Question

Calculate the number of atoms per unit cell of $FCC$ and $BCC$ crystal structure.

Answer 1

In a FCC lattice there rer $8$ atoms at eight corners and $6$ at face centers,
Now each corner contributes to eight cells so per unit cell contribution is $\frac{1}{8}\times 8=1$ atom.
Now similarly each face center contributes to two unit cells so contribution per unit cell by six face centers is equal to $\frac{1}{2}\times 6=3$ atoms. Hence total no. of atoms per unit cell of FCC lattice is $(1+3)=4$ atoms.
For BCC-
A bcc has one atom in center and $\frac{1}{8}th$ part of 8 corner atoms i.e 1 atom.
In total there are $1+1=2$ atoms.