i = i0εαl + qe(eαl − 1).
α

With regard to the minimum energy which must be possessed by a corpuscle to enable it to produce ions by collision, Townsend (loc. cit.) came to the conclusion that to ionize air the corpuscle must possess an amount of energy equal to that acquired by the fall of its charge through a potential difference of about 2 volts. This is also the value arrived at by H. A. Wilson by entirely different considerations. Stark, however, gives 17 volts as the minimum for ionization. The energy depends upon the nature of the gas; recent experiments by Dawes and Gill and Pedduck (Phil. Mag., Aug. 1908) have shown that it is smaller for helium than for air, hydrogen, or carbonic acid gas.

If there is no external source of ionization and no emission of corpuscles from the cathode, then it is evident that even if some corpuscles happened to be present in the gas when the electric field were applied, we could not get a permanent current by the aid of collisions made by these corpuscles. For under the electric field, the corpuscles would be driven from the cathode to the anode, and in a short time all the corpuscles originally present in the gas and those produced by them would be driven from the gas against the anode, and if there was no source from which fresh corpuscles could be introduced into the gas the current would cease. The current, however, could be maintained indefinitely if the positive ions in their journey back to the cathode also produced ions by collisions, for then we should have a kind of regenerative process by which the supply of corpuscles could be continually renewed. To maintain the current it is not necessary that the ionization resulting from the positive ions should be anything like as great as that from the negative, as the investigation given below shows a very small amount of ionization by the positive ions will suffice to maintain the current. The existence of ionization by collision with positive ions has been proved by Townsend. Another method by which the current could be and is maintained is by the anode emitting corpuscles under the impact of the positive ions driven against it by the electric field. J. J. Thomson has shown by direct experiment that positively electrified particles when they strike against a metal plate cause the metal to emit corpuscles (J. J. Thomson, Proc. Camb. Phil. Soc. 13, p. 212; Austin, Phys. Rev. 22, p. 312). If we assume that the number of corpuscles emitted by the plate in one second is proportional to the energy in the positive ions which strike the plate in that second, we can readily find an expression for the difference of potential which will maintain without any external ionization a current of electricity through the gas. As this investigation brings into prominence many of the most important features of the electric discharge, we shall consider it in some detail.

Let us suppose that the electrodes are parallel plates of metal at right angles to the axis of x, and that at the cathode x = 0 and at the anode x = d, d being thus the distance between the plates. Let us also suppose that the current of electricity flowing between the plates is so small that the electrification between the plates due to the accumulation of ions is not sufficient to disturb appreciably the electric field, which we regard as uniform between the plates, the electric force being equal to V/d, where V is the potential difference between the plates. The number of positive ions produced per second in a layer of gas between the planes x and x + dx is αnu·dx. Here n is the number of corpuscles per unit volume, α the coefficient of ionization (for strong electric field α = 1/λ′, where λ′ is the mean free path of a corpuscle), and u the velocity of a corpuscle parallel to x. We have seen that nu = i0εαx, where i0 is the number of corpuscles emitted per second by unit area of the cathode. Thus the number of positive ions produced in the layer is αi0εαx dx. If these went straight to the cathode without a collision, each of them would have received an amount of kinetic energy Vex/d when they struck the cathode, and the energy of the group of ions would be Vex/d·αi0εdx dx. The positive ions will, however, collide with the molecules of the gas through which they are passing, and this will diminish the energy they possess when they reach the cathode.

The diminution in the energy will increase in geometrical proportion with the length of path travelled by the ion and will thus be proportional to ε−βx, β will be proportional to the number of collisions and will thus be proportional to the pressure of the gas. Thus the kinetic energy possessed by the ions when they reach the cathode will be

ε−βx·V(ex/d) · α i0εαx dx,

and E, the total amount of energy in the positive ions which reach the cathode in unit time, will be given by the equation

E = ∫d0 ε−βx · V(ex/d) · α i0εαx dx =Veα i0∫d0 ε−(β−α)x x dx
d
= Veα i0{ 1 − ε−(β−α)d{ 1+ d}}  (1).
d (β − α)²(β − α)² (β − α)

If the number of corpuscles emitted by the cathode in unit time is proportional to this energy we have i0 = kE, where k is a constant; hence by equation (1) we have