# Nuclear fusion : ion vs atom fusion cross section?

I've read critique on ITER project (in Russian), and one of points was that cross section of ion-ion fusion is much lower than ion-atom and atom-atom, and that's the reason why it likely to not work.

Can anyone suggest where I can find how much lower cross sections are? I am mainly interested in D-D, D-T and B11-H reactions.

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Iter attains fusion within a plasma. The very definition of a plasma is that the gas is ionized (fully ionized at Iter energies). It sounds reasonable that atoms accelerated toward each other could have a lower cross-section because Coulomb repulsion (putting aside screening) won't happen until the electron orbitals start interacting, but how would you accelerate them in the first place if they don't have a charge?! –  Alan Rominger Aug 28 '12 at 20:20
Lots of options: Accelerate first then neutralize for example, ITER does have 'neutral beam injectors'. But you surely cannot contain fast neutral atoms in a tokamak. –  BarsMonster Aug 28 '12 at 21:12
I see this question was more sophisticated than I took it for. The Iter parameters are designed to achieve fusion with the ion-ion cross sections, so it seems very weird to cite that as a reason it might not work. It would be an oversight to mess those up. I don't know that the question leads to a good alternative either, an economic reactor design that can slam those nuclei together (with or without their electrons in tow) is the central challenge. Aside from maybe some inertial confinements, all hot-fusion designs I can think of slam ions and not atoms. –  Alan Rominger Aug 28 '12 at 23:37
@AlanSE: In principle, you could slam negatively ionized H-type atom with bare H isotope nuclei, and then the first leg of the coulomb repulsion is actually attraction. The issue is that H- is a very unstable ion, and there is no way you won't eject that electron in a hot environment. –  Ron Maimon Aug 29 '12 at 5:29

I have an answer to establish expectations from first principles. I have not looked up real values, nor am I hopeful of finding such values. All I'm doing here is setting a general expectation for the difference in cross sections of ion-ion fusion and ion-atom or atom-atom fusion. I should note that the one glaring piece of information missing from the question is the energy of the reaction. As such, my answer will be somewhat non-specific on that point, but my position is that it doesn't affect my overall conclusion that no major reaction rate benefit could be obtained in fusion machines by this proposed mechanism.

I used a number of simplifications to craft a simple algebraic form for the difference in cross sections. I should first specify though, I used a "radius" for the cross section defined as $a=\sqrt{\sigma / \pi}$. Yes, it is true that cross-sections aren't really areas, but for the relative scales here it can be treated as such. Now, here are the assumptions I used:

• Fusion cross section << Atomic radius
• Physics are identical to ion-ion interaction after atomic radii meet
• Only the impact of cross section broadening considered, the increase in energy due to atom charge interaction not considered
• Target nuclei doesn't move, you could try to eliminate this with reduced mass or a adjustment to reaction energy, I'll leave that for someone else to try. This assumption isn't really necessary but I just didn't see a quick way around it.

The basic proposition we have is that the ion-ion cross sections of the reactions are already known. This, naturally, depends on the energy.

My proposition is that, whatever the ion-ion reaction trajectory is, it will be different from the ion-atom interaction by the electrostatic force operating at distances greater than the atomic radius. I hope my method id already starting to seem clear, but I'll use an illustration. The proposition is that

1. We know $a_{++}$
2. We know the profile of the $y-y_{++}$ with simple electrostatic physics.

Using the assumption that the atomic radius is small compared to the fusion cross section radius, I can claim $\sin{ \theta} = a/s$. The electrostatic force from the net charges on the atom/ion will operate along the line between the moving atom/ion and the target atom/ion. The distance between these can be approximated to be equal to $s$. If we don't bother with the speedup from this force, then we'll just consider the component inward.

$$F_{in} = \sin{\theta} \frac{ k q_1 q_2 }{ s^2 } = \frac{ k e^2 z_1 z_2 a }{ s^3 }$$

The speed of the incoming atom/ion is roughly constant, so I can give this approximate expression to get the above math in terms of time. This applies from $-\infty$ to $-r/v$. The $v$ follows from the energy of the reaction.

$$s(t) = - v t$$

Then the basic kinematics follow logically.

$$v_{in}(t) = \int_{-\infty}^{t} \frac{F_{in}(t')}{m} dt' = \frac{ k e^2 z_1 z_2 a }{2 v^3 m t^3}$$

$$y(r) - y_{++} (r) = \int_{-\infty}^{-r/v} v_{in}(t) dt$$

$$a - a_{++} = v_{in}(r) \frac{r}{v} + ( y(r) - y_{++} (r) ) = \frac{ k e^2 z_1 z_2 /r }{m v^2} a$$

This gives the difference in cross section radius due to atom-ion or atom-atom interaction beyond that atomic radius. This is a particularly useful form if we introduce $E_c=k e^2 / r$ and the kinetic energy of the incoming atom/ion. The difference in cross sections will be about twice this difference in radii.

$$\sigma - \sigma_{++} = \pi ( a^2 - a_{++}^2 ) \approx - 2 \pi a^2 z_1 z_2 \frac{E_c}{E_k}$$

$$\frac{ \sigma - \sigma_{++} }{ \sigma_{++} } \approx 2 z_1 z_2 \frac{E_c}{E_k}$$

$$E_c = 0.027 keV$$

This makes intuitive sense. The cross section will be affected by an amount proportional to the electrostatic energy of the two touching atoms and inversely proportional to the kinetic energy of the interaction.

The interaction energy in Iter will be about $8 keV$, and the optimal energy for DT fusion is $80 keV$ (previous graph). If you had two neutral atoms interacting, the cross sections will be greater than the ion-ion interaction by 0.6 % at Iter energies. It would be an improvement of 0.06 % at optimal DT energies.

The best you could ever hope for would be the pB interaction with the B fully ionized (+5) and the Hydrogen with (-1). Even this is likely unattainable as some comments have pointed out. If we look at this at Iter energies (impossible but this is best-case), that would increase the cross section by about 5%. Pretty much all other scenarios would have less of an improvement than this.

The bottom line is that the Coulomb attraction at distances beyond the atomic radius is very very small compared with the Coulomb repulsion that the nuclei have to overcome as they get closer than that. Any adjustments to the cross section from this atom-ion or atom-atom interaction will really be <1% for most conceivable fusion reactions.

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I am going to give you the very basic answer: An unionized atom with the electrons attached is about 10^(-10) meters in diameter, so about 10^(-20) square meters cross section. A totally ionized nucleus is about 10^(-15) meters in diameter, so about 10^(-30) meters in cross section, therefore about 10^(-10) times as big as the cross section of an unionized atom.
After that, it varies slightly, by a factor of only a few, with energy, except at resonances, where it jumps up. You can find more details by googling periodic table, table of nuclides, and nuclear resonance.

First off, you need enough energy for the two nucleii to overcome the Coloumb barrier, so the fusion cross section is negligible (read effectively zero) below that energy. Then it climbs rapidly, levels off and declines as it gets too high (but that stage is also irrelevant for a practical fusion reactor). This energy is so high that you can basically ignore the electrons. It doesn't matter much if they are bound to the nucleii or not, or even entirely absent, as far as the fusion cross section is concerned. They do matter as far as energy is concerned, however.
Finally, even at the ideal energy, only about one collision in a thousand results in a fusion. This is why confinement is so critical. The nucleii don't fuse the first time they collide. You have to keep colliding them over and over again.

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This is ridiculous--- he is not asking for the cross section for collision, he is asking for the cross section for fusion. –  Ron Maimon Sep 1 '12 at 7:31
@R Don't be so unforgivable. On one hand Jim's explains very clearly the idea behind the cross section, and it would be easy to follow its steps and compute the surface of large atom's nucleus. And on the other hand, atom-atom fusion in the question is unheard of in the real world. So Jim's answer is perfectly legitimate. –  Shaktyai Sep 1 '12 at 8:35
@Shaktyai I don't see why atom-atom (and atom-ion) fusion is unrelated to real world. Having neutral atoms flying with energy enough for fusion is possible. And I agree with Ron. –  BarsMonster Sep 1 '12 at 13:12
Sorry you didn't like my answer. I think the next thing to say is that indeed, on this oversimplified level, only nucleus nucleus collisions lead to fusion, so you can ignore the neutral-neutral and neutral ion collisions. I will edit my answer. –  Jim Graber Sep 1 '12 at 13:23
@ BarsMonster To fuse nucleus, you need to provide energy. There is no known processes to accelerate a neutral atom to energy high enough to induce fusion. That is why one talks about hot plasmas and hot fusion. At the energy required for fusion, the atoms are stripped out of their electrons. –  Shaktyai Sep 1 '12 at 18:12