p = 



33S H. M. Dadourian — Atmospheric Radio-activity. 



In other words, the number of particles which are transformed 

 is proportional to the number of particles present. Therefore 



X Pdt 



particles are transformed in the differential time dt. If the 

 air be in radio-active equilibrium the number of particles which 

 are deposited on the wire per second will be constant. Denot- 

 ing this number by q, we obtain 



d P = — A Pdt + qdt 



for the net change in the number of particles during the 

 time dt. 



Integrating and remembering the fact that P = for £=0, 

 we get 



The right-hand member approaches the value -f- asymptoti- 



A 



cally as the time increases indefinitely. Denoting this final 

 value by P , 



p=p,(i-«- A ') 



If the disintegration product under consideration is not a ray- 

 less one, the ionization it produces is proportional to the num- 

 ber of its particles, hence 



1 = 1.(1-*-"' 



where 1= value of ionization at any instant of the exposure, 

 and I = the value it would have attained if the exposure were 

 continued for an indefinite length of time. Theoretically an 

 infinitely long exposure is necessary for any of the disintegra- 

 tion products on the wire to reach equilibrium value. But only 

 a comparatively short exposure is necessaiy for the ionization 

 due to any of the products to approach within one one-hun- 

 dredth of the equilibrium value. The last column of the fol- 

 lowing table gives the time of exposure necessary for the 

 ionizations produced by the six disintegration products of 

 radium and thorium to attain 0*99 of their equilibrium values. 

 It takes over three days for thorium A to reach within one one- 

 hundredth of its equilibrium amount. Therefore it is neces- 

 sary to continue the exposure for at least three days in order 

 to make the amounts of the six disintegration products present 

 on the wire proportional to the amounts of the corresponding 

 products present in a unit volume of the air. 



