Using $\delta(R)=\sigma/(\alpha\phi_BR)$, the growth law is
$$
\dot R=\frac{\alpha M\varepsilon}{2\phi_BR}
-\frac{M\sigma}{2\phi_B^2R^2}.
$$
For
$$
F(R)=4\pi\sigma R^2-\frac{4\pi}{3}\Delta R^3,
\qquad
\Delta=2\phi_B\alpha\varepsilon,
$$
one has $F'=8\pi\sigma R-4\pi\Delta R^2$. Therefore, with $\mathcal M(R)=M/(16\pi\phi_B^2R^3)$,
$$
\boxed{\dot R=-\mathcal M(R)\frac{dF}{dR}.}
$$
The first term in $F$ is the positive surface cost and the second is the negative bulk free-energy gain of converting a metastable volume. The nonzero stationary radius and barrier are the <classical nucleation theory> values
$$
\boxed{R^*=\frac{2\sigma}{\Delta}
=\frac{\sigma}{\phi_B\alpha\varepsilon},
\qquad
F^*=F(R^*)=\frac{16\pi\sigma^3}{3\Delta^2}.}
$$
Since $F''(R^*)=-8\pi\sigma<0$, this stationary point is a maximum.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 3 / d / Solution

{wiki}

Classical nucleation theory balances the positive surface free energy of a droplet against its negative bulk free-energy gain. In three dimensions,
$$
F(R)=4\pi\sigma R^2-\frac{4\pi}{3}\Delta R^3
$$
has critical radius $R^*=2\sigma/\Delta$ and barrier $F^*=16\pi\sigma^3/(3\Delta^2)$.


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