radioboy
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Are the Coulomb's barriers for all Hydrogen isotopes the same?
This calculator shows those barriers in MeV, for all nuclides. And while it gives different values for all other nuclides (which is what I expected),
it gives same for all H isotopes. Is it wrong? What are real values for protium, deuterium and tritium? Although it will affect all values between H
and any element and nuclide, here are for PP, DD and TT fusion: 0.389 MeV. You can check here (just select fusion, and those isotopes): nrv.jinr.ru/nrv/webnrv/qcalc/
Also, can somebody explain what MeV means practically? Is it same as applying 1 MV voltage on any element? Or it has to be multiplied with electrons
any atom has? But in fusion all electrons are ionized, so what it means for fusion? How can I set 1 MeV in my fusor? 1 MeV = 1 MV? Or what?
[Edited on 19-4-2016 by radioboy]
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Metacelsus
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All the ions of the hydrogen isotopes have the same charge, so they will feel the same electric repulsion.
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radioboy
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No, but only similar. Nitrogen isotopes also have same charges, but different barriers. For example:
N14+N14: 6.995 MeV.
N14+N15: 6.827 MeV.
N15+N15: 6.659 MeV.
And hundreds of keV is not small difference.
Not only that I care about difference of H isotopes, but also about truth for any! What if person who wrote wrongly that, also wrote wrongly even that
one value (maybe it's wrong for any H isotope). Of course, tunneling probably means values are much less than written ones for all isotopes.
Also for He4+He4 it shows negative barrier! Error only in sign?
[Edited on 19-4-2016 by radioboy]
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Metacelsus
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The difference might be due to a different interaction radius (basically, that some can start fusion at a greater distance than others). The Coulomb
barrier is proportional to the product of the charges of the nuclei, divided by their interaction radius. I would expect the interaction radius to
vary little between isotopes, but for nitrogen the differences seem to be important.
If a negative barrier is shown, then it's wrong.
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Theoretic
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Quote: Originally posted by Metacelsus  | | The difference might be due to a different interaction radius (basically, that some can start fusion at a greater distance than others). The Coulomb
barrier is proportional to the product of the charges of the nuclei, divided by their interaction radius. |
Maybe this is because the closed shell structure of the He-4 makes it so stable, it doesn't release but instead absorbs net energy, (so
correspondingly, Be-8 releases energy when decaying to two He-4).
In the same way as, the thermal neutron absorption cross-section for He-4 is zero because they don't react; but if you tried to derive a number
artificially, it might be negative (depending on the method), because He-5 releases a neutron spontaneously.
Or, the calculator is faulty. It gives negative coulomb barriers for the reaction of He-4 with nearly everything, up to about fluorine. For some
reason it's probably assuming a negative interaction radius.
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radioboy
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I don't care what will happen after overcoming Coulomb's barrier, but only what is it. Does somebody know how to calculate it using formula found on
wikipedia's Coulomb barrier article? Also for p, d and t isotopes.
Another weird thing I found is that B10 has thermal neutron absorption and capture cross section. Isn't that the same? nrv.jinr.ru/nrv/webnrv/map/nucleus.php?q=B10
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Theoretic
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Probably because most of the time B-10 absorbs a neutron, it does *not* capture it to become B-11, but instead disintegrates to Li-7 and He-4. The
small capture cross-section reflects the small probability of B-10 turning into B-11.
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