Emanation in the Lower Regions of the Atmosphere. 27 



with this amount of emanation we multiply the denominator 

 by 



J i.e.---— j or 1 — e A/ , 



using the same notation as before (p. 6). 



The following table gives the values of 1 — e~ kt required 

 in the experiments : — 



Table IX. 



t (hours) . 



l-e- kt 



0223 



14. 



0511 



•099" 



21. 



146 



22. 



•152 



1000 



Thus the amount of radium that would he in radioactive 

 equilibrium with the amount of emanation that mv solution 

 generates in 3 hours is 



3-14xl0- 9 x-0223 gm. 



Five runs of 3 hours each were made with the tubes in 

 parallel, tlie solution being in series with one of the tubes. 

 The results are given in the following table: — 



Table X. — Short Runs. 



Date 

 1909. 



! Mar.2... 

 | July 21. 

 I„ 23. 



| Aug. 27. 



Air alone. 



Air plus Solution. 



I Speed. 



Tabe, r"- Per 



'■ mm. 



Vol. or 



Emanation | 



air. 



caught. 



92 



1-0 



88 



" 7 1 



90 



•5 



90 



11 



90 



1-1 



Tube. 



bpeed. 



lit. per 



min. 



Vol. of 

 air. 



emanation 

 caught. 



B 



•58 



106 



49 



B 



•5 



92 



52 



B 



•5 



90 



51 



B 



•5 



90 



5-8 



A 



•5 



90 



55 



Emanation- 

 generated 

 by solution! 

 in 3 hours. 



39 

 4-5 

 46 

 4-7 

 4-4 



The average amount of emanation caught from the air in 

 3 hours is *8, and the average amount caught from the 

 solution (alone) in 3 hours is £65 (neglecting the reading 

 on Mar. 2 when the speed of the air-stream rose to "58). 



Therefore on the average the emanation in 90 litres of 

 atmospheric air is equal to that which would be in radioactive 

 equilibrium with 



x 3-14 x 10" p x -0223 gm. of radium. 



d'65 



