You are given that mass   37Li=7.0160 u, Mass of    24He=4.0026  u, and mass of    11H=1.0079  uWhen 20g of   37Li is converted into  24He  by proton capture, the energy liberated, (in kWh), is :Mass  of  nucleon=1  GeV/c2

37Li+P→224HeMass defect=△M=Product-Reactant=(7.0160u)+(1.0079u)-(2×4.0026u)⇒△m=0.0187u for 20g of 37Li No. of nucleus=207×6×1023Total mass defect =ΔM=0.0187×207×6×1023 uTotal energy liberated=△E=ΔMc2⇒△E=ΔMc2=0.0187×1207×1023×c2×109×1.6×10−19c2J×1kWhr3.6×106J=1.4×106kWh