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 *q*A*d*. Now if *i* is the current per unit area of
the plate, *e* the charge on an ion, *i*A/*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 α*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 α*n*^{2}·A*d*. 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

*q*A

*d*=

*i*A/

*e*+ α

*n*

^{2}·A

*d*,

*q*=

*i*/

*ed*+ α

*n*

^{2}.

If *u*_{1} and *u*_{2} are the velocities with which the positive and negative
ions move, *nu*_{1}*e* and *nu*_{2}*e* 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*=

*nu*

_{1}

*e*+

*nu*

_{2}

*e*.

If X is the electric force acting on the gas, *k*_{1} and *k*_{2} the velocities
of the positive and negative ions under unit force, u_{1} = *k*_{1}X,
u_{2} = *k*_{2}X; hence

*n*=

*i*/(

*k*

_{1}+

*k*

_{2})X

*e*,

and we have

q = | i | + | αi^{2} | . |

ed | (k_{1} + k_{2})^{2} e^{2}X^{2} |

But *qed* is the saturation current per unit area of the plate; calling
this I, we have

I − i = | dαi^{2} |

e(k_{1} + k_{2})^{2}X^{2} |

or

X^{2} = | i^{2}·dα | . |

e(I − i) (k_{1} + k_{2})^{2} |

Hence if we determine corresponding values of X and *i* we can
deduce the value of α/*e* if we also know (*k*_{1} + *k*_{2}). 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 N_{0} ions if the flash of
Röntgen rays produced N_{0} positive and N_{0} 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
Röntgen rays,

*n*=

*n*

_{0}/(1 + α

*n*

_{0}

*t*).

With the same notation as before, the breadth of either of the outer
layers will in time *dt* increase by X(*k*_{1} + *k*_{2})*dt*, and the number of
ions in it by X(*k*_{1} + *k*_{2})*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(*k*_{1} + *k*_{2}), 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

N = ∫0l/X(k_{1}+k_{2}) |
n_{0}X(k_{1} + k_{2}) |
dt = | X(k_{1} + k_{2}) |
log (1 + | n_{0}αl |
). |

1 + n_{0}αt |
α | X(k_{1} + k_{2}) |

If Q is the charge received by the plate,

Q = Ne = | X | log(1 + | Q_{0}ε | ), |

4πε | 4πX |

where Q_{0} = *n*_{0}le is the charge received by the plate when the electric
force is large enough to prevent recombination, and ε = α4π*e*(R_{1} + R_{2}).
We can from this result deduce the value of ε and hence the value
of α when R_{1} + R_{2} 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
*k*_{1}X, *k*_{2}X represent these velocities respectively; let *n*_{1}, *n*_{2} 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 α*n*_{1}*n*_{2}; then if *e* is the charge on the ion, the volume
density of the electrification is (*n*_{1} − *n*_{2})e, hence

dX | = 4π(n_{1} − n_{2})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,

*n*

_{1}

*ek*

_{1}X +

*n*

_{2}

*ek*

_{2}X = I (2).

From equations (1) and (2) we have

n_{1}e = | 1 | ( | I | + | k_{2} | dX | ) (3), | |

k_{1} + k_{2} | X | 4π | dx |

n_{1}e = | 1 | ( | I | − | k_{2} | dX | ) (4), | |

k_{1} + k_{2} | X | 4π | 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 | (k_{1}n_{1}X) = q − αn_{1}n_{2} (5). |

dx |

− | d | (k_{2}n_{2}X) = q − αn_{1}n_{2} (6). |

dx |

If *k*_{1} and *k*_{2} are constant, we have from (1), (5) and (6)

d²X² | = 8πe(q − αn_{1}n_{2})( |
1 | + | 1 | ) (7). |

dx² | k_{1} | k_{2} |

an equation which is very useful, because it 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* − α*n*_{1}*n*_{2}, which is the excess of ionization over recombination,
is proportional to *d*²X²/*dx*². Thus when the ionization
exceeds the recombination, *i.e.* when *q* − α*n*_{1}*n*_{2} is 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*.

Fig. 11. |

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 *n*_{1}, *n*_{2} given in (3), (4),
we get

d²X² | 8πe[q − |
α | (I + | k_{2} | dX² | )(I − | k_{2} | dX² | )]( | 1 | + | 1 | ) (8). | ||

dx² | e²X²(k_{1} + k_{2})² |
8π | dx | 8π | dx |
k_{1} | k_{2} |

This equation can be solved (see Thomson, *Phil. Mag.* xlvii.
P. 253), when *q* is constant and *k*_{1} = *k*_{2}. From the solution it appears
that if X_{1} be the value of *x* close to one of the plates, and X_{0} the
value midway between them,

X_{1}/X_{0} = | 1 |

β^{2} − 2/β |

where β = 8π*ek*_{1}/α.