Sciencemadness Discussion Board

Are the Coulomb's barriers for all Hydrogen isotopes the same?

radioboy - 19-4-2016 at 03:27

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]

Metacelsus - 19-4-2016 at 05:32

All the ions of the hydrogen isotopes have the same charge, so they will feel the same electric repulsion.

radioboy - 19-4-2016 at 09:22

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]

Metacelsus - 19-4-2016 at 13:02

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.

Theoretic - 21-4-2016 at 00:24

Quote: Originally posted by radioboy  
Also for He4+He4 it shows negative barrier! Error only in sign?
[Edited on 19-4-2016 by radioboy]
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.

radioboy - 21-4-2016 at 02:59

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

Theoretic - 21-4-2016 at 04:14

Quote: Originally posted by radioboy  
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

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.