Since 1.22 cubic centimetres of hydrogen at the temperature 15° C. and pressure 760 mm. of mercury are liberated by the passage through acidulated water of one electromagnetic unit of electricity or 3✕1010 electrostatic units, and since in one cubic centimetre of the gas there are 2.46 N atoms of hydrogen, we have, if E is the charge in electrostatic units, on the atom of hydrogen in the electrolysis of solutions
or
The mean of the values of Ne in the preceding table is 1.24✕1010. Hence we may conclude that the charge of electricity carried by a gaseous ion is equal to the charge carried by the hydrogen atom in the electrolysis of solutions. The values of Ne for the different gases differ more than we should have expected from the probable accuracy of the determination of D and the velocity of the ions: Townsend (Proc. Roy. Soc. 80, p. 207) has shown that when the ionization is produced by Röntgen rays some of the positive ions carry a double charge and that this accounts for the values of Ne being greater for the positive than for the negative ions. Since we know the value of e, viz. 3.5✕10−10, and, also Ne, = 1.24✕1010, we find N the number of molecules in a cubic centimetre of gas at standard temperature and pressure to be equal to 3.5✕1019. This method of obtaining N is the only one which does not involve any assumption as to the shape of the molecules and the forces acting between them.
Another method of determining the charge carried by an ion has been employed by Rutherford (Proc. Roy. Soc. 81, pp. 141, 162), in which the positively electrified particles emitted by radium are made use of. The method consists of: (1) Counting the number of α particles emitted by a given quantity of radium in a known time. (2) Measuring the electric charge emitted by this quantity in the same time. To count the number of the α particles the radium was so arranged that it shot into an ionization chamber a small number of α particles per minute; the interval between the emission of individual particles was several seconds. When an α particle passed into the vessel it ionized the gas inside and so greatly increased its conductivity; thus, if the gas were kept exposed to an electric field, the current through the gas would suddenly increase when an α particle passed into the vessel. Although each α particle produces about thirty thousand ions, this is hardly large enough to produce the conductivity appreciable without the use of very delicate apparatus; to increase the conductivity Rutherford took advantage of the fact that ions, especially negative ones, when exposed to a strong electric field, produce other ions by collision against the molecules of the gas through which they are moving. By suitably choosing the electric field and the pressure in the ionization chamber, the 30,000 ions produced by each α particle can be multiplied to such an extent that an appreciable current passes through the ionization chamber on the arrival of each α particle. An electrometer placed in series with this vessel will show by its deflection when an α particle enters the chamber, and by counting the number of deflections per minute we can determine the number of α particles given out by the radium in that time. Another method of counting this number is to let the particles fall on a phosphorescent screen, and count the number of scintillations on the screen in a certain time. Rutherford has shown that these two methods give concordant results.
The charge of positive electricity given out by the radium was measured by catching the α particles in a Faraday cylinder placed in a very highly exhausted vessel, and measuring the charge per minute received by this cylinder. In this way Rutherford showed that the charge on the α particle was 9.4✕10−10 electrostatic units. Now e/m for the α particle = 5 ✕103, and there is evidence that the α particle is a charged atom of helium; since the atomic weight of helium is 4 and e/m for hydrogen is 104, it follows that the charge on the helium atom is twice that on the hydrogen, so that the charge on the hydrogen atom is 4.7✕10−10 electrostatic units.
Calculation of the Mass of the Ions at Low Pressures.—Although at ordinary pressures the ion seems to have a very complex structure and to be the aggregate of many molecules, yet we have evidence that at very low pressures the structure of the ion, and especially of the negative one, becomes very much simpler. This evidence is afforded by determination of the mass of the atom. We can measure the ratio of the mass of an ion to the charge on the ion by observing the deflections produced by magnetic and electric forces on a moving ion. If an ion carrying a charge e is moving with a velocity v, at a point where the magnetic force is H, a mechanical force acts on the ion, whose direction is at right angles both to the direction of motion of the ion and to the magnetic force, and whose magnitude is evH sin θ, where θ is the angle between v and H. Suppose then that we have an ion moving through a gas whose pressure is so low that the free path of the ion is long compared with the distance through which it moves whilst we are experimenting upon it; in this case the motion of the ion will be free, and will not be affected by the presence of the gas.
Since the force is always at right angles to the direction of motion of the ion, the speed of the ion will not be altered by the action of this force; and if the ion is projected with a velocity v in a direction at right angles to the magnetic force, and if the magnetic force is constant in magnitude and direction, the ion will describe a curve in a plane at right angles to the magnetic force. If ρ is the radius of curvature of this curve, m the mass of the ion, mv2/ρ must equal the normal force acting on the ion, i.e. it must be equal to Hev, or ρ = mv/He. Thus the radius of curvature is constant; the path is therefore a circle, and if we can measure the radius of this circle we know the value of mv/He. In the case of the rapidly moving negative ions projected from the cathode in a highly exhausted tube, which are known as cathode rays, the path of the ions can be readily determined since they make many substances luminous when they impinge against them. Thus by putting a screen of such a substance in the path of the rays the shape of the path will be determined. Let us now suppose that the ion is acted upon by a vertical electric force X and is free from magnetic force, if it be projected with a horizontal velocity v, the vertical deflection y after a time t is 12✕et2/m, or if l is the horizontal distance travelled over by the ion in this time we have since l = vt,
| y = 12 | Xe | l 2 | . | |
| m | v 2 |
Thus if we measure y and l we can deduce e/mv 2. From the effect of the magnetic force we know e/mv. Combining these results we can find both e/m and v.
 |
| Fig. 13. |
The method by which this determination is carried out in practice is illustrated in fig. 13. The cathode rays start from the electrode C in a highly exhausted tube, pass through two small holes in the plugs A and B, the holes being in the same horizontal line. Thus a pencil of rays emerging from B is horizontal and produces a bright spot at the far end of the tube. In the course of their journey to the end of the tube they pass between the horizontal plates E and D, by connecting these plates with an electric battery a vertical electric field is produced between E and D and the phosphorescent spot is deflected. By measuring this deflection we determine e/mv2. The tube is now placed in a uniform magnetic field, the lines of magnetic force being horizontal and at right angles to the plane of the paper. The magnetic force makes the rays describe a circle in the plane of the paper, and by measuring the vertical deflection of the phosphorescent patch at the end of the tube we can determine the radius of this circle, and hence the value of e/mv. From the two observations the value of e/m and v can be calculated.
Another method of finding e/m for the negative ion which is applicable in many cases to which the preceding one is not suitable, is as follows: Let us suppose that the ion starts from rest and moves in a field where the electric and magnetic forces are both uniform, the electric force X being parallel to the axis of x, and the magnetic force Z parallel to the axis of z; then if x, y, are the co-ordinates of the ion at the time t, the equations of motion of the ion are—
| m | d 2x | = Xe − He | dy | , |
| dt 2 | dt |
| m | d 2y | = He | dx | . |
| dt 2 | dt |
The solution of these equations, if x, y, dx/dt, dy/dt all vanish when t = 0, is
| x = | Xm | {1 − cos( | e | Ht)} |
| eH2 | m |
| y = | Xm | { | e | Ht − sin( | e | Ht)}. |
| eH2 | m | m |
These equations show that the path of the ion is a cycloid, the generating circle of which has a diameter equal to 2Xm/eH2, and rolls on the line x = 0.
Suppose now that we have a number of ions starting from the plane x = 0, and moving towards the plane x = a. The particles starting from x = 0 describe cycloids, and the greatest distance they can get from the plane is equal to the diameter of the generating circle of the cycloid, i.e. to 2Xm/eH2. (After reaching this distance they begin to approach the plane.) Hence if a is less than the diameter of the generating circle, all the particles starting from x = 0 will reach the plane x = a, if this is unlimited in extent; while if a is greater than the diameter of the generating circle none of the particles which start from x = 0 will reach the plane x = a. Thus, if x = 0 is a plane illuminated by ultra-violet light, and consequently the seat of a supply of negative ions, and x = a a plane connected with an electrometer, then if a definite electric intensity is established between the planes, i.e. if X be fixed, so that the rate of emission of negative ions from the illuminated plate is given, and if a is less than 2Xm/eH2, all the ions which start from x = 0 will reach x = a. That
