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Re: radioactivity



Doug Craigen wrote:

>
> The explanation implicitely recognizes a neutron:proton ratio in the
> nucleus without providing any reason why. It doesn't tell a student why
> for example you can not have 120 protons with 1000 neutrons. Surely this
> would provide enough spacing between the protons and enough strong force
> that it should be stable? For that matter why can't we have nuclei of
> one proton with ten neutrons, or of ten neutrons with no protons at
> all?

At 05:24 PM 10/27/00 -0500, cliff parker wrote:

I wholeheartedly agree. Any help?

Let me take a spin with this:

Start with ten neutrons and no protons. All such critters are
fermions; you can't have more than one of them in any given state. You
can distinguish neutrons from one another by spin, but there are only two
possible spin-values. After that, you need to distinguish by position,
i.e. by what shell they're in. They go into shells more-or-less analogous
to the way electrons do it. The higher-lying shells have a lot of kinetic
energy.

Now suppose one of the high-lying neutrons gets the bright idea of changing
its isospin label, i.e. to turn into a proton (with the emission of an
electron and a neutrino).
1) This costs nothing from the strong-force point of view (the strong
force is the same between neutrons and protons in any combination,
independent of isospin).
2) For low-Z nuclides, it costs next-to-nothing from the Coulomb point
of view. It's only when you get up around Z=80 does Z^2 get so big that
nucleons get voted off the island (alpha decay) for Coulomb reasons.
3) It's a big win from the kinetic-energy point of view: the proton is
distinguishable from the remaining neutrons, and can plop into the lowest
shell.

At this level of detail, we have an explanation that is a lot better than
nothing (although it is far from complete). It qualitatively predicts:
*) The valley of stability should lie near Z=N, especially for low Z.
*) At higher Z, the valley should lie somewhat below Z=N, for Coulomb
reasons.
*) At some point, the valley will have a northeasternmost end, for
Coulomb reasons. In this region we find lots of alpha emitters.
*) Nuclides that are slightly northwest or southeast of the valley (a
slight Z/N mismatch) will undergo weak decay [i.e. beta emission or
electron capture]
example: http://klbproductions.com/yogi/periodic/K-pg2.html
[Note: ones with a gross mismatch can throw off nucleons as well as betas]

And as I said before:

The Encyclopedia Britannica has a nice writeup on "nuclear models" in the
ATOMS article.
... including a diagram of the valley of stability.