# Find the Energy release in this Nuclear Reaction (Fission)

Plutonium undergoes nuclear fission according to the equation below:

The value of $x=3$. The question is to estimate the energy released in this reaction. I know the Binding Energy per nucleon ($B$) of the all the reactants and the Products. So at first glance, I wrote down the following expression:

$E_{released}=E_{reactants}-E_{products}$

Now $$E_{reactants}=239*B_{pu}+1*B_{n}$$ $$E_{products}=91*B_{sr}+146*B_{ba}+3*B_{n}$$

When I computed the expression I ended up with a negative expression which is definitely incorrect. In the answer, the author has not accounted for the neutrons' Binding Energy. Why is this so?

@Lewis suggested that the neutrons are free particles. How does one go on to determine which particle is free or not?

• Because the original neutron and the final 3 neutrons are not bound. They are free particles. – Lewis Miller Apr 9 '16 at 18:43
• Could you please elaborate.. – model_checker Apr 9 '16 at 18:48
• Binding energy is the difference between the rest energy ($mc^2$) of a composite particle and the sum of the rest energies of its constituents. Although neutrons are composites (made of 3 quarks), they cannot be decomposed into 3 free quarks, s0 when not bound in a nucleus they (neutrons) are free particles. In the fission reaction you describe, the 3 (approximately) neutrons in the final state were part of the 239-Pu nucleus prior to the reaction, but in the final state they are free neutrons. – Lewis Miller Apr 11 '16 at 14:14
• So consider the following reaction: $N^{14}_7+He^{4}_2 ---->O^{17}_8+H^1_1$. Here Nitrogen, Helium, Oxygen can be decomposed and Hydrogen can be decomposed into a proton and an electron. So in this case to find the minimum energy required for the reaction to take place one must account for all the rest energies since there are no free particles. Is my reasoning correct? – model_checker Apr 11 '16 at 17:23
• Technically correct, but the binding energy of the electron to the proton is very small compared to the nuclear binding so you would not be far wrong to treat the proton as a free particle. – Lewis Miller Apr 13 '16 at 12:58