# Volume of BCC and FCC?

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1. Jun 2, 2015

### unscientific

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

BCC

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 $\frac{1}{2}\left(\frac{2\pi}{a}\right)^3$?

2. Jun 2, 2015

### nasu

What is the the magnitude of the unit vectors of the reciprocal lattice? I don't think the size of the reciprocal cube for BCC is 2pi/a.