A fission reaction is given by `""_(92)^(236)U rarr_(54)^(140) Xe + _(38)^(94)St + x + y` , where `x` and `y` are two particle Consider `""_(92)^(236)U` to be at rest , the kinetic energies of the products are deneted by `k_(xe) K _(st) K _(s) (2MeV ) and ` repectively . Let the binding energy per nucleus of `""_(92)^(236)U,` `""_(54)^(140) Xe` and `""_(38)^(94)St` be `7.5` MeV , `8.4` MeV and `8.5` MeV, ` respectively Considering different conservation laws, the correct option (s) is (are)

Options

A. `x = n , y= n , K_(St) = 129 MeV , K_(xe) = 86 MeV `
B. `x = p , y= e^(bar), K_(St) = 129 MeV , K_(xe) = 86 MeV `
C. `x = p, y = n , K_(St) = 129 MeV , K_(xe) = 86 MeV `
D. `x = n , y= n , K_(St) = 86 MeV , K_(xe) = 129 MeV `

Answer

A

Explanation

`_(92)^(236)U rarr _(54)^(140) Xe + _(38)^(94)St + xn + y `
The number of proton in reactants is equal to the products (lesving x and y ) and nujmber of product (leaving x and y ) is two less then reactants
`:. X = p , y = e^(bar) ` is relied out [B] is increases and `x = p , y = n ` is rulled out [C] is increase
`total energy less then = (236 xx 7.5 ) - [ 140 xx 8.5 + 94 xx 8.5] = 219 MeV`
The energies of for and by together is `4 MeV`
The energy remain is disctrthused by sr and Xe which is equal in `219 - 4 = 215 MeV`
`:. `A is the currect option
also momentum is conserved
`:. K.E. prop (1)/(m) therefore K.E. _(m) gt K.E. _(we)`
The energy of ke by togather is `4 MeV`
The energy remain is distthuted by sr and Xe which is equal is `219 - 4 = 215 MeV`
A is the correct option
Also moomentum, is conserved
Therefore `K.E. gt K.E. _(xe)`