 p)2
1 35e 2
4 pe0 rp
2
1 1 Vp
=
35 4p c 2
=
1 1
bp2
35 4p
»
0.01
» 2.3 ´ 10-5 . 
35 ⋅ 4p
If bp = 0.01 ,
10.4 The Neutronic Proton Rotation Energy
72
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
1 Mp
1 c 2 35e 2
2
V =
2 1-b 2 p
2 107 rp
p
Proof:
From the balance between the Centripetal and Electric
forces on the Neutronic Proton in its orbit,
Mp
Vp2
1 35e 2
=
,
4pe0 rp2
1 - bp2 rp
1 Mp
1 1 35e 2
2
v =
.
2 1 - b2 p
2 4pe0 rp
p
Substituting
1
4p
= c 2m0 , and m0 =
,
e0
107
=
1 c 2 65e 2
.
2 107 rp
10.5 The Neutronic Proton Total Energy
U Proton
1 1 2 35e 2
c
» M pc +
rp
2 107
2
Proof:
U Proton
35e 2 1 c 2 35e 2
c
» M pc +
rp
2 107 rp
107
2
= M pc 2 +
1
2
1 1 2 35e 2
c
.
rp
2 107
73
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
11.
The Nucleus Mass-Energy
11.1
The Zinc Nucleus Mass-Energy Balance
rp ö÷
1 1
e2 æ
÷÷
M n - M p - me »
65 ççç 1 +

2 107 rp çè
rNuc ÷ø
Dm
Proof:
30M pc 2 + 35M nc 2 = 65U Proton + 35U Neutronic electron
1 1 2 35e 2
1 1 2 65e 2
2
= 65M pc + 65
c
+ 35m ec + 35
c
.
2 107
rp
2 107 rNuc
2
Dividing by c 2 ,
35(M n - M p - me ) »
rp ö÷
1 1
e2 æ
÷÷ . 
35 ⋅ 65 ççç 1 +
2 107
rp çè
rNuc ø÷
74
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
12.
The Protons’ Orbit Radius
12.1
rp ö÷
1 1 e 2 æç
÷÷
çç 1 +
rp » 65 ⋅
2 107 Dm èç
rNuc ø÷
12.2
rp  1.436178468 ´ 10-13
Proof: Substituting
e = -1.60217733 ´ 10-19 C ,
M N » 1.674 128 6 ´ 10-27 Kg ,
M p » 1.672 623 1 ´ 10-27 Kg ,
m e » 9.109 389 7 ´ 10-31 Kg ,
1 1 e2
1 1 (1.60217733)2 ´ 10-38
»
2 107 Dm
2 107 
5.9456103 ´ 10-31 
0.431742422 ´ 10-7
= 2.15871211 ´ 10-15 .
By 4.3,
rp
rNuc

1
» 0.0223529411
42.5
rp » 65 ´ 2.15871211 ´ 10-15 (1 + 0.0223529411)
= 1.436178468 ´ 10-13 . 
75
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
12.3 The Zinc Proton 1st Orbit Radius is
65´(Neutron Proton 1st Orbit Radius)
Proof: By [Dan5], the Proton Orbit Radius in the Neutron is
rp in n = 2.209505336 ´ 10-15
rp
rp in n
=
1.436178468 ´ 10-13
2.209505336 ´ 10-15
= 64.9999999 . 
By [Dan5, 8.3, or Dan4, p.21]
12.4 The Hydrogen Proton 1st Orbit Radius is
rp in H = 1.221173735 ´ 10-12
12.5 The Hydrogen Proton 1st Orbit Radius is about
8.5´(The Zinc Proton 1st Orbit Radius)
Proof:
rp in H
rp
=
1.221173735 ´ 10-12
1.436178468 ´ 10-13
» 8.502938613 . 
76
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
13.
The Protons’ Speed
bp4
13.1
2
2
æ
1 1 e 2 ÷÷ö 2 æç
1 1 e 2 ö÷÷
ç
+ çç 35
÷ b - ç 35
÷ »0
çè 107 M p rp ÷ø p èçç 107 M p rp ø÷
Proof: From the Force Balance, 8.1,
Mp
1 - bp
bp2
bp4
bp2rp
2
=
35e 2
107
1 e2
1 - bp2
» 35
7 M r
10
p p
1
2
2
æ
1 1 e 2 ö÷÷ 2 æç
1 1 e 2 ö÷÷
ç
+ çç 35
÷ b - ç 35
÷ » 0 .
çè 107 M p rp ø÷ p èçç 107 M p rp ø÷
Substituting
e = -1.60217733 ´ 10-19 C ,
M p » 1.672 623 1 ´ 10-27 Kg ,
rp » 1.436178468 ´ 10-13
35
1 e2
1 (1.60217733)2 ´ 10-38
1
» 35
107 M p rp
107 1.672 623 1 ´ 10-27 1.436178468 ´ 10-13
1
= 3.740095636 ´ 10-4 .
77
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
2
2 ö
æ
çç 35 1 1 e ÷÷ » 1.398831537 ´ 10-7
çç 107 M r ÷÷
è
p pø
13.2
bp4 + 1.398831537 ´ 10-7 bp2 - 1.398831537 ´ 10-7 » 0
Using MAPLE
>
with(RootFinding):
>
>
13.3
bp2 = 3.739396286 ´ 10-4
13.4
bp » 0.01933751868
78
Gauge Institute Journal, Volume 12, No. 2, May 2016
13.5
The Zinc Protons Speed
Vp  5, 801, 257 m/sec .
Proof: Vp = bpc
» 0.01933751868c
» 5, 801, 257 m/sec . 
By [Dan5, 9.5],
13.6 The Neutron Proton Speed
Vp in N  7, 905,145 m/sec .
13.7 The Neutron Proton Speed is about
1.36´( The Zinc Protons Speed)
Proof:
Vp-in-N
Vp

7, 905,145
5, 801, 257
» 1.362660713 . 
By [Dan5, 9.6],
13.8 The Hydrogen Proton Speed
Vp in H  330, 420 m/sec
79
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
13.9
The Zinc Proton Speed is about
17.5 ´(Hydrogen Proton Speed)
Proof:
Vp
Vp in H

5, 801, 257
330,420
» 17.55722111 . 
80
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
14.
The 1st Orbit Proton Frequency
14.1 The 1st Orbit Proton Angular Velocity
Wp =
Proof: Wp =
»
Vp
rp
= 4.039370544 ´ 1019 radians/sec
Vp
rp
5, 801, 257 m/sec
1.436178468 ´ 10-13m
= 4.039370544 ´ 1019 radians/sec . 
14.2 The 1st Orbit Proton Frequency is
Wp
2p
Proof:
= 6.42885789 ´ 1018 cycles/sec
Wp
4.039370544 ´ 1019 radians/sec
=
2p
2p
= 6.42885789 ´ 1018 cycles/sec . 
By [Dan5, 10.7],
81
Gauge Institute Journal, Volume 12, No. 2, May 2016
14.3
The Neutron’s Proton Frequency is
Wp in n
2p
14.4
= 5.69422884 ´ 1020 cycles/sec
The Neutron’s Proton Frequency is
88´(the 1st Orbit Protons Frequency)
Wp in n
Proof:
2p
Wp
=
5.69422884 ´ 1020
6.42885789 ´ 1018
2p
 88.57294632 . 
By [Dan5, 10.8]
14.5 The Hydrogen Proton Frequency is
Wp in H
2p
= 4.30634292 ´ 1016 cycles/sec
14.6 The 1st Orbit Proton Frequency is
149´(Hydrogen Proton’s Frequency)
82
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
Wp
Proof:
2p
Wp in H
=
6.42885789 ´ 1018
4.30634292 ´ 1016
2p
 149.2881082 . 
83
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
15.
The Nucleus Electron Speed
from Radiation Equilibrium
15.1
2
bNuc
 42.5bp2
2
 42.5bp2
Proof: By 8.4, bNuc
65
 0.02951452
35
65
35
= 42.5 ⋅ 3.739396286 ´ 10-4
65
35
= 0.02951452 . 
15.2
bNuc  0.171797906
15.3
vNuc  51, 539, 372 m/sec
Proof: vNuc  0.171797906 ⋅ 300, 000, 000 = 51, 539, 372 m / sec
Next, we solve the Force equation for
estimates to bNuc , and to vNuc .
84
2
bNuc
, and obtain closer
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
16.
The Nucleus Electron Speed
from the Force Equation
16.1
4
bNuc
2
2
2 ö
2 ö
æ
æ
65
e
65
e
÷÷ 2
÷÷
ç
+ ççç
÷÷ bNuc - çç
÷ »0
7
7
çè 42.5rpm e 10 ø
çè 42.5rpme 10 ø÷
Proof: The Force on the Nucleus 1st orbit electron is
me
2
1 - bNuc
2
bNuc
2
bNuc
rNuc
= 65
e2
107
,
65 e 2
2
,
1 - bNuc
=
7
rNucm e 10
Squaring both sides,
4
bNuc
æ 65 e 2 ÷ö2
æ 65 e 2 ÷ö2
2
÷÷ bNuc
÷ »0
+ ççç
- ççç
7
÷
çè rNucm e 107 ÷÷ø
èç rNucm e 10 ø
Since rNuc  42.5rp
4
bNuc
2
2
2 ö
2 ö
æ
æ
65
e
65
e
÷÷ 2
÷÷
ç
+ ççç
÷÷ bNuc - çç
÷ » 0 .
7
7
çè 42.5rpm e 10 ø
çè 42.5rpm e 10 ø÷
85
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
e2
65(1.60217733)2 ´ 10-38

107 (42.5rp )m e
107 (42.5)(1.436178 ´ 10-13 )(9.109 389 7 ´ 10-31 )
65
= 3.0008759 ´ 10-2 .
2
æ 65
ö÷2
e
çç
-4
÷
çç 107 (42.5r )m ÷÷  9.005256346 ´ 10 .
è
p
eø
4
2
bNuc
+ (9.005256346 ´ 10-4 )bNuc
- (9.005256346 ´ 10-4 ) » 0 .
16.2
4
2
bNuc
+ (9.005256346)10-4 bNuc
- (9.005256346)10-4 » 0
2
º x , we seek the zeros of the polynomial
Denoting bNuc
f (x ) = x 2 + (9.005256346 ´ 10-4 )x - (9.005256346 ´ 10-4 )
between x = 0 , and x = 1 .
We used Maple Root-Finding Program.
Maple Input and Output follow:
> with(RootFinding):
> >
>
86
Gauge Institute Journal, Volume 12, No. 2, May 2016
2
bNuc
 0.02956187425
16.3
2
Compare with 15.1 bNuc
 0.02951452
bNuc  0.1719356689
16.4
Compare with 15.2 bNuc  0.171797906
16.5
The Nucleus Electrons Speed
vNuc  51, 560,184 m/sec .
Compare with 15.3 vNuc  51, 539, 372 m/sec
Proof: bNucc  0.1719356689c
» 51, 560,184 m/sec . 
By [Dan5, 11.5],
16.6 The Neutron Electron Speed
ve  51, 558,134 m/sec .
87
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
16.7 Neutron Electron Speed  (Nucleus Electrons Speed)
Proof:
51, 558,134
» 1.
51, 560,184
By [Dan5, 11.6],
16.8 The Hydrogen Electron Speed
vH  a
 c = 2,189, 781 m/sec .
»
1
137
Thus,
16.9
Zinc Nucleus Electron Speed
 23.5´(Hydrogen Electron Speed)
Proof:
51, 560,184
» 23.5458176 . 
2,189, 781
88
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
17.
Nucleus Electron 1st Orbit Radius
17.1 1st Orbit Radius of the Zinc Nucleus Electron
rNuc »
Proof: Substitute
35 2 rp
bNuc
 6.108688998 ´ 10-12
2
65
bp
rp » 1.436178468 ´ 10-13 ,
bp2 = 3.739396286 ´ 10-4 ,
2
bNuc
 0.02953836208 . 
By [Dan5, 12.1],
17.2 Neutron Electron 1st Orbit Radius
RN  9.398741807 ´ 10-14
17.3 Zinc Nucleus Electron 1st Orbit Radius
is  65´ (Neutron Electron Orbit Radius)
Proof:
6.108688998 ´ 10-12
9.398741807 ´ 10-14
= 64.99475274
89
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
By 2.3,
17.4
The Zinc Nucleus Electron Orbit Radius is
 42.5´(The Zinc Proton Orbit Radius)
17.5
The Hydrogen Electron Orbit Radius
rH = 5.29277249 ´ 10-11 m .
17.6 The Hydrogen Electron Orbit Radius is about
8.7´( Zinc Nucleus Electron Orbit Radius)
Proof:
5.29277249 ´ 10-11
6.108688998 ´ 10-12
= 8.664334511 . 
90
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
18.
1st Orbit Nucleus Electron
Frequency
18.1 The 1st Orbit Zinc Nucleus Electron Angular Velocity
wNuc =
Proof:
wNuc =
vNuc
= 8.440466361 ´ 1018 radians/sec
rNuc
vNuc
rNuc
»
51, 560,184 m/sec
6.108688998 ´ 10-12 m
= 8.440466361 ´ 1018 radians/sec . 
18.2 The 1st Orbit Nucleus Electron Frequency
wNuc
= 1.343341943 ´ 1018 cycles/sec
2p
Proof:
wNuc
8.440466361 ´ 1018
=
2p
2p
= 1.343341943 ´ 1018 cycles/sec . 
91
Gauge Institute Journal, Volume 12, No. 2, May 2016
18.3
H. Vic Dannon
1st Orbit Neutron Electron Frequency is
 65´ (1st Orbit Zinc Nucleus Electron Frequency)
Proof:
we in n
20
2p = 5.485642074 ´ 10
wNuc
8.440466361 ´ 1018
2p
 64.99216796 . 
18.4 1st Orbit Hydrogen Electron Frequency
we in H
= 6.58472424 ´ 1015
2p
Proof:
we in H =
v e in H
2,189,781 m/sec
»
re in H
5.2977249 ´ 10-11 m
= 4.1334366 ´ 1016 radians/sec
we in H
4.137304269 ´ 1016 radians/sec
=
2p
2p
= 6.58472424 ´ 1015 . 
18.5
1st Orbit Zinc Nucleus Electron Frequency is
 204´(1st Orbit Hydrogen Electron Frequency)
92
Gauge Institute Journal, Volume 12, No. 2, May 2016
Proof:
wNuc
18
2p = 1.343341943 ´ 10
we in H
6.58472424 ´ 1015
2p
 204.1997296 . 
93
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
19.
Nucleus Electron Quantum of
Angular Momentum
We associated with the Zinc Nucleus Electron
Orbit Radius rNuc  6.108688998 ´ 10-12 m ,
Speed vNuc  51, 560,184 m/sec ,
bNuc  0.1719356689 .
To obtain these, we averaged the 35 electrons orbits into one orbit.
Now, we will consider that one orbit as the 1st orbit of the
electrons in the Nucleus, and construct on it the 2nd, 3rd,
4th,…,orbits.
The Zinc Nucleus Electron nth orbit has angular momentum
me
1-
2
bNuc,
n
2
wNuc,nrNuc,
n.
The Nucleus Electron Quantum of Angular Momentum is the
Angular Momentum of its 1st orbit:
19.1 Zinc Nucleus Electron Quantum of Angular
Momentum
94
Gauge Institute Journal, Volume 12, No. 2, May 2016
me
2
1 - bNuc
H. Vic Dannon
vNucrNuc = (2.761794253)
Proof:
1
2
1 - bNuc
=
1
1 - (0.1719356689)2
» 1.015116939
me
1=
2
bNuc
vNucrNuc =
(1.015116939)(9.1093897)10-31(51, 560,184)(6.108688998)10-12
(1.05457266)10-34

= (2.761794253) . 
19.2 Angular Momentum of the nth orbit Nucleus electron
me
1-
2
bNuc,
n
vNuc,nrNuc,n = n(2.761794253)
95
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
20.
The Nucleus Electron nth Orbit
Radius, Speed, and Frequency
20.1 The Zinc Nucleus Electron 1st Orbit Radius, Speed,
and Frequency
rNuc  6.108688998 ´ 10-12 m
=
1
2
(2.761794253)2 1 - bNuc
rH
65
= (0.110142785)rH
vNuc  51, 560,184 m/sec
=
65
v
2.761794253 H
= (23.53542445)vH
wNuc
= 1.343341943 ´ 1018 cycles/sec
2p
=
652
(13.21701291)3
96
1
2
1 - bNuc-e
vH
2prH
Gauge Institute Journal, Volume 12, No. 2, May 2016
= (203.5958043)
H. Vic Dannon
vH
.
2prH
Proof: The force on the Nucleus 1st orbit electron is
me
2
1 - bNuc
2
vNuc
1 65e 2
=
,
2
4pe0 rNuc
rNuc
2
æ
ö
m
÷
65e
ç
e
= çç
vNucrNuc ÷÷÷
çè 1 - b 2
4pe0
ø÷
Nuc
2
Since
rNuc
me
2
1 - bNuc
1
me
2
1 - bNuc
vNucrNuc = (2.761794253) ,
2 e
1
2h
2
0
=
,
(2.761794253)
1 - bNuc
p mee 2
65

rH
=
1
2
(2.761794253)2 1 - bNuc
rH
65
=
1
(2.761794253)2 1 - (0.1719356689)2 rH
65  
7.627507496
0.985108179
= (0.110142785)rH .
vNuc =
(2.761794253)
me
rNuc
2
1 - bNuc
97
1
rNuc
,
Gauge Institute Journal, Volume 12, No. 2, May 2016
(2.761794253)
=
me
2
1 - bNuc
=
H. Vic Dannon
æ1
ö
çç (2.761794253)2 1 - b 2 r ÷÷
Nuc H ÷
çè 65
ø
65

2.761794253 m erH

vH
=
65
v
2.761794253 H
= (23.53542445)vH
wNuc
v
= Nuc
2p
2prNuc
=
=
65
v
2.761794253 H
æ1
ö
2
2p çç (2.761794253)2 1 - bNuc
rH ÷÷÷
çè 65
ø
652
1
(2.761794253)3
2
1 - bNuc
vH
2prH

We in H
2p
= (203.5958043)
vH
2prH
20.2 The Nucleus Electron nth Orbit Radius, Speed, and
Frequency
98
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
rNuc,n = n 2rNuc ,
 n 2 (6.108688998 ´ 10-12 )m
vNuc,n =
1
v ,
n Nuc

1
51, 560,184 m/sec
n
=
1 wNuc
,
n 3 2p
wNuc,n
2p
=
1
n
3
(1.343341943)1018
Proof: The force on the Nucleus nth orbit electron is
me
2
vNuc,
n
2
r
1 - bNuc,
n Nuc,n
1 65e 2
=
,
2
4pe0 rNuc,
n
æ
ö÷2
ç
me
65e
= çç
vNuc,nrNuc,n ÷÷÷
ç
2
÷÷
4pe0
çè 1 - bNuc,
ø
n
2
Since
me
2
1 - bNuc,
n
rNuc,n =
2
1 - bNuc,
n
vNuc,nrNuc,n = n(2.761794253) ,
1 2
h 2 e0
2
n (2.761794253)2
1 - bNuc,
n ,
2
65
p mee

rH
=
1
me
1 2
2
n (2.761794253)2 1 - bNuc,
n rH
65
99
1
rNuc,n
,
Gauge Institute Journal, Volume 12, No. 2, May 2016
= n2
H. Vic Dannon
1
(2.761794253)2 1 - (0.1719356689)2 rH
65  
7.627507496
0.985108179
= n 2 (0.110142785)rH ,
= n 2rNuc .
vNuc,n =
=
n(2.761794253)
me
rNuc,n
2
1 - bNuc,n
n(2.761794253)
me
2
1 - bNuc,
n
=
æ n2
ö
çç (2.761794253)2 1 - b 2 r ÷÷
Nuc,n H ÷
çè 65
÷ø
1
65

n 2.761794253 m erH

vH
wNuc,n
2p
=
1
65
v
n 2.761794253 H
=
1
(23.53542445)vH
n
=
1
v .
n Nuc
=
v Nuc,n
2prNuc,n
100
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
=
H. Vic Dannon
1
65
v
n 2.761794253 H
æ n2
ö
2
2p çç (2.761794253)2 1 - bNuc
rH ÷÷÷
çè 65
÷ø
1
652
1
n 3 (2.761794253)3
2
1 - bNuc
vH
2prH

We in H
2p
=
=
1
n3
(203.5958043)
vH
2prH
1 wNuc,n
n 3 2p
101
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
21.
The Nucleus Proton Quantum of
Angular Momentum
We associated with the Zinc’s Nucleus Proton
Orbit Radius rp  1.436178468 ´ 10-13 ,
Speed Vp  5, 801, 257 m/sec ,
bp » 0.01933751868 .
To obtain these, we averaged the 65 protons orbits into one orbit.
Now, we will consider that one orbit as the 1st orbit of the protons
in the Nucleus, and construct on it the 2nd, 3rd, 4th,…,orbits.
The Nucleus Electron nth orbit has angular momentum
me
2
1 - bNuc,
n
2
wNuc,nrNuc,
n.
The Nucleus Proton nth orbit has angular momentum
Mp
1-
2
bp,
n
2
Wp,n rp,
n.
The Zinc Nucleus Proton Quantum of Angular Momentum is the
Angular Momentum of its 1st orbit:
102
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
21.2 Quantum of Zinc Nucleus Proton Angular Momentum
Mp
1-
bp2
Vp rp = (13.21701291)
Proof:
1
1 - bp2
=
1
1 - (0.01933751868)2
» 1.000187022
Mp
1 - bp2
=
Vp rp
1
 =

(1.000187022)(1.6726231)10-27 (5, 801, 257)(1.436178468)10-13
(1.05457266)10-34
= (13.21701291) . 
21.3 The Angular Momentum of the nth orbit proton
Mp
1-
2
bp,
n
Vp,n rp,n = n(13.21701291)
103

Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
22.
The Nucleus Proton nth Orbit
Radius, Speed, and Frequency
22.1 The Zinc Nucleus Proton 1st Orbit Radius, Speed, and
Frequency
rp  1.436178468 ´ 10-13m
=
me
1
(13.21701291)2 1 - bp2
(r
)
35
M p p in H
= (0.116456295)rp in H
Vp  5, 801, 257 m/sec
=
1
ö4
æ Mp ÷
35
çç
÷÷ Vp in H
ç
13.21701291 çè me ø÷
= (17.33452434)Vp in H
Wp
2p
= 6.42885789 ´ 1018 cycles/sec
104
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
3
æ M p ö÷4
35
çç
÷÷
=
3ç
ç
m
(13.21701291) è e ø÷
2
1
1-
bp2
(n p in H )
» (148.85)n p in H
Proof: The force on the Nucleus 1st orbit proton is
Mp
Vp2
1 35e 2
=
,
2
2 r
pe
4
r
1 - bp p
0
p
æ M
ö÷2
ç
35e
p
= çç
Vprp ÷÷÷
ç
2
÷
4pe0
èç 1 - bp
ø÷
2
1
Mp
1
,
rp
1 - bp2
Since
rp =
Mp
1-
bp2
(bpc )rp = (13.21701291) ,
m
1
h 2 e0
(13.21701291)2
1 - bp2 e ,
35
p mee 2
Mp


rH
=
me
me
1
,
(13.21701291)2 rH
1 - bp2
35
Mp
Mp

rH
=
me
1
(13.21701291)2 1 - bp2
r
35
Mp H
=
1
1
(13.21701291)2 1 - (0.01933751868)2
rH
35   
1836.152701

174.6894303
0.999813012
105
0.02333703
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
= (0.116456295)rH .
Vp =
(13.21701291)
Mp
r
2 p
1 - bp
=
=
(13.21701291)
æ1
ö
çç (13.21701291)2 r 1 - b 2 m e ÷÷
÷
H
p
ç
M p ø÷
1 - bp2 çè 35
Mp
35

13.21701291 m erH

vH
1
æ M p ÷ö4
35
çç
÷÷
=
ç
13.21701291 èç me ÷ø

1
æ
ö
çç me ÷÷4 v
çç M ÷÷ H
è pø

VH
6.546018057
= (17.33452434)VH
1
ö4
Vp
2prp
æ Mp ÷
35
çç
÷÷ V
13.21701291 ççè me ÷ø H
=
2p
me
1
r
(13.21701291)2 1 - bp2
Mp H
35
3
æ M p ö÷4
35
çç
÷
=
ç
3çm ÷
÷
(13.21701291) è e ø
2
VH
prH
1 - bp2 2
1
Wp in H
2p
106
Gauge Institute Journal, Volume 12, No. 2, May 2016
= (148.8500409)
H. Vic Dannon
Wp in H
2p
22.2 The Zinc Nucleus Proton nth Orbit Radius, Speed, and
Frequency
rp,n = n 2rp ,
 n 2 (1.436178468 ´ 10-13 )m
1
V ,
n p
Vp,n =
1
5, 801, 257 m/sec
n

Wp,n
2p
=
=
1 Wp
,
n 3 2p
1
n3
6.42885789 ´ 1018 cycles/sec
Proof: The force on the Nucleus nth orbit proton is
Mp
Vp,2 n
2 r
1 - bp,
n p,n
1 35e 2
=
,
2
4pe0 rp,
n
æ M
ö2
÷
ç
35e
p
= çç
Vp,n rp,n ÷÷÷
ç
2
÷
4pe0
èç 1 - bp,n
ø÷
2
1
Mp
2
1 - bp,
n
Since
Mp
2
1 - bp,
n
Vp,n rp,n = n(13.21701291) ,
107
1
rp,n
,
Gauge Institute Journal, Volume 12, No. 2, May 2016
rp,n = n 2
H. Vic Dannon
1
h 2 e0
2 me
.
(13.21701291)2
1 - bp,
n
35
Mp
p mee 2

rH
= n 2 rp
Then,
Vp,n =
n(13.21701291)
æ
÷÷ö
çç M p
rp,n ÷÷
çç
2
÷
èç 1 - bp,n
ø÷
n(13.21701291)
=
Mp
2
1 - bp,
n
=
æ1
ö
2 me ÷
÷
ççç n 2 (13.21701291)2 rH 1 - bp,
÷
n
çè 35
M p ÷ø
1
35

n (13.21701291) m erH

vH
1
æ M p ÷ö4
1
35
çç
÷÷
=
n 13.21701291 ççè me ÷ø

1
æ
ö
çç me ÷÷4 v
çç M ÷÷ e in H
è pø

Vp in H
6.546018057
=
1
(17.33452434)Vp in H
n
=
1
V
n p
That is,
the proton’s nth orbit speed is at most its 1st orbit speed,
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Gauge Institute Journal, Volume 12, No. 2, May 2016
bp,n £ bp ,
2
1 - bp,
n » 1,
rp,n » n 2
m
1
(13.21701291)2 rH e
35
Mp
= n 2 rp
Wp,n
2p
=
Vp,n
2prp,n
1
Vp
n
=
2pn 2rp
=
1 Wp
n 3 2p
109
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
23.
Nucleus Radiation Equilibrium is
Equivalent to the Proton
Quantized Angular Momentum
At equilibrium,
23.1
1) the Nucleus is electrodynamically stable if and only if the
inertia moments of the nth orbit electron and the nth orbit
proton are equal:
me
2
1 - bNuc,
n
2
rNuc,
n =
Mp
2
1 - bp,
n
2
rp,
n
2) The Angular velocities of the proton, and the electron are
related by
Wp,n
wNuc,n
1
=
35 æç M p ö÷÷4
ç
÷ = 4.803464029
65 ççè m e ÷ø
23.2 The nth Orbit Radiation Equilibrium is equivalent to
the nth Orbit Quantized Angular Momentum
110
Gauge Institute Journal, Volume 12, No. 2, May 2016
2
m erNuc,
n
2
1 - bNuc,
n

Proof:
=
H. Vic Dannon
2
M p rp,
n
2
1 - bp,
n
2
M p rp,
n Wp,n
2
1 - bp,
n
1
35 æç M p ö÷÷4
= n(2.761794253)
ç
÷
65 ççè m e ÷ø
The nth Orbit Radiation Equilibrium is
2
m erNuc,
n
1-
2
bNuc,
n
=
2
M p rp,
n
2
1 - bp,
n
 the Proton’s n th orbit has Angular Momentum
Mp
2
rp,
n Wp,n =
2
1 - bp,
n


me
2
1-bNuc,
n
2
rNuc,
n
me
2
rNuc,
n wNuc,n
Wp,n
2
wNuc,n
1 - bNuc,
n

  1
n (2.761794253)
35 æç M p ö÷÷4
ç
÷
65 ççè m e ÷÷ø
1
35 æç M p ÷÷ö4
= n(2.761794253)
ç
÷ .
65 ççè m e ÷ø
111
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
IV. ZINC ATOM
24.
Zinc Atom Radiation Equilibrium
The Zinc Nucleus has A=65 Nucleons: Z = 30 Protons, and
A - Z = 35 Neutrons.
The 30 Atomic electrons orbitals balance the 30 atomic protons,
and bond the Zinc Atom
The 30 Atomic electrons orbit the 30 atomic Protons that orbit the
center at radius re .
Since the 30 Atomic electrons do not interact with the interior of
the nucleus,
the 30 Atomic protons appear located at the
boundary of the nucleus, orbiting the center at radius rNuc .
We approximate the 30 atomic electrons by
a charge of 30e,
with mass 30m e
orbiting the center at radius re ,
and speed bec .
We approximate the 30 atomic protons by
a charge of 30e,
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Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
with mass 30M p
orbiting the center at radius rNuc ,
and speed bpNucc .
The Zinc atom is stable because the power radiated by the
accelerating atomic electrons towards the Atomic Protons, equals
the power radiated by the accelerating atomic protons towards the
atomic electrons
ZINC-ATOM RADIAION POWER EQUILIBRIUM
30 Atomic protons with mass
30M p , and radius rNuc
30 Atomic electrons with mass
30m e , and radius re
24.1 The Atomic Electrons Acceleration
The Atomic electrons are attracted to the Atomic protons by the
force
113
Gauge Institute Journal, Volume 12, No. 2, May 2016
30me
1 - be2
ae »
H. Vic Dannon
1 (30e )(30e )
,
4pe0
re2
Thus, the Atomic electrons accelerate at
ae »
1 - be2 1 (30e )2
,
30me 4pe0 re2
and radiate photons into the Atomic Protons fields.
24.2 The Atomic Electrons Radiation Power is
2
æ
ö
2
2
(30e ) 2
(30e ) çç 1 - be 1 (30e ) ÷÷
ae »
÷
ç
6pe0c 3
6pe0c 3 çè 30m e 4pe0 re2 ø÷÷
2
2
24.3 The Atomic Protons Acceleration
The Atomic protons orbit the charge of 30 Atomic Electrons,
located at the center of a circle of Radius rNuc , at speed bpNuc .
The Nucleus mass of 30M p + 35M n , and charge 30e is attracted to
the Atomic electrons by the force
30M p + 35M n
2
1 - bpNuc
Ap&n »
1 (30e )(30e )
,
2
4pe0
rNuc
The Nucleus accelerate towards the Atomic electrons at,
Ap&n »
2
1 - bpNuc
1 (30e )2
,
2
30M p + 35M n 4pe0 rNuc
114
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
and radiates photons into the Atomic electrons fields.
24.4 The Atomic Protons Radiation Power
2
(30e )
6pe0c
3
2
Ap&n
æ
ö2
2
2
1
b
÷
(30e ) çç
1 (30e ) ÷
pNuc
»
ç
÷÷÷
2
6pe0c 3 ççè 30M p + 35M n 4pe0 rNuc
÷ø
2
24.5 Zinc Atom Radiation Equilibrium
At equilibrium, the Power radiated by the Atomic Electrons onto
the Atomic Protons, equals the Power radiated by the Atomic
Protons onto the Atomic electrons.
2
ö2
2
æ
2
2ö
2 æ
2÷
1
b
ç
1
b
(30e ) çç
1 (30e ) ÷÷
(30e ) ç
1 (30e ) ÷
pNuc
e
÷÷
»
÷
ç
ç
÷
2
4pe0 re2 ø÷
6pe0c 3 èç m e
6pe0c 3 èçç 30M p + 35M n 4pe0 rNuc
÷ø÷
2
We use » because we assume one orbit for the Atomic protons,
and one orbit for the Atomic electrons, while there are 30 proton
orbits, and 30 electron orbits.
This Electrodynamics equality leads to a surprisingly purely
mechanical relation between the Relativistic Inertia Moments of
the Atomic electrons and the Nucleons:
24.6 The Zinc Atom Inertia Moments Balance
The Zinc Atom is Electrodynamically stable if and only the Inertia
115
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H. Vic Dannon
Moments of its 30 electrons and 65 Nucleons are related by
30m e
1-
be2
re2 »
30M p + 35M n
1-
116
2
bpNuc
rN2uc
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
25.
re
rNuc
25.1 The Zinc Atom Force Balance
30m e
1-
Proof:
b 2r
2 e e
be
=
(30e )2
10
7
=
30M p + 35M n
1-
2
bpNu
c
The Force on an Atomic electron is
v e2
1 30e 2
=
.
2
2 r
4
pe
r
1 - be e
0
e
me
Dividing both sides by c 2 ,
be2
1 30e 2
=
,
2
2
2 r
4
c
r
pe
1 - be e
0
e
me
=
=
30m e
1-
be2re
be2
117
m0 30e 2
,
4p re2
1 30e 2
107 re2
=
,
(30e )2
107
.
2
bpNuc
rNuc
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
The Force on the Nucleons is
2
30M p + 35M n VpNuc
rNuc
2
1 - bpNuc
1 (30e )2
=
2
4pe0 rNuc
Dividing both sides by c 2 ,
2
30M p + 35M n bpNuc
rNuc
2
1 - bpNuc
=
=
30M p + 35M n
1-
2
bpNuc
(30e )2
1
2
4pe0c 2 rNuc
1 (30e )2
2
107 rNuc
2
bpNuc
rNuc =
,
(30e )2
107
,
.
Therefore,
30m e
1-
be2re
be2
=
(30e )2
10
re »
25.2
7
30M p + 35M n
=
2
1 - bpNuc
rNuc
2
bpNuc
2
bpNuc
rNuc . 
be2
Proof: Divide the Nucleus Inertia Moments Balance 23.3,
30m e
1-
be2
re2 »
30M p + 35M n
1-
by the Force balance 24.1,
118
2
bpNu
c
rN2uc ,
Gauge Institute Journal, Volume 12, No. 2, May 2016
30m e
1 - be2
be2re =
re
25.3
rNuc
H. Vic Dannon
30M p + 35M n
2
1 - bpNuc
30M p + 35M n
»
30m e
4
2
bpNuc
rNuc . 
1 - be2
2
1 - bpNuc
,
Proof: From the Inertia Moments Balance, 23.3,
30m e
1 - be2
re
rNuc
»
re2 »
2
1 - bpNu
c
30M p + 35M n
30m e
re
25.4
30M p + 35M n
»
rNuc
Mp
me
+
4
rN2uc ,
1 - be2
2
1 - bpNuc
.
7 Mn
6 me
 63 .
Proof:
assuming electrons’ speed, and Nucleus speed, much
smaller than light speed,
4
1 - be2
2
1 - bpNuc
Therefore,
119
» 1.
Gauge Institute Journal, Volume 12, No. 2, May 2016
re
rNuc
H. Vic Dannon
Mp
»
me
+
7 Mn
6 me
7
1836.152701 + 1838.683662
6
=
= 63.09741389 . 
re  3.848474069 ´ 10-10 m
25.5
Proof: re  63rNuc
 63(6.108688998 ´ 10-12 m)
= 3.848474069 ´ 10-10 m
25.6
Proof:
WpNuc
we
1
æ M p 7 M ö÷4
n÷
= ççç
+
÷ = 7.937253933
çè me
6 me ø÷
Divide the Force Balance,
30m e
1 - be2
we2re3 =
30M p + 35M n
2
1 - bpNuc
3
W2pNucrNuc
by the Inertia Moments Equilibrium,
30m e
1-
be2
re2 »
30M p + 35M n
1-
120
2
bpNu
c
rN2uc .
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Then,
we2re = W2pNucrNuc
WpNuc
we
=
re
rNuc
1
æM
7 M n ö÷÷4
= ççç p +
÷
çè m e
6 m e ø÷
= 7.937253933 . 
121
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
26.
The Nucleus Speed
26.1 The Nucleus Speed
30
65
bpNuc »
1
Mp
me
+
bNuc
35 M n
30 m e
Proof: Dividing the Atom Force Balance 24.1,
30M p + 35M n
2
bpNuc
1-
2
bpNuc
rNuc
=
(30e )2
107
,
by the Nucleus Force Balance, 8.1,
me
1-
2
bNuc
65e 2
2
bNuc
rNuc =
10
7
,
we have
1 30M p + 35M n
302
2
1 - bpNuc
2
bpNuc
2
bpNuc
rNuc =
me
1
2
bNuc
rNuc
65 1 - b 2
Nuc
30
1
=
65 30M p + 35M n
2
1 - bpNuc
1-
2
bNuc
2
,
bNuc
30m e
bpNuc »
30
65
1
Mp
me
+
4
35 M n
30 me
122
2
1 - bpNuc
1-
2
bNuc
bNuc
Gauge Institute Journal, Volume 12, No. 2, May 2016
Since
4
2
1 - bpNuc
» 1,
2
1 - bNuc
bpNuc »
30
65
1
Mp
me
+
35 M n
30 m e
bNuc . 
bpNuc = 0.001854083888
26.2
Proof:
H. Vic Dannon
bpNuc »
30
65
1
bNuc

M
35 n 0.1719356689
+
30 me
me

Mp
0.01078359
= 0.001854083888 . 
26.3
Proof:
VpNuc  556, 225 m/sec
VpNuc  0.001854083888 ⋅ 300, 000, 000
= 556, 225 m / sec
123
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
27.
The Atomic Electrons Speed
from Radiation Equilibrium
be2 
27.1
Mp
me
+
7 Mn 2
b
6 me pNuc
2
 63bpNuc
Proof: be2 »

re
rNuc
bp2Nuc
Mp
me
+
7 Mn 2
b
6 me pNuc
2
 63bpNuc
.
27.2
27.3
be 
be 
4
Mp
me
30
65
+
7 Mn
b
6 m e pNuc
1
4
Mp
35 M n
+
30 m e
me
124
bNuc
Gauge Institute Journal, Volume 12, No. 2, May 2016
re
Proof: be 
rNuc
30
65
rNuc


4

bpNuc
re
»
Mp
me
+
30
65
1
Mp
me
35 M n
30 m e
1
4
Mp
me
+
35 M n
30 m e
+
35 M n
30 me
be »
30
65

bNuc . 
1
bNuc

Mp
35 M n 0.1719356689
+
30 m e
me


.67936622 4
1
63
= 0.014716334 . 
27.5
bNuc
be = 0.014716334
27.4
Proof:
H. Vic Dannon
v e = 4, 414, 900
Proof: v e = 0.014716334 ⋅ 300, 000, 000
= 4, 414, 900 m/sec . 
125
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
28.
The Atomic Electron Speed
from the Force Equation
28.1
2
2
2 ö
2 ö
æ
æ
e
e
30
30
÷÷ b 2 - ç
÷÷ » 0
be4 + ççç
çç
÷
e
7
7
÷
çè 63rNucm e 10 ÷÷ø
èç 63rNucm e 10 ø
Proof: The Force on the Atomic 1st orbit electron is
me
1be2 =
be2
be2re = 30
e2
107
,
30 e 2
1 - be2 ,
7
rem e 10
Squaring both sides,
be4
æ 30 e 2 ö÷2
æ 30 e 2 ö÷2
2
ç
÷÷ be - çç
÷
+ çç
çèç r m 107 ø÷÷ » 0
7
çè rem e 10 ø÷
e e
where re  3.848474069 ´ 10-10 m . 
30 e 2
30(1.60217733)2 10-38

107 reme
107 (3.848474069)10-10 (9.109 389 7)10-31
= (2.196668758)10-4 .
126
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
æ 30 e 2 ÷ö2
÷  4.825353632 ´ 10-8 .
ççç 7
÷
÷
çè 10 rem e ø
be4 + (4.825353632 ´ 10-8 )be2 - (4.825353632 ´ 10-8 ) » 0 .
28.2
be4 + (4.825353632)10-8 be2 - (4.825353632)10-8 » 0
Denoting be2 º x , we seek the zeros of the polynomial
f (x ) = x 2 + (4.825353632 ´ 10-8 )x - (4.825353632 ´ 10-8 )
between x = 0 , and x = 1 .
We used Maple Root-Finding Program.
Maple Input and Output follow:
> with(RootFinding):
>
>
28.3
be2  0.0002196427503
127
Gauge Institute Journal, Volume 12, No. 2, May 2016
be  0.01482034920
28.4
Compare with 26.4 be  0.014716334
28.5
The Atomic Electrons Speed
v e  4, 446,105 m/sec .
Compare with 26.5 v e = 4, 414, 900 m/sec
Proof: v e = bec
 (0.01482034920)c
» 4, 446,105 m/sec . 
By [Dan5, 11.6],
28.6 The Hydrogen Electron Speed
vH  a
 c = 2,189, 781 m/sec .
»
28.7
1
137
Zinc Atomic Electron Speed
 2´(Hydrogen Electron Speed)
Proof:
4, 446,105
» 2.03038797 . 
2,189, 781
128
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
29.
Atomic Electron 1st Orbit Radius
29.1 1st Orbit Radius of the Atomic Electron
re »
rNuc
2
bpNuc
be2  3.903068097 ´ 10-10 m
compare with 24.5, re  3.848474069 ´ 10-10 m
Proof: Substitute
rNuc » 6.108688998 ´ 10-12 ,
2
bpNuc
 3.437627064 ´ 10-6 ,
be2  0.0002196427503 . 
By [Dan5, 12.1],
29.2 Neutron Electron 1st Orbit Radius
RN  9.398741807 ´ 10-14
29.3 Zinc Atomic Electron 1st Orbit Radius
is  4153´ (Neutron Electron Orbit Radius)
Proof:
3.903068097 ´ 10-10
9.398741807 ´ 10-14
= 4,152.755951
129
Gauge Institute Journal, Volume 12, No. 2, May 2016
29.2
The Hydrogen Electron Orbit Radius
rH = 5.29277249 ´ 10-11 m .
29.3 The Zinc Electron 1st Orbit Radius is about
7´( Hydrogen Electron Orbit Radius)
Proof:
3.903068097 ´ 10-10
5.29277249 ´ 10-11
= 7.367442007 . 
130
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
30.
1st Orbit Atomic Electron
Frequency
30.1 The 1st Orbit Zinc Atomic Electron Angular Velocity
we in Zinc = we =
Proof:
we =
»
ve
= 1.139130779 ´ 1016 radians/sec
re
ve
re
4, 446,105 m/sec
3.903068097 ´ 10-10 m
= 1.139130779 ´ 1016 radians/sec . 
30.2 The 1st Orbit Atomic Electron Frequency
we
= 1.81298294 ´ 1015 cycles/sec
2p
Proof:
we
1.139130779 ´ 1016
=
2p
2p
= 1.81298294 ´ 1015 cycles/sec . 
131
Gauge Institute Journal, Volume 12, No. 2, May 2016
30.3
H. Vic Dannon
1st Orbit Neutron Electron Frequency is
 30,258´ (1st Orbit Zinc Atomic Electron Frequency)
Proof:
we in n
5.485642074 ´ 1020
2p
=
we in Zinc
1.81298294 ´ 1015
2p
 30, 258 . 
30.4 1st Orbit Hydrogen Electron Frequency
we in H
= 6.58472424 ´ 1015
2p
Proof:
we in H =
v e in H
2,189,781 m/sec
»
re in H
5.2977249 ´ 10-11 m
= 4.1334366 ´ 1016 radians/sec
we in H
4.137304269 ´ 1016 radians/sec
=
2p
2p
= 6.58472424 ´ 1015 . 
30.5
1st Orbit Hydrogen Electron Frequency is
 3.6´(1st Orbit Zinc Atomic Electron Frequency)
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Gauge Institute Journal, Volume 12, No. 2, May 2016
Proof:
we in H
6.58472424 ´ 1015
2p
=
we in Zinc
1.81298294 ´ 1015
2p
 3.631983564 . 
30.6
The 1st Orbit Atomic Proton Frequency
WpNuc,n
2p
Proof:
WpNuc,n
2p


»
1
n3
1
n
3
1
n
3

1
n
63
3
(1.439010597)1016 cycles/sec
we
2p
63(1.81298294)1015 cycles/sec
(1.439010597)1016 cycles/sec . 
133
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
31.
Atomic Electron Quantum of
Angular Momentum
We associated with the Zinc Atomic Electron
Orbit Radius re  3.903068097 ´ 10-10 m ,
Speed v e  4, 446,105 m/sec ,
be  0.01482034920 .
To obtain these, we averaged the 30 Atomic electrons orbits into
one orbit.
Now, we will consider that one orbit as the 1st orbit of the
electrons in the Nucleus, and construct on it the 2nd, 3rd,
4th,…,orbits.
The Zinc Atomic Electron nth orbit has Angular Momentum
me
1 - be,2 n
WpNuc,n
we,n
we,nre,2n ,
1
æ M p 7 M ö÷4
n÷
= ççç
+
÷ = 7.937253933 .
çè m e
6 me ø÷
134
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
The Atomic Electron Quantum of Angular Momentum is the
Angular Momentum of its 1st orbit:
31.1 Zinc Atomic Electron Quantum of Angular
Momentum
me
1 - be2
v ere = (14.99154238)
Proof:
1
1 - be2
=
1
1 - (0.01482034920)2
» 1.000109839
me
1 - be2
=
v ere =
(1.000109839)(9.1093897)10-31(4, 446,105)(3.903068097)10-10
(1.05457266)10-34

= (14.99154238) . 
31.2 Angular Momentum of the nth orbit Atomic Electron
me
1 - be,2 n
v e,nre,n = n(14.99154238)
135
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H. Vic Dannon
32.
The Atomic Electron nth Orbit
Radius, Speed, and Frequency
32.1 The Zinc Atomic Electron 1st Orbit Radius, Speed,
and Frequency
re  3.903068097 ´ 10-10 m
=
1
(14.99154238)2 1 - be2rH
30
= (7.490721986)rH
v e  4, 446,105 m/sec
=
30
v
14.99154238 H
= (2.001128319)vH
we
= 1.81298294 ´ 1015 cycles/sec
2p
=
302
(14.99154238)3
136
vH
1 - be2 2prH
1
Gauge Institute Journal, Volume 12, No. 2, May 2016
= (0.267147589)
H. Vic Dannon
vH
.
2prH
Proof: The force on the Nucleus 1st orbit electron is
v e2
1 30e 2
=
,
4pe0 re2
1 - be2 re
me
2
æ
ö
m
÷
30e
ç
e
= çç
v ere ÷÷÷
çè 1 - b 2
4pe0
÷ø
e
2
Since
me
1 - be2
1
me
1 - be2
v ere = (14.99154238) ,
2 e
1
2h
0
re =
(14.99154238)
1 - be2 ,
30
p mee 2


rH
=
1
(14.99154238)2 1 - be2rH
30
=
1
(14.99154238)2 
1 - 0.0002196427503
 rH
30 
0.999890172
224.7463429
= (7.490721986)rH .
ve =
(14.99154238)
me
r
2 e
1 - be
137
1
,
re
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
(14.99154238)
=
æ
ö
çç 1 (14.99154238)2 1 - b 2r ÷÷
e H÷
ø
1 - b 2 çè 30
me
e
=
30

14.99154238 m erH

vH
=
30
v
14.99154238 H
= (2.001128319)vH
we
v
= e
2p
2pre
=
=
30
v
14.99154238 H
æ1
ö
2p çç (14.99154238)2 1 - be2rH ÷÷÷
çè 30
ø
302
vH
2prH
1 - be2 
1
(14.99154238)3
we in H
2p
= (0.267147589)
vH
2prH
32.2 The Nucleus Electron nth Orbit Radius, Speed, and
Frequency
138
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H. Vic Dannon
re,n = n 2re ,
 n 2 (3.903068097 ´ 10-10 )m
1
v ,
n e
v e,n =

1
4, 446,105 m/sec
n
=
1 we
,
n 3 2p
we,n
2p
=
1
n
3
(1.81298294)1015 cycles/sec
Proof: The force on the Nucleus nth orbit electron is
me
v e,2 n
1 - be,2 n re,n
1 30e 2
=
,
4pe0 re,2n
æ
ö÷2
ç
me
30e
= çç
v e,nre,n ÷÷÷
ç
÷÷
4pe0
çè 1 - be,2 n
ø
2
me
Since
1 - be,2 n
re,n =
1 - be,2 n
v e,nre,n = n(14.99154238) ,
1 2
h 2 e0
n (14.99154238)2
1 - be,2 n ,
2
30
p mee


rH
=
1
me
1 2
n (14.99154238)2 1 - be,2 n rH
30
139
1
re,n
,
Gauge Institute Journal, Volume 12, No. 2, May 2016
= n2
1
(14.99154238)2 
1 - 0.0002196427503
 rH
30 
0.999890172
224.7463429
= n 2 (7.490721986)rH ,
= n 2re .
v e,n =
=
n(14.99154238)
me
re,n
2
1 - be,n
n(14.99154238)
me
1 - be,2 n
=
æ 2
ö
çç n (14.99154238)2 1 - b 2 r ÷÷
e,n H ÷
÷ø
çè 30
1
30

n 14.99154238 m erH

vH
we,n
2p
=
1
30
v
n 14.99154238 H
=
1
(2.001128319)vH
n
=
1
v .
n e
=
v e,n
2pre,n
140
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
=
=
1
30
v
n 14.99154238 H
æ n2
ö
2p çç (14.99154238)2 1 - be2rH ÷÷÷
çè 30
ø÷
302
1
vH
prH
1 - be2 2
1
n 3 (14.99154238)3
we in H
2p
=
=
1
n3
(0.267147589)
vH
2prH
1 we,n
n 3 2p
141
H. Vic Dannon
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
33.
The Atomic Proton Quantum of
Angular Momentum
We associated with the Zinc Atomic Proton
Orbit Radius rNuc  6.108688998 ´ 10-12 m ,
bpNuc  0.001854083888 ,
Speed VpNuc = bpNucc  556, 225 m/sec .
To obtain these, we averaged the 30 Atomic protons orbits into one
orbit, at the boundary of the Nucleus.
The 30 atomic electrons react to the net charge of the 30 Atomic
Protons as if the Atomic Protons where orbiting the nucleus at it
boundary at radius rNuc .
We will consider that one orbit as the 1st orbit of the Atomic
protons, and construct on it the 2nd, 3rd, 4th,…,orbits.
The Zinc Atomic Proton nth orbit has Angular Momentum
Mp
2
1 - bpNuc,
n
VpNuc,nrNuc,n .
The Atomic Proton Quantum of Angular Momentum is the
Angular Momentum of its 1st orbit:
142
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H. Vic Dannon
33.1 Zinc Atomic Proton Quantum of Angular Momentum
Mp
1-
2
bpNuc
VpNucrNuc » (53.89157159)
Proof:
1
2
1 - bpNuc
=
1
1 - (0.001854083888)2
» 1.000001719
Mp
2
1 - bpNuc
=
VpNucrNuc =
(1.000001719)(1.6726231)10-27 (556, 225)(6.108688998)10-12
-34
(1.05457266)10

= (53.89157159) . 
33.2 Angular Momentum of the nth orbit Atomic Electron
Mp
2
1 - bpNuc,
n
VpNuc,nrNuc,n » n(53.89157159)
143
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H. Vic Dannon
34.
Atomic Radiation Equilibrium is
Equivalent to the Proton
Quantized Angular Momentum
At equilibrium,
34.1
1) the Nucleus is electrodynamically stable if and only if the
inertia moments of the nth orbit electron and the nth orbit
proton are equal:
30m e
1 - be,2 n
re,2n =
30M p + 35M n
2
1 - bpNuc,
n
2
rNuc,
n
2) The Angular velocities of the proton, and the electron are
related by
WpNuc,n
we,n
1
æ M p 7 M ÷ö4
n÷
= ççç
+
÷ = 7.937253933
çè me
6 me ÷ø
144
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H. Vic Dannon
34.2 The Atomic nth Orbit Radiation Equilibrium is
equivalent to the Proton’s nth Orbit Quantized
Angular Momentum
30m e
1-
be,2 n
re,2n =
30M p + 35M n
1-
2
bpNuc,
n
2
rNuc,
n
1

Proof:
Mp
2
1 - bpNuc,
n
2
rNuc,
n WpNuc,n
æ M p 7 M ö÷4
n÷
= n(14.99154238) ççç
+
÷
çè m e
6 m e ÷ø
The nth Orbit Radiation Equilibrium is
30m e
1-
be,2 n
re,2n =
30M p + 35M n
1-
2
bpNuc,
n
2
rNuc,
n
 the Proton’s n th orbit has Angular Momentum
Mp
2
rNuc,
n WpNuc,n =
2
1 - bpNuc,
n


me
1-be,2 n
me
re,2n we,n
WpNuc,n
we,n
1 - be,2 n


1
n (14.99154238)
re,2n
æ M p 7 M ö÷4
çç
n÷
çç m + 6 m ÷÷÷
è e
e ø
1
æ M p 7 M ÷ö4
n÷
= n(14.99154238) ççç
+
÷ .
çè me
6 me ÷ø
145
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
V. NUCLEAR FORCES
35.
Nucleus Zero Point Energy
The electrical binding energy will exist at temperature zero, where
all thermal motions cease, and is called Zero Point Energy.
If the protons’ motion is neglected, the Zero Point Energy is the
electrons’ total binding energy.
35.1
1 1 (35e )(65e ) 1 æç
a wNuc ö÷
÷
= h ç (35)(65)
2 4pe0
rNuc
2 çè
bNuc 2p ø÷÷

-U Electrons Binding
Proof:
1 1 (35e )(65e ) 1
e2
c
= (35)(65)
w
2 4pe0
rNuc
2
4pe0c wNucrNuc Nuc
 
a
=
1/ bNuc
1 æç
a wNuc ö÷
÷.
h ç (35)(65)
2 çè
bNuc 2p ø÷÷
35.2 The Nucleus electrons’ orbit has photon’s energy
1 æ
a wNuc ö÷
÷.
- h çç (35)(65)
bNuc 2p ø÷÷
2 çè
146
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H. Vic Dannon
35.3 The Nucleus Ground state Energy is Zero Point Energy
1 æç
a wNuc ö÷
÷.
h ç (35)(65)
2 çè
bNuc 2p ø÷÷
35.4
Nucleus Zero Point Energy
1
a
w
 -268, 285 eV
- (35)(65)
2
bNuc Nuc
1
e2
 -268, 498 eV
(35)(65)
8pe0
rNuc
Proof:
=
1
a
(35)(65)
=
w
2
bNuc Nuc
1
1 / 137
é (6.5821220)10-16 eV ù 8.440466361 ´ 1018
(35)(65)
úû
2
(0.1719356689) êë
= 268, 285.5172 eV
1
e2
(35)(65)
=
8pe0
rNuc
1 c2
(1.60217733)2 10-38
eV
(35)(65)
Joul
=
2 107
(6.108688998)10-12
(1.60217733)10-19 Joul
= 268, 498.2722 eV
147
Gauge Institute Journal, Volume 12, No. 2, May 2016
35.5
H. Vic Dannon
The Zero Point Energy Frequency
2
268,498 eV
 1.298451396 ´ 1020 cycles/second
h
(35)(65)
Proof: 2
a wNuc
 (1.297990381)1020 cycles/sec
bNuc 2p
268,498 eV
268,498 eV
=2
h
4.1356692 ´ 10-15 eV
 1.298451396 ´ 1020 cycles/second
(35)(65)
a wNuc
1 / 137
(1.343341943)1018
 (35)(65)
bNuc 2p
(0.171860446)
 (1.297990381)1020 cycles/sec
35.6 Zinc Nucleus Zero Point Energy is
 35´(Neutron’s Zero Point Energy)
Proof:
268, 498
 35 . 
7, 671
35.7 Zinc Nucleus Zero Point Energy is
 19,743´(Hydrogen’s Zero Point Energy)
Proof:
268, 498
 19, 742.5 . 
13.6
148
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H. Vic Dannon
36.
Nucleus Nuclear Energy Binding
Most of binding is due to the Nucleus orbiting protons, and the
Protons’ binding Energy is the actual Zero Point Energy.
36.1 Nucleus Nuclear Energy Binding
1 (35e )(65e ) 1
1 c 2 (35e )(65e ) 1
a
=
= (35)(65) Wp
2 107
rp
2
bp
2 4pe0 rp
Proof:
1 (35e )(65e ) 1
1
e2
c
= (35)(65)
Wp
2 4pe0 rp
2
4pe0c Wp rp
 
a
=
1/ bp
1
a
(35)(65) Wp . 
2
bp
36.2 Nucleus Nuclear Energy Binding is
 42.5(Nucleus Zero Point Energy Binding)
Proof:
1 1
e2
(35)(65)
rp
2 4pe0
1 1
e2
(35)(65)
2 4pe0
rNuc
=
rNuc
 42.5
rp
149
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H. Vic Dannon
36.3 Nucleus Nuclear Energy Binding is
 0.977´(Nucleus Total Energy Binding)
Proof:
36.4
1
 0.977 . 
1 + (1 / 42.5)
1
a
- (35)(65) Wp = -11, 415, 889 eV
bp
2
1
e2
(35)(65)
 -11, 420, 788 eV
8pe0
rp
Proof:
1
a
(35)(65) Wp =
2
bp
1
1 / 137
é (6.5821220)10-16 eV ù (4.039370544)1019
= (35)(65)
úû
2
(0.01933751868) êë
= 11, 415, 888.78 eV
1
e2
(35)(65)

8pe0
rp

1 c2
(1.60217733)210-38
eV
(35)(65)
Joul
2 107
(1.436178468)10-13
(1.60217733)10-19 Joul
= 11, 420, 788.42 eV
150
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H. Vic Dannon
36.5 Hydrogen’s Nuclear Binding Energy
1 e2 1
1 æ a
÷ö 1 æ ac ö
=  ççç
WH ÷÷ =  çç ÷÷÷
÷ø 2 çè rH ÷ø
2 4pe0 rH
2 çè bp in H
Proof:
1 e2 1
1 e2
c
1 æ a
÷ö
=
WH ÷÷ . 
WH =  ççç
÷ø
2 4pe0 rH
2 4pe0c WHrH
2 èç bp in H
 
a
=
36.6
WH =
1/ bp in H
1 æç ac ö÷
 ç ÷.
2 çè rH ø÷÷
1
ö4
vH æç M p ÷
17
ç
÷÷÷ » (2.708286845)10 radians/sec
ç
rH çè me ø
1
Proof:
æ M p ö÷4
÷
WH = wH ççç
 çè m ø÷÷
e
v /r
H
H
1
1
c
137
(1836.152701)4
=
-11
(5.29277249)10
= (2.708286845)1017 . 
36.7
1 æ ac ö
-  çç ÷÷÷ = -590 eV
2 çè rH ø÷
151
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H. Vic Dannon
1 e2
 -590 eV
8pe0 rH
1
c
1 ac
1é
-16
137
ù
= ê (6.5821220)10 eV ú

û (1.221173735)10-12
2 rH
2ë
= 590.1455881 eV . 
1 e2
1 c 2 (1.60217733)2 10-38
eV
Joul
=
2 107 (1.221173735)10-12
8pe0 rH
(1.60217733)10-19 Joul
= 590.3990381 eV . 
36.8 Nucleus Nuclear Energy Binding is
 19349´(Hydrogen’s Nuclear Energy Binding)
Proof:
11, 415, 889 eV
 19, 349 . 
590 eV
152
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H. Vic Dannon
37.
Nuclear Force and Zero Point
Energy Force
37.1 Zinc Nucleus Nuclear Force is
 1836´(Zinc Nucleus Zero Point Energy Force)
Proof:
1 1
e2
(35)(65)
2 4pe0
rp2
1 1
e2
(35)(65) 2
2 4pe0
rNuc
æ r ö÷2
Mp
= ççç Nuc ÷÷ 
» 1836 . 
çè rp ø÷
me
37.2 Zinc Atomic Nuclear Force is
 3969´(Zinc Atomic Zero Point Energy Force)
Proof:
1 1
e2
(30)(30)
2
2 4pe0
rNuc
e2
1 1
(30)(30)
2 4pe0
re2
æ re ö÷2
÷  3969 . 
= çç
çè r ÷÷ø
Nuc

(63)2
153
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H. Vic Dannon
37.3 Zinc Nucleus Nuclear Force is
 7,287,084´(Zinc Atomic Zero Point Energy Force)
Proof:
1 1
e2
(35)(65)
2 4pe0
rp2
e2
1 1
(30)(30)
2 4pe0
re2
2
2
(35)(65) çæ re ÷ö æç rNuc ö÷÷
÷ ç
=
ç
÷  7, 287, 084 . 
(30)(30) çè rNuc ÷÷ø çèç rp ÷ø
 
(63)2
1836
37.4 Zinc Nuclear Force
 694,938,230´(Hydrogen’s Nuclear Force)
Proof:
1 (35e )(65e )
4pe0
rp2
1 e2
4pe0 rH2
æ r ÷ö2
ç
= (35)(65)çç H ÷÷÷ .
çè rp ÷ø
By [Dan4, 4.4], rH » 1.221173735 ´ 10-12
æ 1.221173735 ´ 10-12 ö÷2
÷
= (35)(65)ççç
çè 2.209505336 ´ 10-15 ø÷÷
 694, 938, 230 . 
154
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38.
Gamma Rays Origin is Neutron’s
Protons
Soft X Rays are photons at frequencies
1018 cycles/sec .
The Zinc Nucleus Electron Frequency
wNuc
= 1.343341943 ´ 1018 cycles/sec
2p
is in the range of soft X-rays.
And also the Zinc Nucleus Proton Frequency
Wp
2p
= 6.42885789 ´ 1018 cycles/sec
is in the range of soft X-rays.
Hard X Rays start at
1019 cycles/sec
The Neutron’s Electron Frequency
wne
= 8.73067052 ´ 1019 cycles/sec ,
2p
is in the range of Hard X Rays.
Gamma Rays start at
155
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H. Vic Dannon
1020 cycles/sec
The Neutron’s Proton Frequency
Wnp
2p
 5.69422884 ´ 1020 cycles/second ,
is in the range of Gamma Rays.
Thus, the existence of Gamma rays Radiation proves that the
Neutron is a condensed Hydrogen Atom, composed of an electron,
and a proton.
That is,
38.1
a Neutron’s Proton excited from its orbit into a
higher orbit, returns to a lower Neutron’s Orbit,
and emits a Gamma Ray Photon
156
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39.
The Nuclear Binding Energy
39.1 The Zinc Nuclear Electric Binding Energy
35(M n - M p - m e )c 2 = 11.689515233 MeV
Proof:
From the Nucleus Mass-Energy Equation, 5.1, the source of the
Electric Binding Energy
rp ö÷
1 1 (65e )(35e ) 2 æç
÷÷
c çç 1 +
çè
rp
rNuc ÷ø
2 107
is the Nuclear Binding Energy
35(M n - M p - m e )c 2 = 35(5.9456103 ´ 10-31 )(3 ´ 108 )2 Joul

Dm
= 1.872867245 ´ 10-12 Joul
Since Joule =
1015
MeV ,
160, 2177
= 11.89515233 MeV . 
39.2 The Hydrogen Atom Electric Binding Energy
rp ö÷
1 c 2 e 2 æç
÷÷ = 604.1773234eV
çç 1 +
2 107 rp çè
RH ø÷
157
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H. Vic Dannon
Proof:
Substituting from [Dan4, p.21],
RH
r
= = 42.8503524
rp
r
rp = 1.221173735 ⋅ 10-12
rp ö÷ 1 9 ⋅ 1016 (1.602177331)2 10-38
1 c 2 e 2 æç
1
÷÷ =
çç 1 +
(1 +
)
2 107 rp çè
RH ø÷ 2 107 1.221173735 ⋅ 10-12 
42.8503524
1.02333703
= 9.679990114 ´ 10-17 J
= 9.679990114 ´ 10-17 ´ (6.241507649)1018 eV
= 604.1773234eV . 
39.3 The Zinc Nuclear (Electric) Binding Energy is
 19,688  (Hydrogen Electric Binding Energy)
Proof:
11,895,152.33 eV
= 19, 688.18072 . 
604.1773234 eV
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H. Vic Dannon
40.
The Atomic Binding Energy
40.1 The Zinc Atomic Electric Binding Energy
r ö
1 1 (30e )(30e ) 2 æç
c ç 1 + Nuc ÷÷÷
çè
2 107
rNuc
re ÷ø
Proof: The Zinc Atomic Binding Energy is the Electric energy that
binds the 30 electrons to the 30 protons,
r ö
1 1 (30e )(30e ) 2 æç
c ç 1 + Nuc ÷÷÷ . 
çè
re ÷ø
2 107
rNuc
40.2 The Zinc Nuclear Binding Energy is greater than
 108 ´(The Zinc Atomic Binding Energy)
rp ÷ö
1 1 (65e )(35e ) 2 æç
÷÷
c
1
+
ç
çç
÷ (65)(35) r
rp
r
2 107
è
Nuc ø
Nuc
Proof:
>
= 108.3152778 . 
rNuc ÷ö (30)(30) rp
1 1 (30e )(30e ) 2 æç
c ç1 +
÷

çè
2 107
rNuc
re ÷÷ø
42.85

>1
159
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H. Vic Dannon
41.
Nuclear Forces versus
Atomic Forces
The force between the Nucleus electrons and Protons is enormous
compared to the force between the Atomic electrons and protons:
41.1 Force on a Zinc Nucleus Proton
 2,142  (Force on a Zinc Atomic Proton)
Proof: A Zinc Nucleus Proton is attracted to the 35 Neutronic
Electrons by
1 (e)(35e)
.
4pe0 rp2
A Zinc Atomic Proton is attracted to the 30 Atomic Electrons by
1 (e )(30e )
.
2
4pe0 rNuc
1 (e )(35e )
4pe0 rp2
2
æ
ö
35 ç rNuc ÷÷
çç
=
÷  2142 . 
1 (e )(30e )
30 çè rp ø÷÷

2
4pe0 rNuc
1836
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Gauge Institute Journal, Volume 12, No. 2, May 2016
41.2
H. Vic Dannon
Force on a Zinc Nucleus Electron is
 8845  (Force on a Zinc Atomic Electron)
Proof: A Neutronic Electron is attracted to the 65 Protons by
1 (e )(65e )
.
2
4pe0 rNuc
An Atomic Electron is attracted to the 30 Atomic Protons by
1 (30e )(e )
.
4pe0 re2
1 (65e )(e )
2
4pe0 rNuc
2
65 æç re ö÷÷
=
ç
÷  8600 . 
1 (30e )(e )
30 ççè rNuc ø÷

4pe0 re2
(63)2
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H. Vic Dannon
VI. SUMMARY OF NUMERICAL
RESULTS
Hydrogen Proton nth Orbit
Radius rH,n = n 2rH
me
» n 2 (1.221173735)10-12 m
Mp
1
Speed VH,n
1 c æç m e ÷÷ö4
1
=
çç
÷÷ » 334, 528 m/sec
n 137 çè M p ø
n
1
c
æ
ö
WH,n
1 137 ç M p ÷÷4
1
=
çç
»
Frequency
4.311097293 ´ 1016 cycles/sec
÷
3
3
2p
n 2prH çè m e ÷ø
n
1
æ M p ÷ö4
2
÷÷  = n(6.546018057)
Angular Momentum M p rn Wn = n ççç
çè m e ÷ø
Quantum of Angular Momentum (6.546018057)
Neutron’s Electron nth Orbit
Radius rne,n  n 2 (9.398741807)10-14 m
= n 2 (1.776104665)10-3 rH
Speed vne,n 
=
1
51, 558,134 m/sec
n
vH
1
n 0.043132065
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Gauge Institute Journal, Volume 12, No. 2, May 2016
Frequency
wne,n
2p
H. Vic Dannon
 8.73067052 ´ 1019 cycles/sec
1
=
n3
1.305362686 ´ 104
me
Angular Momentum
1-
2
bne,
n
wH
cycles/sec .
2p
vne,nrne,n = n(0.043132065)
Quantum of Angular Momentum (0.043132065)
Neutron’s Proton nth Orbit
Radius rnp,n  n 2 (2.209505336)10-15 ,
Speed Vnp,n 
Frequency
Wnp
2p
1
7, 905,145 m/sec ,
n
 5.69422884 ´ 1020 cycles/second .
Mp
Angular Momentum
1-
2
bnp,
n
(bnp,nc )rnp,n = n(0.277126027)
Quantum of Angular Momentum (0.277126027)
The Zinc Nucleus
1) is at Radiation Power Equilibrium 
me
1-
2
bNuc
2
rNuc
»
163
Mp
1-
bp2
rp2
Gauge Institute Journal, Volume 12, No. 2, May 2016
2)
Mp
me
1
e2
1
2
bNucrNuc =
=
bp2rp
65 1 - b 2
35 1 - b 2
107
Nuc
p
rNuc
3)
rp
65bp2
Mp
rNuc

rp
me
» 42.5
2
bNuc
 42.5
5)
7)
»
2
35bNuc
4)
6)
H. Vic Dannon
Wp
wNuc
65 2
b
35 p
1
35 æç M p ÷÷ö4
ç
÷ » 4.803464029
65 ççè m e ÷ø

rp ÷ö
1 1
e 2 æç
÷÷
65 çç 1 +
Mass-Energy M n - M p - me »
7

2 10
rp çè
rNuc ÷ø
Dm
Zinc Nucleus Proton nth Orbit
Orbit Radius
Speed
Frequency
rp,n  n 2 (1.436178468)10-13
Vp,n 
Wp,n
2p
=
1
n
3
1
5, 801, 257 m/sec
n
6.42885789 ´ 1018 cycles/sec
164
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H. Vic Dannon
Mp
Angular Momentum
Vp,n rp,n = n(13.21701291)
2
1 - bp,
n
Quantum of Angular Momentum (13.21701291)
Nucleus Radiation Equilibrium  Quantized Angular Momentum
Zinc Nucleus Electron nth Orbit
Orbit Radius
rNuc,n » n 2
rp,n
35 2
bNuc,n
 n 2 (6.108688998)10-12
2
65
bp,n
vNuc,n 
Speed
1
51, 560,184 m/sec
n
 Neutron Electron Speed
Frequency
wNuc,n
2p
Angular Momentum
=
1
n
3
1.343341943 ´ 1018 cycles/sec
me
1-
vNuc,nrNuc,n = n(2.761794253)
2
bNuc,
n
Quantum of Angular Momentum (2.761794253)
The Zinc Atom
1) is at Radiation Power Equilibrium 
30m e
1 - be2
re2 »
30M p + 35M n
1 - bp2Nuc
165
2
rNuc
Gauge Institute Journal, Volume 12, No. 2, May 2016
2)
30me
1-
be2re
be2
=
be2
re
4)
=
107
re
3)
rNuc
»
30M p + 35M n
2
1 - bpNuc
2
bpNuc
rNuc
rNuc
2
bpN
uc
Mp
»
me
+
7 Mn
» 63
6 me
1 30 2
bNuc
63 65
be2 
5)
6)
(30e )2
H. Vic Dannon
1
æ M p 7 M ö÷4
n÷
+
= ççç
÷ »
çè m e
6 m e ø÷
WpNuc
we
63 » 7.937253933
Zinc Atomic Electron nth Orbit
Orbit Radius
1
4, 446,105 m/sec
n
v e,n 
Speed
Frequency
re,n  n 2 (3.903068097)10-10 m
we,n
2p
Angular Momentum
=
1
n
3
1.81298294 ´ 1015 cycles/sec
me
1 - be,2 n
166
v e,nre,n = n(14.99154238)
Gauge Institute Journal, Volume 12, No. 2, May 2016
H. Vic Dannon
Quantum of Angular Momentum (14.99154238)
Zinc Atomic Proton nth Orbit
rNuc,n  n 2 (6.108688998)10-12
Orbit Radius
Speed
Frequency
VpNuc,n 
WpNuc,n

2p
Angular Momentum
1
n
3
1
556, 225 m/sec
n
(1.439010597)1016 cycles/sec
Mp
1-
2
bpNuc,
n
VpNuc,nrNuc,n » n(53.89157159)
Quantum of Angular Momentum (53.89157159)
Atomic Radiation Equilibrium  Quantized Angular Momentum
Zinc Nucleus Zero Point Energy
1
a
- (35)(65)
w
 -268, 285 eV
2
bNuc Nuc
-
1
e2
(35)(65)
 -268, 498 eV
8pe0
rNuc
 19,743´(Hydrogen’s Zero Point Energy)
The Zinc Nucleus Zero Point Energy Frequency
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2
H. Vic Dannon
268,498 eV
 1.298451396 ´ 1020 cycles/sec
h
(35)(65)
a wNuc
 (1.297990381)1020 cycles/sec
bNuc 2p
Zinc Nucleus Nuclear Binding Energy
1
a
- (35)(65) Wp = -11, 415, 889 eV
2
bp
e2
1
(35)(65)
 -11, 420, 788 eV
8pe0
rp
 42.5´(Nucleus Zero Point Energy Binding)
 19349´(Hydrogen’s Nuclear Energy Binding)
Zinc Nucleus Nuclear Force
 1836´(Zinc Nucleus Zero Point Energy Force)
 7,287,084´(Zinc Atomic Zero Point Energy Force)
 694,938,230´(Hydrogen’s Nuclear Force)
Zinc Atomic Nuclear Force
 3969´(Zinc Atomic Zero Point Energy Force)
Zinc Nuclear Binding Energy
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H. Vic Dannon
 19,688  (Hydrogen Binding Energy)
 108 ´(The Zinc Atomic Binding Energy)
Force on a Zinc Nucleus Proton
 2,142  (Force on a Zinc Atomic Proton)
Force on a Zinc Nucleus Electron
 8845  (Force on a Zinc Atomic Electron)
169
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H. Vic Dannon
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176