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Mean Free Paths

  1. Dec 1, 2005 #1
    Question goes as follows:

    "In an ionised gas there are positive ions and electrones moving around in collision with molecules of the gas. Calculate the ratio of the mean free path of the electrons in the gas to the mean free path of the positive ions in the gas."


    As far as I can see I only need
    lamda = (KT)/(4pi root2 r^2 p)

    where everything will cancel except r giving r1:r2 ratio

    I could assume r is roughly 1x10^-10 for the molecules, but I don't think I'm supposed to know or find r for the electron since the compton radius ( (e^2/(me c^2) ) is a little beyond what we've done.


    What have i missed..?

    - Rob
     
  2. jcsd
  3. Dec 2, 2005 #2

    Physics Monkey

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    Hi RichBoxX,

    The r in your mean free path formula refers to the radius of the molecules with which the electrons collide. See here for details: http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/menfre.html

    If the gas is only somewhat ionized, the mean free path is dominated by collisions of electrons with non-ionized molecules and you can ignore the collisions of electrons with positive ions.
     
    Last edited: Dec 2, 2005
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