= Solution
A convex liquid surface has additional surface free energy. The resulting <Kelvin equation> for a pure spherical water droplet is
$$
\boxed{\frac{e_s(r)}{e_s(\infty)}=\exp\left(\frac{2\gamma M_w}{\rho_wRTr}\right)>1.}
$$
Here gamma is <surface tension> and $M_w$ is molar mass. Smaller droplets therefore need a higher environmental vapour pressure to avoid evaporating; forming pure-water embryos by homogeneous condensation has a large nucleation barrier.
An atmospheric droplet normally forms on a <cloud condensation nucleus>. Dissolved salt reduces <water activity> and hence equilibrium vapour pressure, opposing the curvature effect. <Köhler theory> combines them:
$$
\frac{e_{\rm eq}}{e_s(\infty)}=a_w e^{A/r},\qquad
\ln\frac{e_{\rm eq}}{e_s(\infty)}\simeq\frac Ar-\frac B{r^3},\quad A=\frac{2\gamma M_w}{\rho_wRT},\quad B>0.
$$
Using the radius form, with <radius and diameter conventions in Köhler activation> kept distinct, the <Köhler activation radius> is $r_c=\sqrt{3B/A}$, with $\ln S_c=2A/(3r_c)$. Supersaturation above this maximum activates a wet aerosol into a growing cloud droplet. Solubility, particle size and wettability determine the threshold; not every particle activates. Cooling by ascent commonly supplies the necessary modest supersaturation. <A single parameter representation of hygroscopic growth and cloud condensation nucleus activity by M. D. Petters and S. M. Kreidenweis>[2007, curvature/solute competition and critical activation; radius convention explicitly used here]{ref}
\Image[/past-exam-of-the-atmospheric-physics-course-of-the-university-of-cambridge/2012/paper-1-ap2012-kohler-activation.png]
{title=Köhler activation}
{source=paper-1-ap2012-kohler-activation.py}
Back to article page
