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VOLUME

NEUTRINO

7.37249734 x 10^-51 kg

1.41078466754 x 10^-71 m^3

PHOTON

7.37249734 x 10^-51 kg

1.41078466754 x 10^-71 m^3

POSITRON

9.10938356 × 10^-31 kg

1.74315134540 x 10^-51 m^3

ELECTRON

9.10938356 × 10^-31 kg

1.74315134540 x 10^-51 m^3

QUARK

5.57237387 × 10^-28 kg

1.06631704594 x 10^-48 m^3

PROTON

1.63968904 × 10^-27 kg

1.19143405030 x 10^-47 m^3

NEUTRON

1.64059997 × 10^-27 kg

1.60920914103 x 10^-47 m^3

PARTICLE

RELATIVE VOLUME

RELATIVE MASS

NEUTRINO

1 m^3

5.225813345 x 10^20 kg

PHOTON

1 m^3

5.225813345 x 10^20 kg

POSITRON

1.235589941 x 10^20 m^3

6.456962405 x 10^40 kg

ELECTRON

1.235589941 x 10^20 m^3

6.456962405 x 10^40 kg

QUARK

7.558326018 x 10^22 m^3

3.949840112 x 10^43 kg

PROTON

8.445187119 x 10^23 m^3

1.162253232 x 10^44 kg

NEUTRON

1.140648305 x 10^24 m^3

1.162898922 x 10^44 kg

r = ((VOL)*(3/(4 pi))^(1/3)

______________________________________________________________________________________

Universe Today

Space and astronomy news

Posted on July 16, 2008 by Jerry Coffey

Diameter of the Solar System

Defining the diameter of the Solar System is a matter of perspective and characterization. You can look at the Solar System’s diameter as ending at the aphelion of the orbit of the farthest planet, the edge of the heliosphere, or ending at the farthest observable object. To cover all of the objective bases, we will look at all three.

Looking at the aphelion(according to NASA figures) of the orbit of the farthest acknowledged planet, Neptune, the Solar System would have a radius of 4.545 billion km and a 9.09 billion km diameter. This diameter could change if the dwarf planet Eris is promoted after further study.

Sedna is three times farther away from Earth than Pluto, making it the most distant observable object known in the solar system. It is 143.73 billion km from the Sun, thus giving the Solar System a diameter of 287.46 billion km.

Now, that is a lot of zeros, so let’s simplify it into astronomical units. 1 AU(distance from the Earth to the Sun) equals 149,597,870.691 km. Based on that figure, Sedna is nearly 960.78 AU from the Sun and the Solar System is 1,921.56 AU in diameter. Universe Today

A third way to look at the diameter of the Solar System is to assume that it ends at the edge of the heliosphere. The heliosphere is often described as a bubble where the solar wind pushes against the interstellar medium and edge of where the Sun’s gravitational forces are stronger than those of other stars. The heliopause is the term given as the edge of that influence, where the solar wind is stopped and the gravitational force of our Sun fades. That occurs at about 90 AU, giving the Solar System a diameter of 180 AU. If the Sun’s influence ends here, how could Sedna be considered part of the Solar System, you may wonder. While it is beyond the heliopause at aphelion, it falls back within it at perihelion(around 76 AU).

Those determinations of the diameter of the Solar System may seem about as clear as mud, but they give you an idea of what scientists are trying to place a definitive value on. The distances involved are mind boggling and there are too many unknowns to place a absolute figure. Perhaps, an exact number will be determinable as the Voyager probes continue their outward journey.

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https://www.universetoday.com/15585/diameter-of-the-solar-system/

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RADIUS OF SOLAR SYSTEM = 143.73 x 10^12 = 143730000000000

PARTICLE

RELATIVE VOLUME

RELATIVE RADIUS

NEUTRINO

1 m^3

0.62035049089940001 m

PHOTON

1 m^3

0.62035049089940001 m

POSITRON

1.235589941 x 10^20 m^3

76649882644971070353 m

ELECTRON

1.235589941 x 10^20 m^3

76649882644971070353 m

QUARK

7.558326018 x 10^22 m^3

46888112556440073665714 m

PROTON

8.445187119 x 10^23 m^3

523897597500893974559303 m

NEUTRON

1.140648305 x 10^24 m^3

707601735950318554529333 m

WHEN I SUPERSIZE A NEUTRINO TO HAVE A VOLUME OF ONE CUBIC METER AND THEN ADJUST ALL THE

OTHER SUBATOMIC PARTICLES RELATIVE TO THIS FIGURE, I GET THE ABOVE TABLE.

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Milky Way Galaxy

astronomy

Written by Paul W. Hodge

Fact-checked by The Editors of Encyclopaedia Britannica

Last Updated: Jul 2, 2024 Article History

How big is the Milky Way Galaxy?

The first reliable measurement of the size of the Milky Way Galaxy was made in 1917 by American astronomer Harlow Shapley. Assuming that the globular clusters outlined the Galaxy, he determined that it has a diameter of about 100,000 light-years. His values have held up remarkably well over the years.

https://www.britannica.com/place/Milky-Way-Galaxy

___________________________________________________________________________________________________

THE FOLLOWING ARTICLE EXPLAINS THE MODERN DAY VIEW OF GALACTIC FORMATION. THIS IS NOT THE

VIEW THAT THE DYNAMISM MODEL HAS FOR GALACTIC FORMATION.

___________________________________________________________________________________________________

The Ten Largest Galaxies In The Universe

• The largest known galaxy in the universe is IC 1101 at four million light years across

• All of the largest galaxies are either elliptical galaxies or spiral galaxies

• Most galaxies grow to their current size by absorbing other galaxies Galaxies come in a wide variety of shapes and sizes. If we think of galaxies as singular objects, they are some of the largest structures in the universe. Most of the stars and planets in the universe are contained within galaxies.

Astronomers estimate that the universe contains more than 200-billion galaxies. Of all the known galaxies, which ones happen to be the largest and how big are they?

To date, the biggest galaxy that has been discovered is IC 1101. IC 1101 is classified as a supergiant elliptical galaxy, and it looks far different than the Milky Way. As an elliptical galaxy, IC 1101 contains an abundance of low to medium mass red and yellow stars, most of which are quite old. At the center of IC 1101, there exists a supermassive black hole, which happens to be the largest black hole ever discovered. IC 1101 has an estimated diameter of four million light-years. In comparison, the Milky Way is roughly 100,000 light-years in diameter, making IC 1101, 40 times larger than the Milky Way. If you were to place IC 1101 where the Milky Way is, it would completely engulf the Andromeda Galaxy at 2.5 million light-years away. IC 1101 is located roughly one billion light-years away from the Milky Way and likely contains over 100 trillion stars.

https://www.worldatlas.com/space/the-ten-largest-galaxies-in-the-universe.html ___________________________________________________________________________________________________

THE STATEMENT IN THE ABOVE ARTICLE WHICH SAYS; 'Most galaxies grow to their current size by absorbing other galaxies'; IS INACCURATE. GALAXIES ARE ACTUALLY BORN AS SPHERICAL GALAXIES WITH THEIR

ATTENDANT QUASAR CENTER, AND LITTLE TO NO ROTATON, AND THEN SHRINK TO THEIR CURRENT SIZE

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AS THEIR ROTATION SPEED INCREASES.

___________________________________________________________________________________________________

observable universe

astronomy

Written by Karen Sottosanti

Fact-checked by The Editors of Encyclopaedia Britannica

Last Updated: May 20, 2024 Article History

observable universe, the region of space that humans can actually or theoretically observe with the aid of technology. The observable universe, which can be thought of as a bubble with Earth at its centre, is differentiated from the entirety of the universe, which is the whole cosmic system of matter and energy of which Earth, and therefore the human race, is a part. Unlike the observable universe, the universe is possibly infinite and without spatial edges.

The observable universe is approximately 93 billion light-years in diameter. This number is derived from several considerations. A light-year, the distance light can travel in one Earth year, is 9.46 trillion kilometres (5.88 trillion miles). The estimated age of the universe since the big bang is 13.8 billion years, so the light emitted by objects in space that humans can see has been traveling toward Earth for no more than 13.8 billion years. That would seem to indicate that the observable universe is 13.8 billion light-years in any direction from Earth and 27.6 billion light-years in diameter. However, according to Hubble’s law, space has been expanding since the big bang, and thus the observable universe continues to expand as well. Calculations of this expansion show that objects that emitted light 13.8 billion years ago, from a distance of 13.8 billion light-years, are now even farther away from Earth—46 billion light-years away, approximately. This means that the observable universe is more than 46 billion light-years in any direction from Earth and about 93 billion light-years in diameter.

https://www.britannica.com/topic/observable-universe

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HOW FAR IS A LIGHT YEAR?

SPEED OF LIGHT = 299,792,458 m/s 60 SECONDS IN A MINUTE.

60 MINUTES IN AN HOUR.

24 HOURS IN A DAY.

365 DAYS IN A YEAR.

299792458 x 60 x 60 x 24 x 365 = 9454254955488000 m

299792458 x 60 x 60 x 24 x 365 = 9.454254955 x 10^15 m

LIGHT YEAR = 9.454254955 x 10^15 m

PARTICLE

RELATIVE RADIUS

RADIUS IN LIGHT YEARS

NEUTRINO

0.620350490 m

6.561 x 10^-17 ly

PHOTON

0.620350490 m

6.561 x 10^-17 ly

SOLAR SYSTEM

1.4373 x 10^14 m

1.520 x 10^-02 ly

POSITRON

7.6649 x 10^19 m

8,107 ly

ELECTRON

7.6649 x 10^19 m

8,107 ly

MILKY WAY

4.7271 x 10^20 m

50,000 ly

IC 1101

1.8908 x 10^21 m

2,000,000 ly

QUARK

4.6888 x 10^22 m

4,959,460 ly

PROTON

5.2389 x 10^23 m

55,413,144 ly

NEUTRON

7.0760 x 10^23 m

74,844,607 ly

UNIVERSE

4.3489 x 10^26 m

46,000,000,000 ly

NOTE THAT THE ABOVE TABLE HAS BOTH THE RADIUS OF THE SOLAR SYSTEM, THE MILKY WAY GALAXY, IC

1101 WHICH IS THE LARGEST KNOWN GALAXY, AND THE OBSERVABLE UNIVERSE, FOR COMPARISON

PURPOSES.

THE ROLE OF GRAVITY AT THE SUBATOMIC LEVEL

GRAVITY SEEMS INSIGNIFICANT TO THE SUBATOMIC LEVEL BECAUSE IT IS THE WEAKEST FORCE AND THE

SUBATOMIC PARTICLES HAVE INFINITESMAL MASS. THE FORCE OF GRAVITY IS SCORNED BY NUCLEAR

PHYSICIST FOR THESE REASONS.

THE FOLLOWING TABLE SHOWS THE RESULT OF SUPERSIZING ALL OF THE BASIC SUBATOMIC PARTICLES

SUCH THAT THE NEUTRINO HAS A VOLUME OF ONE CUBIC METER, AND EVERYTHING ELSE IS INCREASED

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ACCORDINGLY.

PARTICLE

RELATIVE MASS

RELATIVE RADIUS

NEUTRINO

5.225813345 x 10^20 kg

0.62035049089940001 m

PHOTON

5.225813345 x 10^20 kg

0.62035049089940001 m

POSITRON

6.456962405 x 10^40 kg

76649882644971070353 m

ELECTRON

6.456962405 x 10^40 kg

76649882644971070353 m

QUARK

3.949840112 x 10^43 kg

46888112556440073665714 m

PROTON

1.162253232 x 10^44 kg

523897597500893974559303 m

NEUTRON

1.162898922 x 10^44 kg

707601735950318554529333 m

F = G X M1 X M2

R^2

F = FORCE G = 6.67430 × 10^-11 N·(m/kg)^2 M1 = FIRST MASS M2 = SECOND MASS R = RADIUS

BY MAKING THE NEUTRINO'S MASS EQUAL TO THE SECOND MASS; M2 = 5.225813345 x 10^20 kg FOR

EVERY CASE; I CAN CREATE ANOTHER TABLE SHOWING HOW STRONG THE FORCE OF GRAVITY IS ON AN

OUTSIDE PARTICLE FOR EACH OF THE COMPOSITE PARTICLES.

RADIUS

FORCE OF GRAVITY

NEUTRINO

1.240 × 10^00 m

1.18407 × 10^31 N

PHOTON

NEUTRINO

1.240 × 10^00 m

1.18407 × 10^31 N

POSITRON

NEUTRINO

7.664 × 10^19 m

383,323,007,140 N

ELECTRON

NEUTRINO

7.664 × 10^19 m

383,323,007,140 N

QUARK

NEUTRINO

4.688 × 10^22 m

626,633,532 N

PROTON

NEUTRINO

5.238 × 10^23 m

14,769,560 N

NEUTRON

NEUTRINO

7.076 × 10^23 m

8,100,723 N

1 N = kg x m/s^2

NOTE THAT BY COMPARISON THE FORCE OF GRAVITY AT THE SUN'S SURFACE IS 274 N.

NOTE ALSO THAT WHEN M1 IS A NEUTRINO, OR A PHOTON, THAT THE 'RADIUS' FIGURE IS ACTUALLY JUST

THE DIAMETER OF THE NEUTRINO AS THIS IS THE DISTANCE BETWEEN THE CENTER OF THE TWO

PARTICLES WHEN THEY ARE IN CONTACT WITH EACH OTHER. OBSERVE THAT THE FORCE OF GRAVITY

BETWEEN THESE SMALLEST PARTICLES, NEUTRINO AND NEUTRINO, OR NEUTRINO AND PHOTON, IS

ENORMOUS COMPARED TO THE FORCE EXERTED BY THE LARGER PARTICLES.

THE DYNAMISM MODEL SUGGESTS THAT WHEN ANY OBJECT LOSES MASS IN THE FORM OF

ELECTROMAGNETIC ENERGY, THAT THE ELECTROMAGNETIC RAY IS COMPOSED OF PHOTONS OF

DIFFERENT CHARGES ALTERNATING WITH ONE ANOTHER WITHIN THE RAY. THAT A P-STRING PHOTON

WILL BE FOLLOWED BY AND N-STRING PHOTON, FOLLOWED BYA P-STRING PHOTON, FOLLOWED BY AN NSTRING PHOTON, ... , FOR THE ENTIRE LENGTH OF THE RAY. THAT THE SOURCE OF ALL OF THESE

PHOTONS IS ACTUALLY NEUTRINOS THAT WERE CONVERTED FROM NORMAL MASS INTO THESE PHOTON

PAIRS, WHICH IS THE ENERGY OF THE ELECTROMAGNETIC RAY.

GRAPHICALLY, THE MASS OF ANY PARTICLE, PROTON OR NEUTRON, IS PRIMARILY COMPOSED OF QUARKS, WHICH ARE THEMSELVES MAINLY COMPOSED OF NEUTRINOS. BUT THERE IS ALWAYS THE POSSIBILITY

THAT THERE ARE EXTRA NEUTRINOS SCATTERED ALL THROUGHOUT THE ATOM, NOT JUST IN THE

NUCLEUS BUT IN THE ELECTRON SHELLS ALSO. HOWEVER, THIS IS CONJECTURE. BUT WHEREVER THEY

COME FROM, CLUMPS OF NEUTRINOS THAT ARE LOCKED TOGETHER GRAVITATIONALLY AND INERTIALLY, MAY BE DISTURBED BY SOME OUTSIDE FORCE WHICH CAUSES THEM TO FLY OFF. IN THIS PROCESS, THE

NEUTRINOS WILL SOMEHOW PAIR UP AND THEN TRANSFORM INTO P-STRING AND N-STRING PHOTONS

WHICH THEN GO INTO THE ELECTROMAGNETIC RAY AS DESCRIBED ABOVE.

THE POINT IS THAT THE GRAVITATIONAL FORCE THAT LOCKS THE NEUTRINOS TOGETHER WITH P-STRING

PHOTONS WITHIN A QUARK IS ENORMOUS. BASED ON THE LAW OF INERTIA ALONE, IT WOULD TAKE AN

EVEN MORE MASSIVE OUTSIDE FORCE TO BREAK THESE TIES THAT BIND THEM TOGETHER.

I BELIEVE THAT THIS IS MORE THAN ADEQUATE TO EXPLAIN HOW GRAVITY ALONE IS SUFFICIENT TO KEEP

THE NEUTRINOS, AND PHOTONS, LOCKED INTO THE STRUCTURE OF THE LARGER COMPOSITE PARTICLES

THAT THEY BELONG TO. FOR EXAMPLE, THE P-STRING PHOTONS AT THE OUTER EDGE OF THE POSITRON; AND LIKEWISE THE N-STRING PHOTONS AT THE OUTER EDGE OF THE ELECTRON; EACH HAVE 383 BILLION

NEWTONS OF FORCE TO KEEP THEM LOCKED INSIDE THEIR RESPECTIVE PARTICLES. AT LEAST THAT

WOULD BE THE CASE IF THE LARGER PARTICLE WAS NEUTRALLY CHARGED.

IMAGINE THE RESULTING EXPLOSION THAT WOULD OCCUR IF THE POSITRON WHICH IS COMPOSED OF P-17 of 99

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STRING PHOTONS, WERE TO COME INTO CONTACT WITH AN ELECTRON, WHICH IS COMPOSED OF NSTRING PHOTONS. THE P-STRING PHOTONS WOULD BE IMMENSELY ATTRACTED TO THE NEGATIVE CHARGE

OF THE N-STRING PHOTONS, AND VICE VERSA. UPON CONTACT THE P-STRING PHOTONS WOULD PAIR UP

WITH THEIR N-STRING COUNTERPARTS TO FORM LIGHT RAYS AT EVERY CONCEIVABLE ELECTROMAGNETIC

FREQUENCY. THE POSITRON AND ELECTRON WOULD BE ANIHILATED IN THE PROCESS. ALL THAT WOULD

REMAIN WOULD BE RAYS OF LIGHT.

THE PROTONS STRUCTURE

THE ELECTRON CHARGE TO MASS RATION IS:

e/m = 1.758820 × 10^11 C/kg

THE SUPERSIZED RELATIVE MASS OF THE ELECTRON IS:

ELECTRON MASS = 6.456962405 x 10^40

THEREFORE THE SUPERSIZED ELECTRON'S RELATIVE CHARGE IS:

6.456962405 x 10^40 kg * 1.758820 × 10^11 C/kg = 1.13566346171621 × 10^50 C

F = K X Q1 X Q2

R^2

F = FORCE K = 8.9876 × 10^9 N⋅m^2/C^2 Q1 = FIRST CHARGE Q2 = SECOND CHARGE R = RADIUS

THE PROTON IS COMPOSED OF THREE QUARKS ORBITING AN ELECTRON.

THE RADIUS OF THE SUPERSIZED PROTON IS 5.238 × 10^23 m

A P-STRING PHOTON ON THE OUTER EDGE OF A PROTON HAS THE SAME CHARGE TO MASS RATIO AS THE

ELECTRON AND IT HAS A MASS OF: 5.225813345 x 10^20 kg

5.225813345 x 10^20 kg * 1.758820 × 10^11 C/kg = 9.1912650274529 × 10^31

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (1.13566346171621 × 10^50) * (9.1912650274529 × 10^31)) / (5.238 × 10^23)^2 =

3.42021687816732921929710077391061384912819512 × 10^44 N

THIS IS AT LEAST AS MUCH FORCE THAT THE ELECTRON EXERTS ON EVERY P-STRING PHOTON WITHIN THE

PROTON. BUT THE ELECTRON ALSO EXERTS A PROXIMATE FORCE ON EACH OF THE THREE QUARKS AS

WELL. THIS FORCE IS EXERTED ON THE CENTER OF THE QUARK. SO THE 'R = DISTANCE' WILL BE EQUAL

TO THE RADIUS OF THE QUARK PLUS THE DIAMETER OF THE ELECTRON. THE PROTON'S OVERALL CHARGE

IS EQUAL TO THE CHARGE OF THE ELECTRON. BUT THERE ARE THREE QUARKS. SO AN INDIVIDUAL QUARK

WILL HAVE JUST TWO THIRDS THE CHARGE OF THE ELECTRON.

QUARK'S CHARGE = 2*(1.13566346171621 × 10^50)/3 = 7.5710897447747333 × 10^49 C

QUARK'S RADIUS = 4.688 × 10^22 m

ELECTRON'S DIAMETER = 2*(7.6649 x 10^19 m) = 1.53298 × 10^20 m QUARK'S RADIUS + ELECTRON'S DIAMETER = 4.688 × 10^22 + 1.53298 × 10^20 = 4.7033298 × 10^22 m FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (1.13566346171621 × 10^50) * (7.5710897447747333 × 10^49)) / (4.7033298 ×

10^22)^2 = 3.493340488 × 10^64 N

THE GRAVITATIONAL FORCE CAN THAT KEEPS THE QUARKS WITHIN THE PROTON CAN ALSO BE

CALCULATED THIS WAY. CHANGING THE ELECTRICAL CONSTANT 'K' TO THE GRAVITATION CONSTANT 'G'; AND CHANGING THE CHARGE FIGURES FOR THE QUARK AND THE ELECTRON TO MASS; WHILE THE 'R =

DISTANCE' FIGURE REMAINS THE SAME:

FORCE = (G * M1 * M2) / R^2

FORCE = ((6.67430 × 10^-11) * (3.949840112 x 10^43) * (6.456962405 x 10^40)) / (4.7033298 × 10^22) =

3.619162343747009 x 10^51 N

WITHIN THE PROTON, THE ELECTRICAL BINDING FORCE THAT BINDS THE QUARKS TO THE ELECTRON IS A FORCE OF 3.494273330882157 × 10^64 N, WHILE THE GRAVITATIONAL BINDING FORCE IS A FORCE OF

3.619162343747009 x 10^51 N, WHICH ARE BOTH APPLICABLE TO EACH OF THE THREE QUARKS.

TOTAL FORCE = (3.494273330882157 × 10^64) + (3.619162343747009 x 10^51) =

3.4942733308825189162343747009 x 10^64 N

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THE GRAVITATIONAL FORCE IS SO MUCH SMALLER THAN THE ELECTRICAL ONE THAT IT DOESN'T EVEN

SHOW UP IN THE RESULT. BUT IT IS STILL A FACTOR NEVERTHELESS.

BUT THE FACT STILL REMAINS THAT EVERY P-STRING PHOTON IS REPULSED BY EVERY OTHER P-STRING

PHOTON. LIKEWISE EACH OF THE THE THREE QUARKS IS REPULSED BY THE OTHER TWO QUARKS. THUS

THE ELECTRICAL ATTRACTION THAT THE P-STRING PHOTONS AND THE QUARKS HAVE FOR THE ELECTRON

IS NOT ENOUGH TO KEEP THEM SITUATED STABLY WITHIN THE PROTON.

EACH QUARK IS REPULSED BY THE TWO OTHER QUARKS. IF THE DISTANCE 'R' BETWEEN THE CENTERS OF

THE QUARKS IS ESTIMATED TO BE THREE QUARTERS OF THE DIAMETER OF THE PROTON: PROTON DIAMETER = 2 * PROTON RADIUS = 2 * 5.238 × 10^23 = 1.0476 × 10^24 m r = (3/4) * 1.0476 × 10^24 m = (3/4) * 1.0476 × 10^24 = 7.857 × 10^23 m FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (7.5710897447747333 × 10^49) * (7.5710897447747333 × 10^49)) / (7.857 ×

10^23)^2 = 8.3476258217402632871825893567627411169124361160530015583862548 × 10^61 m THE FORCE IS DOUBLED BECAUSE THERE ARE TWO QUARKS IN A REPULSIVE RELATIONSHIP WITH THE

QUARK:

FORCE = 2 * 8.3476258217402632871825893567627411169124361160530015583862548 × 10^61 =

1.66952516434805265743651787135254822338248722321060031167725096 × 10^62 N

-------------------------------------------------------------------------------------

----------------------------------------------

THE TOTAL ATTRACTIVE FORCES UPON EACH QUARK ARE:

GRAVITATIONAL:

3.619162343747009 x 10^51 N

ELECTRICAL:

3.494273330882157 × 10^64 N

TOTAL:

3.494273330882157 × 10^64 N

-------------------------------------------------------------------------------------

----------------------------------------------

THE TOTAL REPULISVE FORCES UPON EACH QUARK ARE:

ELECTRICAL:

1.669525164348052 × 10^62 N

TOTAL:

1.669525164348052 × 10^62 N

-------------------------------------------------------------------------------------

----------------------------------------------

THE NET FORCES UPON EACH QUARK ARE:

TOTAL ATTRACTION:

3.494273330882157 × 10^64 N

TOTAL REPULSION:

1.669525164348052 × 10^62 N

TOTAL ATTRACTION /

209

TOTAL REPULSION:

-------------------------------------------------------------------------------------

----------------------------------------------

THE ATTRACTIVE FORCES THAT KEEPS AN INDIVIDUAL QUARK WITHIN THE PROTON ARE TWO HUNDRED

AND NINE TIMES GREATER THAN THE REPULSIVE FORCES TRYING TO FORCE THE QUARK TO DEPART.

THE NECESSITY FOR THE MAGNETIC MONOPOLE

___________________________________________________________________________________________________

Coulomb Energy

An alternative derivation in classical form is shown with the magnetic constant and speed of light. This version shows the consistency of energy and mass equations in classical format, as explained further below.

Many of the energy and mass equations are shown with an alternative derivation to show the consistency of the Coulomb energy across all equations (e.g. Electron energy, electron mass, Planck mass, Rydberg energy, etc).

The Coulomb energy is constant across particles, photons and forces. The components of the Coulomb constant from above is found in the next equation as it is expanded to be an energy equation by multiplying amplitude (squared) and dividing by the distance (radius).

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Coulomb Energy Equation

https://energywavetheory.com/physics-constants/coulombs-constant/

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THE ELECTRICAL CONSTANT IS RELATED TO THE MAGNETIC CONSTANT THROUGH THIS EQUATION: ke = (U0*c^2)/(4 pi)

(4 pi * ke) = (U0*c^2)

(4 pi * ke)/(c^2) = U0

U0 = (4 pi * ke)/(c^2)

c = 299 792 458 m/s ke = 8.9876 × 10^9 pi = 3.14159265358979323846264338327

U0 = (4 * 3.14159265358979323846264338327 * 8.9876 × 10^9)/(299792458^2) = 1.2566438025 × 10^-6

U0 = 1.2566438025 × 10^-6 THE MAGNETIC CONSTANT IS: 1.2566438025 × 10^-6 kg/m ___________________________________________________________________________________________________

Units of Magnetism: Ampere, Tesla, Weber, Henry

Magnetism has many different units, such as the Ampere for electric current, Tesla for magnetic field, Weber for magnetic flux and Henry for inductance.

Explanations (1)

Vinitha Ranganeni

Units of Magnetism

Ampere(A):

Unit of measurement for Electric Current

Tesla (T):

Unit of measurement for Magnetic Field

Weber (Wb):

Unit of measurement for Magnetic Flux

Henry (H):

Unit of measurement for Inductance

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https://www.expii.com/t/units-of-magnetism-ampere-tesla-weber-henry-1932

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A P-STRING PHOTON ON THE OUTER EDGE OF A PROTON HAS THE SAME CHARGE TO MASS RATIO AS THE

ELECTRON AND IT HAS A MASS OF: 5.225813345 x 10^20 kg

5.225813345 x 10^20 kg * 1.758820 × 10^11 C/kg = 9.1912650274529 × 10^31

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (1.13566346171621 × 10^50) * (9.1912650274529 × 10^31)) / (5.238 × 10^23)^2 =

3.42021687816732921929710077391061384912819512 × 10^44 N

THIS IS AT LEAST AS MUCH FORCE THAT THE ELECTRON EXERTS ON EVERY P-STRING PHOTON WITHIN THE

PROTON.

THE RELATIVE VOLUME OF A QUARK IS: 7.558326018 x 10^22 m^3

(7.558326018 x 10^22) /(5 × 10^14) = 151166520.36

THE NUMBER OF PHOTONS IN AN ELECTRON = 244470644915512938048

TO GET THE NUMBER OF PHOTONS IN A QUARK I HAVE TO DIVIDE BY THREE: 244470644915512938048 / 3 = 8.1490214971837646016 × 10^19

THE VOLUME OF A QUARK'S SPHERE IS: 7.558326018 x 10^22 m^3

DIVIDING THE VOLUME OF A QUARK'S SPHERE BY THE NUMBER OF PHOTONS IN A QUARK GIVES ME: (7.558326018 x 10^22 m^3) /(8.1490214971837646016 × 10^19) = 927 m THIS MEANS THAT THE AVERAGE DISTANCE BETWEEN RELATIVE PHOTONS WITHIN A RELATIVE QUARK IS

927 m.

THE REPULSION FORCE BETWEEN RELATIVE PHOTONS WOULD BE:

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (9.1912650274529 × 10^31) * (9.1912650274529 × 10^31)) / (927)^2 =

8.837934966885326105 × 10^67 N

THERE IS NO WAY THAT THE ATTRACTIVE FORCE THAT THE ELECTRON HAS ON EACH PHOTON, WOULD BE

ABLE TO KEEP THE QUARKS FROM EXPLODING APART BECAUSE OF THEIR REPULSIVE P-STRING PHOTONS.

R = 1.240 m

FORCE = ((8.9876 × 10^9) * (9.1912650274529 × 10^31) * (9.1912650274529 × 10^31)) / (1.240)^2 =

4.93799838234344647 × 10^73 N

THIS IS THE FORCE OF REPULSION BETWEEN TWO N-STRING PHOTONS THAT ARE IN CONTACT WITH EACH

OTHER WITHIN AN ELECTRON.

----

RADIUS

FORCE OF REPULSION BETWEEN PHOTONS

ELECTRON

7.664 × 10^19 m

4.93799838234344647 × 10^73 N

PROTON

5.238 × 10^23 m

8.83793496688532610 × 10^67 N

MAGNETIC MONOPOLE

F = L X T

R^3

F = FORCE L = MAGNETIC CONSTANT = 1.2566438025 × 10^-6 T = TESLA R = DISTANCE

I WILL HYPOTHESIZE THAT THE FORCE OF THE MAGNETIC FIELD AROUND THE PROTON IS TEN TIMES AS

GREAT AS THE PHOTON REPULSION FORCE:

FORCE = (K * Q1 * Q2) / R^2

FORCE = 10 * 8.837934966885326105 × 10^67 N

FORCE = 8.837934966885326105 × 10^68 N

F = (L * T)/R^3

F * R^3 = (L * T)

(F * R^3)/L = T

T = (F * R^3)/L

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T = ((8.837934966885326105 × 10^68) * (5.238 × 10^23)^3)/(1.2566438025 × 10^-6) T = ((8.837934966885326105 × 10^68) * (5.238 × 10^23)^3)/(1.2566438025 × 10^-6) =

1.01072984557908525741657370 × 10^146

USING THIS FIGURE OF T = 1.01072984557908525741657370 × 10^146 AS THE ELECTRON'S TESLA FIGURE: F = (L * T)/R^3 F = ((1.2566438025 × 10^-6) * (1.01072984557908525741657370 × 10^146))/(7.664 × 10^19)^3

= 2.82150481391573874129832923 × 10^80

----

FORCE OF REPULSION BETWEEN PHOTONS

MAGNETIC FIELD STRENGTH

ELECTRON

4.937998382343446 × 10^73 N

2.821504813915738741 × 10^80 N

PROTON

8.837934966885326 × 10^67 N

8.837934966885326105 × 10^68 N

BOTH THE MAGNETIC FIELD AROUND THE ELECTRON AND THE MAGNETIC FIELD AROUND THE PROTON

ARE HYPOTHESIZED TO BE EQUAL TO THE SAME AMOUNT OF TESLAS WHICH IS: T = 1.01072984557908525741657370 × 10^146

THE MAGNETIC FIELD THAT IS GENERATED AROUND THE ELECTRON AT THE CENTER OF THE PROTON

SERVES AS BOTH A BARRIER TO KEEP THE N-STRING PHOTONS WITHIN THE ELECTRON FROM ESCAPING, AND ALSO TO KEEP THE P-STRING PHOTONS FROM THE QUARK FROM BEING PHYSICALLY PULLED INTO

THE ELECTRON.

THE MAGNETIC FIELD THAT IS GENERATED AROUND THE PROTON SERVES TO KEEP THE P-STRING

PHOTONS IN THE QUARKS FROM ESCAPING AND THUS MAINTAINS THE STRUCTURAL INTEGRITY OF THE

PROTON.

THE NUCLEAR STRUCTURE

THE DYNAMISM MODEL FOR THE NUCLEUS IS BASED ON THE IDEA THAT THE NUCLEUS IS COMPOSED

PRIMARILY OF ALPHA PARTICLES WHICH RESIDE IN NUCLEAR SHELLS.

I HYPOTHESIZE THAT ALL HELIUM PARTICLES ARE BASED ON A SIMILAR TYPE OF STRUCTURE TO THE

DYNAMISM MODEL FOR A PROTON. THE PROTON IS COMPOSED OF THREE ALPHA PARTICLES WHICH ORBIT

AN ELECTRON AT IT CENTER. THIS IS AN EXTREMELY STABLE ARRANGEMENT.

THUS, BY ANALOGICAL EXTENSION I PROPOSE THE FOLLOWING:

• THE DEUTERON (H2) CONSISTS OF A PROTON AND A NEUTRON. I SUBMIT THAT THE STRUCTURE OF A DEUTERON IS ACTUALLY TWO PROTONS ORBITING A SINGLE ELECTRON AT THE CENTER OF THE

DEUTERON IN EXACTLY THE SAME WAY THAT THE THREE QUARKS ORBIT THE CENTRAL ELECTRON

WITHIN A PROTON.

• THE TRITON (H3) CONSISTS OF A PROTON AND TWO NEUTRONS. I HYPOTHESIZE THAT THE

STRUCTURE OF THE TRITON IS ACTUALLY THREE PROTONS ORBITING A DOUBLE ELECTRON CENTER.

THE TRITON DOES NOT EXIST IN NATURES AS STABLY AS THE PROTON OR DEUTERON BECAUSE THIS

STRUCTURE DOES NOT PROVIDE A STRONG ENOUGH MAGNETIC FIELD TO KEEP THE PARTICLE INTACT.

• THE HELIUM THREE PARTICLE (HE3) CONSISTS OF TWO PROTONS AND ONE NEUTRON. I HYPOTHESIZE

THAT THE STRUCTURE OF THE HE3 IS ACTUALLY THREE PROTONS ORBITING A SINGLE ELECTRON

CENTER. THE HE3 DOES NOT EXIST AS STABLY AS THE PROTON OR DEUTERON WITHIN NATURE

BECAUSE THE FUSION PROCESS BY WHICH ALL OF THE NATURAL ELEMENTS ON THE PERIDODIC

TABLE ARE CREATED, NEVER LEAVES AN HE3 PARTICLE AS A STAND ALONE PARTICLE. WITHIN THE

PLASMA WHERE ALL THE ELEMENTS ARE CREATED THROUGH THE FUSION PROCESS, THE HE3 HAS

TOO HIGH A CHARGE TO MASS RATIO TO BE LEFT ALONE, AND THUS WILL ALWAYS ATTRACT MORE

PROTON/ELECTRON PAIRS WITHIN THE PLASMA, IN ORDER TO FUSE INTO A HEAVIER ISOTOPE.

• BOTH THE TRITON (H3), AND THE HELIUM THREE PARTICLE (HE3) LACK THE STRONG MAGNETIC

FIELD NEEDED TO KEEP THESE PARTICLES STABLE OUTSIDE OF THE NUCLEUS. THE MAGNETIC

MONOPOLE, WHICH IS UNDER THE INVERSE CUBE LAW, IS GENERATED BY BOTH THE ELECTRONS AT

THE CENTER AND BY THE PROTONS THAT ORBIT THE ELECTRONS. THE MAGNETIC FIELDS GENERATED

BY BOTH WILL ALIGN TOGETHER TO CREATE ONE MAGNETIC FIELD THAT SURROUNDS THE

ELECTRONS AT THE CENTER, WHICH HAVE A NEGATIVE CHARGE, AND ALSO SURROUNDS THE ENTIRE

PARTICLE, WHICH HAS AN OVERALL POSITIVE CHARGE. WITH THE TRITON, THE OVERALL POSITIVE

CHARGE IS ONLY EQUAL TO ONE AND THIS IS NOT ENOUGH TO GENERATE A STRONG ENOUGH

MAGNETIC FIELD WHICH CAN SURROUND THE ENTIRE PARTICLE AND KEEP THE PROTONS FROM

FLYING AWAY. IRONICALLY, WITH THE HE3 PARTICLE, IT IS THE SINGLE ELECTRON AT THE CENTER, WHICH ONLY HAS A NEGATIVE CHARGE EQUAL TO ONE, THAT CANNOT GENERATE A STRONG ENOUGH

MAGNETIC FIELD TO SURROUND THE ELECTRON AND TO LOCK THE PROTONS INTO ORBIT AROUND

THE ELECTRON. THE LARGER ALPHA PARTICLES, (HE4, HE5, HE6), ALL HAVE AN OVERALL POSITIVE

CHARGE OF TWO, WHICH CREATES STRONG ENOUGH MAGNETIC FIELD TO SURROUND THE PARTICLE

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NEGATIVE CHARGE OF TWO OR MORE, AND THUS THESE ELECTRONS GENERATE A STRONG MAGNETIC

FIELD WHICH SURROUNDS THEM AND LOCKS THE ORBITING PROTONS INTO PLACE. IT IS BECAUSE OF

THE POWER OF THE ALIGNMENT OF THE MAGNETIC FIELDS GENERATED BY BOTH THE ELECTRONS AT

THE CENTER, AND BY THE ORBITING PROTONS, THAT CAUSES ALL ALPHA PARTICLES LARGER THAN

HE3, TO BE SO STABLE INSIDE OR OUTSIDE THE NUCLEUS.

• THE HELIUM FOUR PARTICLE (HE4) CONSISTS OF TWO PROTONS AND TWO NEUTRONS. I HYPOTHESIZE THAT THE STRUCTURE OF THE HE4 IS ACTUALLY FOUR PROTONS ORBITING A DOUBLE

ELECTRON CENTER. THE HE4 IS AN EXTREMELY STABLE PARTICLE BECAUSE IT HAS A PROPER

BALANCE BETWEEN ITS CHARGE AND ITS MASS.

• THE HELIUM FIVE PARTICLE (HE5) CONSISTS OF TWO PROTONS AND THREE NEUTRONS. I HYPOTHESIZE THAT THE STRUCTURE OF THE HE5 IS ACTUALLY FIVE PROTONS ORBITING A TRIPLE

ELECTRON CENTER. THE HE5 IS ALSO AN EXTREMELY STABLE PARTICLE BECAUSE IT HAS A PROPER

BALANCE BETWEEN ITS CHARGE AND ITS MASS.

• THE HELIUM SIX PARTICLE (HE6) CONSISTS OF TWO PROTONS AND FOUR NEUTRONS. I HYPOTHESIZE

THAT THE STRUCTURE OF THE HE6 IS ACTUALLY SIX PROTONS ORBITING A QUADRUPLE ELECTRON

CENTER. THE HE6 IS ALSO AN EXTREMELY STABLE PARTICLE BECAUSE IT HAS A PROPER BALANCE

BETWEEN ITS CHARGE AND ITS MASS.

• THE DOUBLE, TRIPLE, AND QUADRUPLE ELECTRON CENTERS ARE SIMPLE TWO, THREE, OR FOUR

ELECTRONS THAT ARE IN VERY CLOSE PROXIMITY TO EACH OTHER. THEY DO NOT FLY APART WHEN

CONTAINED WITH AN APPROPRIATELY BALANCED HELIUM NUCLEUS, BECAUSE OF THE MAGNETIC

FIELD THAT THEY COLLECTIVELY GENERATE. THE MAGNETIC FIELD GENERATED BY THE ELECTRONS, IS SUBJECT TO THE INVERSE CUBE LAW AND IS EXTREMELY POWERFUL. THE FIELD GENERATED IN

THIS MANNER SERVES TO BOTH KEEP THE ELECTRONS FROM REPELLING EACH OTHER, AND ALSO TO

KEEP THE POSITIVELY CHARGED PARTICLES IN ORBIT AROUND THE ELECTRONS FROM BEING PULLED

INTO THE NEGATIVELY CHARGED ELECTRONS AT THE CENTER.

• THE TRITON DOES NOT EXIST STABLY OUTSIDE OF THE NUCLEUS IN MUCH THE SAME WAY THAT

NEUTRONS DON'T EXIST STABLY OUTSIDE OF THE NUCLEUS, BUT BY ANALOGY, THE NEUTRONS CAN

EXIST STABLY WITHIN A NUCLEUS AND LIKEWISE, SO CAN THE TRITON ALSO EXIST STABLY WITHIN

THE NUCLEUS.

THE HYDROGEN AND HELIUM PARTICLES

THE DYNAMISM MODEL FOR THE HYDROGEN ISOTOPES IS ALWAYS A NEGATIVELY CHARGED PARTICLE

SURROUNDED BY POSITIVELY CHARGED PARTICLES. THIS FRACTAL STRUCTURE IS HYPOTHESIZED AS THE

MOST LIKELY STRUCTURE FOR THE MOST COMMONLY ABUNDANT ISOTOPES IN THE UNIVERSE.

AS PER THE FRACTAL STRUCTURE, I CAN MAKE CERTAIN ANALOGICAL EXTENSIONS IN REGARDS TO THIS

ARRANGEMENT. SINCE THE PROTON IS STABLE AND THE NEUTRON IS NOT; AND BOTH THE DEUTERON

AND THE HE3 ALPHA PARTICLE ARE STABLE, BUT THE TRITON IS NOT; THEN BY ANALOGICAL EXTENSION, THERE MUST BE SOMETHING THAT THE PROTON, THE DEUTERON, AND THE HE3 ALPHA PARTICLE HAVE IN

COMMON, WHICH IS NOT PRESENT IN EITHER THE NEUTRON OR THE TRITON. LIKEWISE, THE NEUTRON

AND THE TRITON MUST HAVE SOMETHING IN COMMON THAT IS NOT PRESENT IN THE PROTON (H1), DEUTERON (H2), OR HE3 PARTICLE.

I SUBMIT THAT IT IS THE RATIO OF POSITIVE TO NEGATIVE COMPONENTS THAT MAKES THE CRITICAL

DIFFERENCE:

• THE PROTON (H1) IS STABLE WITH THREE POSITIVE QUARKS OVER ONE NEGATIVE ELECTRON.(3 : 1)

• THE HE3 IS STABLE WITH THREE POSITIVE PROTONS OVER ONE NEGATIVE ELECTRON.(3 : 1)

• THE DEUTERON (H2) IS STABLE WITH TWO POSITIVE PROTONS OVER ONE NEGATIVE ELECTRON.(2 : 1)

• THE NEUTRON IS UNSTABLE WITH THREE POSITIVE QUARKS OVER TWO NEGATIVE ELECTRONS.(3 : 2)

• THE TRITON (H3) IS UNSTABLE WITH THREE POSITIVE PROTONS OVER TWO NEGATIVE ELECTRONS.(3 : 2)

NOTE THAT THE TERM 'STABLE' IS USED HERE IN REGARDS TO STAND ALONE PARTICLES. WITHIN THE

NUCLEUS ALL PARTICLES ARE 'STABLE'. BUT OUTSIDE OF THE NUCLEUS, BOTH THE NEUTRON AND THE

TRITON ARE 'UNSTABLE' AS THEY WILL BOTH DISINTEGRATE THROUGH BETA PARTICLE RADIATION AS A STAND ALONE PARTICLE.

WHEN THE RATIO FOR THE PARTICLE IS (3 : 1), OR (2 : 1), THEN THE MONOPOLAR MAGNETIC FIELD IS THE

MOST RELEVANT FACTOR THAT MAINTAINS THE INTEGRITY OF THE PARTICLE. BUT WHEN THE RATIO IS (3 : 2), THEN THE MONOPOLAR MAGNETIC FIELD IS MADE TO BE IRRELEVANT, AND THE RESPECTIVE

DIFFERENCES IN POSITVE ELECTRICAL FORCE COMPARED TO NEGATIVE ELECTRICAL FORCE BECOMES

THE MOST SIGNIFICANT FACTOR.

THE DEUTERON

THE DEUTERON, ON THE OTHER HAND, HAS ONLY ONE ELECTRON AT ITS CENTER, BETWEEN TWO

PROTONS. THE 'R' VALUE BETWEEN THE PROTONS IS ASSUMED TO BE THE SAME AS THE DIAMETER OF A PROTON.

THE REPULSIVE FORCE IS BETWEEN THE TWO PROTONS.

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FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (1 * 1.13566346171621 × 10^50)) / (1.0476 ×

10^24)^2 = 1.056214347×10⁶² N

THE ATTRACTIVE FORCE IS BETWEEN THE TWO PROTONS AND THE ELECTRON. HERE THE 'R' VALUE IS THE

ASSUMED TO BE THE RADIUS OF A PROTON.

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (5.238 ×

10^23)^2 = 8.449714778×10⁶² N

THE TOTAL ATTRACTIVE FORCES WITHIN THE DEUTERON:

8.449714778 × 10^62 N

THE TOTAL REPULSIVE FORCES WITHIN THE DEUTERON:

1.056214347 × 10^62 N

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE DEUTERON: 7.393500431 × 10^62 N

EVEN WITHOUT FACTORING IN THE MAGNETIC FIELD OF THE DEUTERON, AND JUST USING THE

ELECTRICAL CHARGES, THE DEUTERON'S NET ATTRACTION/REPULSIVE FORCE IS A LARGE POSITIVE

FIGURE.

THE HE3 ALPHA PARTICLE

THE HE3 ALSO HAS ONLY ONE ELECTRON AT ITS CENTER, BETWEEN THREE PROTONS. THE 'R' VALUE

BETWEEN THE PROTONS IS ASSUMED TO BE THE SAME AS THE DIAMETER OF A PROTON.

THE REPULSIVE FORCE IS BETWEEN THE THREE PROTONS. I CALCULATE THIS TO BE THE TOTAL CHARGE

OF ONE PROTON REPULSING TWO PROTONS ALL MULTIPLIED BY THREE.

FORCE = (K * Q1 * Q2) / R^2

FORCE = 3 * ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (1.0476 ×

10^24)^2 = 6.337286084×10⁶² N

THE ATTRACTIVE FORCE IS BETWEEN THE THREE PROTONS AND THE ELECTRON. HERE THE 'R' VALUE IS

ASSUMED TO BE THE RADIUS OF A PROTON.

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (3 * 1.13566346171621 × 10^50)) / (5.238 ×

10^23)^2 = 1.267457217×10⁶³ N

THE TOTAL ATTRACTIVE FORCES WITHIN THE HE3:

6.337286084 × 10^62 N

THE TOTAL REPULSIVE FORCES WITHIN THE HE3:

1.267457217 × 10^63 N

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE HE3:

6.337286084 × 10^62 N

EVEN WITHOUT FACTORING IN THE MAGNETIC FIELD OF THE HE3 ALPHA PARTICLE, AND JUST USING THE

ELECTRICAL CHARGES, THE HE3'S NET ATTRACTION/REPULSIVE FORCE IS A LARGE POSITIVE FIGURE.

THE NEUTRON

THE NEUTRON IS A NEUTRAL PARTICLE THAT IS ACTUALLY A COMPOSITION OF TWO PARTICLES; ONE

NEGATIVELY CHARGED ELECTRON; AND ONE POSTIVELY CHARGED PROTON. THIS COMPOSITION CAN TAKE

TWO DIFFERENT FORMS IN MY DYNAMISM MODEL. THE FIRST FORM IS A PROTON THAT HAS AN

ELECTRON EMBEDDED WITHIN IT. THIS OCCURS WHEN A NEUTRON IS FORCED TO BE A STAND ALONE

PARTICLE. I HYPOTHESIZE THAT THIS FORM IS RARELY TAKEN BECAUSE THE NEUTRON DOES NOT EXIST

FOR LONG AS A STAND ALONE PARTICLE. THE SECOND FORM IS MUCH MORE PREVALENT AND OCCURS

WHEN A NEUTRON IS EMBEDDED WITHIN A LARGER NUCLEUS. IN THIS CASE, THE NEUTRON TAKES THE

FORM OF A PROTON/ELECTRON PAIR WHERE THE PROTON AND ELECTRON ARE IN ACTUAL CONTACT WITH

EACH OTHER, OR CLOSE ENOUGH TO PHYSICALLY TOUCH, BUT THE ELECTRON PORTION IS PART OF THE

CENTER OF AN ALPHA PARTICLE. BUT REGARDLESS OF WHICH FORM IT TAKES, THE NEUTRON IS ALWAYS A PROTON/ELECTRON PAIR.

THIS GIVES RISE TO A DEFINITE PREDICTION OF THIS DYNAMISM MODEL. THE MODEL PREDICTS THAT IT

IS IMPOSSIBLE FOR NEUTRONS TO BE NATURALLY EMITTED, BY THEMSELVES, AS PARTICLE RADIATION.

THE REASON FOR THE IMPOSSIBILITY IS THAT A NEUTRON IS ALWAYS PART OF A HYDROGEN, OR HELIUM

PARTICLE (MOSTLY HELIUM). THUS ONLY TWO TYPES OF PARTICLE RADIATION ARE POSSIBLE WITHIN THIS

MODEL, AND THAT IS ALPHA PARTICLE, AND BETA PARTICLE RADIATION. ALPHA PARTICLES ARE HELIUM

NUCLEI; AND BETA PARTICLES ARE ELECTRONS.

THE REPULSIVE FORCE THAT THE TWO ELECTRONS HAVE FOR EACH OTHER WITHIN THE STAND ALONE

NEUTRON, IS TOO MUCH FOR THE POSITIVE ELECTRICAL FORCE, WHICH IS ATTRACTIVE TO THE

ELECTRONS, TO BEAR, WITHOUT THE HELP OF A POWERFUL MONOPOLAR MAGNETIC FIELD.

WITHIN EVERY NUCLEUS, THE FORCES OF ATTRACTION, WHICH MAINTAINS THE INTEGRITY OF THE

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NUCLEUS, IS AT WAR WITH THE FORCES OF REPULSION, WHICH WILL DISINTEGRATE THE NUCLEUS TO A MORE STABLE FORMATION WHENEVER THE REPULSIVE FORCES ARE GREATER THAN THE ATTRACTIVE

FORCES.

THE TOTAL REPULSIVE FORCES UPON EACH QUARK WITHIN THE NEUTRON ARE THE SAME AS FOR THE

PROTON:

ELECTRICAL REPULSION EQUALS '1.669525164348052 × 10^62 N'

THE TOTAL REPULSIVE FORCE BETWEEN THE THREE QUARKS WITHIN THE NEUTRON.

FORCE = 3 * 1.669525164348052 × 10^62 = 5.008575493×10⁶² N

THE ELECTRON'S RADIUS IS 7.664 × 10^19 m

THE ELECTRON'S DIAMETER IS 1.5328 × 10^20 m

THEREFORE THE SUPERSIZED ELECTRON'S RELATIVE CHARGE IS 1.13566346171621 × 10^50 C

FORCE = (K * Q1 * Q2) / R^2

THE REPULSIVE FORCE BETWEEN THE TWO ELECTRONS AT THE CENTER OF THE NEUTRON IS

CALCULATED USING THE ELECTRON'S DIAMETER FIGURE FOR 'R'.

R = 1.5328 × 10^20

K = 8.9876 × 10^9

Q1 = Q2 = 1.13566346171621 × 10^50

FORCE = (K * Q1 * Q2) / (R)^2

FORCE = ((8.9876 × 10^9) * (1.13566346171621 × 10^50)^2) / (1.5328 × 10^20)^2

FORCE = 4.93369228 × 10^69 N

THE ATTRACTIVE FORCE BETWEEN THE QUARK AND THE ELECTRONS AT THE CENTER OF THE NEUTRON IS

CALCULATED USING THE PROTON'S RADIUS FIGURE FOR 'R'.

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (2 * 1.13566346171621 × 10^50) * (3 * 7.5710897447747333 × 10^49)) /

(4.7033298 × 10^22)^2 = 2.096004293 × 10^65 N

THE TOTAL ATTRACTIVE FORCES WITHIN THE NEUTRON:

2.09600429 × 10^65 N

THE REPULSIVE FORCES BETWEEN ELECTRONS WITHIN THE NEUTRON: 4.93369228 × 10^69 N

THE TOTAL REPULSIVE FORCES BETWEEN QUARKS WITHIN THE NEUTRON: 5.008575493 × 10^62 N

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE NEUTRON:

-4.93348318 × 10^69 N

THE LACK OF A MAGNETIC FIELD MEANS THAT THE NET ATTRACTION/REPULSION WITHIN THE NEUTRON

IS A LARGE NEGATIVE NUMBER (-4.93348318 × 10^69); WHICH IS WHY THE NEUTRON IS NOT A STABLE

PARTICLE WHEN IT STANDS BY ITSELF OUTSIDE OF THE NUCLEUS OF AN ATOM.

THE TRITON

THE ATTRACTIVE FORCE WITHIN A TRITON IS CALCULATED USING THE DIAMETER OF A PROTON PLUS THE

DIAMETER OF THE ELECTRON AS THE 'R' FIGURE.

PROTON RADIUS = 5.238 × 10^23 m

PROTON DIAMETER = 2 * (5.238 × 10^23) = 1.0476 × 10^24 m

THE ELECTRON'S DIAMETER IS 1.5328 × 10^20 m

PROTON DIAMETER + ELECTRON DIAMETER = 2 * (5.238 × 10^23) = 1.0476 × 10^24 + 1.5328 × 10^20 =

1.04775328 × 10^24 m

THE TOTAL REPULSIVE FORCE BETWEEN THE THREE PROTONS WITHIN THE TRITON.

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (2 * 1.13566346171621 × 10^50) * (3 * 1.13566346171621 × 10^50)) / (1.04775328

× 10^24)^2 = 6.335432006 × 10^62 N

THE ATTRACTIVE FORCES WITHIN THE TRITON:

6.337286084 × 10^62 N

THE REPULSIVE FORCES BETWEEN PROTONS WITHIN THE TRITON: 6.335432006 × 10^62 N

THE REPULSIVE FORCES BETWEEN ELECTRONS WITHIN THE TRITON: 4.93369228 × 10^69 N

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE TRITON:

-4.93369228 × 10^69 N

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THE WEAKNESS OF THE OF THE MAGNETIC FIELD WITHIN THE TRITON BECAUSE IT ONLY HAS A POSITIVE

CHARGE EQUIVALENT TO ONE PROTON; WHILE IT HAS TWO NEGATIVELY CHARGED ELECTRONS AT ITS

CENTER, WHICH REPULSE EACH OTHER; MEANS THAT THE NET ATTRACTION/REPULSION WITHIN THE

TRITON IS ALSO A LARGE NEGATIVE NUMBER (-4.93369228 × 10^69); WHICH IS ALSO WHY THE TRITON IS

NOT A STABLE PARTICLE WHEN IT STANDS BY ITSELF OUTSIDE OF THE NUCLEUS OF AN ATOM.

HOWEVER THE ABOVE HYPOTHESIS FAILS TO ACCOUNT FOR THE MAGNETIC FIELD OF THE TRITON. THE

NEUTRON HAS NO OVERALL CHARGE AND SO CANNOT GENERATE ANY KIND OF MAGNETIC FIELD.

HOWEVER, THE TRITON DOES HAVE AN OVERALL CHARGE AND THUS THE TRITON CAN GENERATE A MAGNETIC FIELD WHICH OBEYS THE INVERSE CUBE LAS FOR MONOPOLAR MAGNETIC FIELDS.

F = (L * T) / R^3

F = ((1.2566438025 × 10^-6) * (1.01072984557908525741657370 × 10^146)) / R^3

F = (1.270127396 × 10^140) / R^3

SO WHAT WOULD THE 'R' FIGURE BE FOR THE TRITON?

THE OUTER 'R' FIGURE WOULD BE THE SAME AS THE ONE USED FOR THE CALCULATION OF THE

ELECTRICAL REPULSION FIGURE BETWEEN THE PROTONS:

R = 1.04775328 × 10^24 m

THE MAGNETIC FORCE FIELD AT THE OUTER EDGE FOR THE TRITON IS: F = (1.270127396 × 10^140) / (1.04775328 × 10^24)^3 = 1.10425709 ×10^68 THE INNER 'R' FIGURE WOULD

BE THE SAME AS THE ONE USED FOR THE CALCULATION OF THE ELECTRICAL REPULSION FIGURE

BETWEEN THE ELECTRONS:

THE ELECTRON'S DIAMETER IS 1.5328 × 10^20 m

THE MAGNETIC FORCE FIELD AT THE CENTER OF THE TRITON IS:

F = (1.270127396 × 10^140) / (1.5328 × 10^20)^3 = 3.526881016 × 10^79 m IF MAGNETISM IS TAKEN INTO ACCOUNT THEN THE TABLE BECOMES: 3.526881016 × 10^79

THE MAGNETIC FORCES WITHIN THE TRITON:

THE ATTRACTIVE FORCES BETWEEN PROTONS AND ELECTRON WITHIN THE 6.337286084 × 10^62

TRITON:

6.337286084 × 10^62

THE REPULSIVE FORCES BETWEEN PROTONS WITHIN THE TRITON: N

4.93369228 × 10^69

THE REPULSIVE FORCES BETWEEN ELECTRONS WITHIN THE TRITON: N

3.526881016 × 10^79

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE TRITON:

THE TRITON HAS ONLY ONE PROTON AS COMPARED TO TWO NEUTRONS, BUT BECAUSE IT HAS SOME

CHARGE, ITS RAPID ROTATIONAL MOVEMENT CONVERTS ITS ELECTRICAL CHARGE INTO A MONOPOLAR

MAGNETIC FIELD WHICH CREATES AN EXTREMELY STRONG BINDING FORCE THAT MAINTAINS THE

STRUCTURAL INTEGRITY OF THE TRITON PARTICLE. BECAUSE THE OTHER FORCES CANCEL OUT, THE ONLY

REPULSIVE FORCE OF NOTE WITHIN THE TRITON, IS THE REPULSION BETWEEN THE TWO ELECTRONS.

THIS REPULSIVE FORCE IS MORE THAN OFFSET BY THE MONOPOLAR MAGNETIC FIELD.

THE HE4 ALPHA PARTICLE

THE HE4 ALSO HAS TWO ELECTRONS AT ITS CENTER, BETWEEN FOUR PROTONS. THE 'R' VALUE BETWEEN

THE PROTONS IS ASSUMED TO BE THE SAME AS THE DIAMETER OF A PROTON.

THE REPULSIVE FORCE IS BETWEEN THE FOUR PROTONS. I CALCULATE THIS TO BE THE TOTAL CHARGE OF

ONE PROTON REPULSING THREE PROTONS ALL MULTIPLIED BY FOUR.

FORCE = (K * Q1 * Q2) / R^2

FORCE = 4 * ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (3 * 1.13566346171621 × 10^50)) / (1.0476 ×

10^24)^2 = 1.267457217×10⁶³ N

THE ATTRACTIVE FORCE IS BETWEEN THE FOUR PROTONS AND THE TWO ELECTRONS. HERE THE 'R' VALUE

IS ASSUMED TO BE THE RADIUS OF A PROTON.

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FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (2 * 1.13566346171621 × 10^50) * (4 * 1.13566346171621 × 10^50)) / (5.238 ×

10^23)^2 = 3.379885911×10⁶³ N

3.526881016 ×

THE MAGNETIC FORCES WITHIN THE HE4:

10^79 N

THE TOTAL ATTRACTIVE FORCES BETWEEN THE PROTONS AND ELECTRONS

3.379885911 ×

WITHIN THE HE4:

10^63 N

4.93369228 × 10^69

THE REPULSIVE FORCES BETWEEN ELECTRONS WITHIN THE HE4:

1.267457217 ×

THE REPULSIVE FORCES BETWEEN PROTONS WITHIN THE HE4:

10^63 N

4.933693547 ×

THE TOTAL REPULSIVE FORCES WITHIN THE HE4:

10^69 N

3.526881016 ×

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE HE4:

10^79 N

THE REPULSIVE FORCE BETWEEN THE TWO ELECTRONS WITHIN THE HE4 PARTICLE IS THE PRIMARY

FORCE THAT WOULD DISINTEGRATE THE PARTICLE. HOWEVER, THE MONOPOLAR MAGNETIC FIELD WHICH

SURROUNDS THE TWO CENTRAL ELECTRONS IS MORE THAN ENOUGH TO RESTRAIN THE TWO REPULSIVE

ELECTRONS. NOTE THAT ONCE THE REPULSIVENESS OF THE ELECTRONS ARE CONTAINED BY THE

MAGNETIC FIELD, THAT THE ATTRACTIVE FORCE BETWEEN PROTONS AND ELECTRONS IS GREATER THAN

THE REPULSIVE FORCE BETWEEN THE PROTONS.

THE HE5 ALPHA PARTICLE

THE HE5 HAS THREE ELECTRONS AT ITS CENTER, BETWEEN FIVE PROTONS. THE 'R' VALUE BETWEEN THE

PROTONS IS ASSUMED TO BE THE SAME AS THE DIAMETER OF A PROTON.

THE REPULSIVE FORCE BETWEEN THE THREE ELECTRONS AT THE CENTER OF THE HE5 IS CALCULATED

USING THE ELECTRON'S DIAMETER FIGURE FOR 'R'. I CALCULATE THIS TO BE THE TOTAL CHARGE OF ONE

ELECTRON REPULSING TWO ELECTRONS ALL MULTIPLIED BY THREE.

FORCE = 3 * ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (1.5328 ×

10^20)^2 = 2.960215368×10⁷⁰ N

THE REPULSIVE FORCE IS BETWEEN THE FIVE PROTONS. I CALCULATE THIS TO BE THE TOTAL CHARGE OF

ONE PROTON REPULSING FOUR PROTONS ALL MULTIPLIED BY FIVE.

FORCE = (K * Q1 * Q2) / R^2

FORCE = 5 * ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (4 * 1.13566346171621 × 10^50)) / (1.0476 ×

10^24)^2 = 2.112428695×10⁶³ N

THE ATTRACTIVE FORCE IS BETWEEN THE FIVE PROTONS AND THE THREE ELECTRONS. HERE THE 'R'

VALUE IS ASSUMED TO BE THE RADIUS OF A PROTON.

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (3 * 1.13566346171621 × 10^50) * (5 * 1.13566346171621 × 10^50)) / (5.238 ×

10^23)^2 = 6.337286084×10⁶³ N

3.526881016 ×

THE MAGNETIC FORCES WITHIN THE HE5:

10^79 N

THE TOTAL ATTRACTIVE FORCES BETWEEN THE PROTONS AND ELECTRONS

6.337286084 ×

WITHIN THE HE5:

10^63 N

2.960215368 ×

THE REPULSIVE FORCES BETWEEN ELECTRONS WITHIN THE HE5:

10^70 N

2.112428695 ×

THE REPULSIVE FORCES BETWEEN PROTONS WITHIN THE HE5:

10^63 N

2.960215368 ×

THE TOTAL REPULSIVE FORCES WITHIN THE HE5:

10^70 N

3.526881013 ×

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE HE5:

10^79 N

THE REPULSIVE FORCE BETWEEN THE THREE ELECTRONS WITHIN THE HE5 PARTICLE IS THE PRIMARY

FORCE THAT WOULD DISINTEGRATE THE PARTICLE. HOWEVER, THE MONOPOLAR MAGNETIC FIELD WHICH

SURROUNDS THE THREE CENTRAL ELECTRONS IS MORE THAN ENOUGH TO RESTRAIN THE THREE

REPULSIVE ELECTRONS. NOTE THAT ONCE THE REPULSIVENESS OF THE ELECTRONS ARE CONTAINED BY

THE MAGNETIC FIELD, THAT THE ATTRACTIVE FORCE BETWEEN PROTONS AND ELECTRONS IS GREATER

THAN THE REPULSIVE FORCE BETWEEN THE PROTONS.

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THE HE6 ALPHA PARTICLE

THE HE6 HAS FOUR ELECTRONS AT ITS CENTER, BETWEEN SIX PROTONS. THE 'R' VALUE BETWEEN THE

PROTONS IS ASSUMED TO BE THE SAME AS THE DIAMETER OF A PROTON.

THE REPULSIVE FORCE BETWEEN THE FOUR ELECTRONS AT THE CENTER OF THE HE6 IS CALCULATED

USING THE ELECTRON'S DIAMETER FIGURE FOR 'R'. I CALCULATE THIS TO BE THE TOTAL CHARGE OF ONE

ELECTRON REPULSING THREE ELECTRONS ALL MULTIPLIED BY FOUR.

FORCE = 4 * ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (3 * 1.13566346171621 × 10^50)) / (1.5328 ×

10^20)^2 = 5.920430736×10⁷⁰ N

THE REPULSIVE FORCE IS BETWEEN THE SIX PROTONS. I CALCULATE THIS TO BE THE TOTAL CHARGE OF

ONE PROTON REPULSING FIVE PROTONS ALL MULTIPLIED BY SIX.

FORCE = (K * Q1 * Q2) / R^2

FORCE = 6 * ((8.9876 × 10^9) * (1 * 1.13566346171621 × 10^50) * (5 * 1.13566346171621 × 10^50)) / (1.0476 ×

10^24)^2 = 3.168643042×10⁶³ N

THE ATTRACTIVE FORCE IS BETWEEN THE SIX PROTONS AND THE FOUR ELECTRONS. HERE THE 'R' VALUE

IS ASSUMED TO BE THE RADIUS OF A PROTON.

FORCE = (K * Q1 * Q2) / R^2

FORCE = ((8.9876 × 10^9) * (4 * 1.13566346171621 × 10^50) * (6 * 1.13566346171621 × 10^50)) / (5.238 ×

10^23)^2 = 1.013965773×10⁶⁴ N

3.526881016 ×

THE MAGNETIC FORCES WITHIN THE HE6:

10^79 N

THE TOTAL ATTRACTIVE FORCES BETWEEN THE PROTONS AND ELECTRONS

1.013965773 ×

WITHIN THE HE6:

10^64 N

5.920430736 ×

THE REPULSIVE FORCES BETWEEN ELECTRONS WITHIN THE HE6:

10^70 N

3.168643042 ×

THE REPULSIVE FORCES BETWEEN PROTONS WITHIN THE HE6:

10^63 N

5.920430736 ×

THE TOTAL REPULSIVE FORCES WITHIN THE HE6:

10^70 N

3.52688101 × 10^79

THE NET ATTRACTION/REPULSIVE FORCES WITHIN THE HE6:

THE REPULSIVE FORCE BETWEEN THE FOUR ELECTRONS WITHIN THE HE6 PARTICLE IS THE PRIMARY

FORCE THAT WOULD DISINTEGRATE THE PARTICLE. HOWEVER, THE MONOPOLAR MAGNETIC FIELD WHICH

SURROUNDS THE FOUR CENTRAL ELECTRONS IS MORE THAN ENOUGH TO RESTRAIN THE FOUR

REPULSIVE ELECTRONS. NOTE THAT ONCE THE REPULSIVENESS OF THE ELECTRONS ARE CONTAINED BY

THE MAGNETIC FIELD, THAT THE ATTRACTIVE FORCE BETWEEN PROTONS AND ELECTRONS IS GREATER

THAN THE REPULSIVE FORCE BETWEEN THE PROTONS.

___________________________________________________________________________________________________

THE NUCLIDES CHART

IN THE DYNAMISM MODEL, THE NUCLEUS OF ATOMS LARGER THAN THE FIRST ELEMENT, OR HYDROGEN

ATOM, IS COMPOSED PRIMARILY OF HELIUM NUCLEI. THE FOLLOWING IS A PRESENTATION OF THE

RESEARCH THAT I HAVE DONE ON THIS SUBJECT IN THE DEVELOPMENT OF THE DYNAMISM MODEL OF

NUCLEAR PHYSICS.

THE NEXT GRAPH IS 'THE KARLSRUHE NUCLIDE CHART':

___________________________________________________________________________________________________

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IN THE ABOVE GRAPH, THE BLACK SQUARES REPRESENT THE STABLE ISOTOPES. THE SQUARES WITH

OTHER COLORS REPRESENT ISOTOPES THAT ARE UNSTABLE FOR VARIOUS REASONS. IN THIS MODEL FOR

THE NUCLEUS, THE UNSTABLE ISOTOPES RADIATE AWAY THREE MAIN TYPES OF PARTICLES.

• ALPHA PARTICLES.

• BETA PARTICLES.

• NEUTRONS.

THE RELATIVE ABUNDANCE OF STABLE ISOTOPES

THE FOLLOWING TABLES REPRESENT EVERY STABLE ISOTOPE FOR ALL THE ELEMENTS OF THE PERIODIC

TABLE, ALONG WITH THEIR NATURAL ABUNDANCE, WITHIN THE EARTH.

___________________________________________________________________________________________________

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http://moltensalt.org/references/static/downloads/pdf/stable-isotopes.pdf ___________________________________________________________________________________________________

THE NEXT TABLE IS TAKEN FROM THE ONE ABOVE, AND REPRESENTS JUST THE SINGLE MOST ABUNDANT

ISOTOPE FOR EACH ELEMENT.

___________________________________________________________________________________________________

Table of Isotopic Masses and Natural Abundances

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NAME

SYMBOL

ATOMIC MASS

ABUNDANCE

1 Hydrogen

1.007825

99.9885

2 Helium

4.002603

99.999863

7 Lithium

7.016004

92.41

9 Beryllium

9.012182

100

11 Boron

11.009305

80.1

12 Carbon

12.000000

98.93

14 Nitrogen

14.003074

99.632

16 Oxygen

15.994915

99.757

19 Fluorine

18.998403

100

20 Neon

19.992440

90.48

23 Sodium

22.989770

100

24 Magnesium

23.985042

78.99

27 Aluminum

26.981538

100

28 Silicon

27.976927

92.2297

31 Phosphorus

30.973762

100

32 Sulphur

31.972071

94.93

35 Chlorine

34.968853

75.78

40 Argon

39.962383

99.6003

39 Potassium

38.963707

93.2581

40 Calcium

39.962591

96.941

45 Scandium

44.955910

100

48 Titanium

47.947947

73.72

51 Vanadium

50.943964

99.750

52 Chromium

51.940512

83.789

55 Manganese

54.938050

100

56 Iron

55.934942

91.754

59 Cobalt

58.933200

100

58 Nickel

57.935348

68.0769

63 Copper

62.929601

69.17

64 Zinc

63.929147

48.63

69 Gallium

68.925581

60.108

74 Germanium

73.921178

36.28

75 Arsenic

74.921596

100

80 Selenium

79.916522

49.61

79 Bromine

78.918338

50.69

84 Krypton

83.911507

57.00

85 Rubidium

84.911789

72.17

88 Strontium

87.905614

82.58

89 Yttrium

88.905848

100

90 Zirconium

89.904704

51.45

93 Niobium

92.906378

100

98 Molybdenum

97.905408

24.13

98 Technetium

97.907216

102 Ruthenium

101.904350

31.55

103 Rhodium

102.905504

100

106 Palladium

105.903483

27.33

108 Silver

106.905093

51.839

114 Cadmium

113.903358

28.73

115 Indium

114.903878

95.71

120 Tin

119.902197

32.58

121 Antimony

120.903818

57.21

130 Tellurium

129.906223

34.08

127 Iodine

126.904468

100

132 Xenon

131.904154

26.89

133 Cesium

132.905447

100

138 Barium

137.905241

71.698

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139 Lanthanum

138.906348

99.910

140 Cerium

139.905434

88.450

141 Praseodymium

140.907648

100

142 Neodymium

141.907719

27.2

145 Promethium

144.912744

152 Samarium

151.919728

26.75

153 Europium

152.921226

52.19

158 Gadolinium

157.924101

24.84

159 Terbium

158.925343

100

164 Dysprosium

163.929171

28.18

165 Holmium

164.930319

100

166 Erbium

165.930290

33.61

169 Thulium

168.934211

100

174 Ytterbium

173.938858

31.83

175 Lutetium

174.940768

97.41

180 Hafnium

179.946549

35.08

181 Tantalum

180.947996

99.988

184 Tungsten

183.950933

30.64

187 Rhenium

186.955751

62.60

192 Osmium

191.961479

40.78

193 Iridium

192.962924

62.7

195 Platinum

194.964774

33.832

197 Gold

196.966552

100

202 Mercury

201.970626

29.86

205 Thallium

204.974412

70.476

208 Lead

207.976636

52.4

209 Bismuth

208.980383

100

209 Polonium

208.982416

210 Astatine

209.987131

222 Radon

222.017570

223 Francium

223.019731

226 Radium

226.025403

227 Actinium

227.027747

232 Thorium

232.038050

100

231 Protactinium

231.035879

100

238 Uranium

238.050783

99.2745

237 Neptunium

237.048167

244 Plutonium

244.064198

243 Americium

243.061373

247 Curium

247.070347

247 Berkelium

247.070299

251 Californium

251.079580

252 Einsteinium

252.082972

10O

257 Fermium

257.095099

101

258 Mendelevium

258.098425

102

259 Nobelium

259.101024

103

262 Lawrencium

262.109692

104

263 Rutherfordium

263.118313

105

262 Dubnium

262.011437

106

266 Seaborgium

266.012238

107

264 Bohrium

264.012496

108

269 Hassium

269.001341

109

268 Meitnerium

268.001388

11O

272 Ununnilium

Uun

272.001463

111

272 Unununium

Uuu

272.001535

112

277 Ununbium

Uub

(277)

114

289 Ununquadium

Uuq

(289)

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116

289 Ununhexium

Uuh

(289)

118

293 Ununoctium

Uuo

(293)

http://moltensalt.org/references/static/downloads/pdf/stable-isotopes.pdf ___________________________________________________________________________________________________

THIS NEXT TABLE IS TAKEN FROM THE ONE ABOVE AND FEATURES THE ATOMS WHICH HAVE JUST ONE

STABLE ISOTOPE, OR ONE HUNDRED PERCENT ABUNDANCE.

Z NAME

SYMBOL

ATOMIC MASS

ABUNDANCE

4 Beryllium

9 Be

9.012182

100

9 Fluorine

19 F

18.998403

100

11 Sodium

23 Na

22.989770

100

13 Aluminum

27 Al

26.981538

100

15 Phosphorus

31 P

30.973762

100

21 Scandium

45 Sc

44.955910

100

25 Manganese

55 Mn

54.938050

100

27 Cobalt

59 Co

58.933200

100

33 Arsenic

75 As

74.921596

100

39 Yttrium

89 Y

88.905848

100

41 Niobium

93 Nb

92.906378

100

45 Rhodium

103 Rh

102.905504

100

53 Iodine

127 I

126.904468

100

55 Cesium

133 Cs

132.905447

100

59 Praseodymium

141 Pr

140.907648

100

65 Terbium

159 Tb

158.925343

100

67 Holmium

165 Ho

164.930319

100

69 Thulium

169 Tm

168.934211

100

79 Gold

197 Au

196.966552

100

83 Bismuth

209 Bi

208.980383

100

9O Thorium

232 Th

232.038050

100

91 Protactinium

231 Pa

231.035879

100

THE FOLLOWING TABLE HAS ALL THE ODD NUMBERED 'Z' FIGURES FROM THE TABLE ABOVE: Z NAME

TRITONS

DEUTERONS

TOTAL

9 Fluorine

3 * 1 = 3

2 * 8 = 16

11 Sodium

3 * 1 = 3

2 * 10 = 20

13 Aluminum

3 * 1 = 3

2 * 12 = 24

15 Phosphorus

3 * 1 = 3

2 * 14 = 28

21 Scandium

3 * 3 = 9

2 * 18 = 36

25 Manganese

3 * 5 = 15

2 * 20 = 40

27 Cobalt

3 * 5 = 15

2 * 22 = 44

33 Arsenic

3 * 9 = 27

2 * 24 = 48

39 Yttrium

3 * 11 = 33

2 * 28 = 56

41 Niobium

3 * 11 = 33

2 * 30 = 60

45 Rhodium

3 * 13 = 39

2 * 32 = 64

103

53 Iodine

3 * 21 = 63

2 * 32 = 64

127

55 Cesium

3 * 23 = 69

2 * 32 = 64

133

59 Praseodymium

3 * 23 = 69

2 * 36 = 72

141

65 Terbium

3 * 29 = 87

2 * 36 = 72

159

67 Holmium

3 * 31 = 93

2 * 36 = 72

165

69 Thulium

3 * 31 = 93

2 * 38 = 76

169

79 Gold

3 * 39 = 117

2 * 40 = 80

197

83 Bismuth

3 * 43 = 129

2 * 40 = 80

209

91 Protactinium

3 * 49 = 147

2 * 42 = 84

231

THE DYNAMISM MODEL FOR THE NUCLEUS OF ATOMS IS BASED ON THE HYPOTHESIS OF THE EXISTENCE

OF A NUCLEAR SHELL STRUCTURE. IN THIS MODEL, THE NUCLEUS OF EVERY ATOM IS CONSTRUCTED

FROM ONLY SEVEN PARTICLES:

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1. NEUTRON

2. PROTON

3. DEUTERON

4. TRITON

5. HE4 ALPHA PARTICLE

6. HE5 ALPHA PARTICLE

7. HE6 ALPHA PARTICLE

THE DEUTERON CONTAINS A PROTON AND A NEUTRON. THE TRITON CONTAINS A PROTON AND TWO

NEUTONS. THE HE4 ALPHA PARTICLE CONTAINS TWO PROTONS AND TWO NEUTRONS. THE HE5 ALPHA PARTICLE CONTAINS TWO PROTONS AND THREE NEUTRONS. THE HE6 ALPHA PARTICLE CONTAINS TWO

PROTONS AND FOUR NEUTRONS.

THE FOLLOWING TABLE SHOWS THE DYNAMISM MODEL'S DISTRIBUTION OF TRITONS AND DEUTERONS IN

THE ODD NUMBERED ELEMENTS WHICH HAVE ONLY ONE STABLE ISOTOPE: NO. OF

NO. OF

TOTAL

Z NAME

TRITONS

DEUTERONS

PARTICLES

NUCLEONS

9 Fluorine

11 Sodium

13 Aluminum

15 Phosphorus

21 Scandium

25 Manganese

27 Cobalt

33 Arsenic

39 Yttrium

41 Niobium

45 Rhodium

103

53 Iodine

127

55 Cesium

133

59 Praseodymium

141

65 Terbium

159

67 Holmium

165

69 Thulium

169

79 Gold

197

83 Bismuth

209

91 Protactinium

231

THE NEXT TABLE EXTENDS THE LOGIC TO ALL OF THE NATURALLY OCCURRING ELEMENTS AND BEYOND

UP TO THE ELEMENT NUMBER 100 WHICH IS FERMIUM.

THE ELEMENT ARGON HAS A DOUBLE ASTERISK BESIDES ITS NAME BECAUSE THE MOST STABLE ISOTOPE

HAS MORE NUCLEONS THAN IS PREDICTED BY THE MODEL. THE MODEL PREDICTS THAT THE STABLE

ISOTOPE FOR ARGON WOULD CONTAIN THIRTY-SEVEN NUCLEONS. BUT THE MOST STABLE ISOTOPE FOR

ARGON CONTAINS 40 NUCLEONS.

NO. OF

TOTAL

Z NAME

TRITONS

DEUTERONS

PARTICLES

NUCLEONS

1 Hydrogen

2 Helium

3 Lithium

4 Beryllium

5 Boron

6 Carbon

7 Nitrogen

8 Oxygen

9 Fluorine

1O Neon

11 Sodium

12 Magnesium

13 Aluminum

14 Silicon

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15 Phosphorus

16 Sulphur

17 Chlorine

18 Argon **

19 Potassium

20 Calcium

21 Scandium

22 Titanium

23 Vanadium

24 Chromium

25 Manganese

26 Iron

27 Cobalt

28 Nickel

29 Copper

30 Zinc

31 Gallium

32 Germanium

33 Arsenic

34 Selenium

35 Bromine

36 Krypton

37 Rubidium

38 Strontium

39 Yttrium

4O Zirconium

41 Niobium

42 Molybdenum

43 Technetium

44 Ruthenium

102

45 Rhodium

103

46 Palladium

106

47 Silver

107

48 Cadmium

114

49 Indium

115

50 Tin

120

51 Antimony

121

52 Tellurium

130

53 Iodine

127

54 Xenon

132

55 Cesium

133

56 Barium

138

57 Lanthanum

139

58 Cerium

140

59 Praseodymium

141

60 Neodymium

142

61 Promethium

145

62 Samarium

152

63 Europium

153

64 Gadolinium

158

65 Terbium

159

66 Dysprosium

164

67 Holmium

165

68 Erbium

166

69 Thulium

169

7O Ytterbium

174

71 Lutetium

175

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72 Hafnium

180

73 Tantalum

181

74 Tungsten

184

75 Rhenium

187

76 Osmium

192

77 Iridium

193

78 Platinum

195

79 Gold

197

80 Mercury

202

81 Thallium

205

82 Lead

208

83 Bismuth

209

84 Polonium

209

85 Astatine

210

86 Radon

222

87 Francium

223

88 Radium

226

89 Actinium

227

9O Thorium

232

91 Protactinium

231

92 Uranium

238

93 Neptunium

237

94 Plutonium

244

95 Americium

243

96 Curium

247

97 Berkelium

247

98 Californium

251

99 Einsteinium

252

100 Fermium

100

257

___________________________________________________________________________________________________

THE FORMULA FOR CALCULATING THE NUMBER OF TRITONS AND DEUTERONS IN EACH NUCLEUS IS TO

FIRST LET THE NUMBER OF NUCLEONS BE EQUAL TO 'U'. LET THE THE ELEMENT NUMBER BE EQUAL TO

'A'.

TRITONS = U - 2 * A

DEUTERONS = A - (U - 2 * A)

FOR EXAMPLE: THE ELEMENT FERMIUM HAS 'U = 257' AND 'A = 100'

TRITONS IN THE NUCLEUS OF FERMIUM = 257 - 2 * 100 = 57

DEUTERONS IN THE NUCLEUS OF FERMIUM = 100 - (257 - 2 * 100) = 43

THERE ARE THREE NUCLEONS IN EACH TRITON: 3 * 57 = 171

THERE ARE 2 NUCLEONS IN EVERY DEUTERON: 2 * 43 = 86

TOTAL NUMBER OF NUCLEONS IN FERMIUM = 171 + 86 = 257

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THE NUCLEAR MAGIC NUMBERS

THE DYNAMISM MODEL HYPOTHESIZES THAT TWO DEUTERONS WILL ALWAYS FORM AN HE4 ALPHA PARTICLE WHENEVER POSSIBLE WITHIN A NUCLEAR SHELL STRUCTURE. LIKEWISE THAT TWO TRITONS

WILL ALWAYS FORM AN HE6 PARTICLE WHENEVER POSSIBLE. HE5 PARTICLES ARE ONLY FORMED WHEN IT

REMAINS THE ONLY OPTION FOR A LEFTOVER DEUTERON AND A LEFTOVER TRITON PARTICLE.

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A Formula for the Nuclear Magic Numbers

applet-magic.com

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Thayer Watkins

Silicon Valley

& Tornado Alley

USA

One of the elements of the physics of nuclei is the matter of magic numbers. They represent a shell being completely filled so additional nucleons have to go into a higher shell. A higher shell involves a greater separation from the other nucleons and lower interaction energy. The conventional magic numbers are {2, 8, 20, 28, 50, 82, 126}. These numbers were found in the case of protons by comparing the number of stable isotopes for different proton numbers. For neutrons the magic numbers were found by comparing the number of stable nuclides with the same neutron numbers.

https://www.sjsu.edu/faculty/watkins/magicnumbers5.htm

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THE FOLLOWING TABLES ARE CALCULATED FROM THE FOLLOWING PREMISES: 1. DEUTERONS WILL FORM HE4 WHENEVER POSSIBLE.

2. TRITONS WILL FORM HE6 WHENEVER POSSIBLE.

3. HELIUM 5 (HE5) WILL ONLY FORM IF THERE IS ONE DEUTERON AND ONE TRITON LEFTOVER AFTER

ALL POSSIBLE HE4 AND HE6 HAVE FORMED.

4. AFTER ALL HELIUM PARTICLES THAT CAN BE FORMED HAVE DONE SO, THEN ANY LEFTOVER

DEUTERONS OR TRITONS WILL COUNT AS A PARTICLE.

5. THE NUCLEAR SHELLS ARE BASED ON THE NEUTRON MAGIC NUMBERS OF: 1. 2

2. 8

3. 28

4. 50

5. 82

6. 126

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

1 Hydrogen

2 Helium

2 MAGIC

NUMBER

FIRST SHELL

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

3 Lithium

4 Beryllium

5 Boron

6 Carbon

7 Nitrogen

8 Oxygen

8 MAGIC

NUMBER

SECOND SHELL

NO.

NO. OF

LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

TRITONS DEUTERONS PARTICLES HYDROGEN

HE4

HE5

HE6

9 Fluorine

1O Neon

11 Sodium

12 Magnesium

13 Aluminum

14 Silicon

15 Phosphorus

16 Sulphur

17 Chlorine

18 Argon **

19 Potassium

20 Calcium

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21 Scandium

22 Titanium

23 Vanadium

24 Chromium

28 MAGIC

NUMBER

THIRD SHELL

NO.

NO. OF

LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

TRITONS DEUTERONS PARTICLES HYDROGEN

HE4

HE5

HE6

25 Manganese

26 Iron

27 Cobalt

28 Nickel

29 Copper

30 Zinc

31 Gallium

32 Germanium

33 Arsenic

34 Selenium

35 Bromine

36 Krypton

37 Rubidium

38 Strontium

39 Yttrium

4O Zirconium

50 MAGIC

NUMBER

FOURTH SHELL

NO.

NO. OF

LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

TRITONS DEUTERONS PARTICLES HYDROGEN

HE4

HE5

HE6

41 Niobium

42 Molybdenum

43 Technetium

44 Ruthenium

45 Rhodium

46 Palladium

47 Silver

48 Cadmium

49 Indium

50 Tin

51 Antimony

52 Tellurium

53 Iodine

54 Xenon

55 Cesium

56 Barium

57 Lanthanum

58 Cerium

59 Praseodymium

60 Neodymium

82 MAGIC

NUMBER

FIFTH SHELL

Z NAME

NO. OF

NO.

LEFTOVER LEFTOVER

TOTAL

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NEUTRONS

TRITONS DEUTERONS PARTICLES HYDROGEN

HE4

HE5

HE6

61 Promethium

62 Samarium

63 Europium

64 Gadolinium

65 Terbium

66 Dysprosium

67 Holmium

68 Erbium

69 Thulium

100

70 Ytterbium

104

71 Lutetium

104

72 Hafnium

108

73 Tantalum

108

74 Tungsten

110

75 Rhenium

112

76 Osmium

116

77 Iridium

116

78 Platinum

117

79 Gold

118

8O Mercury

122

81 Thallium

124

82 Lead

126

83 Bismuth

126

126 MAGIC

NUMBER

SIXTH SHELL

NO.

NO. OF

LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

TRITONS DEUTERONS PARTICLES HYDROGEN

HE4

HE5

HE6

84 Polonium

125

85 Astatine

125

86 Radon

136

87 Francium

136

88 Radium

138

89 Actinium

138

9O Thorium

142

91 Protactinium

140

92 Uranium

146

93 Neptunium

144

94 Plutonium

150

95 Americium

148

96 Curium

151

97 Berkelium

150

98 Californium

153

99 Einsteinium

153

100 Fermium

157

126+ MAGIC

NUMBER

SEVENTH SHELL

THE FORMULA FOR CALCULATING THE NUMBER OF HE4, HE5 AND HE6 IN EACH NUCLEUS IS TO FIRST LET

THE NUMBER OF NUCLEONS BE EQUAL TO 'U'. LET THE THE ELEMENT NUMBER BE EQUAL TO 'A'. LET THE

NUMBER OF TRITONS BE EQUAL TO 'T'. LET THE NUMBER OF DEUTERONS BE EQUAL TO 'D'.

TRITONS = T = U - 2 * A

DEUTERONS = D = A - (U - 2 * A)

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HE4 = INT( D / 2 )

HE6 = INT( T / 2 )

INTERMEDIATE TRITONS = T1 = T - 2 * HE6

INTERMEDIATE DEUTERONS = D1 = D - 2 * HE4

HE5 = D1 * T1

LEFTOVER TRITONS = T2 = (1 - HE5) * T1

LEFTOVER DEUTERONS = D2 = (1 - HE5) * D1

TOTAL PARTICLES = HE4 + HE5 + HE6 + D2 + T2

TOTAL HYDROGEN = D2 + T2

FOR EXAMPLE: THE ELEMENT FERMIUM HAS 'U = 257' AND 'A = 100'

TRITONS IN THE NUCLEUS OF FERMIUM =

T = 257 - 2 * 100 = 57

DEUTERONS IN THE NUCLEUS OF FERMIUM =

D = 100 - (257 - 2 * 100) = 43

THERE ARE THREE NUCLEONS IN EACH TRITON: 3 * 57 = 171

THERE ARE 2 NUCLEONS IN EVERY DEUTERON: 2 * 43 = 86

TOTAL NUMBER OF NUCLEONS IN FERMIUM = 171 + 86 = 257

NO. OF

TOTAL

Z NAME

TRITONS

DEUTERONS

PARTICLES

NUCLEONS

100 Fermium

100

257

HE4 = INT( 43 / 2 ) = INT( 21.5 ) = 21

HE6 = INT( 57 / 2 ) = INT( 28.5 ) = 28

INTERMEDIATE TRITONS = T1 = 57 - 2 * 28 = 57 - 56 = 1

INTERMEDIATE DEUTERONS = D1 = 43 - 2 * 21 = 43 - 42 = 1

HE5 = 1 * 1 = 1

LEFTOVER TRITONS = T2 = (1 - 1) * 1 = 0

LEFTOVER DEUTERONS = D2 = (1 - 1) * 1 = 0

TOTAL PARTICLES = 21 + 1 + 28 + 0 + 0 = 50

TOTAL HYDROGEN = 0 + 0 = 0

NO. OF

NO. OF NO. OF NO. OF LEFTOVER

LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

100 Fermium

157

NOTE THAT ARGON HAS A DOUBLE ASTERISK BECAUSE IT SEEMED TO HAVE MORE NEUTRONS THAN

WOULD OTHERWISE BE EXPECTED. BUT WITHIN THIS TABLE, THE NUMBER OF NEUTRONS THAT IT HAS IS

EXACTLY THE AMOUNT THAT IT WOULD NEED TO FIT THE OVERALL PATTERN; WHICH IS THAT THE TOTAL

NUMBER OF PARTICLES IS AS CLOSE TO EXACTLY HALF OF THE ELEMENT NUMBER AS IT CAN POSSIBLY BE.

THIS IS TRULY AN ASTONISHING RESULT.

ALSO OBSERVE THAT UP UNTIL THE SIXTH SHELL WHERE THE TOTAL NEUTRONS ARE ONE HUNDRED AND

TWENTY-SIX OR LESS, THAT THE NUMBER OF HE4 PARTICLES IS EITHER GREATER THAN OR ROUGHLY THE

SAME AS THE HE6 ALPHA PARTICLES. HOWEVER, THIS CHANGES DRAMATICALLY WITHIN THE SEVENTH

SHELL WHERE THE TOTAL NEUTRONS ARE GREATER THAN ONE HUNDRED AND TWENTY-SIX; FOR IN THIS

SHELL, THE HE6 PARTICLES ARE NOW MUCH GREATER IN TOTAL, FOR EACH ELEMENT, THAN THE HE4

ALPHA PARTICLES.

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THE NUMBER OF HE4 ALPHA PARTICLES NEVER EXCEED TWENTY-TWO WHILE THE HE6 PARTICLES WITHIN

THE SEVENTH SHELL ARE NEVER LESS THAN TWENTY-FOUR. I HYPOTHESIZE THAT THE SEVENTH SHELL

OF ANY NUCLEUS IS ALMOST ENTIRELY COMPOSED OF HE6 PARTICLES, WHICH IMPLIES THAT WHENEVER

ALPHA PARTICLE RADIATION TAKES PLACE, AND AN ALPHA PARTICLE IS EJECTED FROM THE SEVENTH

SHELL, THAT IT IS ALMOST ALWAYS AN HE6 ALPHA PARTICLE THAT IS EJECTED.

THE SHELL STRUCTURE OF EACH ELEMENT UP TO FERMIUM

FIRST SHELL

2 MAGIC

NUMBER

FIRST SHELL

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

1 Hydrogen

2 Helium

SECOND SHELL

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

FIRST SHELL

SECOND SHELL

3 Lithium

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

FIRST SHELL

SECOND SHELL

4 Beryllium

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

FIRST SHELL

SECOND SHELL

5 Boron

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

FIRST SHELL

SECOND SHELL

6 Carbon

DYNAMISM by GEORGE MARTIN WILLIAMS - HTML preview

VOLUME

NEUTRINO

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