Sree DEBASISH DASGUPTA: 2018

Atomic Mass – Energy-Primary Colour

Sree DEBASISH DASGUPTA

ABSTRACT :

This article contains on each and every particle produce by own method, so there are no any man med unit, but the quantity is same; All calculation done without any power for current unit, and equivalent value of Joule to electron Volt or eV to Joule.

We are indebted to those scientist .[The great “Einstein”, sir “Newton”, “Bohr” and lot of others Scientist ] who given new direction to our knowledge by doing hard work relentlessly day by day .In this article making out some investigations after checking out some contributions by them, It is not any kind of self – importance and also have not any mentality to – disrespect to any scientist.

.INTRODUCTION :

(L=0.413334; 1.24 eV/3= 0.413334) The whole calculation is done by keeping the energy of photon =1.24 eV, whole wave length field is known as a triangle, area ½ x base x height, The area of triangle =3 X 0.413334 = 1.24 square unit and 1.602 / 1.24=1.2919 is constant.

All calculation done without any power for current unit,

like one mass = 1.782662 x 10^{^-36} Kg, one energy = 1.6021766208 x 10^{^-19} Joule, here 1.782662 and 1.6021766208.

That means if compare to current unit; calculation should be multiplied (or divided) by 10 or, multiplier of 10 like 1000,10000,10,0000 to convert it from current unit to different unit like- Kilogram, gram, meter, second, etc. As per need.

Each and every Root; square; into; etc. indicate direction of part by part different internal changes.

Subject area of experiment :

Mass and energy of particle, electron; proton; neutron, electron volt; visible lights wave length, frequency;

Experimental future : Hope It will be help for about origin of atomic particles internal action and reaction between charge; energy, magnetic field,etc.

This data article demonstrates is first successful mathematical way for base of atomic system and structure of visible lights.

Experimental Methods about (atomic) Force = Energy X Displacement. Electron Volt = Frequency X 0.413334; Energy = Mass X ( Mass / 10);

1.. Force = Energy X displacement or E = F/s or s = F/E.

3L =1.24 electron Volt. (L=0.413334 ,10L= 4.13334 see later)

Source of ‘7’ ---- (1.24124)^⁹=7 (6.99) [1.24124 to the power 9 = 7].

4.13334 Joule = 2.579828079 x 10^⁺¹⁹ eV,

2.579828079 /2 = 1.2899, 1.2899 x 7 = 9 [Source of 9].

[we know 898,755,178,736,880 m^^2,,s^²= C^², at present calculation here C^²= 8.98755178736880]

9/7 = 1.2857 or 8.98755178736880 / 6.99 = 1.28757.

1.28757 x 4.13334 = 5.322 (see latter) x 4.13334 = 21.99 or 22 [source of 22]

Or 1.28757 x 4.13334^²(square) = 22.

That means one eV to the power 9 = 7 , and (9 divided by 7 ) into 4.13334 square = 22.

10L^²X (9/7) = 22

Or { (1.28757 x 7) / 1.24124^⁹ } x (4.13334)² = 22

Or {(1.28757 x 1.24124^⁹) / (1.24124)^⁹} x (4.13334)²=22

Or 1.28757 x 4.13334^²=22

Or 1.28757 x ( 3.33 x 1.24124)^² = 22 [see latter about 3.33]

So one Pi = (1.28757 x 3.33 x 3.33 x 1.24124 x 1.24124 ) / (1.24124 x 1.24124 x 1.24124 x 1.24124 x 1.24124 x 1.24124 x 1.24124 x 1.24124 x 1.24124)

Or one Pi = (1.28757 x 3.33 x 3.33) / (1.24124)^⁷or (1.28757 x 3.33 x 3.33) / 4.535

Now step by step we can see in which process create one mass of atom

One L=0.413334 x 3 = 1.24 eV, one constant = 1.29219 x 1.24 = 1.601956 or 1.602 unit charge or energy. 1.6019/3= 0.533 (A⁰)unit it is average Radius of 1^st orbit of one Atom’s.

0.53 x2 x 3.14(Pi) = 3.33 (2 Pi x ‘r’)

[Source of 3.33 , it is average circumference when electron revolving around the orbit ]

So 3.33 is “Displacement” of electron’s.= [S]

3.33 x 1.24124 = 4.13334

3.33 x 1.601956 (eng) = 5.334 it is Average “FORCE” for when electron revolving around the orbit. = [F] 10 times of Bohr’ ‘H’ atom’s 1^st orbit.

We know Work = Fs [F= Force & s= Displacement]

If s=0 in that moment W=Fs=0 (because S=0)

Energy = Force / Displacement (s)

Let us see as per this rule “E”=F/s

“E” = 5.334 (F) / 3.33 (s)

or “E” = 1.602 unit

Actually it elastic Collision, Force = Energy x Displacement,

or Displacement = Force / Energy

Any of one “0” means particular in that moment is starting or ending point of new alternative changing period of Energy to another Energy.

Due to Law of “Conservation of Energy” and “Transformation of Energy”

See also as per rule 4/3x ‘Pi’x ‘r’ cube.

4/3x3.14x0.53^{^3} (cube)= 0.623614, 0.623614 eV, it’s equivalent Energy 9.9914x 10^{^-20} Joule. And 0.623614 J0ule to electron Volt = 3.89 x10^{^+18} eV note Red color frequency is 3.8757 unit.

See also as per a^{^2} + b^{^2} = c^{^2} [a=4.13334, b=3.33, c=5.3…]

4.13334^{^2} + 3.33^{^2} = 5.3…or 17.0845 + 11.089 = 28.1735

See also Two dimensional reference frame :-

AC = root over ( AB square + BC square )

[AC = 5.3..AB= 4.13334, BC= 3.33]

5.3…= root over 4.13334^{^2} + 3.33^{^2} = root 28.1765 = 0.53

[See image Force=Energy / Displacement]

One neutron is 1839 time and one proton 1837 times heavier than electron, and one neutron mass value equal energy = 939 MeV. Proton mass value equal energy 938 MeV.

So 939/1839= 0.5106 unit or one proton 938/1837 = 0.5106 energy equal of one mass of electron.(And see later angel 11.25 / 22 = 0.51113 ).

When collusion with one “L” that moment – 0.5106 x 0.413334 = 0.2112796 add one time and space 0.2112796 + 1.296 = 1.5070483404 unit energy.

As per E = m – (m/10). [ E= energy, m= Mass]

Mass = ( ex10) / 9 = (1.5070483404 x 10 ) / 9 = 15.070483404 / 9 = 1.674498 unit mass.

And again 1.5070483404 x 9 = 13.56 unit energy ( as per Scientist ‘Bohr’ later we can see also picture 6)

1837 X 5.334 (Force) = 9798.588.Root = (i) 98.98766 Root = (ii) 9.94925 Root = (iii) 3.15424 Root = (iv) 1.776 Root = (v) 1.332

..(i) nearer 100 (ii) nearer 10,(iii) nearer ‘Pi’. Or ‘h’ bar (iv) mass (v) 4/3,

1837 x 1.602 (energy) = 2942.874 Thirty two Root = 1.283509 x 7 = 8.98456 (C² as per calculation)

1837 x 1.296 (Time & Space) = 2380.75 Twenty seven (3x3x3=27) Root = 1.3337=(4/3) *** That means 1837 times ,Force, energy, time etc. present in a proton.[space 3600 (sec)/1000 =3.6; Space 360 (Degree)/ 1000 =0.36; 3.6 x 0.36 = Time & space as per this calculation]

[ So Total Force x energy x time & space = 9798.588 x 2942.874 x 2380.75 = 68651330478.747, 68651330478.747 nine Root (9) = 16,

And when into total time 1935 , 68651330478.747 x 1935= 1328 4032 4476375.445, if we Root it step by step after 7^th step we will get see 1.289254, 1.289254 x 7 = “9”, ( or Root 128, For Why 128 ?, 0.53 x 100 = 53, & 53 / *0.413334* = 128. & 3 x 3 =”9” , 3 x 3 x 3 = 27) and if 1328 4032 4476375.445 twenty seven (27, 3x3x3 = 27) Root = 3.33, (Displacement) ]

Now ---1837 x 1.11* (see latter) = 1935 Sixteen Root = 1.6047 (energy), Square = 2.5754 Square = 6.63273 (Planck Constant) Square = 43.98 =14 x Pi ,

*1.11 to the power 32** = 28.2056 Root (as per a² +b²=c²= 28.17 & 5.31 is10 times of ‘H’ atoms 1^st orbit radius, 0.53, or it is force or Force for revolving electron around the orbit)

As per formula c = Root over (a² + b² + 2ab) = Root over (a + b)² = a + b

..a + b = (4.13334 + 3.33) = 7.46334, Root 7.46334 = 2.73191 Root = 1.653 Root = 1.285632.

1.285632 x 7 = 8.999 or “9”. (4L= 0.413334x4=1.653336)

*1.11 is Time need for create Displacement to Acceleration. In the time of revolving electron around the orbit. So Time = Root 1.11=1.05365 unit.

We know Force = mass x f (Acceleration) = mass x (changing motion / Time) = Momentum/Time, ie changing of Momentum = Force x Time.

So factor of time & mass, as per view E= F/s or s = F/E. or F = Es [Force=5.334, Energy = 1.602, Displacement (s) = 3.33, Time = 1.05365,Time square= 1.11 ]

Here Acceleration = (3.33 /1.11) = 3 unit.

So as per law F = ma, m=F/a or mass=5.334/3= 1.778 unit. [When Time is 1.11]

We know As per law P=mf or P = mass x (v – u) / time,[Here v= end & u= starting motion, ]

P = 1.782662 mass x (3.33 – 3 = 0.33) / 1.11 = 0.58828 / 1.11 or 0.53 it is radius of “H” atom’s 1^st orbit. And this part total calculation makes from 0.53…unit.

NOW – see also –

(a) 1/3 = 0.33…,

(b) 5.334 / 1.05356 = 5.06283458 , 5.06283458 four Root = 1.5 it is ½ of 3, (some time we use ½ spin ext.)

(d) 3.33 / 0.413334 = 8.056321, 8.056321 eight Root = 1.297 all most 1.296 (Time & Place)

(e) 5.3 + 1.2919 (constant of Visible Light) = 6.5919, 6.5919 sixteen Root = 1.125 x 10 =11.25 angle. [ see later about 11.25 & 1.125 in my article ]

(f) 5.334 + 1.2919 = 6.625 Planck constant. Root = 1.6044 unit energy,(as per E=m-(m/10)

(g) 32** Root = 5.6568, Root = 2.378, Root 1.54221,Root 1.1241,Root = 1.11, Root= 1.055 It is time for produce an atom from Neutron. Four Root 1.24eV= 1.055.,Force= 5.334, Root sixteen 5.334 = 1.11.

2. ATOMIC GRAVITY

Every electron’s radius is same with proton’s radius’s ,proton’s radius = electron’s radius’s Induction area.

As per this view,

F = Es or E = F/s or s = F/E. NOW – As per great scientist Honorable NEWTON’s law of Gravitation,

G₁= F x r ^²/ m^{^2}

G₂= F x r ^²/ m₁m₂

We know one Neutron mass = 1.6749 x 10^{^-27}Kg. and one Neutron 1839 times heavier than Electron.

So as per my view One Electron mass 1.6749 / 1839 = 0.0009 unit.

Hare ‘F’= 5.3…unit, mass= 1.6749, ‘r’= 0.53unit, (as per scientist Bohr ‘H’ atom’s 1^st orbit radius.) Electron mass= 0.000904 unit (as per this calculation electron mass 9.04unit or 0.000904 unit, For details see later)

Gravitation of Neutron --- G = F x r ^²/ m^{^2}

‘G’₁ =(F) 5.3… x (r) 0.53^²… / (1.6749)^{^2} = 0.53..unit That means ‘G’ for an Atom is 0.53…unit,

Again as per G= F x r ^²/ m₁m₂ ‘G’ or Attraction or Repulsion between Neutron and Electron.

‘G’₂ =(F) 5.33 x r (0.533)^² / (N)1.6749 x (e) 0.000904 = 1.5141 / 0.0015141= 1000unit .

As per law “g” = GM / R^{^2}

“g” (neutron) =(0.533 x 1.6749) / 0.533^{^2} =(0.8927217 / 0.284089) = 3.1424 or one Pi

“g” (electron) =(0.533 x 0.000904) / 0.533^{^2} = 0.000481832 / 0.284089 = 0.001696 unit.

Now “g” (neutron) x “g” (electron) = 3.1424 x 0.001696 = 0.00533 unit.

{“g”(n) x “g”(e)} x “G” = (3.1424…x 0.001696) x 1000 = 0.00533 x 1000 = 5.33 unit Force.

[ as per “G”₂ – 1000 root = 31.6227766,root 31.6227766 = 5.6234, 5.624 x 2 = 11.25 (nearer) as per my calculation it is minimum Angel of electron.

Again as per Coulomb rule electric field and intensity of on electric field ---

(a) ‘E’ = Q / k.‘r’^{^2} = “E” = 1.602 / 0.5334 = 5.63 x 2 = 11.25 (nearer) .

Root 5.624 = 2.371

(b) Force = Q₁ X Q₂ / k.’r’² (k= permittivity or dielectric constant) as per Coulomb rule.

F = (1.602 x 1.602) / (0.5334)^² = 9.02 unit

“E’ = 5.63 as per (a) and ‘Torque’ = p x E or “T” = 5.63 x 1.602 = 9.02.

(Torque on an electric dipole in a uniform electric field)

* By ensuring that calculation we get the new rules---

[ Transformation of Energy & Law of Conservation of Energy ]

.1) ‘G’1 = Gravitation of Neutron = 0.53… unit

.2) ‘G’2 = Gravitation between Neutron and Electron = 1000 unit

.3) “g”1 (neutron) = 3.1424 or one Pi

.4) “g”2 (electron) = 0.001696 unit.

.5) “g”1x ‘g’2 x G2 = 5.33 = Force. [see “Atomic Gravity” Image 1; 2; 3 ]

3: primary colour

(L=0.413334) The whole calculation is done by keeping the energy of photon =1.24 eV, whole wave length field is known as a triangle, area ½ x base x height, The area of triangle =3 X 0.413334 = 1.24 square unit and 1.602 / 1.24=1.2919 is constant. The main and primitive triangle is seen as an area of the triangle or ,to calculate the light wave length by multiplying it with 1000000. As ,1.24 x 1000000=1240000.divided by 4 ,each area is calculated, so 1240000/4 =31000 which is shown in picture no.5.

It is seen two Red‘s are inside T1 group, Again if we add up the frequency or quantum no. of “Red” & “Yellow” and divide by 2 we will get frequency of “Orange”. Same T2 group “green” + “Blue” divide by 2 we will get frequency of “Cyan”-color.

Apart from this ,to help geometrical shape drawing we can get same area as by dividing it with 2 to wave length of Red and multiply it with quantum of “Red”, same by dividing it with 2 to wave length of Yellow and multiply it with quantum of Yellow. (This same area consisting triangle met when Orange is created, ) as we add up (both area R&Y) and dividing it with quantum of Orange we can get wave length of Orange.

Same process for (T2 group) Cyan color .So original color is four .

T1 group’s Red, Orange ,Yellow’s addition of frequency is less than of T2 group, which is equal to Red frequency quality,-so, one more “Red” is in T1 group. So T1,group’s total number is four,(Red ,Red ,Orange, Yellow).

From 1^st law, it is seen that each T1,T2, & T3,group’s energy is, Here if added then,T1 + T2 + T3, now,T1,& T2 group’s every addition of mass inside the wave length is , Hare ,

0.413334 which is small to smallest 0.413334 x 4=1.6534 unitary mass which is near to the hydrogen atomic molecular Atomic mass.

Electron which is revolving in an orbit unitary mass is 9 unit [which can see and use in different calculation later]

In pic.5A no [except T3 group ] we will get a list of T1,T2’s energy ,mass. It is proved that if we divide charge to total mass which is equal to total energy.

## If we look at the principle of scientists Bohr & Balmer one ‘H’ atom or atom’s energy is given according to each orbit. In this picture ,the middle part of the two orbits is shown by the energy house

Again Bohr’s orbit is known as A,B,C, ext. so here, first orbit middle part and 7 no house/cell’s centre.[Bhore1^st orbit =A]

Now same process of visible light we can get 1^st & 3^rd quantum or frequency addition is divided by 2 and we will get frequency or quantum no of 2^nd cell’s. same way T2 group’s 4^th & 6^th cell’s quantum or frequency addition is divided by 2 and we will get frequency or quantum no of 5^th cell’s.

In T3 group 7^th cell’s quantum no = 1^st & 6^th cell’s quantum addition is divided by 2 and we will get quantum. And other way for T3 we can see pic.

Every energy status is changed by following one rule. And three group’s are totally separate.

## Four group’s in an atom mass, energy and charge always change among each other and transform to another energy from one as energy universality rule. Every energy is encompassed in a sickle after each and every action or, reaction. Which is shown in images 10; 10A.

If we pick this 0.1335 unit as frequency, then according to 1^st rule, it’s energy =0,1335 x 0.413334=0.055 unit is multiplied by 10 then, it is equal to electron weight by a special unit (0.55amu). Then ,if we calculate this as frequency then , it is equal to unitary unit.

Again ,if we multiply it with 0.1335 x 4 = 1.2919 then all 01.2919unit’s are equal to constant.

So- one neutron is the source of one “H”atom then 2 “electron” means mass or energy is the replacement of one electron mass =9unit (one still mass of “e” =9.1 unit)

# One neutron is heavier than 1839 time’s.9 x 1839 = 16551 unit.

## One proton is heavier than one electron 1837 times. 9 x 1837=16533.unit. [16551 – 16533 = 18]

1 neutron’s mass is equal to 939.7 energy

1 proton’s mass is equal to 938.7 energy

[ 939.7 – 938.7 = 1 , ½ =0.5, we know one energy 0.511 use unit ]

So, it is easily presumed that 2^nd energy or,2^nd mass is the main component of electron and proton of the concerned ENERGY of electron . Which is stored in the “CHAIN’’ of the concerned electron.

IF WE SEE,THE MASS OF “e” FROM ANOTHER VIEW

Mass of ‘e’ 9.1 unit equivalent to energy = 0.511unit

Mass of ,, ,, 1 unit ,, ,,,,,,,,,,,,,,,,,,,,,,,,=0.511/9.1 =.0.0567

1 energy X “L”= 1.24 x 0.413334= 0.51253 or 0.51253/ 0.0567 =9.04 unit (Mass)

1 charge X energy X ‘’L’’ = 1.602 x 13.6 =21.787 or 21.787 x‘’L’’ 0.413334=9 unit (One unit mass)

Or ,1 constant X 7 = 1.2919 x 7=9.04(one unit mass.)

So ,in every aspect 9 unit mass is present.

* By ensuring that calculation we get the new rules---

(A) 1.2919 x 0.413334=0.534 unit “H” atom’s 1^st cell radius.

(B) 1.602 / 0.413334= 3.875 Visible light’s Red frequency .

(1.6028 x 0.413334= 0.6625 x 10 =6.625 )

(D) 0.413334x4=1.6533 mass unit (1 proton is heavier than 1 electron then , 1837 (=1837 X 9 = 16533) so, if we calculate by dividing it with 10000 Then ,we get the used unit

(E) 1 stagnant electron’s mass =9.1 unit we get -,,9.1/7=1.299 or 1.299/0.413334=3.1427=π (one pi )

and we can see also : one pi = 3.141592653589,

pi = 3.141592653589 / 2 = 1.570796326794896,

(1.570796326794896)³ Qube = 3.87578458 it is Red frequency ; 3.87578458 x 0.413334 = 1.602 unit energy,

(F) 1.602 X 13.6 =21.787/7 = 3.11 ,,3.11X0.413334=1.285 (energy) and 9/7 = 1.285.(mass )

(G) Relation with the Pythagoras Theory –

Pythagoras Formula – a² + b² = c²

As per my view – charge²+ “L” ²= mass² (or Weight ² ) (Newton’s Mass= 9 X 1839= 16551,unit /10000=1.6551)

Or (1.602)² + (0.413334)² = (1.6551)²

Or 2.566404 + 0.170845 = 2.73937 Or-- 2.7373249 = 2.74 Or-- 2.74 = 2.74 (PROVED)

Which is the actual relation between Pythagoras main and ancient rules.

Electron Volt = Frequency x 0.413334

Question : (Example) - Electro light action of starting wave length is 7000 Angstrom to any metal transform it to the unit of W0 to eV .

(h=6.60 x 10^-27 erg-s , eV = 1.6 x 10^ -12 erg)

Solve : -

We know Frequency = speed of light / wave length

So here fqn : = 4.2857 x 10^ +14 Hz .

As per my rule (without any power) ev = frequency x 0.413334

eV = 4.2857 x 0.413334 = 1.7714 unit (eV)

and same way frequency = eV / 0.413334

Frequency = 1.7714 / 0.413334 = 4.2857 unit

[ IMAGES 1; 2; 3 ]

4. Induction area of Electron

Every electron’s radius is same with proton’s radius’s ,proton’s radius = electron’s radius’s (include Induction area).

As per this view, if we think (Imagination) proton’s radius = 2 unit, then, we can draw a circle of 2 unit radius. Now ,if we draw same radius circle on it ,we can see ,1st circle or, proton is the centre of 6(six) circles.2nd time, it is 12 circles,3rd time is 18; circles., like this ,every time 6 circles increase each time.

Now, if we divide 6 circles divided by 3,then,(6/3)=2, same, 12 circle (12/3)=4,like this,

As per Bohr’s rule,as n² =orbit, 2n² = number of electron .-

In 1^st orbit’s number of circle 6 , so - 1st orbit = 6/3 = 2 electrons, 2nd orbit =24circles,(24/3)=8 number of electron,3rd orbit =54 circles(54/3)=18 electrons can reside,[shown in black colored electron in the picture--7],Like this, equal number of electrons increase in different orbits.

(Next, if we imagine proton as top point and we draw Equilateral triangle then, in the picture 7A)

example – 1st triangle aPb’s side pa or pb = 2 radius , ( 2 radius = one diameter or one circle ) so we can say one circle ,

2nd triangle cPd’s side pc or pd = (8 radius) 4 circle,

3rd ...9 circles,

so root 1=1,=1^st orbit, root 4=2,=2nd orbirt, root 9=3=3^rd orbit, these are all number of orbit.

And each gap between orbit’s are like this 1^st orbit one circle from middle point of protons (means 2 radius) ,1^st to 2^nd orbit 2circles, 2^nd to 3^rd 4 circles gap, wise it will increase, (in picture7B ASOLSD ).

.If we calculate the perimeter of the triangle, we can see every triangle’s perimeter root 3 =1.73205 unit wise it will increase.

example — [perimeter of triangle 3a, a= (hare number of DIAMETER) x3 (2 radius = 1 Diameter )] 1st—1x3=3 ,root 3 = 1.73205,,2nd 4,-- 4x3=12; root 12 = 3.464101. 3rd—9,,9x3=27,,,root 27=5.196152,

4th-16,,16x3=48,,,root 48=6.926152.

Now two perimeter added,(1+9,2+8, 3+7,,4+6) = 17.3205 which is 10 times of root 3 ,so it can not be more than 17.3205. 5^th number orbit’s perimeter if added to any other perimeter it can not be 17.3205 unit, never. so 5^th number orbit’s perimeter is half of 17.3205 unit,(17.3205 / 2 = 8.66025, 8.66025 / 7 = 1.237 eV of photon) then ,5^th number orbit is the joint place of two groups. so no electron can reside or, stay here. In exam pies -"ASOLSD" picture no. 7, 7a, 7b its been calculated that in every orbit electron and proton's energy, power, induction are control by a cardinal rule.

We know the radius of electron as mass is too negligible than PROTON.

But the induction circle of electron is equal to the radius of proton. In a particular orbit of atom which bears the number of circles (which has the same radius of proton)are divided by 3 equal to the number of electrons of very radius.

If we divide 360 Degree by the number of circles of a particular orbit ,the resultant angle should be drawn from the center of proton and the particular circle (equal to the radius of proton) be placed in that resultant angle is the location of that particular orbit.

In every orbit the distance between the two electrons must be same as two proton's diameter..

[ IMAGE ; 7; 8; 9 ]

As per formula no (i),(ii),(iii)... we can see what is relation between L= 0.413334 and 11.25 Degree angle. [360 degree angle Divided by 32 = 11.25 degree angle

NOW --->> (a) 11.25 Divided by 0.413334 = 27.217698 unit Frequency [ Without energy can't make's any frequency ; So ….]

[Use formula no.-(i) ev = Frequency X 0.413334] (b)--- 27.217698 divided by 2 = 13.60885 unit energy.(as per Bohr electron energy)

(c)--- 13.60885 divided by 2 = 6.804425 unit charge or power .

## - 1.653336 into 32 = 52.906752 mass.

( 32 is max. no of electron in an orbit - 4L = 1.653336 unit mass of a particle)

[Use formula no (ii) - Mass X Charge = Energy or eV ]-->>

Example for 1st orbit as per respected Scientist Bohr --

1.653336 X 2 (no of electron in 1st orbit) = 3.306672 total mass .

3.306672 X 6.8044 = 22.5 .; 22.5/2 = 11.25 either Place , energy or eV. Of one electron.

Example for 2nd orbit --1.653336 X 8 =13.226688 mass; 13.226688 X 6.804425 = 90 unit eng or eV ; 90/8 = 11.25. either place, energy or eV for one.

Example for 3rd orbit -- 1.653336 X 18 = 29.760012 mass.

29.760012 X 6.804425 = 202.5 ; 202.5/ 18 = 11.25 unit either place, eV or enrg for one electron.

**Example for 4th orbit -- 1.653336 X 32 = 52.906752 total max. mass in an orbit. 52.906752 X 6.8044245 = 360 unit Energy or eV or angle either place, !

(1 Hour =3600 Sec.,3600X(1/10)= 360 degree.)

[Use formula No.- (i) eV = Frequency X 0.413334 ] >>

(1) 1st orbit -- eV = 27.217698 X 0.413334 = 11.25.eV either place,

or -- eV = 13.60885 X 0.413334 = 5.625 ; 5.625 X 2 = 11.25.eV either place,

( 2 no of electron in 1st orbit & we know as per Honorable scientist Bohr -- energy label of every orbit that - 1st> 13.6 ; 2nd > 3.4 ; 3rd > 1.51 ; 4th > 0.85 unit. More Please see "ASOLSD" Picture No. 6 & 6 A )

(2) 2nd orbit -- eV = 3.4 X 0.413334 = 1.4053356 X 8 = 11.25.eV either place,

(3) 3rd orbit -- eV = 1.512094 X 0.413334 = 0.625 X 18 = 11.25 .eV either place,

(4) 4th orbit -- eV = 0.85055363 X 0.413334 = 0.3515627;

0.3515627 X 32 = 11.25. eV either place,

If it calculate from back side then get all (Bohr) energy >> 11.25/32 = 0.3515627; 0.3515627/ 0.413334 = 0.85055363 unit of Bohr energy .

** Example for formula no (iii) ‘ L’ square + Charge square = Mass square.

1st orbit --->> (0.413334 X2)square + (1.6009 X 2)square = (1.653336 X2) square.

or (0.826668)square + (3.2016)square = 3.306672 square

or 0.683338 + 10.25152324 = 10.934. ( Same for all orbit)

[ -1.602 eng of electron in 1st orbit of "H" atom,

( Actually mass ,energy, etc. are changeable according to the orbit.

**“mass divided by root over electron volt = energy”

Eg. 1.782662 (mass) / root over 1.24 (eV of photon) = 1.60087 unit energy )

“Every Quantum Number directs a particular unit or, quantity or position change how much”.

Red quantum = 3 and mass 1 unit ; Yellow = quantum. 4 and mass = 1.333,

1.333X3=4 ,is quantum. digit. Green mass 1.416 X 3 = 4.25, 4.25 is quantum. digit of green. Blue mass 1.583 X 3 = 4.75, 4.75 is quantum. of Blue.

Addition of R+O+Y+G+S+B+V = 3+3.5+4+4.25+4.5+4.75+6=30 .total quantum .30.., ]

Total 7 colour Frequency > 30 x 1.2919 = 38.757 unit (10 time's of Red frequency ,)

Average. frequency of one colour --> 38.757 / 7 = 5.5367 unit.

[** 5.5367² (square) =30.65504689 , 30.65504689²(square) = 939.7318 unit. ( nearer of a Neutron's MeV , THAT MEAN'S IT IS "ENERGY" ). and also see that Red Frequency > 3.8757 / 4 = 0.968925 , 0.968925²(square) = 0.9388 or 0.939 unit So it is shown as one complete unit. RED ,(RED is BASIC color) According to that calculation]

Total Energy of 7 colour (as per my 1st law - eV = frequency. X 0.413334)

38.757 X 0.413334 = 16.01958, or 16.02 , for one colour 16.02 / 7 = 2.22857143 . Root Over-> 2.228857143 = 1.5128025,Root -> 1.5128025 = 1.23 unit (energy) of one photon.

[Average. frequency of one colour --> 38.757 / 7 = 5.5367 unit.]

5.5367 (fqn) x 0.413334(L) = 1.24 eV.

[ We know one Neutron's FREQUENCY= 4.285829054907 unit . Here 30 is total of quantum digit - 30 / 7 = 4.2857142847.]

So one color or wave length produce from one neutron. … which is equal to Neutrons MeV.

OR, 1 wavelength of light is structured from 1 Neutron. As per my view maximum 14 electrons equivalent energy including mass, can stay in one orbit , 14 X 9.04 = 126.56 , Root 126.56 = 11.25 unit angle or energy. As per my calculation -- Time 3600 sec Divided by 1000 = 3.6 is atomic time and 360 Degree angle Divided by 1000 = 0.36 is atomic Space. so atomic Time X Space = 3.6 X 0.36 = 1.296,

3.6 X 0.36 = 1.296 ; 1.296 square = 1.679616, unit mass of one particle.

Formula No. (a) -- 3 Mass X atomic Time & space =Frequency.

Example :- 1.679616 X 3 X 1.296 = 6.5303 unit Frequency.(we know electron frequency same unit)

[ 6.5303 X 2 = 13.06 unit Energy .so 2 Freqency = energy ]

Energy = 11.25 / 10 = 1.125, 1.125 square = 1.602 ,

[ Light Rule = 10 times of atomic Rule ]

( 11.25 Degree angle for one electron's as per my opinion)

so Formula No.(b) -- E = 1/10 X space or angle square.

[ 11.25 / 2 = 5.625, Root>> 5.625 = 2.371708,see later about 2.371,

Root >>2.371708 = 1.54 , Root>>1.54 = 1.24098 unit Energy of Photon. That means 3 times square or 8 square of photon Energy = 1/2 of angle]

FORMULA No (C) -- ENERGY = 3 mass / (Divided by) 22/7

or - ENERGY = 3 mass X 7/22.

Example -- 1.679616 X 3 X 7 = 35.271936 / 22 =1.6032698 unit.

Formula No. (d) -- Energy = 3 X Capacity

For example -- 0.534 is radius of 1st orbit of 'H' atom .

we know that radius = capacity . So -- 0.534 X 3 = 1.602 unit.

[ Energy = Light Constant X eV --> 1.2919 X 1.24 = 1.602. ]

** Source of 22/7 and others --> 0.534 X 10 = 5.34 , 5.34 square = 28.512,; 28.512 / 1.296 = 22 , & 7 No. of orbit and Colours.

source of 1.2919 -- five times Root – or 32 root over 3600 = 1.2919.

and already explain in about source of '9'

For example - as per --- E = mc² ,

one electron energy = - 1.6021766208 X 10 to the power -19 J .

mass = eV/c² = 1.782662 X 10 to the power - 36 kg.

so -- c² = eV/m or c²= 1.6021766208 / 1.782662 =0.898755 or '0.9'

mass = Energy x (10/9)

E = m – (m/10) ; E = 1.72662 – 0.172662 or E = 1.6043958 unit

So –

‘m’ - m / 10 = 1.6043958

Or (10 m – m) / 10 = 1.6043958

Or “ 9”m = 1.6043958 x 10

Or “9” m = 16.043958

Or ‘m’ = 16.043958 / 9

Or ‘m’ = 1.782662 unit

As per this view if “E”= 1.6021766208 unit , in that moment à

“m” = (E x 10) / 9

Or ‘m’ = (1.6021766208 x 10 ) / 9

Or ‘m’ = 16.021766208 / 9

Or ‘m’ = 1.7801962453 unit.

[ See also à when “m” = 1.7801962453 unit ;

1.7801962453 / 10 = 0.17801962453 ;

*i* 0.17801962453 x 3 = 0.534057 unit it is all most same of ‘H’ atoms 1st radius ,

*ii* 0.534057 x 3 = 1.60217 unit it is all most same of one mass energy when mass = 1.782662 unit.

*iii* 0.534057 / 4 = 0.13351425

*iv* 0.13351425 square = 0.01782605 & Root over 1.782662 = 1.33516 ;

*v* We know some time we take Radius = Capacity so à

“H” atoms 1^st radius = 0.534 unit as per ‘Hnb’le Bohr

## 1^st radius 0.534 + 2^nd radius (0.534 x 2 ) 1.068 + 3^rd radius (0.534 x 3) 1.602 + 4^th radius (0.534 x 4) 2.136 = 5.34 unit (10 times of 0.534)

Average per orbit 5.34 / 4 = 1.335 ; 1.335 square = 1.782225 & 5.34 x 3 = 178 [ image 6 ]

# (i) -- [Time X Space = 3.6 X 0.36 = 1.296 ] ; 1.296 X 7 = 9.072 ; --> 9.072 X 0.413334 = 3.75 Degree Angle

(ii) -- We know 6.582119 unit frequency of one electron's (without power).

Root over -> 6.582119 = 2.565564 ; 2.565564 / 2 = 1.282782 ; 1.282782 X 7 = 8.98 unit** [ For why use 2 & 7 ? ans :- each & every particle or any others like energy; power; charge ;etc. when create or destroy in the time "MUST" create 2 part - Positive & Negative; or Minus & Plus ...etc..and all are 7 Pair ; that means 7 X 2 = 14 total electrons equal energy or space etc. in one orbit ]

Again --> (1) Root 2.565564 = 1.6017378 unit (energy or charge) ;;Root 1.6017378 = 1.2656 ;; Root ->1.2656 = 1.125 unit (Space);

[ 1.2656 X 3 = 3.7968 all most RED frequency & 1.125 X 10 = 11.25 ;;11.25X 32 ( 32 Max . no of electrons in an atom orbit) =360 Degree; ]

Total 42 electrons are stay in 3 orbit in an atom.

So average equal 14 electrons Energy or Charge and Time & Space are thare ;

(2)-- > Average 14 X 1.24 (Photon eng) X 1.296 (Time & Space) ; = 22.50 ; 22.50 / 2 = 11.25 (eV See ... ) or 7 X 1.24 X 1.296 = 11.25 unit ;

(3) ---> 14 X 1.24 X 1.296 X 1.602 (Charge) = 36.04 ; 36.04 X 10 = 360.4 nearer 360 unit (Space);

(4)---> 32 X 1.24 X 1.602 = 63.567 unit total Energy & Charge in one orbit. (32Max no of elect. in one orbit, 1.24 foton Eng, 1.602 unit Charge)

63.567 / 7 = 9.08,** ;;

next--> 1.296 (Time & Space) Square = 1.68;(mass) 1.68 square = 2.8224;; 2.8224 square = 7.96594; 7.96594 square = 63.546 ; 63.546 / 7 = 9.0779 .**

NOW --> 63.567 / 14 = 4.54 unit Energy either Charge. every part ,

1 st Orbit --- 4.54 / 2 = 2.27 , 2.27 x ( 2 x 0.413334 ) = 1.87653636 for one .electron .

2nd Orbit --- 4.54 / 8 = 0.5675;;0.5675 x (8 x 0.413334) = 1.87653636 for one .electron.

3rd Orbit --- 4.54 / 18 = 0.2522 ;; 0.2522 x (18 x 0.413334) = 1.87653636 for one .electron.

(Same for all Orbit – and Planck constant 6.625 cube root = 1.878 )

Different relation between “10” “9” etc. with Planck constant

(i) Value of Planck const. ‘h’ = 6.626070040(80)x10^-34 Js.

In my view 6.626070040 cube root = 1.878243988,

1.878243988 cube root = 1.2338167707,

1.2338167707 root = 1.11077305

1.11077305 x 9 = “10”

6.626070040 four root = 1.604405 it is energy of a particle’s, when its mass = 1.782662 unit by my rule “energy”= mass – (mass / 10),

*1.782662 electron Volt = 2.85613921732286 Joule,

If we add it as per rule e = m – (m/10) then

1.782662 + 0.1782662 = 3.14175243464572 one pi,

(3.14175243464572 x 7 = 21.99226704252004 or we got 22)

And L= 0.413334 x 10 = 4.13334,,4.13334 cube root = 1.60484 unit

6.626070040 x 0.413334 = 2.7387800339, 2.7387800339 four root

= 1.286439268, 1.286439268 x 7 = “9”

4L= 4x0.413334 = 1.653336, 1.63336 root = 1.285821128, 1.285821128 x 7 = “9”

as per Planck ‘h’ = 4.135667662(25) x10^-15 eV,

4.135667662 cube root = 1.605148475 unit, ½ (pi 3.14…) cube or (1.57079632,)cube = 3.8757 is red light frequency X 0.413334 = 1.601993 unit energy.

(ii)Value of “h” Bar = 1.05457800x10^-34,Js,

1.05457800 root = 1.112134758, 1.112134758 x 9 = “10”

6.582119514(40)x 10^-16 eV,

6.582119514 four root = 1.601737859 energy and angle 11.25 / 10=

1.125 to power 16 = 6.58325

(iii) Value of “hc” = 0.198644568 x 10^-25 Jm.

0.198644568 x 9 = 1.7878011,mass.

1.23984193 eV um,

1.23984193 root =1.11348011 x 9 = “10”

1.23984193 / 0.413334 = 2.999612, 2.999612 Square = “9”

(iv) Value of “hc”Bar = 3.16152649 x 10^-26 Jm,

3.16152649 square = “10”

3.16152649 four root = 1.33344 and 4/3 = 1.3333,

Now à

10 Joule equivalent electron volt = 6.24159…eV

6.24159… x o.413334 = 2.579828…,2.579828…/2=1.289914…x7= “9”

4/3 x (pi) 3.14…= 4.1887, 4.1887x0.413334=1.7313734592, 1.7313734592 square = 2.997654…,2.997654…square = “9”

1.7313734592 eV = 2.77395592x10^-19 Joule.

2.77395592 root = 1.6655227.(mass) 1.6655227root= 1.29 it is constant of visible light’s frequency,

4/3 x (pi) 3.14…x (1.29) cube = 8.99 mass, either energy or

“C” square.

9 Joule = 5.61736x10^-19 eV.

5.61736 cube root =1.778mass, 5.61736 eight root =1.24 photon eV

“5.61736”root = 2.37

[ 2.322 Up to 2.99 is a junction or separation position of Time & Place with various magnetic wave , and create variety of particles , 2.37 – 1.296 (time x place) = 1.078 , 1.078 to the power 32 =“10”

For example one Red frequency 3.8757 divided by 4 = 0.96.… it is a magnetic field or wave . 0.96 / 0.413334 = 2.3227, that means 2.3227 number of 0.413334 is present in 0.96 digit, 2.3227 four root = 1.2345 it is eV, ( that means here present 4 no. of eV or 1.2345 x 1.2345 x 1.2345 x 1.2345 = 2.3227 ), same process, 2.3227 eight root = 1.1111 , is equivalent to 10 / 9 = 1.1111 for easy understanding see Picture no. “1” ]

1.778 power four = “10”

# 4/3 x (pi) 3.14…= 4.18879x1.782662 (mass) =7.467197124

7.467197124 root = 2.73261726629**.;2.73261726629 root=1.653063

1.653063 (4L=1.653336) root = 1.285714974,,1.285714974x7 = “9”

4/3 x (pi) 3.14…x “ r ”cube à

4.18879 x 1.2919 cube (1.2919 is light constant)= “9”

When add place that moment create one mass ( as per rule of globe “4/3 pi ‘r’ cube”, hare “ r ” = 1.296 ,as per my view time 3.6 and space 0.36 3.6x0.36=1.296)

4.18879 x 1.296 cube = 9.1108 mass, 1 stagnant electron’s mass =9.1 unit

That means [(4/3) x pi 3.14 x (4/3) cube] =“10” here “r” = 4/3

And eight root “10” = 1.3335…,1.3335 square = 1.778 mass

That means “4/3” to the power eight = “10” and “4/3” square = mass.

1.60217 Joule = “10” x10^+18 eV,

“10” root = (a) 3.16227766, 3.16227766 root = (b) 1.778, 1.778 x 4 = (c) 7.113,

7.113 / 3 = (d) 2.371***, 2.371 root = (e) 1.5398, 1.5298 root = 1.2408 eV of photon.

Now we can see what is activity each part of root “10”

Here (a) is nearest of pi, (b)is mass, (d) “4/3”,

(c)=7.113, 7.113 root = 2.660702, 2.660702 / 2 = 1.3335,

1.3335 x 4 = 5.334 it is 10 times of ‘H’ atoms radius,

(d) = 2.371, 2.371 cube root = 1.33345, (4/3=1.3333)

Root 2.371 = 1.5398, 1.5398 + 1.60217 energy = 3.14104 (pi)

1.5398 + 1.60217 energy + 1.2919 constant of light fqn. + 1.24 photon eV = 5.67297, 5.67297 cube root = 1.783 mass.

Equivalent energy 5.67297 eV = “9.089” x 10^-19 Joule. [image 1 & 2]

Relation with joule – eV. Light’s frequency, energy, electron volt, and mass –

[ For easy understanding please follow image 4 & 5 ]

As per rule eV = frequency x 0.413334

Red frequency 3.8757 unit

eV = 3.8757 x 0.413334 = 1.602 unit

Orange frequency 4.5216 unit

eV = 4.5216 x 0.413334 = 1.8689 unit

Yellow frequency = 5.1676 unit

eV = 5.1676 x 0.413334 = 2.1359 unit

Green frequency = 5.4905 unit

eV = 5.4905 x 0.413334 = 2.2694 unit

Cyan frequency = 5.8135 unit

eV = 5.8135 x 0.413334 = 2.4029 unit

Blue frequency = 6.1365 unit

eV = 6.1365 x 0.413334 = 2.5364 unit

As per formula energy (electron Volt) x Charge = Either mass power .

RED à 1.602 / 1.602 = “1” unit

Orange à 1.868 / 1.602 = 1.166

Yellow à 2.1359 / 1.602 = 1.333

Green à 2.2649 / 1.602 = 1.413

Cyan à 2.4029 / 1.602 = 1.499

Blue à 2.5364 / 1.602 = 1.583

(see picture No.1 & 5 , 5A )

Red fqn.= 3.8757 ,

3.8757 Joule equivalent eV = 2.419021829011563 eV.

2.419021829011563 x 0.413334 = 1 unit,

Orange fqn. = 4.5216, 4.5216 Joule = 2.822160926299426 eV.

2.822160926299426 x 0.413334 = 1.1665

Yellow fqn. = 5.1676, 5.1676 joule = 3.22536243868208 eV

3.22536243868208 x 0.413334 = 1.3331

Green fqn. = 5.4905, 5.4905 joule = 3.426900779778618 eV

3.426900779778618 x 0.413334 = 1.4164

Cyan fqn.= 5.8135 , 5.8135 joule = 3.628501535969947 eV

3.628501535969947 x 0.413334 = 1.4997

Blue fqn.= 6.1365 ,6.1365 Joule = 3.830102292161276 eV

3.830102292161276 x 0.413334 = 1.5831

If it calculate as per identical result ,the quantum no. of red = 3 , orange =3.5,yellow = 4,etc divided by resultant eV we get 1.24 that is ev of photon.

See

3/2.4190218….=1.24

3.5/2.822160…=1.24

4/3.3353624….=1.24

That means all color produce from equal energy or element and also every quantum digit indicate that maintain a particular rule and process for create each and every color.

Total frequency of four PRIMARY COLOUR (Red + Yellow = 9.0433) + (Green + Blue = 11.627) ) = 20.6703 unit

20.6703 Joule = 1.29014 x 10^+20 eV.

All most visible light constant 1.2919

10eV = 1.60217…x 10^ -18 Joule

10x 10^+18 eV = 1.60217 Joule

1.60217 eV = 2.566959…x10^ -19 Joule

Root over 2.56695…= 1.60217…unit energy

As calculation of the mass of total electrons is ->in every orbit total electron energy not more than 16.978 unit and less than 16.472754 unit.

When one electron goes up from 1^st to last orbit have no total energy maximum 20.253366 unit.

As per Bohr’s energy of electron’s –Total energy of every orbit
(-)ev	orbit	energy	charge	Energy X no of e	Total energy
13.6	X 1/ 1^2	=13.6	/1.602=	8.489x2 =	16.978 unit
13.6	X 1/ 2^2	=3.4	/1.602=	2.1223x8 =	16.978 unit
13.6	X 1/ 3^2	=1.511	/1.602=	0.9432x18=	16.978 unit
Same way for others orbit= PROVE THAT equal energy in every orbit

[Calculation done by formula no 2 in this data article or (ii) Mass x Charge =Energy and electrons energy are taken from picture no -6 (For Example 2^nd orbit………) ]

10eV = 1.60217…x 10^ -18 Joule

10x 10^+18 eV = 1.60217 Joule

1.60217 eV = 2.566959…x10^ -19 Joule

Root over 2.56695…= 1.60217…unit energy

16.978 Joule = 2.7201753…x 10^+18 eV

2.7201753 / 0.413334 = 6.581058… it is Frequency either “h”bar 6.581058…

2.7201753 four root =1.2842489 x 7 = “9”or 3²or C²

16.978 Joule = 1.05968…x10^+20 eV.

1.05968…/ 0.41334= 2.563746…,Root 2.563746…=1.60117… (energy)

2.563746…/2=1.28187…x7= 8.97311…mass either , “C²”

Conclusion

(i) All colors produce from equal quantity of element or energy.[1.2401707unit]

(ii) Equal total energy in every orbit.

(iii) Visible light constant 1.2919

(iv) These two energy “Time”& “Space”

(v) Any Time is not necessary “C^2”for each calculation of “energy”

(vi) Rresultant angle should be drawn from the center of proton and the particular circle (equal to the radius of proton) be placed in that resultant angle is the location of that particular orbit . .

(vii) In every orbit the distance between the two electrons must be same as two proton's diameter

(viii) induction circle of electron is equal to the radius of proton

(ix) Primary or original color is four “Red” “Yellow” & “Green” “Blue”

(x) eV = 0.413334 x Frequency

(xi) Force = E x s

(xii) Mass = Energy x (10/9)

Images: