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Beta decay - derive an expression relating source angle to energy

  1. Oct 15, 2005 #1
    Hi guys,
    Need a spot of help as i cant seem to find where to go next :(
    Ok, so there is a source which emits beta particles (currently unknown whether they are positrons or electrons). It can move at an angle +/- 90 degrees to the GM tube. I need to find an expression for the energy of the particles related to the angle the apparatus is at. Here are the equations given to me:
    [tex]E = m_{0}c^2 [\sqrt{1+(\frac{p}{m_{0}c})^2} - 1][/tex] (1)
    [tex] r = \frac{R}{\tan(\frac{\Theta}{2})}[/tex] (2)

    Where

    r = the radius of the electron/positron circle.
    R = the radius of the circle used for the source/GM tube setup.

    I also have the following equation:

    [tex] F = \frac{mv^2}{r} = qvB[/tex] (3)

    Combining (2) and (3) gives me:

    [tex] \frac{R}{\tan(\frac{\Theta}{2})} = \frac{mv}{qB}[/tex]

    I cant see where to go next...any ideas? Cheers :)
     
  2. jcsd
  3. Oct 15, 2005 #2

    Chi Meson

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    I can't picture the setup of the apparatus. Could you provide a sketch?
    Until then, have you noticed that momentum (mv) is in both formulas (the formulas at the top and bottom of your post)?
     
  4. Oct 15, 2005 #3
    oops, sure :)

    http://img427.imageshack.us/img427/6121/scan1lu.jpg

    I didnt notice that momentum was in both, silly mistake. Ok so if i do substitute in momentum it still seems like a bit of a hefty equation..i dunno if i can simplify it much?

    Out of interest, could anyone break down the energy equation for me or hint towards what part is what? Apparently its both kinetic and potential energies in one but i cant really see how they got that. Thanks :)
     
  5. Oct 16, 2005 #4

    Chi Meson

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    Yes it is hefty. This is where physics gets fun. Just take a deep breath, get all your values together and thank god you don't need to use a slide-rule anymore.

    And the first formula is the kinetic energy formula for objects travelling at relativistic speeds. it has conveniently been put in terms of the object's momentum rather than speed.
     
  6. Oct 16, 2005 #5
    ok ive now got this:

    [tex]E = m_{0}c^2 [\sqrt{1+(\frac{RqB}{m_{0}c\tan{\frac{\Theta}{2}}})^2} - 1][/tex]

    I take it that cant be simplified any further. Ack this is gonna be hard to work out errors with!

    Thanks for your help!
     
  7. Oct 16, 2005 #6

    Chi Meson

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    I've seen worse.
     
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