Lattice constant of a simple cubic structure

The density of a crystal equals the mass of the atoms in one unit cell divided by the volume of the cell.

Let

• $N$ = number of atoms per unit cell,
• $M$ = atomic weight (mass of one mole of atoms, kg/mol),
• $N_A$ = Avogadro's number,
• $a$ = lattice constant, $\rho$ = density.

Proof

Mass of one atom:

$$m_{atom} = \frac{M}{N_A}$$

Mass of atoms in one unit cell:

$$m_{cell} = N\cdot\frac{M}{N_A}$$

The unit cell is a cube of side $a$, so its volume is $V_{cell} = a^3$.

Density:

$$\rho = \frac{m_{cell}}{V_{cell}} = \frac{NM}{N_A\,a^3}$$

Rearranging:

$$a^3 = \frac{NM}{\rho N_A} \qquad\Rightarrow\qquad a = \left[\frac{NM}{\rho N_A}\right]^{1/3}$$

Hence proved. For a simple cubic cell, $N = 8\times\frac{1}{8} = 1$; the same relation holds for BCC ($N = 2$) and FCC ($N = 4$) with the proper $N$.

Example

Copper (FCC): $N = 4$, $M = 63.55\times10^{-3}$ kg/mol, $\rho = 8960$ kg/m³:

$$a = \left[\frac{4\times63.55\times10^{-3}}{8960\times6.022\times10^{23}}\right]^{1/3} = 3.61\times10^{-10}\ \text{m} = 0.361\ \text{nm}$$

which agrees with the measured lattice constant of copper (0.3615 nm).