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huey910
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As we all know, the nuclear stability graph strays away from the y=x line from mass no. = 20 onwards. Why is that?
jambaugh said: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.)
Bill_K said:An empirical fit to the Valley of Stability is Z = A/(1.98 + 0.015 A2/3). Which as A gets large is increasingly less than Z = A/2.
No. They tried that for the solar system once, and it didn't work there either.Has anyone observed or discovered a correlation between concentric icosahedral shells and proton/neutron ratios?
randomedesign said:Has anyone observed or discovered a correlation between concentric icosahedral shells and proton/neutron ratios?
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.randomedesign said:Is there an equation which predicts the P/N curve if this one doesn't do that?
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.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 said: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.
mfb said: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.
mfb said: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 said:Am I correct in thinking the magic numbers have suggested an unknown order to nuclear scientist for a long time?
The Nuclear Stability Belt is a concept in nuclear physics that refers to the region of stable isotopes on a graph of protons versus neutrons.
This deviation occurs due to the presence of extra neutrons, which help to stabilize the nucleus by balancing out the repulsive forces between protons.
These elements are considered unstable or radioactive, meaning their nuclei will eventually undergo decay to reach a more stable state.
The Nuclear Stability Belt helps to predict the stability of different elements and their likelihood of undergoing nuclear reactions, which is important in fields such as nuclear energy and medicine.
Yes, there are some elements that fall within the Nuclear Stability Belt but are still considered radioactive due to their high proton or neutron numbers, making them more prone to decay.