90 Dr. L. Vegard on 



The distances l x and Z 2 (figs. 4, 5) from the Zr and Si 

 atoms to the oxygen atoms will be 



k = e x a \/2 = 2*71 x 10~ 8 cm. 



l 2 = e 2 a \/2 = 1-08 xlO" 8 cm. 



The distance from the Zr atom to one of the oxygen 

 atoms associated with it is more than twice the corresponding 

 distance for Si, a fact which may be due to the greater 

 affinity between the Si and atoms. 



§ 9. Rutile and Kassiterite. 



If the lattices of (Ti0 2 ) 2 and (Sn0 2 ) 2 belong to the type 

 given in Table IV., the spectra should be derived from the 

 formulae (12) by inserting the corresponding values of the 

 atomic numbers. 



In both cases Ni = N 2 = N, and for the (111) face we get 



f/^n) =N(l + (-l)«)+ 2 (N + 2N.) cosn| 

 I 111 ) \ +2N 3 (cos ^+(-1)* cos no,). 



l/iM=o. 



Now the Rontgen ray analysis shows that spectra of 

 uneven order with regard to n disappear ; consequently we 

 have for all values of q : 



/ 1 (2gr-l)=0 = 2N, (cos (2q-l)cc 1 - cos {2q-l)a 2 \ 



which gives 



This is an important result as it shows tbat in each of the 

 two minerals all the metal atoms are identical as regards 

 their relation to the oxygen atoms, and it is impossible to 

 consider rutile say — as a titanum — titanate. 



The amplitudes for the four cases considered will be 



A ni = N + (-l)*(N + 2N,) + 2N t cob 2n«, \ 



Ano = N-+ N 8 + N 8 cos 2na, 

 A 101 = N-f2E" 3 cos not, 



A 100 = N _|_ 9^ 3 cos 2) ia . , 



The expression for A m , A 110 , and A 100 will not be altered 

 if we substitute it— ol for a. Which of these two is the 

 right value can be decided by means of the (101) spectrum. 



