Calculating Ratio of Electron and Positive Ion Mean Free Paths in an Ionised Gas

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SUMMARY

The discussion focuses on calculating the ratio of the mean free paths of electrons and positive ions in an ionized gas using the formula λ = (KT)/(4π√2 r² p). The key takeaway is that the radius (r) in the formula represents the molecular radius of the gas, which is approximately 1x10^-10 meters. It is established that in a partially ionized gas, the mean free path for electrons is primarily influenced by their collisions with non-ionized molecules, allowing for the simplification of the calculations by ignoring collisions with positive ions.

PREREQUISITES
  • Understanding of kinetic theory of gases
  • Familiarity with the concept of mean free path
  • Knowledge of ionization processes in gases
  • Basic grasp of molecular radii and their significance in collision theory
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  • Study the effects of ionization on gas properties
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RichRobX
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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
 
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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.
 
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