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Nuclear fission in an atomic bomb

  1. Jan 10, 2015 #1
    A person said:

    "However uranium-235 (U235) makes up only 0.72% of normal uranium metal and has to be separated from the remainder (mostlyuranium-238) in special factories which makes uranium-235 (U-235) a little more expensive. An uranium-235 (U-235) atomic has three neutrons less than an uranium-238 atomic. To, e.g., obtain 61 kg of pure uranium-235 (U-235) metal you need about 8 400 kg or 8.4 ton of uranium to separate the uranium-235 (U-235) from. How to separate uranium-235 (U-235) from uranium-238?

    It is, however 2012 - 67 years later, still TOP SECRET, what US factory managed to separate U-235 from 10 tons of U-238 by gas separation or whatever - magnetism? - and then making it, the U-235, a 72 kg metal slab again and what workshop manufactured and drilled the U-235 metal target rings and projectile ringsin 1945! Reason apparently being that no such workshop or technology existed at that time, 1942-1945, and no rings were ever manufactured. Of course, there was Oak Ridge, TN, with 75 000 people but they didn't know what they were doing."

    "It is the speed of the neutron, when it hits the nucleus that has a lot to do with how likely a fission is to occur. One might think, intuitively, that if the neutron is going really fast that it has a better chance of “shattering” the nucleus, but that’s not really how it works. Actually, for the fissile nuclei such U-235 the SLOWER the neutron is going, the more probable fission is.

    So slowed-down neutrons to maximize fission are an absolute requirement. And then from fission comes more neutrons, which continue the reaction. Well, mostly right. Actually, the neutrons born from fission are going really fast. Really, really fast. And they have to slow down to have a good chance of causing fission. That’s where the moderator comes in.

    The moderator in a nuclear reactor is the material whose job it is to slow down neutrons without absorbing them. This slowing-down is done by neutrons bouncing off the nuclei of the atomics in the moderating material. For most reactors, moderation takes place in the water that also cools the reactor. For a high-temperature reactor like the liquid-fluoride reactor, graphite (carbon) is used as the moderator. This was not really known in the 1940's when the atomic bomb was said to have been invented."
     
  2. jcsd
  3. Jan 10, 2015 #2

    mfb

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    What do you want to discuss? Who is "a person"?

    The US tried many different techniques of isotope separation. They are well-known know, and gas centrifuges became standard, while laser separation is a relatively new technique that might become cheaper and easier in the future (in 1945 lasers did not even exist).
     
  4. Jan 10, 2015 #3

    Astronuc

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    There are two aspects being discussed in the OP. One concerns the enrichment of uranium, from natural enrichment 0.72 U-235 to very high levels, and the other concerns the physics of an atomic bomb. Both technologies are mature and well-known to the practitioners.

    An atomic bomb works with prompt neutrons, where there is essentially no moderation, and the system is prompt super-critical until it disperses within a few microseconds. Most modern systems use Pu-239 rather than U-235.

    Many commercial nuclear reactors use moderation, and enrichments typically less than 5% U-235. Fast reactors have very little moderation, and use enrichments on the order of 20%, and typically use Pu rather than U.

    The OP is somewhat confusing. Other than the question "How to separate uranium-235 (U-235) from uranium-238?", it's not clear what is to be discussed. mfb answered the question about enrichment.
     
  5. Jan 10, 2015 #4
    Sorry about my last post. I just want to know if the person, who said the following, is correct or not?

    "So a 52 kg sphere of uranium-235 is a critical mass like 61 kg of rings of uranium-235? What?

    The temperature caused by exponential chain reaction will simply melt the uranium-235 (U-235) metal assembly that will flow or evaporate away.

    It is the speed of the neutron, when it hits the nucleus that has a lot to do with how likely a fission is to occur. One might think, intuitively, that if the neutron is going really fast that it has a better chance of “shattering” the nucleus, but that’s not really how it works. Actually, for the fissile nuclei such U-235 the SLOWER the neutron is going, the more probable fission is.

    So slowed-down neutrons to maximize fission are an absolute requirement. And then from fission comes more neutrons, which continue the reaction. Well, mostly right. Actually, the neutrons born from fission are going really fast. Really, really fast. And they have to slow down to have a good chance of causing fission. That’s where the moderator comes in.

    The moderator in a nuclear reactor is the material whose job it is to slow down neutrons without absorbing them. This slowing-down is done by neutrons bouncing off the nuclei of the atomics in the moderating material. For most reactors, moderation takes place in the water that also cools the reactor. For a high-temperature reactor like the liquid-fluoride reactor, graphite (carbon) is used as the moderator. This was not really known in the 1940's when the atomic bomb was said to have been invented.

    98.5% of the nuclei do not fission as the free neutrons miss them and fly away. 98.5%? Why not 100%?

    When the first (and only?) nucleus of 4x1026 U-235 nuclei in the above bomb fissions (it is split by free neutrons from somewhere - the Polonium-Beryllium initiators?), it will only heat the surrounding as it is not cooled and the 2.5 free neutrons will just fly away and produce nothing with their 2 MeV energy each. They cannot possibly collide with and fission anything. They must be moderated to produce further fission like in an atomic power plant."
     
  6. Jan 10, 2015 #5

    mfb

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    The critical mass of rings will depend on the ring geometry, but it is always larger than the critical mass of a sphere (at normal density).
    52kg is a realistic value for a hypothetical sphere of pure U-235.
    In a bomb, it will evaporate.

    Slowing down neutrons is very useful for reactors because it allows to go with lower enrichment rates and gives more control over the reaction. In principle you could do the same for nuclear weapons, but working with fast neutrons is easier there (you do not want to control the reaction anyway).
    That specific number suggests some specific origin where it might apply, it is impossible to tell what is meant here without context. A typical bomb will use more than 1.5% of the nuclei for fission.

    That is wrong.
     
  7. Jan 10, 2015 #6
    I meant "Little Boy."


    Here is another one, is this correct or not?:

    "Now, as for the claim that each fission results in on average 2.5 neutrons being released, this is true, but those diagrams illustrating a nuclear chain reaction are very misleading. They have the nuclei the size of marbles, and lined up like bowling pins, so that you see the two neutrons from each fission hitting the NEXT nuclei ! But the fact is, that if the nuclei of U-235 were the size of a marble ( about a half inch ), then the whole U-235 atom, with its outer electron clouds, would be the size of Yankee stadium ! And the neutrons would be the size of mustard seeds. One of these things emitted is just NOT going to hit the "next nucleus" ----- it will have to travel very far, passing by millions of other nuclei before hitting another nucleus. So, a chain reaction is a very sparse process."

    Also from this site: http://www.laradioactivite.com/en/site/pages/FastNeutrons.htm

    "The drawback of fast neutrons in reactors is that the probabilities of their capture by nuclei are comparatively small. Travelling in matter, neutrons see nuclei as targets. The apparent cross-section of these targets is much more smaller for fast neutrons than it is for slow neutrons. As a result, an intense neutron flux and a fuel rich in fissile elements are both needed to compensate for this lower probability."
     
    Last edited: Jan 10, 2015
  8. Jan 10, 2015 #7

    Astronuc

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    Well clearly, atomic bombs have been successfully demonstrated at Hiroshima and Nagasaki, as well as numerous tests by the nations that possess such weapons. As to exactly how they work, such details are not for discussion in the public domain.

    The fission chain reaction is a sparse process on the atomic level, but remember that there are on the order of 6.023E23 atoms per 235 gm of U-235.
     
  9. Jan 10, 2015 #8

    SteamKing

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    I don't think anyone has ever claimed that an A-bomb fissions all of the U-235 in its core, and the first bombs, while making a spectacular explosion, were very inefficient in this regard.

    The first U-235 device, Little Boy, which was dropped on Hiroshima, converted only about 600-860 mg (milligrams) of its enriched 64 kg U-235 core into energy. Assuming an enrichment of 80% U-235 in the core, the efficiency of the Hiroshima blast was at most about 0.000860/(0.80*64) = 0.0017% of U-235 fissioned, with the rest of the uranium being vaporized.

    http://en.wikipedia.org/wiki/Little_Boy

    This device used a crude gun-type mechanism to shoot a small uranium bullet into a larger fixed mass of uranium to produce a critical mass. All subsequent uranium bombs used an implosion mechanism, like that required for detonating plutonium cores.

    You don't have to take what "a person said" about these devices at face value. There are many sources of information about nuclear physics and nuclear weapons on the web and in published articles and books. These sources contain information about nuclear devices that has been de-classified over the years.
     
  10. Jan 11, 2015 #9

    mfb

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    I think you mixed the mass-energy (where 600-860mg is a good estimate) with the amount of uranium fissioned. Something like 2/3 kg of U-235 were used, which is roughly 1% of the uranium. 1.5% sounds reasonable. It was a very easy (they did not test this design before) and inefficient design.
     
  11. Jan 11, 2015 #10

    SteamKing

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    The early estimates of the yield of Little Boy varied quite a bit. Although the yield was initially announced as the equivalent of 20 KT, revised estimates put the yield as low as 13 KT, which finally became the accepted figure derived from data collected from examining the damage produced in other atomic explosions.

    The mechanism of Little Boy was tested by scientists at Los Alamos using mockups and simulations years after the real article was used in the war.

    According to more accurate calculations made after the war, 1 pound (0.45 kg) of U-235 undergoing complete fission has a yield of 8 KT. Since the final yield figure, as determined from calculations and simulations run after the war, was no more than 16 KT, the actual device used up no more than 2 lbs of U-235 (about 0.91 kg), out of a total of 141 lbs. (64 kg) of the uranium fuel with which the bomb was equipped, or about 1.5% fission efficiency.

    Sorry for not making this distinction clear in my earlier post.
     
  12. May 20, 2015 #11
    The Energy Released by Little Boy and Fat Man.

    Energy released E =

    (Mt+Mf )/2{[( Ro Kcom^2/3 - Rc )( 1 - ( Rc/Ro Kcom^2/3)^2 )( n - 1 )] /4t}^2

    Rc = ( pi Aw / 2 p Av ) { 1/ [ 3(n-1) Gf Gs] }^1/2

    Mt = mass of tamper/reflector Aw = atomic weight
    Mf = mass of fission material Av = Avogadro constant
    R = radius of fission material Gf = fission cross-section
    Rc = critical mass radius Gs = scatter cross-section
    Kcom = compression factor n = neutron yield
    p = density of fission material t = fission time
    Mc = critical mass pi = 3.14

    Hiroshima yield estimate.

    Mf = 64 Kg, Mt = 311 Kg (Tungsten carbide), U235 enrichment factor x = 0.8.
    Gf Gs
    U235 1.235 4.566 x 10^-24
    U238 0.3 4.804 ||
    n = 2.637, t = 10^-8 sec. p = 18.95 gm/cm3 and Kcom = 1.

    Over the range of 80 to 100% enrichment, a simplified relationship of the sum of the proportionate relative parameters is utilized. i.e. Gf*= Gf235x + Gf238(1-x)

    Gf* = 1.043x10^-24, Gs* = 4.613x10^-24, Mc = 22.38 Kg,

    Rc = 6.65 cm, Ro = 9.30 cm, E = 5.28x10^20 ergs

    TNT 4.200 jouls/gm E /4.2 x 10^16 ergs/ton = 12,600 tons TNT.

    For an enrichment of 85% , Mc = 21.8 Kg and the yield would be 15,300 tons TNT.
    Nagasaki yield estimate.

    Using the following parameters :-

    Pu 239, Mt ~ 300 Kg ( Uranium ), Mf = 6.2 Kg, p = 15.6 gm/cm3, n = 3,
    t = 8.5 nano-seconds, Mc = 10.5 Kg and Kcom = 2.25 & 2.4.

    Thus Ro = 4.56 cm, Rc = 5.44 cm, E = 18,800 tons TNT at Kcom = 2.25
    20,800 2.4

    It is estimated that 10 to 20% of the yield was due to the fission of the Uranium tamper.
     
  13. May 21, 2015 #12
    I forgot to add that approximately 55 milli-gram of fissile material per equivalent ton TNT undergoes fission.
     
  14. May 23, 2015 #13
    The Critical Mass of Plutonium 239 with regard to Fat Man.

    Jezebel 1950 ~55 ; Pu239 critical radius of a bare sphere = 6.285 cm at density 15.6 gm/cm3
    Mc bare = 16.22 Kg.
    For a reflector of 9.5 inches or 24.13 cm of Tungsten Mc = 5.8 Kg.

    Los Alamos 2013 Pu239 Mc bare = 16.624 Kg.

    Mragheb ; Pu 239 n = 2.98, Gs = 1.85 b, Gs = 6.8 b, p = 15.6 gm/cm3.

    For an infinite reflector where diffusion coef. Di = Do

    Rc = pi Aw/2 p Av [ 1/ 3 ( n -1) Gf Gs ]^½ = 4.62 cm , Mc = 6.4 Kg.

    Bare critical mass U 235 = 50 Kg and with reflector 17.5 Kg, ratio approximately 2.86 or 2^3/2. For Pu239 assume bare critical mass = 16.5 Kg then with reflector
    16.5 / 2^3/2 = 5.83 Kg or about 6 Kg.

    Fat man uranium reflector approximately 6 cm thick ( ~2.3” ), from ‘Los Alamos 1955’ for a 2” reflector the critical mass of U235 rises from 17.5 Kg to 26 Kg. Applying this to Pu 239 gives a figure of about 8.5 Kg.
    For Fat Man the Pu239 mass is quoted as 6.2 Kg and the criticality factor is then
    ( Mo/Mc )^1/3 = 0.9 giving a safety margin of 10%.

    Energy released ** E =
    ( Mt+Mf )/2{[( Ro Kcom^2/3 - Rc )( 1 - ( Rc/Ro Kcom^2/3)^2 )( n - 1 )] /4t}^2

    ( Mt + Mf ) = 300 Kg , n = 3, t = 8.5 nano-seconds, Rc = 5.067 cm, R = 4.56 cm and setting Kcom = 2.1, gives a energy release of 8.78 x 10^20 ergs, equivalent to 21,000 tons TNT.

    ** The estimate of yield is based on a lecture given by Werner Heisenberg 1945.
     
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