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A third question

  1. Nov 20, 2004 #1
    What are all of the particles, radioactive or otherwise, left over from a nuclear explosion, and what are their characteristics? Scientists who have been involved in monitoring underground detonations would be best qualified to answer this, however, I would welcome all input and information.
     
  2. jcsd
  3. Nov 21, 2004 #2

    DaveC426913

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    A short list:
    Alpha particles - the common helium nucleus, stripped of its electrons (He^2+)
    Beta radiation - high-energy electrons or positrons ejected from a nucleus
    Gamma rays - an energetic form of electromagnetic radiation (above X-rays)

    I found some good descriptions on Wikipedia: http://en.wikipedia.org/wiki/Alpha_particle
    Google them to taste, for all the info you can read.
     
  4. Dec 1, 2004 #3
    Spectral Analysis

    Has femto-second or, at least nano-second spectral analysis ever been done during a nuclear detonation? And, are there particles that have been created then destroyed during the blast? Are alpha particles the only particles that exist without electrons? Are there other particles that occur naturally or otherwise that are without electrons? What is meant by 'fast-moving particles' in reference to particle radiation? Do alpha and beta particles recombine with water when it rains to become non-radioactive? How long do the alpha and beta particles last and what (other than rain or water) makes them become non-radioactive? Are there different types of radioactivity with different types of bombs (Other than alpha, beta, gamma and neutron exitation)? Or is there just more of these particles with different types of nuclear bombs? You didn't mention x-rays. What is the full spectrum of radiant energy emitted by a nuclear detonation and do all types of bombs emit the same spectum of energy? Is there different intensities at different frequencies for different bombs? What are the intensities of radiant energies? Does the spectrum change during the duration of the fireball?
     
    Last edited: Dec 1, 2004
  5. Dec 7, 2004 #4
    Also, about positrons... Do they only annialate with electrons? Can they annialate with protons or nuetrons? If they're anti-matter shouldn't they be able to annialate with any form of matter? How long do they last after a nuclear detonation? My interest is purely scientific and I will return the favors with another 'interesting conclusion' if my questions are answered, otherwise, this forum is of no further use to me and I will leave. None of these questions should be covered by any secrets act, and if they are, someone is really paranoid. Just about anybody in the world can make a nuclear bomb and if they had any brains, they would know that they don't need fissionable materials to do it. :surprised
     
    Last edited: Dec 7, 2004
  6. Dec 8, 2004 #5
    Holy crap...you dont want much lol
     
  7. Dec 8, 2004 #6

    Morbius

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    Dockwood,

    In terms of radiation, you get alpha, beta, gamma, and neutrons.

    The alpha particles are short range - slow down, pick up stray electrons, and
    become neutral helium-4.

    Beta are electrons - they also are slowed down and captured and become part
    of the background material.

    Gammas are Compton scattered and eventually absorbed, for example, by
    pair production - which produces a positron / electron pair. The positron
    eventually annihilates with another electron. A particle of anti-matter only
    annihilates with its matter counterpart.

    There are conservation laws - conservation of lepton and baryon numbers -
    for example. An electron is a lepton, a positron an anti-lepton. A proton is
    a baryon. The positron has a lepton number of -1. The proton has a baryon
    number of +1. If you annihilated a positron with a proton - the number of
    leptons would increase by 1, and the number of baryons would decrease by 1.
    The conservation laws say that can't happen.

    As far as making a nuclear bomb - you don't need "fissionable" material - but
    you do need "fissile" material. It's not as easy as you think!

    Dr. Gregory Greenman
    Physicist
     
  8. Dec 8, 2004 #7
    Nuclear fission fragments

    He said, "left over," Morbius, not "immediate emissions." Here is a fairly concise answer to his question, in terms of fission (and not fusion)...
    http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/fisfrag.html
    ...except that that article does not deal directly with fission explosions. Fission explosions would also at least involve releasing unfissioned uranium or plutonium (since fission bombs do not burn their nuclear fuel with 100% efficiency).
     
  9. Dec 8, 2004 #8

    Morbius

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    hittssquad,

    "Immediate" or "left over" - depends on your time scale. Since Dockwood
    had mentioned nanosecond time scales, I assumed he was talking about
    what is left over after the nuclear event - but still on the short time
    scale.

    Nothing is 100% efficient - but you'd be surprised what the efficiency is.

    Dr. Gregory Greenman
    Physicist
     
  10. Dec 8, 2004 #9
    What is a lepton number? And what about combinding a lepton and baryon? Are those just different names for protons and electrons?
     
  11. Dec 9, 2004 #10

    Morbius

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    Candyman,

    The lepton number is the number of leptons you have.

    Leptons and baryons are not just another name for electrons and protons -
    they are names of a class or category.

    Leptons include electrons and neutrinos

    Baryons include protons and neutrons.

    Let me take another example of this - beta minus decay. Essentially you
    have a radioisotope that has too many neutrons to be stable. One of the
    neutrons decays to a proton, an electron, and an anti-neutrino:

    n -> p + e + nu-bar [ "nu-bar" - a nu with a bar over it is the symbol
    for the anti-neutrino ]

    Let's look at the baryon number. Before the reaction, we had 1 baryon;
    the neutron - and after the reaction we have 1 baryon - the proton.
    So baryon number is conserved - +1 before, +1 after.

    Let's look at lepton number. We had 0 leptons before. Afterwards we
    have an electron [ lepton number +1 ] and an anti-neutrino [ lepton
    number -1 ]. [ The number of the anti-particle is the negative of the
    particle's number ]. The lepton number after the reaction is +1 + (-1) = 0
    So we started with 0, and ended with 0 - lepton number was conserved.

    Let's look at charge. Before the reaction, we had a chargeless neutron -
    so the charge was 0. Afterwards we have a proton [ charge +1 ], and an
    electron [ charge -1 ]. Total charge afterwards is +1 + (-1) = 0
    We started with 0 net charge and ended with 0 net charge.

    So the number of baryons was conserved, the number of leptons was
    conserved, and the net charge was conserved.

    Dr. Gregory Greenman
    Physicist
     
  12. Dec 9, 2004 #11
    Thank you for the examples with that reply, Dr. Greenman. But, why does net charge have to be conserved? Is charge not a form of energy, and is there not a small change in the total mass between a neutron changeing into a proton/electron/anti-neutrino?
     
  13. Dec 9, 2004 #12

    Morbius

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    Candyman,

    Just like conservation of energy - the laws of physics dictate conservation
    of charge too. It comes out of Maxwell's equations. Courtesy of
    Georgia State University Dept of Physics:

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxsup.html

    Courtesy of Clark Atlantic University:

    http://www.hbcumi.cau.edu/tqp/451/451 Module I/451-03/451-03.html

    Since total energy is conserved - if in the beta decay reaction - you
    take the energy equivalent of the mass of the parent radioisotope -
    and compare that to the sum of the energy equivalent of the daughter
    isotope, plus the energy equivalent of the electron, plus the energy
    equivalent of the anti-neutrino plus the kinetic energies - you will find
    that the energy equation balances too - total energy is conserved.

    Dr. Gregory Greenman
    Physicist
     
    Last edited: Dec 9, 2004
  14. Dec 9, 2004 #13
    What is a neutrino's mass in relation to electron? About the same size? The same for anti-neutrinos?

    Also, I thought neutrions moved at 99 percent the speed of light. Would this not require alot of energy, to attain this speed? Or is that speed only the ones coming from the sun as their source?

    Sorry to bother you with so many questions.
     
  15. Dec 9, 2004 #14

    Morbius

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    Candyman,

    Actually it's a very good question.

    At one time, physicists thought that neutrinos moved at the speed of light
    like photons. In order to do that - they would have to have a rest mass of
    zero like the photon.

    More recently, it's been suggested that the neutrino moves somewhat
    slower than the speed of light. Therefore, it has a small mass - smaller
    than the electron.

    It's understandable that this is not a settled issue - it's very hard to
    study neutrinos - they interact so weakly with matter. Our experiments
    determine the properties of neutrinos [ or any other nuclear particle
    for that matter ] by how it iteracts with other matter. The neutrino
    interacts so weakly with matter, that most of the time, neutrinos
    stream right through matter without "doing anything" - so how do you
    study them?

    There is a concept called the "mean free path" - which is on average -
    how far will a particle "free stream", i.e. propagate without interacting
    before it eventually interacts. You can think of it as the average
    distance the particle goes before it "collides" with something.

    You can determine the mean free path of practically any particle in
    any material if you know the nuclear "cross-section" or interaction
    probability.

    If you calculate the mean free path of neutrinos in solid lead - you would
    measure that distance in "light-years". It's that long!!!

    Scientists use huge underground vats of liquid [ I believe a certain
    cleaning fluid has the right properties, and is cheap enough to make a
    huge underground detector.] There are photo-detectors surrounding the
    vat of fluid that watch for the almost infinitesimal flashes of a neutrino
    interaction. They build it underground for shielding - nothing but
    neutrinos can get all the way down underground.

    http://antwrp.gsfc.nasa.gov/apod/ap990623.html

    [Notice that man standing near the bottom left corner - it gives you
    a sense of the size of the detector.]

    Even with these large detectors - a neutrino interaction "event" is rare.

    It's amazing that we can study them at all.

    Dr. Gregory Greenman
    Physicist
     
  16. Dec 9, 2004 #15
    It is huge! And more than a mile underground!

    Light-year distance through solid lead? I am guessing that is because they have mass, unlike photons, which implies they do not have wavelike properties which is why photons do not go throught solid objects?
     
  17. Dec 10, 2004 #16

    Morbius

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    Candyman,

    Nope - even when we thought that neutrinos had zero rest mass - the
    mean free path in lead was still many light-years.

    The reason the mean free path is so long is that the force with which
    neutrinos interact with matter is the "weak" force. [ which is a certain
    manifestation of the "electro-weak" force - there's been a unification
    between the weak force and the Coulomb force].

    Quantum mechanically, neutrinos have wavelike properties - just like
    any particle does - even electrons, protons, neutrons....

    Photons don't go through solid matter because they react with the
    electrons that surround atoms. If a photon has an energy that is
    equal to the difference in energy between a lower "orbit" that has an
    electron in it, and an unpopulated higher "orbit" - the electron in the
    low orbit can absorb the photon and use its energy to move to the
    higher orbit. If the photon has an energy in excess of 1.02 MeV - then
    you can have "pair production" - if the electron passes close enough to
    an atom - its energy can be converted to an electron / positron pair.
    There are other reactions that scatter or deflect the photon instead
    of absorbing it.

    Photons interact with electrons via the Coulomb force.

    However, neutrinos interact with the weak force - and it's just not as
    "strong" as the Coulomb force. With such a meager force causing
    interactions - neutrinos just flat out don't "do anything" when passing
    through matter [ which is mostly empty space ]. THATS why neutrinos
    have such a long mean free path.

    Dr. Gregory Greenman
    Physicist
     
  18. Jan 11, 2005 #17
    I appreciate the concise responses of Dr. Greenman (Moribus) and the questions and comments of the Candyman, even though only part of my questions were answered. It takes me awhile to play catchup, meaning that some of your responses require a great deal of research and investigation on my part, followed by a time of brainwracking to thoroughly understand the issues and 'test' the theories and, sometimes, developing my own. You are very good at communicating Dr. and it is obvious that you have a great deal of understanding. I cannot express enough my gratitude for your responses. I still have several unanswered questions and, doubtless, I'll have more when these are answered.

    First unanswered question: What is the largest particle extant (natural or synthetic) without electrons, not counting the 'free' particles of an accelerator. (I stated my reasons for not believing that accelerator particles are free of electrons in my 'Another question' which I posted under the Quantum physics forum, which forum was truncated to three pages about the time my thread hit page four.)

    Second: Has femto, pico, or nano second spectral analysis ever been performed during a nuclear detonation? The reason that I ask is that I assume that the greatest changes in the fireball will likely happen in the first few femto seconds after detonation, and that elemental changes will also occur during this time.

    Thirdly: What is the bandwidth and intensity of propagative emmisions during the entire length of a detonation. I am assuming that certain frequencies will dominate during the first femto seconds of the detonation while other frequencies will dominate at later intervals. Most likely from gamma to lower frequencies.

    F.Y.I., quoting from Sarbacher's 'Encyclopedic Dictionary of Electronics and Nuclear Engineering (Published in 1959): "Fissile. In nucleonics, fissionable (q.v.). Fissile is used more in England and Canada than in the the United States."

    I know that much of the material in the book is obsolete, but I use it mostly to weed through some of the current mis-information floating around on the internet and sometimes this forum. Plus I use it as a comparison of old and new. That being said, I'll use fissile henceforth, assuming that fissionable is obsolete. I bought the book in Japan in 1968.
     
    Last edited: Jan 11, 2005
  19. Jan 11, 2005 #18

    Morbius

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    Nuclear tests were heavily diagnosed. The purpose of a test was not to
    determine simply whether the device went "boom" or not - but the
    details of exactly how the process happened.

    It's certainly very complex. At detonation, the device starts out
    extremely hot, and cools as it radiates and expands. The big fireball
    one sees in films of nuclear detonations is a state that has expanded
    and cooled a great deal from when all that energy was contained in
    only the mass of the nuclear device, itself.

    The accepted definitions of "fissile" and "fissionable" by scientists are
    as follows.

    "Fissionable" means that the nuclide will fission - but there is a threshold
    energy that the neutron must have. That is - if one were to introduce
    a low energy neutron into the nuclide - the nuclide would not fission.
    The neutron has to carry with it an additional amount of kinetic energy
    in order to make the "fissionable" nuclide unstable and subsequently
    fission. U-238, for example, is "fissionable". U-238 will fission, but only
    if the triggering neutron has an energy above ~1 MeV.

    "Fissile" means that the nuclide will fission with low energy neutrons.
    The mere introduction of a neutron into the nuclide will cause the
    nuclide to be unstable with respect to fission. The neutron need not
    bring any additional kinetic energy with it - the mere fact that the
    neutron falls into the nuclide's nuclear potential well is enough to cause
    the nuclide to fission. U-235 is "fissile".

    In a light water moderated nuclear reactor - most of the neutrons are
    low energy "thermal" neutrons - their temperature is in equilibrium
    with the temperature of the materials of the reactor. Such neutrons
    have energies that are a small fraction of an eV.

    If the reactor has enough U-235 in it - one can have a self-sustaining
    chain reaction. However, if the reactor had nothing but U-238 in it -
    it would not support a chain reaction - even though U-238 is "fissionable".
    [There aren't enough high energy neutrons is a water moderated,
    "thermal reactor".]

    A thermal reactor runs on thermal neutrons fissioning U-235. As a result
    of fission - there are fast neutrons produced [ which will later slow down
    after scattering off the hydrogen in the water ]. Before these neutrons
    slow down - they are above the threshold of U-238 fission, and will cause
    some fissioning of U-238 - but this is a small percentage of the fissions.

    A very early reactor calculational model was Enrico Fermi's "four factor
    formula". One of the four terms was the "fast fission factor". You might
    search for information on those terms.

    Dr. Gregory Greenman
    Physicist
     
  20. Jan 17, 2005 #19
    Forces

    When you were a kid you must have been very good at dodgeball Dr.. I guess you could consider me as a sort of technical no spin zone. Obviously, my questions are not going to be answered. I'm not really surprised, though, and we both know why. That's not to say that you don't provide really good information. I have a great admiration for people of your capability. I have one last question, or at least an attempt at one. How would you compare the different forces in the core? In other words, if I gave the attractive forces of the particles a weight of one per particle, what would the repulsive forces be (per repulsive particle)?
     
  21. Jan 17, 2005 #20

    Astronuc

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    First question - do mean largest particle or largest stable particle without electrons? Excluding atomic nuclei, the largest stable particle would be a proton. On the other hand, since protons and electrons really dig each other, they don't stay very far away from each other. Charge neutrality is the local gig. You did, man? :biggrin:

    Quantum mechanically speaking, in a hydrogen atom, the electron orbits some distance from the proton. Furthermore, protons like to pair up with their electron partners to form diatomic hydrogen molecules. :biggrin:

    Second question - femto-spectral analysis? Get real! No way! The resolution of the best electronics is going to be nano-seconds, and by the time one processes the data - more like 10's or 100's of nano-seconds. And the measurements will not give one a direct observation of what is happening at the atomic/nuclear level. One will get a prompt gamma-burst, but that will involve some degradation of the gammas by the Compton effect.

    And what is meant by 'elemental changes'? The fission process is well known and understood. No surprises. Dr. Greenman has answered your questions about subatomic particles - quite adequately.

    Third question - the fission process takes place over nano-seconds. Why be concerned with femto- or pico-seconds. Firstly, the fission process releases gamma-rays (prompt gammas, with energies of 1 to 7 MeV), with fission products and neutrons. Some fission products may decay by beta-decay within a few milliseconds - but most have half-lives of seconds or more, so they do not release beta-particles or anti-neutrinos until after the CM has dissociated.

    The fissile CM (e.g. Pu-239 (+240, 241)) becomes a plasma of unfissioned Pu and fission products. The prompt gammas will scatter on the electrons (Compton effect) and the other atoms will collide causing ionization, from which recombination will produce photons of wavelengths characteristic of the particular atoms.

    So, the detonated CM will produce a broad spectrum of electromagnetic radiation consisting of gamma rays, X-rays, UV, visible and infrared.

    Just curious - what is the point of your questions?
     
    Last edited: Jan 18, 2005
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