While the exponential law, in some cases, approximately represents the variation of the ionization with distance, in others the divergence from it is wide. The ionization, due to a plane surface of polonium, for example, falls off more rapidly than the exponential law indicates. The α rays from an active substance like radium are highly complex; the law of variation of the ionization due to them is by no means simple and depends upon a variety of conditions. The distribution of ionization is quite different according as a thick layer or a very thick film of radio-active matter is employed. The question is fully considered at the end of chapter IV., but for simplicity, the exponential law is assumed in the following calculations. Consider two parallel plates placed as in Fig. 1, one of which is covered with a uniform layer of radio-active matter. If the distance _d_ between the plates is small compared with the dimensions of the plates, the ionization near the centre of the plates will be sensibly uniform over any plane parallel to the plates and lying between them. If _q_ be the rate of production of ions at any distance _x_ and _q₀_ that at the surface, then _q_ = _q₀__e_^{–λ_x_}. The saturation current _i_ per unit area is given by $$ i = \int₀^d qe' dx $$, where _e´_ is the charge on an ion, $$ = q₀e' \int₀^d e^{–λ x} dx = \frac{q₀e'}{λ} (1 - e^{–λ d}) $$ hence, when λ_d_ is small, _i.e._ when the ionization between the plates is nearly constant, _i_ = _q₀e´d_. The current is thus proportional to the distance between the plates. When λ_d_ is large, the saturation current _i₀_ is equal to _q₀e´_/λ, and is independent of further increase in the value of _d_. In such a case the radiation is completely absorbed in producing ions between the plates, and $$ \frac {i}{i₀} = 1 − e^{–λ d} $$ For example, in the case of a thin layer of uranium oxide spread over a large plate, the ionization is mostly produced by rays the intensity of which is reduced to half value in passing through 4·3 mms. of air, _i.e._ the value of λ is 1·6. The following table is an example of the variation of _i_ with the distance between the plates. Distance Saturation Current 2·5 mms. 32 5 „ 55 7·5 „ 72 10 „ 85 12·5 „ 96 15 „ 100 Thus the increase of current for equal increments of distance between the plates decreases rapidly with the distance traversed by the radiation. The distance of 15 mms. was not sufficient to completely absorb all the radiation, so that the current had not reached its limiting value. When more than one type of radiation is present, the saturation current between parallel plates is given by $$ i = A (1 − e^{λ d}) + A_1 (1 − e^{–λ_1 d}) $$ &c.