B Question about the s-process and r-process in stellar evolution

AI Thread Summary
The discussion focuses on the s-process and r-process in stellar evolution, highlighting the tendency for these processes to produce elements with even numbers of protons and neutrons. This preference is linked to the stability of even-numbered isotopes, as even elements generally have more stable isotopes compared to odd ones. The conversation also touches on the role of alpha particles in the formation of heavier elements and the significance of neutron/proton ratios in nuclear stability. Participants express a need for more detailed data on isotopic abundances rather than just elemental counts. Overall, the discussion underscores the complexities of nuclear stability and elemental formation in stars.
TheCelt
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Abundances relating to Z number
Hello

I have been reading about the s-process and r-process in star evolution. And one thing i briefly saw mentioned was that both processes generally create higher abundances of elements with even number of protons than odd.

It however did not explain why there is a slight bias towards even number of protons. What is the explanation for this ?
 
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TheCelt said:
slight bias towards even number of protons
It's one or two orders of magnitude. Wikipedia does not give much of a theoretical explanation of the Oddo–Harkins rule
 
It is both a bias toward even numbers of protons and even numbers of neutrons.

You get a similar effect in chemistry with electron orbitals. They have a "spin" and they are more stable when two occupy the same space in pairs.
 
I understood that it is because fusion of hydrogen to helium creates a large number of helium nuclei (alpha particles). Heavier elements are then built up from these alpha particles, which then have an even number of protons.
 
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phyzguy said:
I understood that it is because fusion of hydrogen to helium creates a large number of helium nuclei (alpha particles). Heavier elements are then built up from these alpha particles, which then have an even number of protons.

The s-process adds one neutron at a time. The elements beta decay to increase proton number.

There is also an alpha process.
 
stefan r said:
There is also an alpha process.
Yes. That it what I was referring to. Here is a Wikipedia link. And this process is very significant for building up medium weight elements.
 
Keith_McClary said:
It's one or two orders of magnitude. Wikipedia does not give much of a theoretical explanation of the Oddo–Harkins rule
Indeed.
One obvious Duh! is that even elements have more stable isotopes. From O to P all even elements (O, Ne, Mg, Si) have 3 stable isotopes, and all odd elements (F, Na, Al, P) have 1. From S on, the numbers are less regular, but even elements generally have more.
But this accounts for half an order of magnitude, not whole or two.
Does anyone have abundances of nuclei by mass rather than proton count?
 
snorkack said:
Indeed.
One obvious Duh! is that even elements have more stable isotopes. From O to P all even elements (O, Ne, Mg, Si) have 3 stable isotopes, and all odd elements (F, Na, Al, P) have 1. From S on, the numbers are less regular, but even elements generally have more.
But this accounts for half an order of magnitude, not whole or two.
Does anyone have abundances of nuclei by mass rather than proton count?

Abundance in the Universe for all the elements in the Periodic Table
 
snorkack said:
Indeed.
One obvious Duh! is that even elements have more stable isotopes. From O to P all even elements (O, Ne, Mg, Si) have 3 stable isotopes, and all odd elements (F, Na, Al, P) have 1. From S on, the numbers are less regular, but even elements generally have more.
But this accounts for half an order of magnitude, not whole or two.
Does anyone have abundances of nuclei by mass rather than proton count?

But why is an odd number of protons with an odd number of neutrons less stable than an even number of protons with an even number of neutrons they would both be neutral atoms yet the even version is more prevalent? I'm not sure i totally understand why that would make any difference if they balance out in the nucleus ?
 
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TheCelt said:
But why is an odd number of protons with an odd number of neutrons less stable than an even number of protons with an even number of neutrons they would both be neutral atoms yet the even version is more prevalent? I'm not sure i totally understand why that would make any difference if they balance out in the nucleus ?

They only balance when the other particle is identical. Note that there is "an uncertainty in the position of a particle". Where is the proton? It is "in the nucleus".

Wikipedia has a nice article about orbital shapes. I only saw this in p-chem class. We did not have the animated drum model. For each "shell" you are using the same drum but making it resonate differently. They are resonating around the same central point. On average all of the electrons (or particles) would be at the central point over a period of time. At any instantaneous measurement none of them are at the center (the position/momentum of the center is also uncertain though). I'm not sure about the nuclear physics the orbitals fill in slightly differently but it looks similar.

My impression is that the neutron/proton ratio has nothing to do with orbitals or spin. Rather, the balance prevents various forms of radioactive decay. If you have "to many" protons you can get positron emission, electron capture, alpha decay, proton emission, or fission. If you have "too many" neutrons you get beta radiation, neutron emission, or fission.

Calcium isotopes verify that neutrons and protons are filling their own shells. 41Ca electron captures because there are too many protons. 47Ca beta decays. 40Ca and 48Ca are stable within the lifetime of the universe. Then numbers 20 and 28 are magic numbers because nuclear orbitals are filled so both 40Ca and 48Ca are doubly magic.
 
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stefan r said:
That´ s precisely what I have seen before (abundance of elements, by proton count), not what I have been missing and just asked for (abundance of isotopes, compared not just against other isotopes of same element, but all isotopes of all elements, or rather, combining isobars into one entry?)
Do you have the underlying table (abundances, not the periodic table)?
 
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