n 235U 88Kr 146Ba 2 n a Calculate the energy released in

n + 235U 88Kr + 146Ba + 2 n

(a) Calculate the energy released in the neutron-induced fission reaction above, given m(88Kr) = 87.914447 u and m(146Ba) = 145.930107 u. Please show the math.

Solution

m(n)=1.008664u, m(235U)=235.043929u

So mass of reactants=236.052593u

mass of products= (87.914447+145.930107+2.017328)u=235.861882u

So lost mass=0.190711u

Energy released=0.190711*1..66*10^(-27)*c^2=2.8492*10^(-11)Kg m^2*s^(-2)=2.849*10^(-11) Joules