Nucleus having only neutrons ?

  • Thread starter Sourabh N
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Main Question or Discussion Point

Can we have a nucleus containing neutrons alone?

Thanks in advance.
 

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  • #2
jtbell
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http://paul.merton.ox.ac.uk/science/administratium.html [Broken] :biggrin:
 
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  • #3
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No electron would like to be "around" such an electrically neutral nucleus. Then it cannot be an atom nucleus. But a big (very, very big) such an assembly exist as neutron stars, but it cannot be said that they are nuclei.
 
  • #4
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Thanks Ifpr. I too thought so but i don't think that's a correct answer to my question : What i know is that strong forces (short ranged) are dominant in nucleus ; in neutron stars i hope gravitational forces are dominant. Why can't i have a "something" of few fm long that has only neutrons held by strong forces ?
 
  • #5
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May be that the word "nucleus" in your post was a bad choice. I understand that your question was "are there linked states between neutrons?". I don't know the answer for sure, but I think that if there are such states, they are not stable. Even a single neutron outside a nucleus is not stable.
 
  • #6
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I think that if there are such states, they are not stable. Even a single neutron outside a nucleus is not stable.
You are correct. The ultimate reason why there is no such thing as a neutronic nucleus is that neutrons would turn into protons, which are lighter.

Also, as you have said, a neutron star is structurally different from a nucleus, since it is hold together by gravity and does not collapse thanks to the fact that neutrons are fermions. In fact, in the process of creation of a neutron star, inverse beta decay occurs where protons capture electrons and emit neutrinos, thus turning into neutrons.

The neutron star matter is very interesting for nuclear physicists, but is not a giant nucleus. :smile:
 
  • #7
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can u explain me why a neutron is so unstable outside nucleus?
 
  • #8
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can u explain me why a neutron is so unstable outside nucleus?
Because the neutron itself is heavier than the sum of the rest masses of the proton, the electron (and the (anti-)neutrino). Now of course, you could insist "but why is it heavier ?", since this is the true reason for the instability. Should I take the risk of going into the inifinite chain of whys ? :biggrin:
 
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  • #9
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No. I think thats enough.
 
  • #10
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No. I think thats enough.
You know, it is actually a very difficult (and thus interesting) question when you think about it. I'll let you figure out the numbers exactly, but the difference of rest mass between the proton and the neutron is about 1.3 MeV. The rest mass of the (anti-)neutrino is negligible, and the rest mass of the electron is about 0.5 MeV. If you think in terms of quark masses, [tex]m_d-m_u[/tex] could be amywhere between 0 and 7 MeV (or even -3 to 13 MeV...). It is remarquable that, although quark masses are so small compared to nucleons masses (about 1 GeV compared to a few MeV) the difference of the mass are not that off. Those considerations are the basis for chiral symmetry, and isospin (flavor) symmetries, which are historically of uttermost importance in the establishment of QCD.

Ultimately, a really satisfactory answer would involve masses, charges, and mix the electroweak with the strong sectors in a non-trivial (beyond the standard model) scheme. Such an "holly-graal" would possibly also either provide hints, or solve totally, the problems of (light quarks) confinement and chiral symmetry breaking.
 

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