Chapter III: Electric Conduction Through Gases (2)

+----------------------+----------------------+
| 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).

From equations (1) and (2) we have

1 / I k2 dX \
n1e = ------- ( - + ----- -- ) (3),
k1 + k2 \ X 4[pi] dx /

1 / I k1 dX \
n2e = ------- ( - - ----- -- ) (4),
k1 + k2 \ X 4[pi] dx /

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

d
--(k1n1X) = q - [alpha]n1n2 (5).
dx

d
- --(k2n2X) = q - [alpha]n1n2 (6).
dx

If k1 and K2 are constant, we have from (1), (5) and (6)

d^2X^2 / 1 1 \
------ = 8[pi]e(q - [alpha]n1n2)( --- + --- ) (7),
dx^2 \ k1 k2 /

an equation which is very useful, because it enables us, if we know
the distribution of X^2, 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 - [alpha]n1n2, which is the excess of ionization over
recombination, is proportional to d^2X^2/dx^2. Thus when the ionization
exceeds the recombination, i.e. when q - [alpha]n1n2 is positive, the
curve for X^2 is convex to the axis of x, while when the recombination
exceeds the ionization the curve for X^2 will be concave to the axis of
x. Thus, for example, fig. 11 represents the curve for X^2 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, n2 given in (3), (4),
we get
_ _
d^2X^2 | [alpha] / k^2 dX^2\ / k2 dX^2\ | / 1 1 \
------ = 8[pi]e |q - ----------------- (1 + ----- ---- ) (1 - ----- ---- )| ( --- + --- ) (8).
dx^2 |_ e^2X^2(k1 + k2)^2 \ 8[pi] dx / \ 8[pi] dx /_| \k1 k2 /

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
X1 be the value of x close to one of the plates, and X0 the value
midway between them,

1
X1/X0 = -------------------
[beta]^2 - 2/[beta]

where [beta] = 8[pi]ek1/[alpha].

Since e = 4 X 10^-10, [alpha] = 2X10^-6, and k1 for air at atmospheric
pressure = 450, [beta] is about 2.3 for air at atmospheric pressure
and it becomes much greater at lower pressures.

Thus X1/X0 is always greater than unity, and the value of the ratio
increases from unity to infinity as [beta] increases from zero to
infinity. As [beta] does not involve either q or I, the ratio of X1 to
X0 is independent of the strength of the current and of the intensity
of the ionization.

No general solution of equation (8) has been found when k1 is 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 / [alpha])^1/2

k1
k1n1Xe = ------- I,
k1 + k2

k2
k2n2Xe = ------- I,
k1 + k2

/[alpha]\^1/2 I
X = ( ------- ) ----------.
\ q / e(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 [lambda] is the distance
of P from this plate, this region cannot furnish more than q[lambda]
positive ions, and only this number if there are no recombinations.
Hence the solution cannot hold when q[lambda] is less than k1I/(k1 +
k2)e, or where [lambda] 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

/[alpha]\^1/2 I
X0 = ( ------- ) ----------
\ q / 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 X1 and
X2 are the values of X at the positive and negative plates
respectively, and X0 the value of X outside the layer,

/ k1 I \^1/2 / k2 I \^1/2
X1 = X0 ( I + -- --------- ) , X2 = X0 ( I + -- --------- ) ,
\ k2 [epsilon]/ \ k1 [epsilon]/

where [epsilon] = [alpha]/4[pi]e(k1 + k2). Langevin found that for air
at a pressure of 152 mm. [epsilon] = 0.01, at 375 mm. [epsilon] =
0.06, and at 760 mm. [epsilon] = 0.27. Thus at fairly low pressures
1/[epsilon] is large, and we have approximately

/k1\^1/2 I /k2\^1/2 I
X1 = X0 ( -- ) ---------------, X2 = X0 ( -- ) ---------------.
\k2/ [root][epsilon] \k1/ [root][epsilon]

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, V2 at the two layers when 1/[epsilon] is
large can be shown to be given by the equations

/[epsilon]\^3/2 /k1\^1/2
V1 = 8[pi]^2( --------- ) k1 ( -- ) i^2,
\q [alpha]/ \k2/

/[epsilon]\^3/2 /k2\^1/2
V2 = 8[pi]^2( --------- ) k2 ( -- ) i^2,
\q [alpha]/ \k1/

hence V1/V2 = k1^2/k2^2,

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 + Bi^2.

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

d
--(n1k1X) = q,
dx

d
--(n2k2X) = -q.
dx

The solutions of which if q is constant are

n1k1X = qx,

n2k2X = I/e - qx = q(l - x),

if l is the distance between the plates, and x = 0 at the positive
electrode. Since

dX/dx = 4[pi](n1 - n2)e,

we get

1 dX^2 / 1 1 \ l
----- ---- = qx ( -- + -- ) - q --,
8[pi] d^2x \k1 k2 / k2

or

X^2 x^2 / 1 1 \ lx
----- = q --- ( -- + -- ) - q -- + C,
8[pi] 2 \k1 k2 / k2

where C is a quantity to be determined by the condition that

_
/ l
| Xdx = V,
_/0

where V is the given potential difference between the plates. When the
force is a minimum dX/dx = 0, hence at this point

lk1 lk2
x = -------, l - x = -------.
k1 + k2 k1 + k2

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,

d
--(n2k2X) = q - [alpha]n1n2,
dx

dX
-- = 4[pi](n1 - n2)e.
dx

In this case practically all the current is carried by the negative
ions so that i = n2k2Xe, and therefore q = [alpha]n1n2.

Thus

n2 = i/k2Xe, n1 = qk2Xe/[alpha]i.

Thus

dX 4[pi]e^2k2qX 4[pi]i
-- = ------------ - ------,
dx [alpha]i k2X

or

dX^2 8[pi]e^2k2qX^2 8[pi]i
---- - -------------- = - ------.
dx [alpha]i k2

The solution of this equation is

[alpha] i^2
X^2 = ------- ------- + C[epsilon]^(8[pi]e^2k2qx/[alpha]i)
q k2^2e^2

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)

[alpha] i^2
X^2 = ------- ------- + C^1[epsilon]^(8[pi]e^2k2qx/[alpha]i)
q k2^2e^2

To find the value of C^1 we see by equation (7) that

d^2X^2 k1k2 1
------ ------- ------ = q - [alpha]n1n2;
dX^2 k1 + k2 8[pi]e

hence
_ _ _
| dX^2 k1k2 1 |^x1 / x1
| ---- ------- ------ | = | (q - [alpha]n1n2)dx.
|_ dX k1 + k2 8[pi]e_| _/0

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 area of the
cathode, as hot metal cathodes emit large quantities of negative
electricity i0 may in some cases be considerable, thus the right hand
side of equation is (i - i0)/e. When x1 is large dX^2/dx = 0; hence we
have from equation

[alpha]i(i - i0) k1 + k2
C^1 = ---------------- -------,
qk1k2e^2 k2

and since k1 is small compared with k2, we have

[alpha]i^2 / k2 i - i0 \
X^2 = ---------- (1 + -- ------ [epsilon]^{-8[pi]e^2k2.qx/[alpha].i}).
qk2^2e^2 \ k1 i /

From the values which have been found for k2 and [alpha], we know that
8[pi]ek2/[alpha] 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.[root]([alpha]/q), while
when i0 = 0, the force at the cathode itself bears to the uniform
force the ratio of (k1 + k2)^1/2 to k1^1/2. As k1 is many thousand
times k2 the 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.[root]([alpha]/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.[root]([alpha]l/q), where l is the
distance between the electrodes. Close to the cathode, the electric
force when i0 is not nearly equal to i is approximately given by the
equation

i /[alpha]\^1/2
X = --------- (---------) [epsilon]^{-4[pi]e^2k2qx/[alpha]i},
e(k1k2)^1/2 \ q / ,

and the fall of potential at the cathode is equal approximately to

_[oo]
/
| X dx,
_/0

that is to

i /[alpha]\^1/2 [alpha]i
--------- (---------) -----------.
e(k1k2)^1/2 \ q / 4[pi]e^2k2q

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

/[alpha]\^1/2 i / i[alpha]^2 1 \
V = (---------) --- ( il + ---------- ------------ ).
\ q / ek2 \ 4[pi]e^2q [root](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 form

V = Ai + Bi^2,

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 q^(3/2), 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
i0 is the stream of negative electricity which comes from the cathode
itself, thus as i0 increases 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 = 4[pi]ne, and if u is the velocity of the ion neu
= i. But u = kX, hence we have kX/4[pi] . dX/dx = i, and since the
right hand side of this equation does not depend upon x, we get
kX^2/8[pi] = 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,
_ _____ _ _
/ l 1 /8[pi] | |
V = | Xdx = --- / ----- | ( il + C )^3/2 - C^3/2 |.
_/0 i \/ k |_ _|

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

V^2 = 8[pi]l^3i/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 n0 be the value of n, then
we have q = [alpha]n0^2 + i/ed. If X0 be the value of X when x = 0,
kX0n0e = i, and,

kX0^2 i^2 [alpha] i^2
C = ----- = ------------ = --------- . -------- (2).
8[pi] n0^2ke.8[pi] 8[pi]ke^2 q + i/ed

Since [alpha]/8[pi]ke 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
I0 is 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 therefore _a fortiori_ compared 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 current to 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.

_Methods of counting the Number of Ions._--The detection of the ions and the estimation of their number in a given volume is much facilitated by the property they possess of promoting the condensation of water-drops in dust-free air supersaturated with water vapour. If such air contains no ions, then it requires about an eightfold supersaturation before any water-drops are formed; if, however, ions are present C. T. R. Wilson (_Phil. Trans._ 189, p. 265) has shown that a sixfold supersaturation is sufficient to cause the water vapour to condense round the ions and to fall down as raindrops. The absence of the drops when no ions are present is due to the curvature of the drop combined with the surface tension causing, as Lord Kelvin showed, the evaporation from a small drop to be exceeding rapid, so that even if a drop of water were formed the evaporation would be so great in its early stages that it would rapidly evaporate and disappear. It has been shown, however (J. J. Thomson, _Application of Dynamics to Physics and Chemistry_, p. 164; _Conduction of Electricity through Gases_, 2nd ed. p. 179), that if a drop of water is charged with electricity the effect of the charge is to diminish the evaporation; if the drop is below a certain size the effect the charge has in promoting condensation more than counterbalances the effect of the surface tension in promoting evaporation. Thus the electric charge protects the drop in the most critical period of its growth. The effect is easily shown experimentally by taking a bulb connected with a piston arranged so as to move with great rapidity. When the piston moves so as to increase the volume of the air contained in the bulb the air is cooled by expansion, and if it was saturated with water vapour before it is supersaturated after the expansion. By altering the throw of the piston the amount of supersaturation can be adjusted within very wide limits. Let it be adjusted so that the expansion produces about a sixfold supersaturation; then if the gas is not exposed to any ionizing agents very few drops (and these probably due to the small amount of ionization which we have seen is always present in gases) are formed. If, however, the bulb is exposed to strong Rontgen rays expansion produces a dense cloud which gradually falls down and disappears. If the gas in the bulb at the time of its exposure to the Rontgen rays is subject to a strong electric field hardly any cloud is formed when the gas is suddenly expanded. The electric field removes the charged ions from the gas as soon as they are formed so that the number of ions present is greatly reduced. This experiment furnishes a very direct proof that the drops of water which form the cloud are only formed round the ions.

This method gives us an exceedingly delicate test for the presence of ions, for there is no difficulty in detecting ten or so raindrops per cubic centimetre; we are thus able to detect the presence of this number of ions. This result illustrates the enormous difference between the delicacy of the methods of detecting ions and those for detecting uncharged molecules; we have seen that we can easily detect ten ions per cubic centimetre, but there is no known method, spectroscopic or chemical, which would enable us to detect a billion (10^12) times this number of uncharged molecules. The formation of the water-drops round the charged ions gives us a means of counting the number of ions present in a cubic centimetre of gas; we cool the gas by sudden expansion until the supersaturation produced by the cooling is sufficient to cause a cloud to be formed round the ions, and the problem of finding the number of ions per cubic centimetre of gas is thus reduced to that of finding the number of drops per cubic centimetre in the cloud. Unless the drops are very few and far between we cannot do this by direct counting; we can, however, arrive at the result in the following way. From the amount of expansion of the gas we can calculate the lowering produced in its temperature and hence the total quantity of water precipitated. The water is precipitated as drops, and if all the drops are the same size the number per cubic centimetre will be equal to the volume of water deposited per cubic centimetre, divided by the volume of one of the drops. Hence we can calculate the number of drops if we know their size, and this can be determined by measuring the velocity with which they fall under gravity through the air.

The theory of the fall of a heavy drop of water through a viscous
fluid shows that v = (2/9)ga^2/[mu], where a is the radius of the drop,
g the acceleration due to gravity, and [mu] the coefficient of
viscosity of the gas through which the drop falls. Hence if we know v
we can deduce the value of a and hence the volume of each drop and the
number of drops.

_Charge on Ion._--By this method we can determine the number of ions
per unit volume of an ionized gas. Knowing this number we can proceed
to determine the charge on an ion. To do this let us apply an electric
force so as to send a current of electricity through the gas, taking
care that the current is only a small fraction of the saturating
current. Then if u is the sum of the velocities of the positive and
negative ions produced in the electric field applied to the gas, the
current through unit area of the gas is neu, where n is the number of
positive or negative ions per cubic centimetre, and e the charge on an
ion. We can easily measure the current through the gas and thus
determine neu; we can determine n by the method just described, and u,
the velocity of the ions under the given electric field, is known from
the experiments of Zeleny and others. Thus since the product neu, and
two of the factors n, u are known, we can determine the other factor
e, the charge on the ion. This method was used by J. J. Thomson, and
details of the method will be found in _Phil. Mag._ [5], 46, p. 528;
[5], 48, p. 547; [6], 5, p. 346. The result of these measurements
shows that the charge on the ion is the same whether the ionization is
by Rontgen rays or by the influence of ultra-violet light on a metal
plate. It is the same whether the gas ionized is hydrogen, air or
carbonic acid, and thus is presumably independent of the nature of the
gas. The value of e formed by this method was 3.4 X 10^-10
electrostatic units.

H. A. Wilson (_Phil. Mag._ [6], 5, p. 429) used another method. Drops
of water, as we have seen, condense more easily on negative than on
positive ions. It is possible, therefore, to adjust the expansion so
that a cloud is formed on the negative but not on the positive ions.
Wilson arranged the experiments so that such a cloud was formed
between two horizontal plates which could be maintained at different
potentials. The charged drops between the plates were acted upon by a
uniform vertical force which affected their rate of fall. Let X be the
vertical electric force, e the charge on the drop, v1 the rate of fall
of the drop when this force acts, and v the rate of fall due to
gravity alone. Then since the rate of fall is proportionate to the
force on the drop, if a is the radius of the drop, and [rho] its
density, then

Xe + (4/3)[pi][rho]ga^3 v1
---------------------- = ---,
4/3[pi][rho]ga^3 v

or Xe = (4/3)[pi][rho]ga^3(v1 - v)/v.

But v = 2/9ga^2[rho]/[mu],

so that

/ / [mu]^3 v^(3/2)(v1 - v)
Xe = \/ 2.9[pi] - / ------ . ---------------.
\/ g[rho] v

Thus if X, v, v1 are known e can be determined. Wilson by this method
found that e was 3.1 X 10^-10 electrostatic units. A few of the ions
carried charges 2e or 3e.

Townsend has used the following method to compare the charge carried
by a gaseous ion with that carried by an atom of hydrogen in the
electrolysis of solution. We have

u/D = Ne/[Pi],

where D is the coefficient of diffusion of the ions through the gas, u
the velocity of the ion in the same gas when acted on by unit electric
force, N the number of molecules in a cubic centimetre of the gas when
the pressure is [Pi] dynes per square centimetre, and e the charge in
electrostatic units. This relation is obtained on the hypothesis that
N ions in a cubic centimetre produce the same pressure as N uncharged
molecules.

We know the value of D from Townsend's experiments and the values of u
from those of Zeleny. We get the following values for Ne X 10^-10:--

+---------------+---------------------+---------------------+
| | Moist Gas. | Moist Gas. |
+---------------+----------+----------+----------+----------+
| Gas. | Positive | Negative | Positive | Negative |
| | Ions. | Ions. | Ions. | Ions. |
+---------------+----------+----------+----------+----------+
| Air | 1.28 | 1.29 | 1.46 | 1.31 |
| Oxygen | 1.34 | 1.27 | 1.63 | 1.36 |
| Carbonic acid | 1.01 | .87 | .99 | .93 |
| Hydrogen | 1.24 | 1.18 | 1.63 | 1.25 |
+---------------+----------+----------+----------+----------+
| Mean | 1.22 | 1.15 | 1.43 | 1.21 |
+---------------+----------+----------+----------+----------+

Since 1.22 cubic centimetres of hydrogen at the temperature 15 deg. C.
and pressure 760 mm. of mercury are liberated by the passage through
acidulated water of one electromagnetic unit of electricity or 3 X
10^10 electrostatic units, and since in one cubic centimetre of the
gas there are 2.46 N atoms of hydrogen, we have, if E is the charge in
electrostatic units, on the atom of hydrogen in the electrolysis of
solutions

2.46NE = 3 X 10^10,

or

NE = 1.22 X 10^10.

The mean of the values of Ne in the preceding table is 1.24 X 10^10.
Hence we may conclude that the charge of electricity carried by a
gaseous ion is equal to the charge carried by the hydrogen atom in the
electrolysis of solutions. The values of Ne for the different gases
differ more than we should have expected from the probable accuracy of
the determination of D and the velocity of the ions: Townsend (_Proc.
Roy. Soc._ 80, p. 207) has shown that when the ionization is produced
by Rontgen rays some of the positive ions carry a double charge and
that this accounts for the values of Ne being greater for the positive
than for the negative ions. Since we know the value of e, viz. 3.5 X
10^-10, and, also Ne, = 1.24 X 10^10, we find N the number of
molecules in a cubic centimetre of gas at standard temperature and
pressure to be equal to 3.5 X 10^19. This method of obtaining N is the
only one which does not involve any assumption as to the shape of the
molecules and the forces acting between them.

Another method of determining the charge carried by an ion has been
employed by Rutherford (_Proc. Roy. Soc._ 81, pp. 141, 162), in which
the positively electrified particles emitted by radium are made use
of. The method consists of: (1) Counting the number of [alpha]
particles emitted by a given quantity of radium in a known time. (2)
Measuring the electric charge emitted by this quantity in the same
time. To count the number of the [alpha] particles the radium was so
arranged that it shot into an ionization chamber a small number of
[alpha] particles per minute; the interval between the emission of
individual particles was several seconds. When an [alpha] particle
passed into the vessel it ionized the gas inside and so greatly
increased its conductivity; thus, if the gas were kept exposed to an
electric field, the current through the gas would suddenly increase
when an [alpha] particle passed into the vessel. Although each [alpha]
particle produces about thirty thousand ions, this is hardly large
enough to produce the conductivity appreciable without the use of very
delicate apparatus; to increase the conductivity Rutherford took
advantage of the fact that ions, especially negative ones, when
exposed to a strong electric field, produce other ions by collision
against the molecules of the gas through which they are moving. By
suitably choosing the electric field and the pressure in the
ionization chamber, the 30,000 ions produced by each [alpha] particle
can be multiplied to such an extent that an appreciable current passes
through the ionization chamber on the arrival of each [alpha]
particle. An electrometer placed in series with this vessel will show
by its deflection when an [alpha] particle enters the chamber, and by
counting the number of deflections per minute we can determine the
number of [alpha] particles given out by the radium in that time.
Another method of counting this number is to let the particles fall on
a phosphorescent screen, and count the number of scintillations on the
screen in a certain time. Rutherford has shown that these two methods
give concordant results.

The charge of positive electricity given out by the radium was
measured by catching the [alpha] particles in a Faraday cylinder
placed in a very highly exhausted vessel, and measuring the charge per
minute received by this cylinder. In this way Rutherford showed that
the charge on the [alpha] particle was 9.4 X 10^-10 electrostatic
units. Now e/m for the [alpha] particle = 5 X 10^3, and there is
evidence that the [alpha] particle is a charged atom of helium; since
the atomic weight of helium is 4 and e/m for hydrogen is 10^4, it
follows that the charge on the helium atom is twice that on the
hydrogen, so that the charge on the hydrogen atom is 4.7 X 10^-10
electrostatic units.