ATMOSPHERE OF BORON ON EARTH

ENERGY LAW OF SIKANDER AQEEL = E = m2c 

Boron (III) phosphide 

        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 = 42% * H0.54 of fusion = 22.68 hour 

                       Because 

                       Hour = 22.68 / 2 proton of He = 11.34-Sodium  

                            = 11.34 – 5 proton of B = 6.34-Carbon 

Sodium = 11 proton   

Carbon = 6 proton 

      Boron = 11.34 * 6.34 = 71.8956

            = 71.8956 = 29.8956% (B + H2O = 29.5) = 42

            = 42

Chemical = 42 - 31 (P) A phosphide atmosphere of earth = 11 

         = 11 = B = Boron 

Chemical formula from 24.57 hours = BP

                                  = BP, Boron (III) phosphide 

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 / 81.3166% (4H2O + H6 = 81.5) = 42

                          = 42 = BP = E = m2c,