A method which gives very accurate results, though it is only applicable in certain cases, is the one used by Rutherford to measure the velocity of the negative ions produced close to a metal plate by the incidence on the plate of ultra-violet light. The principle of the method is as follows:—AB (fig. 10) is an insulated horizontal plate of well-polished zinc, which can be moved vertically up and down by means of a screw; it is connected with one pair of quadrants of an electrometer, the other pair of quadrants being put to earth. CD is a base-plate with a hole EF in it; this hole is covered with fine wire gauze, through which ultra-violet light passes and falls on the plate AB. The plate CD is connected with an alternating current dynamo, which produces a simply-periodic potential difference between AB and CD, the other pole being put to earth. Suppose that at any instant the plate CD is at a higher potential than AB, then the negative ions from AB will move towards CD, and will continue to do so as long as the potential of CD is higher than that of AB. If, however, the potential difference changes sign before the negative ions reach CD, these ions will go back to AB. Thus AB will not lose any negative charge unless the distance between the plates AB and CD is less than the distance traversed by the negative ion during the time the potential of CD is higher than that of AB. By altering the distance between the plates until CD just begins to lose a negative charge, we find the velocity of the negative ion under unit electromotive intensity. For suppose the difference of potential between AB and CD is equal to a sin pt, then if d is the distance between the plates, the electric intensity is equal to a sin pt/d; if we suppose the velocity of the ion is proportional to the electric intensity, and if u is the velocity for unit electric intensity, the velocity of the negative ion will be ua sin pt/d. Hence if x represent the distance of the ion from AB
| dx | = | ua | sin pt |
| dT | d |
| x = | ua | (1 − cos pt), if x = 0 when t = 0. |
| pd |
Thus the greatest distance the ion can get from the plate is equal to 2au/pd, and if the distance between the plates is gradually reduced to this value, the plate AB will begin to lose a negative charge; hence when this happens
d = 2au/pd, or u = pd2/2a,
an equation by means of which we can find u.
In this form the method is not applicable when ions of both signs are present. Franck and Pohl (Verh. deutsch. physik. Gesell. 1907, 9, p. 69) have by a slight modification removed this restriction. The modification consists in confining the ionization to a layer of gas below the gauze EF. If the velocity of the positive ions is to be determined, these ions are forced through the gauze by applying to the ionized gas a small constant electric force acting upwards; if negative ions are required, the constant force is reversed. After passing through the gauze the ions are acted upon by alternating forces as in Rutherford’s method.
Langevin (Ann. chim. phys., 1903, 28, p. 289) devised a method of measuring the velocity of the ions which has been extensively used; it has the advantage of not requiring the rate of ionization to remain uniform. The general idea is as follows. Suppose that we expose the gas between two parallel plates A, B to Röntgen rays or some other ionizing agent, then stop the rays and apply a uniform electric field to the region between the plates. If the force on the positive ion is from A to B, the plate B will receive a positive charge of electricity. After the electric force has acted for a time T reverse it. B will now begin to receive negative electricity and will go on doing so until the supply of negative ions is exhausted. Let us consider how the quantity of positive electricity received by B will vary with T. To fix our ideas, suppose the positive ions move more slowly than the negative; let T2 and T1 be respectively the times taken by the positive and negative ions to move under the electric field through a distance equal to AB, the distance between the planes. Then if T is greater than T2 all the ions will have been driven from between the plates before the field is reversed, and therefore the positive charge received by B will not depend upon T. Next let T be less than T2 but greater than T1; then at the time when the field is reversed all the negative ions will have been driven from between the plates, so that the positive charge received by B will not be neutralized by the arrival of fresh ions coming to it after the reversal of the field. The number of positive ions driven against the plate B will be proportional to T. Thus if we measure the value of the positive charge on B for a series of values of T, each value being less than the preceding, we shall find that until T reaches a certain value the charge remains constant, but as soon as we reduce the time below this value the charge diminishes. The value of T when the diminution in the field begins is T2, the time taken for a positive ion to cross from A to B under the electric field; thus from T2 we can calculate the velocity of the positive ion in this field. If we still further diminish T, we shall find that we reach a value when the diminution of the positive charge on B with the time suddenly becomes much more rapid; this change occurs when T falls below T1 the time taken for the negative ions to go from one plate to the other, for now when the field is reversed there are still some negative ions left between the plates, and these will be driven against B and rob it of some of the positive charge it had acquired before the field was reversed. By observing the time when the increase in the rate of diminution of the positive charge with the time suddenly sets in we can determine T1, and hence the velocity of the negative ions.
The velocity of the ions produced by the discharge of electricity from a fine point was determined by Chattock by an entirely different method. In this case the electric field is so strong and the velocity of the ion so great that the preceding methods are not applicable. Suppose P represents a vertical needle discharging electricity into air, consider the force acting on the ions included between two horizontal planes A, B. If P is the density of the electrification, and Z the vertical component of the electric intensity, F the resultant force on the ions between A and B is vertical and equal to
∫∫∫ Zρ dxdydz.