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Re: tesla and fusion



Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Jim,

I have no problem with using keV. My text book uses both almost interchangably and I tried to quote the book where ever I could. It also talks about fusion at almost any average temperature because the temperature of individual particles have a distribution (a Maxwell distribution, I think I've heard). There is a statistical probability that two particles will have the required energy to break the coulumb barrier. Of course, the lower the average temperature the lower the probability of this occuring.

I can believe the 40 keV number for deuterium. The smaller the energy, the smaller the fusion cross section and a more perfect head on collision is required. My text book (of 33 years ago) says 144 keV for hydrogen based particles, but I dont know what cross section that equation is assuming and it will be different for deuterium vs tritium.

I'm a little confused as to your 100 million K is only 10 keVish statement. It seems like if 11000K = 1 keV, then 100 million K = 9000 keV; and a billion K = 90000 keV

The easy part, I believe, is to get two particles to fuse. The hard part is to get this to happen in high numbers.

Gerry



Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>

But, working in kelvins is misleading.. a better way is to talk about keV (1 kiloelectronvolt = 11000 K)..

You get detectable fusion in Deuterium at 40 keV.. and that's easy to come by..

Google for "fusor" and you'll find a fair number of amateurs have done tabletop fusion for low budgets.

100 million kelvin is only 10 keVish.. and while there are some reactions which "go" at that energy, they're pretty low cross section. A billion kelvin is 100 keV, and for DT the cross section is pretty good.






At 06:28 PM 5/2/2007, Tesla list wrote:
Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Steve,

You may be interested to know that in order to get two interacting particles to fuse, the nuclei need to approach within an order of 10^-14 m. This is about 1/10000 of the diameter of a hydrogen atom. The nuclear radius of a particle is generally taken to be 1.5*10^(-15)*A^(1/3) where A is the atomic weight of the particle (A needs to be less than 20 for fusion to occur). The obstacle to having this occur is the coulomb barrier. In order to have enough kinetic energy to over come this barrier and get to about 10^-14 m, the kinetic energy needs to be:

Ue = 1.44 * 10^5 * Z1*Z2 ev

where Z1 and Z2 are the atomic numbers of the two interacting particles

The temperature needed to get great numbers of fusion reactions is about 1 billion degrees Kelvin. The temperature needed to get some reactions can be as low as 40 million degrees Kelvin but generally it is thought that 100 million degrees Kelvin is needed for fusion.


Gerry R.