Saved Bookmarks
| 1. |
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) Xeand _(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) |
|
Answer» x=n, y=n, `K_"Sr"`=129 MEV, `K_"Xe"`=86 MeV `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 . |
|