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THIRD SHELL

FOURTH SHELL

FIFTH SHELL

SIXTH SHELL

SEVENTH SHELL

99 Einsteinium

153

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

THIRD SHELL

FOURTH SHELL

FIFTH SHELL

SIXTH SHELL

SEVENTH SHELL

100 Fermium

157

ESTABLISHMENT OF NEW SHELLS

THE DYNAMISM MODEL IS PREDICATED ON THE IDEA THAT THE INNER SHELLS HAVE THE SAME

COMPOSITIONAL STRUCTURE FOR ALL ELEMENTS THAT ARE IN A GIVEN SHELL. THIS IMPLIES THAT UPON

THE ESTABLISHMENT OF A NEW SHELL, THAT A TRANSFORMATION ALWAYS OCCURS IN THE PREVIOUS

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SHELL THAT CONVERTS THE PARTICLES TO SOME STANDARD FOR THAT INNER SHELL.

Z NAME

NEUTRONS HE4 HE5 HE6 TRITONS

DEUTERONS

PARTICLES

HYDROGEN

2 Helium

FIRST SHELL

SECOND SHELL

3 Lithium

DIFFERENCE

IT IS IMPORTANT TO REALIZE THAT THESE NUCLEAR SHELLS WERE FORMED WITHIN THE PLASMA THAT

OCCUPIED ALL OF THE SPACE WITHIN THE UNIVERSE BILLIONS OF YEARS AGO. THE PLASMA DOES NOT

HAVE AN ELECTRON SHELL BECAUSE THE MAGNETIC FIELDS THAT PERMEATE THE PLASMA ARE TOO

POWERFUL TO ALLOW ELECTRON SHELLS TO FORM. THE FIRST SHELL CONTAINS TWO NEUTRONS (TWO IS

A MAGIC NUMBER), WHICH ARE THE TWO NEUTRONS IN A HELIUM FOUR (HE4) PARTICLE. THE ONLY

NUCLEUS THAT IS NOT CENTERED BY AN HE4 PARTICLE IS THE HYDROGEN NUCLEUS WHICH IS SIMPLY A PROTON. ALL OTHER SHELLS HAVE AN HE4 PARTICLE AS THE SOLE PARTICLE IN ITS FIRST SHELL. THE

LITHIUM NUCLEUS HAS FOUR NEUTRONS WHICH FORCES THE CREATION OF A NEW SHELL. WHEN A NEW

SHELL IS FORMED AND THE FIRST SHELL BECOMES THE INNER SHELL OF THE TWO SHELL NUCLEUS, THERE IS NO DIFFERENCE BETWEEN THE NUMBER OF PARTICLES BETWEEN THE FIRST SHELL OF THE

HELIUM NUCLEUS (WHICH IS THE ONLY SHELL) AND THE FIRST SHELL OF THE LITHIUM NUCLEUS (WHICH

HAS TWO SHELLS).

Z NAME

NEUTRONS HE4 HE5 HE6 TRITONS

DEUTERONS

PARTICLES

HYDROGEN

8 Oxygen

SECOND SHELL

THIRD SHELL

9 Fluorine

DIFFERENCE

THE OXYGEN NUCLEUS HAS EIGHT NEUTRONS (EIGHT IS A MAGIC NUMBER), AND IS THE LAST NUCLEUS

TO HAVE ONLY TWO SHELLS. THE FLUORINE NUCLEUS HAS TEN NEUTRONS WHICH FORCES THE

CREATION OF A THIRD SHELL. THE SECOND SHELL OF THE OXYGEN NUCLEUS IS THE SAME AS THE

SECOND SHELL OF THE FLUORINE NUCLEUS BECAUSE THE EIGHT NEUTRONS OF THE OXYGEN NUCLEUS

IS EXACTLY THE SAME AS THE MAGIC NUMBER EIGHT WHICH EXACTLY FILLS OUT THIS SECOND SHELL.

THE HELIUM PARTICLE IN THE FLUOURINE NUCLEUS'S THIRD SHELL IS CREATED AS A COMPLETELY NEW

PARTICLE DURING THE FUSION PROCESS.

Z NAME

NEUTRONS HE4 HE5 HE6 TRITONS

DEUTERONS

PARTICLES

HYDROGEN

24 Chromium

THIRD SHELL

FOURTH SHELL

25 Manganese

DIFFERENCE

NOTE THE DIFFERENCE BETWEEN THE CHROMIUM NUCLEUS'S THIRD SHELL AND THE MANGANESE

NUCLEUS'S THIRD SHELL. THERE IS A DIFFERENCE OF FOUR NEUTRONS WHICH IS DUE TO A HELIUM SIX

(HE6) PARTICLE BEING TAKEN OUT OF THE THE THIRD SHELL OF THE MANGANESE NUCLEUS AND PLACED

WITHIN THE FOURTH SHELL OF THE NUCLEUS. THIS IS SURPRISING BECAUSE THE CHROMIUM NUCLEUS

HAS EXACTLY TWENTY-EIGHT NEUTRONS AND TWENTY-EIGHT IS A MAGIC NUMBER. SINCE THE TOTAL

NUMBER OF NEUTRONS IN THE CHROMIUM NUCLEUS IS EXACTLY THE SAME AS THE MAGIC NUMBER, I EXPECTED THAT A BRAND NEW SHELL WOULD BE STARTED WITH A COMPLETELY NEW PARTICLE FOR THE

FOURTH SHELL OF THE MANGANESE NEUCLEUS. BUT INSTEAD THE MANGANESE NUCLEUS TRANSFERRED

AN HE6 PARTICLE FROM ITS THIRD SHELL TO ITS FOURTH SHELL. THE TOTAL DIFFERENCE IN NEUTRON

NUMBERS BETWEEN THE CHROMIUM NUCLEUS AND THE MANGANESE NUCLEUS IS TWO NEUTRONS, WHICH IS ACCOUNTED FOR BY A SINGLE TRITON PARTICLE. THERE SEEMS TO BE AN OVERALL PATTERN

WHERE ALPHA PARTICLES HAVE TO BE THE MAJORITY OF THE PARTICLES IN EVERY SHELL. WHENEVER IT

IS POSSIBLE FOR A SINGLE TRITON, OR A SINGLE DEUTERON, (WHICH ARE HYDROGEN PARTICLES), TO BE

THE ONLY PARTICLE IN A NEW SHELL, THIS NEVER HAPPENS. WHAT ACTUALLY HAPPENS IS THAT ONE, OR

MORE, ALPHA PARTICLES IS PULLED OUT OF THE INNER SHELL AND PLACED IN THE NEW OUTER SHELL

BESIDE THE HYDROGEN PARTICLE.

Z NAME

NEUTRONS HE4 HE5 HE6 TRITONS

DEUTERONS

PARTICLES

HYDROGEN

4O Zirconium

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FOURTH SHELL

FIFTH SHELL

41 Niobium

DIFFERENCE

THE ZIRCONIUM NUCLEUS HAS FIFTY NEUTRONS WHICH IS A MAGIC NUMBER FOR NEUTRONS. AGAIN I EXPECTED THAT NEW SHELL WOULD BE FORMED FOR THE NIOBIUM PARTICLE THAT WOULD HAVE A BRAND NEW CREATED PARTICLE, FOR THIS WAS ANOTHER OPPORTUNITY FOR A HYDROGEN PARTICLE TO

EXIST BY ITSELF IN A NEW SHELL. HOWEVER, THIS IS NOT TO BE, FOR THERE IS A DIFFERENCE OF FOUR

NEUTRONS BETWEEN THE ZIRCONIUM FOURTH SHELL AND THE NIOBIUM FOURTH SHELL WHICH IS DUE

TO TWO OF THE HE4 PARTICLES BEING REMOVED FROM THE NIOBIUM NUCLEUS'S FOURTH SHELL AND

PLACED IN ITS FIFTH SHELL, RIGHT BESIDE THE TRITON HYDROGEN PARTICLE. IT BECOMES MORE AND

MORE APPARENT THAT FOR SOME REASON, THERE IS A CLEAR BIAS AGAINST THE HYDROGEN PARTICLES

OCCUPYING ANY SHELL, (EXCEPT THE FIRST SHELL), BY THEMSELVES.

Z NAME

NEUTRONS HE4 HE5 HE6 TRITONS

DEUTERONS

PARTICLES

HYDROGEN

60 Neodymium

FIFTH SHELL

SIXTH SHELL

61 Promethium

DIFFERENCE

THE EXACT SAME THING HAPPENS HERE AGAIN. THE NEODYMIUM NUCLEUS HAS EXACTLY EIGHTY-TWO

NEUTRONS, (WHICH IS A MAGIC NUMBER FOR NEUTRONS), WHILE THE PROMETHIUM NUCLEUS HAS

EIGHTY-FOUR NEUTRONS, WHICH PRESENTS AN OPPORTUNITY FOR A TRITON NUCLEUS TO OCCUPY THE

NEW SIXTH SHELL ALL BY ITSELF. BUT THIS IS NOT TO BE, BECAUSE TWO HE4 PARTICLES ARE REMOVED

FROM THE FIFTH SHELL OF THE NEODYMIUM NUCLEUS AND PLACED IN THE SIXTH SHELL BESIDE THE

TRITON HYDROGEN PARTICLE, JUST AS WHAT OCCURRED IN THE PREVIOUS CASE.

Z NAME

NEUTRONS HE4 HE5 HE6 TRITONS DEUTERONS

PARTICLES

HYDROGEN

83 Bismuth

126

SIXTH SHELL

SEVENTH SHELL

84 Polonium

125

DIFFERENCE

IN THE CASE OF THE CREATION OF THE NEW SEVENTH SHELL, THE DIFFERENCE BETWEEN THE BISMUTH

NUCLEUS'S SIXTH SHELL, AND THE POLONIUM NUCLEUS'S SIXTH SHELL IS TEN NEUTRONS. TWO HE4, ONE HE6, AND ONE TRITON ARE TAKEN OUT OF THE BISMUTH NUCLEUS'S SIXTH SHELL AND FIND THEIR

WAY INTO POLONIUM'S SEVENTH SHELL. APPARENTLY, SINCE POLONIUM HAS ONE MORE PROTON THAN

BISMUTH HAS, AND HAS ONE LESS NUETRON THAN BISMUTH HAS, THAT MOST LIKELY A BISMUTH

NEUTRON EMITTED AN ELECTRON THROUGH BETA PARTICLE RADIATION, TO BECOME A POLONIUM

PROTON.

THIS WOULD INFER THAT ALL POLONIUM NUCLEI ARE ACTUALLY FORMED FROM THE BETA DECAY OF A BISMUTH NUCLEI. MATHEMATICALLY SPEAKING, THIS PROCESS REQUIRES THAT THE AN HE6 PARTICLE

FROM THE BISMUTH SIXTH SHELL, SPONTANEOUSLY FISSIONS INTO AN HE4 PARTICLE PLUS TWO

NEUTRONS. THE TWO NEUTRONS WOULD THEN BETA DECAY, BY EMITTING AN ELECTRON, TO BECOME A NEUTRON PLUS A PROTON, WHICH IS A DEUTERON PARTICLE. THE DEUTERON WHICH HAS A PROTON AND

A NEUTRON, WOULD THEN COMBINE WITH THE TRITON, WHICH HAS A PROTON AND TWO NEUTRONS, TO

FORM AN HE5 PARTICLE WHICH HAS TWO PROTONS AND FIVE NEUTRONS.

THE NET CHANGE IS THAT THE POLONIUM SEVENTH SHELL GAINS AN HE4 AND AN HE5 ALPHA PARTICLE, BUT HAS ONE LESS HE6 PARTICLE, AND ONE LESS NEUTRON THAN THE DIFFERENCE BETWEEN THE SIXTH

SHELLS OF THE BISMUTH AND POLONIUM NUCLEI.

Z NAME

NEUTRONS HE4 HE5 HE6 TRITONS DEUTERONS

PARTICLES

HYDROGEN

83 Bismuth

126

DIFFERENCE

SEVENTH SHELL

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84 Polonium

125

NO. OF

NO. OF NO. OF NO. OF LEFTOVER

LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS

DEUTERONS PARTICLES HYDROGEN

82 Lead

126

83 Bismuth

126

84 Polonium

125

85 Astatine

125

86 Radon

136

87 Francium

136

88 Radium

138

NOTE THAT THE TOTAL NEUTRON NUMBERS FOR POLONIUM AND ASTATINE, (125), ARE LOWER THAN THE

TWO NUCLEI THAT PRECEDED THEM, WHICH ARE LEAD AND BISMUTH, (126), WHICH MAKES NO LOGICAL

SENSE UNTIL YOU LOOK AT THE TOTAL PARTICLE NUMBERS FOR ALL FOUR NUCLEI. FOR SOME REASON, THE NEUTRON NUMBERS ARE FORCED TO CHANGE IN ORDER TO PRESERVE THE SEQUENTIAL PATTERN

ESTABLISHED BY THE TOTAL PARTICLE NUMBERS FOR ALL OF THESE NUCLEI.

THE MAIN THING THAT I TOOK AWAY FROM STUDYING THESE TABLES, IS THAT, AS PER THE PREMISE THAT

THE INNER SHELLS ARE IDENTICAL FOR ALL OF THE NUCLEI AT A CERTAIN SHELL LEVEL, THERE IS A FLEXIBILITY IN THE PARTICLES THAT COMPOSE THE VARIOUS SHELLS, SUCH THAT THEY WILL

TRANSFORM, AS IS NECESSARY, FROM ONE TYPE OF PARTICLE TO ANOTHER, IN ORDER TO MAINTAIN SOME

TYPE OF OVERALL PATTERN. THIS IS A VERY SURPRISING FINDING FOR ME. I DID NOT EVEN SUSPECT THAT

AS A POSSIBILITY!

THE DYNAMISM NUCLEAR SHELL MODEL

THE NEUTRINO AND THE PHOTON ARE THE MOST FUNDAMENTAL PARTICLES BECAUSE THEY CANNOT BE

BROKEN DOWN INTO ANY SMALLER PARTICLE. THE PHOTONS ARE THE SOLE COMPONENTS OF

ELECTRONS AND POSITRONS, WITH P-STRING PHOTONS BEING THE SOLE COMPONENTS OF THE

POSITIVELY CHARGED POSITRON, AND N-STRING PHOTONS BEING THE SOLE COMPONENTS OF THE

NEGATIVELY CHARGED ELECTRONS. HOWEVER, THE OVERALL CHARGE OF THE UNIVERSE IS NEUTRAL

WHICH MEANS THAT THERE ARE EXACTLY AS MANY POSITIVE CHARGES AS THERE ARE NEGATIVE

CHARGES. SINCE THE OVERWHELMINGLY VAST MAJORITY OF THE ELECTRON/POSITRON PARTICLES ARE

THE NEGATIVELY CHARGED ELECTRONS, THIS FORCES THE POSITIVE CHARGES TO ACCUMULATE

SOMEWHERE ELSE IN ORDER TO BALANCE OUT THIS COSMIC EQUATION.

THE QUARKS ARE WHERE THE OVERWHELMINGLY VAST MAJORITY OF THE POSITIVE CHARGES

ACCUMULATE. THERE IS ONLY ONE TYPE OF QUARK, AND THAT IS THE UPQUARK WHICH HAS POSITIVE

TWO-THIRDS CHARGE. EACH UPQUARK HAS TWO-THIRDS OF THE P-STRING PHOTONS WHICH COMPRISE A POSITRON; THE REST OF THE MASS OF THE QUARK IS MADE UP OF NEUTRINOS. THE OVERWHELMINGLY

VAST MAJORITY OF THE NEUTRINOS THAT ARE WITHIN THE UNIVERSE ARE WITHIN THESE UPQUARKS.

THE UNIVERSE HAS A SIMPLE WAY TO BALANCE THIS PHYSICS EQUATION; AND THAT IS TO TAKE TWO

ELECTRONS AND TWO POSITRONS AND SOME NEUTRINOS AND TO MAKE THE EQUIVALENT OF A NEUTRON, WHICH IS A NEUTRALLY CHARGED PARTICLE. BUT THE UNIVERSE IS EXTREMELY EFFICIENT, AND

EXTREMELY CLEVER IN THE NUMBER OF WAYS THAT IT CREATES THESE NEUTRONS. THE HYDROGEN

ATOM IS NEUTRALLY CHARGED AND HAS ALL THE SAME COMPONENTS AS A NEUTRON. WITHIN THE PURE

PLASMA WHICH IS EFFECTIVELY DARK MATTER, THERE EXISTS A PROTON/ELECTRON PAIR WHICH IS NON-ATOMIC BUT HAS THE SAME COMPONENTS AS A NEUTRON. EVEN A POSITRON/ELECTRON PAIR CAN BE

CONSIDERED TO BE A 'NEUTRON' IN THE SENSE THAT THE TWO TOGETHER ARE NEUTRALLY CHARGED.

BUT THE BASIC IDEA IS A PROTON AND AN ELECTRON NEUTRALIZING EACH OTHER'S CHARGE.

THE STRUCTURE OF THE PARTICLES THAT ARE LARGER THAN AN ELECTRON ARE HYPOTHESIZED TO BE

RATHER FRACTAL IN NATURE, WITHIN THE DYNAMISM MODEL. BY 'FRACTAL', IT IS MEANT THAT THEY ARE

ALL SELF-SIMILAR. THE BASIC FEATURES ARE REPEATED AT EVERY LEVEL: 1. THE NEGATIVELY CHARGED ELECTRON, OR ELECTRONS, SURROUNDED BY POSITIVELY CHARGED

PARTICLES, IS A BASIC FEATURE.

2. THE PROTON CONSISTS OF A SINGLE ELECTRON SURROUNDED BY THREE POSITIVELY CHARGED

UPQUARKS.

3. THE NEUTRON CONSISTS OF A PAIR OF ELECTRONS (DOUBLE ELECTRON) SURROUNDED BY THREE

UPQUARKS.

4. THE DEUTERON IS A SINGLE ELECTRON SURROUNDED BY TWO PROTONS.

5. THE TRITON IS A DOUBLE ELECTRON PAIR SURROUNDED BY THREE PROTONS.

6. THE NUMBER OF ELECTRONS AT THE CENTER OF AN ALPHA PARTICLE IS THE SAME AS THE NUMBER

OF NEUTRONS WITHIN THE PARTICLE.

7. THE HE3 IS A SINGLE ELECTRON (ONE ELECTRON = ONE NEUTRON) SURROUNDED BY THREE

PROTONS.

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8. THE HE4 IS A DOUBLE ELECTRON (TWO ELECTRONS = TWO NEUTRONS) SURROUNDED BY FOUR

PROTONS.

9. THE HE5 IS A TRIPLE ELECTRON (THREE ELECTRONS = THREE NEUTRONS) SURROUNDED BY FIVE

PROTONS.

10. THE HE6 IS A QUADRUPLE ELECTRON (FOUR ELECTRONS = FOUR NEUTRONS) SURROUNDED BY SIX

PROTONS.

11. NOTE THAT THE OVERALL PATTERN FOR HYDROGEN PARTICLES IS: 1. THERE IS A SINGLE ELECTRON SURROUNDED BY QUARKS AT THE CENTER OF THE H1 PARTICLE.

2. THERE IS A SINGLE ELECTRON SURROUNDED BY TWO PROTONS AT THE CENTER OF THE H2

PARTICLE (DEUTERON).

3. THERE IS A DOUBLE ELECTRON SURROUNDED BY THREE PROTONS AT THE CENTER OF THE H3

PARTICLE (TRITON).

12. NOTE THAT THE OVERALL PATTERN FOR HELIUM PARTICLES IS: 1. THERE IS A SINGLE ELECTRON SURROUNDED BY THREE PROTONS AT THE CENTER OF THE HE3

PARTICLE.

2. THERE IS A DOUBLE ELECTRON SURROUNDED BY FOUR PROTONS AT THE CENTER OF THE HE4

PARTICLE.

3. THERE IS A TRIPLE ELECTRON SURROUNDED BY FIVE PROTONS AT THE CENTER OF THE HE5

PARTICLE.

4. THERE IS A QUADRUPLE ELECTRON SURROUNDED BY SIX PROTONS AT THE CENTER OF THE HE6

PARTICLE.

13. THE FIRST ELEMENT IS ALWAYS A HYDROGEN PARTICLE IN THE FIRST SHELL AND THESE ALWAYS

HAVE AN ELECTRON AT THE CENTER.

14. ALL OTHER ELEMENTS HAVE AN HE4 PARTICLE IN THEIR FIRST SHELL, AND THERE IS ALWAYS AN

ELECTRON AT THE CENTER OF THE HE4 PARTICLE.

15. FOR THE THIRD ELEMENT AND HIGHER, THE FIRST SHELL IS THE CENTER OF THE NUCLEI, AND IT IS

ALWAYS A HE4 PARTICLE, AND IT ALWAYS HAS AN ELECTRON AT ITS CENTER.

16. ATOMIC NUCLEI ARE LIKE ONIONS AND ONIONS HAVE LAYERS.

17. THE INNERMOST LAYER AT THE CENTER IS AN HE4 ALPHA PARTICLE WHICH IS SURROUNDED BY AS

MANY AS SIX OTHER LAYERS.

18. MY CONCLUSION, AS PER THIS HYPOTHETICAL DYNAMISM MODEL FOR THE NUCLEAR STRUCTURE, IS

THAT ALL NUCLEI HAVE ELECTRONS AT THEIR CORE.

19. THE MONOPOLAR MAGNETIC FIELD EMANATING FROM THESE CORE ELECTRONS IS ADDED TO BY ALL

THE OTHER CHARGED PARTICLES THAT SURROUND THEM, AND THIS CREATES THE OVERALL

MAGNETIC FIELD THAT PROVIDE THE ORBITAL PATHS FOR ALL THE CHARGED PARTICLES IN SHELL

THREE AND HIGHER.

20. THE FIRST ONE HUNDRED ELEMENTS ARE IN SEVEN SHELLS. THE SHELLS ARE AS FOLLOWS

1. FIRST SHELL: MAGIC NUMBER FOR NEUTRONS IS TWO (2).

2. SECOND SHELL: MAGIC NUMBER FOR NEUTRONS IS EIGHT (8).

3. THIRD SHELL: MAGIC NUMBER FOR NEUTRONS IS TWENTY-EIGHT (28).

4. FOURTH SHELL: MAGIC NUMBER FOR NEUTRONS IS FIFTY (50).

5. FIFTH SHELL: MAGIC NUMBER FOR NEUTRONS IS EIGHTY-TWO (82).

6. SIXTH SHELL: MAGIC NUMBER FOR NEUTRONS IS ONE HUNDRED AND TWENTY-SIX (126).

7. SEVENTH SHELL: MAGIC NUMBER FOR NEUTRONS IS ONE HUNDRED AND TWENTY-SIX PLUS

(126+).

FIRST

SHELL

1 H

2 He

SECOND

SHELL

3 Li

4 Be

5 B

6 C

7 N

8 O

THIRD

SHELL

9 F

1O Ne

11 Na

12 Mg

13 Al

14 Si

15 P

16 S

17 Cl

18 Ar

19 K

20 Ca

21 Sc

22 Ti

23 V

24 Cr

FOURTH

SHELL

25 Mn

26 Fe

27 Co

28 Ni

29 Cu

30 Zn

31 Ga

32 Ge

33 As

34 Se

35 Br

36 Kr

37 Rb

38 Sr

39 Y

40 Zr

FIFTH

SHELL

41 Nb

42 Mo

43 Tc

44 Ru

45 Rh

46 Pd

47 Ag

48 Cd

49 In

50 Sn

51 Sb

52 Te

53 I

54 Xe

55 Cs

56 Ba

57 La

58 Ce

59 Pr

60 Nd

SIXTH

SHELL

61 Pm

62 Sm

63 Eu

64 Gd

65 Tb

66 Dy

67 Ho

68 Er

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69 Tm

7O Yb

71 Lu

72 Hf

73 Ta

74 W

75 Re

76 Os

77 Ir

78 Pt

79 Au

8O Hg

81 Tl

82 Pb

83 Bi

SEVENTH

SHELL

84 Po

85 At

86 Rn

87 Fr

88 Ra

89 Ac

90 Th

91 Pa

92 U

93 Np

94 Pu

95 Am

96 Cm

97 Bk

98 Cf

99 Es

100 Fm

DYNAMISM MODEL FOR ALPHA PARTICLE RADIATION

IN THE DYNAMISM MODEL, THE REASON FOR ALPHA PARTICLE RADIATION FROM THE NUCLEUS OF ATOMS, IS THE CONVERSE OF THE REASON FOR THE INTEGRITY OF THE STRUCTURE OF THE NUCLEUS. THE

STRENGTH OF THE MONOPOLAR MAGNETIC FIELD AT SHORTER DISTANCES FROM THE CENTER OF THE

NUCLEUS IS WHAT MAINTAINS THE INTEGRITY OF THE NUCLEAR STRUCTURE. THIS IS TRUE FOR THE

SIXTH SHELL, THE FIFTH SHELL, THE FOURTH SHELL, ALL THE WAY DOWN TO THE FIRST SHELL. BUT THE

SEVENTH SHELL OF THE LARGEST NUCLEI IS FAR AWAY FROM THE CENTER OF THE NUCLEUS OF THE

ATOM, AND THIS DISTANCE SEVERELY WEAKENS THE STRENGTH OF THE MONOPOLAR MAGNETIC FIELD

BECAUSE THE FORCE OF THE MAGNETIC FIELD OBEYS THE INVERSE CUBE LAW.

MAGNETIC MONOPOLE

F = L X T

R^3

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

THE ELECTRICALLY CHARGED PARTICLES WITHIN THE NUCLEAR SHELLS, WILL ALL ORBIT THE CENTER OF

THE NUCLEUS BY FOLLOWING THE MAGNETIC FIELD LINES WHICH ARE GENERATED BY THE MOVEMENT

OF THE POSITIVELY CHARGED NUCLEUS. THE FIELD LINES ARE STRONGEST NEAR THE CENTER AND THEN

TAPER OFF WITH INCREASING DISTANCE AWAY FROM THE CENTER. THE FIRST SIX SHELLS ARE CLOSE

ENOUGH TO THE CENTER OF THE NUCLEUS TO KEEP THE CHARGED PARTICLES SECURELY WITHIN THEIR

RESPECTIVE SHELLS. BUT THE SEVENTH SHELL IS FAR ENOUGH AWAY FROM THE CENTER TO WEAKEN

THE MAGNETIC FIELD ENOUGH TO CREATE THE OPPORTUNITY FOR CHARGED PARTICLES TO ESCAPE.

I HYPOTHESIZE THAT EACH NUCLEAR SHELL WOULD HAVE A WIDTH OF TWO TIMES THE DIAMETER OF A HELIUM FOUR PARTICLE. IF THE RADIUS OF A HELIUM FOUR PARTICLE IS EQUAL TO THE DIAMETER OF A PROTON PLUS THE DIAMETER OF AN ELECTRON THEN THE SHELL WIDTH WOULD BE EQUAL TO THIS

RADIUS TIME TWO AND THEN TIME ONE AND A HALF; WHICH WOULD BE EQUAL TO THIS RADIUS TIME

FOUR:

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

ELECTRON DIAMETER = 153299765289942140706 m

PROTON DIAMETER = 1047795195001787949118606 m

HE4 RADIUS = 153299765289942140706 + 1047795195001787949118606 = 1047948494767077891259312 m HE4 DIAMETER = 2 * 1047948494767077891259312 = 2095896989534155782518624 m NUCLEUS SHELL WIDTH = 2 * 2095896989534155782518624 = 4191793979068311565037248 m THE FOLLOWING TABLE IS BASED ON THE FERMIUM NUCLEUS BECAUSE IT HAS ALL SEVEN SHELLS. IT IS

BASED ON THE PREMISE THAT THE FORCE OF THE MAGNETIC MONOPOLE THAT EMANATES FROM THE

CENTER IS:

T = 1.01072984557908525741657370 × 10^146

WHICH IS THE ELECTRON'S TESLA FIGURE:

THE ESTIMATE FOR THE WIDTH OF THE SHELL IS CALCULATED UNDER THE PREMISE THAT THE SEVENTH

SHELL WILL HAVE A WEAKER MAGNETIC FIELD THAN THE FORCE OF ELECTRICAL REPULSION OF THE

PARTICLES FOR EACH OTHER. THIS MEANS THAT THE SIXTH AND LOWER SHELLS WILL HAVE A STRONG

ENOUGH MAGNETIC FIELD TO KEEP THE PARTICLES SECURELY IN PLACE, BUT THE SEVENTH SHELL WILL

NOT HAVE A STRONG ENOUGH MAGNETIC FIELD TO KEEP THE SEVENTH SHELL PARTICLES SECURELY IN

PLACE. THIS IS MY HYPOTHETICAL MODEL FOR THE REASON FOR THE ALPHA RADIATION THAT OCCURS IN

THE SEVENTH SHELL, BUT WHICH DOES NOT OCCUR IN THE SIXTH AND LOWER SHELLS.

THIS IS HOW I CALCULATE THE WIDTH OF EACH SHELL:

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83 Bismuth

F = (8.9876 × 10^9) * (81* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50) / R^2

F = (1.87783771185448471935877616505407592 × 10^112) / R^2

F = (1.27012739644873951235165450088949425 × 10^140) / R^3

(1.87783771185448471935877616505407592 × 10^112) / R^2 = (1.27012739644873951235165450088949425 ×

10^140) / R^3

R * (1.87783771185448471935877616505407592 × 10^112) = (1.27012739644873951235165450088949425 ×

10^140)

R = (1.27012739644873951235165450088949425 × 10^140) / (1.87783771185448471935877616505407592 ×

10^112)

R = 6.763776168891653354401924872 × 10^27

SHELL WIDTH = R / 6

SHELL WIDTH = (6.763776168891653354401924872 × 10^27) / 6

SHELL WIDTH = 1.127296028148608892400320812 × 10^27

SHELL #

SHELL DIAMETER

DIAMETER^2

DIAMETER^3

1.12729602 × 10^27

1.270796335 × 10^54

1.43256366 × 10^81

2.25459205 × 10^27

5.083185340 × 10^54

1.14605092 × 10^82

3.38188808 × 10^27

1.143716701 × 10^55

3.86792188 × 10^82

4.50918411 × 10^27

2.033274136 × 10^55

9.16840743 × 10^82

5.63648014 × 10^27

3.176990837 × 10^55

1.79070457 × 10^83

6.76377616 × 10^27

4.574866806 × 10^55

3.09433750 × 10^83

7.89107219 × 10^27

6.226902041 × 10^55

4.91369335 × 10^83

F = (L * T) / R^3

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

F = (1.270127396 × 10^140) / R^3

F = (1.270127396 × 10^140) / (1.432563661 × 10^81) = 8.866114861 × 10^58

F = (1.270127396 × 10^140) / (1.146050929 × 10^82) = 1.108264358 × 10^58

F = (1.270127396 × 10^140) / (3.867921885 × 10^82) = 3.283746245 × 10^57

F = (1.270127396 × 10^140) / (9.168407431 × 10^82) = 1.385330447 × 10^57

F = (1.270127396 × 10^140) / (1.790704576 × 10^83) = 7.092891888 × 10^56

F = (1.270127396 × 10^140) / (3.094337508 × 10^83) = 4.104682806 × 10^56

F = (1.270127396 × 10^140) / (4.913693358 × 10^83) = 2.584873137 × 10^56

SHELL #

SHELL DIAMETER

DIAMETER^3

MAGNETIC FIELD

1.12729602 × 10^27

1.43256366 × 10^81

8.866114860 × 10^58

2.25459205 × 10^27

1.14605092 × 10^82

1.108264357 × 10^58

3.38188808 × 10^27

3.86792188 × 10^82

3.283746244 × 10^57

4.50918411 × 10^27

9.16840743 × 10^82

1.385330446 × 10^57

5.63648014 × 10^27

1.79070457 × 10^83

7.092891888 × 10^56

6.76377616 × 10^27

3.09433750 × 10^83

4.104682805 × 10^56

7.89107219 × 10^27

4.91369335 × 10^83

2.584873137 × 10^56

COULOMB'S LAW

F = K X Q1 X Q2

R^2

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

THE ELECTRON CHARGE TO MASS RATIO 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

THIS IS THE SAME CHARGE AS FOR THE SUPERSIZED PROTON.

8 OXYGEN:

F = (8.9876 × 10^9 * (6 * 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((5.083185340 ×

10^54)) = 2.736455204031550580853895960840607869710294686992467601033 × 10^56

24 Chromium:

F = (8.9876 × 10^9 * (22 * 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((1.143716701 ×

10^55)) = 4.459408482593533300478725658663849833910924065451764352613 × 10^56

4O Zirconium:

F = (8.9876 × 10^9 * (38* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((2.033274136 ×

10^55)) = 4.332720739716621753018668604664295793707966587738073701636 × 10^56

60 Neodymium:

F = (8.9876 × 10^9 * (58* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((3.176990837 ×

10^55)) = 4.232384049568231070222685382277413726138486813646419087862 × 10^56

83 Bismuth:

F = (8.9876 × 10^9 * (81* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 4.104682806047325871280843941260911804565442030488701401550 × 10^56

100 Fermium:

F = (8.9876 × 10^9 * (98* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.648606939001705389084801154181078532884535544598910127611 × 10^56

SHELL #

SHELL DIAMETER

DIAMETER^2

ELECTRICAL FORCE

4.19179397 × 10^24

1.270796335 × 10^54

2 -- 8 Oxygen

8.38358795 × 10^24

5.083185340 × 10^54

2.736455204 × 10^56

3 -- 24 Chromium

1.25753819 × 10^25

1.143716701 × 10^55

4.459408483 × 10^56

4 -- 4O Zirconium

1.67671759 × 10^25

2.033274136 × 10^55

4.332720740 × 10^56

5 -- 60 Neodymium

2.09589698 × 10^25

3.176990837 × 10^55

4.232384050 × 10^56

6 -- 83 Bismuth

2.51507638 × 10^25

4.574866806 × 10^55

4.104682806 × 10^56

7 -- 100 Fermium

2.93425578 × 10^25

6.226902041 × 10^55

3.648606939 × 10^56

MAGNETIC -

SHELL #

MAGNETIC FIELD

ELECTRICAL FORCE

ELECTRICAL

8.866114860 × 10^58

2 -- 8 Oxygen

1.108264357 × 10^58

2.736455204 × 10^56

1.080899805 × 10^58

3 -- 24 Chromium

3.283746244 × 10^57

4.459408483 × 10^56

2.837805396 × 10^57

4 -- 40 Zirconium

1.385330446 × 10^57

4.332720740 × 10^56

9.520583720 × 10^56

5 -- 60 Neodymium

7.092891888 × 10^56

4.232384050 × 10^56

2.860507838 × 10^56

6 -- 83 Bismuth

4.104682805 × 10^56

4.104682806 × 10^56

−1.00000000 × 10^47

7 -- 100 Fermium

2.584873137 × 10^56

3.648606939 × 10^56

−1.06373380 × 10^56

SIXTH AND SEVENTH SHELL ODD NUMBERED ELEMENTS

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61 Promethium:

F = (8.9876 × 10^9 * (59* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 2.989830686 × 10^56

63 Europium:

F = (8.9876 × 10^9 * (61* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.091180879 × 10^56

65 Terbium:

F = (8.9876 × 10^9 * (63* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.192531071 × 10^56

67 Holmium:

F = (8.9876 × 10^9 * (65* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.293881264 × 10^56

69 Thulium :

F = (8.9876 × 10^9 * (67* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.395231457 × 10^56

71 Lutetium :

F = (8.9876 × 10^9 * (69* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.49658165 × 10^56

73 Tantalum :

F = (8.9876 × 10^9 * (71* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.597931842 × 10^56

75 Rhenium :

F = (8.9876 × 10^9 * (73* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.699282035 × 10^56

77 Iridium:

F = (8.9876 × 10^9 * (75* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.800632228 × 10^56

79 Gold:

F = (8.9876 × 10^9 * (77* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.901982421 × 10^56

81 Thallium:

F = (8.9876 × 10^9 * (79* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 4.003332613 × 10^56

83 Bismuth:

F = (8.9876 × 10^9 * (81* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 4.104682806 × 10^56

MAGNETIC -

SHELL # 6

SHELL DIAMETER

MAGNETIC FIELD

ELECTRICAL FORCE

ELECTRICAL

7.89107219 ×

4.104682805 ×

2.989830686 ×

1.114852119 ×

Promethium

10^27

10^56

7.89107219 ×

4.104682805 ×

3.091180879 ×

1.013501926 ×

63 Europium

10^27

10^56

7.89107219 ×

4.104682805 ×

3.192531071 ×

65 Terbium

9.12151734 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

3.293881264 ×

67 Holmium

8.10801541 × 10^55

10^27

10^56

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7.89107219 ×

4.104682805 ×

3.395231457 ×

69 Thulium

7.09451348 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

3.496581650 ×

71 Lutetium

6.08101155 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

3.597931842 ×

73 Tantalum

5.06750963 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

3.699282035 ×

75 Rhenium

4.05400770 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

3.800632228 ×

77 Iridium

3.04050577 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

3.901982421 ×

79 Gold

2.02700384 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

4.003332613 ×

81 Thallium

1.01350192 × 10^55

10^27

10^56

7.89107219 ×

4.104682805 ×

4.104682806 ×

83 Bismuth

−1.0000000 × 10^47

10^27

10^56

FOR THE ODD NUMBERED ELEMENTS IN THE SIXTH SHELL, THE ONLY ELEMENT FOR WHICH THE FORCE

OF ELECTRICAL REPULSION IS GREATER THAN THE MAGNETIC FIELD, IS BISMUTH. THIS IS THE ONLY

PARTICLE IN THE SIXTH SHELL THAT IS RADIOACTIVE WITHIN THE SIXTH SHELL.

85 Astatine:

F = (8.9876 × 10^9 * (83* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.090146693 × 10^56

87 Francium:

F = (8.9876 × 10^9 * (85* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.164608059 × 10^56

89 Actinium:

F = (8.9876 × 10^9 * (87* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.239069425 × 10^56

91 Protactinium:

F = (8.9876 × 10^9 * (89* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.313530792 × 10^56

93 Neptunium:

F = (8.9876 × 10^9 * (91* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.387992158 × 10^56

95 Americium:

F = (8.9876 × 10^9 * (93* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.462453524 × 10^56

97 Berkelium:

F = (8.9876 × 10^9 * (95* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.53691489 × 10^56

99 Einsteinium:

F = (8.9876 × 10^9 * (97* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.611376256 × 10^56

MAGNETIC -

SHELL # 7

SHELL DIAMETER

MAGNETIC FIELD

ELECTRICAL FORCE

ELECTRICAL

7.89107219 ×

2.584873137 ×

3.090146693 ×

−5.05273556 ×

85 Astatine

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.164608059 ×

−5.79734922 ×

87 Francium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.239069425 ×

−6.54196288 ×

89 Actinium

10^27

10^56

10^55

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7.89107219 ×

2.584873137 ×

3.313530792 ×

−7.28657655 ×

Protactinium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.387992158 ×

−8.03119021 ×

93 Neptunium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.462453524 ×

−8.77580387 ×

95 Americium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.536914890 ×

−9.52041753 ×

97 Berkelium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.611376256 ×

−1.02650311 ×

99 Einsteinium

10^27

10^56

FOR THESE ODD NUMBERED ELEMENTS IN THE SEVENTH SHELL, EVERY ELEMENT HAS AN ELECTRICAL

REPULSION FORCE WHICH EXCEEDS THE MAGNETIC FIELD STRENGTH OF THE SEVENTH SHELL, AND

EVERY ELEMENT WITHIN THIS SEVENTH SHELL IS RADIOACTIVE.

NO. OF

NO. OF NO. OF NO. OF LEFTOVER LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

85 Astatine

87 Francium

89 Actinium

91 Protactinium

93 Neptunium

95 Americium

97 Berkelium

99 Einsteinium

• EINSTEINIUM - HE6 = BERKELIUM

• BERKELIUM - HE4 = AMERICIUM

• AMERICIUM - HE6 = NEPTUNIUM

• NEPTUNIUM - HE6 = PROTACTINIUM

• PROTACTINIUM - HE4 = ACTINIUM

• ACTINIUM - HE4 = FRANCIUM

FOR BOTH THE ODD NUMBERED, AND EVEN NUMBERED ELEMENTS IN THE SEVENTH SHELL, IT IS

POSSIBLE TO MOVE FROM ONE ELEMENT TO ANOTHER USING THE ADDITION AND SUBTRACTION OF

ALPHA PARTICLES.

SIXTH AND SEVENTH SHELL EVEN NUMBERED ELEMENTS

62 Samarium:

F = (8.9876 × 10^9 * (60* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.040505782×10⁵⁶

64 Gadolinium:

F = (8.9876 × 10^9 * (62* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.141855975×10⁵⁶

66 Dysprosium:

F = (8.9876 × 10^9 * (64* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.243206168×10⁵⁶

68 Erbium:

F = (8.9876 × 10^9 * (66* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.34455636×10⁵⁶

70 Ytterbium:

F = (8.9876 × 10^9 * (68* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.445906553×10⁵⁶

72 Hafnium:

F = (8.9876 × 10^9 * (70* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.547256746×10⁵⁶

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74 Tungsten :

F = (8.9876 × 10^9 * (72* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.648606939×10⁵⁶

76 Osmium:

F = (8.9876 × 10^9 * (74* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.749957131×10⁵⁶

78 Platinum:

F = (8.9876 × 10^9 * (76* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.851307324×10⁵⁶

80 Mercury:

F = (8.9876 × 10^9 * (78* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 3.952657517×10⁵⁶

82 Lead :

F = (8.9876 × 10^9 * (80* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((4.574866806 ×

10^55)) = 4.05400771×10⁵⁶

MAGNETIC -

SHELL # 6

SHELL DIAMETER

MAGNETIC FIELD

ELECTRICAL FORCE

ELECTRICAL

62 Samarium 7.89107219 × 10^27 4.104682805 × 10^56 3.040505782 × 10^56 1.06417702 × 10^56

64 Gadolinium 7.89107219 × 10^27 4.104682805 × 10^56 3.141855975 × 10^56 9.62826830 × 10^55

66 Dysprosium 7.89107219 × 10^27 4.104682805 × 10^56 3.243206168 × 10^56 8.61476637 × 10^55

68 Erbium

7.89107219 × 10^27 4.104682805 × 10^56 3.344556360 × 10^56 7.60126445 × 10^55

70 Ytterbium 7.89107219 × 10^27 4.104682805 × 10^56 3.445906553 × 10^56 6.58776252 × 10^55

72 Hafnium

7.89107219 × 10^27 4.104682805 × 10^56 3.547256746 × 10^56 5.57426059 × 10^55

74 Tungsten 7.89107219 × 10^27 4.104682805 × 10^56 3.648606939 × 10^56 4.56075866 × 10^55

76 Osmium

7.89107219 × 10^27 4.104682805 × 10^56 3.749957131 × 10^56 3.54725674 × 10^55

78 Platinum

7.89107219 × 10^27 4.104682805 × 10^56 3.851307324 × 10^56 2.53375481 × 10^55

80 Mercury

7.89107219 × 10^27 4.104682805 × 10^56 3.952657517 × 10^56 1.52025288 × 10^55

82 Lead

7.89107219 × 10^27 4.104682805 × 10^56 4.054007710 × 10^56 5.06750950 × 10^54

FOR THE EVEN NUMBERED ELEMENTS IN THE SIXTH SHELL, NO ELEMENT HAS A FORCE OF ELECTRICAL

REPULSION WHICH IS GREATER THAN THE MAGNETIC FIELD. CONSEQUENTLY, THERE ARE NO

RADIOACTIVE ELEMENTS WITHIN THIS SIXTH SHELL.

84 Polonium:

F = (8.9876 × 10^9 * (82* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.05291601 × 10^56

86 Radon:

F = (8.9876 × 10^9 * (84* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.127377376 × 10^56

88 Radium:

F = (8.9876 × 10^9 * (86* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.201838742 × 10^56

9O Thorium:

F = (8.9876 × 10^9 * (88* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.276300108 × 10^56

92 Uranium:

F = (8.9876 × 10^9 * (90* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.350761475 × 10^56

94 Plutonium:

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F = (8.9876 × 10^9 * (92* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.425222841 × 10^56

96 Curium:

F = (8.9876 × 10^9 * (94* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.499684207 × 10^56

98 Californium:

F = (8.9876 × 10^9 * (96* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / (6.226902041 ×

10^55) = 3.574145573 × 10^56

100 Fermium:

F = (8.9876 × 10^9 * (98* 1.13566346171621 × 10^50) * (2 * 1.13566346171621 × 10^50)) / ((6.226902041 ×

10^55)) = 3.648606939×10⁵⁶

MAGNETIC -

SHELL # 7

SHELL DIAMETER

MAGNETIC FIELD

ELECTRICAL FORCE

ELECTRICAL

7.89107219 ×

2.584873137 ×

3.052916010 ×

−4.68042873 ×

84 Polonium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.127377376 ×

−5.42504230 ×

86 Radon

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.201838742 ×

−6.16965605 ×

88 Radium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.276300108 ×

−6.91426971 ×

9O Thorium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.350761475 ×

−7.65888338 ×

92 Uranium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.425222841 ×

−8.40349704 ×

94 Plutonium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.499684207 ×

−9.14811070 ×

96 Curium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.574145573 ×

−9.89272436 ×

Californium

10^27

10^56

10^55

7.89107219 ×

2.584873137 ×

3.648606939 ×

−1.06373380 ×

100 Fermium

10^27

10^56

FOR THESE EVEN NUMBERED ELEMENTS IN THE SEVENTH SHELL, EVERY ELEMENT HAS AN ELECTRICAL

REPULSION FORCE WHICH EXCEEDS THE MAGNETIC FIELD STRENGTH OF THE SEVENTH SHELL, AND

EVERY ELEMENT WITHIN THIS SEVENTH SHELL IS RADIOACTIVE.

NO. OF

NO. OF NO. OF NO. OF LEFTOVER

LEFTOVER

TOTAL

Z NAME

NEUTRONS

HE4

HE5

HE6

TRITONS DEUTERONS PARTICLES HYDROGEN

84 Polonium

86 Radon

88 Radium

9O Thorium

92 Uranium

94 Plutonium

96 Curium

98 Californium

100 Fermium

• FERMIUM - HE6 = CALIFORNIUM

• CALIFORNIUM - HE6 = CURIUM

• CURIUM + HE6 - HE5 - HE4 = PLUTONIUM

• PLUTONIUM - HE6 = URANIUM

• URANIUM - HE6 = THORIUM

• THORIUM - HE6 = RADIUM

• RADIUM - HE4 = RADON

PARTICLE RADIATION

THERE ARE TWO TYPES OF PARTICLE RADIATION:

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1. ALPHA PARTICLE RADIATION.

2. BETA PARTICLE RADIATION.

FROM THE HYPOTHETICAL DYNAMISM NUCLEAR SHELL MODEL THAT I HAVE DESCRIBED ABOVE, I SUBMIT

THAT THE ABILITY OF THE NUCLEUS TO HOLD ON TIGHTLY TO ITS POSITIVELY CHARGED PARTICLES, IS

COMPLETELY DUE TO THE COMBINED ATTRACTIVE FORCES OF THE MAGNETIC MONOPOLE, AND GRAVITY.

THE MODEL CLEARLY SHOWS THAT FOR ALL BUT ONE PARTICLE, (BISMUTH), THAT THE STRENGTH OF THE

MAGNETIC FIELD FOR SHELLS ONE THROUGH SIX IS CLEARLY STRONGER THAN THE REPULSIVE

ELECTRICAL FORCE THAT THE ALPHA PARTICLES HAVE WITHIN THE NUCLEUS. THIS MAGNETIC FIELD IS

OBVIOUSLY THE MAJOR FACTOR THAT IS HOLDING THESE POSITIVELY CHARGED ALPHA PARTICLES IN

PLACE. THE LONE EXCEPTION TO THIS, WHICH IS THE BISMUTH ELEMENT, HAS AN ELECTRICAL

REPULSION WHICH EXCEEDS THE STRENGTH OF THE MAGNETIC FIELD OF THE SIXTH SHELL, AND IS THUS

THE ONLY RADIOACTIVE PARTICLE. I SUBMIT THAT THIS LONE EXCEPTION PROVES THE RULE.

THEREFORE, WHEN THE NUCLEUS SIZE IS TOO LARGE FOR THE MAGNETIC FIELD TO KEEP THE ALPHA PARTICLES SECURELY WITHIN THE SHELL, THEN AN ALPHA PARTICLE MAY BREAK LOOSE FROM THE

OUTER SEVENTH SHELL AND, HAVING REACHED ESCAPE VELOCITY DUE TO THE WEAKENING HOLD OF THE

MAGNETIC FIELD, FLY OFF AS ALPHA PARTICLE RADIATION. IN THIS DYNAMISM MODEL FOR SHELL

STRUCTURE; NEUTRONS MAY HAVE TWO ELECTRONS AT THEIR CORE, SURROUNDED BY THREE UPQUARKS.

SIMILARLY, TRITONS HAVE TWO ELECTRONS AT THEIR CORE SURROUNDED BY THREE PROTONS. BOTH OF

THESE PARTICLES ARE CAPABLE OF EMITTING ONE OF THE ELECTRONS AT THEIR CORE BECAUSE THEIR

STRUCTURE IS INHERENTLY UNSTABLE. WHEN A NEUTRON EMITS AN ELECTRON AS BETA PARTICLE

RADIATION, IT BECOMES A PROTON. WHEN A TRITON EMITS AN ELECTRON AS BETA PARTICLE RADIATION, IT BECOMES AN HE3 ALPHA PARTICLE.

IN THE DYNAMISM MODEL, THE REASON FOR THE BETA PARTICLE RADIATION, WHICH IS THE EMISSION OF

AN ELECTRON, FROM THE NEUTRON AND FROM THE TRITON, IS DUE TO THE LACK OF ADEQUATE

AMOUNTS OF POSITIVE CHARGE IN THE PARTICLES THAT SURROUND THE ELECTRONS. IN BOTH THE

NEUTRON AND THE TRITON, THERE ARE TWO ELECTRONS AT THE CORE OF EACH PARTICLE. THE NEUTRON

AND THE TRITON HAVE TWO ELECTRONS AT THEIR CENTER, WHICH ARE SURROUNDED BY THREE

POSITIVELY CHARGED PARTICLES. IN THE NEUTRON'S CASE, THE TWO ELECTRONS ARE SURROUNDED BY

THREE UPQUARKS, WHICH EACH HAVE POSITIVE TWO-THIRDS ELECTRIAL CHARGE. IN THE TRITON'S CASE, THE TWO ELECTRONS ARE SURROUNDED BY THREE PROTONS WHICH EACH HAVE POSITIVE ONE

ELECTRICAL CHARGE. THE TOTAL POSITIVE CHARGE IS NOT ENOUGH TO ATTRACT BOTH ELECTRONS AT

THE CENTER OF EACH PARTICLE. THEREFORE, ONE OF THE ELECTRONS IS INVARIABLY EMITTED AS A BETA PARTICLE. THIS CONVERTS THE NEUTRON INTO A PROTON, AND CONVERTS THE TRITON INTO AN HE3

ALPHA PARTICLE. THE NEWLY FORMED PROTON NOW HAS JUST ONE ELECTRON AT ITS CENTER, SURROUNDED BY THREE UPQUARKS, WHICH IS AN EXTREMELY STABLE FORMATION. THE NEWLY FORMED

HE3 PARTICLE ALSO HAS AN ELECTRON AT ITS CENTER, BUT THIS ELECTRON IS SURROUNDED BY THREE

PROTONS. BUT THE HE3 PARTICLE IS AN UNBALANCED ALPHA PARTICLE, AND ITS HIGH POSITIVE CHARGE

TO MASS RATIO MEANS THAT IT WILL ATTRACT A PROTON/ELECTRON PAIR, (A NEUTRON), FROM WITHIN

THE PLASMA, TO BECOME AN HE4 ALPHA PARTICLE, WHICH IS AN EXTREMELY STABLE PARTICLE; MUCH

LIKE THE PROTON. THE HE4 HAS TWO ELECTRONS AT ITS CENTER, SURROUNDED BY FOUR PROTONS, AND

IS AN EXTREMELY BALANCED PARTICLE; WITH APPROPRIATE CHARGE TO MASS RATIO, TO KEEP THE

ELECTRONS WITHIN IT FROM BEING EJECTED.

THE EVOLUTION OF ATOMIC MATTER

BELOW IS A TABLE THAT GIVES THE FIRST TEN ELEMENTS OF THE PERIODIC TABLE AND THEIR NATURAL

ABUNDANCE IN THE EARTH'S SURFACE ENVIRONMENT.

___________________________________________________________________________________________________

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

THIS IS JUST THE FIRST TEN ELEMENTS; BUT ALREADY A PATTERN IS READILY PERCEPTIBLE THAT THE

NUMBER OF NUCLEONS IN THE STABLE ISOTOPES FOR EACH ELEMENT, IS UNIQUE FOR EACH ELEMENT.

FOR EXAMPLE, HE3 HAS A SLIGHT ABUNDANCE OF 0.000137 PER CENT OF ALL STABLE HELIUM PARTICLES.

HELIUM THREE HAS THREE NUCLEONS. OBSERVE THAT FOR H3 (TRITIUM) WHICH IS THE HYDROGEN

ISOTOPE WITH THREE NUCLEONS, THAT THE ABUNDANCE IS ZERO PERCENT OF ALL STABLE HYDROGEN

PARTICLES.

THIS OCCURS AGAIN WITH N14 (NITROGEN FOURTEEN) WHICH HAS A MASSIVE ABUNDANCE OF 99.632 PER

CENT OF ALL STABLE NITROGEN PARTICLES. N14 HAS FOURTEEN NUCLEONS. OBSERVE THAT FOR C14

(CARBON FOURTEEN) WHICH IS THE CARBON ISOTOPE WITH FOURTEEN NUCLEONS, THAT THE

ABUNDANCE IS ZERO PERCENT OF ALL STABLE CARBON PARTICLES.

FURTHER OBSERVATION SHOWS THAT EVEN WHEN A NUCLEON HAS STABLE ISOTOPES IN TWO

SUCCESSIVE ELEMENTS, THAT THE OVERWHELMING PERCENTAGE IS IN ONE OF THE ELEMENTS, AND THE

OTHER ELEMENT HAS A MINISCULE PERCENTAGE OF STABLE ISOTOPES WITH THAT NUCLEON NUMBER.

THE FOLLOWING TABLE SHOWS THAT ONLY FOR THE NUCLEON NUMBER FORTY IS THERE AN ABUNDANCE

OF OVER FIFTY PER CENT FOR MORE THAN ONE ELEMENT; THE ELEMENTS BEING CALCIUM AND ARGON.

IN EVERY OTHER CASE, WHEN THE NUCLEON NUMBER HAS AN ISOTOPE IN MORE THAN ONE ELEMENT, THE ISOTOPES ARE USUALLY WELL UNDER FIFTY PER CENT ABUNDANCE.

___________________________________________________________________________________________________

Table of Isotopic Masses and Natural Abundances

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THE APPORTIONING OF NUCLEONS AMONG THE ELEMENTS

NAME

SYMBOL

ATOMIC MASS

ABUNDANCE

Tritium

3.016049

Helium

3He

3.016029

0.000137

# 3

------

---

--------

Carbon

14C

14.003242

Nitrogen

14N

14.003074

99.632

# 14

------

---

--------

Sulphur

36S

35.967081

0.02

Argon

36Ar

35.967546

0.3365

# 36

------

---

--------

Argon

40Ar

39.962383

99.6003

Potassium

40K

39.963999

0.0117

Calcium

40Ca

39.962591

96.941

# 40

------

---

--------

Calcium

46Ca

45.953693

0.004

Titanium

46Ti

45.952629

8.25

# 46

------

---

--------

Calcium

48Ca

47.952534

0.187

Titanium

48Ti

47.947947

73.72

# 48

------

---

--------

Titanium

50Ti

49.944792

5.18

Vanadium

50V

49.947163

0.250

Chromium

50Cr

49.946050

4.345

# 50

------

---

--------

Chromium

54Cr

53.938885

2.365

Iron

54Fe

53.939615

5.845

# 54

------

---

--------

Iron

58Fe

57.933280

0.282

Nickel

58Ni

57.935348

68.0769

# 58

------

---

--------

Nickel

64Ni

63.927970

0.9256

Zinc

64Zn

63.929147

48.63

# 64

------

---

--------

Zinc

70Zn

69.925325

0.62

Germanium

70Ge

69.924250

20.84

# 70

------

---

--------

Germanium

74Ge

73.921178

36.28

Selenium

74Se

73.922477

0.89

# 74

------

---

--------

Selenium

78Se

77.917310

23.77

Krypton

78Kr

77.920386

0.35

# 78

------

---

--------

Selenium

80Se

79.916522

49.61

Krypton

80Kr

79.916378

2.28

# 80

------

---

--------

Krypton

84Kr

83.911507

57.00

Strontium

84Sr

83.913425

0.56

# 84

------

---

--------

Krypton

86Kr

85.910610

17.30

Strontium

86Sr

85.909262

9.86

# 86

------

---

--------

Zirconium

92Zr

91.905040

17.15

Molybdenum

92Mo

91.906810

14.84

------

---

--------

Zirconium

94Zr

93.906316

17.38

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Molybdenum

94Mo

93.905088

9.25

# 94

------

---

--------

Molybdenum

96Mo

95.904679

16.68

Ruthenium

96Ru

95.907598

5.54

# 96

------

---

--------

Molybdenum

98Mo

97.905408

24.13

Technetium

98Tc

97.907216

Ruthenium

98Ru

97.905287

1.87

# 98

------

---

--------

Molybdenum

100Mo

99.907477

9.63

Ruthenium

100Ru

99.904220

12.60

# 100

------

---

--------

Ruthenium

102Ru

101.904350

31.55

Palladium

102Pd

101.905608

1.02

# 102

------

---

--------

Palladium

106Pd

105.903483

0.187

Cadmium

106Cd

105.906458

1.25

# 106

------

---

--------

Palladium

108Pd

107.903894

27.33

Cadmium

108Cd

107.904183

0.89

# 108

------

---

--------

Palladium

110Pd

109.905152

11.72

Cadmium

110Cd

109.903006

12.49

# 110

------

---

--------

Cadmium

112Cd

111.902757

24.13

Tin

112Sn

111.904821

0.97

# 112

------

---

--------

Cadmium

114Cd

113.903358

28.73

Tin

114Sn

113.902782

0.66

# 114

------

---

--------

Cadmium

116Cd

115.904755

7.49

Tin

116Sn

115.901744

14.54

# 116

------

---

--------

Tin

120Sn

119.902197

32.58

Tellurium

120Te

119.904020

0.09

# 120

------

---

--------

Tin

122Sn

121.903440

4.63

Tellurium

122Te

121.903047

2.55

# 122

------

---

--------

Tin

124Sn

123.905275

5.79

Tellurium

124Te

123.902819

4.74

Xenon

124Xe

123.905896

0.09

# 124

------

---

--------

Xenon

132Xe

131.904154

26.89

Barium

132Ba

131.905056

0.101

# 132

------

---

--------

Xenon

136Xe

135.907220

8.87

Barium

136Ba

135.904570

7.854

# 136

------

---

--------

Barium

138Ba

137.905241

71.698

Lanthanum

138La

137.907107

0.090

Cerium

138Ce

137.905986

0.251

# 138

------

---

--------

Cerium

142Ce

141.909240

11.114

Neodymium

142Nd

141.907719

27.2

# 142

------

---

--------

Neodymium

144Nd

143.910083

23.8

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Samarium

144Sm

143.911995

3.07

# 144

------

---

--------

Neodymium

148Nd

147.916889

5.7

Samarium

148Sm

147.914818

11.24

# 148

------

---

--------

Neodymium

150Nd

149.920887

5.6

Samarium

150Sm

149.917271

7.38

# 150

------

---

--------

Samarium

152Sm

151.919728

26.75

Gadolinium

152Gd

151.919788

0.20

# 152

------

---

--------

Samarium

154Sm

153.922205

22.75

Gadolinium

154Gd

153.920862

2.18

# 154

------

---

--------

Gadolinium

156Gd

155.922120

20.47

Dysprosium

156Dy

155.924278

0.06

# 156

------

---

--------

Dysprosium

162Dy

161.926795

25.51

Erbium

162Er

161.928775

0.14

# 162

------

---

--------

Dysprosium

164Dy

163.929171

28.18

Erbium

164Er

163.929197

1.61

# 164

------

---

--------

Erbium

168Er

167.932368

26.78

Ytterbium

168Yb

167.933894

0.13

# 168

------

---

--------

Erbium

170Er

169.935460

14.93

Ytterbium

170Yb

169.934759

3.04

# 170

------

---

--------

Ytterbium

174Yb

173.938858

31.83

Hafnium

174Hf

173.940040

0.16

# 174

------

---

--------

Hafnium

180Hf

179.946549

35.08

Tantalum

180Ta

179.947466

0.012

Tungsten

180W

179.946706

0.12

# 180

------

---

--------

Tungsten

184W

183.950933

30.64

Osmium

184Os

183.952491

0.02

# 184

------

---

--------

Osmium

190Os

189.958445

26.26

Platinum

190Pt

189.959930

0.014

# 190

------

---

--------

Osmium

192Os

191.961479

40.78

Platinum

192Pt

191.961035

0.782

# 192

------

---

--------

Platinum

196Pt

195.964935

25.242

Mercury

196Hg

195.965815

0.15

# 196

------

---

--------

Platinum

198Pt

197.967876

7.163

Mercury

198Hg

197.966752

9.97

# 198

------

---

--------

Mercury

204Hg

203.973476

6.87

Lead

204Pb

203.973029

1.4

# 204

------

---

--------

http://moltensalt.org/references/static/downloads/pdf/stable-isotopes.pdf https://euro-fusion.org/fusion/fusion-vs-fission/

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___________________________________________________________________________________________________

THE FOLLOWING TABLE SHOWS ALL THE ISOTOPES THAT ARE AT OR ABOVE FIFTY PERCENT ABUNDANCE.

IT IS CLEAR THAT ONLY 40Ar AND 40Ca BOTH HAVE OVER FIFTY PERCENT ABUNDANCE AND ALSO HAVE

THE SAME NUMBER OF NUCLEONS.

NAME

SYMBOL

ATOMIC MASS

ABUNDANCE

Nitrogen

14N

14.003074

99.632

Argon

40Ar

39.962383

99.6003

Calcium

40Ca

39.962591

96.941

Titanium

48Ti

47.947947

73.72

Nickel

58Ni

57.935348

68.0769

Selenium

80Se

79.916522

49.61

Krypton

84Kr

83.911507

57.00

Barium

138Ba

137.905241

71.698

40ARGON AND 40CALCIUM ARE BOTH OVER NINETY-FIVE PERCENT ABUNDANCE. 40CALCIUM HAS TWENTY

NEUTRONS WHICH MAKES IT A MAGIC NUMBER ISOTOPE. THIS MAY PLAY A ROLE IN WHY THIS ANOMALY

OCCURS.

A NEWLY DISCOVERED HYPOTHESIS

I ORIGINALLY ASSUMED THAT THE FORMATION OF NUCLEI, WITHIN THE PLASMA, OCCURRED

SIMULTANEOUSLY WITH THE FORMATION OF THE HYDROGEN PARTICLES, THE HELIUM PARTICLES, AND

THE NUCLEAR SHELL STRUCTURE. BUT UPON CLOSER EXAMINATION OF THE DATA ABOVE, I DISCOVERED

THAT ALMOST ALL OF THE NUCLEONS ARE ASSOCIATED WITH ONLY ONE ATOMIC ELEMENT NUMBER

WITHIN THE PERIODIC TABLE. THE EXCEPTION BEING 40Ca OR THE ISOTOPE OF CALCIUM WHICH HAS

TWENTY NEUTRONS AND TWENTY PROTONS WHICH MAKES IT NUMBER TWENTY ON THE PERIODIC TABLE

OF ELEMENTS. APPARENTLY, THE FACT THAT TWENTY IS A MAGIC NUMBER IS THE REASON WHY THIS

PARTICULAR ISOTOPE OF CALCIUM IS SO STABLE. 40Ca IS DOUBLY MAGIC AS IT HAS TWENTY PROTONS

AND TWENTY NEUTRONS, WHICH ACCOUNTS FOR WHY IT HAS OVER NINETY-FIVE PERCENT ABUNDANCE

FOR ALL ISOTOPES OF CALCIUM.

CONSIDERING THAT THE MAGIC NUMBER OF TWENTY IS WHAT CAUSES 40Ca TO BE ANOMALOUS, THEN

THE REST OF THE HYPOTHESIS HOLDS TRUE. THIS HYPOTHESIS BEING THAT IF A NUCLEON NUMBER IS

SHARED BY TWO OR MORE ELEMENTS ON THE PERIODIC TABLE, THAT ONLY ONE OF THEM HAS A CHANCE

OF BEING OVER FIFTY PERCENT ABUNDANT FOR ITS PARTICULAR ELEMENT.

THIS IS ACTUALLY EXTREMELY SIGNIFICANT IN ITS IMPLICATIONS AS TO HOW THE NUCLEAR SHELL

STRUCTURE IS FORMED. IT IMPLIES THAT THERE ISN'T ANY FORMATION OF HELIUM PARTICLES, OR

NUCLEAR SHELLS WITHIN THE PLASMA.

THUS I HYPOTHESIZE AS A POSTDICTION, THAT THERE ARE NO SHELLS OF ANY KIND WITHIN NUCLEI THAT

EXIST WITHIN THE PURE PLASMA; THE PURE PLASMA BEING THE THE ORIGINAL PLASMA THAT FILLED UP

ALL THE SPACE WITHIN THE EARLY UNIVERSE THAT GOD CREATED.

WHICH DIRECTLY IMPLIES THAT THE ONLY NUCLEAR PROCESSES THAT OCCUR WITHIN THE PLASMA, ARE

FUSION PROCESSES. SIMPLY PUT, A PROTON ADDS A NEUTRON TO BECOME A DEUTERON (AN H2

HYDROGEN NUCLEUS), THEN ADDS ANOTHER NEUTRON TO BECOME A TRITON (AN H3 HYDROGEN

NUCLEUS), AND THEN ANOTHER NEUTRON IS ADDED TO MAKE IT AN H4 HYDROGEN NUCLEUS, THEN

ANOTHER NEUTRON IS ADDED TO MAKE IT AN H5 HYDROGEN NUCLEUS, ... , AND SO ON UNTILE IT

BECOME AN HN HYDROGEN NUCLEUS WHERE 'N' CAN BE OVER TWO HUNDRED AND FIFTY NUCLEONS ('N

> 250'), IN EXTREME CASES.

IMAGINE THAT! THIS IS BASICALLY A HYDROGEN ISOTOPE WITH A MASSIVE AMOUNT OF NEUTRONS

ATTACHED TO IT. THE SIGNIFICANCE OF THIS IS THE STRUCTURE OF THIS MASSIVE BLOB OF NEUTRONS; BECAUSE THE NEUTRONS ARE COMPOSED OF PROTON/ELECTRON PAIRS WITHIN THIS OVERALL

STRUCTURE. THIS IS VERY IMPORTANT TO UNDERSTAND BECAUSE THE PROTONS ARE ALL POSITIVELY

CHARGED AND ARE REPULSIVE TO EACH OTHER; AND SIMILARLY, THE ELECTRONS ARE ALL NEGATIVELY

CHARGED AND ARE REPULSIVE TO EACH OTHER; WHICH MEANS THAT SOME TYPE OF GRID STRUCTURE

HAS TO OCCUR WHERE THE PROTONS ARE ON A PROTON GRID, AND THE ELECTRONS ARE ON AN

ELECTRON GRID, AND ALL OF THIS IS OCCURRING WITHIN THE NUCLEUS.

THE PURE PLASMA THAT THIS NUCLEUS INHABITS, ALSO FEATURE A PROTON GRID STRUCTURE

OVERLAPPING AN ELECTRON GRID STRUCTURE. THE MAIN DIFFERENCE IS THE DENSITY FACTOR. THE

NUCLEUS IS MUCH DENSER THAN THE PLASMA WITHIN WHICH IT IS CONTAINED.

___________________________________________________________________________________________________

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EUROfusion

Home » What is Fusion? » Fusion vs Fission

In fission, energy is gained by splitting heavy atoms, for example uranium, into smaller atoms such as iodine, caesium, strontium, xenon and barium, to name just a few. However, fusion is combining light atoms, for example two hydrogen isotopes, deuterium and tritium, to form the heavier helium. Both reactions release energy which, in a power plant, would be used to boil water to drive a steam generator, thus producing electricity.

Fission and chain reactions

Fission is the nuclear process that is currently run in nuclear power plants. It is triggered by uranium absorbing a

neutron, which renders the nucleus unstable. The result of the instability is the nucleus breaking up, in any one of many different ways, and producing more neutrons, which in turn hit more uranium atoms and make them unstable and so on. This chain reaction is the key to fission reactions, but it can lead to a runaway process resulting in nuclear accidents. In conventional nuclear power stations today, there are systems in place to moderate the chain reactions to prevent accident scenarios and stringent security measures to deal with proliferation issues.

Fusion: inherently safe but challenging

Unlike nuclear fission, the nuclear fusion reaction in a tokamak is an inherently safe reaction. The reasons that have made fusion so difficult to achieve to date are the same ones that make it safe: it is a finely balanced reaction which

is very sensitive to the conditions – the reaction will die if the plasma is too cold or too hot, or if there is too much

fuel or not enough, or too many contaminants, or if the magnetic fields are not set up just right to control the

turbulence of the hot plasma. This is why fusion is still in the research and development phase – and fission is already making electricity.

Left: fusion, when light atoms fuse and release energy. Right: fission, when heavy atoms split and release energy.

Binding energy

The key to why some atoms split and release energy while others fuse to do the same lies in how tightly the protons and neutrons are held together. If a nuclear reaction produces nuclei that are more tightly bound than the originals then energy will be produced by fusion, and for fission the opposite is true.

It turns out that the most tightly bound atomic nuclei are around the size of iron, which has 26 protons in the nucleus. So, one can release energy either by splitting very large nuclei, like uranium with 92 protons, to get

smaller products, or fusing very light nuclei, like hydrogen, with just one proton to get bigger products.

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Smaller nuclei fuse and release energy until at iron no more energy is released by fusion.

https://euro-fusion.org/fusion/fusion-vs-fission/

___________________________________________________________________________________________________

THE FORMATION OF THE SHELL STRUCTURE

THE MAGNETIC FIELD WITHIN A PLASMA IS WEAKENED BY THE FUSION PROCESSES THAT GO ON WITHIN

THE PLASMA. FUSION ONLY TAKES PLACE WITHIN THE STATIC PLASMA THAT FEATURES A STATIONARY

ELECTRON GRID OVERLAPPING WITH A STATIONARY PROTON GRID. IT IS WITHIN THESE STATIONARY GRIDS

THAT THESE MASSIVE HYDROGEN ISOTOPES DEVELOP OVER TIME. AS HAS BEEN EXPLAINED ABOVE, THESE HYDROGEN ISOTOPES HAVE A GRID-LIKE STRUCTURE FOR BOTH THE PROTONS AND THE

ELECTRONS AS WELL. BUT AS MORE FUSION TAKES PLACE, THESE NUCLEI CONTINUE TO GROW WITHIN

THE PLASMA, AND THE PLASMA LOSES BOTH ELECTRONS AND PROTONS AS THEY ARE SWALLOWED UP BY

THE GROWING NUCLEI. THE MORE FUSION, THE LESS OVERALL CHARGE WITHIN THE PLASMA, AND THE

WEAKER THE MAGNETIC FIELD GETS; BECAUSE THE MAGNETIC FIELD IS GENERATED BY MOVING

ELECTRICAL CHARGES. THERE ARE TWO FORCES AT WORK WITHIN THE PLASMA, AND THESE FORCES ARE

OPPOSED TO EACH OTHER. THE MAGNETIC FIELD GENERATED BY THE SPINNING ELECTRICAL CHARGES

WITHIN THE PLASMA, LOCKS THE PROTONS AND ELECTRONS INTO PLACE WITHIN THE GRID STRUCTURE, AND HAS THE OVERALL EFFECT OF FIGHTING OFF THE FORCE OF GRAVITY; WHICH SEEKS TO MAKE THE

PLASMA CONTRACT INTO A DENSER OBJECT.

THEREFORE, WHEN ENOUGH FUSION HAS OCCURRED, THIS IS PRECISELY WHAT HAPPENS; THE MAGNETIC

FIELD WEAKENS ENOUGH AGAINST GRAVITY THAT THE GRAVITATIONAL FORCE CAUSES THE PLASMA TO

COLLAPSE INTO A VERY DENSE BALL. THIS GRAVITATIONAL COLLAPSE OF THE PLASMA RESULTS IN A QUASAR, WHICH IS THE BIRTH OF A SUPERMASSIVE BLACK HOLE, WHICH IS THE CENTER OF A NEWLY

BORN SPHERICAL GALAXY. THE DYNAMISM MODEL POSTDICTS THAT GALAXIES ARE BORN OUT OF THE

ORIGINAL PLASMA THAT COMPRISED THE UNIVERSAL MATTER, IN PRECISELY THIS WAY, FOR EVERY

GALAXY THAT HAS EVER BEEN CREATED, AND PREDICTS THAT IT WILL DO SO FOR EVERY FUTURE GALAXY

THAT WILL EVER BE CREATED.

THIS QUITE LITERALLY MEANS THAT THE FUSION PROCESS IS HOW THE ORIGINAL PLASMA THAT HAS

EXISTED SINCE GOD'S CREATION OF THE UNIVERSE, IS PHYSICALLY EVOLVED INTO THE UNIVERSE THAT IS

REPLETE WITH GALAXIES THAT WE SEE TODAY. THUS DYNAMISM HYPOTHESIZES THAT THE ORIGINAL

PLASMA, WHICH IS DARK MATTER, BECAUSE IT DOES NOT HAVE ANY SHELL STRUCTURE, NUCLEAR OR

ELECTRON; AND IS THEREFORE NOT ATOMIC MATTER, IS TRANSFORMED INTO ATOMIC MATTER THROUGH

THE FUSION PROCESS; WHICH IS PRIMARILY RESPONSIBLE FOR THE CREATION OF GALAXIES AND SOLAR

SYSTEMS WHICH ARE PRIMARILY ATOMIC MATTER BECAUSE THE SUBSTANCE OF THESE SOLAR SYSTEMS

DOES HAVE NUCLEAR AND ELECTRON SHELL STRUCTURES, WHICH IS WHAT MAKES LIFE POSSIBLE.

DURING THE EXOTHERMIC PROCESS WHICH CREATES THE GALAXIES AND THE SOLAR SYSTEMS, THE

NUCLEAR AND ELECTRON SHELLS ARE CREATED AS THE NUCLEI ARE CONVERTED FROM A VERY DENSE

NUCLEUS, DUE TO THE PRESENCE OF THE POWERFUL MAGNETIC FIELDS WITHIN THE PLASMA, TO A MUCH MORE EXPANSIVE NUCLEUS, DUE TO THE WEAKENING OF THE MAGNETIC FIELD WITHIN WHICH

THE NUCLEUS EXISTS.

THIS TABLE 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

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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

NOTE THE NUCLEON NUMBERS: 9, 19, 23, 27, 31, 45, 55, 59, 75, 89, 93, 103, 127, 133, 141, 159, 165, 169, 197, 209, 232, 231. THE HYDROGEN ISOTOPES THAT HAVE THESE NUCLEON NUMBERS WHEN THE EXOTHERMIC

REACTION OF GRAVITATIONAL COLLAPSE BEGINS, ARE HYPOTHESIZED, IN THE DYNAMISM MODEL, TO

CONVERT DIRECTLY INTO THE ISOTOPE FOR THE ELEMENTS IN THE ABOVE TABLE, WITH THEIR

ATTENDANT NUCLEAR AND ELECTRON SHELLS. NOTE THAT MOST OF THE ABOVE ISOTOPES ARE ODD

NUMBERED WHICH MEANS THAT THESE ODD NUMBERED ONES WOULD HAVE AT LEAST ONE HYDROGEN

PARTICLE, EITHER A DEUTERIUM OR A TRITIUM PARTICLE AS PART OF THE NUCLEUS. BERYLLIUM AND

THORIUM ARE THE EVEN NUMBERED ELEMENTS.

IN REFERENCE TO THE TABLE WHICH IS TITLED "THE APPORTIONING OF NUCLEONS AMONG THE

ELEMENTS" NOTE THAT EXCEPT FOR TRITIUM WHICH HAS THREE NUCLEONS AS DOES HELIUM THREE

(HE3), ALL OF THE REST OF THE NUCLEONS THAT ARE SPREAD OVER MORE THAN ONE ATOMIC ELEMENT

ARE EVEN NUMBERED. WHAT COULD POSSIBLE ACCOUNT FOR THIS DIFFERENCE BETWEEN ODD

NUMBERED NUCLEONS HAVING ONE HUNDRED PERCENT ABUNDANCE FOR A PARTICULAR ATOMIC

ELEMENT, AND EVEN NUMBERED NUCLEONS BEING SPREAD OUT OVER MULITIPLE ATOMIC ELEMENTS?

THERE ARE SOME ODD NUMBERED ELEMENTS THAT SHARE A NUCLEON NUMBER WITH OTHER

ELEMENTS, BUT THE OTHER ELEMENTS ARE ALWAYS EVEN NUMBERED ELEMENTS. SO THESE ODD

NUMBERED ELEMENTS ARE ALWAYS IN THE MIDDLE OF A GROUP OF THREE ELEMENTS; SANDWICHED

BETWEEN TWO EVEN NUMBERED ELEMENTS. THESE ISOTOPES ARE:

1. 19 Potassium 40K 0.0117

2. 23 Vanadium 50V 0.250

3. 43 Technetium 98Tc *

4. 57 Lanthanum 138La 0.090

5. 73 Tantalum 180Ta 0.012

ALSO NOTEWORTHY IS THAT THEIR ABUNDANCE IS ALWAYS LESS THAN, OR EQUAL TO, ONE QUARTER OF A PERCENTAGE POINT.

SINCE THE PERCENTAGE OF ABUNDANCE FOR THESE ANOMALOUS ODD NUMBERED ELEMENTS ARE SO

INSIGNIFICANTLY SMALL, WHY EVEN BOTHER TO POINT IT OUT?

I BELIEVE THAT IN THESE CASES, IT IS THE EXCEPTIONS THAT PROVE THE RULE. FOR INSTANCE, WHAT

WOULD HAPPEN IF I SIMPLY DISCARDED THESE ANOMALOUS RESULTS AS INSIGNIFICANT. LET'S SAY THAT

I JUST IGNORED THEM AS IF THEY WEREN'T THERE. THEN THE OVERWHELMING EVIDENCE WOULD

SUPPORT THE IDEA THAT THE DISTRIUBUTION OF EVEN NUMBERED NUCLEONS WOULD BE ENTIRELY

AMONGST THE EVEN NUMBERED ELEMENTS. SO WHAT IS THE SIGNIFICANCE OF THAT CONCLUSION?

TAKING INTO ACCOUNT THAT ODD NUMBERED ELEMENTS HAVE TO HAVE A HYDROGEN PARTICLE INSIDE

THE SHELL SOMEWHERE, EITHER A DEUTERON OR A TRITON, AND EVEN NUMBERED ELEMENTS ARE

ABSENT ANY HYDROGEN PARTICLES; THEN THE CONCLUSION ABOVE WOULD IMPLY THAT THE PRESENCE

OF THE HYDROGEN PARTICLES IS EXTREMELY IMPORTANT, NOT ONLY TO THE FORMATION OF THE SHELLS, BUT ALSO TO THE STABILITY OF THE ISOTOPES THAT ARE FORMED FROM THESE SHELLS.

OUT OF FORTY-NINE POSSIBLE HYDROGEN PARTICLES ONLY THESE FOUR ELEMENT HAVE A DEUTERON IN

THEIR OUTER SHELL:

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1. 7 Nitrogen

2. 43 Technetium

3. 85 Astatine

4. 99 Einsteinium

THE OTHER FORTY-FIVE ODD NUMBERED ELEMENTS ALL HAVE A TRITON IN THEIR OUTER SHELL.

EVERY ONE OF THE TWENTY ODD NUMBERED ELEMENTS THAT HAVE ONE HUNDRED PERCENT

ABUNDANCE HAS A TRITON HYDROGEN PARTICLE IN ITS OUTER SHELL. THIS IMPLIES THAT THERE IS

SOMETHING ABOUT HAVING A TRITON PARTICLE IN ITS OUTER SHELL, COMBINED WITH HAVING AN ODD

NUMBER OF NUCLEONS, THAT LENDS ITSELF TO THAT ISOTOPE OF THAT ELEMENT BEING EXTREMELY

STABLE.

BY THE WAY, THE FOUR ODD NUMBERED ELEMENTS IN THE ABOVE LIST, WHICH ALL HAVE DEUTERONS IN

THEIR OUTER SHELL, ARE NOT ONE HUNDRED PERCENT ABUNDANCE.

ONE HUNDRED PERCENT ABUNDANCE FOR ANY ISOTOPE OF AN ELEMENT IMPLIES THAT THE OTHER

ISOTOPES FOR THAT ELEMENT ARE ALL RELATIVELY UNSTABLE. EXCEPT FOR BERYLLIUM AND THORIUM, WHICH ARE EVEN NUMBERED ELEMENTS AND DON'T HAVE A HYDROGEN PARTICLE IN ITS SHELL, ALL THE

OTHER ONE HUNDRED PERCENT ABUNDANCE ELEMENTS ARE ODD NUMBERED AND HAVE A TRITIUM

PARTICLE IN ITS OUTER SHELL. IT IS ALSO REMARKABLE THAT THESE TWENTY ODD NUMBERED ELEMENTS

HAVE EXCLUSIVE ASSOCIATION WITH THEIR RESPECTIVE NUCLEON NUMBER ISOTOPE. THIS MEANS THAT

EVERY NUCLEUS WITH THAT PARTICULAR NUCLEON WILL FORM A SHELL STRUCTURE THAT HAS ONLY

THAT ISOTOPE NUMBER, AND NO OTHER.

THUS I HYPOTHESIZE THAT THE TRITIUM HYDROGEN PARTICLE MAY HAVE A ROLE WITHIN THE OUTER

SHELL OF THESE ODD NUMBERED ELEMENTS, OF A SHEPHERD PARTICLE. THAT SOMEHOW, THE

PRESENCE OF THE SMALLER TRITIUM PARTICLE KEEPS THE LARGER HELIUM PARTICLES IN CHECK; SUCH

THAT WHEN THE NUCLEAR SHELL IS BEING FORMED FOR ODD NUMBERED ELEMENTS, THAT THE

PRESENCE OF TRITIUM WILL ACT TO STABILIZE AND CONCENTRATE THE ODD NUMBERED NUCLEONS AND, WHENEVER POSSIBLE, MAKE THE ODD NUMBERED NUCLEON EXCLUSIVE TO THE ODD NUMBERED

ELEMENT SUCH THAT IT HAS ONE HUNDRED PERCENT ABUNDANCE FOR THAT PARICULAR ISOTOPE. MIND

YOU, THIS IS MERELY A HYPOTHESIS; AN EDUCATED GUESS, AND THERE IS A HIGH PROBABILITY THAT IT IS

WRONG TO SOME EXTENT.

BUT WHAT I MEAN WHEN I SAY "STABILIZE AND CONCENTRATE' IS THAT THE TRITIUM HYDROGEN

PARTICLE, SOMEHOW PREVENTS BETA DECAY FROM HAPPENNING IN THE ODD NUMBERED ELEMENTS.

THIS WOULD EXPLAIN THE EXCLUSIVITY OF THOSE PARTICULAR NUCLEONS FOR THOSE PARTICULAR

ELEMENTS. IN THE ABSENCE OF THE HYDROGEN PARTICLES IN THE OUTER SHELL, THE EVEN NUMBERED

NUCLEONS ARE EXPERIENCING BETA DECAY AFTER THE SHELL HAS BEEN FORMED, AND THIS GIVES RISE

TO THE SAME NUCLEON BEING SHARED AMONGST MULTIPLE EVEN NUMBERED ELEMENTS. FOR

INSTANCE, TAKE A LOOK AT THIS TABLE:

NAME

SYMBOL

ATOMIC MASS

ABUNDANCE

Hafnium

180Hf

179.946549

35.08

Tantalum

180Ta

179.947466

0.012

Tungsten

180W

179.946706

0.12

# 180

------

---

--------

THE HYPOTHESIS IS THAT THE ORIGINAL SHELL FORMATION WAS ENTIRELY HAFNIUM WHICH IS AN EVEN

NUMBERED ELEMENT. HOWEVER, IN THE ABSENCE OF THE SHEPHERDING PARTICLE TRITIUM, 180Hf WILL

TEND TO BETA DECAY TO 180Ta, WHICH IS A TANTALUM ISOTOPE; BUT IT IS UNSTABLE BECAUSE THAT

PARTICULAR ISOTOPE WAS NOT ORIGINALLY CREATED BUT WAS ARRIVED AT THROUGH BETA DECAY. BUT

DUE TO ITS INSTABILITY, IT WILL FURTHER BETA DECAY TO 180W, WHICH IS TUNGSTEN ISOTOPE. THIS

TUNGSTEN ISOTOPE IS ALSO UNSTABLE FOR THE SAME REASON.

I HYPOTHESIZE THAT ALL OF THE OTHER INSTANCES OF THE SAME NUCLEON BEING DISTRIBUTE

AMONGST TWO OR MORE ELEMENTS IS DUE TO THE SAME REASON. THAT THE LOWER ONE IS ANALOGOUS

TO 180Hf, AND THE HIGHER ONES ARE DUE TO BETA DECAY. THIS IMPLIES THAT ALL OF THE NUCLEONS

WERE TURNED INTO JUST ONE ISOTOPE OF ONE ELEMENT IN AN EXCLUSIVE MANNER, WHEN THE

NUCLEAR SHELL WAS FIRST FORMED. BUT THEN THROUGH THE PROCESS OF BETA DECAY, THE OTHER

ISOTOPES OF THE OTHER ELEMENTS CAME TO SHARE THE SAME NUCLEON NUMBER. HOWEVER WHEN

THE TRITIUM PARTICLE IS PRESENT, THE BETA DECAY IS SOMEHOW HALTED BECAUSE IT PLAYS THE ROLE

OF A SHEPHERD PARTICLE, WHICH IS WHY THOSE PARTICULAR ODD NUMBERED ELEMENTS HAVE ONE

HUNDRED PERCENT ABUNDANCE WHEN THEY HAVE A TRITIUM PARTICLE IN THEIR OUTER SHELL.

THUS THE HYPOTHESIS IMPLIES THAT ALL OF THE ISOTOPES THAT HAVE THE SAME NUCLEON NUMBER

ARE INDEED CONNECTED THROUGH BETA DECAY. FROM THE ABOVE TABLES, IT IS CLEAR THAT NONE OF

THE ODD NUMBERED ELEMENTS ARE CONNECTED TO EACH OTHER BY BETA DECAY. THUS ALL OF THE

BETA DECAY CONNECTED ELEMENTS ARE PRIMARILY EVEN NUMBERED ELEMENTS WITH SOME ODD

NUMBERED ELEMENT SANDWICHED BETWEEN THEM OCCASIONALLY.

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FISSION DOESN'T OCCUR UNTIL THE NUCLEAR SHELL HAS BEEN FORMED. ACCORDING TO THE DYNAMISM

MODEL, THE ONLY NUCLEAR PROCESS THAT OCCURS WITHIN THE ORIGINAL PLASMA ARE FUSION

REACTIONS WHICH CAN BUILD UP NUCLEI UP TO TWO HUNDRED AND FIFTY NUCLEONS IN EXTREME

CASES. THESE NUCLEI THAT EXIST WITHIN THE PLASMA ARE EXTREMELY DENSE AND REMAIN SO UNTIL

THE TRANSFORMATION INTO NUCLEAR SHELLS OCCUR. THIS TRANSFORMATION ONLY OCCURS AFTER

THE MAGNETIC FIELD HAS WEAKENED ENOUGH TO ALLOW IT. ONCE THE NUCLEAR SHELL HAS FORMED

THEN FISSION BECOMES POSSIBLE. EVERY FORM OF PARTICLE RADIATION IS A FISSION REACTION. ALPHA RADIATION, BETA RADIATON AND NEUTRON RADIATION ARE ALL NUCLEAR FISSION REACTIONS, ACCORDING TO THE DYNAMISM MODEL.

BETA PARTICLE RADIATION IS A NATURAL NUCLEAR FISSION REACTION THAT LEAVES THE TOTAL

NUCLEON NUMBER UNCHANGED BUT WHICH CHANGES THE ELEMENT NUMBER OF THE ATOMIC

NUCLEUS. THE ELEMENT NUMBER RISES BY ONE, WHICH MEANS THAT IF THE ELEMENT WAS EVEN

NUMBERED, IT WILL BE RAISED BY ONE TO BECOME ODD NUMBERED; AND CONVERSELY, IF THE ELEMENT

WAS ODD NUMBERED, IT WOULD BE RAISED BY ONE TO BECOME AN EVEN NUMBERED ELEMENT.

ASSUMING THAT THE NUCLEAR SHELLS ONLY CONTAIN HYDROGEN AND HELIUM PARTICLES, THEN THE

SHELL FROM WHICH THE BETA DECAY OCCURRED (THE OUTER MOST SHELL) WILL HAVE TO UNDERGO A TRANSFORMATION OF SOME TYPE TO RENORMALIZE.

THE CALCULATION PROCESS FOR THE NEW STRUCTURE OF THE OUTER SHELL IS TO: 1. FIRST FIND THE TOTAL NUMBER OF PROTONS AND NEUTRONS IN THE SHELL.

2. CONVERT THESE TO TRITONS AND DEUTERONS AS ALREADY EXPLAINED ABOVE.

3. CONVERT THESE TO HE4, HE5, AND HE6 ALPHA PARTICLES, WITH WHATEVER DEUTERONS OR TRITONS

THAT ARE LEFT OVER AS EXPLAINED ABOVE.

4. COMBINE THIS SHELL INFORMATION WITH THE INNER SHELL DATA TO ARRIVE AT THE NEWLY

COMPLETED SHELL STRUCTURE.

THIS ENTIRE RENORMALIZING PROCESS IS DONE AUTOMATICALLY WITHIN THE OUTER SHELL, EVERY

TIME BETA DECAY OCCURS. THE IMPORTANT POINT TO MAKE IS THAT NUCLEAR FISSION ONLY OCCURS IN

ATOMIC MATTER WHICH IS TECHNICALLY CALLED 'BARYONIC' MATTER. THIS ATOMIC MATTER EXISTS

WITHIN A MUCH WEAKER MAGNETIC FIELD THAN WHICH EXISTED IN THE PLASMA MATTER. BECAUSE OF

THE MUCH WEAKER FIELD THE ATOMIC NUCLEI ARE FAR LESS DENSE AND THIS ALLOWS BOTH NUCLEAR

AND ELECTRON SHELLS TO FORM. WITHIN THE NUCLEAR SHELL, ALPHA PARTICLE DECAY, BETA PARTICLE

DECAY, AND PERHAPS EVEN NEUTRON PARTICLE DECAY BECOMES POSSIBLE.

THE ARRANGEMENT OF ELECTRONS

THE POSITIVIVELY CHARGED PARTICLE'S ORBITAL PATHS WITHIN THEIR RESPECTIVE SHELLS, IS IN

ALIGNMENT WITH THE MAGNETIC FIELD. I POSTULATE THAT THIS SAME MAGNETIC FIELD WHICH

GOVERNS THE ORIBTAL PATHS OF THE PARTICLES WITHIN THE NUCLEUS IS EXTENDED BEYOND THE

NUCLEUS TO THE ELECTRONS IN THE OUTER SHELL, AND GOVERNS THE ORBITAL PATHS OF THESE

ELECTRONS AS WELL. THE MAGNETIC FIELD ALSO SERVES THE PURPOSE OF PREVENTING THE

NEGATIVELY CHARGED ELECTRONS FROM APPROACHING THE POSITIVELY CHARGED NUCLEUS, BECAUSE

ELECTRICALLY CHARGED PARTICLES HAVE GREAT DIFFICULTY PASSING ACROSS MAGNETIC FIELD LINES

OF FORCE.

AS PER THIS DYNAMISM MODEL, THE INITIAL PLASMA THAT WAS FORMED FROM THE VOID AT THE BIRTH

OF THE UNIVERSE, CONSISTED OF JUST PROTONS AND ELECTRONS. BUT THE REASON WHY THE PLASMA IS NOT ATOMIC MATTER IS BECAUSE THE ELECTRONS WITHIN THE PLASMA WERE NOT ORBITING THE

PROTONS AS THEY WOULD IN ATOMIC MATTER, BUT INSTEAD THE ELECTRONS WERE FIXED IN PLACE

EQUIDISTANT FROM ONE ANOTHER, AS THERE WERE NO ELECTRON SHELLS FOR THEM TO ORBIT AROUND

THE PROTONS.

THIS IS THE PUREST TYPE OF PLASMA BECAUSE OF THE POWERFUL EXTERNAL MAGNETIC FIELDS WHICH

SURROUND THE PROTONS. THE ELECTRONS DO NOT ORBIT THE PROTONS, BUT INSTEAD ARE FIXED INTO

PLACE EQUIDISTANT FROM ONE ANOTHER. IT IS THE ABSENCE OF THIS POWERFUL EXTERNAL MAGNETIC

FIELD WHICH MAKES THE ELECTRON SHELL WHICH SURROUNDS THE NUCLEUS POSSIBLE.

___________________________________________________________________________________________________

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https://chem.libretexts.org/Courses/University_of_Illinois_Springfield /

CHE_124%3A_General_Chemistry_for_the_Health_Professions_(Morsch_and_Andrews)

/02%3A_Elements%2C_Atoms%2C_and_the_Periodic_Table/2.6%3A_Arrangements_of_Electrons ___________________________________________________________________________________________________

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THE ARTICLE SEGMENT ABOVE STATES THAT 'THE MODERN THEORY OF ELECTRON BEHAVIOR IS CALLED

QUANTUM MECHANICS.'

THIS MEANS THAT THE ELECTRON SUBSHELL CONFIGURATIONS OF 'S, P, D AND F'; ARE ALL DERIVED

FROM QUANTUM MECHANICS. IN OTHER WORDS, IT IS ALL THEORETICAL.

___________________________________________________________________________________________________

https://byjus.com/physics/quantum-mechanics/

___________________________________________________________________________________________________

THIS CALCULATIONS FOR THE ELECTRON SUBSHELL CONFIGURATIONS OF 'S, P, D AND F'; ARE ALL

DERIVED FROM QUANTUM MECHANICS EQUATIONS SIMILAR TO THOSE ABOVE. IN OTHER WORDS, IT IS

ALL THEORETICAL; AND FOR THE MOST PART, IN MY OPINION, UNPROVEN.

MY MAIN REASON FOR BUILDING THIS DYNAMISM MODEL OF THE SUBATOMIC WORLD IS TO PROVIDE AN

ALTERNATIVE TO THE THEORY WHICH MAINTAINS THAT A SO-CALLED 'STRONG FORCE' WHICH IS CARRIED

BY A 'GLUON' IS THE REASON WHY THE POSITIVELY CHARGED PARTICLES (WITHIN A PROTON, THE

POSITIVELY CHARGED UPQUARKS), (WITHIN THE NUCLEUS, THE POSITIVELY CHARGED PROTONS); ARE

HELD WITHIN THE NUCLEUS DESPITE ALL OF THEIR REPULSION TO EACH OTHER. THE FORCE OF THE

MAGNETIC MONOPOLE CREATED BY A SINGLE POINT SOURCE OF ELECTRICAL CHARGE, RAPIDLY

ROTATING, IS THE WHAT I SUBMIT AS THE REAL REASON WHY THE NUCLEUS DOESN'T DISINTEGRATED

DUE TO ALL OF THOSE POSITIVELY CHARGED PARTICLES REPELLING EACH OTHER.

BECAUSE THE MAGNETIC MONOPOLE OBEYS THE INVERSE CUBE LAW, THE STRENGTH OF ITS MAGNETIC

FIELD RAPIDLY DECLINES WITH DISTANCE FROM THE SOURCE OF THE FIELD. I SUBMIT THAT IT IS THIS

RAPID DECLINE THAT IS THE REAL REASON FOR ALPHA PARTICLE RADIATION FROM THE SEVENTH SHELL

OF THE LARGER NUCLEI. AT THE POINT WHERE THE REPULSIVE FORCES OF THE POSITIVELY CHARGED

PARTICLES, IS GREATER THAN THE STRENGTH OF THE MAGNETIC FIELD, WHICH OCCURS WITHIN THE

SEVENTH SHELL, THIS IS WHERE THE NUCLEUS BREAKS DOWN AND ALPHA PARTICLE RADIATION OCCURS.

THE FOLLOWING ARTICLE IS A BRIEF SUMMARY OF THE QUANTUM THEORY EXPLANATION OF ALPHA PARTICLE DECAY.

___________________________________________________________________________________________________

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https://www.sciencedirect.com/topics/earth-and-planetary-sciences/alpha-decay ___________________________________________________________________________________________________

"The maximum potential energy of the barrier can be calculated as 30MeV but alpha particles with energy of 4.18

MeV are emitted from 238U by tunnelling as illustrated by Fig. 2.3. This cannot be explained by classical physics."

PERHAPS NO ONE WITHIN CLASSICAL PHYSICS IS GOING TO TRY TO EXPLAIN 'TUNNELLING' WITH

CLASSICAL PHYSICS BECAUSE THAT WOULD BE RUBBING TOO MANY PEOPLE IN THE QUANTUM

THEORETICAL FIELDS THE WRONG WAY! WE LIVE IN A THEORETICAL AGE WHERE THEORIES THAT CAN GET

FUNDING FROM DEEP POCKETED PEOPLE ARE ALL THAT MATTER. A CLASSICAL EXPLANATION THAT

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DIRECTLY OPPOSES SOME WELL ACCEPTED THEORY IS CAREER SUICIDE FOR A PHYSICIST.

THE DYNAMISM MODEL FOR THE ARRANGEMENT OF ELECTRONS IS IS BASED ON THE NOTION OF

MAGNETIC LINES OF FORCE THAT ARE GENERATED BY A MONOPOLE MAGNETIC SOURCE. THE

MONOPOLAR MAGNETIC FIELDS ARE THEMSELVES GENERATED BY THE MOVEMENT OF THE POSITIVELY

CHARGED NUCLEUS, AND OBEYS THE INVERSE CUBE LAW.

THE IMPORTANCE OF THE INVERSE CUBE LAW TO THE MAGNETIC FIELDS CAN NOT BE OVERSTATED. ALL

OF THE POSITIVELY CHARGED SUBATOMIC PARTICLES ARE INFLUENCED BY THIS FIELD. THE MAGNETIC

LINES OF FORCE ARE MOST DENSELY PACKED AT THE CENTER OF THE NUCLEUS AND THEN START TO

THIN OUT AS THE DISTANCE FROM THE CENTER INCREASES. THE ALPHA PARTICLES WITHIN THE

NUCLEUS ARE AFFECTED BY THE MAGNETIC FIELD, AS ARE THE ELECTRONS IN THE SHELL. BUT THE

ELECTRONS ARE MUCH FARTHER AWAY FROM THE CENTER OF THE NUCLEUS THAN THE ALPHA PARTICLES

WITHIN THE NUCLEUS. THIS MEANS THAT THE FURTHER AWAY FROM THE NUCLEUS THAT AN ELECTRON

IS, THE LESS RESTRAINT THAT THE MAGNETIC FIELD EMANATING FROM THE NUCLEUS HAS ON IT. THIS

MEANS THAT THE ELECTRONS HAVE MORE AND MORE FREEDOM OF MOVEMENT AS THEY ORBIT FARTHER

AND FARTHER AWAY FROM THE NUCLEUS. THIS IS ALSO WHY THE ELECTRONS AT THE OUTER SHELLS OF

THE ATOMS ARE FAR MORE ENERGETIC THAN THE ELECTRONS CLOSER TO THE NUCLEUS IN THE INNER

SHELLS.

NEWTONIAN PHYSICS VS LEIBNIZIAN MATHEMATICS

THE NAVIER STOKES MILLENNIUM PROBLEM REQUIRES THAT THE READER HAVE A MINIMUM AMOUNT OF

CALCULUS KNOWLEDGE. CALCULUS IS THE MATHEMATICS OF MOTION. THE WAY THAT NEWTON

ORIGINALLY DESCRIBED IT, ALL OF HIS CALCULUS FUNCTONS WERE BASED ON TIME AS THE

INDEPENDENT VARIABLE. BUT THE WAY THAT CALCULUS WAS DEVELOPED OVER THE CENTURIES, WAS

BASED ON LEIBNIZ'S VIEW THAT THE INDEPENDANT VARIABLE WOULD BE A SPATIAL VARIABLE 'X'.

SO IN NEWTON'S CALCULUS; 'X' IS A SPATIAL DIRECTION AND 'T' IS A TEMPORAL DIMENSION. THIS WOULD

BE STATED TODAY AS A FUNCTION WHERE THE SPATIAL COORDINATE OF 'X', WHICH IS THE DEPENDANT

VARIABLE; WOULD BE BASED ON WHERE IT WAS AT TIME 'T', WHICH IS THE INDEPENDANT VARIABLE.

THIS GIVE US THE FUNCTION:

'X(T)'

EXAMPLE:

X = T^2 + T

SO A SIMPLE DIAGONAL LINE IN NEWTON'S CALCULUS IS:

X = T

LEIBNIZ HAS A DIFFERENT NOMENCLATURE. IN THE CALCULUS OF LEIBNIZ, THE INDEPENDANT VARIABLE

IS 'X' AND THE DEPENDANT VARIABLE IS 'Y'. NOTE THAT THESE ARE BOTH SPATIAL DIMENSIONS. THUS

LEIBNIZ DISREGARDED THE TEMPORAL DIRECTION OF TIME 'T'. THIS GIVE US THE FUNCTION:

'Y(X)'

EXAMPLE:

Y = X^2 + X

SO A SIMPLE DIAGONAL LINE IN LEIBNIZ'S CALCULUS IS:

Y = X

YOU MAY BE FAMILIAR WITH THE SIMPLE EQUATION 'Y = X'.

A TYPICAL STRAIGHT LINE IS WRITTEN AS A FUNCTION:

F(X) = X

WHICH IS THE SAME AS:

Y = X

IN MATHEMATICS ALL LINES HAVE A SLOPE AND AN X-INTERCEPT:

Y = MX + B

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WHEN 'M = 1' AND 'B = 0'

Y = 1X + 0

Y = X

F(X) = X

FOR ANY STRAIGHT LINE (OTHER THAN A VERTICAL LINE) THE 'M' COEFFICIENT IS THE SLOPE OF THE

LINE AND IS ALSO THE DERIVATIVE. THE TECHNICAL DEFINITION OF 'THE DERIVATIVE' IS THE SLOPE OF

THE GRAPH AT ANY PARTICULAR POINT. BUT FOR A GRAPH THAT IS JUST A NONVERTICAL STRAIGHT LINE, THE SLOPE IS THE SAME AT EVERY POINT AND IS EQUAL TO 'M'.

THE FIRST FOUR EXPONENTS ARE:

1. X^0 = 1; CONSTANT FUNCTION.

2. X^1 = X; LINE FUNCTION.

3. X^2 = X^2; QUADRATIC FUNCTION.

4. X^3 = X^3; CUBIC FUNCTION.

THE 'EXPONENT' OF 'X' IS ALSO THE 'POWER' OF 'X'.

THIS IS TO SAY THAT FOR A 'CUBIC FUNCTION' THE LARGEST EXPONENT IN THE FUNCTION, IS 'X' RAISED

TO THE POWER OF '3'.

FOR A QUADRATIC FUNCTION, THE LARGEST EXPONENT IN THE FUNCTION, IS 'X' RAISED TO THE POWER

OF '2'.

FOR A LINE FUNCTION, THE LARGEST EXPONENT IN THE FUNCTION, IS 'X' RAISED TO THE POWER OF '1'.

FOR A CONSTANT FUNCTION, THE LARGEST EXPONENT IN THE FUNCTION, IS 'X' RAISED TO THE POWER OF

'0'.

THE 'COEFFICIENT' IS THE NUMBER JUST TO THE LEFT OF 'X'. FOR EXAMPLE: FOR '1X'; 'X' HAS A COEFFICIENT OF 1.

FOR '2X'; 'X' HAS A COEFFICIENT OF 2.

FOR '3X'; 'X' HAS A COEFFICIENT OF 3.

FOR '4X'; 'X' HAS A COEFFICIENT OF 4.

THE DERIVATIVE OF THE CALCULUS

THE SIMPLEST WAY TO CALCULATE THE COEFFICIENT OF THE DERIVATIVE IS TO MULTIPLY THE

COEFFICIENT OF THE ORIGINAL BY THE EXPONENT OF THE ORIGINAL: FOR EXAMPLE, FOR A COEFFICIENT OF ONE:

1. FOR X^1 COEFFICIENT IS ONE; EXPONENT IS ONE; DERIVATIVE COEFFICIENT = (1 * 1 = 1) 2. FOR X^2 COEFFICIENT IS ONE; EXPONENT IS TWO; DERIVATIVE COEFFICIENT = (1 * 2 = 2) 3. FOR X^3 COEFFICIENT IS ONE; EXPONENT IS THREE; DERIVATIVE COEFFICIENT = (1 * 3 = 3) 4. FOR X^4 COEFFICIENT IS ONE; EXPONENT IS FOUR; DERIVATIVE COEFFICIENT = (1 * 4 = 4) FOR EXAMPLE, FOR A COEFFICIENT OF TWO:

1. FOR 2X^1 COEFFICIENT IS TWO; EXPONENT IS ONE; DERIVATIVE COEFFICIENT = (2 * 1 = 2) 2. FOR 2X^2 COEFFICIENT IS TWO; EXPONENT IS TWO; DERIVATIVE COEFFICIENT = (2 * 2 = 4) 3. FOR 2X^3 COEFFICIENT IS TWO; EXPONENT IS THREE; DERIVATIVE COEFFICIENT = (2 * 3 = 6) 4. FOR 2X^4 COEFFICIENT IS TWO; EXPONENT IS FOUR; DERIVATIVE COEFFICIENT = (2 * 4 = 8) FOR EXAMPLE, FOR A COEFFICIENT OF THREE:

1. FOR 3X^1 COEFFICIENT IS THREE; EXPONENT IS ONE; DERIVATIVE COEFFICIENT = (3 * 1 = 3) 2. FOR 3X^2 COEFFICIENT IS THREE; EXPONENT IS TWO; DERIVATIVE COEFFICIENT = (3 * 2 = 6) 3. FOR 3X^3 COEFFICIENT IS THREE; EXPONENT IS THREE; DERIVATIVE COEFFICIENT = (3 * 3 = 9) 4. FOR 3X^4 COEFFICIENT IS THREE; EXPONENT IS FOUR; DERIVATIVE COEFFICIENT = (3 * 4 = 12) FOR EXAMPLE, FOR A COEFFICIENT OF FOUR:

1. FOR 4X^1 COEFFICIENT IS FOUR; EXPONENT IS ONE; DERIVATIVE COEFFICIENT = (4 * 1 = 4) 2. FOR 4X^2 COEFFICIENT IS FOUR; EXPONENT IS TWO; DERIVATIVE COEFFICIENT = (4 * 2 = 8) 3. FOR 4X^3 COEFFICIENT IS FOUR; EXPONENT IS THREE; DERIVATIVE COEFFICIENT = (4 * 3 = 12) 4. FOR 4X^4 COEFFICIENT IS FOUR; EXPONENT IS FOUR; DERIVATIVE COEFFICIENT = (4 * 4 = 16) THE SIMPLEST WAY TO CALCULATE THE EXPONENT OF THE DERIVATIVE IS TO SUBTRACT ONE FROM THE

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EXPONENT OF THE ORIGINAL:

FOR EXAMPLE, FOR AN EXPONENT OF ONE:

1. FOR 1X^1 COEFFICIENT IS ONE; EXPONENT IS ONE; DERIVATIVE EXPONENT = (1 - 1 = 0) 2. FOR 2X^1 COEFFICIENT IS TWO; EXPONENT IS ONE; DERIVATIVE EXPONENT = (1 - 1 = 0) 3. FOR 3X^1 COEFFICIENT IS THREE; EXPONENT IS ONE; DERIVATIVE EXPONENT = (1 - 1 = 0) 4. FOR 4X^1 COEFFICIENT IS FOUR; EXPONENT IS ONE; DERIVATIVE EXPONENT = (1 - 1 = 0) FOR EXAMPLE, FOR AN EXPONENT OF TWO:

1. FOR 1X^2 COEFFICIENT IS ONE; EXPONENT IS TWO; DERIVATIVE EXPONENT = (2 - 1 = 1) 2. FOR 2X^2 COEFFICIENT IS TWO; EXPONENT IS TWO; DERIVATIVE EXPONENT = (2 - 1 = 1) 3. FOR 3X^2 COEFFICIENT IS THREE; EXPONENT IS TWO; DERIVATIVE EXPONENT = (2 - 1 = 1) 4. FOR 4X^2 COEFFICIENT IS FOUR; EXPONENT IS TWO; DERIVATIVE EXPONENT = (2 - 1 = 1) FOR EXAMPLE, FOR AN EXPONENT OF THREE:

1. FOR 1X^3 COEFFICIENT IS ONE; EXPONENT IS THREE; DERIVATIVE EXPONENT = (3 - 1 = 2) 2. FOR 2X^3 COEFFICIENT IS TWO; EXPONENT IS THREE; DERIVATIVE EXPONENT = (3 - 1 = 2) 3. FOR 3X^3 COEFFICIENT IS THREE; EXPONENT IS THREE; DERIVATIVE EXPONENT = (3 - 1 = 2) 4. FOR 4X^3 COEFFICIENT IS FOUR; EXPONENT IS THREE; DERIVATIVE EXPONENT = (3 - 1 = 2) FOR EXAMPLE, FOR AN EXPONENT OF FOUR:

1. FOR 1X^4 COEFFICIENT IS ONE; EXPONENT IS FOUR; DERIVATIVE EXPONENT = (4 - 1 = 3) 2. FOR 2X^4 COEFFICIENT IS TWO; EXPONENT IS FOUR; DERIVATIVE EXPONENT = (4 - 1 = 3) 3. FOR 3X^4 COEFFICIENT IS THREE; EXPONENT IS FOUR; DERIVATIVE EXPONENT = (4 - 1 = 3) 4. FOR 4X^4 COEFFICIENT IS FOUR; EXPONENT IS FOUR; DERIVATIVE EXPONENT = (4 - 1 = 3) COMBINING BOTH OPERATIONS WILL GIVE THE DERIVATIVE:

FIRST SET OF EXAMPLES HAS CONSTANT COEFFICIENTS:

FOR EXAMPLE, FOR A COEFFICIENT OF ONE:

1. FOR X^1 COEFFICIENT IS ONE; EXPONENT IS ONE; DERIVATIVE = 1X^0

2. FOR X^2 COEFFICIENT IS ONE; EXPONENT IS TWO; DERIVATIVE = 2X^1

3. FOR X^3 COEFFICIENT IS ONE; EXPONENT IS THREE; DERIVATIVE = 3X^2

4. FOR X^4 COEFFICIENT IS ONE; EXPONENT IS FOUR; DERIVATIVE = 4X^3

FOR EXAMPLE, FOR A COEFFICIENT OF TWO:

1. FOR 2X^1 COEFFICIENT IS TWO; EXPONENT IS ONE; DERIVATIVE = 2X^0

2. FOR 2X^2 COEFFICIENT IS TWO; EXPONENT IS TWO; DERIVATIVE = 4X^1

3. FOR 2X^3 COEFFICIENT IS TWO; EXPONENT IS THREE; DERIVATIVE = 6X^2

4. FOR 2X^4 COEFFICIENT IS TWO; EXPONENT IS FOUR; DERIVATIVE = 8X^3

FOR EXAMPLE, FOR A COEFFICIENT OF THREE:

1. FOR 3X^1 COEFFICIENT IS THREE; EXPONENT IS ONE; DERIVATIVE = 3X^0

2. FOR 3X^2 COEFFICIENT IS THREE; EXPONENT IS TWO; DERIVATIVE = 6X^1

3. FOR 3X^3 COEFFICIENT IS THREE; EXPONENT IS THREE; DERIVATIVE = 9X^2

4. FOR 3X^4 COEFFICIENT IS THREE; EXPONENT IS THREE; DERIVATIVE = 12X^3

FOR EXAMPLE, FOR A COEFFICIENT OF FOUR:

1. FOR 4X^1 COEFFICIENT IS FOUR; EXPONENT IS ONE; DERIVATIVE = 4X^0

2. FOR 4X^2 COEFFICIENT IS FOUR; EXPONENT IS TWO; DERIVATIVE = 8X^1

3. FOR 4X^3 COEFFICIENT IS FOUR; EXPONENT IS THREE; DERIVATIVE = 12X^2

4. FOR 4X^4 COEFFICIENT IS FOUR; EXPONENT IS FOUR; DERIVATIVE = 16X^3

SECOND SET OF EXAMPLES HAS CONSTANT EXPONENTS:

FOR EXAMPLE, FOR AN EXPONENT OF ONE:

1. FOR 1X^1 COEFFICIENT IS ONE; EXPONENT IS ONE; DERIVATIVE = 1X^0

2. FOR 2X^1 COEFFICIENT IS TWO; EXPONENT IS ONE; DERIVATIVE = 2X^0

3. FOR 3X^1 COEFFICIENT IS THREE; EXPONENT IS ONE; DERIVATIVE = 3X^0

4. FOR 4X^1 COEFFICIENT IS FOUR; EXPONENT IS ONE; DERIVATIVE = 4X^0

FOR EXAMPLE, FOR AN EXPONENT OF TWO:

1. FOR 1X^2 COEFFICIENT IS ONE; EXPONENT IS TWO; DERIVATIVE = 2X^1

2. FOR 2X^2 COEFFICIENT IS TWO; EXPONENT IS TWO; DERIVATIVE = 4X^1

3. FOR 3X^2 COEFFICIENT IS THREE; EXPONENT IS TWO; DERIVATIVE = 6X^1

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4. FOR 4X^2 COEFFICIENT IS FOUR; EXPONENT IS TWO; DERIVATIVE = 8X^1

FOR EXAMPLE, FOR AN EXPONENT OF THREE:

1. FOR 1X^3 COEFFICIENT IS ONE; EXPONENT IS THREE; DERIVATIVE = 3X^2

2. FOR 2X^3 COEFFICIENT IS TWO; EXPONENT IS THREE; DERIVATIVE = 6X^2

3. FOR 3X^3 COEFFICIENT IS THREE; EXPONENT IS THREE; DERIVATIVE = 9X^2

4. FOR 4X^3 COEFFICIENT IS FOUR; EXPONENT IS THREE; DERIVATIVE = 12X^2

FOR EXAMPLE, FOR AN EXPONENT OF FOUR:

1. FOR 1X^4 COEFFICIENT IS ONE; EXPONENT IS FOUR; DERIVATIVE = 4X^3

2. FOR 2X^4 COEFFICIENT IS TWO; EXPONENT IS FOUR; DERIVATIVE = 8X^3

3. FOR 3X^4 COEFFICIENT IS THREE; EXPONENT IS FOUR; DERIVATIVE = 12X^3

4. FOR 4X^4 COEFFICIENT IS FOUR; EXPONENT IS FOUR; DERIVATIVE = 16X^3

HOPEFULLY, THE ABOVE EXAMPLES ARE HELPFUL IN SHOW YOU, THE READER HOW TO CALCULATE THE

DERIVATIVE OF A FUNCTION.

THE DERIVATIVE OF A FUNCTION IS EQUIVALENT TO THE SLOPE

OF THE FUNCTION

MODERN DAY MATH IS CURRENTLY BASED ON THE CONCEPT OF A FUNCTION. EXAMPLES OF FUNCTION

NOTATION ARE:

• F(X)

• G(X)

• H(X)

• J(X)

THE 'X' REPRESENTS THE HORIZONTAL AXIS IN A TWO-DIMENSIONAL CARTESIAN PLANE. THE 'Y'

REPRESENTS THE VERTICAL AXIS. THE 'X' IS THE INDEPENDANT VARIABLE, AND THE 'Y' IS THE

DEPENDANT VARIABLE. FOR EXAMPLE:

• Y = F(X)

• Y = G(X)

• Y = H(X)

• Y = J(X)

THIS CAN BE ORALLY STATED AS:

• 'Y' IS EQUAL TO THE FUNCTION F(X).

• 'Y' IS EQUAL TO THE FUNCTION G(X).

• 'Y' IS EQUAL TO THE FUNCTION H(X).

• 'Y' IS EQUAL TO THE FUNCTION J(X).

THE NOTATION FOR THE FIRST DERIVATIVES OF THE ABOVE FUNCTIONS ARE:

• Y = F'(X)

• Y = G'(X)

• Y = H'(X)

• Y = J'(X)

THE NOTATION FOR THE SECOND DERIVATIVES OF THE ABOVE FUNCTIONS ARE:

• Y = F''(X)

• Y = G''(X)

• Y = H''(X)

• Y = J''(X)

THIS TYPE OF NOTATION PROCEEDED DIRECTLY FROM LEIBNIZIAN CALCULUS WHERE BOTH 'X' AND 'Y' ARE

SPATIAL DIMENSIONS WITHIN A CARTESIAN GRAPH. THIS TYPE OF NOTATION IS THE BASIS FOR ALL KINDS

OF MATHEMATICS INCLUDING ALGEBRA, GEOMETRY, TRIGONOMETRY, AND A HOST OF OTHERS.

APPLIED IN THE CONTEXT OF A THREE DIMENSIONAL CARTESIAN PLANE, THIS ALSO HAS TO ACCOUNT

FOR THE DIMENSION OF TIME:

• 'X' IS EQUAL TO THE HORIZONTAL AXIS.

• 'Y' IS EQUAL TO THE VERTICAL AXIS.

• 'Z' IS EQUAL TO THE DEPTH AXIS.

• 'T' IS EQUAL TO THE TIME AXIS.

LEIBNIZ WAS A GREAT MATHEMATICIAN. BUT HE WAS NOT A PHYSICISTS.

WHEN YOU TRY TO CALCULATE A SPATIAL COORDINATE 'Y' FROM ANOTHER SPATIAL COORDINATE 'X'; YOU

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EFFECTIVELY END UP WITH THESE TYPES OF FUNCTIONS:

• X = X(Z) + X(Y)

• Y = Y(X) + Y(Z)

• Z = Z(X) + Z(Y)

WHEN DEALING WITH THE PHYSICS OF MOTION, THE ABOVE NOTATION IS WRONG FOR TWO REASONS.

1. THE FUNCTIONS END UP LOOKING LIKE THIS:

◦ X(Z) = Z^3 + Z

◦ X(Y) = Y^2 + 1

◦ Y(X) = X^2 + 1

◦ Y(Z) = Z^3 + 2Z

◦ Z(X) = X^2 + 3X

◦ Z(Y) = Y^4 + 3Y^3 + Y

◦ WHICH REQUIRES A TECHNIQUE CALLED 'PARTIAL DIFFERENTIATION' TO FIND THE DERIVATIVES.

2. THERE IS NO TIME DIMENSION WHATSOEVER. HOW CAN ANYONE DESCRIBE THE MOTION OF AN

OBJECT WITHOUT ANY REFERENCE TO THE TIME DIMENSION?

NEWTONIAN CALCULUS IS THE CORRECT MATHEMATICS FOR DEALING WITH OBJECTS IN MOTION. LOOK

HOW MUCH THE ABOVE FUNCTIONS WILL SIMPLIFY IF YOU ADD THE TIME DIMENSION IN THE RIGHT WAY.

FOR EXAMPLE:

• X = T

• Y = 2T

• Z = 3T

SUBSTITUTE INTO THE FOLLOWING:

• X(Z) = Z^3 + Z

X(Z) = (3T)^3 + (3T)

X(Z) = 3T^3 + 3T

• X(Y) = Y^2 + 1

X(Y) = (2T)^2 + 1

X(Y) = 2T^2 + 1

• Y(X) = X^2 + 1

Y(X) = (T)^2 + 1

Y(X) = T^2 + 1

• Y(Z) = Z^3 + 2Z

Y(Z) = (3T)^3 + 2(3T)

Y(Z) = 3T^3 + 6T

• Z(X) = X^2 + 3X

Z(X) = (T)^2 + 3(T)

Z(X) = T^2 + 3T

• Z(Y) = Y^4 + 3Y^3 + Y

Z(Y) = 2T^4 + 6T^3 + 2T

___________________________________________________________________________________________________

THE MILLENNIUM PROBLEMS

BY KEITH DEVLIN

NAVIER STOKE EQUATIONS

"...The resulting equations are known as the Navier-Stokes equations.

Though these equations can be solved in the hypothetical two-dimensional case of an infinitely thin planar film of fluid, it is not known whether there is a solution in the (more realistic) three-dimensional case. Notice that the issue is not Do we know what the solution is? It's more basic than that. We don't even know whether there is a solution!

Let's begin with Euler's equations for fluid motion, the equations that govern flow in a (hypothetical) frictionless fluid that extends to infinity in all directions.

We assume that each point P = (x,y,z) in the fluid is subject to a force that varies with time. We can specify the force at P at time t by giving its values in each of the three axial directions: fx(x,y,z,t), fy(x,y,z,t), fz(x,y,z,t). (To win the Clay Prize, it is enough to solve the problem in the case where there is no externally applied force, that is, when 92 of 99

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each of fx, fy, fz is zero at all locations and all times. But historically, the problem was formulated in the way I am presenting it.)

let p(x, y, z, t) be the pressure in the fluid at the point P at time t.

The motion of the fluid at point P at time t can be specified by giving its velocity in the three axial directions. Let ux(x, y, z, t) be the velocity of the fluid at P in the x-direction, uy(x, y, z, t) be the velocity of the fluid at P in the y-direction, uz(x, y, z, t) be the velocity of the fluid at P in the z-direction.

We assume that the fluid is incompressible. That is, when a force is applied to it, it may flow in some direction but it can not be compressed; nor can it expand. This is expressed by the following equation: The problem assumes that we know how the fluid is moving at the start, i.e., when t = 0. That is, we know ux(x, y, z, t), uy(x, y, z, t), uz(x, y, z, t) (as functions of x, y, and z).

Moreover, thesee initial functions are assumed to be well-behaved ones. (Exactly what this means is technical, but we don't require a definition in order to obtain an overall understanding of the problem. The precise formulation of the restirction is, however, relevant to the Millennium Prize statement of the Navier-Stokes Problem, so would-be problem solvers would need to know the exact statement.)

By applying Newton's law

to each point P in the fluid, Euler produced the flollowing equations, which when combined wih the incompressiblility equation (1) above describe the motion of the fluid: Equations (1) through (4) are Euler's equations for fluid motion. To allow for viscosity, Navier and Stokes introduced a positive constant v, the viscosity, which measures the frictioal forces within the fluid, and added and additional force--the viscous force--to the right-hand side of quations (2), (3), and (4). The term to be added to the right-hand side of equation (2) is

with entirely similar terms (with ux, replaced by uy, and uz respectively) added to equations (3) and (4).

Here, the notation

denotes the second partial derivative, obtained by first differentiating the result again with respect to x, i.e.,

with analogous definitions in the y and z cases.

Unless you are a calculus whiz, chances are that the above formulas look pretty daunting. To be honest, mathematicians find them a bit overwhelming as well. The problem is that when we try to capture the motion of the fluid at any point in terms of its motion in each of the x-, y-, and z-directions, we make life unnecessarily complicated for ourselves. As you can see, there is relativley little difference among equations (2), (3), and (4), and the three additional viscosity terms we add are all variations on a single theme, one for each axial direction.

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During the nineteenth century, mathematicians developed a notation and a method to handle directional motion in a simpler fashion. The idea was to introduce a new kind of quantity called a vector. Whereas a number simply has quantity, a vector has both quantity and direction. "Vector calculus" is the method you get when you develop calculus for vectors and vector functions instead of number-variables and number-variable functions. Using vectors, mathematicians can write the Navier-Stokes equations more compactly: here, the quantities f and u are vector functions and the symbols/terms

, grad, and div denote operations of

vector calculus. (If you want to know more about this, consult one of the references given at the end of the chapter.) So little progress has been made toward solving the Navier-Stokes equations that the Clay Institute will award the $1 million prize for the solution to any one of several variations of the problem. The simplest version to state, though not necessarily the easiest to solve, assumes that you make the force functions fx, fy, and fz all zero. Can you then find functions p(x, y, z, t), ux(x, y, z, t), uy(x, y, z, t), uz(x, y, z, t) that satisfy the modified versions of equation (1) through 4) (i.e., the versions that include the viscosity terms, with v > 0) and are sufficiently "well-behaved" that they could plausibly correspond to physical reality?

Let me mention that the analoous problem where the viscosity is 0 (i.e., for the Euler equations) has also not been solved but that version is not a Millennium Problem.

If the Navier-Stokes problem is reduced to two dimensions (by setting all z terms equal to 0), it can be solved. This is an old result, but it provides no clue to what happens in three dimensions.

The full three-dimensional problem can also be solved in a highy restricted way. Given the various initial conditions, it is always possible to find a posiitive number T. such that the equations can be solved for all times In

general, the number T is fairly small, so this answer is not particularly useful in real life. The number T is called the

"blowup" time for the particular system.

Will the Navier-Stokes Problem Be Solved?

Will anyone win the $1 million prize for solving the Navier-Stokes equations? By choosing this challenge as a Millennium Problem, the Clay Institute has shined the mathematical spotlight onto a part of mathematics that goes back over two hundred years; the calculus of fluid flow. Given the length of time that mathematicains have been trying to solve these equations, it is hard to deny the thought that they may simply be unsolvable. At the very least, it seems likely that a solution will require some genuine new techniques. The equations may look like a problem in a typical student textbook. But they are definitely far more difficult than that."

THE MILLENNIUM PROBLEMS BY KEITH DEVLIN [PAGE 152 TO 155]

___________________________________________________________________________________________________

NEWTONIAN CALCULUS VS LEIBNIZIAN CALCULUS

THE HEART OF THE PROBLEM IS THE PARTIAL DERIVATIVES!

USING LEIBNIZIAN CALCULUS TO TRY TO SOLVE A PROBLEM IN FLUID DYNAMICS YIELDS FOUR VARIABLES

THAT ARE ALL INDEPENDANT VARIABLES AT SOME POINT WITHIN THESE EQUATIONS; (x, y, z, t). THIS

RESULTS IN EXTREMELY COMPLICATED PARTIAL DIFFERENTIAL EQUATIONS WHICH HAVE PROVEN TO BE

UNSOLVABLE FOR HUNDREDS OF YEARS. IT IS QUITE OBVIOUS THAT LEIBNIZIAN CALCULUS IS NOT THE

RIGHT TOOL FOR THE JOB!

THE RIGHT KIND OF TOOL FOR PROBLEMS OF THIS KIND, IS THE NEWTONIAN CALCULUS. WITH

NEWTONIAN CALCULUS, THERE IS ONLY ONE INDEPENDANT VARIABLE, WHICH IS THE TIME VARIABLE 't', AND THIS ELIMINATES ALL OF THE PAINFUL PARTIAL DIFFERENTIAL MATHEMATICS THAT COMES WITH

THE LEIBNIZIAN APPROACH.

" Let ux(x, y, z, t) be the velocity of the fluid at P in the x-direction, uy(x, y, z, t) be the velocity of the fluid at P in the y-direction, uz(x, y, z, t) be the velocity of the fluid at P in the z-direction."

"The problem assumes that we know how the fluid is moving at the start, i.e., when t = 0. That is, we know ux(x, y, z, t), uy(x, y, z, t), uz(x, y, z, t) (as functions of x, y, and z)."

I WILL PRESUME THAT THESE VELOCITY EQUATIONS ARE THE EQUIVALENT OF MOMENTUM EQUATIONS

WHERE 'm = 1' EFFECTIVELY NEUTRALIZING THE MASS VARIABLE WHICH LEAVES ONLY THE VELOCITY

FUNCTIONS.

USING NEWTONIAN CALCULUS, THE VELOCITY FUNCTIONS IN EACH OF THE THREE AXIAL DIRECTIONS

ARE:

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1. U[t] = U[X(t), Y(t), Z(t)]

2. ux(x, y, z, t) = X(t)

3. uy(x, y, z, t) = Y(t)

4. uz(x, y, z, t) = Z(t)

IN KEEPING WITH THE PRESUMPTION OF 'm = 1', WHICH NEUTRALIZES THE MASS VARIABLE, THE FORCE

FUNCTIONS ARE REDUCED TO ACCELERATION FUNCTIONS WHERE ACCELERATION IS THE DERIVATIVE OF

VELOCITY.

THE DERIVATIVES OF THESE VELOCITY EQUATIONS ARE THE FORCE/ACCELERATION EQUATIONS [WHICH

ARE MASSLESS BECAUSE 'm = 1'] IN EACH OF THE THREE AXIAL DIRECTIONS: 1. F[t] = F[X'(t), Y'(t), Z'(t)]

2. fx(x,y,z,t) = X'(t)

3. fy(x,y,z,t) = Y'(t)

4. fz(x,y,z,t) = Z'(t)

WHICH ARE ALSO THE EQUIVALENT OF THE PARTIAL DERIVATIVE VERSIONS OF THE FORCE/ACCELERATION

EQUATIONS:

SO THE INCOMPRESSIBILITY EQUATIONS TRANSLATES TO:

X'(t) + Y'(t) + Z'(t) = 0

THE VELOCITY FUNCTION IN THE X AXIAL DIRECTION:

[NOTE THAT I PRESUME THAT AN INTEGRAL SIGN

WAS MISSING FROM THE FIRST TERM]

TRANSLATES TO:

= X(t)

THE VELOCITY FUNCTION IN THE Y AXIAL DIRECTION:

[NOTE THAT I PRESUME THAT AN INTEGRAL SIGN

WAS MISSING FROM THE FIRST TERM]

TRANSLATES TO:

= Y(t)

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THE VELOCITY FUNCTION IN THE Z AXIAL DIRECTION:

[NOTE THAT I PRESUME THAT AN INTEGRAL SIGN

WAS MISSING FROM THE FIRST TERM]

TRANSLATES TO:

= Z(t)

THE MISSING INTEGRAL SIGN IS SIGNIFICANT BECAUSE THE INTEGRAL IS THE ANTIDERIVATIVE WHICH

MEANS THAT WHEN AN INTEGRAL AND A DERIVATIVE IS IN THE SAME TERM, THE OPERATIONS CANCEL

EACH OTHER OUT, WHICH JUST LEAVES THE ORIGINAL FUNCTION.

THE COMPLETE PRESSURE EQUATIONS

THE PRESSURE EQUATIONS IN EACH OF THE THREE AXIAL DIRECTIONS ARE: 1.

THEREFORE:

TRANSLATES TO:

X(t) = X'(t) - PX(t)

TRANSLATES TO:

Y(t) = Y'(t) - PY(t)

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TRANSLATES TO:

Z(t) = Z'(t) - PZ(t)

NOTICE THE DIFFERENCE BETWEEN THE LEIBNIZIAN EQUATIONS:

AND THE NEWTONIAN EQUATIONS:

1. X(t) = X'(t) - PX(t) (2)

2. Y(t) = Y'(t) - PY(t) (3)

3. Z(t) = Z'(t) - PZ(t) (4)

THIS MEANS THAT THE VISCOSITY EQUATIONS ARE THE SECOND DERIVATIVES OF THE VELOCITY

EQUATIONS AND ARE ALSO MASS NEUTRAL 'm = 1':

"The term to be added to the right-hand side of equation (2) is with entirely similar terms (with ux, replaced by uy, and uz respectively) added to equations (3) and (4)."

WHICH GIVES:

1. X(t) = X'(t) - PX(t) + X''(t) (2)

2. Y(t) = Y'(t) - PY(t) + Y''(t) (3)

3. Z(t) = Z'(t) - PZ(t) + Z''(t) (4)

REARRANGING THE EQUATIONS GIVES:

1. PX(t) = X'(t) - X(t) + X''(t) (2)

2. PY(t) = Y'(t) - Y(t) + Y''(t) (3)

3. PZ(t) = Z'(t) - Z(t) + Z''(t) (4)

WHICH ARE THE MASS NEUTRAL ('m = 1') COMPLETE PRESSURE EQUATIONS; INCLUDING THE NAVIER-STOKES VISCOSITY ADDITIONS; FOR DYNAMIC FLUID FLOW.

EXAMPLE OF APPLICATION OF GENERAL SOLUTION OF

NAVIER-STOKES EQUATION TO A SPECIFIC CASE

"Can you then find functions p(x, y, z, t), ux(x, y, z, t), uy(x, y, z, t), uz(x, y, z, t) that satisfy the modified versions of equation (1) through 4) (i.e., the versions that include the viscosity terms, with v > 0) and are sufficiently "well-behaved" that they could plausibly correspond to physical reality?"

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NOTICE THAT THE VISCOSITY TERMS ARE THE SECOND DERIVATIVE OF THE VELOCITY TERMS. ALL THAT IS

NECESSARY FOR THE VISCOSITY TERMS TO BE GREATER THAN ZERO; (v > 0); IS THE VELOCITY OF THE

FLUID IN EACH OF THE THREE AXIAL DIRECTIONS; HAS TO BE A FUNCTION OF 't' SUCH THAT: 1. X(t) = at^2 + d

2. Y(t) = bt^2 + d

3. Z(t) = ct^2 + d

4. {a, b, c} > 0

THEN THE FIRST DERIVATIVES OF THE VELOCITY FUNCTIONS ARE THE FORCE FUNCTIONS IN EACH OF

THE THREE AXIAL DIRECTIONS:

1. X'(t) = 2at

2. Y'(t) = 2bt

3. Z'(t) = 2ct

4. {a, b, c} > 0

THEN THE SECOND DERIVATIVES OF THE VELOCITY FUNCTIONS ARE THE VISCOSITY FUNCTIONS IN EACH

OF THE THREE AXIAL DIRECTIONS:

1. X''(t) = 2a

2. Y''(t) = 2b

3. Z''(t) = 2c

4. {a, b, c} > 0

IF THESE ARE AN ACTUAL EXAMPLE, THEN THE PRESSURE EQUATIONS IN EACH OF THE THREE AXIAL

DIRECTIONS ARE:

1. PX(t) = 2at - (at^2 + d) + 2a (2)

2. PY(t) = 2bt - (bt^2 + d) + 2b (3)

3. PZ(t) = 2ct - (ct^2 + d) + 2c (4)

4. {a, b, c} > 0

IF a = 1; b = 2; c = 3; d = 4:

1. PX(t) = 2(1)t - ((1)t^2 + (4)) + 2(1) (2)

= 2t - t^2 + 4 + 2

= 2t - t^2 + 6

2. PY(t) = 2(2)t - ((2)t^2 + (4)) + 2(2) (3)

= 4t - 2t^2 + 4 + 4

= 4t - 2t^2 + 8

3. PZ(t) = 2(3)t - ((3)t^2 + (4)) + 2(3) (4)

= 6t - 3t^2 + 4 + 6

= 6t - 3t^2 + 10

AS LONG AS THE HIGHEST POWER OF THE VELOCITY FUNCTIONS; IN EACH OF THE THREE AXIAL

DIRECTIONS; IS TWO OR GREATER, THEN THE VISCOSITY FUNCTION IN THE THREE AXES WILL ALL BE

GREATER THAN ZERO [v > 0]; WHICH WILL SATISFY THE REQUIREMENTS OF THE NAVIER-STOKES

EQUATIONS.

I THEREFORE SUBMIT THIS CHAPTER AS A SOLUTION TO THE "NAVIER-STOKES EQUATIONS" WHICH IS ONE

OF THE SEVEN MILLENNIUM PROBLEMS.

BREAKING THE THIRD SEAL

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Revelation 6:5-6

King James Version

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5 And when he had opened the third seal, I heard the third beast say, Come and see. And I beheld, and lo a black horse; and he that sat on him had a pair of balances in his hand.

6 And I heard a voice in the midst of the four beasts say, A measure of wheat for a penny, and three measures of barley for a penny; and see thou hurt not the oil and the wine.

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Genesis 1:9-13

King James Version

9 And God said, Let the waters under the heaven be gathered together unto one place, and let the dry land appear: and it was so.

10 And God called the dry land Earth; and the gathering together of the waters called he Seas: and God saw that it was good.

11 And God said, Let the earth bring forth grass, the herb yielding seed, and the fruit tree yielding fruit after his kind, whose seed is in itself, upon the earth: and it was so.

12 And the earth brought forth grass, and herb yielding seed after his kind, and the tree yielding fruit, whose seed was in itself, after his kind: and God saw that it was good.

13 And the evening and the morning were the third day.

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