How large is the collision cross section compared to nuclear fusion cross section

  • #1
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Main Question or Discussion Point

Most Fusion reactors, and the leading ones like JET, use high temp. plasma and confine it. So, the plasma would approximate the Maxwell- Boltzmann distribution. This means that only a small portion of the plasma has enough energy to fuse. But, collisions are much more often, right? Since not all collisions result in fusion, the cross section of collision is much higher than the cross section of fusion. Is this true?

  • But how much smaller is the cross section of fusion compared to the cross section of collisions. And do the electrons possess higher energy than the ions, in which case thee nuclear cross section would be orders of magnitude lower.
Am I right? If approximately all the collisions resulted in fusion, how much more energy would we be able to produce?

And what if there were 2 beam travelling toward each other. But these not just follow straight paths. There exists a HELICAL magnetic field and the particle spiral around the field lines and approach each other. Here, I found out that the particle spiral direction would be in opposite directions( one clock wise and the other anticlockwise.) If the beams interact when they meet, how would the collision cross section be affected? (Note: The particle possess reasonable energies at least higher than 25 Kev)

How would I calculate the collision cross section? I seriously need help. Thanks
 

Answers and Replies

  • #2
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This means that only a small portion of the plasma has enough energy to fuse. But, collisions are much more often, right? Since not all collisions result in fusion, the cross section of collision is much higher than the cross section of fusion. Is this true?
Right.

DT fusion has a maximal cross section of 5*10-28 m2 at about 60 keV center-of-mass energy.
Something like a collision happens if the nuclei come closer than ~200 fm, which corresponds to a cross section of 4*10-26 m2.

At a temperature of 10 keV, only a small fraction of nuclei collisions will lead to 60 keV center-of-mass energy, which reduces the effective fusion cross section even more.
Am I right? If approximately all the collisions resulted in fusion, how much more energy would we be able to produce?
The plasma would be too short-living then, and we would have to run it at a lower temperature.
If the beams interact when they meet, how would the collision cross section be affected?
The magnetic field is only relevant for beam steering, the actual collision is completely independent of that.
How would I calculate the collision cross section?
See the papers doing that.
 
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  • #3
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For the 4*10^-26 m squared cross section, what is the cross section for? and how did we end up with it? and an other thing I was wondering about was that I said we had a helical magnetic field. I've figured out that the particles spin in opposite directions during collision. Doesn't this mean that almost all of these are head on collisions? I think this would increase the fusion rate, wouldn't it? And these particle beams are not in thermal equilibrium and hence all have approximately the same energy, so all collisions would result in fusion right? So.. according to what you just said the cross section wouldn't change. So, am I safe in assuming these assumptions are accurate?
 
  • #4
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And approximately how many more collisions do we have compared to fusion?
 
  • #5
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For the 4*10^-26 m squared cross section, what is the cross section for?
For any collision. Those elastic collisions don't have a well-defined total cross section - larger impact parameters will just lead to smaller deflections. See Rutherford scattering. I used a distance where the deflection will be small (few degrees).
Doesn't this mean that almost all of these are head on collisions?
If you have two well-defined beams, you don't even need magnetic fields for that. The fusion rate will be low due to the low achievable particle density in the beams.
Even at the ideal energy for fusion, most collisions won't lead to fusion, see the comparison of the numbers in post 2. Most particles would be scattered out of the beam without fusing.
 
  • #6
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What if low density beams bombard a high density plasma?
 
  • #7
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Then you still lose most of the particles due to elastic scattering.

You can get fusion with accelerator systems, and accelerator-based fusion is a convenient source of neutrons. You just do not get enough to use it as power source because you have to put in much more energy than you get out.
 
  • #8
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But then, why doesn't it happen in plasmas. I know it happens, but I think the effect is much smaller. Hence, they don't loose energy through coulomb scattering. Is this right?
 
  • #9
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But then, why doesn't it happen in plasmas. I know it happens, but I think the effect is much smaller.
Which effect is smaller?
The energy is not lost in either case. In a plasma, collisions between atoms just redistribute the energy within the plasma particles - a completely normal thermal process. In beams of cold atoms, it means some particles get scattered out of the beam and the beam gets hotter.
 
  • #10
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Got it. I will think about this one more time.
 
  • #11
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Recently, I have learn't about the Gamows factor, and I have plotted it. Then I realized that the probability of Deuterium and tritium fusing at a temperature of 150 million kelvin when they collide, was about 0.00048. But, I know that the nuclear cross section for fusion is extremely small (about 1-1000 barns). But why is this so? I f the probability of fusion is 0.00048 when they collide, then why is the nuclear cross section so small? Is it because of the collisions? Are the collisions really that rare? Is the collision cross section really that small? Why is the nuclear cross section so small?

OO0Ef.png



nSWUb.png


  • Here the x-axis is the temperature and the y-axis is the probability of fusion when a collision occurs.
 
  • #12
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Why do you expect a large cross section? To actually hit the nuclei in a classical sense you would need about 2 MeV - but then you have to hit exactly, which is incredibly unlikely as nuclei are tiny (smaller than 1 barn). Everything below that has the nuclei approach each other, and then you have to hope that they tunnel to fuse before they fly apart again. That process is not very frequent either.
 
  • #13
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Does the ion-ion collision rate or collision frequency increase or decrease with an increase in temperature? I've read that most plasmas could be considered collisionless or have little collisions taking place.
 
  • #14
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For a constant density, the collision rate increases with temperature, simply because the particles move faster.
 
  • #15
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Now, for a plasma put under a constant pressure, the molar density multiplied by the temperature is always a constant which equals Pressure/Gas constant. According to the ideal has equation. So the collision rate is a constant for a constant pressure right?
 
  • #16
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At constant pressure: double the temperature and you double the path length between collisions, but the speed only grows with a factor sqrt(2) - collisions get less frequent. The chance of fusion increases more up to some point, so increasing temperature is still advisable.
 
  • #17
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So thats why the cross section drops after 150 million kelvin for D-T fusion? The collisions become less frequent?
 
  • #18
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The velocity you considered is the average velocity isn't it?
 
  • #19
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So thats why the cross section drops after 150 million kelvin for D-T fusion? The collisions become less frequent?
The cross section itself decreases as well if the energy gets too high, this is a more important effect.
The velocity you considered is the average velocity isn't it?
It is true for all speeds. Average, median, top 0.01%, ...
 
  • #20
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But why does the temperature become too high? Isn't it going to help us overcome the coulomb barrier? The reason I think it can get too high is through losses. Is there a quantum mechanical effect that affects the cross section at extremely high temperatures?
 
  • #21
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The reason I think it can get too high is through losses.
The cross section has nothing to do with losses. It has nothing to do with temperature either. For the cross section, you just have two nuclei colliding with a given center of mass energy.
Faster nuclei lead to a shorter collision time, that could contribute.
 
  • #22
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Oh. So time matters?
 
  • #23
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An other question: If I accelerates particles, and make them hit a plasma target, the portion of the particles that actually collide with the nucleus' described by the collision cross section either fuse or lose their energy in elastic collisions. So, the percentage of particles that fuse are described Gamow equation. Right? The rest just lose their energy through elastic collisions and braking radiation. Is this right? ....So, in this case, what exactly is a collision? How close are particles supposed to get in order for the interaction to be considered a collision?
 
  • #24
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Oh. So time matters?
Sure.

If you use an accelerator, there is no need to have plasma target. A regular target (solid, liquid, gas) will work as well.

So, the percentage of particles that fuse are described Gamow equation.
That just describes the fusion cross section, not how the energy of the particles evolves if they don't fuse.

Some fraction will fuse with the first nucleus they get close to. Others will scatter once and fuse with the second one. Others will scatter twice and fuse with the third one. And so on, but fusion gets less likely over time as the nuclei lose energy. Most will just scatter and never fuse.
So, in this case, what exactly is a collision? How close are particles supposed to get in order for the interaction to be considered a collision?
As mentioned before, the lower cutoff for calling something collision is a bit arbitrary. That doesn't change the physics, just the names we assign to things.
 
  • #25
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Yeah. I understand now. Thanks
 

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