Volume change in iron allotropy

Method: compare volume per atom in the two structures.

BCC has 2 atoms per cell and FCC has 4 atoms per cell.

Volume per atom

$$V_{BCC} = \frac{a^3}{2} = \frac{(0.2903)^3}{2} = \frac{0.02446}{2} = 0.012232\ \text{nm}^3$$ $$V_{FCC} = \frac{a^3}{4} = \frac{(0.3647)^3}{4} = \frac{0.04851}{4} = 0.012127\ \text{nm}^3$$

Volume change

$$\Delta V\% = \frac{V_{FCC} - V_{BCC}}{V_{BCC}}\times100 = \frac{0.012127 - 0.012232}{0.012232}\times 100 = -0.86\%$$

The negative sign means a contraction of about 0.86% on heating through 912 °C (and an expansion on cooling).

Atomic radius

$$R_{BCC} = \frac{\sqrt3\, a}{4} = \frac{1.732 \times 0.2903}{4} = 0.1257\ \text{nm}$$ $$R_{FCC} = \frac{\sqrt2\, a}{4} = \frac{1.414 \times 0.3647}{4} = 0.1289\ \text{nm}$$

The FCC structure is denser (APF 0.74 against 0.68) but the atom appears larger because of coordination 12 against 8; the net effect is a small contraction.

Practical effect

When steel cools through the $\gamma\rightarrow\alpha$ change, the volume change causes distortion and internal stresses, and in quenching can lead to cracking. Also, since FCC iron dissolves much more carbon (2.14% against 0.022%), steel is austenitised before hardening.

Answer: volume contraction of 0.86% on $\alpha\rightarrow\gamma$; $R_{BCC} = 0.1257$ nm, $R_{FCC} = 0.1289$ nm.