The nucleus radius above which growth becomes the more stable path in nucleation. For a spherical nucleus, r* = −2γ/ΔGv, which becomes smaller as the undercooling increases.
Japanese name: 臨界核半径
Critical nucleus radius appears in 2 practice questions across 1 topic, including Thin films, surfaces and nanomaterials.
A small cluster of atoms (an embryo) that forms in a liquid or a supersaturated parent phase lowers its energy in proportion to its volume, but raises it by the interfacial energy in proportion to its surface area. For a sphere, the free energy change ΔG = (4/3)πr³ΔGv + 4πr²γ first increases at small r, where the surface term wins, reaches a maximum at a certain radius and then decreases. The position of this maximum is the critical nucleus radius r*. An embryo smaller than r* lowers its energy by shrinking, and a nucleus larger than r* by growing, so r* marks the boundary between survival and disappearance.
In solidification, the free energy change per unit volume can be approximated as ΔGv ≈ −ΔHf ΔT/Tm (ΔHf is the heat of fusion per unit volume and ΔT the undercooling), so r* = 2γTm/(ΔHf ΔT), inversely proportional to the undercooling. Taking gold as an example, with γ = 0.132 J/m², ΔHf = 1.16×10⁹ J/m³ and Tm = 1337 K, r* is about 3.0 nm at ΔT = 100 K and about 1.5 nm at 200 K. Because the nucleation barrier ΔG* is inversely proportional to ΔT², a small increase in undercooling raises the nucleation rate by orders of magnitude.
In heterogeneous nucleation on a mold wall or an impurity particle, the nucleus is a spherical cap with contact angle θ. Its radius of curvature at the critical state is the same r* as in homogeneous nucleation; what decreases is the volume of the nucleus and the barrier. The barrier is multiplied by f(θ) = (2 + cosθ)(1 − cosθ)²/4, which is about 0.16 at θ = 60° and about 0.013 at 30°. Also, r* is the minimum size at which a nucleus becomes stable and is different from the size of grains or precipitates after transformation. The final grain size is set by the number of nuclei formed and how fast they then grow.
Adding an inoculant in casting refines the grains because it supplies substrates with a small contact angle, lowering the heterogeneous nucleation barrier so that many nuclei form even at small undercooling. In precipitation heat treatment, lowering the aging temperature increases the supersaturation (driving force), which makes r* smaller and gives fine precipitates at high density. The same idea applies to thin-film growth: lowering the substrate temperature or raising the deposition rate increases the supersaturation and shrinks the critical nucleus, so films tend to form with a high island density and a fine microstructure.
Answer: C. About 1.0 nm
Differentiating ΔG with respect to r and setting it to zero gives r* = −2γ/ΔGv, so r* = 2 × 0.10/(2.0×10⁸) = 1.0×10⁻⁹ m = 1.0 nm. About 0.5 nm drops the factor of 2, giving −γ/ΔGv, and about 2.0 nm mistakes the diameter of the critical nucleus (2r*) for its radius. Embryos smaller than r* lower their energy by shrinking and tend to disappear, and only nuclei larger than r* keep growing. The greater the undercooling, the larger |ΔGv|, so r* becomes smaller and the nucleation barrier ΔG* = 16πγ³/(3ΔGv²) also drops sharply, making nucleation easier.
Answer: C. It is the same
A heterogeneous nucleus has the shape of a spherical cap (part of a sphere) sitting on the substrate with contact angle θ. Its radius of curvature at the critical state is determined only by the interfacial energy and the free energy change per unit volume, r* = −2γ/ΔGv, the same as for homogeneous nucleation. The difference is that, at the same radius of curvature, the spherical cap has a smaller volume than a sphere, so the number of atoms needed and the energy barrier are reduced by the factor f(θ). It is easy to confuse the lowered barrier with a smaller radius, so be careful. Nor does the presence of a substrate make the critical radius larger.
Find the radius at which the free energy change of a spherical nucleus, ΔG = (4/3)πr³ΔGv + 4πr²γ, is a maximum. Setting dΔG/dr = 0 gives r* = −2γ/ΔGv, and substituting this back into ΔG gives the barrier height ΔG* = 16πγ³/(3ΔGv²).
It depends on the material and the undercooling. For gold at ΔT = 100 K, r* is about 3 nm, and a sphere of that size holds about 7000 atoms. At ΔT = 200 K the radius halves and the number of atoms drops to about 1/8, a little under 1000.
As ΔT approaches 0, ΔGv also approaches 0, so r* becomes infinitely large and no nucleus can form. This is why solidification does not start exactly at the melting point and some undercooling is required.
Ostwald ripening Quantum size effect Stranski–Krastanov growth Thermal stress