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 _(x) (2MeV ) and `Ky(2MeV) 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"`=129 MeV, `K_"Xe"`=86 MeV
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

`._92^236U to ._54^140Xe + ._38^94Sr + x+ y`
`K_x`=2 MeV , `K_y`=2 MeV , `K_"Xe"`=? , `K_"Sr"`=?
By conservation of charge number and mass number , `x-=y-=n`
B.E. per nucleon of `._92^236U`=7.5 MeV
B.E. per nucleon of `._54^140Xe`or `._38^94Sr`=8.5 MeV
Q value of reaction,
Q=Net Kinetic energy gained in the process `=K_(Xe)+K_(Sr)+2+2-0=K_(Xe)+K_(Sr)+4`...(i)
As number of nucleons is conserved in a reaction , so Q = Difference of binding energies of the nuclei
`=140xx8.5+9.4xx8.5-236xx7.5`=219 MeV ...(ii)
From Eqns.(i) and (ii)
`K_"Xe"+K_"Sr"`=219-4=215 MeV
Xe and Sr have momentum of same magnitude but in opposite directions.
Hence, lighter body has larger kinetic energy
So, from options,
`K_"Sr"`=129 MeV , and `K_"Xe"` =86 MeV
Hence, option (a) is correct .