Diamond with Theoretical Carbon Atoms. 2G' 



radii, %m 2 for the carbon atom is, therefore, 

 ' 12 



6-735 2 x 8 =363 

 2-338 2 x 4= 21*9 



2m 2 = 384-9 (13) 



12 



This is to be compared with this same quantity determined 

 from the diamond in (11) above, namely, 378. The 

 agreement is surprisingly close. 



To find the Length of the Edge of the Elementary 

 Tetrahedron in Diamond. 



It may be of interest to give the calculation of the length 

 of the edge of the elementary tetrahedron in diamond, as 

 the method differs somewhat from the usual but gives sub- 

 stantially the same value. 



I = distance between corners of tetrahedron. 

 — y/6l= „ „ nearest planes of atoms. 



a = the edge f a cube of the crystal containing one 



No. planes of atoms in distance "a"= -jy/6. 



No. atoms in one plane = -^ ~%— • 



i" O 



No. atoms in cube of edge a = ... 2y/2 (14) 



d = density. 



a 3 = volume per gram = cu. cm (15j 



No. electrons in a gram = 6 x 10". 



No. carbon atoms in a gram = — x 10 23 = \*> x 10 23 . (1C) 



Replacing a 3 in (14) by - and equating to (16) we find 



P=~ X 10" 23 cu. cm (17) 



Taking d = 3'5 for diamond, we get 



Z 3 = 16-16 xl0~ 24 cu. cm. and I = 2-528 x 10- 8 cm. . (18) 



