Question

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).

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.

Answer 1

Correct Answer: 4

Step 1: Recall where the critical nucleus radius formula comes from.
When a spherical solid nucleus of radius \(r\) forms inside an undercooled liquid, the total change in Gibbs free energy has two competing parts: a volume term that favors solidification (negative, since \(\Delta G_v\) is negative below the melting point) and a surface term that opposes it (always positive, since creating a new solid-liquid interface costs energy).
\[ \Delta G(r) = \frac{4}{3}\pi r^3 \Delta G_v + 4\pi r^2 \gamma \]
Here \(\Delta G_v\) is the free energy change per unit volume of solidification and \(\gamma\) is the solid-liquid interfacial energy per unit area.

Step 2: Find the critical radius by maximizing \(\Delta G(r)\).
The curve \(\Delta G(r)\) rises to a maximum and then falls, because the negative volume term (growing as \(r^3\)) eventually overtakes the positive surface term (growing as \(r^2\)). A nucleus becomes stable, meaning it keeps growing rather than shrinking back, only once it crosses this maximum. That radius is the critical radius \(r^*\), found by setting the derivative of \(\Delta G(r)\) to zero:
\[ \frac{d(\Delta G)}{dr} = 4\pi r^2 \Delta G_v + 8\pi r \gamma = 0 \]
Taking the non-zero root,
\[ r^* = \frac{-2\gamma}{\Delta G_v} \]

Step 3: Substitute the given values.
\(\gamma = 0.1\) J/m\(^2\) and \(\Delta G_v = -0.5\times10^{8}\) J/m\(^3\). Since \(\Delta G_v\) is negative, the ratio \(-2\gamma/\Delta G_v\) comes out positive, as a radius should.
\[ r^* = \frac{-2(0.1)}{-0.5\times10^{8}} = \frac{0.2}{0.5\times10^{8}} = 4\times10^{-9} \text{ m} \]

Step 4: Convert to nanometers.
\[ r^* = 4\times10^{-9} \text{ m} = 4 \text{ nm} \]

Step 5: Final Answer.
The critical nucleus radius is \(4\) nm. If a solver instead reports the critical nucleus size as the diameter (\(2r^*\)) rather than the radius, the value doubles to \(8\) nm; the official answer key accepted both \(4\) and \(8\) because of this radius-versus-diameter reading of "nucleus size". Working from the standard definition of the critical radius, the direct answer is \(4\) nm.
\[ \boxed{r^* = 4 \text{ nm}} \]