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Nuclear explosion in space x-rays hit atmosphere.

  1. Oct 20, 2009 #1
    Suppose we set off a nuclear weapon 400 km above the surface of the Earth. See:

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

    When the x-rays hit the atmosphere what gets the greatest velocity boost, electrons or ions?

    Can we assume the averaged velocity of all boosted particles is in the direction opposite the blast?

    Thank you for any help.
     
  2. jcsd
  3. Oct 20, 2009 #2

    Astronuc

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    Staff: Mentor

    Electrons being lighter would scatter gamma/x-ray photons and receive more energy than ions. The blast wave would apply some pressure on the ions.

    However, realize that the blast front density rapidly decreases with 1/r3.
     
  4. Oct 20, 2009 #3
    Thanks, though I'm not clear on the last sentence. The blast front is well(?)..... the front of the blast? X-rays lead the race followed by electrons and ions? And if I double my distance from the blast I should halve my x-ray dose? What is it that goes as 1/r^3?

    Thanks again!
     
  5. Oct 21, 2009 #4
    If I double my distance from a source (point like) I should reduce my dose by 1/4.
     
  6. Oct 21, 2009 #5
    1) Electrons are lighter and easier to displace.

    2) The averaged - yes but keep in mind that an X-ray is a transversal electromagnetic field. It makes charges oscillate in the transversal to the propagation direction. As a result, the charges will obtain transversal and longitudinal vectors of velocity. The averaged transversal components give zero, the longitudinal remain.
     
    Last edited: Oct 21, 2009
  7. Oct 21, 2009 #6
    The plots of the absorption cross section of photons below 100-200 KeV show a large increase above the Thomson -scattering cross section, due primarily to deep-core photoejection of bound atomic electrons. There is a plot in Evans, "The Atomic Nucleus", (page 696) showing the directional distribution of photoejected electrons as a function of photon energy varying from about ~60 degrees max at 20 KeV to ~15 degrees at 511 KeV. On the same page it states that the "photoelectrons tend to be ejected in the direction of the electric vector of the incident radiation."
    Bob S
     
  8. Oct 21, 2009 #7
    Which should average to zero(?) perpendicular to the beam as the x-rays have equal probability for all polarizations(?).

    If this "beam" of x-rays scatter into all directions then the initial momentum of the x-ray beam is greater then the final summed momentum of the scattered x-rays, this momentum difference shows up in the momentum of the electrons and ions with the average momentum in the direction of the beam, as you stated above?.

    Thank you for your time!
     
  9. Oct 21, 2009 #8
    Posted by Bob S
    The plots of the absorption cross section of photons below 100-200 KeV show a large increase above the Thomson -scattering cross section, due primarily to deep-core photoejection of bound atomic electrons. There is a plot in Evans, "The Atomic Nucleus", (page 696) showing the directional distribution of photoejected electrons as a function of photon energy varying from about ~60 degrees max at 20 KeV to ~15 degrees at 511 KeV. On the same page it states that the "photoelectrons tend to be ejected in the direction of the electric vector of the incident radiation."

    Probably not. Low energy x-rays come from electrons hitting high-Z targets and creating bremsstrahlung, and the x-rays are largely polarized at 90 degrees at low energies. See Heitler, "Quantum Theory of Radiation", 3rd edition, pgs 244-245.

    Deep-core photoejection and the photoelectric effect absorb the entire energy of the incident photon. There are no scattered x-rays. Compton scattering (and Thomson scattering), usually dominant above ~100 KeV, both have scattered x-rays.
    Bob S
     
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