Question

7. What is the energy released in the fission reaction    0¹n + 92²³⁵U ? 55¹⁴¹Cs + 37⁹²Rb + 3  0¹n? (The atomic mass of ¹⁴¹Cs is 140.920046 u and that of ⁹²Rb is 91.919729 u)   MeV

Answer 1

  0¹n + 92²³⁵U ? 55¹⁴¹Cs + 37⁹²Rb + 3  0¹n

The atomic mass of ¹⁴¹Cs is m =140.920046 u

The atomic mass of ⁹²Rb is m ' = 91.919729 u

Atomic mass of 235 U is M = 235.0439299 u

Mass of neutron M ' = 1.008665 u

Mass defect m = ( M + M ' ) - (m+ m' + 3 M ' )

                         = M - (m + m ' + 2 M ' )

                        = 0.1868249 u

We know 1 u = 931.5 MeV

So, energy released in the fission reaction E = m x 931.5 MeV

                                                                 = 174.027 MeV