The energy released per fission of nucleus of X 240 is 200 MeV . The energy released if all the atoms in 120 g of pure X 240 undergo fission is _____ × 10 25 MeV . (Given N A = 6 × 10 23 )

Step-by-step solution

Energy released per fission is 200   MeV . The number of atoms n in 120   g : n = 120 240 × 6 × 10 23   atoms . Total energy released, E = n × 200 × 10 6 = 6 × 10 25   MeV