that the positive (and the same is true for the negative) ions in a
mixture of gases are all of the same kind. This conclusion is one of
considerable importance, as it would not be true if the ions consisted
of single molecules of the gas from which they are produced.
Recombination.—Several methods enable us to deduce the coefficient of recombination of the ions when we know their velocities. Perhaps the simplest of these consists in determining the relation between the current passing between two parallel plates immersed in ionized gas and the potential difference between the plates. For let q be the amount of ionization, i.e. the number of ions produced per second per unit volume of the gas, A the area of one of the plates, and d the distance between them; then if the ionization is constant through the volume, the number of ions of one sign produced per second in the gas is qAd. Now if i is the current per unit area of the plate, e the charge on an ion, iA/e ions of each sign are driven out of the gas by the current per second. In addition to this source of loss of ions there is the loss due to the recombination; if n is the number of positive or negative ions per unit volume, then the number which recombine per second is αn2 per cubic centimetre, and if n is constant through the volume of the gas, as will approximately be the case if the current through the gas is only a small fraction of the saturation current, the number of ions which disappear per second through recombination is αn2·Ad. Hence, since when the gas is in a steady state the number of ions produced must be equal to the number which disappear, we have
q = i/ed + αn2.
If u1 and u2 are the velocities with which the positive and negative ions move, nu1e and nu2e are respectively the quantities of positive electricity passing in one direction through unit area of the gas per second, and of negative in the opposite direction, hence
If X is the electric force acting on the gas, k1 and k2 the velocities of the positive and negative ions under unit force, u1 = k1X, u2 = k2X; hence
and we have
| q = | i | + | αi 2 | . |
| ed | (k1 + k2)2 e2X2 |
But qed is the saturation current per unit area of the plate; calling this I, we have
| I − i = | dαi 2 |
| e(k1 + k2)2X2 |
or
| X2 = | i 2·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,
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
| dXdx = 4π(n1 − n2)e | (1). |
If I is the current through unit area of the gas and if we neglect any diffusion except that caused by the electric field,
| n1ek1X + n2ek2X = I | (2). |
From equations (1) and (2) we have
| n1e = 1k1 + k2 ( IX+k24π dXdx) | (3). |
| n2e = 1k1 + k2 ( IX−k14π dXdx) | (4). |
and from these equations we can, if we know the distribution of electric intensity between the plates, calculate the number of positive and negative ions.
In a steady state the number of positive and negative ions in unit volume at a given place remains constant, hence neglecting the loss by diffusion, we have
| ddx(k1n1X) = q − αn1n2 | (5). |
| −ddx(k2n2X) = q − αn1n2 | (6). |
If k1 and k2 are constant, we have from (1), (5) and (6)
| d2X2dx2 = 8πe(q − αn1n2)(1k1+1k2) | (7). |
an equation which is very useful, because it enables us, if we know the distribution of X2, to find whether at any point in the gas the ionization is greater or less than the recombination of the ions. We see that q − αn1n2, which is the excess of ionization over recombination, is proportional to d2X2/dx2. Thus when the ionization exceeds the recombination, i.e. when q − αn1n2 is positive, the curve for X2 is convex to the axis of x, while when the recombination exceeds the ionization the curve for X2 will be concave to the axis of x.
 |
| Fig. 11. |
Thus, for example, fig. 11 represents the curve for X2 observed by Graham (Wied. Ann. 64, p. 49) in a tube through which a steady current is passing. Interpreting it by equation (7), we infer that ionization was much in excess of recombination at A and B, slightly so along C, while along D the recombination exceeded the ionization. Substituting in equation (7) the values of n1, n2 given in (3), (4), we get
| d2X2 | 8πe[q − | α | (I + | k2 | dX2 | )(I − | k2 | dX2 | )]( | 1 | + | 1 | ) | ||
| dx2 | e2X2(k1 + k2)2 | 8π | dx | 8π | dx | k1 | k2 |
This equation can be solved (see Thomson, Phil. Mag. xlvii.
P. 253), when q is constant and k1 = k2. From the solution it appears
that if X1 be the value of x close to one of the plates, and X0 the
value midway between them,
| X1/X0 = | 1 |
| β2 − 2/β |
where β = 8πek1/α.
