Why is Deuterium fusion so much easier than Hydrogen fusion?

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SUMMARY

Deuterium fusion is significantly easier than hydrogen fusion due to the lower temperature requirements for the reactions. Hydrogen fusion (H+H->D+e+) necessitates temperatures exceeding 10 million K, while deuterium fusion (H+D->3He) can occur at around 1 million K. The presence of a neutron in deuterium creates a deeper potential well, stabilizing the resulting helium-3 nucleus against spontaneous separation of protons, which is not the case with the unstable diproton state formed during hydrogen fusion. Consequently, the probability of successful fusion reactions is vastly higher for deuterium at lower temperatures.

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Buzz Bloom
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This question occurred to me as a result of the discussion in the thread
Apparently
(1) H+H->D+e+
requires a temperature of >10,000,000 K
while
(2) H+D->3He requires only about 1,000,000 K.
I confess that the references for these temperatures are not high quality, but they are the best i could find.
Ref (1)
The threshold temperature for hydrogen fusion, sometimes referred to as the proton-proton chain, is 10-14 million K (Kelvin).
http://www.answers.com/Q/What_temperature_is_required_for_hydrogen_fusion
Ref (2)
Deuterium is the most easily fused nucleus available to accreting protostars,[1] and such fusion in the center of protostars can proceed when temperatures exceed 10^6 K.
https://en.wikipedia.org/wiki/Deuterium_fusion

The role of a high temperature is to give the interacting nuclei sufficient energy/momentum that that they can overcome their positive charge repulsion to approach each other close enough for the strong force to have it's effect to complete the process of combining them together. What is puzzling is that the repulsive force is the same for (1) and (2).

My question is: Why does the presence of the neutron in D influence the strong force so much that only about 1/10 of the energy/momentum of the pair of interacting nuclei sufficient to allow for the fusion to happen?
 
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I think it's just the matter of whether there is or isn't a need for a weak interaction to occur.

In the first case, once the parent nuclei are momentarily joined together, one of the protons must almost immediately undergo a beta plus decay into a neutron for the daughter nucleus to remain stable. The probability of such decay is low, so most p-p collisions end up in the two protons flying apart ag.

Higher temperature simply means that the rate of collisions goes up sufficiently, so that the number of beta plus decays occurring is significant despite low individual probability.

In the second case, once the electric potential barrier is broken and the two nuclei join, they tend to stay joined.
 
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Bandersnatch said:
In the first case, once the parent nuclei are momentarily joined together, one of the protons must almost immediately undergo a beta plus decay into a neutron for the daughter nucleus to remain stable. The probability of such decay is low, so most p-p collisions end up in the two protons flying apart ag.
To put this slightly differently, the diproton state is unstable and quickly disintegrates back into two protons. Had the diproton state been stable to disintegration, there would have been plenty of time for undergoing the beta decay, but as it is only a tiny fraction of diprotons actually have time to decay via beta decay to deuterium.
Buzz Bloom said:
Why does the presence of the neutron in D influence the strong force so much that only about 1/10 of the energy/momentum of the pair of interacting nuclei sufficient to allow for the fusion to happen?
The neutron provides a deeper potential well, enough to make the He-3 state stable against spontaneous separation of the charged protons.
 
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Orodruin said:
The neutron provides a deeper potential well, enough to make the He-3 state stable against spontaneous separation of the charged protons.

Can you elaborate a bit? How does that relate back to the temperature? My first thought was that the addition of the neutron meant that the strength of the strong force was higher at a given distance, so the incoming proton didn't have to get as close in order to fuse. However, since this is the quantum scale, I know things are never that simple.
 
The volume of a nucleus is roughly proportional to its mass number, He-3 has a larger nucleus than He-2 would have. A larger nucleus means the lowest energy states have a lower energy. In He-3 you have two protons and one neutron in a low-energetic state, while in He-2 you have two protons with a higher energy. Too high to be bound. To get a reaction, you need the weak interaction, and that makes the reaction very unlikely.Both fusion reactions happen at both temperatures, but the rates are different by many orders of magnitude.
pp at 106 K is negligible.
pD at 106 K is important -fast enough to support a small star.

pp at 107 K is important - fast enough to support a larger star.
pD at 107 K is extremely fast - so fast that deuterium nuclei just survive a few seconds.
 
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