ATMOSPHERE OF BORON ON EARTH

ENERGY LAW OF SIKANDER AQEEL = E = m2c 

Boron (III) orthophosphate 

        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 = 106% * H0.54 of fusion = 57.24 hour 

                       Because 

                       Hour = 57.24 / 6 proton of C = 9.54-Neon  

                            = 9.54 – 5 proton of B = 4.54-Boron 

Neon = 10 proton   

Boron = 5 proton 

      Boron = 9.54 * 4.54 = 43.3116

            = 43.3116 + 62.6884 (O3 + CH2 = 62.5) = 106

            = 106

Chemical = 106 - 31 (P) A phosphoric atmosphere of earth = 75 

         = 75 - 64 (O4) = 11 

         = 11 = B = Boron  

Chemical formula from 57.24 hours = BPO4

                                  = BPO4, Boron (III) orthophosphate 

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 / 32.2198 (O2 = 32) = 106

                          = 106 = BPO4 = E = m2c,