The average energy released per fission for the nucleus of 23592U is 190 MeV. When all the atoms of 47 g pure 23592U undergo fission process, the energy released is 1023 MeV. The value of is . (Avogadro Number = 6 1023 per mole)
Answer
228
Solution
Number of moles = 47235 = 0.2
Number of nuclei = 0.2 6 1023 = 1.2 1023
Energy = 1.2 1023 190 MeV = 228 1023 MeV
= 228
Answer: 228