A star initially has 3×1020  deuterons. It produces energy via the process   1H2+1H2→1H3+p and  1H2+1H3→2He4+n . Find the total energy released during the deuteron supply of star is exhausted (mass of  2He4=4.001amu, 1H1=1.007amu, 1H2=2.014amu, 0n1=1.008amu )

1H2+1H2→1H3+pplus 1H2+1H3→2He4+ngives31H2→2He4+n+pMass defect =ΔM=mass  of   2He4+n+P−mass 31H2 ⇒ΔM=4.001+1.008+1.007−32.014⇒ΔM=−0.026 amu∴Energy released =ΔE=ΔM931MeV=0.026×931×106×1.6×10−19J =3.87×10−12J No. of reactions =N3=3×10203=1020∴Total energy released =E=N3ΔE=3.87×10−12×1020=3.87×108J