| X2 = | i2·dα | . |
| e(I − i) (k1 + k2)2 |
Hence if we determine corresponding values of X and i we can deduce the value of α/e if we also know (k1 + k2). The value of I is easily determined, as it is the current when X is very large. The preceding result only applies when i is small compared with I, as it is only in this case that the values of n and X are uniform throughout the volume of the gas. Another method which answers the same purpose is due to Langevin (Ann. Chim. Phys., 1903, 28, p. 289); it is as follows. Let A and B be two parallel planes immersed in a gas, and let a slab of the gas bounded by the planes a, b parallel to A and B be ionized by an instantaneous flash of Röntgen rays. If A and B are at different electric potentials, then all the positive ions produced by the rays will be attracted by the negative plate and all the negative ions by the positive, if the electric field were exceedingly large they would reach these plates before they had time to recombine, so that each plate would receive N0 ions if the flash of Röntgen rays produced N0 positive and N0 negative ions. With weaker fields the number of ions received by the plates will be less as some of them will recombine before they can reach the plates. We can find the number of ions which reach the plates in this case in the following way:—In consequence of the movement of the ions the slab of ionized gas will broaden out and will consist of three portions, one in which there are nothing but positive ions,—this is on the side of the negative plate,—another on the side of the positive plate in which there are nothing but negative ions, and a portion between these in which there are both positive and negative ions; it is in this layer that recombination takes place, and here if n is the number of positive or negative ions at the time t after the flash of Röntgen rays,
n = n0/(1 + αn0t).
With the same notation as before, the breadth of either of the outer layers will in time dt increase by X(k1 + k2)dt, and the number of ions in it by X(k1 + k2)ndt; these ions will reach the plate, the outer layers will receive fresh ions until the middle one disappears, which it will do after a time l/X(k1 + k2), where l is the thickness of the slab ab of ionized gas; hence N, the number of ions reaching either plate, is given by the equation
| N = ∫0l/X(k1+k2) | n0X(k1 + k2) | dt = | X(k1 + k2) | log (1 + | n0αl | ). |
| 1 + n0αt | α | X(k1 + k2) |
If Q is the charge received by the plate,
| Q = Ne = | X | log(1 + | Q0ε | ), |
| 4πε | 4πX |
where Q0 = n0le is the charge received by the plate when the electric force is large enough to prevent recombination, and ε = α4πe(R1 + R2). We can from this result deduce the value of ε and hence the value of α when R1 + R2 is known.
Distribution of Electric Force when a Current is passing through an Ionized Gas.—Let the two plates be at right angles to the axis of x; then we may suppose that between the plates the electric intensity X is everywhere parallel to the axis of x. The velocities of both the positive and negative ions are assumed to be proportional to X. Let k1X, k2X represent these velocities respectively; let n1, n2 be respectively the number of positive and negative ions per unit volume at a point fixed by the co-ordinate x; let q be the number of positive or negative ions produced in unit time per unit volume at this point; and let the number of ions which recombine in unit volume in unit time be αn1n2; then if e is the charge on the ion, the volume density of the electrification is (n1 − n2)e, hence
| dX | = 4π(n1 − n2)e (1). |
| dx |