308 Prof. W. H. Bragg on the Structure of 



Let the carbon atoms of one face-centred lattice be re- 

 placed by oxygen tetrahedra of one orientation, and the 

 atoms of the other lattice by the other orientation. It then 

 follows that when the process of reflexion and shifting is 

 carried out as in the case of the diamond, we still obtain 

 the same absolute coincidence. From the crystallographic 

 point of view the symmetry about the (100) plane is 

 complete. 



As regards the (110) planes the introduction of the oxygen 

 tetrahedra in place of the carbon atoms does not disturb the 

 simple mirror symmetry which already exists. Any (110) 

 plane parsing through a tetrahedron contains some one edge 

 and bisects the opposite edge at right angles, thus dividing- 

 it into two halves which are the reflexions of each other in 

 the plane. 



A tetrahedron of one kind does not reflect into a tetra- 

 hedron of the other kind over a (111) plane, but the crystal 

 of course possesses no symmetry of that kind ; so the- 

 introduction of tetrahedra creates no difficulty. 



Tims all the symmetries are maintained in full. 



Next consider the effects of the substitution on the spectra. 



As regards the (100) planes the spacing in the case of the 

 diamond is one quarter of the edge of the cube shown in 

 fig. 2, a cube which contains a volume associated with eight 

 atoms of carbon. The (100) planes contain alternately all 

 blacks and all whites; and when both blacks and whites are 

 carbon atoms they are necessarily exactly similar. When 

 the blacks represent oxygen tetrahedra of the one orientation 

 and the whites tetrahedra of the other, the planes are still 

 effectively identical. A (100) plane divides a tetrahedron 

 so that two oxygen atoms lie on each side of it. All four 

 are at the same distance from the plane, though the pair on 

 one side is not the reflexion of the pair on the other. Each 

 (100) plane of carbon atoms becomes, therefore, a pair of 

 oxygen planes, and whether the original plane contained 

 blacks or whites these sets of two are indistinguishable from 

 an X-ray point of view. If they were not we should have a 

 new spacing double as great as the old, and the (100) spectra 

 would be interleaved by new ones bisecting the spaces 

 between those of the diamond. We have nothing of this 

 sort to account for, and the tetrahedra therefore still satisfy 

 the conditions required. 



As regards the (110) planes no difficulty arises because 

 each (110) plane contains blacks and whites in equal 

 numbers, and the spacing of the (110) planes does not 



