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) = 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)^(236)U to ._(54)^(140)X_(e) +._(38)^(a_(4))Sr +._(0)^(1)x+._(0)^(1)y`
x = y = n , `Q = 236 xx 7.5 -(140 xx 8.5 +94 xx 8.5)`
= `1770 -(1190 +799) = 219 ` MeV
In A and D and 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 conservation of momentum will not hold .