Mean Free Path of an Electron -

1. dekoi

0
Mean Free Path of an Electron -- URGENT

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:
$$\lambda = \frac{1}{\sqrt{2} \pi \frac{N}{V} r^2}$$

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 $$\lambda = \frac{L}{N_{coll}}$$
$$N_{coll} = \frac{L}{\lambda } = \frac{3}{50} = 0.06 collisions$$
I then know that

$$N_{coll} = \frac{N}{V}V_{cyl} = \frac{N}{V}\sqrt{2} \pi r^2 L$$

However, I don't know how to continue from here to find the pressure, because I know neither the number density $\frac{N}{V}$ or the radius $r$.

2. Physics Monkey

1,362
Clearly you can't treat electrons as point masses in your current model since a point mass corresponds to $$r = 0$$ 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.phy-astr.gsu.edu/hbase/kinetic/menfre.html

3. Gokul43201

11,046
Staff Emeritus
PM, the OP is talking about electron-molecule collisions, not electron-electron collisions. The molecules having a non-zero diameter gives the electron a finite mean-free path.

Last edited: Nov 23, 2005
4. Gokul43201

11,046
Staff Emeritus
You can solve this from the equation above for the mean free path.
What relation tells you the number density of an ideal gas from the pressure and temperature ? As for the radius, I think you may assume that the gas in the SLAC is mostly N2 and use the size of an N2 molecule as the diameter.

5. Physics Monkey

1,362
Right Gokul, thanks. The more I thought about it, something did seem amiss. Heh, I should probably go to sleep.

Last edited: Nov 23, 2005