Question

Consider the fission 23892U of by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are 14058C and 9944Ru. Calculate Q for this fission process. The relevant atomic and particle masses are m(23892U) =238.05079 u m(14058) =139.90543 u m(9944Ru) = 98.90594 u

Answer 1

Dear Student,

◆ Answer -

Q-value = 230.97 MeV

◆ Explaination -

Fusion of 92U238 takes place aa follows -

92U238 + 0n1 --> 58Ce140 + 44Ru99 + Q

Mass defect is calculated by -

∆m = [m(92U238) + m(n)] - [m(58Ce140) + m(44Ru99)]

∆m = (238.05079 + 1.00867) - (139.90543 + 98.90594)

∆m = 0.24809 amu

Q-value is calculated by -

Q-value = ∆m × 931 MeV

Q-value = 0.24809 × 931 MeV

Q-value = 230.97 MeV

This process is energetically allowed as Q-value for this reaction is positive.

Thanks dear...