Question

In diamond carbon atom occupy fcc lattice points as well as alternate tetrahedral voids if as length of the unit cell is 3:56 p.m. then the diameter of carbon atom is

Answer 1

Number of atoms presented  by the atoms at corner position  to an unit cell is calculated as  1/8×8 =1 , similarly the  number of atoms contributed by the facial and central atoms to the unit cell is  calculated as 1/2 × 6 = 3 and atoms inside the structure, as per the problem is = 4.

Therefore, total number of atoms presented  in a diamond cubic unit cell is 1 + 3 + 4 = 8.

Again, since each carbon atom is surrounded by four more carbon atoms, the co-ordination number is estimated at 4.From triangle WXY

XY2=(a/4)2+(a/4)2

=(a2)/8

From triangle XYZ

XZ2=(a2)/8 + (a/4)2

=(3a2)/16

But XZ=2r

So (2r)2 = (3a2)/16

Atomic radius r =( a(3)1/2)/8

So, diameter of carbon atom is , D=2( a(3)1/2)/8