`U-235` is decayed by bombardment by neutron as according to the equation:
`._(92)U^(235) + ._(0)n^(1) rarr ._(42)Mo^(98) + ._(54)Xe^(136) + x ._(-1)e^(0) + y ._(0)n^(1)`
Calculate the value of `x` and `y` and the energy released per uranium atom fragmented (neglect the mass of electron). Given masses (amu) `U-235 = 235.044`,
`Xe = 135.907, Mo = 97.90, e = 5.5 xx 10^(-4), n = 1.0086`.

Explanation

The nuclear reaction is:
`._(92)U^(235) + ._(0)n^(1) rarr ._(42)Mo^(98) + ._(54)Xe^(136) + x_(-1)e^(0) + y_(0) n^(1)`….(i)
Equating the mass of the sides, we get
`235 + 1 = 98 + 136 + x 0 + y + 1`
`235 = 234 + y`
`236 = 234 + y`
`:. y = 236 - 234 = 2`
Similarly equating the atomic number of both sides,
we get
`92 + 0 = 42 + 53 + x xx (-1) + y xx 0`
`92 = 96 - x`
`:. x = 96 - 92 = 4`
Mass defect of Eq (i) = Masses o f`RHS -` Masses of `LHS`
`= 236.0526 - 235.8264`
`= 0.2262 "amu"`
`:.` Energy released `= mc^(2)`
or `E = m xx 931.48 MeV`
`= 0.2262 xx 931.48 = 210.7 MeV`