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 /
[Illustration: Fig. 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
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.
[Illustration: FIG. 12.]
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