Why Can't We Construct a Particle Consisting Only of Neutrons?

[Hak]

Why can't we construct a particle consisting of only neutrons? Why are there nuclei of $_2^3He$ (2 protons and 1 neutron) and $_2^4He$ (2 protons and 2 neutrons) but there is no $_2^{50}He$?

Maybe a first approach might be that neutrons and protons interact with each other not through the electromagnetic force, since neutrons are neutral, but through the strong nuclear force.

What ideas do you have on this? Can you provide some detailed and in-depth information?

Feel free to move the thread to the more appropriate forum if this one is not. Thank you.

 

[Vanadium 50]

Why can't we construct a particle consisting of only neutrons?

Because such a state isn't bound.
We can't make what doesn't exist.

 

[Hak]

Because such a state isn't bound.
We can't make what doesn't exist.

Thanks. Why isn't it bound? I thought there was a more specific answer to my question related to Quantum Mechanics, the strong nuclear force, etc...

 

[OmCheeto]

I liked your question and upon googling I found a brief synopsis:

Four neutrons together momentarily​

Momentarily in their case being ≈10 ^-22 seconds.

As best I can tell, the following is the answer to your question:

Nuclear forces are essentially identical between all nucleons, whether they are protons or neutrons. So it might seem strange that the tetraneutron is not bound but that the α-particle of two protons and two neutrons is strongly bound, despite the additional electrical repulsion between protons. The explanation is based on the Pauli exclusion principle, which forbids two identical nucleons from occupying the same quantum state.

 

[snorkack]

I am not sure whether there is a simple first principles proof but there might be.
Starting from the smallest nuclei: two nucleons. The only bound state of dinucleon is paradeuteron - no bound excited states for that. Orthodeuteron is a virtual state, which is easily observed by neutron scattering. Orthodineutron and orthodiproton are unbound for the same reasons that orthodeuteron is, orthodiproton also for Coulomb repulsion. And paradineutron and paradiproton are banned by Pauli rule.
The thing is that when you have too many neutrons, they have to occupy higher energy states - and unless there are enough protons, these high energy states are higher than the unbound state.
It also helps if the neutrons occupy states of opposite spins - but only to a limit.
So, dineutron is accounted for (unbound), tetraneutron also unbound. What are the observed and predicted lifetimes for tri-, penta- and hexaneutrons?
For H, there are 4 isotopes known beyond T. All are unbound... and yet after H-5 their lifetimes increase:
H-4 139 ys
H-5 86 ys
H-6 294 ys
H-7 652 ys
He has a pattern of paired isotopes:
He-5 602 ys (758 keV)
He-6 bound (807 ms)
He-7 2510 ys (182 keV)
He-8 bound (120 ms)
He-9 2500 ys
He-10 260 ys (1760 keV)
So an alpha particle will not bind one neutron (it scatters off) but it will bind one pair, or two pairs. But not three pairs.
Li-9 binds 1 pair - Li-11 8,7 ms, Li-13 confirmed unbound
Be-12 binds 1 pair - Be-14 4,5 ms, Be-16 confirmed unbound
Already for B, the lifetimes of the less stable isotopes - B-16, B-18, B-20 and B-21 are given as bounds only... wrong magnitude (for B-18), wrong sign (B-16, B-20, B-21).
The neutron dripline is generally expected to go on. But how good are first principles calculations at predicting the exact position of dripline in terms of widths and cross-sections of nuclei on both sides?

 

[Hornbein]

Going the other way there is short lived nitrogen-9 with two neutrons (usual is nitrogen-14 with seven).

https://physics.aps.org/articles/v16/186

 

[berkeman]

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