Calculate the number of particles per unit cell in fcc.

Explanation

(a)
The number of atoms per unit cell in bec structures is two. Each atom is considered as one sphere.
So, the volume of two atoms (two spheres) =` 2xx(4)/(3)pir^(3)`
From the figure it is clear that there are three spheres touching each other along the body diagonal. Let a be the edge length of the cube and r be the radius of the sphere. The body diagonal AF = c and face diagonal FD=b, then AF-c=4r.
  `DeltaEFD`
` FD^(2)=EF^(2)+ED^(2)`
`b^(2)=a^(2)+a^(2)=2a^(2)`             `b=sqrt2a`
In `DeltaAFD`
`AF^(2)=AD^(2)+FD^(2)c^(2)=a^(2)+b^(2)=a^(2)+2a^(2)=3a^(2)`.
` c=sqrt3a`
But c=4r 
Thereforr, ` sqrt=4r`
`a=(4r)/sqrt3` 
volume of the unit cell =`a^(3)=((4r)/(sqrt3))^(3)`
Therefore Packing efficiency= `("Volume of two atoms (two spheres) in unit cell")/("volume of the unit cell")xx100`
Packing efficiency = `(2xx(4)/(3)pir^(3))/((4r)/sqrt3)^(3)xx100%`=`((8)/(3)pir^(3))/(64/(3sqrt3)r^(3))xx100%=(sqrt3pi)/(8)xx100=68%`
Therefore 68% of unit cell is occupid by atoms and the rest 32% is empty space in bcc structure.
The packing fraction is bcc structures =0.68.
  The fraction of empty space in bcc structures =0.32.
(b) Total no. of practice due to corners =`(1)/(8)xx8=1`.
Total no. of practice due to faces `=6xx(1)/(2)=2`.
Total no. of practice  unit cells =3+1=4.