LAST ELECTRON OF IODINE WITH CROSS-TRON

WITH ENERGY LAW  ‘‘E = m + m x C1’’

Iodine 53 last electron of Iodine Perhaps for this reason his says atom number 53. Or` atomic number 53 meaning last electron of  Iodine So Electron number 53 have also one orbit. and his connect was also with sector 9000L like our earth. and` we 

shall solve motion of last electron with planets motion. we would get here four information. (1) Electron energy. (2) Electron orbit. around the atom (3) micro meter distance between Electron and Iodine level till. (4) micro seconds distance 

between Electron and Iodine But before the begin. I want some tell you, about Atomic number. with a simple method. Our earth have 365 days, But` if sees to his, like a atomic theory. Then sun is like a atom and Earth is a Electron and his number 

is 365. Perfectly like this` Iodine is a atom and his last electron is his atomic number 53 Last electron of Iodine that in last Electron is turn around with micro meter speed. after of all.

EARTH HOT AND COLD PERIOD, 

So E = m + m x C1. Earth was hot in begin and Earth has cold now. m365 hot + m365 cold = 730 x C1 300000000 = 2.19000000000 x .008 photon = 1752000000 / 1000m / 100m /10mm = 1752 seconds distance between Earth and Sun. Now` we 

shall make his distance in meters and orbit. 1752 x 300000000 = 525600000000m and` it is distance in meters. 525600000000 x 6 = 3153600000000m and` it is orbit of Earth. Now` to this orbit` we shall do shift in atom number. (1) electron has not 

moon, (2) electron has not days months and year. from this reason` orbit of electron we shall divide with 24 hours and 360 degrees. 3153600000000m / 24 / 360 degrees = 365000000 / 1000m / 100m / 10mm = 365 atomic number. Or` 365 atomic number of Earth. now

IODINE  ELECTRON AND, E = m + m x C1. 

Iodine atomic number 53. So m53 hot + m53 cold = 106 x C1 300000000 = 31800000000 x .008 one photon = 254400000 / 1000m / 100m /10mm = 254.4 is micro seconds distance between last electron. and 31800000000  volt energy of last 

electron of Iodine Now` we shall make his distance in micro meter 254.4 x 300000000 =  76320000000 / 1000mc / 100mc /10mc = 76320 it is the distance in micro meters between last electron and  Iodine surface till. to the proof 76320  / 24 hours / 60 minute 

= 53 electron of Iodine Now` we shall make to his orbit. 76320 x 6 = 457920 so it is orbit of last electron. in the micro meter measurement. (  now to the atomic number ) we shall make his atomic number. 457920 orbit / 24 hours = 19080 / 360 degrees = 53 atomic Number of Iodine

DETAILS OF ELECTRON OF IODINE  

(1) electron energy 31800000000  volt. (2) electron orbit. 457920 micro meter. 

(3) 76320 is micro meter distance between last electron and Iodine level till. 

(4) 254.4 micro seconds distance between last electron and Iodine

PROOF OF DETAILS.  

1. one electron charge 14.4 and one proton charge 25 percent excessive from electron. 

1. 31800000000 volt / 14.4 = 2208333333 x .008 = 1766666667 / 1000000 = 17 + 8 = 25 

percent excessive charge of proton from one electron.

2. we shall compare between hydrogen size and electron charge. with micro meter distance 76320 

2.   76320 / 1.25 hydrogen size = 61056 / 60 seconds hydrogen time = 1017.6 / 100 area 

= 10.176 + 4.224 = 14.4 electron charge.

3. orbit, we shall to know percentage of electron charge here. with photon quantity. in one proton happens 5 photons. 

3. 457920 / 5 = 91584 / 360 degrees = 254.4 / 25 proton percentage = 10.176 +  4.224 = 

14.4 active charge.  

4. we shall to know percentage of charge with proton charge 28.8. 

4. 254.4 micro meter distance. So 254.4 / 28.8 = 8 + 4.5 = 12.5 Sun hydrogen atom size  

 LAST ELECTRON OF IODINE  AND CROSS-TRON.

If the electron charge value would be right. then will be solved automatically time of mass and weight of mass through the new energy law. So Iodine atomic seconds distance 254.4 

and one electron charge 14.4 and energy law E = m + m x C1. Now, m14.4 + m254.4 = 268.8 x 300000000 = 80640000000 / 24 hours / 60 minute / 1000m / 100cm light measure = 560 / 254.54 light code = 2.2 Cross-tron of Iodine Now to the work condition  Iodine will have need 254.54 volt 

energy for one atom and you will be get 31800000000 volt energy from one atom. But how` when we shall be reached at Hydrogen atom till then you will understood. and Every math is possible! If you have right atomic energy law. You know very well that i test at earth first of all. and second at atom. and third at human body.