Amazing Random Posted August 1, 2019 Posted August 1, 2019 Well we know that most cores of atoms are said to be stable or unstable by doing experiments . But I think I have found a way to predict if a core is stable or not without the known proton/neutron ratios : There are 3 interactions which contribute to the stability of a core . 1) Electrostatic repulsive forces 2) Pauli repulsion 3) Mass of the neutron being bigger than the mass of the proton When a core is unstable we say it has a big weak interaction potential energy . The weak interaction potential energy is mass . This is how it works : Let's take Hydrogen-2 and Helium-2 : There can be only 1 stable core with a certain amount of nucleons . So which is it in this case? Hydrogen-2 is made from 1 proton and 1 neutron . The mass of a neutron is bigger than the mass of a neutron and this gives the core a relative big mass. Helium-2 is made from 2 protons. The electrostatic force between the two protons is relatively big (because they are close) and this gives the core a relative big mass. It turns out that the electric potential energy in a helium-2 core gives the core more mass than the excess mass of a neutron. Helium-2 has bigger mass than Hydrogen-2 and it decays to Hydrogen-2. P.S. It is energetically favourable a core with domain: Proton-Neutron-Proton-Neutron...... A core with two protons next to each other is unstable 100%.For a core to be stable at any axis two protons can never be next to each other. Hydrogen-4 is a synthesized product by firing high energetically Hydrogen-2 atoms to a Hydrogen-3 atom. It is highly unstable . Hydrogen-4 has 1 proton and 3 neutrons . Well due to the neutrons being close to each very close to each other , and since two fermions in a quantum system cannot occupy the same quantum state , the pauli repulsion becomes very big which means lot of potential energy which is converted to mass . Glad to be helpful!Comments and corrections appreciated.
Strange Posted August 1, 2019 Posted August 1, 2019 How many atoms and isotopes have you tested this with? Can you show us the data? ! Moderator Note Moved to Speculations. Note the special rules for this forum, in particular the need to provide evidence 1
swansont Posted August 1, 2019 Posted August 1, 2019 I don't see where you're predicting anything, other than relative stability of isobars (atoms with the same number of nucleons). You haven't described a method of showing if a particular isotope is stable, since you could have isobars where they are both unstable. 9 hours ago, Amazing Random said: A core with two protons next to each other is unstable 100%.For a core to be stable at any axis two protons can never be next to each other. "Next to" kinda loses meaning at this scale.
Amazing Random Posted August 2, 2019 Author Posted August 2, 2019 On 8/1/2019 at 1:29 PM, swansont said: I don't see where you're predicting anything, other than relative stability of isobars (atoms with the same number of nucleons). You haven't described a method of showing if a particular isotope is stable, since you could have isobars where they are both unstable. "Next to" kinda loses meaning at this scale. Next to I mean proton and the nearest nucleon is a proton. I have already given examples. On 8/1/2019 at 10:29 AM, Strange said: How many atoms and isotopes have you tested this with? Can you show us the data? ! Moderator Note Moved to Speculations. Note the special rules for this forum, in particular the need to provide evidence Helium-2 vs Hydrogen-2 Helium-3 vs Hydrogen-3 Hydrogen-4&more Helium-5&more Beryllium-7 vs Lithium-7
Strange Posted August 2, 2019 Posted August 2, 2019 9 minutes ago, Amazing Random said: Next to I mean proton and the nearest nucleon is a proton. How are you assuming they are arranged? Are you taking into account the shell structure? 10 minutes ago, Amazing Random said: Helium-2 vs Hydrogen-2 Helium-3 vs Hydrogen-3 Hydrogen-4&more Helium-5&more Beryllium-7 vs Lithium-7 So one or two percent (of stable isotopes), then. I would need a lot more data to find this convincing.
Amazing Random Posted August 2, 2019 Author Posted August 2, 2019 23 minutes ago, Strange said: How are you assuming they are arranged? Are you taking into account the shell structure? So one or two percent (of stable isotopes), then. I would need a lot more data to find this convincing. Those cores are easy to do because they have few nucleons . I have to make lot of calculations for all isotopes.....Also I am not working with numbers , I work with facts....so someone must teach me how to represent those things mathematically . I know calculus and derivatives but I dont have a mathematical mind....
swansont Posted August 2, 2019 Posted August 2, 2019 1 hour ago, Amazing Random said: Next to I mean proton and the nearest nucleon is a proton. I have already given examples. You gave examples — simple ones — where that argument might hold up. I'm saying that there are hundreds of ones where it is not clear you can treat a nucleus this way. 1 hour ago, Amazing Random said: Helium-2 vs Hydrogen-2 Helium-3 vs Hydrogen-3 Hydrogen-4&more Helium-5&more Beryllium-7 vs Lithium-7 What about e.g. K-40 vs Ca-40 vs Sc-40? Or Kr-86 vs Rb-86 vs Sr-86?
Amazing Random Posted August 2, 2019 Author Posted August 2, 2019 1 hour ago, swansont said: You gave examples — simple ones — where that argument might hold up. I'm saying that there are hundreds of ones where it is not clear you can treat a nucleus this way. What about e.g. K-40 vs Ca-40 vs Sc-40? Or Kr-86 vs Rb-86 vs Sr-86? It is very hard to check because there are many interactions between many nucleons.
swansont Posted August 2, 2019 Posted August 2, 2019 28 minutes ago, Amazing Random said: It is very hard to check because there are many interactions between many nucleons. That's part of the point of asking. Can you show your toy model doesn't break? Even slightly simpler, with 13 nucleons, which in this view is one central ball surrounded by 12 others, close-packed. C-13 is stable, N-13 is not. But you don't explain how big of an effect this has. N-13 would have all of its protons being next to another proton. Would that make it massively unstable, i.e. having a super-short half-life? Is that more or less of an issue as compared to e.g. Be-8, which would have fewer protons near each other, but has a much, much shorter half-life? Once you add a new layer onto it, can you say that no protons will be adjacent? Can you even say how the nucleons would arrange themselves — would they tend to go toward the gaps in the previous layer, to minimize r (close-packing)? Does something with ≠ 13 nucleons still try to be spherical? You should be able to look at the electric quadrupole moment to see. (zero meaning the charge distribution is, indeed spherical)
Strange Posted August 2, 2019 Posted August 2, 2019 4 hours ago, Amazing Random said: It is very hard to check because there are many interactions between many nucleons. So it is not a very useful model, is it?
Amazing Random Posted August 2, 2019 Author Posted August 2, 2019 25 minutes ago, Strange said: So it is not a very useful model, is it? If you have the time , you can continue working with it.
Strange Posted August 2, 2019 Posted August 2, 2019 2 hours ago, Amazing Random said: If you have the time , you can continue working with it. Why? If you don't think it is worth spending the effort on, why would anyone else?
Amazing Random Posted August 3, 2019 Author Posted August 3, 2019 11 hours ago, Strange said: Why? If you don't think it is worth spending the effort on, why would anyone else? It is worth spending the effort on , but I am currently trying to prove another theory ( gravity doesnt exist in the microscopic world ) so I cant work with two theories at the same . I just found those examples because they are easy and dont require effort.
Strange Posted August 3, 2019 Posted August 3, 2019 54 minutes ago, Amazing Random said: It is worth spending the effort on , but I am currently trying to prove another theory ( gravity doesnt exist in the microscopic world ) so I cant work with two theories at the same . I just found those examples because they are easy and dont require effort. It is so worth spending time on that you can't be bothered. Right.
Amazing Random Posted August 3, 2019 Author Posted August 3, 2019 (edited) 2 minutes ago, Strange said: It is so worth spending time on that you can't be bothered. Right. I dont like your attitude. Edited August 3, 2019 by Amazing Random
Strange Posted August 3, 2019 Posted August 3, 2019 3 minutes ago, Amazing Random said: I dont like your attitude. I’m not terribly impressed by yours :-|
Amazing Random Posted August 3, 2019 Author Posted August 3, 2019 1 minute ago, Strange said: I’m not terribly impressed by yours :-| Hahahaha
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