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