the Absorption of the Gamma Rays. 133 



The total amount, S 1? of the scattered rays which enter 

 the electroscope E, therefore, is 



8 l = 27rI ae-^ +a V ( ' i V+^ 8ecfl - 1 >/(^) sin e de dx > ( 2 ) 



Jo Jo 



where t is the thickness of the plate and 1 is half the angle 

 subtended at the point P by the side of the electroscope E. 

 The intensity, I 1} of the beam which reaches the electro- 

 scope is 



I 1 = V-^+^ + Sj (3) 



If the plate be close to the electroscope then the upper 

 limit of the integral with respect to 6 in (2) is approxi- 

 mately 7r/2, and therefore the amount, S 2 , of the scattered 

 rays entering the electroscope in this case is 



S 5 = 2irW-<*+ 



*>< ^ ( 'V+^W" 6 - l \t\e) sin 6 d6 dx, . (4) 



which is the total amount of the emergent scattered rays. 

 The total intensity, I 2 , of the beam which reaches the 

 electroscope is given by 



I 2 = I„«-<"+^ + S 2 (5) 



The difference between (3) and (5) is 

 I 2 — Ix = S 2 — Sj 



= 2ttI c76-^+^ \ ]eb+°)*(«*o-i)f(0) s in<9 dB dx 



Jo J 0i 



= S, say (6) 



Hence, if this integral can be evaluated, we can find the value 

 of a from the observed values of S and (/jl + <t). Strictly 

 speaking, 6 is a function of x ; but if the plate be very thin, 

 6 can be considered to be approximately independent of a?, 

 and we have for (6) 



27rLo- f* ir/2 ^fci + ffXsec 0-1)* _1 



fi+<r Jfc sec<9-l yv J K ' 



To integrate (6') with respect to 0, the expression f(6) 

 must be known. According to Sir J. J. Thomson's theory 

 of scattering of X rays, the intensity of scattered radiation 

 in any direction is proportional to l-f-cos 2 #, and may be 

 written 



l 6 = 1^(1 + cos 2 0). 



