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A convex liquid surface has additional surface free energy. The resulting Kelvin equation for a pure spherical water droplet isHere 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: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. ^[ref]

For $A,B>0$, differentiating the simplified Köhler theory curve givesThe derivative is positive below $r_c$ and negative above it, so $r_c$ is the Köhler activation radius, the unique maximum. At that point $B/r_c^3=A/(3r_c)$, givingThe curve crosses $S=1$ at $r=\sqrt{B/A}=r_c/\sqrt{3}$, reaches $S_c$ at $r_c$, and approaches $1$ from above as $r\to \infty$. Its formal divergence to negative values as $r\to 0$ is outside the dilute-chemical solution approximation; a finite dry solute core and composition-dependent properties must replace that extrapolation.

At the Köhler activation radius, the two coefficients satisfyConverting the atmospheric supersaturation to a fraction gives $S_c-1=0.004$, not $0.4$. Henceand the supplied curvature coefficient gives the temperatureThe coefficient and the stated radius consistently use a radius convention; replacing radius by diameter without changing the coefficient would introduce a factor-of-two error.