In the fusion reaction  12H+12H⟶23He+01n, the masses of deuteron, helium and neutron expressed in amu are 2.015, 3.017 and 1.009, respectively. If 1g of deuterium undergoes complete fusion, then the amount of total energy released is 9 x 10n kJ. Find the value of n.

Δm=2(2.015)−(3.017+1.009)=0.004amu∴  Energy released =(30.004×931.5)MeV =3.726MeVEnergy released per deuteron                        =3.7262=1.863MeVNumber of deuterons in 1g                         =6.02×10232=3.01×1023∴Energy released per g of deuterium fusion=3.01×1023×1.863=5.6×1023MeV≈9.0×1010J≃9×109kJ≃9×10nkJ (given) ∴n=9