ATMOSPHERE OF BORON ON EARTH

ENERGY LAW OF SIKANDER AQEEL = E = m2c 

Boron (III) oxide 

        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 = 70% * H0.54 of fusion = 37.8 hour 

                       Because 

                       Hour = 37.8 / 4 proton of Be = 9.45-Neon  

                            = 9.45 – 5 proton of B = 4.45-Boron 

Neon = 10 proton   

Boron = 5 proton 

      Boron = 9.45 * 4.45 = 42.0525

            = 42.0525 + 27.9475 (BN + H2 = 27.5) = 70

            = 70

Chemical = 70 – 48 (O3) A radical atmosphere of earth = 22 

         = 22 = B2 = Boron  

Chemical formula from 37.8 hours = B2O3

                                  = B2O3, Boron (III) oxide 

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 / 48.79 (O3 + 0.79 = 48.79) = 70

                          = 70 = B2O3 = E = m2c,