What would be the energy required to dissociate completely 1 gram of Ca-40 into its constituent particles?Mass of proton = 1.007277amuMass of neutron — 1.00866 amuMass of Ca-40 = 39.97545amu (Take one amu = 931 MeV)

Mass defect∆m=20(1.007277+1.00866)-39.97545        =40.31874-39.97545        =0.34329 amu∴Binding energy =0.34329 x 931 = 319.6 MeVWhen one atom of Ca-40 completely dissociates, the energy to be supplied =319.6 MeV.Number of atoms in 1 g of Ca-40 = 6.023 x 102340                                                     =1.506 x 1022The emergy required for the dissociation of 1 g of Ca-40=319.6 x 1.506 x 1022 =4.813 x 1024 MeV