686 Prof. B. M. Sen on the Kinetic Theory of Solids 

 4. Face-centred Cubic Crystals. 



This arrangement consists in having a molecule placed at 

 the eight corners of a cube and one at the centre of each of 

 the six faces. To find the arrangement of the molecules 

 about any individual one, we note that there are two types, 

 viz. those at the corners and those at the centres of the 

 faces. For both the types it will be found that the following 

 arrangement gives the requisite number of molecules situated 

 nearest to each individual molecule 0. 



With O as centre describe a sphere of radius I and draw 

 a system of three great circles intersecting one another at 

 right angles. Each great circle is divided into four quad- 

 rants by the other two. At the middle point of each of 

 these twelve quadrants place a molecule. With a series of 

 such models we can build up a body of the face-centred 

 cubic crystal type, the edge of the cube being V /2Z. It may 

 be noted that there are twelve molecules surrounding each 

 individual molecule instead of the fourteen in the isotropic 

 arrangement, and that the spherical distance between any 

 two adjacent molecules not on the same co-ordinate plane is 

 60°. In the isotropic model the distance between any two 

 on the same great circle is 60°. 



Let us calculate the potential energy of the displacement 

 of the central molecule. Take the planes of the great 

 circles for the co-ordinate planes. Take two straight lines 

 in the .ry-plane, the x 1 and the y 1 axes, bisecting the angles 

 between the x and the n axes. Place four molecules at the 

 extremities of the a\ and the^ axes. Put similarly four at the 

 extremities of the y 2 and z 2 axes which bisect the angles 

 between the a, z axes in the //z-plane, and also four at the 

 extremities of the z z and x z axes which bisect the angles 

 between the z; ,v axes in the zx-piane. Let (x, y, z) be the 

 most general displacement of the central molecule. It can 

 also be written in the form (#j, y x , z), (as, y 2 , z 2 ), (a? 3 , y, z B ), 

 so that 



x 2 + y 2 + z 2 = x 1 2 +y l 2 + z 2 = x 2 + y 2 2 + z 2 2 = x 3 2 + y 2 + z 3 2 = a 2 



( sa y)- 



For the four molecules on the x 1} y 1 axes, the potential 

 energy due to the displacement x\ 



=/(Z + *i)+/G-*i) + 2/(z+ l^f), 



=4/(0+v{^ + /"(0}- 



