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Is it possible for two neutrons to 'bond'?

  1. Jan 18, 2010 #1
    I'm just wondering why it's important for atomic structures to contain protons. Hydrogen-3 has two neutrons and one proton. Could you have a particle with two or more neutrons held together by nothing but the strong nuclear force? Is the reason I don't know about this because such particles aren't interesting in chemical or nuclear reactions?
     
  2. jcsd
  3. Jan 18, 2010 #2
    Well, neutrons can be very important in atomic reactions.
    But you do bring up a good question.
     
  4. Jan 19, 2010 #3

    Vanadium 50

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    The di-neutron is known to be unbound.
     
  5. Jan 19, 2010 #4
    For light atoms or 'atoms', I believe protons tend to increase the binding energy. I don't know why this is or if it's even true. Deuterium is stable, Tritium is unstable. The binding energies for Deuterium and Helium-4 are both on the order of an MeV or two, I think, which compared to the masses of a free proton and neutron (and more importantly, their mass difference), means that these isotopes just barely form as is. I suspect that if one were to calculate the binding energy of two neutrons, it'd work out to be positive a few MeV or something like that, and thus wouldn't form (or would be incredibly unstable at reasonable temperatures). Take all of this with a grain of salt, cause I'm just guessing :)
     
  6. Jan 19, 2010 #5
    I just asked the question(which got deleted), why can't there be clusters of neutron atoms? I don't see anywhere in any current field theory that would prevent a stable neutron nucleus from forming.

    If there are no neutron clusters formed by decay, stars, accelerators..etc, then that would imply the binding energies/gluons would be a result of some difference between the neutron/proton relationship and the magic number quantum states.

    Is that already established or am I just behind the curve?
     
  7. Jan 19, 2010 #6
    I don't know about your question, TheNerf, but something else I forgot to point out is that free neutrons decay into protons by emitting an electron and an antineutrino. The halflife is short enough (~15 minutes) that I imagine most things comprised of just neutrons would eventually decay their way into the stable nuclei we know of. I know what's special about the stable nuclei is the binding energy is just enough that it means the neutron no longer decays (it's energetically favorable to remain in the bound state).
     
  8. Jan 19, 2010 #7

    bcrowell

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    It's fairly easy to show based on basic ideas of quantum mechanics that the most stable form of nuclear matter (for small A) has N and Z approximately equal. The reason is simply the Pauli exclusion principle. If I have 8 protons and 8 neutrons, I can put the 8 protons in the lowest 8 energy levels, and the 8 neutrons in the lowest 8 energy levels. If I have 16 neutrons and no protons, then the first 8 neutrons can go in the lowest 8 energy levels, and the next 8 have to go in the next 8 levels, with higher energies. That's enough to show that N=16, Z=0 isn't going to be a stable system. At best, it might beta decay to 16O.

    There's also the question of whether these systems would even be bound with respect to particle emission, in which case they could only be observed as resonances. Since there is no Coulomb barrier for neutrons, they don't have to tunnel out. If they've got enough energy to escape, they just do. Then it becomes a question of the strength of the nuclear force. The best experimental and theoretical work so far seems to show that N=2 is unbound, and N=6, 8, 10, ... are also pretty clearly unbound. There is still at least some question whether N=4 is bound.

    These people claimed to have detected the N=4 system, with a lifetime of at least ~100 ns, which means it would have to be bound, although not stable with respect to beta decay:

    http://arxiv.org/abs/nucl-ex/0111001

    Whether they're right is a whole different question. I wouldn't bet a six-pack on it.
     
  9. Jan 19, 2010 #8

    Vanadium 50

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    I'll go further - it's almost certainly wrong.

    The most straightforward thing to do is something called double charge exchange: you get a liquid or solid helium target and look for the reaction pi- + 4He -> (4n) + pi+. No evidence for a bound tetraneutron has been seen.
     
  10. Jan 19, 2010 #9
    No doubt this is the case experimentally. Two, four (or any number larger than one) of neutrons do not bind. The usual explanation is isospin: two neutrons will be in isospin=1 state, which does not bind, while the isospin= 0 does. This of course is a good rule of thumb that is consistent with all observed data, but it is no explanation. It is just a selection rule. In the end it should come from some fundamental computations in QCD. Maybe it does, but I am not aware of any published work on that. (I hope I am wrong, I will be the first to check and read it, if it exists.) Worse yet, there are no QCD calculations that I know of which explains why there are no six quark color singlet states. Of course, one might consider Deutorium a six quark color singlet, but it does not cut the muster. Because it is not really a six quark color singlet; it is a bound state of two more-or-less spatially separated three-quark color singlets. Why should that be more energetically favorable relative to one completely-fused six quark color singlet should fall out of QCD calculations, but it does not to my knowledge. We have no data that contradicts QCD when QCD is able to make a prediction, yet we have plenty of data which QCD cannot explain very easily and quantitatively. I think something as simple as a Deutorium or di-neutronium falls into that class.
     
  11. Sep 24, 2011 #10
    Two neutrons must have opposite spins according to pauli exclusion principle, therefore a particle which consists of two bound neutrons has 0 spin. The deuteron has spin-1 (because protons and neutrons are different and they can have the same spin). That means that in deuteron there is also nuclear tensor force (which works when the system has spin). The nuclear tensor force is much stronger than the strong nuclear force in the case of two nucleons. This is why dueteron is stable and a system with two protons or two neutrons is not.
     
  12. Sep 24, 2011 #11

    Vanadium 50

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    This thread is almost two years old.
     
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