He-4 & Gamma Rays, H-3 & Proton, He-3 & Neutron Fusion: Answers

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In summary: You'll have to decide what energy range you're interested in though. If you're interested in temperature averaged cross sections (energy is not the same as temperature, after all), S-factors are also available in EXFOR for a lot or reactions.
  • #1
TESL@
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A basic question for which I could not find any answers on the internet:

What factors do determine whether the fusion results with a He-4 and gamma rays, H-3 and proton, or He-3 and neutron? If it is random, are there any methods to predict the rations of these reactions?

Thank you.
 
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  • #2
It is random, and quantum mechanics allows to calculate the probabilities (at least in theory).
 
  • #3
I thought it would be closely related to temperature. So what affects those probabilities?
 
  • #4
The kinetic energy of the D-D pair in their center of mass frame (which is related to temperature, of course= will influence the rates at high energies, but apart from that they are fixed, given by the strength of the relevant interactions and the properties of the collision partners and final particles.
 
  • #5
TESL@ said:
A basic question for which I could not find any answers on the internet:

What factors do determine whether the fusion results with a He-4 and gamma rays, H-3 and proton, or He-3 and neutron? If it is random, are there any methods to predict the rations of these reactions?

Thank you.
From what I can gather (Wikipedia) there is no He-4 reaction. The H-3 and He-3 reactions are equally likely.
 
  • #6
Well, it is an electromagnetic process, so it is significantly weaker. It should still occur sometimes.
 
  • #7
In stellar cores, like those of the sun, He-4 only results at the end of a chain of different fusion reactions, starting with two separate H-H fusion chains.
Eventually, you get He3-He3 fusion, which produces He-4 + 2p, where the 2p are then available to start another H-H fusion reaction.

http://en.wikipedia.org/wiki/Nuclear_fusion

Even given the conditions at the sun's core, the rates of these reactions are "notoriously slow" to occur.
 
  • #8
He3 + n and t + p should be roughly equal, because of isospin. He4 + gamma should be about 1% of those, because it's electromagnetic.
 
  • #9
Vanadium 50 said:
He3 + n and t + p should be roughly equal, because of isospin.

I don't understand why "roughly". Isn't the interaction force between both nucleons equal?

I am working on an inertial electrostatic confinement fusion reactor, that I need more precise percentages in order to observe changes in fusion rate by using a neutron detector. Can you give me a range (...%-...%) or any equations?

Thank you.
 
  • #10
TESL@ said:
I don't understand why "roughly". Isn't the interaction force between both nucleons equal?

I am working on an inertial electrostatic confinement fusion reactor, that I need more precise percentages in order to observe changes in fusion rate by using a neutron detector. Can you give me a range (...%-...%) or any equations?

Thank you.

If we knew all the answers, we'd have working fusion reactors and flying cars by now. (Sorry.)

It's one thing to be working on the design of a fusion reactor; it's quite another to be looking for answers to theoretical physics questions on a website, even a reputable one like PF.

Don't you have access to a technical library somewhere, you know where all the papers on the latest research can be accessed?
 
  • #11
OK you are right. Thank you all by the way.
 
  • #12
Here I found what I was looking for:

source: http://en.wikipedia.org/wiki/Cold_fusion#Lack_of_expected_reaction_products

Conventional deuteron fusion is a two-step process, in which an unstable high energy intermediary is formed:

D + D → 4He* + 24 MeV
Experiments have observed only three decay pathways for this excited-state nucleus, with the branching ratio showing the probability that any given intermediate follows a particular pathway. The products formed via these decay pathways are:

4He* → n + 3He + 3.3 MeV (ratio=50%)
4He* → p + 3H + 4.0 MeV (ratio=50%)
4He* → 4He + γ + 24 MeV (ratio=10-6)
 
  • #13
TESL@ said:
I don't understand why "roughly". Isn't the interaction force between both nucleons equal?

Yes, but the mass of the decay products is not, so you have phase space effects slightly favoring p+t. These are at the fraction of a percent level, but so is the He4 channel.
 
  • #14
NNDC to the rescue!

What you can do is to compare the relative probabilities ("cross sections") of each of the reactions.

In general, you can look at Sigma (Evaluated Nuclear Data File retrieval and plotting) from NNDC for a lot of cross sections, but that doesn't give me the d-d fusion probability, it seems. A different database - EXFOR (Experimental Nuclear Reaction Data) gives me a whole lot of hits for that reaction. You'll have to decide what energy range you're interested in though.

If you're interested in temperature averaged cross sections (energy is not the same as temperature, after all), S-factors are also available in EXFOR for a lot or reactions.
 
  • #15
e.bar.goum said:
NNDC to the rescue!

What you can do is to compare the relative probabilities ("cross sections") of each of the reactions.

In general, you can look at Sigma (Evaluated Nuclear Data File retrieval and plotting) from NNDC for a lot of cross sections, but that doesn't give me the d-d fusion probability, it seems. A different database - EXFOR (Experimental Nuclear Reaction Data) gives me a whole lot of hits for that reaction. You'll have to decide what energy range you're interested in though.

If you're interested in temperature averaged cross sections (energy is not the same as temperature, after all), S-factors are also available in EXFOR for a lot or reactions.

Thank you e.bar.goum!
 

1. What is the process of He-4 & Gamma Ray, H-3 & Proton, He-3 & Neutron Fusion?

In He-4 & Gamma Ray fusion, a helium-4 nucleus combines with a gamma ray photon to form a beryllium-8 nucleus. In H-3 & Proton fusion, a hydrogen-3 nucleus combines with a proton to form a helium-4 nucleus. In He-3 & Neutron fusion, two helium-3 nuclei combine to form a helium-4 nucleus and two protons.

2. What are the advantages of He-4 & Gamma Ray, H-3 & Proton, He-3 & Neutron Fusion?

The advantages of these fusion reactions are that they release a large amount of energy, they produce very little radioactive waste, and the fuel sources (helium-4, hydrogen-3, and helium-3) are abundant in nature.

3. How does He-4 & Gamma Ray, H-3 & Proton, He-3 & Neutron Fusion differ from other fusion reactions?

Unlike other fusion reactions, He-4 & Gamma Ray, H-3 & Proton, He-3 & Neutron Fusion do not require extreme temperatures and pressures to initiate the reaction. This makes them more feasible for practical energy production.

4. Can He-4 & Gamma Ray, H-3 & Proton, He-3 & Neutron Fusion be used for energy production?

Yes, these fusion reactions have the potential to be used for energy production. However, more research and development is needed to make the process efficient and cost-effective.

5. What are the potential applications of He-4 & Gamma Ray, H-3 & Proton, He-3 & Neutron Fusion?

Aside from energy production, these fusion reactions have potential applications in medical imaging and cancer treatment, as well as in space propulsion for long-distance space travel.

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