A parent nucleus X of mass number 240 , initially at rest, undergoes asymmetric fission into two fragments Y and Z of mass numbers 80 and 160 respectively. The binding energy per nucleon for the parent nucleus X is 7.5 MeV , while for the fragments Y and Z it is 8.4 MeV and 8.1 MeV respectively. Assuming all the released energy appears as the kinetic energy of the fragments, the kinetic energy of the lighter fragment

Step-by-step solution

The total initial binding energy of nucleus X is, E_ X = 240 7.5 = 1800 MeV The total final binding energy of fragments Y and Z is, E_ Y + E_ Z = (80 8.4) + (160 8.1) = 672 + 1296 = 1968 MeV The total energy released ( Q -value) is, Q = 1968 - 1800 = 168 MeV By conservation of linear momentum, the two fragments move in opposite directions with equal magnitude of momentum p . The kinetic energy of a fragment of mass m is given by K = p^2 2m . Thus, the kinetic energy is inversely proportional to the mass. The ratio