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Industrial grade positron formation

  1. Jun 30, 2011 #1
    Is there any way to economically produce positrons using particle accelerators or any other method?
     
  2. jcsd
  3. Jun 30, 2011 #2
    Sure. Just irradiate some material ina particle accelerator that would become a positron emitter. Look up the process for creating F-18.
     
  4. Jun 30, 2011 #3

    Bill_K

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    The accelerators typically produce positrons by letting a beam of gamma rays strike a target. This creates electron-positron pairs, which are then separated and thermalized.
     
  5. Jul 2, 2011 #4
    ok so is it possible to form positrons by bombarding electrons on a material (tungsten or lead - because of its high atomic no.)? because electrons on colliding with other electrons basically produce x rays which are basically low energy gamma raysso these gamma rays can produce positron electron pairs
     
  6. Jul 2, 2011 #5

    Bill_K

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    Here's a report on a positron source that does it that way:

    "Positrons are generated via pair creation when energetic electrons are stopped in a high-Z target, the electron-positron converter. These positrons are then moderated, i.e. slowed to thermal energies, via interaction with a suitable material such as tungsten, and a slow positron beam produced."

    http://arxiv.org/abs/1102.1220
     
  7. Jul 5, 2011 #6
    what would be the efficiency of this method?
     
  8. Jul 5, 2011 #7

    Vanadium 50

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    It's in the paper that Bill_K pointed you to. It's even in the abstract. We can point you to papers, but you have to read them yourself.
     
  9. Jul 5, 2011 #8

    jambaugh

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    Reading the thread title, my first question was
    So positrons come in different grades?

    I'll take 36 industrial grade positrons and 204 consumer grade positrons if you please! ;)
     
  10. Jul 5, 2011 #9

    Vanadium 50

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    Prime, choice, select, standard, commercial, cutter and canner. (Mmmm...utility grade positrons)

    And nuclei come in sizes small, medium, large, extra large, jumbo, giant, colossal and super colossal.
     
  11. Jul 5, 2011 #10

    jambaugh

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    So that's the REAL reason for the "non commuting c-numbers"! It's a promotional gimmick!

    SUPER-SIZE ME!
     
  12. Jul 7, 2011 #11
    can any one express the efficiency in billi_k's link in terms of percentage?
     
  13. Jul 7, 2011 #12

    Vanadium 50

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    I'm sure they can. Where are you getting stuck in the calculation?
     
  14. Jul 7, 2011 #13
    ummm.. at the beginning the calculation was a bit complicated
     
  15. Jul 7, 2011 #14

    Vanadium 50

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    Show us how far you got, please.
     
  16. Jul 8, 2011 #15
    ok so the abstract said "For an electron beam energy of 5 MeV (10 MeV) and current 240 $\mu$A (30 $\mu$A) production of a slow positron beam of intensity 5 $\times$ 10$^{6}$ s$^{-1}$ is predicted. The simulation also calculates the average energy deposited in the converter per electron, allowing an estimate of the beam heating at a given electron energy and current. For low energy, high-current operation the maximum obtainable positron beam intensity will be limited by this beam heating."

    however i could not get the SI unit so can anyone tell me the SI unit as well as the formula for converting this unit into percentage
     
  17. Jul 18, 2011 #16
    Can any one tell me how many grams of positrons would be generated in one hour?
     
  18. Jul 18, 2011 #17

    xts

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    Skullcracker, if you know how many positrons are created in one second, then you know how many seconds are in one hour, and you know what is positron mass (if you don't - wiki it), then you should be able to answer your question yourself. If not, ask your 3rd grade primary school ma'am for help.
     
  19. Jul 23, 2011 #18
    xts u dint get the question i wanted to know what 'production of a slow positron beam of intensity 5 $\times$ 10$^{6}$ s$^{-1}$ is predicted.'meant and i dont think third grade teachers would be teaching particle physics
     
  20. Jul 23, 2011 #19

    Vanadium 50

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    We're done here.
     
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