A heavy nucleus X of mass number 240 undergoes symmetric fission, splitting into two identical fragments Y . If the binding energy per nucleon of the parent nucleus X is 7.6 MeV and that of the product nucleus Y is 8.5 MeV , the total energy released in this fission event is :

Step-by-step solution

Total binding energy of the parent nucleus X is E_X = 240 7.6 MeV . Since X splits into two identical fragments Y , the mass number of each fragment is 240 2 = 120 . Total binding energy of the two product fragments is E_Y = 2 (120 8.5) = 240 8.5 MeV . The energy released ( Q ) is the difference between the total binding energy of the products and the reactant: Q = E_Y - E_X Q = 240 8.5 - 240 7.6 Q = 240 (8.5 - 7.6) = 240 0.9 = 216 MeV . Answer: 216 MeV