=106. Absorption of the γ rays=. In an examination of the active substances by the electrical method, the writer[169] found that both uranium and thorium gave out γ rays in amount roughly proportional to their activity. An electroscope of the type shown in Fig. 12 was employed. This was placed on a large lead plate ·65 cm. thick, the active substance being placed in a closed vessel beneath. The discharge due to the natural ionization of the air in the electroscope was first observed. The additional ionization due to the active substance must be that produced by rays which have passed through the lead plate and the walls of the electroscope. The following table shows that the discharge due to these rays decreases approximately according to an exponential law with the thickness of lead traversed. Thickness of lead Rate of discharge ·62 cms. 100 „ + ·64 cms. 67 „ + 2·86 „ 23 „ + 5·08 „ 8 Using 100 grs. of uranium and thorium, the discharge due to the rays through 1 cm. of lead was quite appreciable, and readily measured. The results showed that the amount of γ rays was about the same for equal weights of thorium and uranium oxides. The penetrating power was also about the same as for the radium rays. [Illustration: Fig. 44.] The writer showed that the absorption of the γ rays from radium was approximately proportional to the density of the substance traversed. A more detailed examination of the absorption of these rays in various substances has been recently made by McClelland[170]. The curve (Fig. 44) shows the decrease of the ionization current in a testing vessel due to the β and γ rays with successive layers of lead. It is seen that the β rays are almost completely stopped by 4 mms. of lead; the ionization is then due entirely to the γ rays. In order to leave no doubt that all the β rays were absorbed, the radium was covered with a thickness of 8 mms. of lead, and measurements of the coefficient of absorption λ were made for additional thicknesses. The average value of λ was calculated from the usual equation $$ \frac {I} {I₀} = e^{–λ d} $$, where _d_ is the thickness of matter traversed. The following table shows the value of λ, (I) for the first 2·5 mms. of matter traversed (after initially passing through 8 mms. of lead), (II) for the thickness 2·5 to 5 mms., (III) for 5 to 10 mms., (IV) 10 to 15 mms. TABLE A. Substance I II III IV