﻿Problems in Radioactivity, 



219 



These results follow from the solutions of the problems 

 already considered by making use of the reciprocal theorem 

 stated' at the end of § i of the present paper. This 

 theorem explains how the two different types of absorption 

 problems lead to the same mathematical expressions. 



It is perhaps not without interest to notice that the in- 

 tensity at a point on the axis of a circular slab containing a dis- 

 tribution of radioactive matter can be evaluated. The result 

 is easily obtained from (5) by taking the integrals from 



to <fi instead of from to - , </> being the angle subtended by 



the radius a of the plate at P. Then by (7) we obtain for 

 the ionization at P, 



27rQ/?n r 



^'0 



f{\s)-f(\c + llK) — CbB 



4Kc^)~K^?)j] 



. . . (13) 



This result neglects the contribution of the ring generated 

 by the revolution of the shaded area in the figure around 

 the axis of the slab. This will be negligible if the radius 

 be laro-e compared with h, and P be not too close to the 

 plate. 



If we have a screen of absorbing matter, thickness b, 

 coefficient of absorption //,, interposed between P and the 

 circular plate, the intensity at P is given by the formula 



n = ~ 7r %U ° f(fib + X:) —/(/Jib + \z + h/c) 

 Phil Mag. S. 6. Vol. 23. No. 134. Feb. 1912. S 



