---
title: "Critical nucleus radius for copper"
subject: "Material Science"
chapter: 5
key: "semester_3/material_science:critical-nucleus-radius-copper"
source: https://www.bctengineer.com/q/semester_3/material_science/critical-nucleus-radius-copper
site: BCT Engineering
---

# Critical nucleus radius for copper

Material Science · Chapter 5: Solidification, Phase Relations and Strengthening Mechanism

- Asked in: year not given, 5 marks
- Video: [Materials - Chapter 4 - Critical radius](https://www.youtube.com/watch?v=PCRzFRSBSAM) — iylia elena
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/critical-nucleus-radius-copper
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-5

## Question

Derive an expression for the critical radius of a nucleus in homogeneous nucleation. For pure copper with undercooling of 236 K, melting point 1356 K, latent heat of fusion $1.628\times10^{9}$ J/m$^3$ and solid-liquid surface energy 0.177 J/m$^2$, calculate the critical radius, the critical free-energy barrier and the number of atoms in the critical nucleus (FCC, a = 0.3615 nm).

## Short answer

r* = 2γTm/(ΔHf ΔT) = 1.25 nm; ΔG* = 16πγ^3 Tm^2/(3 ΔHf^2 ΔT^2) = 1.16×10^{-18} J. Nucleus volume 8.17 nm^3; FCC Cu (a=0.3615 nm) volume/atom = a^3/4 = 0.0118 nm^3, so about 690 atoms.

## Answer

#### Derivation
For a spherical nucleus of radius $r$, the total free-energy change is volume term (negative) plus surface term (positive):

$$
\Delta G = \frac{4}{3}\pi r^3 \Delta G_v + 4\pi r^2 \gamma
$$

where $\Delta G_v = -\dfrac{\Delta H_f\,\Delta T}{T_m}$ (per unit volume) and $\gamma$ is the solid-liquid surface energy. $\Delta G$ is maximum at $r^*$:

$$
\frac{d\Delta G}{dr} = 4\pi r^2 \Delta G_v + 8\pi r\gamma = 0
$$

$$
r^{*} = \frac{-2\gamma}{\Delta G_v} = \frac{2\gamma T_m}{\Delta H_f\,\Delta T}
$$

Substituting back gives the barrier

$$
\Delta G^{*} = \frac{16\pi\gamma^3 T_m^2}{3\,\Delta H_f^2\,\Delta T^2}
$$

A bigger undercooling lowers both $r^*$ and $\Delta G^*$.

#### Numerical
$$
r^{*} = \frac{2 \times 0.177 \times 1356}{(1.628\times10^{9})(236)} = \frac{480.0}{3.842\times10^{11}} = 1.249\times10^{-9}\ \text{m} = 1.25\ \text{nm}
$$

$$
\Delta G^{*} = \frac{16\pi (0.177)^3 (1356)^2}{3 (1.628\times10^{9})^2 (236)^2} = 1.16\times10^{-18}\ \text{J}
$$

Volume of nucleus:

$$
V = \frac{4}{3}\pi (1.249)^3 = 8.17\ \text{nm}^3
$$

Volume per atom in FCC copper $= \dfrac{a^3}{4} = \dfrac{(0.3615)^3}{4} = 0.0118\ \text{nm}^3$.

$$
N = \frac{8.17}{0.0118} \approx 692\ \text{atoms}
$$

**Answer: $r^* = 1.25$ nm, $\Delta G^* = 1.16\times10^{-18}$ J, about 690 atoms in the critical nucleus.**

## Check yourself

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

1. Critical radius r* =
   A. 2 gamma Tm / (dHf dT)
   B. gamma / dT
   C. 16 pi gamma / 3
   D. dHf dT / (2 gamma Tm)
   Answer: A. Setting dDG/dr = 0 gives r* = -2 gamma/DGv. (Key point: Critical radius)

2. Larger undercooling does what?
   A. Raises r* and barrier
   B. Lowers both r* and barrier
   C. Raises r* only
   D. Nothing
   Answer: B. Both r* and DG* fall with dT. (Key point: Undercooling effect)

3. r* for copper is approx
   A. 0.125 nm
   B. 12.5 nm
   C. 1.25 nm
   D. 125 nm
   Answer: C. 2(0.177)(1356)/(1.628e9 x 236) = 1.25 nm. (Key point: Value)

4. Which nuclei are stable?
   A. r < r*
   B. r = 0
   C. Any r
   D. r > r*
   Answer: D. Clusters larger than r* lower their free energy by growing. (Key point: Stability)

## Related in Material Science

- [Solidification: nucleation and grain growth](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Asolidification-nucleation-and-grain-growth) — Material Science, chapter 5

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