Lithium hydrogen fusion

  1. why doesn't background radiation cause lithium hydride to explode in a nuclear chain reaction
    Li7 + H = Be8 + energy
    Be8 = 2He4 + energy
  2. jcsd
  3. SteamKing

    SteamKing 10,978
    Staff Emeritus
    Science Advisor
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    Because fusion reactions are not self-sustaining like fission reactions. In order for fusion to initially occur, certain conditions must be present, like extremely high temperature or high pressure. In a hydrogen bomb, these conditions are obtained by first detonating an atomic (fission) bomb.
  4. does that mean that initially the particle struck by the radiation will fuse and maybe a few after that the energy will be to dispersed or that none of the particles will fuse
  5. SteamKing

    SteamKing 10,978
    Staff Emeritus
    Science Advisor
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    I'm afraid I don't understand what you are trying to ask.

    However, fusion reactions are not like fission reactions, where a fissile nucleus, such as U-235, undergoes fission after absorbing a thermal neutron. In most circumstances, bombarding a hydrogen nucleus, or a lithium nucleus, with various forms of radiation does not initiate a fusion reaction, even for a few lucky nuclei.

    The following article explains the fusion process:

    Each nucleus is positively charged because of the protons present, so in order to fuse, first the electrostatic forces tending to repel nuclei must be overcome, which takes energy. However, once two nuclei are brought close enough, nuclear forces, which normally bind protons and neutrons together, become strong enough to overcome electrostatic repulsion, and fusion may take place.

    Certain particles called muons can serve as catalysts for fusion reactions, but these particles are unstable and exist only for about 2 microseconds before they decay into neutrinos, which don't interact much with other forms of matter. The only naturally known source of muons is produced by the interaction of cosmic rays striking the upper atmosphere; producing artificial muons takes a lot of energy, which is why muon-catalyzed fusion has only been studied in the laboratory.
  6. mfb

    Staff: Mentor

    A high-energetic proton, hitting a Li7 nucleus, can lead to fusion and then the production of two He4. Those He4 will heat their environment a bit and nothing else happens. There is nothing that could produce a chain reaction.
  7. why don't the alpha particles produces excite protons again to continue the reaction though
    Last edited: May 6, 2014
  8. mfb

    Staff: Mentor

    They'll lose their energy via to ionization. Direct hits of other nuclei are really unlikely, and even then it is very unlikely that those will give a proton enough speed for fusion - and even then, it is very unlikely that this proton will indeed hit another lithium nucleus and fuse with it.
    Nuclear reactions of slow (nonrelativistic) charged objects are really, really rare.
  9. Now, lithium 6 deuteride is another matter!

    Step 1: with a slow neutron, like stray neutron from cosmic rays or from spontaneous fission of a nearby U nucleus
    n+6Li=α+t+4,78 MeV, of which about 2,73 MeV are to triton
    This is overwhelmingly the only fate of a slow neutron. There are possible reactions for chain loss:
    β decay
    but all of these are very low probability, because 6Li is a very good neutron absorber
    Step 2: with the said fast triton
    t+d=α+n+17,5 MeV. If triton was relatively slow, the breakdown of energy is 14 MeV for neutron and 3,5 MeV for α, but fast tritons allow a but wider spread of neutron energy.
    Now THAT step does include a significant probability of chain loss. The triton may collide with electrons and lithium 6 nuclei, lose energy to ionization or undergo elastic or braking radiation with deuterons without fusing. So the said 2,73 MeV triton, or faster triton if a faster neutron was involved in step 1, may thermalize without fusion, and in cold deuteride this loses the chain.

    No matter how low the probability of chain loss, chain reaction cannot happen unless there is some way for chain to branch.

    Branching step seems to be
    Step 3: with a fast neutron most likely
    d+n+2,2 MeV=p+2n
    The same might happen to α which also has enough energy. But I suspect that it would have less cross-sections even at same energy, and α starts at quarter the energy of a fission neutron.

    So... if the lithium 6 deuteride is cold then a thermalized triton means chain loss. If the deuteride is hot then a thermalized triton eventually still finds a deuteron and propagates chain.

    What is the neutron multiplication factor in cold lithium deuteride?
  10. mfb

    Staff: Mentor

    I would be surprised if the tritium nuclei has any relevant probability (let's say larger than .1) to do a fusion reaction before it thermalizes.
    The same applies to the deuteron splitting reaction. Heavy water is used in some nuclear reactors and I don't remember any notion of this reaction as a relevant contribution to neutron flux.
  11. There are thousands of kilograms of Li6D in existing thermonuclear warheads in storage. They even have some non-negligible spontaneous neutron sources nearby.

    I'm sure I didn't miss any news about some of them spontaneously exploding...
  12. Spontaneous explosion of fission warheads is also rare. Though not unknown if mishandled. As Daghlian and Slotin found out. And what happened to Palomares and other warheads whose chemical explosives were spontaneously set off by impact, heat or sparks?

    But if the infinite multiplication factor of cold lithium 6 deuteride were above unity then an assembly of cold lithium 6 deuteride could be made critical by reaching critical mass. And if cold lithium 6 deuteride could be made to explode by assembly of critical mass alone, without employing a fission explosion to heat it, then pure fusion bombs would be more convenient than fission-fusion ones.

    This argument shows that the infinite multiplication factor of cold lithium 6 deuteride must be below unity. But it does not show what it is within the range 0 to 1.
  13. Astronuc

    Staff: Mentor

    There is much more to a 'chain reaction' than the simple reactions. One must assemble a critical mass, and for fusion, that requires high temperature and density, which together produce high pressure. In addition to the technical constraints of high pressure, there is the matter of energy losses from the system, which increase with Z. In addition, there is a lot more scattering than fusion in a fusion plasma.

    Stars represent a controlled chain reaction, but the densities and pressure are well beyond what can be achieved in a terrestrial, man-made system.
  14. Why density? In small mass, yes - for a given mass the losses via escape outwards decrease with increasing density. In large volumes, though - the density does not matter, because the reactive particles are not slowed or lost save through collisions.
  15. Astronuc

    Staff: Mentor

    Radiative energy losses and losses due to neutrals leaking out of the plasma at the surface increase with density.

    Density also determines power density, and density with temperature determines pressure, and currently one of the major constraints on magnetic confinement is the magnetic fields which provide the 'magnetic pressure' that retains the plasma.

    Large volumes require large confinement systems, and that introduces challenges to confinement and power conversion systems.

    If it was so easy, we'd have commercial fusion plants online by now.
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