The arrangement used by Zeleny is represented in fig. 8. P and Q are square brass plates. They are bored through their centres, and to the openings the tubes R and S are attached, the space between the plates being covered in so as to form a closed box. K is a piece of wire gauze completely covering the opening in Q; T is an insulated piece of wire gauze nearly but not quite filling the opening in the plate P, and connected with one pair of quadrants of an electrometer E. A plug of glass wool G filters out the dust from a stream of gas which enters the vessel by the tube D and leaves it by F; this plug also makes the velocity of the flow of the gas uniform across the section of the tube. The Röntgen rays to ionize the gas were produced by a bulb at O, the bulb and coil being in a lead-covered box, with an aluminium window through which the rays passed. Q is connected with one pole of a battery of cells, P and the other pole of the battery are put to earth. The changes in the potential of T are due to ions giving up their charges to it. With a given velocity of air-blast the potential of T was found not to change unless the difference of potential between P and Q exceeded a critical value. The field corresponding to this critical value thus made the ions move with the known velocity of the blast.Fig. 8.Fig. 9.Another method which has been employed by Rutherford and McClelland is based on the action of an electric field in destroying the conductivity of gas streaming through it. Suppose that BAB, DCD (fig. 9) are a system of parallel plates boxed in so that a stream of gas, after flowing between BB, passes between DD without any loss of gas in the interval. Suppose the plates DD are insulated, and connected with one pair of quadrants of an electrometer, by charging up C to a sufficiently high potential we can drive all the positive ions which enter the system DCD against the plates D; this will cause a deflexion of the electrometer, which in one second will be proportional to the number of positive ions which have entered the system in that time. If we charge A up to a high potential, B being put to earth, we shall find that the deflexion of the electrometer connected with DD is less than it was when A and B were at the same potential, because some of the positive ions in their passage through BAB are driven against the plates B. If u is the velocity along the lines of force in the uniform electric field between A and B, and t the time it takes for the gas to pass through BAB, then all the positive ions within a distance ut of the plates B will be driven up against these plates, and thus if the positive ions are equally distributed through the gas, the number of positive ions which emerge from the system when the electric field is on will bear to the number which emerge when the field is off the ratio of 1 − ut/l to unity, where l is the distance between A and B. This ratio is equal to the ratio of the deflexions in one second of the electrometer attached to D, hence the observations of this instrument give 1 − ut/l. If we know the velocity of the gas and the length of the plates A and B, we can determine t, and since l can be easily measured, we can find u, the velocity of the positive ion in a field of given strength. By charging A and C negatively instead of positively we can arrive at the velocity of the negative ion. In practice it is more convenient to use cylindrical tubes with coaxial wires instead of the systems of parallel plates, though in this case the calculation of the velocity of the ions from the observations is a little more complicated, inasmuch as the electric field is not uniform between the tubes.Fig. 10.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 notlose 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 ABdx=uasin ptdTdx =ua(1 − cos pt), if x = 0 when t = 0.pdThus 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 happensd = 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 T2and T1be 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 T2all 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 T2but 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 T2we 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 T1the 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.Let us suppose that the velocity of the ion is proportional to the electric intensity, so that if w is the vertical velocity of the ions, which are supposed all to be of one sign, w = RZ.Substituting this value of Z, the vertical force on the ions between A and B is equal to1∫∫∫wρ dxdydz.RBut ∫∫ wρdxdy = ι, where ι is the current streaming from the point. This current, which can be easily measured by putting a galvanometer in series with the discharging point, is independent of z, the vertical distance of a plane between A and B below the charging point. Hence we haveF =ι∫dz =ι· z.RRThis force must be counterbalanced by the difference of gaseous pressures over the planes A and B; hence if pBand pAdenote respectively the pressures over B and A, we havepB− pA=ιz.RHence by the measurement of these pressures we can determine R, and hence the velocity with which an ion moves under a given electric intensity.There are other methods of determining the velocities of the ions, but as these depend on the theory of the conduction of electricity through a gas containing charged ions, we shall consider them in our discussion of that theory.By the use of these methods it has been shown that the velocities of the ions in a given gas are the same whether the ionization is produced by Röntgen rays, radioactive substances, ultra-violet light, or by the discharge of electricity from points. When the ionization is produced by chemical action the ions are very much less mobile, moving in the same electric field with a velocity less than one-thousandth part of the velocity of the first kind of ions. On the other hand, as we shall see later, the velocity of the negative ions in flames is enormously greater than that of even the first kind of ion under similar electric fields and at the same pressure. But when these negative ions get into the cold part of the flame, they move sluggishly with velocities of the order of those possessed by the second kind. The results of the various determinations of the velocities of the ions are given in the following table. The velocities are in centimetres per second under an electric force of one volt per centimetre, the pressure of the gas being 1 atmosphere. V+ denotes the velocity of the positive ion, V- that of the negative. V is the mean velocity of the positive and negative ions.Velocities of Ions.—Ions produced by Röntgen Rays.Gas.V+.V-.V.Observer.Air....1.6RutherfordAir (dry)1.361.87..Zeleny”1.601.70..Langevin”1.391.78..Phillips”1.541.78..WellischAir (moist)1.371.81..ZelenyOxygen (dry)1.361.80..”Oxygen (moist)1.291.52..”Carbonic acid (dry)0.760.81..”” ”0.860.90..Langevin” ”0.810.85..WellischCarbonic acid (moist)0.820.75..ZelenyHydrogen (dry)6.707.95..”Nitrogen....1.6RutherfordSulphur dioxide0.440.41..WellischHydrochloric acid....1.27RutherfordChlorine....1.0”Helium (dry)5.096.31..Franck and PohlCarbon monoxide1.101.14..WellischNitrous oxide0.820.90..”Ammonia0.740.80..”Aldehyde0.310.30..”Ethyl alcohol0.340.27..”Acetone0.310.29..”Ethyl chloride0.330.31..”Pentane0.360.35..”Methyl acetate0.330.36..”Ethyl formate0.300.31..”Ethyl ether0.290.31..”Ethyl acetate0.310.28..”Methyl bromide0.290.28..”Methyl iodide0.210.22..”Carbon tetrachloride0.300.31..”Ethyl iodide0.170.16..”Ions produced by Ultra-Violet Light.Air1.4RutherfordHydrogen3.9RutherfordCarbonic acid0.78RutherfordIons in Gases sucked from Flames.Velocities varying from .04 to .23McClellandIons in Flames containing Salts.Negative ions12.9 cm./sec.Gold+ions for salts of Li, Na, K, Rb, Cs62H. A. Wilson”200Marx”80MoreauIons liberated by Chemical Action.Velocities of the order of 0.0005 cm./sec.BlochIons from Point Discharge.Hydrogen5.47.436.41ChattockCarbonic acid0.830.9250.88ChattockAir1.321.801.55ChattockOxygen1.301.851.57ChattockIt will be seen from this table that the greater mobility of the negative ions is very much more marked in the case of the lighter and simpler gases than in that of the heavier and more complicated ones; with the vapours of organic substances there seems but little difference between the mobilities of the positive and negative ions, indeed in one or two cases the positive one seems slightly but very slightly the more mobile of the two. In the case of the simple gases the difference is much greater when the gases are dry than when they are moist. It has been shown by direct experiment that the velocities are directly proportional to the electric force.Variation of Velocities with Pressure.—Until the pressure gets low the velocities of the ions, negative as well as positive, vary inversely as the pressure. Langevin (loc. cit.) was the first to show that at very low pressures the velocity of the negative ions increases more rapidly as the pressure is diminished than this law indicates. If the nature of the ion did not change with the pressure, the kinetic theory of gases indicates that the velocity would vary inversely as the pressure, so that Langevin’s results indicate a change in the nature of the negative ion when the pressure is diminished below a certain value. Langevin’s results are given in the following table, where p represents the pressure measured in centimetres of mercury, V+ and V- the velocities of the positive and negative ions in air under unit electrostatic force,i.e.300 volts per centimetre:—Negative Ions.Positive Ions.p.V-.pV-/76.p.V+.pV+/76.7.565606477.5443043720.0220458020.0163443041.599453041.578242776.051051076.0480420142.0270505142.0225425The increase in the case of pV- indicates that the structure of the negative ion gets simpler as the pressure is reduced. Wallisch in some experiments made at the Cavendish Laboratory found that the diminution in the value of pV- at low pressures is much more marked in some gases than in others, and in some gases he failed to detect it; but it must be remembered that it is difficult to get measurements at pressures of only a few millimetres, as the amount of ionization is so exceedingly small at such pressures that the quantities to be observed are hardly large enough to admit of accurate measurements by the methods available at higher pressures.Effect of Temperature on the Velocity of the Ions.—Phillips (Proc. Roy. Soc., 1906, 78, p. 167) investigated, using Langevin’s method, the velocities of the + and − ions through air at atmospheric pressure at temperatures ranging from that of boiling liquid air to 411° C.; R1and R2are the velocities of the + and − ions respectively when the force is a volt per centimetre.R1.R2.Temperature Absolute.2.002.495411°1.952.40399°1.852.30383°1.812.21373°1.672.125348°1.602.00333°1.391.785285°0.9451.23209°0.2350.23594°We see that except in the case of the lowest temperature, that of liquid air, where there is a great drop in the velocity, the velocities of the ions are proportional to the absolute temperature. On the hypothesis of an ion of constant size we should, from the kinetic theory of gases, expect the velocity to be proportional to the square root of the absolute temperature, if the charge on the ion did not affect the number of collisions between the ion and the molecules of the gas through which it is moving. If the collisions were brought about by the electrical attraction between the ions and the molecules, the velocity would be proportional to the absolute temperature. H. A. Wilson (Phil. Trans.192, p. 499), in his experiments on the conduction of flames and hot gases into which salts had been put, found that the velocity of the positive ions in flames at a temperature of 2000° C. containing the salts of the alkali metals was 62 cm./sec. under an electric force of one volt per centimetre, while the velocity of the positive ions in a stream of hot air at 1000° C. containing the same salts was only 7 cm./sec. under the same force. The great effect of temperature is also shown in some experiments of McClelland (Phil. Mag.[5], 46, p. 29) on the velocities of the ions in gases drawn from Bunsen flames and arcs; he found that these depended upon the distance the gas had travelled from the flame. Thus, the velocity of the ions at a distance of 5.5 cm. from the Bunsen flame when the temperature was 230° C. was .23 cm./sec. for a volt per centimetre; at a distance of 10 cm. from the flame when the temperature was 160° C. the velocity was .21 cm./sec; while at a distance of 14.5 cm. from the flame when the temperature was 105° C. the velocity was only .04 cm./sec. If the temperature of the gas at this distance from the flame was raised by external means, the velocity of the ions increased.We can derive some information as to the constitution of the ions by calculating the velocity with which a molecule of the gas would move in the electric field if it carried the same charge as the ion. From the theory of the diffusion of gases, as developed by Maxwell, we know that if the particles of a gas A are surrounded by a gas B, then, if the partial pressure of A is small, the velocity u with which its particles will move when acted upon by a force Xe is given by the equationu =XeD,(p1/N1)where D represents the coefficient of inter-diffusion of A into B, and N1the number of particles of A per cubic centimetre when the pressure due to A is p1. Let us calculate by this equation the velocity with which a molecule of hydrogen would move through hydrogen if it carried the charge carried by an ion, which we shall prove shortly to be equal to the charge carried by an atom of hydrogen in the electrolysis of solutions. Since p1/N1is independent of the pressure, it is equal to Π/N, where Π is the atmospheric pressure and N the number of molecules in a cubic centimetre of gas at atmospheric pressure. Now Ne= 1.22×1010, if e is measured in electrostatic units; Π = 106and D in this case is the coefficient of diffusion of hydrogen into itself, and is equal to 1.7. Substituting these values we findu = 1.97×104X.If the potential gradient is 1 volt per centimetre, X =1⁄300. Substituting this value for X, we find u = 66 cm./sec, for the velocity of a hydrogen molecule. We have seen that the velocity of the ion in hydrogen is only about 5 cm./sec, so that the ion moves more slowly than it would if it were a single molecule. One way of explaining this is to suppose that the ion is bigger than the molecule, and is in fact an aggregation of molecules, the charged ion acting as a nucleus around which molecules collect like dust round a charged body. This view is supported by the effect produced by moisture in diminishing the velocity of the negative ion, for, as C. T. R. Wilson (Phil. Trans.193, p. 289) has shown, moisture tends to collect round the ions, and condenses more easily on the negative than on the positive ion. In connexion with the velocities of ions in the gases drawn from flames, we find other instances which suggest that condensation takes place round the ions. An increase in the size of the system is not, however, the only way by which the velocity might fall below that calculated for the hydrogen molecule, for we must remember that the hydrogen molecule, whose coefficient of diffusion is 1.7, is not charged, while the ion is. The forces exerted by the ion on the other molecules of hydrogen are not the same as those which would be exerted by a molecule of hydrogen, and as the coefficient of diffusion depends on the forces between the molecules, the coefficient of diffusion of a charged molecule into hydrogen might be very different from that of an uncharged one.Wellisch (loc. cit.) has shown that the effect of the charge on the ion is sufficient in many cases to explain the small velocity of the ions, even if there were no aggregation.Mixture of Gases.—The ionization of a mixture of gases raises some very interesting questions. If we ionize a mixture of two very different gases, say hydrogen and carbonic acid, and investigate the nature of the ions by measuring their velocities, the question arises, shall we find two kinds of positive and two kinds of negative ions moving with different velocities, as we should do if some of the positive ions were positively charged hydrogen molecules, while others were positively charged molecules of carbonic acid; or shall we find only one velocity for the positive ions and one for the negative? Many experiments have been made on the velocity of ions in mixtures of two gases, but as yet no evidence has been found of the existence of two different kinds of either positive or negative ions in such mixtures, although some of the methods for determining the velocities of the ions, especially Langevin’s, ought to give evidence of this effect, if it existed. The experiments seem to showthat the positive (and the same is true for the negative) ions in a mixture of gases are all of the same kind. This conclusion is one of considerable importance, as it would not be true if the ions consisted of single molecules of the gas from which they are produced.Recombination.—Several methods enable us to deduce the coefficient of recombination of the ions when we know their velocities. Perhaps the simplest of these consists in determining the relation between the current passing between two parallel plates immersed in ionized gas and the potential difference between the plates. For let q be the amount of ionization,i.e.the number of ions produced per second per unit volume of the gas, A the area of one of the plates, and d the distance between them; then if the ionization is constant through the volume, the number of ions of one sign produced per second in the gas is qAd. Now if i is the current per unit area of the plate, e the charge on an ion, iA/e ions of each sign are driven out of the gas by the current per second. In addition to this source of loss of ions there is the loss due to the recombination; if n is the number of positive or negative ions per unit volume, then the number which recombine per second is αn2per cubic centimetre, and if n is constant through the volume of the gas, as will approximately be the case if the current through the gas is only a small fraction of the saturation current, the number of ions which disappear per second through recombination is αn2·Ad. Hence, since when the gas is in a steady state the number of ions produced must be equal to the number which disappear, we haveqAd = iA/e + αn2·Ad,q = i/ed + αn2.If u1and u2are 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, hencei = nu1e + nu2e.If X is the electric force acting on the gas, k1and k2the velocities of the positive and negative ions under unit force, u1= k1X, u2= k2X; hencen = i/(k1+ k2)Xe,and we haveq =i+αi2.ed(k1+ k2)2e2X2But qed is the saturation current per unit area of the plate; calling this I, we haveI − i =dαi2e(k1+ k2)2X2orX2=i2·dα.e(I − i) (k1+ k2)2Hence if we determine corresponding values of X and i we can deduce the value of α/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 N0ions if the flash of Röntgen rays produced N0positive and N0negative 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 + αn0t).With the same notation as before, the breadth of either of the outer layers will in time dt increase by X(k1+ k2)dt, and the number of ions in it by X(k1+ k2)ndt; these ions will reach the plate, the outer layers will receive fresh ions until the middle one disappears, which it will do after a time l/X(k1+ k2), where l is the thickness of the slab ab of ionized gas; hence N, the number of ions reaching either plate, is given by the equationN =∫0l/X(k1+k2)n0X(k1+ k2)dt =X(k1+ k2)log(1 +n0αl).1 + n0αtαX(k1+ k2)If Q is the charge received by the plate,Q = Ne =Xlog(1 +Q0ε),4πε4πXwhere Q0= n0le is the charge received by the plate when the electric force is large enough to prevent recombination, and ε = α4πe(R1+ R2). We can from this result deduce the value of ε and hence the value of α when R1+ R2is known.Distribution of Electric Force when a Current is passing through an Ionized Gas.—Let the two plates be at right angles to the axis of x; then we may suppose that between the plates the electric intensity X is everywhere parallel to the axis of x. The velocities of both the positive and negative ions are assumed to be proportional to X. Let k1X, k2X represent these velocities respectively; let n1, n2be respectively the number of positive and negative ions per unit volume at a point fixed by the co-ordinate x; let q be the number of positive or negative ions produced in unit time per unit volume at this point; and let the number of ions which recombine in unit volume in unit time be αn1n2; then if e is the charge on the ion, the volume density of the electrification is (n1− n2)e, hencedX= 4π(n1− n2)e (1).dxIf I is the current through unit area of the gas and if we neglect any diffusion except that caused by the electric field,n1ek1X + n2ek2X = I (2).From equations (1) and (2) we haven1e =1(I+k2dX)(3),k1+ k2X4πdxn1e =1(I-k2dX)(4),k1+ k2X4πdxand from these equations we can, if we know the distribution of electric intensity between the plates, calculate the number of positive and negative ions.In a steady state the number of positive and negative ions in unit volume at a given place remains constant, hence neglecting the loss by diffusion, we haved(k1n1X) = q − αn1n2(5).dx-d(k2n2X) = q − αn1n2(6).dxIf k1and k2are constant, we have from (1), (5) and (6)d²X²= 8πe(q − αn1n2)(1+1)(7).dx²k1k2an equation which is very useful,becauseit enables us, if we know the distribution of X², to find whether at any point in the gas the ionization is greater or less than the recombination of the ions. We see that q − αn1n2, which is the excess of ionization over recombination, is proportional to d²X²/dx². Thus when the ionization exceeds the recombination,i.e.when q − αn1n2is positive, the curve for X² is convex to the axis of x, while when the recombination exceeds the ionization the curve for X² will be concave to the axis of x. Thus, for example, fig. 11 represents the curve for X² observed by Graham (Wied. Ann.64, p. 49) in a tube through which a steady current is passing. Interpreting it by equation (7), we infer that ionization was much in excess of recombination at A and B, slightly so along C, while along D the recombination exceeded the ionization. Substituting in equation (7) the values of n1, n2given in (3), (4), we getd²X²8πe[q -α(I +k2dX²)(I -k2dX²)](1+1)(8).dx²e²X²(k1+ k2)²8πdx8πdxk1k2Fig. 11.This equation can be solved (see Thomson,Phil. Mag.xlvii. P. 253), when q is constant and k1= k2. From the solution it appears that if X1be the value of x close to one of the plates, and X0the value midway between them,X1/X0=1β2− 2/βwhere β = 8πek1/α.Since e = 4×10-10, α = 2×10-6, and k1for air at atmospheric pressure = 450, β is about 2.3 for air at atmospheric pressure and it becomes much greater at lower pressures.Thus X1/X0is always greater than unity, and the value of the ratio increases from unity to infinity as β increases from zero to infinity. As β does not involve either q or I, the ratio of X1to X0is independent of the strength of the current and of the intensity of the ionization.No general solution of equation (8) has been found when k1is not equal to k2, but we can get an approximation to the solution when q is constant. The equations (1), (2), (3), (4) are satisfied by the values—n1= n2= (q/α)1/2k1n1Xe =k1I,k1+ k2k2n2Xe =k2I,k1+ k2X =(α)1/2I.qe(k1+ k2)These solutions cannot, however, hold right up to the surface of the plates, for across each unit of area, at a point P, k1I/(k1+ k2)e positive ions pass in unit time, and these must all come from the region between P and the positive plate. If λ is the distance of P from this plate, this region cannot furnish more than qλ positive ions, and only this number if there are no recombinations. Hence the solution cannot hold when qλ is less than k1I/(k1+ k2)e, or where λ is less than k1I/(k1+ k2)qe.Similarly the solution cannot hold nearer to the negative plate than the distance k2I/(k1+ k2)qe.The force in these layers will be greater than that in the middle of the gas, and so the loss of ions by recombination will be smaller in comparison with the loss due to the removal of the ions by the current. If we assume that in these layers the loss of ions by recombination can be neglected, we can by the method of the next article find an expression for the value of the electric force at any point in the layer. This, in conjunction with the value X0= (α/q)1/2· I/e(k1+ k2) for the gas outside the layer, will give the value of X at any point between the plates. It follows from this investigation that if X1and X2are the values of X at the positive and negative plates respectively, and X0the value of X outside the layer,X1= X0(1 +k11)1/2,X2= X0(1 +k21)1/2,k2εk1εwhere ε = α/4πe(k1+ k2). Langevin found that for air at a pressure of 152 mm. ε = 0.01, at 375 mm. ε = 0.06, and at 760 mm. ε = 0.27. Thus at fairly low pressures 1/ε is large, and we have approximatelyX1= X0(k1)1/21,X2= X0(k2)1/21.k2√εk1√εThereforeX1/X2= k1/k2,Fig. 12.or the force at the positive plate is to that at the negative plate as the velocity of the positive ion is to that of the negative ion. Thus the force at the negative plate is greater than that at the positive. The falls of potential V1, V2at the two layers when 1/ε is large can be shown to be given by the equationsV1= 8π²(ε)3/2k1(k1)1/2i²,qαk2V2= 8π²(ε)3/2k2(k2)1/2i²,qαk1henceV1/V2= k1²/k2²,so that the potential falls at the electrodes are proportional to the squares of the velocities of the ions. The change in potential across the layers is proportional to the square of the current, while the potential change between the layers is proportional to the current, the total potential difference between the plates is the sum of these changes, hence the relation between V and i will be of the formV = Ai + Bi².Mie (Ann. der. Phys., 1904, 13, P. 857) has by the method of successive approximations obtained solutions of equation (8) (i.) when the current is only a small fraction of the saturation current, (ii.) when the current is nearly saturated. The results of his investigations are represented in fig. 12, which represents the distribution of electric force along the path of the current for various values of the current expressed as fractions of the saturation current. It will be seen that until the current amounts to about one-fifth of the maximum current, the type of solution is the one just indicated,i.e.the electric force is constant except in the neighbourhood of the electrodes when it increases rapidly.Though we are unable to obtain a general solution of the equation (8), there are some very important special cases in which that equation can be solved without difficulty. We shall consider two of these, the first being that when the current is saturated. In this case there is no loss of ions by recombination, so that using the same notation as before we haved(n1k1X) = q,dxd(n2k2X) = -q.dxThe solutions of which if q is constant aren1k1X = qx,n2k2X = I/e − qx = q(l − x),if l is the distance between the plates, and x = 0 at the positive electrode. SincedX/dx = 4π(n1− n2)e,we get1dX²= qx{1+1}- ql,8πd²xk1k2k2orX²= qx²(1+1)- qlx+ C,8π2k1k2k2where C is a quantity to be determined by the condition that∫l0Xdx = V, where V is the given potential difference between the plates. When the force is a minimum dX/dx = 0, hence at this pointx =l k1, l − x =l k2.k1+ k2k1+ k2Hence the ratio of the distances of this point from the positive and negative plates respectively is equal to the ratio of the velocities of the positive and negative ions.The other case we shall consider is the very important one in which the velocity of the negative ion is exceedingly large compared with the positive; this is the case in flames where, as Gold (Proc. Roy. Soc.97, p. 43) has shown, the velocity of the negative ion is many thousand times the velocity of the positive; it is also very probably the case in all gases when the pressure is low. We may get the solution of this case either by putting k1/k2= 0 in equation (8), or independently as follows:—Using the same notation as before, we havei = n1k1Xe + n2k2Xe,d(n2k2X) = q − αn1n2,dxdX= 4π (n1− n2)e.dxIn this case practically all the current is carried by the negative ions so that i = n2k2Xe, and therefore q = αn1n2.Thusn2= i/k2Xe, n1= qk2Xe/αi.ThusdX=4πe²k2qX-4πi,dxαik2XordX²-8πe²k2qX²= -8πi.dxαik2The solution of this equation isX² =αi²+ Cε8πe² k2qx/αi.qk²2e²Here x is measured from the positive electrode; it is more convenient in this case, however, to measure it from the negative electrode. If x be the distance from the negative electrode at which the electric force is X, we have from equation (7)X² =αi²+ C¹ε8πe² k2qx/αi.qk²2e²To find the value of C¹ we see by equation (7) thatd²X²k1k21= q − αn1n2;dX²k1+ k28πehence[dX²k1k21]x1=∫x10(q − αn1n2)dx.dXk1+ k28πeThe 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 areaof the cathode, as hot metal cathodes emit large quantities of negative electricity i0may in some cases be considerable, thus the right hand side of equation is (i − i0)/e. When x1is large dX²/dx = 0; hence we have from equationC¹ =αi(i − i0)k1+ k2,qk1k2e²k2and since k1is small compared with k2, we haveX² =α i²(1 +k2i − i0ε-8πe² k2 · qx/α · i).qk²2e²k1iFrom the values which have been found for k2and 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)1/2to k11/2. As k1is many thousand times k2the 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 i0is not nearly equal to i is approximately given by the equationX =i(α)1/2ε−4π e²k2 qx / αi,e(k1k2)1/2qand the fall of potential at the cathode is equal approximately to∫∞0Xdx, that is toi(α)1/2α i.e(k1k2)1/2q4πe²k2qThe potential difference between the plates is the sum of the fall of potential in the uniform part of the discharge plus the fall at the cathode, henceV =(α)1/2i(il +i α²1).qek24πe²q√(k1k2)The fall of potential at the cathode is proportional to the square of the current, while the fall in the rest of the circuit is directly proportional to the current. In the case of flames or hot gases, the fall of potential at the cathode is much greater than that in the rest of the circuit, so that in such cases the current through the gas varies nearly as the square root of the potential difference. The equation we have just obtained is of the formV = Ai + Bi²,and H. A. Wilson has shown that a relation of this form represents the results of his experiments on the conduction of electricity through flames.The expression for the fall of potential at the cathode is inversely proportional to q3/2, q being the number of ions produced per cubic centimetre per second close to the cathode; thus any increase in the ionization at the cathode will diminish the potential fall at the cathode, and as practically the whole potential difference between the electrodes occurs at the cathode, a diminution in the potential fall there will be much more important than a diminution in the electric force in the uniform part of the discharge, when the force is comparatively insignificant. This consideration explains a very striking phenomenon discovered many years ago by Hittorf, who found that if he put a wire carrying a bead of a volatile salt into the flame, it produced little effect upon the current, unless it were placed close to the cathode where it gave rise to an enormous increase in the current, sometimes increasing the current more than a hundredfold. The introduction of the salt increases very largely the number of ions produced, so that q is much greater for a salted flame than for a plain one. Thus Hittorf’s result coincides with the conclusions we have drawn from the theory of this class of conduction.The fall of potential at the cathode is proportional to i − i0, where i0is the stream of negative electricity which comes from the cathode itself, thus as i0increases the fall of potential at the cathode diminishes and the current sent by a given potential difference through the gas increases. Now all metals give out negative particles when heated, at a rate which increases very rapidly with the temperature, but at the same temperature some metals give out more than others. If the cathode is made of a metal which emits large quantities of negative particles, (i − i0) will for a given value of i be smaller than if the metal only emitted a small number of particles; thus the cathode fall will be smaller for the metal with the greater emissitivity, and the relation between the potential difference and the current will be different in the two cases. These considerations are confirmed by experience, for it has been found that the current between electrodes immersed in a flame depends to a great extent upon the metal of which the electrodes are made. Thus Pettinelli (Acc. dei Lincei[5], v. p. 118) found that,ceteris paribus, the current between two carbon electrodes was about 500 times that between two iron ones. If one electrode was carbon and the other iron, the current when the carbon was cathode and the iron anode was more than 100 times the current when the electrodes were reversed. The emission of negative particles by some metallic oxides, notably those of calcium and barium, has been shown by Wehnelt (Ann. der Phys.11, p. 425) to be far greater than that of any known metal, and the increase of current produced by coating the cathodes with these oxides is exceedingly large; in some cases investigated by Tufts and Stark (Physik. Zeits., 1908, 5, p. 248) the current was increased many thousand times by coating the cathode with lime. No appreciable effect is produced by putting lime on the anode.Conduction when all the Ions are of one Sign.—There are many important cases in which the ions producing the current come from one electrode or from a thin layer of gas close to the electrode, no ionization occurring in the body of the gas or at the other electrode. Among such cases may be mentioned those where one of the electrodes is raised to incandescence while the other is cold, or when the negative electrode is exposed to ultra-violet light. In such cases if the electrode at which the ionization occurs is the positive electrode, all the ions will be positively charged, while if it is the negative electrode the ions will all be charged negatively. The theory of this case is exceedingly simple. Suppose the electrodes are parallel planes at right angles to the axis of x; let X be the electric force at a distance x from the electrode where the ionization occurs, n the number of ions (all of which are of one sign) at this place per cubic centimetre, k the velocity of the ion under unit electric force, e the charge on an ion, and i the current per unit area of the electrode. Then we have dX/dx = 4πne, and if u is the velocity of the ion neu = i. But u = kX, hence we have kX/4π · dX/dx = i, and since the right hand side of this equation does not depend upon x, we get kX²/8π = ix + C, where C is a constant to be determined. If l is the distance between the plates, and V the potential difference between them,V =∫l0Xdx =1√8π[(il + C)3/2− C3/2].ikWe shall show that when the current is far below the saturation value, C is very small compared with il, so that the preceding equation becomesV² = 8πl³ i/k (1).To show that for small currents C is small compared with il, consider the case when the ionization is confined to a thin layer, thickness d close to the electrode, in that layer let n0be the value of n, then we have q = αn0² + i/ed. If X0be the value of X when x = 0, kX0n0e = i, and,C =kX0²=i²=α·i²(2).8πn0²ke·8π8πke²q + i/edSince α/8πke is, as we have seen, less than unity, C will be small compared with il, if i/(eq + i/d) is small compared with l. If I0is the saturation current, q = I0/ed, so that the former expression = id/(I0+ i), if i is small compared with I0, this expression is small compared with d, and thereforea fortioricompared with l, so that we are justified in this case in using equation (1).From equation (2) we see that the current increases as the square of the potential difference. Here an increase in the potential difference produces a much greater percentage increase than in conduction through metals, where the current is proportional to the potential difference. When the ionization is distributed through the gas, we have seen that the current is approximately proportional to the square root of the potential, and so increases more slowly with the potential difference than currents through metals. From equation (1) the current is inversely proportional to the cube of the distance between the electrodes, so that it falls off with great rapidity as this distance is increased. We may note that for a given potential difference the expression for the current does not involve q, the rate of production of the ions at the electrode, in other words, if we vary the ionization the current will not begin to be affected by the strength of the ionization until this falls so low that the current is a considerable fraction of the saturation current. For the same potential difference the current is proportional to k, the velocity under unit electric force of the ion which carries the current. As the velocity of the negative ion is greater than that of the positive, the current when the ionization is confined to the neighbourhood of one of the electrodes will be greater when that electrode is made cathode than when it is anode. Thus the current will appear to pass more easily in one direction than in the opposite.Since the ions which carry the current have to travel all the way from one electrode to the other, any obstacle which is impervious to these ions will, if placed between the electrodes, stop the currentto the electrode where there is no ionization. A plate of metal will be as effectual as one made of a non-conductor, and thus we get the remarkable result that by interposing a plate of an excellent conductor like copper or silver between the electrode, we can entirely stop the current. This experiment can easily be tried by using a hot plate as the electrode at which the ionization takes place: then if the other electrode is cold the current which passes when the hot plate is cathode can be entirely stopped by interposing a cold metal plate between the electrodes.
The arrangement used by Zeleny is represented in fig. 8. P and Q are square brass plates. They are bored through their centres, and to the openings the tubes R and S are attached, the space between the plates being covered in so as to form a closed box. K is a piece of wire gauze completely covering the opening in Q; T is an insulated piece of wire gauze nearly but not quite filling the opening in the plate P, and connected with one pair of quadrants of an electrometer E. A plug of glass wool G filters out the dust from a stream of gas which enters the vessel by the tube D and leaves it by F; this plug also makes the velocity of the flow of the gas uniform across the section of the tube. The Röntgen rays to ionize the gas were produced by a bulb at O, the bulb and coil being in a lead-covered box, with an aluminium window through which the rays passed. Q is connected with one pole of a battery of cells, P and the other pole of the battery are put to earth. The changes in the potential of T are due to ions giving up their charges to it. With a given velocity of air-blast the potential of T was found not to change unless the difference of potential between P and Q exceeded a critical value. The field corresponding to this critical value thus made the ions move with the known velocity of the blast.
Another method which has been employed by Rutherford and McClelland is based on the action of an electric field in destroying the conductivity of gas streaming through it. Suppose that BAB, DCD (fig. 9) are a system of parallel plates boxed in so that a stream of gas, after flowing between BB, passes between DD without any loss of gas in the interval. Suppose the plates DD are insulated, and connected with one pair of quadrants of an electrometer, by charging up C to a sufficiently high potential we can drive all the positive ions which enter the system DCD against the plates D; this will cause a deflexion of the electrometer, which in one second will be proportional to the number of positive ions which have entered the system in that time. If we charge A up to a high potential, B being put to earth, we shall find that the deflexion of the electrometer connected with DD is less than it was when A and B were at the same potential, because some of the positive ions in their passage through BAB are driven against the plates B. If u is the velocity along the lines of force in the uniform electric field between A and B, and t the time it takes for the gas to pass through BAB, then all the positive ions within a distance ut of the plates B will be driven up against these plates, and thus if the positive ions are equally distributed through the gas, the number of positive ions which emerge from the system when the electric field is on will bear to the number which emerge when the field is off the ratio of 1 − ut/l to unity, where l is the distance between A and B. This ratio is equal to the ratio of the deflexions in one second of the electrometer attached to D, hence the observations of this instrument give 1 − ut/l. If we know the velocity of the gas and the length of the plates A and B, we can determine t, and since l can be easily measured, we can find u, the velocity of the positive ion in a field of given strength. By charging A and C negatively instead of positively we can arrive at the velocity of the negative ion. In practice it is more convenient to use cylindrical tubes with coaxial wires instead of the systems of parallel plates, though in this case the calculation of the velocity of the ions from the observations is a little more complicated, inasmuch as the electric field is not uniform between the tubes.
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 notlose 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
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 T2and T1be 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 T2all 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 T2but 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 T2we 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 T1the 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.
Let us suppose that the velocity of the ion is proportional to the electric intensity, so that if w is the vertical velocity of the ions, which are supposed all to be of one sign, w = RZ.
Substituting this value of Z, the vertical force on the ions between A and B is equal to
But ∫∫ wρdxdy = ι, where ι is the current streaming from the point. This current, which can be easily measured by putting a galvanometer in series with the discharging point, is independent of z, the vertical distance of a plane between A and B below the charging point. Hence we have
This force must be counterbalanced by the difference of gaseous pressures over the planes A and B; hence if pBand pAdenote respectively the pressures over B and A, we have
Hence by the measurement of these pressures we can determine R, and hence the velocity with which an ion moves under a given electric intensity.
There are other methods of determining the velocities of the ions, but as these depend on the theory of the conduction of electricity through a gas containing charged ions, we shall consider them in our discussion of that theory.
By the use of these methods it has been shown that the velocities of the ions in a given gas are the same whether the ionization is produced by Röntgen rays, radioactive substances, ultra-violet light, or by the discharge of electricity from points. When the ionization is produced by chemical action the ions are very much less mobile, moving in the same electric field with a velocity less than one-thousandth part of the velocity of the first kind of ions. On the other hand, as we shall see later, the velocity of the negative ions in flames is enormously greater than that of even the first kind of ion under similar electric fields and at the same pressure. But when these negative ions get into the cold part of the flame, they move sluggishly with velocities of the order of those possessed by the second kind. The results of the various determinations of the velocities of the ions are given in the following table. The velocities are in centimetres per second under an electric force of one volt per centimetre, the pressure of the gas being 1 atmosphere. V+ denotes the velocity of the positive ion, V- that of the negative. V is the mean velocity of the positive and negative ions.
Velocities of Ions.—Ions produced by Röntgen Rays.
Ions from Point Discharge.
It will be seen from this table that the greater mobility of the negative ions is very much more marked in the case of the lighter and simpler gases than in that of the heavier and more complicated ones; with the vapours of organic substances there seems but little difference between the mobilities of the positive and negative ions, indeed in one or two cases the positive one seems slightly but very slightly the more mobile of the two. In the case of the simple gases the difference is much greater when the gases are dry than when they are moist. It has been shown by direct experiment that the velocities are directly proportional to the electric force.
Variation of Velocities with Pressure.—Until the pressure gets low the velocities of the ions, negative as well as positive, vary inversely as the pressure. Langevin (loc. cit.) was the first to show that at very low pressures the velocity of the negative ions increases more rapidly as the pressure is diminished than this law indicates. If the nature of the ion did not change with the pressure, the kinetic theory of gases indicates that the velocity would vary inversely as the pressure, so that Langevin’s results indicate a change in the nature of the negative ion when the pressure is diminished below a certain value. Langevin’s results are given in the following table, where p represents the pressure measured in centimetres of mercury, V+ and V- the velocities of the positive and negative ions in air under unit electrostatic force,i.e.300 volts per centimetre:—
The increase in the case of pV- indicates that the structure of the negative ion gets simpler as the pressure is reduced. Wallisch in some experiments made at the Cavendish Laboratory found that the diminution in the value of pV- at low pressures is much more marked in some gases than in others, and in some gases he failed to detect it; but it must be remembered that it is difficult to get measurements at pressures of only a few millimetres, as the amount of ionization is so exceedingly small at such pressures that the quantities to be observed are hardly large enough to admit of accurate measurements by the methods available at higher pressures.
Effect of Temperature on the Velocity of the Ions.—Phillips (Proc. Roy. Soc., 1906, 78, p. 167) investigated, using Langevin’s method, the velocities of the + and − ions through air at atmospheric pressure at temperatures ranging from that of boiling liquid air to 411° C.; R1and R2are the velocities of the + and − ions respectively when the force is a volt per centimetre.
We see that except in the case of the lowest temperature, that of liquid air, where there is a great drop in the velocity, the velocities of the ions are proportional to the absolute temperature. On the hypothesis of an ion of constant size we should, from the kinetic theory of gases, expect the velocity to be proportional to the square root of the absolute temperature, if the charge on the ion did not affect the number of collisions between the ion and the molecules of the gas through which it is moving. If the collisions were brought about by the electrical attraction between the ions and the molecules, the velocity would be proportional to the absolute temperature. H. A. Wilson (Phil. Trans.192, p. 499), in his experiments on the conduction of flames and hot gases into which salts had been put, found that the velocity of the positive ions in flames at a temperature of 2000° C. containing the salts of the alkali metals was 62 cm./sec. under an electric force of one volt per centimetre, while the velocity of the positive ions in a stream of hot air at 1000° C. containing the same salts was only 7 cm./sec. under the same force. The great effect of temperature is also shown in some experiments of McClelland (Phil. Mag.[5], 46, p. 29) on the velocities of the ions in gases drawn from Bunsen flames and arcs; he found that these depended upon the distance the gas had travelled from the flame. Thus, the velocity of the ions at a distance of 5.5 cm. from the Bunsen flame when the temperature was 230° C. was .23 cm./sec. for a volt per centimetre; at a distance of 10 cm. from the flame when the temperature was 160° C. the velocity was .21 cm./sec; while at a distance of 14.5 cm. from the flame when the temperature was 105° C. the velocity was only .04 cm./sec. If the temperature of the gas at this distance from the flame was raised by external means, the velocity of the ions increased.
We can derive some information as to the constitution of the ions by calculating the velocity with which a molecule of the gas would move in the electric field if it carried the same charge as the ion. From the theory of the diffusion of gases, as developed by Maxwell, we know that if the particles of a gas A are surrounded by a gas B, then, if the partial pressure of A is small, the velocity u with which its particles will move when acted upon by a force Xe is given by the equation
where D represents the coefficient of inter-diffusion of A into B, and N1the number of particles of A per cubic centimetre when the pressure due to A is p1. Let us calculate by this equation the velocity with which a molecule of hydrogen would move through hydrogen if it carried the charge carried by an ion, which we shall prove shortly to be equal to the charge carried by an atom of hydrogen in the electrolysis of solutions. Since p1/N1is independent of the pressure, it is equal to Π/N, where Π is the atmospheric pressure and N the number of molecules in a cubic centimetre of gas at atmospheric pressure. Now Ne= 1.22×1010, if e is measured in electrostatic units; Π = 106and D in this case is the coefficient of diffusion of hydrogen into itself, and is equal to 1.7. Substituting these values we find
u = 1.97×104X.
If the potential gradient is 1 volt per centimetre, X =1⁄300. Substituting this value for X, we find u = 66 cm./sec, for the velocity of a hydrogen molecule. We have seen that the velocity of the ion in hydrogen is only about 5 cm./sec, so that the ion moves more slowly than it would if it were a single molecule. One way of explaining this is to suppose that the ion is bigger than the molecule, and is in fact an aggregation of molecules, the charged ion acting as a nucleus around which molecules collect like dust round a charged body. This view is supported by the effect produced by moisture in diminishing the velocity of the negative ion, for, as C. T. R. Wilson (Phil. Trans.193, p. 289) has shown, moisture tends to collect round the ions, and condenses more easily on the negative than on the positive ion. In connexion with the velocities of ions in the gases drawn from flames, we find other instances which suggest that condensation takes place round the ions. An increase in the size of the system is not, however, the only way by which the velocity might fall below that calculated for the hydrogen molecule, for we must remember that the hydrogen molecule, whose coefficient of diffusion is 1.7, is not charged, while the ion is. The forces exerted by the ion on the other molecules of hydrogen are not the same as those which would be exerted by a molecule of hydrogen, and as the coefficient of diffusion depends on the forces between the molecules, the coefficient of diffusion of a charged molecule into hydrogen might be very different from that of an uncharged one.
Wellisch (loc. cit.) has shown that the effect of the charge on the ion is sufficient in many cases to explain the small velocity of the ions, even if there were no aggregation.
Mixture of Gases.—The ionization of a mixture of gases raises some very interesting questions. If we ionize a mixture of two very different gases, say hydrogen and carbonic acid, and investigate the nature of the ions by measuring their velocities, the question arises, shall we find two kinds of positive and two kinds of negative ions moving with different velocities, as we should do if some of the positive ions were positively charged hydrogen molecules, while others were positively charged molecules of carbonic acid; or shall we find only one velocity for the positive ions and one for the negative? Many experiments have been made on the velocity of ions in mixtures of two gases, but as yet no evidence has been found of the existence of two different kinds of either positive or negative ions in such mixtures, although some of the methods for determining the velocities of the ions, especially Langevin’s, ought to give evidence of this effect, if it existed. The experiments seem to showthat the positive (and the same is true for the negative) ions in a mixture of gases are all of the same kind. This conclusion is one of considerable importance, as it would not be true if the ions consisted of single molecules of the gas from which they are produced.
Recombination.—Several methods enable us to deduce the coefficient of recombination of the ions when we know their velocities. Perhaps the simplest of these consists in determining the relation between the current passing between two parallel plates immersed in ionized gas and the potential difference between the plates. For let q be the amount of ionization,i.e.the number of ions produced per second per unit volume of the gas, A the area of one of the plates, and d the distance between them; then if the ionization is constant through the volume, the number of ions of one sign produced per second in the gas is qAd. Now if i is the current per unit area of the plate, e the charge on an ion, iA/e ions of each sign are driven out of the gas by the current per second. In addition to this source of loss of ions there is the loss due to the recombination; if n is the number of positive or negative ions per unit volume, then the number which recombine per second is αn2per cubic centimetre, and if n is constant through the volume of the gas, as will approximately be the case if the current through the gas is only a small fraction of the saturation current, the number of ions which disappear per second through recombination is αn2·Ad. Hence, since when the gas is in a steady state the number of ions produced must be equal to the number which disappear, we have
qAd = iA/e + αn2·Ad,q = i/ed + αn2.
If u1and u2are 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, k1and k2the velocities of the positive and negative ions under unit force, u1= k1X, u2= k2X; hence
n = i/(k1+ k2)Xe,
and we have
But qed is the saturation current per unit area of the plate; calling this I, we have
or
Hence if we determine corresponding values of X and i we can deduce the value of α/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 N0ions if the flash of Röntgen rays produced N0positive and N0negative 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 + αn0t).
With the same notation as before, the breadth of either of the outer layers will in time dt increase by X(k1+ k2)dt, and the number of ions in it by X(k1+ k2)ndt; these ions will reach the plate, the outer layers will receive fresh ions until the middle one disappears, which it will do after a time l/X(k1+ k2), where l is the thickness of the slab ab of ionized gas; hence N, the number of ions reaching either plate, is given by the equation
If Q is the charge received by the plate,
where Q0= n0le is the charge received by the plate when the electric force is large enough to prevent recombination, and ε = α4πe(R1+ R2). We can from this result deduce the value of ε and hence the value of α when R1+ R2is known.
Distribution of Electric Force when a Current is passing through an Ionized Gas.—Let the two plates be at right angles to the axis of x; then we may suppose that between the plates the electric intensity X is everywhere parallel to the axis of x. The velocities of both the positive and negative ions are assumed to be proportional to X. Let k1X, k2X represent these velocities respectively; let n1, n2be respectively the number of positive and negative ions per unit volume at a point fixed by the co-ordinate x; let q be the number of positive or negative ions produced in unit time per unit volume at this point; and let the number of ions which recombine in unit volume in unit time be αn1n2; then if e is the charge on the ion, the volume density of the electrification is (n1− n2)e, hence
If I is the current through unit area of the gas and if we neglect any diffusion except that caused by the electric field,
n1ek1X + n2ek2X = I (2).
From equations (1) and (2) we have
and from these equations we can, if we know the distribution of electric intensity between the plates, calculate the number of positive and negative ions.
In a steady state the number of positive and negative ions in unit volume at a given place remains constant, hence neglecting the loss by diffusion, we have
If k1and k2are constant, we have from (1), (5) and (6)
an equation which is very useful,becauseit enables us, if we know the distribution of X², to find whether at any point in the gas the ionization is greater or less than the recombination of the ions. We see that q − αn1n2, which is the excess of ionization over recombination, is proportional to d²X²/dx². Thus when the ionization exceeds the recombination,i.e.when q − αn1n2is positive, the curve for X² is convex to the axis of x, while when the recombination exceeds the ionization the curve for X² will be concave to the axis of x. Thus, for example, fig. 11 represents the curve for X² observed by Graham (Wied. Ann.64, p. 49) in a tube through which a steady current is passing. Interpreting it by equation (7), we infer that ionization was much in excess of recombination at A and B, slightly so along C, while along D the recombination exceeded the ionization. Substituting in equation (7) the values of n1, n2given in (3), (4), we get
This equation can be solved (see Thomson,Phil. Mag.xlvii. P. 253), when q is constant and k1= k2. From the solution it appears that if X1be the value of x close to one of the plates, and X0the value midway between them,
where β = 8πek1/α.
Since e = 4×10-10, α = 2×10-6, and k1for air at atmospheric pressure = 450, β is about 2.3 for air at atmospheric pressure and it becomes much greater at lower pressures.
Thus X1/X0is always greater than unity, and the value of the ratio increases from unity to infinity as β increases from zero to infinity. As β does not involve either q or I, the ratio of X1to X0is independent of the strength of the current and of the intensity of the ionization.
No general solution of equation (8) has been found when k1is not equal to k2, but we can get an approximation to the solution when q is constant. The equations (1), (2), (3), (4) are satisfied by the values—
n1= n2= (q/α)1/2
These solutions cannot, however, hold right up to the surface of the plates, for across each unit of area, at a point P, k1I/(k1+ k2)e positive ions pass in unit time, and these must all come from the region between P and the positive plate. If λ is the distance of P from this plate, this region cannot furnish more than qλ positive ions, and only this number if there are no recombinations. Hence the solution cannot hold when qλ is less than k1I/(k1+ k2)e, or where λ is less than k1I/(k1+ k2)qe.
Similarly the solution cannot hold nearer to the negative plate than the distance k2I/(k1+ k2)qe.
The force in these layers will be greater than that in the middle of the gas, and so the loss of ions by recombination will be smaller in comparison with the loss due to the removal of the ions by the current. If we assume that in these layers the loss of ions by recombination can be neglected, we can by the method of the next article find an expression for the value of the electric force at any point in the layer. This, in conjunction with the value X0= (α/q)1/2· I/e(k1+ k2) for the gas outside the layer, will give the value of X at any point between the plates. It follows from this investigation that if X1and X2are the values of X at the positive and negative plates respectively, and X0the value of X outside the layer,
where ε = α/4πe(k1+ k2). Langevin found that for air at a pressure of 152 mm. ε = 0.01, at 375 mm. ε = 0.06, and at 760 mm. ε = 0.27. Thus at fairly low pressures 1/ε is large, and we have approximately
Therefore
X1/X2= k1/k2,
or the force at the positive plate is to that at the negative plate as the velocity of the positive ion is to that of the negative ion. Thus the force at the negative plate is greater than that at the positive. The falls of potential V1, V2at the two layers when 1/ε is large can be shown to be given by the equations
hence
V1/V2= k1²/k2²,
so that the potential falls at the electrodes are proportional to the squares of the velocities of the ions. The change in potential across the layers is proportional to the square of the current, while the potential change between the layers is proportional to the current, the total potential difference between the plates is the sum of these changes, hence the relation between V and i will be of the form
V = Ai + Bi².
Mie (Ann. der. Phys., 1904, 13, P. 857) has by the method of successive approximations obtained solutions of equation (8) (i.) when the current is only a small fraction of the saturation current, (ii.) when the current is nearly saturated. The results of his investigations are represented in fig. 12, which represents the distribution of electric force along the path of the current for various values of the current expressed as fractions of the saturation current. It will be seen that until the current amounts to about one-fifth of the maximum current, the type of solution is the one just indicated,i.e.the electric force is constant except in the neighbourhood of the electrodes when it increases rapidly.
Though we are unable to obtain a general solution of the equation (8), there are some very important special cases in which that equation can be solved without difficulty. We shall consider two of these, the first being that when the current is saturated. In this case there is no loss of ions by recombination, so that using the same notation as before we have
The solutions of which if q is constant are
if l is the distance between the plates, and x = 0 at the positive electrode. Since
dX/dx = 4π(n1− n2)e,
we get
or
where C is a quantity to be determined by the condition that∫l0Xdx = V, where V is the given potential difference between the plates. When the force is a minimum dX/dx = 0, hence at this point
Hence the ratio of the distances of this point from the positive and negative plates respectively is equal to the ratio of the velocities of the positive and negative ions.
The other case we shall consider is the very important one in which the velocity of the negative ion is exceedingly large compared with the positive; this is the case in flames where, as Gold (Proc. Roy. Soc.97, p. 43) has shown, the velocity of the negative ion is many thousand times the velocity of the positive; it is also very probably the case in all gases when the pressure is low. We may get the solution of this case either by putting k1/k2= 0 in equation (8), or independently as follows:—Using the same notation as before, we have
i = n1k1Xe + n2k2Xe,
In this case practically all the current is carried by the negative ions so that i = n2k2Xe, and therefore q = αn1n2.
Thus
n2= i/k2Xe, n1= qk2Xe/αi.
Thus
or
The solution of this equation is
Here x is measured from the positive electrode; it is more convenient in this case, however, to measure it from the negative electrode. If x be the distance from the negative electrode at which the electric force is X, we have from equation (7)
To find the value of C¹ we see by equation (7) that
hence
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 areaof the cathode, as hot metal cathodes emit large quantities of negative electricity i0may in some cases be considerable, thus the right hand side of equation is (i − i0)/e. When x1is large dX²/dx = 0; hence we have from equation
and since k1is small compared with k2, we have
From the values which have been found for k2and 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)1/2to k11/2. As k1is many thousand times k2the 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 i0is not nearly equal to i is approximately given by the equation
and the fall of potential at the cathode is equal approximately to∫∞0Xdx, that is to
The potential difference between the plates is the sum of the fall of potential in the uniform part of the discharge plus the fall at the cathode, hence
The fall of potential at the cathode is proportional to the square of the current, while the fall in the rest of the circuit is directly proportional to the current. In the case of flames or hot gases, the fall of potential at the cathode is much greater than that in the rest of the circuit, so that in such cases the current through the gas varies nearly as the square root of the potential difference. The equation we have just obtained is of the form
V = Ai + Bi²,
and H. A. Wilson has shown that a relation of this form represents the results of his experiments on the conduction of electricity through flames.
The expression for the fall of potential at the cathode is inversely proportional to q3/2, q being the number of ions produced per cubic centimetre per second close to the cathode; thus any increase in the ionization at the cathode will diminish the potential fall at the cathode, and as practically the whole potential difference between the electrodes occurs at the cathode, a diminution in the potential fall there will be much more important than a diminution in the electric force in the uniform part of the discharge, when the force is comparatively insignificant. This consideration explains a very striking phenomenon discovered many years ago by Hittorf, who found that if he put a wire carrying a bead of a volatile salt into the flame, it produced little effect upon the current, unless it were placed close to the cathode where it gave rise to an enormous increase in the current, sometimes increasing the current more than a hundredfold. The introduction of the salt increases very largely the number of ions produced, so that q is much greater for a salted flame than for a plain one. Thus Hittorf’s result coincides with the conclusions we have drawn from the theory of this class of conduction.
The fall of potential at the cathode is proportional to i − i0, where i0is the stream of negative electricity which comes from the cathode itself, thus as i0increases the fall of potential at the cathode diminishes and the current sent by a given potential difference through the gas increases. Now all metals give out negative particles when heated, at a rate which increases very rapidly with the temperature, but at the same temperature some metals give out more than others. If the cathode is made of a metal which emits large quantities of negative particles, (i − i0) will for a given value of i be smaller than if the metal only emitted a small number of particles; thus the cathode fall will be smaller for the metal with the greater emissitivity, and the relation between the potential difference and the current will be different in the two cases. These considerations are confirmed by experience, for it has been found that the current between electrodes immersed in a flame depends to a great extent upon the metal of which the electrodes are made. Thus Pettinelli (Acc. dei Lincei[5], v. p. 118) found that,ceteris paribus, the current between two carbon electrodes was about 500 times that between two iron ones. If one electrode was carbon and the other iron, the current when the carbon was cathode and the iron anode was more than 100 times the current when the electrodes were reversed. The emission of negative particles by some metallic oxides, notably those of calcium and barium, has been shown by Wehnelt (Ann. der Phys.11, p. 425) to be far greater than that of any known metal, and the increase of current produced by coating the cathodes with these oxides is exceedingly large; in some cases investigated by Tufts and Stark (Physik. Zeits., 1908, 5, p. 248) the current was increased many thousand times by coating the cathode with lime. No appreciable effect is produced by putting lime on the anode.
Conduction when all the Ions are of one Sign.—There are many important cases in which the ions producing the current come from one electrode or from a thin layer of gas close to the electrode, no ionization occurring in the body of the gas or at the other electrode. Among such cases may be mentioned those where one of the electrodes is raised to incandescence while the other is cold, or when the negative electrode is exposed to ultra-violet light. In such cases if the electrode at which the ionization occurs is the positive electrode, all the ions will be positively charged, while if it is the negative electrode the ions will all be charged negatively. The theory of this case is exceedingly simple. Suppose the electrodes are parallel planes at right angles to the axis of x; let X be the electric force at a distance x from the electrode where the ionization occurs, n the number of ions (all of which are of one sign) at this place per cubic centimetre, k the velocity of the ion under unit electric force, e the charge on an ion, and i the current per unit area of the electrode. Then we have dX/dx = 4πne, and if u is the velocity of the ion neu = i. But u = kX, hence we have kX/4π · dX/dx = i, and since the right hand side of this equation does not depend upon x, we get kX²/8π = ix + C, where C is a constant to be determined. If l is the distance between the plates, and V the potential difference between them,
We shall show that when the current is far below the saturation value, C is very small compared with il, so that the preceding equation becomes
V² = 8πl³ i/k (1).
To show that for small currents C is small compared with il, consider the case when the ionization is confined to a thin layer, thickness d close to the electrode, in that layer let n0be the value of n, then we have q = αn0² + i/ed. If X0be the value of X when x = 0, kX0n0e = i, and,
Since α/8πke is, as we have seen, less than unity, C will be small compared with il, if i/(eq + i/d) is small compared with l. If I0is the saturation current, q = I0/ed, so that the former expression = id/(I0+ i), if i is small compared with I0, this expression is small compared with d, and thereforea fortioricompared with l, so that we are justified in this case in using equation (1).
From equation (2) we see that the current increases as the square of the potential difference. Here an increase in the potential difference produces a much greater percentage increase than in conduction through metals, where the current is proportional to the potential difference. When the ionization is distributed through the gas, we have seen that the current is approximately proportional to the square root of the potential, and so increases more slowly with the potential difference than currents through metals. From equation (1) the current is inversely proportional to the cube of the distance between the electrodes, so that it falls off with great rapidity as this distance is increased. We may note that for a given potential difference the expression for the current does not involve q, the rate of production of the ions at the electrode, in other words, if we vary the ionization the current will not begin to be affected by the strength of the ionization until this falls so low that the current is a considerable fraction of the saturation current. For the same potential difference the current is proportional to k, the velocity under unit electric force of the ion which carries the current. As the velocity of the negative ion is greater than that of the positive, the current when the ionization is confined to the neighbourhood of one of the electrodes will be greater when that electrode is made cathode than when it is anode. Thus the current will appear to pass more easily in one direction than in the opposite.
Since the ions which carry the current have to travel all the way from one electrode to the other, any obstacle which is impervious to these ions will, if placed between the electrodes, stop the currentto the electrode where there is no ionization. A plate of metal will be as effectual as one made of a non-conductor, and thus we get the remarkable result that by interposing a plate of an excellent conductor like copper or silver between the electrode, we can entirely stop the current. This experiment can easily be tried by using a hot plate as the electrode at which the ionization takes place: then if the other electrode is cold the current which passes when the hot plate is cathode can be entirely stopped by interposing a cold metal plate between the electrodes.