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only 0.5 metres to transmute thorium to uranium 233

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  • only 0.5 metres to transmute thorium to uranium 233



    The rest mass of helium is 6.64*10^-27 kg

    The energy necessary to overcome the Coulomb barrier is 1,96 Mev.

    So applying the laws of physics, we can calculate the distance between the two beryllium target.
    signature...
    (sorry for my poor English, but I am italien)

  • #2
    what does this do? Have you built it?
    Atoms move for free. It's all about resonance and phase. Make the circuit open and build a generator.

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    • #3
      it does the famous rubbiatron.

      Energy amplifier - Wikipedia, the free encyclopedia

      No !
      I have not build yet, because I don't find a peace of thorium


      Mr. Rubbia tease people by saying you need an accelerator as big as a cathedral to the spallation of thorium by protons. (1 GeV)
      The reason is that protons have electric charge, so is natural it needs 1 Gev, instead we need is a little artificial neutrons generator.
      signature...
      (sorry for my poor English, but I am italien)

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      • #4
        The problem with this kind of experiment is 'luck'. Firstly you need a big operation and big energy to start; Secondly you need to generate and bombard the thorium with a LOT of neutrons. Conditions has to be perfect, else no fusion. Particles are very particular when it comes to them selecting a mate . . .

        Thorium is available in some normal household products. Same you can also use Amercium (Americanium) in stead of Thorium.

        Within the controlled USA it might be easier to look for Monazite and by simple methods you can extract Thorium 232.

        Overall in theory this does makes perfect value and sense; but in practice you will still use much more input than result achieved. Unless you do go VERY big.

        @The italian:
        Here is another forum specializing in this this kind of theory and experiments:
        Focus Fusion Society Forum | Could pB11 focus fusion device be modified to use thorium?
        Enjoy!
        Last edited by Aromaz; 06-16-2010, 01:11 AM.
        Therefore we need to find NEW ways, NEW experiments and NEW lines of thoughts.

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        • #5
          And the enormous costs to professionally get rid of the uranium without getting in trouble which you would be in anyway for political reasons doing stuff like this
          As Aromaz said, nothing for a hobbyist OU home-researcher.

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          • #6
            Originally posted by Aromaz View Post
            you need to generate and bombard the thorium with a LOT of neutrons.
            You would be right if the neutrons source is simply powder of radium mixed with powdered of beryllium.
            But the neutrons source is artificial, so neutrons flow depends on the level of vacuum helium.

            The neutron flux is almost inversely proportional to the vacuum level

            flux_neutrons = 1 / (rarefied_helium)

            Power electrical consumption from the source of neutrons depends on the level of vacuum.

            many neutrons = much electricity

            It does not sense to say that the neutrons souce produces few neutrons, because the number of neutrons depends on the electrical power and the vacuum level.

            The collision of one million helium atoms against to beryllium traget, makes 30 neutrons, which is not little.

            Instead powder of radium mixed with powdered of beryllium, makes practically nothing.

            Unfortunately exists the political to make false information, I do not know why of this one.
            Last edited by The_italian; 06-17-2010, 02:58 PM.
            signature...
            (sorry for my poor English, but I am italien)

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            • #7
              Vacuum will have no effect on neutrons or radiation quantity.

              The only issues are spin speed, temperature and quantity.

              But then: If you really start building this device, I will be the first in support, contribute and working along with physical duplications.
              Therefore we need to find NEW ways, NEW experiments and NEW lines of thoughts.

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              • #8
                You might find this page quite interesting; and best still: A real working model has already been build. However cost . . .

                New Page 1
                Therefore we need to find NEW ways, NEW experiments and NEW lines of thoughts.

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                • #9
                  suppose that the generator can produce an electric current 1 mA

                  We already know that 1 amp = 6.24 * 10 ^ 18 electrons

                  then
                  1 amp / 6.24 * 10 ^ 18 electrons = 1 * 10 ^ -3 amp / X:

                  X = (6.24 * 10^18*10^-3) / 1
                  X = (6.24 * 10^(18-3)) / 1
                  X = 6.24 * 10 ^ 15 electrons

                  a neutral helium atom has 2 electrons

                  so if 6.24 * 10 ^ 15 electrons go from a party, 3.24 * 10 ^ 15 helium nucleus must be to opposite side.


                  We need one million collisions to produce 30 neutrons.

                  one million = 1* 10^6

                  30 / 10 ^ 6 = X / 3.24 * 1 * 10^15

                  X = (3.24 * 10^15 * 30) / 10^6
                  X = (97,2*10^15) / 1* 10^6
                  X = 9,72*10^(16-6)
                  X = 9,72*10^10 neutrons

                  9,72*10^10 netrons are many, much more than powder of radium mixed with powdered of beryllium
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                  (sorry for my poor English, but I am italien)

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