Earlier I wrote:
>
> A capsule of LiD with a small hole, in a differentially-pumped Electron Beam Welder capable
> of 100 Kev at a few milliamperes in a millimeter diameter beam (the LiD has a vapor pressure of
> about 60 Torr at it's ~688 deg C melting point) should tell the story.
> of 100 Kev at a few milliamperes in a millimeter diameter beam (the LiD has a vapor pressure of
> about 60 Torr at it's ~688 deg C melting point) should tell the story.
>
In order to make the soccer game more interesting and the Deuteron target easier to come by, reacting
lithium (or other alkali metals) with D2O to get solid lithium hydroxide (LiOD) or NaOD-KOD might be expedient.
My "calculations" suggest that the energy gain of the deuteron is a function
of the square of the number of Electon-Beam electron-deuteron collisions.
IOW. a 100 Kev electron has a momentum of ~ 1.54e-22 kg-meter/sec and can impart up
to 20 eV to a deuteron. Hence 100 Kev/20 = 5,000
Hence, (5,000)^1/2 = 71. Thus 71* 1.54e-22 = 1.1e-20 kg-meter/sec which gives the free-space deuteron velocity
of 3.3e6 meters/sec as compared with the free-space beam electron velocity of 1.7e8 meters/sec.
A deuteron with 3.3e6 meters/sec has a kinetic energy of 113 Kev.
Might do something?
Frederick
Trying to determine the terminal energy/momentum of the deuteron "soccer ball" is a bit cumbersome.
Frederick
