Some questions: What is the field inside a 12cm diameter, vertical, meter long positively charged pipe that is closed at the top except for a 1cm diameter hole in the center for a gas inlet tube and open at the other end with a 4mm gap between the rim of this open end and the closed, but for a funnel shaped hole in the middle, end of a slightly larger 12.8cm diameter 1 meter long grounded pipe that extends upward and surrounds the first pipe and is open at the opposite(upper) end? The funnel shaped hole is 1 to 2 cm in the narrowest part. A potential difference of 20kV between the pipes produces a vertical axial field and a radial field on H2 molecules entering the inner pipe in a 10^-6 Torr vacuum at 1840m/s and so spending 1/1840 sec before exiting the tube or hitting the sides of the funnel shaped hole. Rutherford used such an apparatus to produce and accelerate protons in 1934 before lineacs and cyclatrons etc were used for proton beams etc. and reported that after running for some time this apparatus produced a 1 milliamp beam of protons into a Faraday cylinder and
19millamp current of electrons moving toward and through the pipe. And this proton beam occurred with a voltage of 20kV and changes in this voltage produced no better ratio of protons to electrons. (perhaps because protons recombined and remained with electrons on the surface of the funnel shaped hole?) Since V=Ed and d =1 meter from the closed end of the positive pipe to the almost closed end of the grounded pipe, wouldn't the vertical tangential acceleration given to an electron in an H2 molecule from this field be (x)( eE)(10^-3)/9(10^-31) = [(2)(x) ((1.602)(10^-19 )(10^(4-3))]/(9)(10^-31) for some small fraction,x, of the 10^-3 seconds that the molecule moves the length of the pipe, when this force is unopposed, twice every 10^-15 sec about, by the nuclei holding the electrons in their orbits around the proton nuclei? Thus x is about 10^-6 and this produces evidently an elliptical extension of the bound orbits enough to cause ejection of the electrons. Note mv^2/r =9(10^9)e^2/r^2 where r then is twice the .5 (10^-10) Hydrogen atom radius we can estimate the speed,v, of the figure eight or circular orbiting molecular electrons (9(10^9)e^2/rm)^1/2= 10^6 m/s . and the escape velocity is 1.4 times this. If the funnel shaped hole was widened and the voltage increased so the field remained the same, would this increase the ratio of protons and electrons in the total current produced? If the length of the pipes was increased and the voltage so that the field remained the same would this increase the ratio of protons to electrons? Further acceleration of the beam by a negatively charged 2cm long pipe section of the same 12 cm diameter at a gap of 4mm below the funnel shaped hole could be focused into a narrower beam by a 1 to 2cm long pipe section at a gap of 4mm below this and charged to a higher potential that would slightly retard the accelerated protons. The beam could be directed thusly through a small hole in a pipe section at a gap of 4mm below this and at a more negative potential. What should the focusing potential etc be relative to the preceding accelerating potential? And how best to determine the number of protons that got through the hole compared to those that didn't?- posted
18 years ago