408 



Dr. L. Vegard on 

 Table IV. 



j 



(001). 



(101). 



(HI). 



From 

 lattice. 



From 

 obs. 



From From 

 lattice. obs. 



From 



i lattice. 



100 

 44 

 19 

 62 



From 

 obs. 



100 



48 



(?) 

 72 



A, ...... 



A 2 



A 3 



^ 



28 



45 



9 



100 



21 



30 







100 



50 52 



36 44 



100 100 



69 72 



Of course, we cannot prove that these three angles are 

 exactly equal, but the difference cannot be more than a few 

 degrees. Thus, if we put B = 30° and = 40°, or vice versa, 

 we find intensities which are very different from those 

 observed. 



Inserting the values of A, B, and C from equation (11) 

 equations (13) give 



A =B = C 



8 = 0, 

 = 0. 



1 



(14) 



J 



Thus we find that two of the parameters introduced are 

 nearly equal to zero. From equation (9 a) we see that 

 /' = 0and '/ = 0. 



The disappearance of the first parameter V means that in 

 the molecular element the plane of the C atoms cuts the 

 N atom, and l =Q means thai the centres of the molecular axes 

 are arranged in a "prism-centred " lattice. Or we might say 

 that the N and 1 atoms are arranged in such a icay that each 

 elementary I lattice has an N lattice at its centre, and vice versa. 



Thus the simple values found for V and 7 will considerably 

 simplify the lattice. 



The arrangement of the I, N, and C atoms is illustrated 

 in fig. 4. We notice that the lattice found leads to a very 

 simple arrangement of the C atoms. They appear in groups 

 of four atoms belonging to four different molecular elements. 

 These four C atoms are placed at the corners of a sphenoid 

 (deformed tetrahedron) like OPQR. Two of the atoms of 

 the group belong to the one elementary lattice of molecular 



