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I Why is Deuterium fusion so much easier than Hydrogen fusion?

  1. Jul 14, 2017 #1
    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?
     
    Last edited: Jul 14, 2017
  2. jcsd
  3. Jul 14, 2017 #2

    Bandersnatch

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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.
     
  4. Jul 14, 2017 #3

    Orodruin

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    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.
    The neutron provides a deeper potential well, enough to make the He-3 state stable against spontaneous separation of the charged protons.
     
  5. Jul 14, 2017 #4

    Drakkith

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    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.
     
  6. Jul 14, 2017 #5

    mfb

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    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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