1.

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

Answer»

Solution :(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`


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