For One For All

Does a \(Ne\) nucleus exist?  Yes, the paired orbits all without opposite particles, each generates a weak \(g\) field that sums to the resultant represented by the dotted arrow.

At the negative end of this weak field, a \(g^{+}\) particle spins as the nucleus spins.  This is the start of the nucleus cyclic set (\(g^{+}\), \(T^{+}\), \(p^{+}\)), as the next layer to the nucleus.

All \(n\) solutions can have a resultant weak field extending beyond the nucleus.