If the binding energy per nucleon in ₃ Li ^7 and ₂ He ^4 nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of p in the reaction p+ ₃ L i^7 2 ₂ H e^4 is

Step-by-step solution

The given nuclear reaction is p + ₃Li⁷ 2₂He⁴ . The total binding energy of the reactants is: Binding energy of p = 0 MeV (since it is a single nucleon). Binding energy of ₃Li⁷ = 7 5.60 MeV = 39.20 MeV . Total binding energy of reactants = 0 + 39.20 = 39.20 MeV . The total binding energy of the products is: Binding energy of 2₂He⁴ = 2 (4 7.06 MeV ) = 2 28.24 MeV = 56.48 MeV . The energy released ( Q -value) in the reaction is the difference between the total binding energy of the products and the total binding energ