A heavy nucleus X of mass number 240 undergoes symmetric fission into two identical nuclei Y of mass number 120 . The energy released per fission is 240 MeV . If the binding energy per nucleon of the product nucleus Y is 8.5 MeV , the binding energy per nucleon of the parent nucleus X is:

Step-by-step solution

The energy released ( Q -value) in a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactants. Q = Total BE of products - Total BE of reactant Let the binding energy per nucleon of nucleus X be E_X . Total binding energy of the reactant X = 240 E_X Since two identical nuclei Y of mass number 120 are formed, the total binding energy of the products is: Total BE of products = 2 (120 8.5) = 2040 MeV Substitute the known values into the Q