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 = 98% * H0.54 of fusion = 52.92 hour 

                       Because 

                       Hour = 52.92 / 6 proton of C = 8.82-Fluorine  

                            = 8.82 – 5 proton of B = 3.82-Beryllium 

Fluorine = 9 proton   

Beryllium = 4 proton 

      Boron = 8.82 * 3.82 = 33.6924

            = 33.6924 + 64.3076 (B2H6)2 + H4 = 64) = 98

            = 98

Chemical = 98 - 76 (F4) A hydrofluoric atmosphere of earth = 22 

         = 22 = B2 = Boron  

Chemical formula from 52.92 hours = B2F4

                                  = B2F4, 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 / 34.85 (H2O2 = 34.5) = 98

                          = 98 = B2F4 = E = m2c,