A heavy nucleus of mass number 240 undergoes symmetric fission into two identical fragments. The binding energy per nucleon of the parent nucleus is 7.6 MeV . If the total energy released in the fission process is 192 MeV , the binding energy per nucleon of the product fragments is:

Step-by-step solution

The total initial binding energy of the parent nucleus is, E_ i = 240 7.6 = 1824 MeV Let the binding energy per nucleon of the product fragments be E_ f . Since the fission is symmetric, the total mass number of the products is 240 . The total final binding energy is, E_ final = 240 E_ f The energy released ( Q -value) is the difference between the final and initial total binding energies: Q = E_ final - E_ i 192 = 240 E_ f - 1824 240 E_ f = 2016 E_ f = 2016 240 = 8.4 MeV Answer: 8.4 MeV