_Ions from Point Discharge._
+---------------+------+-------+------+----------+
| Hydrogen | 5.4 | 7.43 | 6.41 | Chattock |
| Carbonic acid | 0.83 | 0.925 | 0.88 | Chattock |
| Air | 1.32 | 1.80 | 1.55 | Chattock |
| Oxygen | 1.30 | 1.85 | 1.57 | Chattock |
+---------------+------+-------+------+----------+
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:--
+----------------------+----------------------+
| Negative Ions. | Positive Ions. |
+-------+------+-------+-------+------+-------+
| p. | V-. |pV-/76.| p. | V+. |pV+/76.|
+-------+------+-------+-------+------+-------+
| 7.5 | 6560 | 647 | 7.5 | 4430 | 437 |
| 20.0 | 2204 | 580 | 20.0 | 1634 | 430 |
| 41.5 | 994 | 530 | 41.5 | 782 | 427 |
| 76.0 | 510 | 510 | 76.0 | 480 | 420 |
| 142.0 | 270 | 505 | 142.0 | 225 | 425 |
+-------+------+-------+-------+------+-------+
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 deg.
C.; R1 and R2 are the velocities of the + and - ions respectively when
the
force is a volt per centimetre.
+-------+-------+---------------------+
| R1. | R2. |Temperature Absolute.|
+-------+-------+---------------------+
| 2.00 | 2.495 | 411 deg. |
| 1.95 | 2.40 | 399 deg. |
| 1.85 | 2.30 | 383 deg. |
| 1.81 | 2.21 | 373 deg. |
| 1.67 | 2.125 | 348 deg. |
| 1.60 | 2.00 | 333 deg. |
| 1.39 | 1.785 | 285 deg. |
| 0.945 | 1.23 | 209 deg. |
| 0.235 | 0.235 | 94 deg. |
+-------+-------+---------------------+
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 deg. 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
deg. 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 deg. C. was .23
cm./sec. for a volt per centimetre; at a distance of 10 cm. from the
flame when the temperature was 160 deg. C. the velocity was .21
cm./sec; while at a distance of 14.5 cm. from the flame when the
temperature was 105 deg. 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
Xe
u = ------- D,
(p1/N1)
where D represents the coefficient of inter-diffusion of A into B, and
N1 the 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/N1 is independent of the pressure, it is equal to
[Pi]/N, where [Pi] is the atmospheric pressure and N the number of
molecules in a cubic centimetre of gas at atmospheric pressure. Now Ne
= 1.22 X 10^10, if e is measured in electrostatic units; [Pi] = 10^6
and 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 X 10^4X.
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 show that 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 [alpha]n^2 per 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 [alpha]n^2.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 + [alpha]n^2.Ad,
q = i/ed + [alpha]^n2.
If u1 and u2 are the velocities with which the positive and negative
ions move, nu1e and nu2e are respectively the quantities of positive
electricity passing in one direction through unit area of the gas per
second, and of negative in the opposite direction, hence
i = nu1e + nu2e.
If X is the electric force acting on the gas, k1 and k2 the velocities
of the positive and negative ions under unit force, u1 = k1X, u2 =
k2X; hence
n = i/(k1 + k2)Xe,
and we have
i [alpha]i^2
q = -- + -----------------.
ed (k1 + k2)^2e^2X^2
But qed is the saturation current per unit area of the plate; calling
this I, we have
d[alpha]i^2
I - i = ---------------
e(k1 + k2)^2X^2
or
i^2.d[alpha]
X^2 = -------------------.
e(I - i)(k1 + k2)^2
Hence if we determine corresponding values of X and i we can deduce
the value of [alpha]/e if we also know (k1 + k2). The value of I is
easily determined, as it is the current when X is very large. The
preceding result only applies when i is small compared with I, as it
is only in this case that the values of n and X are uniform throughout
the volume of the gas. Another method which answers the same purpose
is due to Langevin (_Ann. Chim. Phys._, 1903, 28, p. 289); it is as
follows. Let A and B be two parallel planes immersed in a gas, and let
a slab of the gas bounded by the planes a, b parallel to A and B be
ionized by an instantaneous flash of Rontgen rays. If A and B are at
different electric potentials, then all the positive ions produced by
the rays will be attracted by the negative plate and all the negative
ions by the positive, if the electric field were exceedingly large
they would reach these plates before they had time to recombine, so
that each plate would receive N0 ions if the flash of Rontgen rays
produced N0 positive and N0 negative ions. With weaker fields the
number of ions received by the plates will be less as some of them
will recombine before they can reach the plates. We can find the
number of ions which reach the plates in this case in the following
way:--In consequence of the movement of the ions the slab of ionized
gas will broaden out and will consist of three portions, one in which
there are nothing but positive ions,--this is on the side of the
negative plate,--another on the side of the positive plate in which
there are nothing but negative ions, and a portion between these in
which there are both positive and negative ions; it is in this layer
that recombination takes place, and here if n is the number of
positive or negative ions at the time t after the flash of Rontgen
rays,
n = n0/(1 + [alpha]n0t).
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
_
/ l/X(k1+k2) n0X(k1 + k2) X(k1 + k2) / n0[alpha]l \
N = | --------------dt = ---------- log( 1 + ---------- ).
_/ 0 1 + n0[alpha]t [alpha] \ X(k1 + k2) /
If Q is the charge received by the plate,
X / Q0[epsilon]\
Q = Ne = -------------- log ( 1 + ----------- ),
4[pi][epsilon] \ 4[pi]X /
where Q0 = n0le is the charge received by the plate when the electric
force is large enough to prevent recombination, and [epsilon] =
[alpha]4[pi]e(R1 + R2). We can from this result deduce the value of
[epsilon] and hence the value of [alpha] when R1+R2 is 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, n2 be
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 [alpha]n1n2; then if e is the charge on the ion, the volume density
of the electrification is (n1 - n2)e, hence
dX
-- = 4[pi](n1 - n2)e (1).
dx
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).