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Mean Free Path of an Electron  
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#1
Nov2305, 08:32 PM

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An electron can be thought of as a point particle with zero radius.
Find an expression for the mean free path of an electron through a gas. The mean free path of an electron can be represented by: [tex]\lambda = \frac{1}{\sqrt{2} \pi \frac{N}{V} r^2}[/tex] Electrons travel through L = 3 km in the Stanford Linear Accelerator. In order for scattering losses to be negligible, the pressure inside the accerlator tube must be reduced to the point where the mean free path is at least 50 km. What is the maximum possible pressure inside the accelerator tube, assuming T = 293 K? By using the equation [tex] \lambda = \frac{L}{N_{coll}}[/tex] [tex] N_{coll} = \frac{L}{\lambda } = \frac{3}{50} = 0.06 collisions[/tex] I then know that [tex]N_{coll} = \frac{N}{V}V_{cyl} = \frac{N}{V}\sqrt{2} \pi r^2 L[/tex] However, I don't know how to continue from here to find the pressure, because I know neither the number density [itex]\frac{N}{V}[/itex] or the radius [itex]r[/itex]. 


#2
Nov2305, 09:24 PM

Sci Advisor
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P: 1,333

Clearly you can't treat electrons as point masses in your current model since a point mass corresponds to [tex] r = 0 [/tex] i.e. infinite mean free path. I would suggest trying to examine your assumptions again.
Here is some information about the model you are currently using: http://hyperphysics.phyastr.gsu.edu...ic/menfre.html 


#3
Nov2305, 09:41 PM

Emeritus
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PF Gold
P: 11,155




#4
Nov2305, 09:53 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

Mean Free Path of an Electron 



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