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 i = nu1e + nu2e. 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 n = i/(k1 + k2)Xe, and we have i [alpha]i² q = -- + --------------. ed (k1 + k2)²e²X² But qed is the saturation current per unit area of the plate; calling this I, we have d[alpha]i² I - i = ------------- e(k1 + k2)²X² or i²·d[alpha] X² = ------------------. e(I - i)(k1 + k2)² Hence if we determine corresponding values of X and i we can deduce the value of [alpha]/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 + [alpha]n0t).