Critical nucleus radius for copper

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.