Consider the fission reaction _ 92 ^ 236 U X ^ 117 + Y ^ 117 + n + n i.e. two nuclei of…
Physics · JEE Main · NTA Exams — Atoms and Nuclei
Consider the fission reaction \({ }_{92}^{236} \mathrm{U} \longrightarrow \mathrm{X}^{117}+\mathrm{Y}^{117}+\mathrm{n}+\mathrm{n}\)i.e. two nuclei of same mass number 117 are found plus two neutrons. The binding energy per nucleon of X and Y is 8.5 MeV whereas of U236 is 7.6 MeV. The total energy liberated is
- 2000 MeV
- 200 MeV
- 20 MeV
- 2 MeV
Answer
(B) 200 MeV
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