Page:EB1911 - Volume 06.djvu/897

GASES] of the cathode, as hot metal cathodes emit large quantities of negative electricity i0 may in some cases be considerable, thus the right hand side of equation is (i − i0)/e. When x1 is large dX2/dx = 0; hence we have from equation

and since k1 is small compared with k2, we have

From the values which have been found for k2 and, we know that 8ek2/ is a large quantity, hence the second term inside the bracket will be very small when eqx is equal to or greater than i; thus this term will be very small outside a layer of gas next the cathode of such thickness that the number of ions produced on it would be sufficient, if they were all utilized for the purpose, to carry the current; in the case of flames this layer is exceedingly thin unless the current is very large. The value of the electric force in the uniform part of the field is equal to $i⁄k_{2}e$ · &radic; $undefined⁄q$, while when i0 = 0, the force at the cathode itself bears to the uniform force the ratio of (k1 + k2)$1⁄2$ to k1$1⁄2$. As k1 is many thousand times k2 the force increases with great rapidity as we approach the cathode; this is a very characteristic feature of the passage of electricity through flames and hot gases. Thus in an experiment made by H. A. Wilson with a flame 18 cm. long, the drop of potential within 1 centimetre of the cathode was about five times the drop in the other 17 cm. of the tube. The relation between the current and the potential difference when the velocity of the negative ion is much greater than the positive is very easily obtained. Since the force is uniform and equal to $i⁄k_{2}e$&radic; $undefined⁄q$, until we get close to the cathode the fall of potential in this part of the discharge will be very approximately equal to $i⁄k_{2}e$&radic; $undefined⁄q$l, where l is the distance between the electrodes. Close to the cathode, the electric force when i0 is not nearly equal to i is approximately given by the equation

and the fall of potential at the cathode is equal approximately to $\int_0^{\infty}$Xdx, that is to

The potential difference between the plates is the sum of the fall of potential in the uniform part of the discharge plus the fall at the cathode, hence

The fall of potential at the cathode is proportional to the square of the current, while the fall in the rest of the circuit is directly proportional to the current. In the case of flames or hot gases, the fall of potential at the cathode is much greater than that in the rest of the circuit, so that in such cases the current through the gas varies nearly as the square root of the potential difference. The equation we have just obtained is of the form

V = Ai + Bi 2,

and H. A. Wilson has shown that a relation of this form represents the results of his experiments on the conduction of electricity through flames.

The expression for the fall of potential at the cathode is inversely proportional to q3/2, q being the number of ions produced per cubic centimetre per second close to the cathode; thus any increase in the ionization at the cathode will diminish the potential fall at the cathode, and as practically the whole potential difference between the electrodes occurs at the cathode, a diminution in the potential fall there will be much more important than a diminution in the electric force in the uniform part of the discharge, when the force is comparatively insignificant. This consideration explains a very striking phenomenon discovered many years ago by Hittorf, who found that if he put a wire carrying a bead of a volatile salt into the flame, it produced little effect upon the current, unless it were placed close to the cathode where it gave rise to an enormous increase in the current, sometimes increasing the current more than a hundredfold. The introduction of the salt increases very largely the number of ions produced, so that q is much greater for a salted flame than for a plain one. Thus Hittorf’s result coincides with the conclusions we have drawn from the theory of this class of conduction.

The fall of potential at the cathode is proportional to i − i0, where i0 is the stream of negative electricity which comes from the cathode itself, thus as i0 increases the fall of potential at the cathode diminishes and the current sent by a given potential difference through the gas increases. Now all metals give out negative particles when heated, at a rate which increases very rapidly with the temperature, but at the same temperature some metals give out more than others. If the cathode is made of a metal which emits large quantities of negative particles, (i − i0) will for a given value of i be smaller than if the metal only emitted a small number of particles; thus the cathode fall will be smaller for the metal with the greater emissitivity, and the relation between the potential difference and the current will be different in the two cases. These considerations are confirmed by experience, for it has been found that the current between electrodes immersed in a flame depends to a great extent upon the metal of which the electrodes are made. Thus Pettinelli (Acc. dei Lincei [5], v. p. 118) found that, ceteris paribus, the current between two carbon electrodes was about 500 times that between two iron ones. If one electrode was carbon and the other iron, the current when the carbon was cathode and the iron anode was more than 100 times the current when the electrodes were reversed. The emission of negative particles by some metallic oxides, notably those of calcium and barium, has been shown by Wehnelt (Ann. der Phys. 11, p. 425) to be far greater than that of any known metal, and the increase of current produced by coating the cathodes with these oxides is exceedingly large; in some cases investigated by Tufts and Stark (Physik. Zeits., 1908, 5, p. 248) the current was increased many thousand times by coating the cathode with lime. No appreciable effect is produced by putting lime on the anode.

Conduction when all the Ions are of one Sign.—There are many important cases in which the ions producing the current come from one electrode or from a thin layer of gas close to the electrode, no ionization occurring in the body of the gas or at the other electrode. Among such cases may be mentioned those where one of the electrodes is raised to incandescence while the other is cold, or when the negative electrode is exposed to ultra-violet light. In such cases if the electrode at which the ionization occurs is the positive electrode, all the ions will be positively charged, while if it is the negative electrode the ions will all be charged negatively. The theory of this case is exceedingly simple. Suppose the electrodes are parallel planes at right angles to the axis of x; let X be the electric force at a distance x from the electrode where the ionization occurs, n the number of ions (all of which are of one sign) at this place per cubic centimetre, k the velocity of the ion under unit electric force, e the charge on an ion, and i the current per unit area of the electrode. Then we have and if u is the velocity of the ion neu = i. But u = kX, hence we have $1⁄2$ $1⁄2$ = i, and since the right hand side of this equation does not depend upon x, we get kX2/8 = ix + C, where C is a constant to be determined. If l is the distance between the plates, and V the potential difference between them,

We shall show that when the current is far below the saturation value, C is very small compared with il, so that the preceding equation becomes

To show that for small currents C is small compared with il, consider the case when the ionization is confined to a thin layer, thickness d close to the electrode, in that layer let n0 be the value of n, then we have q = n02 + i/ed. If X0 be the value of X when x = 0, kX0n0e = i, and,

(2).

Since /8ke is, as we have seen, less than unity, C will be small compared with il, if i/(eq + i/d) is small compared with l. If I0 is the saturation current, q = I0/ed, so that the former expression = id/(I0 + i), if i is small compared with I0, this expression is small compared with d, and therefore a fortiori compared with l, so that we are justified in this case in using equation (1).

From equation (2) we see that the current increases as the square of the potential difference. Here an increase in the potential difference produces a much greater percentage increase than in conduction through metals, where the current is proportional to the potential difference. When the ionization is distributed through the gas, we have seen that the current is approximately proportional to the square root of the potential, and so increases more slowly with the potential difference than currents through metals. From equation (1) the current is inversely proportional to the cube of the distance between the electrodes, so that it falls off with great rapidity as this distance is increased. We may note that for a given potential difference the expression for the current does not involve q, the rate of production of the ions at the electrode, in other words, if we vary the ionization the current will not begin to be affected by the strength of the ionization until this falls so low that the current is a considerable fraction of the saturation current. For the same potential difference the current is proportional to k, the velocity under unit electric force of the ion which carries the current. As the velocity of the negative ion is greater than that of the positive, the current when the ionization is confined to the neighbourhood of one of the electrodes will be greater when that electrode is made cathode than when it is anode. Thus the current will appear to pass more easily in one direction than in the opposite.

Since the ions which carry the current have to travel all the way from one electrode to the other, any obstacle which is impervious to these ions will, if placed between the electrodes, stop the current