Periodic Table and strong fields at high Z There is a known "phenomenon" recalled by Greiner et al. in several of his books related to hypothetical elements with Z>172: in some point arount this Z, the nuclear field is strong enough to pop out an electron-positron pair from vacuum. What does it mean? Are elements beyond that Z "possible" or are they unstable? How important are quantum mechanics, special relativity and quantum electrodynamics there?
In particular, does it mean that superheavy elements emit positrons at no cost from the continuum sea of negative energy?
 A: The idea that a sufficiently strong DC electric field can cause electron-positron creation goes all the way back to the early papers on antimatter by Dirac.  The original interpretation of the Dirac equation was that free electrons can occupy states with both positive and negative energies, and that to achieve the vacuum we observe there must be a Dirac sea of electrons filling up all the negative-energy states everywhere.  In this model a uniform electric field makes the zero-energy surface of the sea vary with position.  It's a pretty straightforward one-dimensional QM problem to find the rate at which negative-energy electrons begin to tunnel out of the sea towards the positive electrode, leaving positively-charged "holes" to move towards the negative electrode, for a given field strength; the result agrees with modern approaches for calculating field emission.  Unfortunately I haven't looked at that for ages and don't have a reference handy.
So electron-positron pair creation at strong electric fields is not unique to the problem of superheavy nuclei.
Now, if you did have a nucleus whose surface field were this strong, would it emit electrons and positrons?  Suppose that it did.  The positrons would be repelled from the nucleus and fly away to infinity.  The electrons would try their hardest to stay inside the nucleus and reduce the total field strength.  But the wavefunction for an electron that overlaps with a nucleus is an atomic $s$ orbital; those are big, and there aren't very many of them.  I think that once the atom was neutral, the nucleus couldn't emit pairs any more, since there would be no final state for the electrons to occupy.
In real heavy nuclei, like uranium, the electromagnetic repulsion wins out over the strong force.  The attractive part of the potential for two nucleons goes like 
$$
V = \frac{\alpha}{r}  e^{-m_\pi r }
$$
where $m_\pi$ is the mass of the pion.  This is essentially a contact force, since the pion's range is only about a femtometer.  A nucleus with $A\approx 240$ has a diameter of about 7 fm, so a substantial fraction of the protons see each other's electric charge but see no strong attraction to each other at all.  This means in practice that the heavy nuclei have nonspherical ground states ($^{238}$U is cigar-shaped) and have finite probabilities to emit alpha particles or to spontaneously fission.  That would be the fate of your nucleus with $Z=170$.
In super-super heavy nuclei, where gravity is strong enough to prevent particle escape, the charge density is reduced when electrons and protons interact weakly to form neutrons.  This is of course why we talk about neutron stars.
Whether there is relatively stable nuclear matter between the island of stability around $Z=90$ and the neutron stars is an open research question.
