When slow neutrons are incident on a target containing `._(92)U^(235)` , a possitle fission reactionis
`._(92)U^(235)+n rarr ._(56)Ba^(141)+_(36)Kr^(92)+3n+Q`
Estimate the amount of energy released using the following data :
`M[._(92)U^(235)]=235.04u,`
`M[._(56)Ba^(141)]=140.91u,`
`M[._(36)Kr^(92)]=91.926u,`
`M_(n)=1.0087u.`
Take `1u=1.661xx10^(-27)kg.`
`1MeV=1.602xx10^(-13)J`

In equation ._(92)U^(235) + ._(0)n^1 to ._(56)Ba^(144) + ._(36)Kr^(89) + X : X is-

The reaction ._(92)U^(235) + ._(0)n^(1) rarr ._(56)Ba^(140) + ._(36)Kr^(93) + 3 ._(0)n^(1) represents

Complete the reaction n + ._92^235 U rarr ._56^144+….+3 n .

When slow neutrons are incident on a target containing underset(92)overset(235).U , a possible fission reaction is underset(92)overset(235).U+underset(0)overset(1).nrarr underset(56)overset(141).Ba+underset(36)overset(92).Kr+3_(0)^(1)n Estimate the amount of energy released using the following data Given, mass of underset(92)overset(235).U=235.04 amu , mass of underset(0)overset(1).n=1.0087 amu , mass of underset(56)overset(141).Ba=140.91 amu , mass of underset(36)overset(92).Kr=91.926 amu , and energy equivalent to 1 amu=931 MeV.

In the nuclear chain ractio: ._(92)^(235)U + ._(0)^(1)n rarr ._(56)^(141) Ba + ._(36)^(92)Kr + 3 ._(0)^(1)n + E The number fo neutrons and energy relaesed in nth step is:

Estimate the amount of energy released in the nuclear fusion reaction: _(1)H^(2)+._(1)H^(2)rarr._(2)He^(2)+._(0)n^(1) Given that M(._(1)H^(2))=2.0141u, M(._(2)He^(3))=3.0160u m_(n)=1.0087u , where 1u=1.661xx10^(-27)kg . Express your answer in units of MeV.