Question

a) Calculate the energy released in the neutron-induced fission

n + ²³⁵U --> ⁹²Kr + ¹⁴²Ba + 2n

given the neutron is 1.008665 u, the mass of is ²³⁵ U 235.043924 u, the mass of ⁹²Kr is 91.926269 u, and the mass of ¹⁴²Ba is 141.916361 u. b) Confirm that the total number of nucleons and total charge are conserved in this reaction. The atomic mass is 1.66053904 X 10^-27 kg.

Answer 1

first, we find the mass defect

m = 141.916361 + 91.926269 + 1.008665 - 235.043924

m = - 0.1927 u

so,

energy released = m * 931.5 MeV

energy released = 179.53 MeV ( energy released per fission - you can put negative sign to indicate fission)

or

energy release = 7.4e10 J

---------------------------------------------------------

yes, charge is conserved

as we can see in the reaction

U (92) = Ba( 56) + Kr (36)

92 = 92