In order to test the amount of absorption of the α rays for different thicknesses of matter, an apparatus similar to that shown in Fig. 17, p. 98, was employed[158]. A thin layer of the active material was spread uniformly over an area of about 30 sq. cms., and the saturation current observed between two plates 3·5 cms. apart. With a thin layer[159] of active material, the ionization between the plates is due almost entirely to the α rays. The ionization due to the β and γ rays is generally less than 1% of the total. The following table shows the variation of the saturation current between the plates due to the α rays from radium and polonium, with successive layers of aluminium foil interposed, each ·00034 cm. in thickness. In order to get rid of the ionization due to the β rays from radium, the radium chloride employed was dissolved in water and evaporated. This renders the active compound, for the time, nearly free from β rays. The initial current with 1 layer of aluminium over the active material is taken as 100. It will be observed that the current due _Polonium._ _Radium._ Layers of Current Ratio of Layers of Current Ratio of aluminium decrease aluminium decrease for each for each layer layer 0 100 0 100 ·41 ·48 1 41 1 48 ·31 ·48 2 12·6 2 23 ·17 ·60 3 2·1 3 13·6 ·067 ·47 4 ·14 4 6·4 ·39 5 0 5 2·5 ·36 6 ·9 7 0 to the radium rays decreases very nearly by half its value for each additional thickness until the current is reduced to about 6% of the maximum. It then decays more rapidly to zero. Thus, for radium, over a wide range, the current decreases approximately according to an exponential law with the thickness of the screen, or $$ \frac {i} {i₀} = e^{λ d} $$,