Why can we not produce a "giant" nucleus?

[A. Neumaier]

The $O(r^{-6})$ part of the van der Waals potential is the attractive portion of it (induced dipole-induced dipole). The repulsive portion is far more complicated, but an exponential works pretty well as an approximation.

The repulsive force between neutral atoms is the exchange force due to the Pauli principle. It grows for $r\to 0$ only like $r^{-1}$. An exponential is a traditional but incorrect approximation.

But I just realized that all this onöy applies to atoms and not to nuclei. It is still relevant for the formation of neutron stars since gravitation must overcome the repulsive force between atoms...

 

[Xforce]

But why in the Earth we only see a finite number of neutrons in atoms?What is the condition to exist a "giant" atom?

Actually, when gravity is holding them together instead of strong nuclear force (I’m talking about neutron stars) that’s the condition of having a giant atom... because gravity decays much slower than nuclear forces over distances. However, try not to make it too big because it will collapse and becomes a black hole.

 

[fxdung]

What is an exchange force as Prf Neumaier said?

 

[Vanadium 50]

https://en.wikipedia.org/wiki/Exchange_interaction

 

[Nik_2213]

Reminds me of an article on the Russian 'heavy isotope' factory, and their extraordinary work to push further and further beyond known nuclei, into 'super-heavy country'. Part of the problem is the lower periodic table's 'magic numbers' no longer hold. To put it politely, the nucleus becomes too big for neat 'shells' to hold sway. The hoped-for 'Island of Stability' may prove nothing more than a 'shoal'...

IMHO, chemical analogy may be gold, mercury and lead, whose nuclei are so massive they significantly alter the behaviour of their inner electron 'shells', the relativistic effects causing remarkable changes in properties. Metals, yes, but anomalous colour, melting point etc. IIRC, copper almost qualifies...

And, I learned about the 'drip line', beyond which lesser nuclei shed surplus neutrons they cannot stabilise...

 

[wacki]

Apart from the giant nucleus. Why is it not possible to have a 2 neutron bound state. (I'm not sure if that should be asked in a separate thread, but it is so closely related). For the 2 neutron bound state there is no electric repulsion and if they have opposite spin they could both sit in the ground state. Shouldn't that be extremely stable?

 

[Nik_2213]

Seems not. Nature can be inconvenient...
FWIW, see...
https://en.wikipedia.org/wiki/Tetraquark
https://en.wikipedia.org/wiki/Pentaquark

Could be said that, without a deutron's proton to supply the relevant 'glue', any di-neutron is a 'hexa-quark' which promptly falls apart to two single neutrons, which then decay in the usual way...

 

[vanhees71]

Already the nucleon-nucleon potential is a quite complicated beast. Particularly it's spin-dependent. Model calculations, e.g., based on effective low-energy effective descriptions of the nuclear two-body force (the most modern version is based on a systematic expansion using chiral effective theory; historically there are effective one-boson exchange models like the Walecka model too), show that the two-nucleon system has only one bound state, namely the deuteron with $S=1$ and total angular momentum $J=1$. That implies that the orbital angular momentum must be $L=0$ or $L=2$, and indeed the bound state is a mixture of both, dominated by $L=0$ but with a small admixture of the $L=2$ state to account for the non-vanishing electrical quadrupole moment.

Now consider two protons or two neutrons. Since they are identical fermions the only bound state possible for the pn system (the deuteron) is not allowed for these systems, because for them the two-body wave function must be antisymmetric. The spatial part and spin of the bound state is symmetric, and this cannot be for pp or nn. Thus neither the pp nor the nn system have any bound state.