Why are synthetic elements unstable?

So far 20 synthetic elements have been synthesized. All are unstable, decaying with half-lives between years and milliseconds.

Why is that?

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It's kind of the other way around...they are only available after synthesis because they are sufficiently unstable. If they were stable we might be able to lay hands on some naturally occurring samples... –  dmckee Oct 15 '12 at 20:29
@dmckee You should post as an answer - I was going to post, but I don't know that I could add much more. –  Chris White Oct 15 '12 at 21:31
@Chris Well, there is room for another kind of answer: one detailing why heavy isotopes tend towards every lower binding energies. I would like to see one on the site because it's an interesting story. Alas, I'd have to do some serious boning up to write it. –  dmckee Oct 15 '12 at 23:27
Thanks everyone. –  SparKot ॐ Oct 24 '12 at 6:58

Protons are positively charged, and neutrons are neutral, so large nuclei are highly positively charged. A postively charged sphere will energetically prefer to break up into two separate charged droplets which move far apart, this reduces the electrostatic energy, since the electrostatic field does work during this process.

This thing, spontaneous fission, is usually phase-space unlikely, since you need to have a large chunk of the nucleus tunnel away from another large chunk, and it's unlikely for all those particles to tunnel out together. But at large atomic numbers, you are unstable even just to shooting out an alpha-particle, and this doesn't require a conspiracy, so large Z nuclei are alpha unstable, usually with long half-lives.

The positive charge on nuclei puts a limit to the stable ones. The reason is simply that the electrostatic force is long range, while the cohesion force is short range. The same phenomenon causes the instability of water droplets, so that if you charge one up, it will break into a fine mist. The cohesion of the droplets is local, while the electrostatic repulsion is long range.

The scale at which you get a fission instability directly can be estimated from surface-tension considerations. If you break a sphere into two adjacent spheres of same total volume, the radius is reduced by the cube-root of two, so that the surface area is decreases by the square of this, and you multiply by 2 (since there are two spheres) so the net factor is the cube-root of 2, which is around 1.3. So the extra surface tension energy is increased by a factor of 1.3, or 30%.

But in separating the two spheres, you have taken one ball of charge, with an energy of $Q^2\over R$ and separated it into two adjacent balls of reduced radius and half the charge. Adding up the electrostatic energy, it is about 80% of the original electrostatic energy in the single sphere.

So spontaneous droplet fission will happen when you have a charged ball for which 30% of the surface tension energy is less than 20% of the charge energy. Since charge goes up almost as the volume (not quite, but close) while the surface tension goes up as the area, there is a crossover, and charged droplets will spontaneously separate when they are too big.

The surface tension can be found from the binding energy curve of nuclei, and these simple considerations limit stable nuclear size to about that of Uranium. The U nucleus can spontaneously fission at an extremely low rate, but the transuranics become progressively more unstable because their electrostatic energy is increasing as the volume to a power greater than 2/3, while their surface tension energy is increasing as the surface area, which grows as the 2/3 power of the volume.

These considerations, in much more sophisticated form, are due to Niels Bohr in the seminal liquid drop model of the 1940s. This model explained the nuclear binding energy curve quantitatively, and accounted well for fission phenomena. The only major thing left out of this was the shell model and magic numbers, which was supplied by Mayer.

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Hi there Ron; what would happen if the unstable nuclei in question would have sufficiently high spin so that the self-attraction of the charges due to its magnetic momenta would exceed or partially cancel the electrostatic self-repulsion enough to keep the nucleus stable? –  lurscher Oct 16 '12 at 15:45
@lurscher: This is not possible, as magnetism from rotation only equals electrostatic repulsion when things are moving at the speed of light. Besides, the angular momentum of large nuclei isn't that huge. –  Ron Maimon Oct 16 '12 at 20:49
magnetism from rotation only equals electrostatic repulsion when things are moving at c - what if there is already a background magnetic field, the net magnetic field will be due to inner rotation plus background, so the rotation speed doesn't need to be unphysical to exceed electrostatic repulsion. If you have an answer to that, please consider contributing it here: physics.stackexchange.com/q/38728/955 There was some people that worked in studying high spin excitation phenomena to nuclei books.google.com.pa/… –  lurscher Oct 19 '12 at 1:54
i realise now how moronic this idea was. For a moment i thought the antipodal part of a charged ring will generate an attractive force, so i expected an equilibrium be found somewhere, now i realise that such force is actually repulsive, just like the electrostatic component –  lurscher Oct 21 '12 at 14:53
@lurscher: oh yeah, they are antiparallel, aren't they. I didn't notice that either, but of course you are right. I didn't answer not because it's moronic in this way but because I just didn't care about highly spun up nuclei, and you found a detailed reference about that anyway. –  Ron Maimon Oct 21 '12 at 15:26