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)=129MeV, K_(Xe)=86MeV`
C. `x=p, y=n, K_(Sr)=129MeV, K_(Xe)=86MeV`
D. `x=n, y=n, K_(Sr)=86MeV, K_(Xe)=129MeV`

Answer

A

Explanation

`._(92)^(236)Uto._(54)^(140)Xe+._(38)^(94)Sn+._(0)^(1)X+._(1)^(0)Y`
`x=y=n`
`Q=(140xx8.5+9.4xx8.5)-236xx7.5`
`=(1190+799)-1770=219MeV`
As `Q=K_(Xe)+K_(Sr)+K_(x)+K_(y)`
`=129+86+4=219MeV`.
In option (a) and (d), energy and charge conservation are followed.
In option (d), `p_(Xe)gt(p_(Sr)+p_(x)+p_(y))`
Therefore, conservation of momentum is not followed. Hence option 'a' is correct.