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How can I calculate the mean kinetic energy of an emitted neutron in a nuclear fission. Take for example the fission of U-235 to Ba-141 and Kr-92.

A calculation which just shows that the mean energy will be in the range of "fast" neutrons (> 1MeV) would be enough for me.

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    $\begingroup$ Use their mass excesses, calculate what energy you have left over and see if this fits within the range you want. (I would add more but I'm not sure if this is a homework question or not...it sounds suspiciously like one). $\endgroup$ – Lio Elbammalf May 24 '17 at 13:07
  • $\begingroup$ No this is not a homework question. I don't see how to do this using mass excesses. That approach would just give me the total energy released which is distributed to kinetic energy of the fragments, kinetic enregy of the neutron and gamma emission. But I don't see of how to get the energy of the neutrons. So it would be great, if you could post a detailed answer. $\endgroup$ – Julia May 24 '17 at 13:24
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    $\begingroup$ The final products consist of a number of particles; the daughter nuclei and a few neutrons. You can evaluate the total kinetic energy. How that kinetic energy is distributed has no unique answer as applying momentum conservation does not yield only one possible answer. You could assume that the daughters go off with the same speed in the same direction and assume that all (two or three or . . . ) neutrons go off in the opposite direction with the same speed but of course this is a highly unlikely event. So then try other scenarios. $\endgroup$ – Farcher May 24 '17 at 13:58
  • $\begingroup$ I know how the neutron spectrum looks like from measurements. I am just looking for a way to get the mean value from a theoretical estimation. $\endgroup$ – Julia May 24 '17 at 19:17
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This is an interesting problem because the naïve approach gives the wrong answer.

If you model the fission as a four- or five-body decay,

$$ \rm ^{236}U \to Ba + Kr + (\text{2 or 3})n, $$

where the initial $\rm^{236} U$ is at rest, the total momentum of all the decay products has to add up to zero and the total energy has to add up to the $Q$-value of the fission reaction. The most likely way to add up to zero momentum is for all the fragment momenta $p_\mathrm{Ba}$, $p_\mathrm{Kr}$, $p_\mathrm n$ to have roughly the same magnitude but to be directed in random directions in space. (Compare to a two-body decay, where the final momenta must have the same magnitude and opposite directions, or to a three-body decay, where the phase space for two products to have nearly parallel momenta is much smaller than the phase space for the angles between all the momenta to be large.)

If all the momenta have more or less the same magnitude, that predicts that the kinetic energies

$$ E_i = p_i^2 / 2m_i $$

should mostly be carried by the least massive decay products, the neutrons. The reality, however, goes the other way. That's a strong argument that the five-body decay model is wrong.

A more likely approach might be a series of two-body decays, such as \begin{align} \rm^{236}U &\to \rm {}^{142}Ba^* + {}^{92}Kr^* \\ \rm ^{142}Ba^* &\to \rm{}^{141}Ba + n \\ \rm^{92}Kr^* &\to \rm{}^{91}Kr + n \end{align} where the time it takes for the fission fragments to decay by neutron emission is comparable to the time required for the fission framgents to (partially) thermalize by interacting with the surrounding medium. In the literature (for example) this is known as "neutron evaporation" from the excited fission fragments. It's easy to check that the fission fragments will be born traveling at 3--5% of the speed of light, which gives them plenty of time for many thermalizing interactions with other nuclei in the fuel during the $\sim10^{-15}\rm\,s$ before the "prompt" neutrons are emitted.

So now your question becomes: why do neutrons which evaporate from an excited nucleus at rest have typical energies of a few MeV? That answer is much more boring: it's because a mega-eV is the typical spacing between nuclear energy levels. Alpha and beta decay energies are typically a few MeV for the same reason.

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I still won't plug the numbers in for you but I'll explain the principle.

You start off with your $^{235}$U and all the energy that is wrapped up in there will be our total (we can see how much we have to play with from the mass excess). Then, through fission, you get your $^{141}$Ba and $^{92}$Kr and two neutrons $^\dagger$ (this depends on whether you have spontaneous fission or neutron induced fission but you can adapt this as fits your situation). Now check how much energy you have left to play with from your original uranium using the basic rule:

$$ Q = \Delta(A) + \Delta(B) - \Delta(C) -\Delta(D) $$

Where your $\Delta(A)$ is the mass excess of some particle $A$ in the reaction $A+B \rightarrow C + D$. Your energy $Q$ is left to be distributed among your remaining products. This energy is then distributed as a Maxwell-Boltzmann PDF, the neutrons(being lightest) could take the majority of that energy and conserve momentum (since momentum depends on $v$ and kinetic energy on $v^{2}$) so you can approximate the maximum kinetic energy of the neutron as the $Q$ value.

$\dagger$ $235-92-141=2$

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  • $\begingroup$ Thanks, but it doesn't really help: Essentially you explain how to get the total energy (which is clear) and say that the neutrons take most of the energy which is wrong (measurements show thats about 2 MeV per neutron and about 170 MeV for the fragments). $\endgroup$ – Julia May 24 '17 at 19:16
  • $\begingroup$ @Julia I said the maximum is the approximately the $Q$ value, the MB distribution wouldn't indicate this is common. To work out the average you want to fit the distribution which has a maximum energy of your calculated $Q$ value and find where it peaks. $\endgroup$ – Lio Elbammalf May 24 '17 at 20:39

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