nuclear stability belt

huey910

As we all know, the nuclear stability graph strays away from the y=x line from mass no. = 20 onwards. Why is that?

jambaugh

The proton-proton repulsion is a long ranged force while the strong force is not. As the size of the nucleus grows electric repulsion begins to kick in and higher proton count becomes a disadvantage.

The equal number of protons and neutrons is favored by the weak force interactions (the n + p can couple into an isospin singlet and be much less likely to interact with the weak force.)

You will note in the picture which shows the types of decays, as the atomic numbers grow you move into predominantly beta- decay and alpha decay.

(Image from wikipedia)
The -beta is due to the neutron excess [edit: a neutron becomes a proton emitting a -W boson which decays into an electron and antineutrino or more rarely muon or talon and corresponding anti-neutrinos] and the alpha due to the protonic repulsion. (tunneling w.r.t. the strong force potential.) Alpha decay, proton emission, and fission are all driven mostly by proton repulsion.

I'm not quite clear on the causes of neutron emission. But that's a rare case an less involved in your question.

randomedesign

Is there a formula which predicts the P/N ratio?

Bill_K

An empirical fit to the Valley of Stability is Z = A/(1.98 + 0.015 A^2/3). Which as A gets large is increasingly less than Z = A/2.

bcrowell

This is incorrect. First off, the weak force is irrelevant. The shape of the line of stability is determined solely by the shape of the potential energy surface as a function of N and Z, which in turn is determined only by the strong force. The weak force is irrelevant.

The reason that N=Z is favored, when the electrical repulsion is negligible, is the Pauli exclusion principle. Here is a FAQ entry I wrote up on this:

FAQ: Why does the line of stability have the average over-all shape it does?

For light nuclei, the line of stability hugs the N=Z line, and this is because of the Pauli exclusion principle. If you have N=8 and Z=8 (16O), you can put the 8 neutrons in the 8 lowest energy states, and the 8 protons in the 8 lowest energy states. With N=10 and Z=6 (16C), the exclusion principle forces you to put those last few neutrons in high-energy states that weren't occupied in 16O.

For heavy nuclei, the mutual electrical repulsion of the protons breaks the symmetry in the way the strong nuclear force treats neutrons and protons. This effect favors higher N/Z ratios, so the line of stability bends away from N=Z.

The line of stability also has little wiggles superimposed on top of its broad over-all curve. These are caused by quantum mechanical shell effects, the nuclear analogs of the ones in atomic physics that make the noble gases so chemically stable. These shell effects have nothing to do with the over-all shape of the line of stability. For example, the nucleus 100Sn (N=50, Z=50) has two closed shells, but it is very far from the line of stability.

randomedesign

Has anyone observed or discovered a correlation between concentric icosahedral shells and proton/neutron ratios?

randomedesign

Is this equation derived from the curve?

Is there an equation which predicts the P/N curve if this one doesn't do that?

Bill_K

No. They tried that for the solar system once, and it didn't work there either.

bcrowell

Presumably this question is motivated by your having heard the term "shell model." Don't take it too literally. The neutrons and protons in different shells are not physically confined to concentric spheres. They are also not located at specific points. They are waves that overlap one another. (The same is true for electrons in atomic physics.)

randomedesign

Why do the ratios of these equations closely correlate to the proton/neutron ratios of the nuclides in the 'valley of stability'?

p+=(10)F^2+2 and n0=(10/3)F^3+(5)F^2+(11/3)F+1

mfb

It does. Divide both sides by A and you get P/(P+N) = (something depending on A), which you can solve for P/N, too, if you want.


I would expect that those values are the result of some fitting procedure. In this case: The prefactors are chosen to give the valley of stability. There is nothing magical going on here. The valley is quite smooth, so polynomials and similar functions can be used to approximate it.

randomedesign

Could the stability of a nuclide be partly due to its geometry?
The equations are not from a fitting procedure. The first one is a hollow icosahedral shell formula. The second is centered icosahedral numbers.
http://oeis.org/A005901
http://oeis.org/A005902

mfb

Ah, I see.
Well, there are many series which can be combined somehow to give a relation close to the nuclear stability belt. In addition, your series can be used for 4 nuclei only, and one of them (U239) is highly unstable.
Bethe-Weizsäcker formula allows to calculate a relation based on a physical model (nuclei as "liquids"), with this shape as result. It does not use any discrete, geometric objects. As bcrowell already said, there are no solid shells anywhere. Nuclei are not a pile of solid balls.

randomedesign

Yes, I understand there are no balls. Is it possibly useful to think of them as balls just like it can be useful to think of nuclei as drops?

The series can be used for all of the nuclei, not just 4 of them. For example:

F=((n-2)/10)^1/2

Theoretical A.W. Actual A.W. Correlation
12.162277660168 12.01070000 0.990776011424475Carbon
14.271236166328 14.00670000 0.992298016093952Nitrogen
16.389381125702 15.99940000 0.994244822575429Oxygen
18.519960159205 18.99840320 0.995406989439364Fluorine
20.664705543000 20.17970000 0.996416539217974Neon
22.824555320337 22.98976928 0.997157206240969Sodium (Natrium)
25.000000000000 24.30500000 0.997676269632494Magnesium
27.191264886581 26.98153860 0.998141309120224Aluminium
29.398412548413 28.08550000 0.998252146435306Silicon
31.621403400793 30.97376200 0.998572224990406Phosphorus
33.860132971833 32.06500000 0.998607120703053Sulfur
36.114455552060 35.45300000 0.998767356287702Chlorine
38.384199576606 39.94800000 0.997521153821952Argon
40.669177823045 39.09830000 0.997773870642422Potassium (Kalium)

mfb

Be careful with those "Actual A.W." values - they depend on the isotopic composition, and this depends on fusion processes, geology and so on. The values refer to the earth's crust, you will get different values on the moon, on venus or elsewhere.

How do you define a correlation for one formula result and one experimental value?

As I said, the valley is quite smooth, if you use some polynomial function or something similar, you can get a good approximation. But in this case, you lose your geometric interpretation of the values, and it is just coincidence.

Some other examples:
Hydrogen: Your formula gives 2 (using the exact values (1,1)), the real value is ~1. Ok, deuterium is stable.
Helium: Using your formula I get 3, while the real value is ~4. Ok, He-3 is stable.
Lithium: Formula gives 5.8, real value is ~7. Ok, Li-6 is stable.
Beryllium: Formula gives 7.9, real value is 9 (and all other isotopes are unstable)
Boron: Formula gives 10,1, real value is 10.8 due to the isotopic composition (80% B-11, 20% B-10)

What about heavier elements?
Xenon: Formula gives 128.9, real value is 131.3
Lead: Formula gives 209, real value is 207.2.
Uranium: Formula gives 239, real value is 238.03 and slowly rising towards 238.05 as the heaviest isotope has the longest lifetime. It was closer to 236.5 some billion years ago.

randomedesign

Thanks, mfb. I will do more homework. Maybe instead of using A.W. I will stick with just the stable nuclides. I just thought there might be a connection to geometry and stability, or that someone might wonder why no one is aware of what seems to be a close approximation of nature purely based on icosahedral numbers. Rarely an amateur stumbles across an idea that others can make use of. Am I correct in thinking the magic numbers have suggested an unknown order to nuclear scientist for a long time?

bcrowell

There is no "unknown order." It's very well understood why the magic numbers are what they are.