(a) Calculate the number of particles in BCC lattice.
(b) Calculate the number of particles per unit cell in FCC.

Explanation

(a) The number of atoms per unit cell in bcc structures is two. Each atom is considered as one sphere.
So, the volume of two atoms (two spheres) =? `2xx4/3pir^(3)`
From the figure it is clear that there are three spheres touching each other along the L. diagonal. Let a be the edge length of the cube and r be the radius of the sphere. The diagonal AF = c and face diagonal FD = b , then AF - c = 4r.
In AEFD
`FD^(2)= EF^(2) + ED^(2)`
`b^(2) – a^(2) + a^(2) = 2a^(2)` `b = V2a `
In `Delta AFD`
`AF^(2)=AD^(2)+FD^(2)c^(2)=a^(2)+b^(2)=a^(2)+2a^(2)=3a^(2)`.
`c= sqrt(3)a`
but c=4r
therefore `sqrt(3)a=4r`
a`=(4r)/sqrt(3)`
volume of unit cell=`a^(3)=((4r)/sqrt(3))^(3)`
Therefore Packing efficiency = `("Volume of two atoms (two spheres) in unit cell")/("Volume of the unit cell")xx100`
Packing efficiency `= (8/3pir^(3))/(64/(3sqrt(3))r^(3))xx100% =sqrt(3) pi xx 100= 68%`
Therefore 68% of unit cell is occupied by atoms and the rest 32% is empty space in bcc structures.
The packing fraction in bcc structures = 0.68
The fraction of empty space in bcc structures = 0.32.
(b) Number of particles per unit cell of FCC =
`(1)/(82)`(No. of comer particles) + `(1)/(82)`(No. of facial particle) = `1/8 + 1/6 = l + 3= 4`