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I Fission of 238U with high energy neutrons & other questions

  1. Jan 31, 2017 #1
    Hi,
    I have a question related to 238U.
    I know of course that it can fission with neutrons above 1 MeV, but there is a elevated risk it would absorb the neutron and produce 239Pu instead.
    Let's assume you have a ram of metallic 238U, with perhaps 0,7% 235U, and that you collide it very very quickly (perhaps 1800 meters per second) with a target.
    Wouldn't the compression allow for an increase in the density of the metal, that would thus increase the probability of a nuclear fission ? Of course that nuclear fission would never be sustained.
    Second, wouldn't the presence of 0,7% 235U also provide for some fuel for a non-sustained chain reaction ? Neutrons would be slowed down by the 238U...
    The source of neutrons for the beginning of the non-sustained chain reactions would be the spontaneous fission of 238U (approx. 28 neutrons per second in a 2 kg rod of 238U).
    So these non-sustained chain reactions would create some heat (perhaps a few thousands of degrees C), and some light, but no explosion at all ?

    Thanks a lot in advance,
     
  2. jcsd
  3. Jan 31, 2017 #2
    Solids are negligibly compressed by pressure ... You maybe thinking of a gun type fission bomb , where an explosive fires a chunk of U235 into another piece to make a sphere of the metal .. this shape has minimal neutron loss so creates criticality with the least mass .... compression is not a factor .

    it would create some heat , perhaps a 0.000000000001C increase in temp ....28 neutrons a second is nothing.
     
  4. Jan 31, 2017 #3

    mfb

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    If your material is prompt critical, it is a bomb.

    If your material is critical only including delayed neutrons, it can give a stable reaction with some additional regulation - something you can use in a nuclear reactor.

    With just 0.7% U-235, your material is not critical - the density doesn't matter, the U-238 will absorb too many neutrons. Fission will be completely negligible. Even if you get 0.99 new neutrons per initial neutron, that just gives a multiplication factor of 100. Your 28 neutrons per second from spontaneous fission lead to 1000 fission reactions per second, a power of some nanowatts. Completely negligible.

    You just need more pressure. Fuel in nuclear weapons does get compressed. Plutonium for example has different phases with large differences in density - apply enough pressure and it goes to a lattice structure with a 23% higher density.
     
  5. Jan 31, 2017 #4
    Ok thanks a lot.
    Another possibility is the contamination by 238Pu and 240Pu. My question was actually related to the use of uranium in so-called "depleted" uranium weapons, which are not made with depleted uranium as :
    - Asaf Durakovic found 0,7% 235U in several samples taken in Irak, as well as 236U (0,005%), indicating this is not natural uranium from the ground
    - The US army itself acknowledged actual contamination with plutonium, neptunium and americium (during a press conference)
    The presence of 236U confirms that there could be isotopes of, for instance, plutonium, together with the uranium.
    My idea is that the "tip" of the rod, when penetrating into the target, gets compressed and that it would increase the probability that a limited reaction happens there (the 0,7% of 235U could react with the neutrons.from the spontaneous fissions of 238Pu and 240Pu).
    This would explain actually why is there always, at the impact of a "depleted" uranium shell, a "flash" of light before the fireball.
    The metal oxydises (the fire can reach 5000°C), it is "pyrophoric" like most metals, but someone told me there would need to be a "firestarter" somewhere, and suggested me some fission.
     
  6. Jan 31, 2017 #5
    Actually, if there were traces of 250Cm or 252Cf, undetectable levels could be enough to provide a good chunk of neutrons to the 235U...
     
  7. Jan 31, 2017 #6

    mfb

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    You would need weapon-grade uranium (enriched to ~90% - something that certainly doesn't get shot away) to have any chance to get criticality. Forget nuclear reactions. Everything that happens is mechanical or chemical.
     
  8. Jan 31, 2017 #7
    I thought of subcriticality, not criticality. I also thought that the spontaneous fission could feed 238U with neutrons > 1 MeV. If there's even only trace amounts of 250Cm and 252Cf, these would get closer to 238U atoms when compressed... The neutrons get closer to 238U locally during a few milliseconds...
     
  9. Jan 31, 2017 #8

    mfb

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    Subcritical is equivalent to "nothing nuclear happens" - see above, even if you are very close to criticality the energy released in fission processes is completely negligible. Unless you have a massive particle accelerator as spallation source nearby, but that is a different topic (accelerator-driven fission).
     
  10. Jan 31, 2017 #9
    Ok thanks a lot.
     
  11. Jan 31, 2017 #10
    Something important to keep track of is infinite multiplication factor.
    It depends only on composition and is independent on size, density or presence of neutron reflectors.
    For U-238, it is below unity. And you need appreciable amounts of fissile isotopes to reach unity.
     
  12. Jan 31, 2017 #11
    In other words, do kinetic antitank projectiles made of DU or natural uranium have nuclear reactions during impact? Answer is "no".
     
  13. Feb 1, 2017 #12
    Nonsense. Such element would cost astronomical money to manufacture (billions per gram).

    Uranium is better because tungsten is not pyrophoric, and mushrooms on impact, reducing penetration. Also, not sure tungsten is cheaper. DU is basically free since supply is vastly bigger than demand.
     
  14. Feb 1, 2017 #13
    Right.
    What could be feasible is put highly enriched U-235 instead of depleted U, and thus build a fission bomb which is deformed into criticality not by internal high explosive, but by kinetic impact into target.
    However, such a nuclear bomb needs to contain high critical mass of HEU, would be low yield and expensive.
     
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