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a) Calculate the energy released in the neutron-induced fission n + 235U --> 92Kr + 142Ba...

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.

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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

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