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Consider the case of bombardment of U^(2...

Consider the case of bombardment of `U^(235)` nucleus with a thermal neutron. The fission products are `Mo^(95)` & `La^(139)` and two neutrons. Calculate the energy released by one `U^(235)` nucleus. (Rest masses of the nuclides are `U^(235)=235.0439 u, ._(0)^(1)n=1.0087u, Mo^(95)=94.9058 u, La^(139).9061u)`.

Text Solution

Verified by Experts

The correct Answer is:
`[M_(U)+m_(n)-M_(La)-2m_(n)]931=207.9 MeV`
`U^(235)+n rarrM_(0)^(95)+La^(139)+2n+Q`
`Q=(m_(u)+m_(n)-m_(M_(0))-m_(La)-2m_(n)).c^(2)`
`(235.0439+1.0087-94.9058-138.9061-2xx1.0087)xx931 MeV.`
`207.9 MeV`.
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