cannot be greater than [epsilon]^{V/60}, when V
is measured in volts. This result is based on Townsend's experiments
with very weak currents; we must remember, however, that when the
collisions are so frequent that the effects of collisions can
accumulate, [alpha] may have much larger values than when the current
is small. In some experiments made by J. J. Thomson with intense
currents from cathodes covered with hot lime, the increase in the
current when the potential difference was 60 volts, instead of being e
times the current when there was no ionization, as the preceding theory
indicates, was several hundred times that value, thus indicating a
great increase in [alpha] with the strength of the current.

Townsend has shown that we can deduce from the values of [alpha] the
mean free path of a corpuscle. For if the ionization is due to the
collisions with the corpuscles, then unless one collision detaches
more than one corpuscle the maximum number of corpuscles produced will
be equal to the number of collisions. When each collision results in
the production of a corpuscle, [alpha] = 1/[lambda] and is independent
of the strength of the electric field. Hence we see that the value of
[alpha], when it is independent of the electric field, is equal to the
reciprocal of the free path. Thus from the table we infer that at a
pressure of 17 mm. the mean free path is 1/325 cm.; hence at 1 mm. the
mean free path of a corpuscle is 1/19 cm. Townsend has shown that this
value of the mean free path agrees well with the value 1/21 cm.
deduced from the kinetic theory of gases for a corpuscle moving
through air. By measuring the values of [alpha] for hydrogen and
carbonic acid gas Townsend and Kirby (_Phil. Mag._ [6], 1, p. 630)
showed that the mean free paths for corpuscles in these gases are
respectively 1/11.5 and 1/29 cm. at a pressure of 1 mm. These results
again agree well with the values given by the kinetic theory of gases.

If the number of positive ions per unit volume is m and v is the
velocity, we have nue+mve = i, where i is the current through unit
area of the gas. Since nue = i0[epsilon]^nx and i = i0[epsilon]^nl,
when l is the distance between the plates, we see that

nu / mv = [epsilon]^(nx) / ([epsilon]^(nl) - [epsilon]^(nx)),

n v [epsilon]^(nx)
-- = -- . -------------------------------.
m u [epsilon]^(ne) - [epsilo