A heavy nucleus of mass number 240 undergoes fission into two identical fragments, each of mass number 120 . If the binding energy per nucleon of the parent nucleus is 7.6 MeV and the energy released per fission is 216 MeV, the binding energy per nucleon of the fragment nuclei is:

Step-by-step solution

Let the binding energy per nucleon of the fragment nuclei be E_b . The total binding energy of the parent nucleus is: BE _ parent = 240 7.6 = 1824 MeV The total binding energy of the two product nuclei is: BE _ products = 2 (120 E_b) = 240 E_b The energy released in the fission process ( Q -value) is the difference between the total binding energy of the products and the reactants: Q = BE _ products - BE _ parent 216 = 240 E_b - 1824 240 E_b = 1824 + 216 = 2040 MeV E_b = 2040 240 = 8.5 MeV Answer: 8.5 MeV