When the ions are not removed from the gas, they will increase until the number of ions of one sign which combine with ions of the opposite sign in any time is equal to the number produced by the ionizing agent in that time. We can easily calculate the number of free ions at any time after the ionizing agent has commenced to act.

Let *q* be the number of ions (positive or negative) produced in
one cubic centimetre of the gas per second by the ionizing agent,
*n*_{1}, *n*_{2}, the number of free positive and negative ions respectively per
cubic centimetre of the gas. The number of collisions between
positive and negative ions per second in one cubic centimetre of the
gas is proportional to *n*_{1}*n*_{2}. If a certain fraction of the collisions
between the positive and negative ions result in the formation of an
electrically neutral system, the number of ions which disappear per
second on a cubic centimetre will be equal to α*n*_{1} *n*_{2}, where α is a
quantity which is independent of *n*_{1}, *n*_{2}; hence if *t* is the time since
the ionizing agent was applied to the gas, we have

*dn*

_{1}/

*dt*=

*q*− α

*n*

_{1}

*n*

_{2},

*dn*

_{2}/

*dt*=

*q*− α

*n*

_{1}

*n*

_{2}.

Thus *n*_{1} − *n*_{2} is constant, so if the gas is uncharged to begin with, *n*_{1}
will always equal *n*_{2}. Putting *n*_{1} = *n*_{2} = n we have

*dn*/

*dt*=

*q*− α

*n*

^{2}(1),

the solution of which is, since *n* = 0 when *t* = 0,

n = | k(ε^{2kαt} − 1) |
(2), |

ε^{2kαt} + 1 |

if *k*^{2} = *q*/α. Now the number of ions when the gas has reached a
steady state is got by putting *t* equal to infinity in the preceding
equation, and is therefore given by the equation

*n*

_{0}=

*k*= √ (

*q*/α).

We see from equation (1) that the gas will not approximate to its
steady state until 2kα*t* is large, that is until *t* is large compared with
½kα or with ½√ (*q*α). We may thus take ½√ (*q*α) as a measure of
the time taken by the gas to reach a steady state when exposed to
an ionizing agent; as this time varies inversely as √*q* we see that
when the ionization is feeble it may take a very considerable time for
the gas to reach a steady state. Thus in the case of our atmosphere
where the production of ions is only at the rate of about 30 per cubic
centimetre per second, and where, as we shall see, α is about 10^{−6},
it would take some minutes for the ionization in the air to get into
a steady state if the ionizing agent were suddenly applied.

We may use equation (1) to determine the rate at which the ions
disappear when the ionizing agent is removed. Putting *q*=0 in
that equation we get *dn*/α*t* = −α*n*^{2}.

Hence

*n*=

*n*

_{0}/(1 +

*n*

_{0}α

*t*) (3),

where *n*_{0} is the number of ions when *t* = 0. Thus the number of ions
falls to one-half its initial value in the time 1/*n*_{0}α. The quantity α is
called the *coefficient of recombination*, and its value for different gases
has been determined by Rutherford (*Phil. Mag.* 1897 [5], 44, p. 422),
Townsend (*Phil. Trans.*, 1900, 193, p. 129), McClung (*Phil. Mag.*,
1902 [6], 3, p. 283), Langevin (*Ann. chim. phys.* [7], 28, p. 289),
Retschinsky (*Ann. d. Phys.*, 1905, 17, p. 518), Hendred (*Phys. Rev.*,
1905, 21, p. 314). The values of α/*e*, *e* being the charge on an ion in
electrostatic measure as determined by these observers for different
gases, is given in the following table:—

Townsend. | McClung. | Langevin. | Retschinsky. | Hendred. | |

Air | 3420 | 3380 | 3200 | 4140 | 3500 |

O_{2} | 3380 | ||||

CO_{2} | 3500 | 3490 | 3400 | ||

H_{2} | 3020 | 2940 |

The gases in these experiments were carefully dried and free from
dust; the apparent value of α is much increased when dust or small
drops of water are present in the gas, for then the ions get caught
by the dust particles, the mass of a particle is so great compared
with that of an ion that they are practically immovable under the
action of the electric field, and so the ions clinging to them escape
detection when electrical methods are used. Taking *e* as 3.5×10^{−10},
we see that α is about 1.2×10^{−6}, so that the number of recombinations
in unit time between *n* positive and *n* negative ions in unit
volume is 1.2×10^{−6}*n*^{2}. The kinetic theory of gases shows that
if we have *n* molecules of air per cubic centimetre, the number of
collisions per second is 1.2×10^{−10}*n*^{2} at a temperature of 0° C. Thus
we see that the number of recombinations between oppositely
charged ions is enormously greater than the number of collisions
between the same number of neutral molecules. We shall see that
the difference in size between the ion and the molecule is not nearly
sufficient to account for the difference between the collisions in the
two cases; the difference is due to the force between the oppositely
charged ions, which drags ions into collisions which but for this force
would have missed each other.

Several methods have been used to measure α. In one method
air, exposed to some ionizing agent at one end of a long tube, is
slowly sucked through the tube and the saturation current measured
at different points along the tube. These currents are proportional
to the values of *n* at the place of observation: if we know the
distance of this place from the end of the tube when the gas was
ionized and the velocity of the stream of gas, we can find *t* in equation
(3), and knowing the value of *n* we can deduce the value of α from
the equation

*n*

_{1}− 1/

*n*

_{2}= α(

*t*

_{1}−

*t*

_{2}),

where *n*_{1}, *n*_{2} are the values of *n* at the times *t*_{1}, *t*_{2} respectively. In this
method the tubes ought to be so wide that the loss of ions by diffusion
to the sides of the tube is negligible. There are other methods which
involve the knowledge of the speed with which the ions move under
the action of known electric forces; we shall defer the consideration
of these methods until we have discussed the question of these
speeds.

In measuring the value of α it should be remembered that the theory of the methods supposes that the ionization is uniform throughout the gas. If the total ionization throughout a gas remains constant, but instead of being uniformly distributed is concentrated in patches, it is evident that the ions will recombine more quickly in the second case than in the first, and that the value of α will be different in the two cases. This probably explains the large values of α obtained by Retschinsky, who ionized the gas by the α rays from radium, a method which produces very patchy ionization.

*Variation of* α *with the Pressure of the Gas.*—All observers agree
that there is little variation in α with the pressures for pressures of
between 5 and 1 atmospheres; at lower pressures, however, the
value of α seems to diminish with the pressure: thus Langevin
(*Ann. chim. phys.*, 1903, 28, p. 287) found that at a pressure of 15
of an atmosphere the value of α was about 15 of its value at
atmospheric pressure.

*Variation of* α *with the Temperature.*—Erikson (*Phil. Mag.*, Aug.
1909) has shown that the value of α for air increases as the temperature
diminishes, and that at the temperature of liquid air −180° C.,
it is more than twice as great as at +12° C.

Since, as we have seen, the recombination is due to the coming
together of the positive and negative ions under the influence of the
electrical attraction between them, it follows that a large electric
force sufficient to overcome this attraction would keep the ions apart
and hence diminish the coefficient of recombination. Simple considerations,
however, will show that it would require exceedingly
strong electric fields to produce an appreciable effect. The value of
α indicates that for two oppositely charged ions to unite they must
come within a distance of about 1.5×10^{−6} centimetres; at this
distance the attraction between them is *e*^{2}×10^{12}/2.25, and if X is the
external electric force, the force tending to pull them apart cannot
be greater than X*e*; if this is to be comparable with the attraction,
X must be comparable with *e*×10^{12}/2.25, or putting *e* = 4×10^{−10},
with 1.8×10^{2}; this is 54,000 volts per centimetre, a force which
could not be applied to gas at atmospheric pressure without producing
a spark.

*Diffusion of the Ions.*—The ionized gas acts like a mixture of gases,
the ions corresponding to two different gases, the non-ionized gas
to a third. If the concentration of the ions is not uniform, they will
diffuse through the non-ionized gas in such a way as to produce a
more uniform distribution. A very valuable series of determinations
of the coefficient of diffusion of ions through various gases has been
made by Townsend (*Phil. Trans.*, 1900, A, 193, p. 129). The method
used was to suck the ionized gas through narrow tubes; by measuring
the loss of both the positive and negative ions after the gases
had passed through a known length of tube, and allowing for the loss
by recombination, the loss by diffusion and hence the coefficient of
diffusion could be determined. The following tables give the values
of the coefficients of diffusion D on the C.G.S. system of units as
determined by Townsend:—

*Coefficients of Diffusion (D) in Dry Gases.*

Gas. | D for +ions. | D for −ions. | Mean Value of D. | Ratio of D for − to D for +ions. |

Air | .028 | .043 | .0347 | 1.54 |

O_{2} | .025 | .0396 | .0323 | 1.58 |

CO_{2} | .023 | .026 | .0245 | 1.13 |

H_{2} | .123 | .190 | .156 | 1.54 |

*Coefficients of Diffusion in Moist Gases.*

Gas. | D for +ions. | D for −ions. | Mean Value of D. | Ratio of D for − to D for +ions. |

Air | .032 | .037 | .0335 | 1.09 |

O_{2} | .0288 | .0358 | .0323 | 1.24 |

CO_{2} | .0245 | .0255 | .025 | 1.04 |

H_{2} | .128 | .142 | .135 | 1.11 |