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_(Sr)=129MeV, K_(xe)=86MeV`
B. `x=p, y=e^(-), K_(Sr)=129 MeV, K_(xe)=86 MeV`
C. `x=p,y=n,K_(Sr)=129 MeV, K_(xe)=86 MeV`
D. `x=n, y=n,K_(Sr)= 86 MeV, K_(xe)=129 MeV`

Answer

A

Explanation

`._(92)^(236)Urarr._(54)^(140)x+._(38)^(a_(4))Sr+._(0)^(1)y`
`x=y=n`
`Q=236xx7.5-(140xx8.5+94xx8.5)=1770-(1190+799)=219MeV`
In `A` and `D` energy and charge conservation is followed
So `Q=K_(xe)+K_(Sr)+K_(x)+K_(y)=129+86+4=219`
In `D`,
`p_(xe) gt p_(Sr)+p_(x)+p_(y)`
so convervation of momentum will not hold