I used this formula to calculate the mean free path length of an electron in high vacuum.

$$\lambda = \frac{k T}{\sqrt{2}\cdot4\pi r^2 \cdot p}$$

where k is the Boltzmann konstant, T the temperature in Kelvin, r the radius of the particle in question, in m and p the pressure in pascal.

I used the following numbers:

$$k = 1.38*10^{-23}, T = 300, r \approx3*10^{-15}, p = 10^{-6}$$

Am I crazy for getting a result of

$$\lambda \approx 2.6*10^{13}m$$

I mean am I missing something? Maybe the radius of the electron changes when it moves? (Actually it moves with 50 keV but I have neglected that so far.)

I am sorry but this result seems just so....unreal. And I have to be sure because I have an exam about this stuff soon!

  • $\begingroup$ You better use the size of the atoms, which is bigger, not the electron which is much smaller, so $10^{-10}$ instead of $3 ~ 10^{-15}$. $\endgroup$ – Kostas Oct 5 '19 at 20:32
  • $\begingroup$ The formula you are using appears to be one that might apply to the molecules of a gas. When you speak of a high vacuum, are you referring to a low density of molecules (as in outer space) or a plasma of electrons? In either case you need a collision cross section based on the probability that the electron wave packet will interact with the other particles in your system. $\endgroup$ – R.W. Bird Oct 5 '19 at 23:36

The presence of $r$ in the formula you used implies that it is based a contact scattering assumption.

That kind of assumption makes sense for modeling the atoms and molecules of a neutral gas or neutron diffusion, but it doesn’t make sense for modeling the interactions of charged particles unless you tunes the radius you use to express the median impact parameter at which the scattering angle is non-trivial.

There is no particular reason to assume that the so-called “classical electron radius” is the right value.

  • $\begingroup$ Ah! I see! So would you have a guess what would be the correct mean free path for electrons in such a vacuum? $\endgroup$ – Caito Oct 6 '19 at 14:44
  • $\begingroup$ Hard to say. In contact scatting scenarios the mean free path is insensitive to temperature. But electromagnetic scattering is long range, so we set a threshold scattering angle to threat as a “collision”. But the impact parameter corresponding to that scattering angle is velocity (I.e. temperature) dependent. $\endgroup$ – dmckee --- ex-moderator kitten Oct 6 '19 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.