In the fusion reaction ₁^2 H + ₁^2 H ₂^3 He + ₀^1 n , the masses of deuteron, helium and neutron expressed in amu are 2.015,3.017 and 1.009 , respectively. If 1 kg of deuterium undergoes complete fusion, find the amount of total energy released. (1 amu =931.5 MeV ) .

Step-by-step solution

m =2(2.015)-(3.017+1.009)=0.004 amu Energy released =(0.004 931.5) MeV =3.726 MeV Energy released per deuteron = 3.726 2 =1.863 MeV Number of deuterons in 1 kg = 6.02 10²⁶ 2 =3.01 10²⁶ Energy released per kg of deuterium fusion aligned & = (3.01 10²⁶ 1.863 ) & =5.6 10²⁶ MeV 9.0 10¹³ ~J aligned