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## Main Question or Discussion Point

Taken from http://dao.mit.edu/8.231/BZandRL.pdf

In real space, it has a simple cubic lattice with one basis in the centre. Total number of atoms per unit cell = 2. Volume of primitive unit cell is then ##\frac{1}{2}a^3##.

In reciprocal space, BCC becomes an FCC structure. It has a simple cubic lattice of length ##\frac{2\pi}{a}## with 4 atoms in total. Why is the total volume then ##4 \left(\frac{2\pi}{a}\right)^3## and not ##\frac{1}{4}\left(\frac{2\pi}{a}\right)^3##?

In real space, it has a simple cubic lattic with 3 basis. Total number of atoms per unit cell = 4. Volume of primitive unit cell is then ##\frac{1}{4}a^3##.

In reciprocal space, FCC becomes a BCC structure. It has a simple cubic lattice of length ##\frac{2\pi}{a}## with 2 atoms in total. Why is the total volume then ##2 \left(\frac{2\pi}{a}\right)^3## and not ##\frac{1}{2}\left(\frac{2\pi}{a}\right)^3##?

BCCBCC

In real space, it has a simple cubic lattice with one basis in the centre. Total number of atoms per unit cell = 2. Volume of primitive unit cell is then ##\frac{1}{2}a^3##.

In reciprocal space, BCC becomes an FCC structure. It has a simple cubic lattice of length ##\frac{2\pi}{a}## with 4 atoms in total. Why is the total volume then ##4 \left(\frac{2\pi}{a}\right)^3## and not ##\frac{1}{4}\left(\frac{2\pi}{a}\right)^3##?

__FCC__In real space, it has a simple cubic lattic with 3 basis. Total number of atoms per unit cell = 4. Volume of primitive unit cell is then ##\frac{1}{4}a^3##.

In reciprocal space, FCC becomes a BCC structure. It has a simple cubic lattice of length ##\frac{2\pi}{a}## with 2 atoms in total. Why is the total volume then ##2 \left(\frac{2\pi}{a}\right)^3## and not ##\frac{1}{2}\left(\frac{2\pi}{a}\right)^3##?