Fusion Message Board

In this space, visitors are invited to post any comments, questions, or skeptical observations about Philo T. Farnsworth's contributions to the field of Nuclear Fusion research.

Subject: Mean Free Path
Date: Apr 14, 12:24 pm
Poster: Richard Hull

On Apr 14, 12:24 pm, Richard Hull wrote:


I am readying a long paper on fusor operation. As part of that effort I did a lot of poking around various references. One of the key items concerned mean free path.

For those in the dark on the matter, Mean free path in vacuum work means... the path length over which half of any moving particle group can move with out bunmping into something else> (usually taken to mean a brother particle.)

There are lookup tables, equations, etc., for molecules of gas in their own species. Since we use deuterium in the fusor, the mean free path for deuterons is realtively simple. At the one micron level, at room temperature, the mean free path for an un accelerated deuteron is 100 mm. this will increase slightly with energy.

Electrons, however are a different animal and due to their minute size, they can travel farther than the large gas atoms. In the same pressure range as our deuterons, electrons of 10ev or less can travel about 3-4 times the gas molecule's mean free path or ~400mm. However, under acceleration or at high energy, the electron path length grows tremendously.

For accelerated electrons, collisions with gas molecules are very important as that is how we generate deuterons. Due to the longer mean free path of an accelerated electron, we are less likely to ionize a lot of deuterium atoms.

There are a couple of great equations I found in Spangenberg's book "Electorn Tubes" Mc Graw Hill, 1946 which give useful info on the electron mean free path.

The first is the distance L in cm between ionizing collisions by a single electron in a gas at pressure P in mm or torr with the electron at an energy Ekv in kilovolts is found by the equation

L= Ekv/60P

Thus, we see that a 20kev electron at 1 micron in deuterium is going to travel, on average, 333 cm or 3.3 meters between ionizations!!!!

Now this means that in a 3.3meter sphere, 100% of the electrons will create an ionization. This does not mean that all electrons will need to travel 3.3 meters to ionize a gas molecule or that a 6" chamber is hopelessly too small to allow any ionization of gas by electrons.

There is another equation which might be more useful and that is the one which gives the number of ionizations/cm or travel for a given electron current. Here, the countless billions of electrons are averaged over that long path to give a reasonable number of ionizsations to be expected per cm of path. This is normally taken as a beam, but can be looked at over a spherical isotropic sourced volume as well.

The number N of ions formed per second per cm of path by an electron current Ie in amps accelerated to a voltage E in volts at pressure P in mm is given by the equation.

N= 3.75x10e23 X I X P / E

For a standard 10ma fusor run with 20kev electrons near the other shell, we are ionizing about 2x1014 deuterium atoms per second per cm of chamber radius right at the outer chamber wall.

One must cogitate and excersize the senses a little to appreciate the numbers and how a seemingly long mean free path can produce so many ionizations/cm in a 6" fusor.

Of course, one would have to integrate this over the entire chamber volume. Example, at 1cm out from the inner grid, we are at a 2.5kev level and the ionizations are much greater per cm at this point, being about an order of magnitude greater. All these deuterons are lost to us as they will not fall through a geat enough potential to achieve fusion energy.

As in my former postings, we now are producing the least number of deuterons in the area they are needed the most. (near the outer shell of the chamber)

Both the deuteron and electron mean free paths as well as there production levels at key areas are antagonistic. Still, inspite of all these problems, the thing fuses just fine.

Richard Hull