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econd by unit area of the cathode. Thus the number of positive ions produced in the layer is [alpha]i0[epsilon]^[alpha]x 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.[alpha]i0[epsilon]^dx 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 [epsilon]^-[beta]x, [beta] 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 [epsilon]^{-[beta]x} . V(ex/d) . [alpha]i0[epsilon]^{[alpha]x} 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 _ /d E = | [epsilon]^{-[beta]x} . V(ex/d) . [alpha]i0[epsilon]^{[alpha]x} dx _/0 _ Ve[alpha]i0 /d = ----------- | [epsilon]^{-([beta]-[alpha])x}.x.dx d _/0 Ve[alpha]i0 / 1 / 1 d \ \ = ----------- { ------------------ - [epsilon]^{-([beta]-[alpha])d} { ------------------ + ---------------- } } (1). d \([beta]-[alpha])^2 \([beta]-[alpha])^2 ([beta]-[alpha])/ / 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 ([beta]-[alpha])^2 d V = ------------------ . --, ke[alpha] I where I = 1 - [epsilon]^{-([beta]-[alpha])d} (1 + d([beta] - [alpha])). Since both [beta] and [alpha] are proportional to the pressure, I and ([beta] - [alpha])^2d/[alpha] are both functions of pd, the product of the pressure and the spark length, hence we see that V is expressed by an equation of the form 1 V = -- [int](pd) (2), ke where [int](pd) denotes a f
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