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I Is nuclear fusion induced by gamma photons possible?

  1. May 8, 2016 #1
    Hi folks,

    IMO, it should be possible to initiate a nuclear fusion of H1-H2 and H2-H2 in crystals of Lithium Hydride induced by gamma photons emitted form decay of Na24.

    Could anyone please verify if the following is correct?

    By using nuclear photodisintegration effect, we need a gamma photon with energy high enough to overcame a threshold energy, which is specific to the reaction[1][4]:
    H2 -> n0 + H1 - 2,2246 [MeV] *
    Li6 -> He3 + H3 - 15,794 [MeV]
    Li6 -> He4 + H2 - 1,4738 [MeV] *
    Li6 -> He5 + H1 - 4,5883 [MeV]
    Li7 -> He4 + H3 - 2,4676 [MeV] *
    Li7 -> He5 + H2 - 9,6148 [MeV]
    Li7 -> He6 + H1 - 9,9740 [MeV]

    Gamma emission energy for radioactive decay of chosen radioisotopes is[2]:
    Na24: 1.368626 [MeV] @ 99.99 [%]
    Na24: 2.754007 [MeV] @ 99.85 [%] *
    Y88: 1.836063 [MeV] @ 99.20 [%]

    Therefore, for mentioned radioisotopes and photodisintegration channels of H2, Li6 and Li7 marked above with asterisks, we will gain some release of kinetic energy[1][4]:
    Na24: (gamma @ 2,754 [MeV]) + H2 -> n0 + H1 + 0,5294 [MeV]
    Na24: (gamma @ 2,754 [MeV]) + Li6 -> He4 + H2 + 1,2802 [MeV] *
    Y88: (gamma @ 1,836 [MeV]) + Li6 -> He4 + H2 + 0,3622 [MeV]
    Na24: (gamma @ 2,754 [MeV]) + Li7 -> He4 + H3 + 0,2864 [MeV]

    According to the law of conservation of energy and momentum we should expect energy distribution like this[4]:
    Na24: (gamma @ 2,754 [MeV]) + H2 -> ( n0 @ 0,2646 [MeV]) + (H1 @ 0,2648 [MeV])
    Na24: (gamma @ 2,754 [MeV]) + Li6 -> (He4 @ 0,4285 [MeV]) + (H2 @ 0,8516 [MeV]) *
    Y88: (gamma @ 1,836 [MeV]) + Li6 -> (He4 @ 0,1212 [MeV]) + (H2 @ 0,2410 [MeV])
    Na24: (gamma @ 2,754 [MeV]) + Li7 -> (He4 @ 0,1231 [MeV]) + (H3 @ 0,1633 [MeV])

    H1-H2 and H2-H2 fusion requires to exceed a threshold energy of 0,389 [MeV]. When achieved, the reaction will release extra energy[1][3][5]:
    (H2 @ 0,389 [MeV]) + H1 -> He3 + 5,4935 [MeV] *
    (H2 @ 0,389 [MeV]) + H2 -> He4 + 23,8465 [MeV] *

    Therefore a two stage reaction should lead to nuclear fusion of H1-H2 and H2-H2 between stationary H1 or H2 in Lithium Hydride crystals with H2 @ 0,8516 [MeV] released in photodisintegration reaction of Li6:
    1) Na24: (gamma @ 2,754 [MeV]) + Li6 -> He4 + (H2 @ 0,8516 [MeV])
    2a) (H2 @ 0,8516 [MeV]) + H1 -> He3 + 5,96 [MeV]
    2b) (H2 @ 0,8516 [MeV]) + H2 -> He4 + 24,31 [MeV]

    The reactions if correct, have a very small cross sections and will occur very rarely. However I am just curious if my understanding of above is correct.

    References:
    [1] http://nrv.jinr.ru/nrv/webnrv/qcalc/
    [2] https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html
    [3] R. Bass, Nuclear Reactions with heavy ions, Springer-Verlag, NY, 1980
    [4] Photodisintegration of Lithium Isotopes, Ward Andrew Wurtz, 2010
    [5] Hydrogen Properties for Fusion Energy, P. Clark Souers, 1985

    Many Thanks,
    Toreno
     
  2. jcsd
  3. May 8, 2016 #2

    Simon Bridge

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    You are thinking that ##\text{H}_1^1 + \text{H}_1^1 +\gamma \to \text{He}_2^2## ?
    Or are you thinking of some multi-stage process like ##\gamma + X_a^z \to X_a^{z-1} + n_0^1## and the neutron goes to fuse with something else?
     
  4. May 8, 2016 #3
    None of those. I think to use gamma ray to disintegrate 6Li nucleus and create products like 2H or 1H with kinetic energy enough to cause 1H-2H or 2H-2H fusion with stationary hydrogen in Lithium Hydride/Deuteride.
     
  5. May 8, 2016 #4

    Simon Bridge

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    OK - so a fission-fusion approach.
    But certainly you can, in principle, induce a fission whose daughter nuclei have a vanishing but non-zero probability to go on to fuse.
    In a way this is how the fusion bomb works - you stimulate Li to produce tritium starting from a neutron source.
     
  6. May 8, 2016 #5

    mfb

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    If that would work, then accelerator-driven fusion power plants would be a thing. Way easier than the indirect approach with photons, where most photons would do pair production instead of photodisintegration.
    Fast hydrogen atoms have a negligible chance to fuse before they lose their energy in a material. You get a few fusion reactions, but far away from anything that would be relevant.
     
  7. May 11, 2016 #6

    Simon Bridge

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    Depends what you mean by "work" ... "negligible chance to fuse" is not zero chance - although saying: "this is not a practical path to fusion power stations" is something of an understatement.

    rereading the detail in light of above:
    Is the idea that hydrogen in lithium hydride will be doing the fusion?
    Is the idea to use LiH as a salt bridge in an electrolytic cell to transport H+ to high densities, then initiating fusion in the salt by a timely gamma-ray burst?
    What is the overarching idea here?
     
  8. May 12, 2016 #7
    > Is the idea that hydrogen in lithium hydride will be doing the fusion?
    1) Yes, but hydrogen/deuteron in LiH is used as a target to collide with
    daughter deuteron produced in Li6 fission induced by gamma photon.
    2) If we use Na24 as source of gamma photons, the kinetic energy of this daughter H2
    is enough to fuse with stationary target H1 and/or H2 in cristals of LiH.

    > Is the idea to use LiH as a salt bridge in an electrolytic cell to transport H+ to high densities,
    > then initiating fusion in the salt by a timely gamma-ray burst?
    No, this is not related to this process
     
  9. May 12, 2016 #8
    Furthermore: If we use Lithium-6 Hydride-1 with Sodium-24, then the fusion is clean i.e. no radioactive isotopes like tritium nor neutrons are produced because only the reactions mentioned above are permitted for those isotopes and 2,754 [MeV] photons.
     
  10. May 12, 2016 #9

    mfb

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    Yes, but where is the point in having a few reactions per second where you need some sextillions?

    Did you check the sodium and its decay product for possible neutron spallation?
     
  11. May 13, 2016 #10
    > Yes, but where is the point in having a few reactions per second where you need some sextillions?
    I have never claimed this is the invention which will revolutionize energy industry. As I have written above: "The reactions if correct, have a very small cross sections and will occur very rarely. However I am just curious if my understanding of above is correct". Unfortunately I didn't found the answer yet :)

    > Did you check the sodium and its decay product for possible neutron spallation?
    No, but actually why?
    Na24 decays with beta- @ 100% to Mg24,
    Na24m decays with IT to Na24 and also
    Na24m decays with beta- to Mg24.
    However T1/2 for Na24m is about 10 ms.
     
  12. May 13, 2016 #11

    mfb

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    To check that the photons cannot kick out neutrons or other radioactive parts there. For neutrons the energy is not sufficient, checked that.
    I don't see which question would be open.
    Yes you get fusion, but at a completely negligible rate.
     
  13. May 13, 2016 #12
    Ah, this way. Yes I have also checked this. There is only way if you use Lithium-7 and/or Deuteride as mentioned above (marked with asterisk):
    Ok, so yes is the answer. Many thanks :)
     
  14. May 13, 2016 #13

    mfb

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    Then the only neutrons you get will be from splitting D -> p+n.
     
  15. May 13, 2016 #14
    Yes, exactly. Therefore, if we use Lithium-6 Hydride-1, no radioactive isotopes are produced.
     
  16. May 13, 2016 #15

    mfb

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    Li6 + gamma -> He4 + H2
    H2 + gamma -> p + n
    There is your neutron.
     
  17. May 13, 2016 #16
    Sure, but according to above rare reactions this is extremely negligible :)
     
  18. May 13, 2016 #17

    mfb

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    It is a second order effect, but fusion itself is so unlikely I would expect a significant neutron to fusion ratio.
     
  19. May 14, 2016 #18
    Actually you might be right. However it is difficult to tell that without performing an experiment. In fact experiment should be easy to perform: Just observe amount of He3, He4 and neutrons produced in vacuum with Lithium Hydride near to gamma source (like Na24). This should tell enough about all reactions occurring at mentioned conditions.
     
  20. May 14, 2016 #19

    Vanadium 50

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    Then why don't you do it?
     
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