536 Miss Jadwiga Szuridt on Distribution of Energy in 



unit thickness close to the source, and //, 2 is the absorption 

 coefficient of the hard rays in air. In order to find the 

 relative amounts of energy the knowledge of! the absolute 

 values of n x and n 2 is not necessary, but simply of the ratio 



— ; this can be obtained by examining an absorption curve, 

 n 2 



say in aluminium, of the 7 rays and extrapolating the part 



of the curve where it follows an exponential law right to the 



beginning, as shown in fig. 3, which illustrates the case of 











Fig. 3. 





































































































....... 





















■w> 



I 1-2 1-4 



Mms. of Aluminium. 



radium D. The ratio of the ordinates (OB — OA) : OA (after a 

 small correction has been applied for the comparatively 

 larger absorption of the soft rays in the space between source 

 and ionization-chamber, and in the face wall of the latter) is 



then equal to — . To calculate — l = -^- 2 we have then 

 x n 2 JN 2 ^2/^1 



only to know the values of fi x and ^. A similar method can 



be applied when more types of radiations are present. 



y Rays from Radium D. — The source of rays consisted of 

 the same preparation of radium D on filter-paper which had 

 been used for the absorption and ionization experiments de- 

 scribed in Part I. The experimental arrangement was very 

 similar to the one shown in fig. 1, the cylinder A being- 

 removed and the source placed nearer the electroscope. An 

 examination of the absorption curve with air in the ionization 

 chamber gives n x : n 2 equal to 106 : 7, after all necessary 

 corrections are applied. 



The value of ^ for air is taken from Table I. It is difficult 

 to decide what value to assume for /i 2 , as it could not be 

 obtained by direct measurement. Supposing the density law 

 to hold in this case } for which the mass absorption coefficient 



