﻿Lead and the End Product of Ti 



675 



of m, the amount of original lead is assumed as above, and 

 the values of k and m are obtained by solution of two equa- 

 tions obtained as follows. The results of any two of the 

 four analyses Nos. 3, 6, 7, 12 are substituted in the above 

 equation, and the two equations obtained by different com- 

 binations of the analyses give on solution the required values 

 -of k and m. In the following table is given the series of 

 results so obtained : — 



Table VI. a. 



Combination of 

 Analyses. 



Value of 7c. 



Value of m. 



Nos. 3 : 6 



0043 



5.10~ 5 



„ 3: 7 



0044 



4.10- 5 



„ 3:12 



0-042 



7.10" 5 



„ 6: 7 



0-045 



1.10" 5 



„ 6:12 



0042 



8.10-5 



„ 7:12 



0-042 



8.10 



Mean = 6.lO -5 



The value obtained from 6 : 7 is undoubtedly too low, and 

 is due to these two analyses being almost alike. 



(6) In the second method we assume Pb = 0*0005, and 

 the value of k to be 0'042 for Devonian minerals, and find 

 the value of m from each of the four analyses Nos. 3, 6, 7, 

 and 12. Before passing on, there is one point to be men- 

 tioned in connexion with the value of k above stated. The 

 Pbi/U m ratio for No. 12 expressed in terms of the time- 

 average value of uranium is 0*041. In this discussion the 

 actual amount of uranium present is being used in the cal- 

 culations, and hence we must use the ratio between the 

 present lead and uranium contents of the minerals Pb/U. 

 For mineral No. 12 this is equal to 0042. As in the previous 

 case, we require the equation Pb^ = Pb 4-& .U* + »i . Th< for 

 the purpose of calculating m, which is the only unknown 

 present. The following table gives the results obtained by 

 this method : — 



Table VI. b. 



No. of Analysis... 



3. 



6. 



7. 





7.10- 5 



io. i(T 5 



9.10" 5 







Mea 



10"' 



2X2 



