A nucleus undergoes fission according to the simplified reaction The binding energy…

PHYSICS · Multi-part · A Level · Physics · Nuclear physics · Nuclear physics

Problem

A 236 ⁣U^{236}\!\mathrm{U} nucleus undergoes fission according to the simplified reaction 236 ⁣U→92 ⁣Kr+144 ⁣Ba^{236}\!\mathrm{U} \rightarrow {}^{92}\!\mathrm{Kr} + {}^{144}\!\mathrm{Ba} The binding energy per nucleon values are: Ebn(236 ⁣U)=7.6 MeVE_{bn}(^{236}\!\mathrm{U}) = 7.6\,\text{MeV} Ebn(92 ⁣Kr)=8.5 MeVE_{bn}(^{92}\!\mathrm{Kr}) = 8.5\,\text{MeV} Ebn(144 ⁣Ba)=8.3 MeVE_{bn}(^{144}\!\mathrm{Ba}) = 8.3\,\text{MeV} (i) Compute the total binding energy of the parent nucleus and of the two product nuclei. (ii) Hence determine the energy released in one fission event. (iii) Discuss what the calculated energy release indicates about the relative stability of the parent and product nuclei.

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