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Suppose you had a room full of neutrons at standard pressure and temperature. Would it be just like a room full of hydrogen?

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    $\begingroup$ Related: physics.stackexchange.com/q/17573 $\endgroup$ – dmckee Oct 28 '18 at 0:13
  • $\begingroup$ Putting aside the answers' observations that the walls of the room matter, if you could neglect that the neutrons would decay viz. $n\to p+e^-$, resulting in hydrogen that would quickly become diatomic. $\endgroup$ – J.G. Oct 31 '18 at 20:12
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Neutrons interact with matter through the strong nuclear force. The neutrons you're talking about are called thermal neutrons (because they're at thermal energies rather than the MeV energies typical of nuclear processes), and the cross-section for thermal neutrons to interact with many forms of matter are quite high. The interaction can be through elastic or inelastic scattering, or absorption. Therefore most of the neutrons would probably be absorbed or scattered by the walls within a fraction of a second (basically some small multiple of the time of flight across the room).

Since several of the answers seem to either overstate or understate the transparency of matter to neutrons, it may be helpful to give an example from a lab where I worked. We did accelerator experiments that produced quite a lot of neutrons (enough that it would show up on a dosimeter badge and would cause us to minimize time spent in the room when there was beam on target). For this reason, the target rooms had walls constructed from (IIRC) several feet of concrete, and the entrance hallways had twists and turns in them so that neutrons and gammas would have no direct line of flight out into the other areas, where people were working. This amount of shielding (along with the $1/r^2$ factor, I suppose) was enough to make the radiation dose negligible in the adjoining rooms. So yes, it requires pretty bulky shielding to shield against neutrons, but no, it's not true that they just pass through matter. Also, the neutrons in this lab were MeV neutrons, but thermal neutrons tend to interact more strongly.

Returning to your example, at least some of the absorbed or scattered neutrons would probably induce nuclear reactions that would release large amounts of energy. ("Probably" because it depends on the material.) Some of these reactions would probably be effectively instantaneous, e.g., reactions in which the neutron was absorbed by the nucleus but a proton was ejected. In any such instantaneous reaction, you would essentially be taking ~1 eV of energy from the thermal neutron, and immediately converting it to ~1 MeV (the nuclear energy scale). Thus the result would be an extremely energetic explosion, perhaps comparable to a small nuclear bomb. The exact result would depend sensitively on the material that the wall was built out of, because the cross-sections for such processes depend on the isotope.

The products of the explosion would also include a lot of material with delayed radioactivity, primarily $\beta^-$ and $\gamma$. So your room is essentially a "dirty bomb."

The basic insight to have here is that radioactivity is not something we normally interact with in macroscopic quantities such as Avogadro's number, and to do so would be extremely unhealthy.

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  • $\begingroup$ Wow! I figured it'd be extremely unhealthy, but I wasn't expecting it to cause an explosion. How does the danger from the radioactivity of free neutron decay compare to that of the absorption & scattering of the neutrons by the ambient matter in the room? I assume from the info on the page dmckee linked that with the right materials you could make a room that would act as a neutron bottle. But you can't stop that $\beta^-$ decay. $\endgroup$ – PM 2Ring Oct 28 '18 at 14:44
  • $\begingroup$ So dear Ben, you agree that they will not pass through. So my answer states basically the same. Though I will correct the EM interaction part, but I do not understand the downvotes. $\endgroup$ – Árpád Szendrei Oct 28 '18 at 23:06
  • $\begingroup$ A useful order-of-magnitude number in the context of this answer: a Hiroshima-type nuclear bomb generates roughly a mole of free neutrons. $\endgroup$ – rob Nov 23 '18 at 6:32
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A major problem with that scenario is that neutrons have a half-life of 11 minutes, so the heat generated would be colossal! - Possibly gigawatts ... or more even. And it's extremely hypothetical by reason also that neutrons diffuse into solid substances, and are absorbed by them - irreversibly as well, as they stick to the nuclei forming new isotopes, and are not ejected again ... rendering fabulous any notion of containing a room's volume of them as a gas.

Rate of temperature rise of gas of neutrons in K/s =

1.6 e-19 (electronic charge in C)

× 7.8 e5 (decay energy of neutron in eV)

× 6 e23 (number of particles in one mole)

/(8.3 ×1.5 (heat capacity of ideal gas with three degrees of freedom at constant volume in J/molK)

× 9 e2 (mean life of neutron in s ))

≈ 6.7 e6!!

So the temperature of this roomful of neutrons would rise at 6.7 million Kelvin per second ... hotter than the sun in a millisecond - basically a nuclear explosion!

I once noticed how in footage of nuclear explosions the fireball seems not to cool down for a good few tens of seconds. It requires really a very small amount of fission products (& fusion byproducts) having a half-life in the region of a minute to pump sufficient heat into it.

I found that interesting about ultracold neutrons, though - how below a certain - extremely low energy - they do not pass into matter & therefore might be contained by material walls.

There's also the question of neutron-neutron, neutron-proton, & neutron-electron scattering cross section. This might actually put the mean free path up by 10 or more orders of magnitude - to greater than the dimensions of a typical room - so that the products of the decays would instantly impinge on the walls, heating them, rather than the gas heating itself. It would be no less a nuclear explosion, though.

And actually maybe you wouldn't get a nuclear explosion - or atmost a much smaller one - by reason that the substance of the walls would be ablated off & enter the space of the room & mop all the neutrons up before this process would be able much to proceed. But that would depend on what the walls were made of - if it were a substance that has a high capture cross-section, and on capturing a neutron turns into a nuclide with higher decay energy and-or shorter half-life than that of free neutron, it could make the explosion very much bigger! If the cross-section were very small, it wouldn't even 'mop them up' atall.

And that would make containing these ultra cold neutrons difficult, and impose a requirement that it be exceedingly tenuous, coz the walls would be heated, & the neutrons are interacting with the walls. And it wouldn't need to be heated much atall before ceasing to be an ultracold gas, as the coldness requirement is extremely stiff (kT = 300 neV, apparantly).

Anyway, speculation seems to be spiralling here.

On balance, I think a roomful of neutrons as a gas is not atall practicable!

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The most obvious difference would be that the neutrons would tend to go straight through the walls of the room, unlike hydrogen atoms or molecules. Atoms have no overall charge, but they are negatively charged on the "outside" and positively charged on the "inside" (apologies for the over-simplification!) and therefore tend to repel each other when they are close together. Neutrons have no electric charge at all.

If the neutrons did interact with the walls of the room, the isotopes created either by fission or by neutron absorption would typically be unstable and radioactive.

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    $\begingroup$ "neutrons would tend to go straight through the walls of the room" Might be too strong. Neutrons interact by scattering off atoms when they come within residual strong force range of the nuclei, which makes the cross-sections smalls but not trivial. Moderately energetic neutrons are sufficiently penetrating that shielding them calls for more areal mass density than a typical interior wall, but not not by so much that you can simply ignore structure like that. And thermal neutrons get their penetrating power from the statistics of random walks: they hit stuff in the wall over and over again. $\endgroup$ – dmckee Oct 28 '18 at 0:27
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pressure and temperature

First thing to notice is that pressure doesn't apply for neutron since, as other people already explain, they do not interact (or very weakly) with each with the wall. (We need to remind that pressure exist because particle of gaz are bouncing against the wall of the room).

Also for the situation you describe temperature does not exist as the usual thermodynamic temperature since neutron are not interacting with each other and then will not converge toward Boltzmann distribution. But as temperature is related to kinetic energy we usually said that the kinetic energy of neutron are its temperature. But through the moderation process with a second particle (usually hydrogen) they can follow Boltzmann distribution. So for some special case we can say than the neutron gas has a define thermodynamic temperature (but hey are still not interacting with each other or this is very very...very unlikely so negligible)

Standard temperature is 290K which is 25 meV. This is the thermal neutron range. Thermal neutron are relatively slow and interact "a lot" with material since the cross section evolve as the inverse of the neutron velocity (but usual free neutron are in the "fast neutron" range ( = MeV range) since they are usually produce through nuclear process in nuclear reactor or proton beam on target) One can guide thermal neutron in guides of super-mirror (see Neutron Optic).

We can also mention the existence of Ultra Cold Neutron with energy below 300neV. Ultra Cold Neutron does not penetrate material (to be exact it depend of the so call Fermi potential of the material / 300nev is a high Fermi potential). Consequence : they can be store in material trap !!

Also free neutron is not stable with a lifetime of ~880s so a you will loose all your neutrons in few hours even if they are perfectly trapped in your room.

edit : neutron with high energy go through the wall "relatively" easily. What we are doing in nuclear reactor to shield against neutron is to surround the reactor with water which contain a lot of hydrogen. Neutron are slow down by elastics scattering on the hydrogen and then, since the cross section follow $1/v$, they are capture by the wall. But if they are note thermalise by the hydrogen (in the water) they can go through concrete easily since elastic collision on heavy atom does not slow them a lot (think about a ping pong ball on steel ball...). One can also add some high capture cross section element in concrete (Bore for exemple) to reduce the needed concrete.

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    $\begingroup$ We need to remind that pressure exist because particle of gaz are bouncing on each other and against the wall of the room The pressure of an ideal gas actually has nothing to do with collisions between particles. $\endgroup$ – Ben Crowell Oct 27 '18 at 23:04
  • $\begingroup$ Oups yes you are right I was of course thinking about pressure and temperature. I will correct it $\endgroup$ – Dalnor Oct 27 '18 at 23:11
  • $\begingroup$ I think the statement "The pressure of an ideal gas actually has nothing to do with collisions between particles." is wrong .atmo.arizona.edu/students/courselinks/fall11/atmo551a/… quote: : "Pressure arises from the multiple collisions the molecules of a gas have with the walls that contain the gas" in 2.a origin of pressure $\endgroup$ – anna v Oct 31 '18 at 16:45
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First, H atoms consist of a proton and an electron. A neutron does not have EM charge, is neutral.

The reason H atoms cannot go through thick walls is because of the electron could around their nucleus, and the wall's atoms have the same electron cloud around themselves. Now these electron clouds of the H atoms and the wall's atoms, will repel. The H atoms will pop off the atoms of the wall.

Now in the case of neutrons, you do not have a negative EM charge. The quarks building up the neutron do have EM charge. -1/3 and 2/3 that of the electron's EM charge.

And though the neutron itself is EM neutral, its constituent quarks do have EM charge, but the Em interaction is negligible here. Neutrons interact through the strong force, and when they reach the atoms of the wall, they will too pop off. They will not go through the wall (most of them will not).

It is a common misconception that the neutron is like a neutrino. Neutrinos do pass through the wall. Because the weakly interact with normal matter. Not because they are EM neutral. And neutrinos are elementary, pointlike. They usually do not even see the atoms of the wall, just the space inbetween.

Now neutrons are not elementary. They consist of quarks, and neutrons do have a size. Now neutrons are the size of order of 10^-15m, orders of magnitude smaller then the atoms. But what really matters is the spacing between the atoms. The spacing between the atoms of the wall might be much bigger then the neutrons, but the strong interaction will not let neutrons pass through.

It is all probabilities. As per QM, some of the neutrons will pass through, they will tunnel through. It is because of the Heisenberg uncertainty principle. But most of the will not.

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  • $\begingroup$ So I have two contradictory answers. Which is correct? $\endgroup$ – Stephen Montgomery-Smith Oct 27 '18 at 22:36
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    $\begingroup$ And though the neutron itself is EM neutral, its constituent quarks do have EM charge, and when they reach the atoms of the wall, they will too pop off. No, this is wrong. The interaction of neutrons with matter is due to the strong force. The electromagnetic interaction (such as magnetic dipole interactions) is negligible. $\endgroup$ – Ben Crowell Oct 27 '18 at 23:00
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    $\begingroup$ Now neutrons are not elementary. They consist of quarks, and neutrons do have a size. The spacing between the atoms of the wall and the EM charge of the electron clouds will not let neutrons pass through. No, the size of a neutron is ~10^-15 m, which is orders of magnitude smaller than an atom, and in any case it's not physical size that's relevant here. $\endgroup$ – Ben Crowell Oct 27 '18 at 23:00
  • $\begingroup$ @BenCrowell I edited what you commented. I still say that after your correct comments and the edits I made, my answer states the same as yours, that the neutrons will not pass through the wall (most of them will not). I do not understand the downvotes. $\endgroup$ – Árpád Szendrei Oct 28 '18 at 23:11

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