| X² = | α | i² | + C¹ε 8πe² k2qx/αi. | |
| q | k²2e² |
To find the value of C¹ we see by equation (7) that
| d²X² | k1k2 | 1 | = q − αn1n2; | ||
| dX² | k1 + k2 | 8πe |
hence
| [ | dX² | k1k2 | 1 | ]x1 = ∫x10 (q − αn1n2)dx. | ||
| dX | k1 + k2 | 8πe |
The right hand side of this equation is the excess of ionization over recombination in the region extending from the cathode to x1; it must therefore, when things are in a steady state, equal the excess of the number of negative ions which leave this region over those which enter it. The number which leave is i/e and the number which enter is i0/e, if it is the current of negative ions coming from unit area 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
| C¹ = | αi(i − i0) | k1 + k2 | , | |
| qk1k2e² | k2 |
and since k1 is small compared with k2, we have
| X² = | α i² | (1 + | k2 | i − i0 | ε -8πe² k2 · qx/α · i). | |
| qk²2 e² | k1 | i |
From the values which have been found for k2 and α, we know that 8πek2/α 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/k2e · √a/q, while when i0 = 0, the force at the cathode itself bears to the uniform force the ratio of (k1 + k2)1/2 to k11/2. 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/k2e · √a/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/k2e · √(a/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