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If a neutron is alone byitself, will it be able to bind electron(s)?


Cordor

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Yes it would. Both particles have mass, therefore would be attracted to each other by the gravitational force. This cannot be observed experimentally due to the Earth's magnetic field; but probably does occur naturally in deep space.

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there are magnetic fields everywhere

 

 

Including near a neutron, which has a magnetic moment.

 

It is left as an exercise to find the gravitational energy levels of an electron and a neutron in a Bohr-atom-like system to quantify the validity of the claim that the system wouldn't form.

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I heard a neutron will attract another neutron. Is this true?

 

Yes, but there's some fine print. (but I'll use regular-size print to try and explain what I know)

 

Nucleons attract via the strong nucelar force, but two neutrons together are unstable, as there is an empty proton shell, and so you would get beta-minus decay if you could form the system. And there is an issue that in order to form a bound system you have to release energy, and I think that there are no states available to the di-neutron system. The neutrons would have to have anti-aligned spins, since the Pauli exclusion principle forbids them having the same spin in the same energy state. I think the situation is that the energy of the antialigned state is in the continuum. (very vague recollections here, so take this with a few units of NaCl) Anyway, it's true thet you don't get n-n (nor p-p) bound states forming. So you can't even get the system to the point where it would decay anyway.

 

However, once you get to systems that have both neutrons and protons so you can get around some of these issues and have multi-particle attractions, neutrons attract each other just fine. This is why you tend to get more neutrons than protons in stable nuclei heavier than the first couple of rows of the periodic table - the attraction of the neutrons, without that inconvenient repulsion of the protons.

 

And if Severian shows up to give a more detailed explanation, rather than my "approximately correct" summary, maybe he can explain isospin to you.

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Free neutrons only have a mean life of about 15 minutes before they decay into protons. So to your question I say yes, if you wait a little while.

 

True only if there is an external magnetic field present. Two neutrons alone in infinity are not in a magnetic field, but mass distorts space creating gravitation with an infinite range. Weak it may be, but it is the only external force and will therefore draw the neutrons together.

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Free neutrons only have a mean life of about 15 minutes before they decay into protons. So to your question I say yes' date=' if you wait a little while.

 

True only if there is an external magnetic field present. Two neutrons alone in infinity are not in a magnetic field, but mass distorts space creating gravitation with an infinite range. Weak it may be, but it is the only external force and will therefore draw the neutrons together.[/quote']

 

All the neutron would need would be an ever so slight nudge and it will have "escape velocity", meaning that although gravity may always slow it down, it will limit towards a particular speed that ensures it will head away forever.

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(quoted from TM) Free neutrons only have a mean life of about 15 minutes before they decay into protons. So to your question I say yes' date=' if you wait a little while.

 

---

 

True only if there is an external magnetic field present. Two neutrons alone in infinity are not in a magnetic field, but mass distorts space creating gravitation with an infinite range. Weak it may be, but it is the only external force and will therefore draw the neutrons together.[/quote']

 

Why does an external magnetic field need to be present for the neutron to decay?

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Why does an external magnetic field need to be present for the neutron to decay?

 

Within an atom neutrons can be said to be in a neutral field created by the presence of equal numbers of electron and protons. If neutron nuclear stability is attributed to the nuclear strong force it is necessary to explain why free neutrons decay and free protons do not. But if nuclear neutron stability is attributed to neutral charge, then it can be said that neutral particles (with mass) decay in a charged field and charged particles do not. Keep in mind that all our experiments are conducted within the Earth's magnetic field. and that it is questionable whether any portion of the universal field is completely free of magnetic charge.

Charged particles within an atom cannot primarily be considered to be in a neutral field but are primarily paired with an opposite charged particle to create a neutral field.

I cannot explain the case for neutral particles without mass, without proposing a new interpretation; this is not the proper place to do that.

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Within an atom neutrons can be said to be in a neutral field created by the presence of equal numbers of electron and protons. If neutron nuclear stability is attributed to the nuclear strong force it is necessary to explain why free neutrons decay and free protons do not. But if nuclear neutron stability is attributed to neutral charge' date=' then it can be said that neutral particles (with mass) decay in a charged field and charged particles do not. Keep in mind that all our experiments are conducted within the Earth's magnetic field. and that it is questionable whether any portion of the universal field is completely free of magnetic charge.

Charged particles within an atom cannot [b']primarily[/b] be considered to be in a neutral field but are primarily paired with an opposite charged particle to create a neutral field.

I cannot explain the case for neutral particles without mass, without proposing a new interpretation; this is not the proper place to do that.

 

Neutron decay is adequately explained by the weak interaction. The only external effect that appears to change nuclear decay rates is pressure in the case of electron capture.

 

I'm sure you could go through the literature and find measurements made in different magnetic fields.

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