Published online by Cambridge University Press: 24 October 2008
The results of experiments on the diffusion of hydrogen through metals from a pressure p on one side to a vacuum on the other show that at high pressures the amount diffusing varies linearly with p½ but that at low pressures it varies more rapidly. The difficulty usually encountered when diffusion from an adsorbed layer into the solid is considered theoretically is that the theory indicates that saturation should be reached. In this paper it is shown that this difficulty is due to the omission of an important process at the surface and that by including this process the experimental results can be explained.
* (I) Proc. Roy. Soc. A, 150 (1935), 172Google Scholar; (II) ibid. 152 (1935), 706; (III)ibid. 155 (1936), 195.
† Richardson, , Nicol, , and Parnell, , Phil. Mag. 8 (1904), 1.CrossRefGoogle Scholar
‡ Cf. Smithells and Ransley, loc. cit. (I), p. 192.
§ Loc. cit. (I), p. 193; see also Melville, and Rideal, , Proc. Roy. Soc. A, 153 (1935), 89.CrossRefGoogle Scholar
* Fowler, , Proc. Camb. Phil. Soc. 31 (1935), 260.CrossRefGoogle Scholar
† See papers of Smithells and Ransley; see also Braaten, and Clark, , Proc. Roy. Soc. A, 153 (1936), 504.CrossRefGoogle Scholar
‡ See section 2 below. Cf. also Melville and Rideal, loc. cit., in particular p. 100.
§ I am indebted to these authors for letting me see their manuscript before publication.
∥ In the work referred to in the preceding footnote Smithells and Ransley also measured the diffusion of oxygen through nickel and they found the phenomenon of saturation, but on account of the fact that oxygen disintegrates the crystal lattice the phenomena with oxygen are much more complicated.
* In this connection processes involving the pressure of atoms in the gas phase can be neglected, cf. Roberts, , Proc. Camb. Phil. Soc. 32 (1936), 152.CrossRefGoogle Scholar
* It is interesting to consider the relations between θ, ν and p in the equilibrium state. If detailed balancing is assumed, we have
Similarly
where we have dropped the suffix 1, because in the case of no diffusion the conditions on both surfaces are the same. Solving these equations, we get
(this equation is the same as (2) with)
and a relation
* The reasoning which leads to this result may be outlined as follows. We first show that n → ∞ as p → ∞. For if n remains finite, then by (15)
p (1 − θ1) → const.,
and consequently p (1 − θ1)2 → 0,
which is contradictory to (13), the left-hand side of which remains greater than a positive constant. The right-hand side of (14) remains finite, therefore n (1 − θ2) → const., and, since n → ∞ as p → ∞, we must have θ2 → 1. To show that for large p, we use (15) and get
n ˜ k′p (1 − θ1).
Substituting in (13), we obtain p (1 − θ1)2 → const.