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The .(92)U^(235) absorbs a slow neturon ...

The `._(92)U^(235)` absorbs a slow neturon (thermal neutron) & undergoes a fission represented by `._(92)U^(235)+._(0)n^(1)rarr._(92)U^(236)rarr._(56)Ba^(141)+_(36)Kr^(92)+3_(0)n^(1)+E`. Calculate:
The energy released when `1 g` of `._(92)U^(235)` undergoes complete fission in `N` if `m=[N]` then find `(m-2)/(5)`. `[N]` greatest integer
Given
`._(92)U^(235)=235.1175"amu (atom)"`,
`._(56)Ba^(141)=140.9577 "amu (atom)"`,
`._(36)r^(92)=91.9263 "amu(atom)" , ._(0)n^(1)=1.00898 "amu", 1 "amu"=931 MeV//C^(2)`

Text Solution

Verified by Experts

The correct Answer is:
`4`
(i) `E=[M_(U)+m_(n)-M_(Ba)-M_(kr)-3m_(n)]931=200.57 MeV` (ii) `(N_(A))/(235)xxE=22.84 MWh`
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