Question

how do you calculate the TOTAL energy that is released during thermal fission, of the reaction 235U+1n--->140Ba+98Kr+2n. Emphasis on total, meaning prompt and delayed neutron energy should be taken into account.

Answer 1

²³⁵U + ¹n ¹⁴⁰Ba + ⁹⁸Kr + ²n

atomic mass of ²³⁵U = 235.0439299 u

atomic mass of ¹n = 1.008665 u

atomic mass of ¹⁴⁰Ba = 139.910581 u

atomic mass of ⁹⁸Kr = 97.95191 u

atomic mass of ²n = 2.01733 u

The total mass of a product which will be given as -

m_products = m(¹⁴⁰Ba) + m(⁹⁸Kr) + m(²n)

m_products = [(139.910581 u) + (97.95191 u) + (2.01733 u)]

m_products = 239.879821 u

The total mass of a reactants which will be given as -

m_reactants = m(²³⁵U) + m(¹n)

m_reactants = [(235.0439299 u) + (1.008665 u)]

m_reactants = 236.0525949 u

The mass defect will be given by -

m = m_reactants - m_products

m = [(236.0525949 u) - (239.879821 u)]

m = 3.8272261 u

Therefore, the total energy released during thermal fission which will be given by =

E = m c²

E = (3.8272261 u) [(931.5 MeV/c²) / (1 u)] c²

E = 3565.06 MeV

Converting MeV to GeV :

E = 3.56 GeV