sree DEBASISH DASGUPTA: The relationship of Energy with Atoms and Time, Space ©

The relationship of Energy with Atoms and Time, Space ©

sree Debasish Dasgupta

In each of the previous articles, electrons, protons, etc., the relationship of ‘L’ with particle and light has been shown. It also provides detailed information on why each number used in physics is being used, and what sources are used to find the source.

In the present article, too, are electrons, protons, Pi, time, space, how they come together, and what is the source of ‘C’²? (in the great formula e=mC² . I respect sir Einstein’s contribution) .This has been discussed.

I look at each article from different angles and from different perspectives; I have tried to prove their authenticity by calculating the atomic and light constants, quantum etc. by mathematics. All of these articles were later published in Mendely Data. Which has taken a place in the court of the world (scientist community) today through social media.

And how the properties of matter/particles are hidden in all these combinations of Atom

Therefore, I urge all readers to read each of the previous articles once to understand this article better or to know the details.

INTRODUCTION :

This article discusses relation between atomic mass and energy with electric charge of electron and Proton’s. Here, it is found that, by comparing with e=mc² .

Which gives a new process of calculation about atomic structure, ‘Pi’; electron mass, and also C^{^2} = Speed of Light Square. in great equation e=mc² .

In present essay, have shown the digit or number ‘L’ = 0.413334 used with Proton mass, electron mass, energy, Pi, etc.

And also have shown that how to involve Time 3600 second with Proton mass, electron mass & energy,

1.01

As per reference articles, ‘L’ = (photon’s eV 1.24 unit / 3) = 0.413334,

(0.413334 *10) ^{^2} = 17.084449955556 , either [L x (Time /space)] = [0.413334 * (3600/360)]^{^2} =17.084449955556 [3600 second & 360 Degree ]

9.109390 e-31/17.084449955556 = 5.3319773382798041975309487724255e-32

(Electron mass) / (0.413334 *10) ^{^2} = 5.3319773382798041975309487724255e-32 ---- (1a)

1.02

22/7 = 3.1428571428571428571428571428571 [Pi]

1.2985270578788836 /0.413334 = 3.141592653589793242268964082316

[1.2985270578788836 is constant unit and ‘L’ = 0.413334, see reference articles]

(3.1428571428571428571428571428571)^{^9} = 2.9917256660402129603928590571842e+4

--- (2a)

1.03

2.9917256660402129603928590571842e+4 / 5.3319773382798041975309487724255e-32

= 5.6109121930465663243853382177961e+35 [2a / 1a] --- (3a)

1/5.6109121930465663243853382177961e+35 = 1.7822414e-36 Unit Mass.

As per (e=mc^{^2}) equation. (Mass = 1.782662e-36 Kg. When energy = 1.60217653e-19 J. )

Either

5.3319773382798041975309487724255e-32 / 2.9917256660402129603928590571842e+4

— (1a)/(2a) .

= 1.7822414e-36 Unit mass.

So [ (Electron mass) / (0.413334 *10) ^{^2} ] divided by (Pi)^{^9} = Mass. --- (i)

[As per (e=mc^{^2}) equation. Mass = 1.782662e-36 Unit. When energy = 1.60217653e-19 Unit ]

2.01

5.3319773382798041975309487724255e-32 * 2.9917256660402129603928590571842e+4

= 1.5951813453676469100727741096844e-27 = (1a)x(2a).

1.5951813453676469100727741096844e-27 * (3.1428571428571428571428571428571 /3)

= 1.6711423618137253343619538291932e-27 Unit. Proton’s mass.

So, [(Electron mass) / ( ‘L’ x 10 )^{^2} ] x [(Pi) / 3 ] *(Pi)^{^9} = Proton’s mass. …(ii)

As per (e=mc^{^2}) equation. Proton’s mass = 1.672623e-27 Kg.

2.02 About ( Pi) with charge and energy,

[We know, 1.5032784196545335010226279575152e-10 Unit = Proton’s energy As per “e=mc^{^2} .”

And 1.60217653e-19 Unit electric charge of a Proton then (-1.60217653e-19) Unit electric charge of an electron]

1.60217653e-19 <root = 4.0027197378782342122581519961685e-10

4.0027197378782342122581519961685e-10 / 1.5032784196545335010226279575152e-10 = 2.6626602800551704435263034591073 /2 = (1.3313301400275852217631517295536)^{^2}= (1.7724399417458696743044966223555) ^{^2} = 3.1415433470961018845922239897721 =[Pi].

So. {[(Root> charge of a Proton) / (Proton’s energy in Joule)] / 2}^{^4} = [Pi].

[Nb. Please see, Rule:- 3, {[(root > Energy) / Proton energy ] / 2 }^{^4} = Pi

(here Energy = 1.60217653e-19 Unit) In reference article ‘Mass Energy in Pythagoras formula’]

Back calculation

3.1415433470961018845922239897721 <4root = 1.3313353638003897127975349179503

1.3313353638003897127975349179503 *2 = 2.6626707276007794255950698359006

1.5032784196545335010226279575152e-10 * 2.6626707276007794255950698359006

= 4.0027354434480865514832910049719e-10 square.

= 1.6021891030235550091508407813388e-19 , Unit ]

So, {[4Root over > (Pi)] x (Proton’s energy as per e=mc^{^2}) x 2}^{^2} = Proton’s electric charge. …(iii)

Or,

[ x (1.50327841965…5152e-10) x 2]^{^2} = Proton’s electric charge =1.602189e-19 , Unit

[Nb. Rule:-2 [root> (1.60217653e-19 / 2)] / 4root> 4Pi = Proton Energy.

In reference article ‘Mass Energy in Pythagoras formula’

As per this view: we can put in place of Proton’s electric charge =1.602189e-19 ,

[root> {[4Root over > (Pi)] x (Proton’s energy as per e=mc^{^2}) x 2}^{^2} / 2 x (1 / 4root> 4Pi) = Proton Energy. in Joule. In this process we can make a lot of new rule. For example Please see image 1 ]

3.01

Next >

(root 3)^{^3} = 5.1961524227066318805823390245176 * (10)^{^2} = 5.1961524227066318805823390245176e+2

(5.1961524227066318805823390245176e+2)^{^4} = 7.29e+10

0.413334 / 7.29e+10 = 5.6698765432098765432098765432099e-12 ……(1b)

5.6698765432098765432098765432099e-12 * 1.60217653e-19

= 9.0841431255283950617283950617284e-31 electron mass.

9.0841431255283950617283950617284e-31 *[ (10/9)^{^2} * 0.413334*3600 (second)] =

1.6687934287320679526610673400061e-27 Unit. Mass.

3.02

## We know ,

Proton mass = 1.672623e-27 Unit & energy = 1.5032784196545335010226279575152e-10

(Energy = 1.60217653e-19 Unit, Mass = 1.782662e-36 Unit as per e=mc²)

Electron mass = 9.109390 x 10^{^-31} Kg. or 9.109390 e-31 Unit.

Neutron mass = 1.6749274171e-27 Kg or 1.6749e-27 Unit.

Electron energy in MeV. Unit =0.51099895 MeV.

0.51099895 MeV = 0.51099895 * 1000000 = 5.1099895e+5 electron Volt.

[10 /9=1.1111111111111111111111111111111 root = 1.2345679012345679012345679012346*0.413334=0.51028888888888878683111111111112

As per my (sree Debasish Dasgupta) formula ‘eV=frequency * 0.413334’

1.2345679012345679 is ‘frequency’ and 0.51028888 is ‘eV’

Almost same Digits (0.51…) of electron’s energy in MeV. Unit. For why used (9 & 10) in my (sree Debasish Dasgupta) formula, e=m–(m/10)]

Now as per unitary process:

When energy is 1.60217653e-19 Unit in that moment electron’s mass = 9.109390 e-31 Unit.

When energy is 1 Unit ,, ,, ,, ,, mass = 9.109390 e-31/1.60217653e-19 = 5.6856344038443753760392433160908e-12 Unit. …… (2b)

When energy (Proton) is 1.5032784196545335010226279575152e-10 Unit ,, ,, ,, mass = 8.5470915013446183292177653419436e-22 Unit. …… (3b)

(Proton energy) x (2b) =

1.5032784196545335010226279575152e-10 * 5.6856344038443753760392433160908e-12

= 8.5470915013446183292177653419436e-22

8.5470915013446183292177653419436e-22 / 5.1099928085077260860443538932226e+5

= 1.672623e-27 Unit [Energy of m_e = 0.51099895000(15)MeV. Unit]

[Nb. Almost same value 3.01 equation. ……(1b) & (2b) ]

So, [(Proton energy) x (1b)] / electron energy (eV,) = Proton mass.

Or, [(Proton energy) x (electron mass/electron’s or, Proton charge)] / electron energy (eV,) = Proton mass. ……….…(iv)

Example: (electron’s energy 0.51099895 Mev = (0.51099895 * 1000000) = 510998.95 eV.)

8.5470915013446183292177653419436e-22 / 510998.95 = 1.6726240829544988946098158013717e-27 Unit Proton mass.

[Nb, 8.5470915013446183292177653419436e-22 / 0.5334 (H, Atom’s 1^st orbit radius 0.53 Unit)

1.6023793590822306578960939898657e-21 *100 = 1.6023793590822306578960939898657e-19 unit charge of Proton.]

## 510998.95 eV / 5.6856344038443753760392433160908e-12 =

8.9875449897813519895404631923762e+16 = SPEED of LIGHT Square. [As per ‘e=mc^{^2}’ ]

So we can write here

(electron energy) / (electron mass/electron’s or, Proton’s charge) = SPEED of LIGHT Square. ……………(v)

4.01

Electron mass X electron MeV. X 3600 Second =

9.109390 e-31 * 0.51 *3600 = 1.672484004e-27 Unit Proton Mass.

So, we can write here, Proton Mass. / (electron MeV. X 3600 Second) = Electron mass ….(vi)

Example, 1.672623e-27 / (0.51099895 *3600) =9.0923376652731e-31 = Electron mass

Or Neutron mass 1.6749274171e-27 / (0.51099895 *3600) =9.1048644201932530607178568783969e-31Unit = Electron mass

Or, (0.51099895 *3600) = 1.6749274171e-27 / 9.1048644201932530607178568783969e-31

Or, (0.51099895 *3600) = 1839.59622

Or, 0.51099895 = (1839.59622 / 3600) = 0.51099895

Either number of Proton Mass divided by (electron MeV. X 3600 Second) = number of Electron mass.

For example = 3 * 1.672623e-27 / (0.51099895 *3600)

= 2.7277012995819267339003338460872e-30 = 3 number of Electron mass.

(2.7277012995819267339003338460872e-30 / 9.0923376652731e-31 = 3 )

Either number of Proton Mass divided by {[ (10/9)^{^2} * 0.413334 ] * 3600} = number of Electron mass …………. (vii)

For example = 3 * 1.672623e-27 / (0.510288889 * 3600) = 2.7314968639264258015227919258105e-30 = 3 number of Electron mass.

[ 2.7314968639264258015227919258105e-30 / 3 = 9.105e-31 one Electron mass.]

Same method for energy value

For example

Proton energy = 1.5032784196545335010226279575152e-10 Unit.

1.5032784196545335010226279575152e-10 / (0.510288889 * 3600) = 8.1831556162479166287762448959421e-14 Unit energy of electron.

So Proton energy = electron energy * (0.510288889 * 3600) or electron energy *1837.

For example 8.1831556162479166287762448959421e-14 * 1837 = 1.5032456867047422847061961873846e-10 Unit.

So in MeV. Unit,

Proton energy = {[ (10/9)^{^2} * 0.413334 ] * 3600} * electron’s energy ……….(viii)

(0.51099895 Mev is electron’s energy mass energy)

For example , {[ (10/9)^{^2} * 0.413334 ] * 3600} * 0.51099895

Or, 1837.04 * 0.51099895 = 938.725511108 Mev , Proton energy.

As per formula ‘eV=frequency * 0.413334’

1.2345679012345679 is ‘frequency’ and 0.51028888 is ‘eV’ …. (3.02)

SO.

Photon (1.2345679012345679 )‘frequency’ and ‘L’ (= 0.413334) and 3600 second present in the digit of 1837.04’

And (electron energy in MeV.) x (Time in Second) = {[ (10/9)^{^2} * 0.413334 ] * 3600}

Ie. (Photon frequency) x (L) x (Time) x (electron energy MeV) = Proton energy ….(ix)

For example

(1.2345679012345679) * (0.413334) * (3600) * (0.51099695) =938.721837028 MeV.

Conclusion:

· [ (Electron mass) / (0.413334 *10) ^{^2} ] divided by (Pi)^{^9} = Mass.

· {[(Root> charge of a Proton) / (Proton’s energy in Joule)] / 2}^{^4} = [Pi].

· [(Electron mass) / ( ‘L’ x 10 )^{^2} ] x [(Pi) / 3 ] *(Pi)^{^9} = Proton’s mass.

· {[4Root over > (Pi)] x (Proton’s energy as per e=mc^{^2}) x 2}^{^2} = Proton’s electric charge

· [(Proton energy) x (electron mass/electron’s or, Proton charge)] / electron energy (eV,) = Proton mass.

· (electron energy) / (electron mass/electron’s or, Proton’s charge) = SPEED of LIGHT Square

· Proton Mass. / (electron MeV. X 3600 Second) = Electron mass

· Number of Proton Mass divided by {[ (10/9)^{^2} * 0.413334 ] * 3600} = Number of Electron mass

· Proton energy = {[ (10/9)^{^2} * 0.413334 ] * 3600} * electron’s energy

· (Photon frequency) x (L) x (Time) x (electron energy MeV) = Proton energy

Reference:

1) “Atomic Mass Energy Primary Colour”

2) ‘Mass Energy in Pythagoras formula’

3) ‘Atom to Nucleus’

4) “Atomic Time Space”

5) ‘Atomic Mass Energy and Constant’

sree DEBASISH DASGUPTA

মঙ্গলবার, ৬ জুলাই, ২০২১

The relationship of Energy with Atoms and Time, Space ©

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