ATMOSPHERE OF BORON ON EARTH

ENERGY LAW OF SIKANDER AQEEL = E = m2c 

Boron Tri-fluoride 

        Boron = 11-B  

        Boron = 11 

        Boron = 5 proton 

        Boron = 6 neutron 

        a2 + b2 = c2  

        a2 (neutron)     +     b2 (proton) = c2  

        a2 (6)           +     b2 (5)   

        a2 (36)          + b2 (25) = c2

        a2 + b2 = 61 = c2 

        a2 + b2 = 61 / 11-B = 5.54

        a2 + b2 = 5.54 = 5 proton + H0.54

Boron energy atmosphere = 68% * H0.54 of fusion = 36.72 hour 

                       Because 

                       Hour = 36.72 / 6 proton of C = 6.12-Carbon  

                            = 6.12 – 5 proton of B = 1.12-Hydrogen 

Carbon = 6 proton  

Hydrogen = 1.25 proton 

      Boron = 6.12 * 1.12 = 6.8544

            = 6.8544 + 61.1456% (C2(H2O)2 = 61) = 68 

            = 68

Chemical = 68 - 57 (F3) A powerful oxidizing atmosphere of earth = 11 

         = 11 = B = Boron 

The earth is most large planet in solar system for combination of oxygen 

Chemical formula from 36.72 hours = BF3  

                                  = BF3, Boron Tri-fluoride  

ENERGY IN ACT BY C1     

Light speed meter seconds = 300000000 / 61 (m2) = 4918032.78

                          = 4918032.78 / 24 hour = 204918.0325

                          = 204918.0325 / 60 minute = 3415.30

                          = 3415.30 / 50.225% (H2O)2 + CH2 = 50.25) = 68

                          = 68 = BF3 = E = m2c,