Let's say that, for some reason, producing a certain type of field causes all free neutrons with a kinetic energy greater than 100 kEV to spontaneously emit visible-spectrum photons and lose the corresponding amount of kinetic energy. Never mind how that field is produced; that's the one big lie of this fictional setting.

Would this emission of photons and loss of kinetic energy actually stop neutron-induced fission nuclear reactions - both neutron-induced nuclear fission detonations and neutron-induced nuclear fission reactions used for power - from occurring? Why?

It only effects free neutrons - that is, neutrons not bound to other particles.

  • $\begingroup$ basically all normal matter is largely composed of neutrons. $\endgroup$
    – John
    Commented Sep 8, 2021 at 19:30
  • $\begingroup$ @John Well, normal matter is made of neutrons and protons. Also, this field effects free neutrons - i.e. those that are not bound to normal matter but are flying around on their own due to a nuclear reaction. $\endgroup$
    Commented Sep 8, 2021 at 19:33

2 Answers 2


It would have next to zero impact on any present day nuclear fission technology

To oversimplify, in order for a neutron to cause a fissile atom to fission, it needs to be whizzing past the nucleus of that atom close enough to be captured via quantum mechanical processes. We measure the probability of capture by a value called "cross section" where the higher the cross section of an isotope, the more likely it will capture the neutron. Here is the fission cross section for selected actinides:

enter image description here

As you can see, the slower the neutron, the more likely it is for the nucleus to capture it and cause fission. An analogy is to imagine throwing a ping-pong ball (the neutron) at a slightly sticky bowling ball (the uranium nucleus). If you throw the ping-pong ball fast, it is more likely to bounce off, but if you gently toss it, it is more likely to stick.

Neutrons from fission are generally produced with energies around a couple of MeV. Because they are unlikely to interact with fuel atoms at this energy, we design reactors such that the neutrons will scatter off light atoms (usually hydrogen in water or carbon in graphite), hence slowing down enough to the point where they are likely to cause fission. These slowed neutrons generally have less than a couple of hundredths of an eV, well below the 100 keV limit of the field you propose. This is easily achieved with a few tonnes of water or graphite.

Some reactors are designed to use faster neutrons, perhaps in the 200-500 keV region. However, the change in cross section by slowing these down to 100 keV is pretty negligible. Consequently even nuclear reactors (or weapons) that rely on fast neutron fission probably won't be affected much.

In summary, if you brought your "neutron slowing field" into any modern nuclear powerplant, they probably wouldn't even be able to tell if you had turned it on without sensitive equipment.


Nope! In fact, it would be very useful for getting them to happen more controllably and reliably!

It would inhibit some types of nuclear reactions, because some types of nuclear reactions do in fact require fast neutrons to happen efficiently. But different isotopes have different neutron interactions cross section curves, with peaks at different energies, and getting a reactor to run is often a matter of slowing down neutrons to maximize their interaction potential. That's what a neutron moderator does.

So, you've just made it a lot easier to precisely control power-producing nuclear reactors. And if you intended it to be a defense, you've just put some new design constraints on the fuel structure for weapons intended to get past it.

  • $\begingroup$ This may be accurate only if the neutron-slowing field is uniform and creates, effectively, the new physics. However, if this field is applied only intermittently, all existing nuclear fission designs would stop working when it's applied. $\endgroup$
    – Alexander
    Commented Sep 8, 2021 at 20:05

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