# Why is the energy taken to free neutrons irrelevant to the energy released in a fission reaction?

Our teacher is currently teaching nuclear physics and I am confused on the process of calculating the energy released in a fission reaction. An example given in class is as follows:

Uranium-235 + 1n ---> Krypton-92 + Barium-141 + 3n

I understand that nuclear binding energy is the amount of energy needed to free all the nucleons in the nucleus, and the method taught to us is to take the binding energy of the krypton and barium nuclei and subtract the binding energy of the uranium nucleus to find the energy released.

My question is, why is the energy needed to free the 2 new free neutrons not also subtracted because they were previously bound in the uranium nucleus (at about 7.59 MeV)?

• Because there is no binding energy of one free neutron. It is just, well, a neutron. Apr 16, 2019 at 2:29
• Yes, of course, but were the neutrons not previously bound in the nucleus? Where is the energy needed to "unbind" them coming from? Apr 16, 2019 at 2:33
• The binding energy is what would be released if you started with all the protons and all the neutrons at infinity, then brought them together to make a particular nucleus. So, thought experiment: Start with 236 nucleons that make (U-235 + n) at infinity. Instead of (U-235 + n) make (Kr-92 + Ba-141 + 3n) instead. Each of Kr-92 and Ba-141 has its own binding energy (again derived from all the nucleons at infinity). Now see that an isolated neutron has no binding energy - it is already at the reference position (free neutron). Apr 16, 2019 at 12:40

In this kind of reaction you have two states: the one on the left, and the one on the right. The energy of a state doesn't depend on how it got there; it only depends on the state (this is analogous to how the gravitational potential energy of something doesn't depend on the path it took to get there; it only depends on the height).

So since free particles don't have binding energy, there's no further calculation involved.

• I see. So does that mean the question of where the neutrons get the energy is a nonsensical one? Apr 16, 2019 at 2:41
• I don't understand your question. The free neutrons in this reaction got their energy from the reaction, of course. But if I just gave you a neutron and ask you where it got its energy from, it'd be a nonsensical question. Apr 16, 2019 at 2:45
• Well, by taking the binding energy of Uranium-235, it is about 1784MeV. The combined binding energy of Krypton-92 and Barium-141 is 1957MeV. This means that about 173MeV is unaccounted for. The original state has 1 free neutron, while the new has 3, 2 of which were previously within the U-235 nucleus, with about 7.6MeV of binding energy each. My question is that the answer that is correct is 173MeV, but why is the energy needed to free those 2 electrons counted as "released" energy when it is used to free the neutrons and seemingly converted to mass? Apr 16, 2019 at 3:45
• I still don't understand I'm afraid. Why would it not be counted as "released" energy? The neutrons are flying away at high speed, why wouldn't that energy not be "released"? I don't see where the "converted to mass" is coming from either; mass is being converted to energy here. Apr 16, 2019 at 4:08

My question is that the answer that is correct is 173MeV, but why is the energy needed to free those 2 electrons counted as "released" energy when it is used to free the neutrons and seemingly converted to mass?

This 173Mev is released as the kinetic energy of the daughter nuclei as they fly apart. Usually in your case the fission is reached by adding a high energy neutron, thus first creating a peanut nucleus, at that point the EM repulsion is strong enough to make the two daughter nuclei fly apart at 0.3c.

AS the daughter nuclei fly apart, energy is released as their kinetic energy, and it is that 173Mev.

The daughter nuclei's rest mass is more then the original nucleus' rest mass. This is because to get the daughter nuclei as separate entities, you need to add extra energy to the original nucleus. This extra energy is needed to tear it apart.

To see that the separation energies of the free neutrons are accounted for, compute also the binding energy of the alternative final state

$$\to\rm Kr ^{93}+Ba^{131}+2n$$

and similar hypothetical states where some of the final neutrons are bound.