---
title: "Volume change in iron allotropy"
subject: "Material Science"
chapter: 6
key: "semester_3/material_science:volume-change-bcc-to-fcc-iron"
source: https://www.bctengineer.com/q/semester_3/material_science/volume-change-bcc-to-fcc-iron
site: BCT Engineering
---

# Volume change in iron allotropy

Material Science · Chapter 6: Iron – Iron Carbide diagram and Heat Treatment of Steels

- Asked in: year not given, 5 marks
- Video: [BCC to FCC transformation in Iron (Problem discussion)](https://www.youtube.com/watch?v=dwfoCMmS5Bo) — C Patel Metallurgy & Chemistry [IISc Bangalore]
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/volume-change-bcc-to-fcc-iron
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-6

## Question

Iron changes from BCC (alpha) to FCC (gamma) at 912 °C. The lattice constants at this temperature are 0.2903 nm for BCC and 0.3647 nm for FCC. Calculate the percentage volume change on transformation and the atomic radius in each structure. What is the practical effect of this change?

## Short answer

Volume per atom in BCC is 0.012232 nm³, in FCC 0.012127 nm³, giving a 0.86% contraction on heating. Atomic radii: BCC 0.1257 nm, FCC 0.1289 nm. This volume change causes distortion and cracking during cooling.

## Answer

**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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the volume change on BCC to FCC transformation found?
   A. Compare unit-cell volumes a^3 directly
   B. Compare volume per atom: a^3/2 for BCC and a^3/4 for FCC
   C. Compare atomic radii only
   D. Compare a^3/4 for BCC and a^3/2 for FCC
   Answer: B. BCC has 2 atoms and FCC has 4 atoms per cell, so volume per atom is a^3/2 and a^3/4, respectively. (Key point: Method)

2. Which formulas give the atomic radius in BCC and FCC?
   A. BCC: sqrt(2)a/4; FCC: sqrt(3)a/4
   B. BCC: a/2; FCC: a/4
   C. BCC: sqrt(3)a/4; FCC: sqrt(2)a/4
   D. BCC: sqrt(3)a/2; FCC: sqrt(2)a/2
   Answer: C. Atoms touch along the body diagonal in BCC (4R = sqrt3 a) and along the face diagonal in FCC (4R = sqrt2 a). (Key point: Atomic radius formulas)

3. What is the result on heating through 912 °C (alpha to gamma)?
   A. About 0.86% contraction
   B. About 0.86% expansion
   C. About 8.6% contraction
   D. No change in volume
   Answer: A. V per atom falls from 0.012232 to 0.012127 nm^3, a contraction of 0.86% (an expansion on cooling). (Key point: Sign and size of change)

## Related in Material Science

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