if k2 = q/α. Now the number of ions when the gas has reached a steady state is got by putting t equal to infinity in the preceding equation, and is therefore given by the equation
n0 = k = √ (q/α).
We see from equation (1) that the gas will not approximate to its steady state until 2kαt is large, that is until t is large compared with ½kα or with ½√ (qα). We may thus take ½√ (qα) as a measure of the time taken by the gas to reach a steady state when exposed to an ionizing agent; as this time varies inversely as √q we see that when the ionization is feeble it may take a very considerable time for the gas to reach a steady state. Thus in the case of our atmosphere where the production of ions is only at the rate of about 30 per cubic centimetre per second, and where, as we shall see, α is about 10-6, it would take some minutes for the ionization in the air to get into a steady state if the ionizing agent were suddenly applied.
We may use equation (1) to determine the rate at which the ions disappear when the ionizing agent is removed. Putting q=0 in that equation we get dn/αt = -αn2.
Hence
n = n0/(1 + n0 αt) (3),
where n0 is the number of ions when t = 0. Thus the number of ions falls to one-half its initial value in the time 1/n0α. The quantity α is called the coefficient of recombination, and its value for different gases has been determined by Rutherford (Phil. Mag. 1897 [5], 44, p. 422), Townsend (Phil. Trans., 1900, 193, p. 129), McClung (Phil. Mag., 1902 [6], 3, p. 283), Langevin (Ann. chim. phys. [7], 28, p. 289), Retschinsky (Ann. d. Phys., 1905, 17, p. 518), Hendred (Phys. Rev., 1905, 21, p. 314). The values of α/e, e being the charge on an ion in electrostatic measure as determined by these observers for different gases, is given in the following table:—
| Townsend. | McClung. | Langevin. | Retschinsky. | Hendred. | |
| Air | 3420 | 3380 | 3200 | 4140 | 3500 |
| O2 | 3380 | ||||
| CO2 | 3500 | 3490 | 3400 | ||
| H2 | 3020 | 2940 |
The gases in these experiments were carefully dried and free from dust; the apparent value of α is much increased when dust or small drops of water are present in the gas, for then the ions get caught by the dust particles, the mass of a particle is so great compared with that of an ion that they are practically immovable under the action of the electric field, and so the ions clinging to them escape detection when electrical methods are used. Taking e as 3.5×10-10, we see that α is about 1.2×10-6, so that the number of recombinations in unit time between n positive and n negative ions in unit volume is 1.2×10-6n2. The kinetic theory of gases shows that if we have n molecules of air per cubic centimetre, the number of collisions per second is 1.2×10-10n2 at a temperature of 0° C. Thus we see that the number of recombinations between oppositely charged ions is enormously greater than the number of collisions between the same number of neutral molecules. We shall see that the difference in size between the ion and the molecule is not nearly sufficient to account for the difference between the collisions in the two cases; the difference is due to the force between the oppositely charged ions, which drags ions into collisions which but for this force would have missed each other.
Several methods have been used to measure α. In one method air, exposed to some ionizing agent at one end of a long tube, is slowly sucked through the tube and the saturation current measured at different points along the tube. These currents are proportional to the values of n at the place of observation: if we know the distance of this place from the end of the tube when the gas was ionized and the velocity of the stream of gas, we can find t in equation (3), and knowing the value of n we can deduce the value of α from the equation