The energy released per fission of nucleus of ²⁴⁰X is 200 MeV. The energy released if all the atoms in 120g of pure ²⁴⁰X undergo fission i s _____ × 10²⁵ MeV. (Given Nₐ = 6 × 10²³)

Question

G1 Physics · Accepted Answer

Number of moles = 120/240 = 1/2. Number of nuclei = (1/2)Nₐ = 3 × 10²³. Total energy = (3 × 10²³)(200 MeV) = 6 × 10²⁵ MeV. Hence the required value is 6.