A fission reaction is given by `._(92)^(236)U rarr ._(54)^(140)Xe + ._(54)^(140)Xe + ._(38)^(94)Sr + x +y`, where `x` and `y` are two particles. Consi
A fission reaction is given by `._(92)^(236)U rarr ._(54)^(140)Xe + ._(54)^(140)Xe + ._(38)^(94)Sr + x +y`, where `x` and `y` are two particles. Considering `._(92)^(236)U` to be at rest, the kinetic energies of the products are denoted by `K_(xe),K_(Sr),K_(x)(2MeV)` and `K_(y)(2Me V)`, respectively. Let the binding energies per nucleon of `._(92)^(236)U, ._(54)^(140)Xe` and `._(38)^(94)Sr` be `7.5 MeV, 8.5 Me V` and `8.5 MeV`, respectively. Considering different conservation laws, the correct options `(s)` is (are)
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
1 Answers
Correct Answer - a
`._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 .