A given volume of liquid is undercooled just below the melting temperature to form a spherical solid nucleus (consider homogeneous nucleation). The Gibbs free energy of solidification (\(\Delta G_v\)) is \((-0.5 \times 10^{8})\) J/m\(^3\). The solid-liquid interfacial energy (\(\gamma\)) is isotropic and its value is \(0.1\) J/m\(^2\).
The critical nucleus size for a stable nucleus is _______ nm (answer in integer).
Show Hint
Use the critical nucleus radius formula for homogeneous nucleation, \(r^* = -2\gamma/\Delta G_v\), where \(\gamma\) is the interfacial energy and \(\Delta G_v\) is the volumetric free energy of solidification.
Step 1: Use a mechanical equilibrium argument instead of calculus.
There is a quicker way to reach the critical radius without differentiating the free energy curve: think of the balance between the driving force pushing the interface to grow and the resisting pressure from surface tension, similar to the Young-Laplace pressure across a curved interface.
Step 2: Write the two pressures.
The volumetric driving force for solidification, per unit volume, is $|\Delta G_v|$ (we use the magnitude since $\Delta G_v$ is negative and favors the solid). The resisting curvature pressure from the interfacial energy $\gamma$ on a sphere of radius $r$ is the Laplace pressure,
\[
\Delta P = \frac{2\gamma}{r}
\]
Step 3: Set the two equal at the critical radius.
At the critical radius $r^*$, the driving force per unit volume exactly balances the resisting curvature pressure, the same condition reached by setting the derivative of $\Delta G(r)$ to zero:
\[
|\Delta G_v| = \frac{2\gamma}{r^*} \quad\Rightarrow\quad r^* = \frac{2\gamma}{|\Delta G_v|}
\]
This matches the usual formula $r^*=-2\gamma/\Delta G_v$ once we account for the sign of $\Delta G_v$.
Step 4: Plug in the numbers.
$\gamma=0.1$ J/m$^2$ and $|\Delta G_v|=0.5\times10^{8}$ J/m$^3$.
\[
r^* = \frac{2(0.1)}{0.5\times10^{8}} = \frac{0.2}{0.5\times10^{8}} = 4\times10^{-9}\text{ m} = 4\text{ nm}
\]
Step 5: Note the alternate reading.
If "nucleus size" is taken to mean the diameter rather than the radius, the number to report becomes $2r^*=8$ nm, which is why the official key marked both $4$ and $8$ as correct. Using the standard radius definition, the answer is $4$ nm.
\[
\boxed{r^*=4\text{ nm}}
\]
