Consider the nuclear reaction  3Li7+1H1→2 2He4 . The relevant atomic masses are  7Li=7.018 amu, 1H=1.0080, 4He=4.004 amu . 0.5 gms of  1H1  completely consumed in the reaction along with sufficient amount of  7Li . Find energy released.

Mass Defect =Reactants - Products ⇒ΔM=7.018+1.008 -2(4.004) ⇒ΔM=0.018 amu Energy released per atom= ΔMc2=0.018×1.67×10−27×3×1082=2.7×10−12J  Now, in reaction hydrogen is limiting reagent. ∴Moles of hydrogen consumed =massMolar mass=0.51=0.5 mol Number of atoms of hydrogen consumed =moles×NA =0.5 ×6.023×1023= 3.0115×1023  Hence, energy released in reaction =  3.0115×1023 ×2.7×10−12=8.1×1011J