A stationary nucleus of mass number 200 decays into two fragments of mass numbers 160 and 40 . The binding energy per nucleon for the parent nucleus is 7.6 MeV, while for the heavier and lighter fragments it is 8.1 MeV and 8.6 MeV, respectively. The kinetic energy of the lighter fragment is:

Step-by-step solution

First, we calculate the Q -value of the decay process. Total binding energy of the parent nucleus: BE _ parent = 200 7.6 = 1520 MeV Total binding energy of the fragments: BE _ fragments = (160 8.1) + (40 8.6) = 1296 + 344 = 1640 MeV The energy released ( Q -value) is: Q = 1640 - 1520 = 120 MeV This energy is shared as kinetic energy between the two fragments. By conservation of linear momentum, the momenta of the two fragments are equal in magnitude ( p ). The kinetic energy of a fragment of mass m is given by K =