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An aerosol particle becomes a freely growing cloud droplet after crossing its Köhler theory growth barrier under the parcel's atmospheric supersaturation history. Counting particles whose critical supersaturation is below a specified peak gives an equilibrium-threshold estimate; calculating that peak requires cooling, water vapor depletion and condensational-growth kinetics. Neither total aerosol number nor dry size alone determines activation.

The first-order combination of the Kelvin equation and Raoult's law for a chemical solution droplet with $A/r\ll 1$ and $B/r^3\ll 1$. Its maximum is at $r_c=\sqrt{3B/A}$, with $S_c-1=2A/(3r_c)$. For ambient saturation ratio $1<S<S_c$, the smaller equilibrium radius is stable and the larger is unstable: the sign of condensational growth is the sign of $S-S_{\mathrm{eq}}(r)$. In particular, exceeding $r_c$ alone need not activate continued growth at subcritical atmospheric supersaturation.

A particle count per stated volume and logarithmic size interval. Its integral over $d\ln D$ gives a number density, unlike an aerosol mass distribution. In a cloud-droplet activation estimate, integrating the dry-size distribution above the size-dependent Köhler theory threshold gives activated number for a specified peak atmospheric supersaturation and composition.

Below cloud base, an initially unsaturated closed parcel expands as ambient pressure falls. In the absence of heat exchange its temperature falls approximately at the dry adiabatic lapse rate $g/c_{p,d}$. Its vapour mixing ratio is conserved, but its saturation mixing ratio decreases strongly as it cools. Hence relative humidity rises until the lifting condensation level is reached. The density normally decreases through expansion; the environmental profile determines the parcel's buoyancy and thus whether ascent can continue.

Near saturation, slight atmospheric supersaturation permits suitable cloud condensation nuclei to activate. Köhler theory describes the competition between curvature, which raises equilibrium vapour pressure, and dissolved solute, which lowers it. Cooling generates supersaturation, while condensation onto the new droplets consumes it. Their balance normally keeps in-cloud relative humidity close to, but not identically at, saturation. The activated droplet number depends on the aerosol population and the maximum supersaturation produced by the ascent.

Above cloud base, condensation lowers the vapour content and raises the liquid-water content. The latent heat released partly offsets expansion cooling, so the moist adiabat has a smaller lapse-rate magnitude than the dry one. In a dilute approximation,A closed adiabatic parcel conserves total water $q_t=q_v+q_l$ and approximately its moist entropy; retaining condensate makes this a reversible-cloud idealization, rather than a precipitating pseudoadiabat. In a shallow cloud, the saturation-mixing-ratio decrease and slow density variation yield nearly constant condensation per height, so liquid-water content is approximately linear above base. Once activation is complete, assuming nearly fixed droplet number per volume gives increasing droplet size as condensate accumulates. Entrainment, continued activation, collision-coalescence and atmospheric precipitation can all violate that simplified description; they are excluded from the stated retrieval model.

$S=e/e_s^l$ is the ambient saturation ratio over a flat liquid-water surface; $S-1$ is the liquid atmospheric supersaturation. Curvature increases the surface equilibrium vapour pressure, producing the Kelvin penalty $A/r$, while dissolved solute lowers it and gives the opposing Raoult contribution $B/r^3$. Thus the dilute Köhler equilibrium curve is $S_{\mathrm{eq}}-1=A/r-B/r^3$, and the numerator is the atmospheric supersaturation available to drive growth beyond surface equilibrium.

$F_d^l$ describes the resistance of vapour diffusion through the surrounding air. $F_k^l$ describes the thermal-conduction resistance to removing latent heat released by condensation: a warmed droplet surface has a higher equilibrium vapour pressure and hence a weaker mass-transfer drive. Common continuum approximations are $F_d^l\simeq R_{v}T/(D_v{e}_s^l)$ and $F_k^l\simeq L_v[k_{\mathrm{air}}T{]}^{-1}[L_v/(R_{v}T)-1]$, showing the roles of diffusivity, conductivity, saturation pressure and latent heat. Both resistances have the supplied units $\mathrm{s}\mathrm{m}\mathrm{k}{\mathrm{g}}^{-1}$ and add for this coupled transport problem.

For a spherical droplet of constant density, $m=4\pi \rho r^3/3$. Multiplying the radius equation by $4\pi \rho r$ givesAt sufficiently large radius, curvature and solute corrections are negligible, soThis limit also requires sustained positive atmospheric supersaturation. It need not describe the earliest submicrometre stage or an unactivated haze particle.

Use the large-radius condensational-growth approximation from part (a), with fixed ambient atmospheric supersaturation and transport resistances. It givesTaking $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $S-1=0.01$, and negligible initial $r_0^2$ givesThe squared-radius law makes condensation relatively fast for reaching cloud-droplet sizes but far too slow for producing millimetre rain within a typical evolving cloud. Collision–coalescence, or ice growth followed by aggregation/riming and melting, bypasses this bottleneck. In reality atmospheric supersaturation is shared by many particles and changes with ascent and vapour depletion, so even the constant-$1\%$ estimate is idealized.

These are growth times after activation, with negligible activation delay. The statement $r_0\ll 1\mu \mathrm{m}$ alone does not ensure that a seed grows: for pure water with $B=0$, a seed with $r_0<A/(S-1)$ has a negative growth rate and shrinks. For soluble seeds the Köhler barrier depends on $A,B$ and the environmental atmospheric supersaturation. The literal full-model time iswhich cannot be assigned a unique initial-stage numerical value without those microphysical parameters. The two boxed estimates are the intended continuum condensation scales for activated droplets, rather than a claim that every arbitrarily small seed reaches those sizes.

The target liquid-water mass isThe thin-plate ice-growth time uses the prescribed ice mass law, which therefore requires $r_{i\ast }=(m_{\ast }/\alpha {)}^{1/3}=45.31\mu \mathrm{m}$, not the same radius as the water droplet. With $C=2r_i/\pi$ and $F_i=F_k^i+F_d^i$,Hence, for negligible seed mass and constant conditions,This is about $8.24$ times faster than the liquid estimate to the same mass. Both laws have $dm/dt\propto m^{1/3}$ under the given shape assumptions. Ice benefits here from five times the atmospheric supersaturation and a larger geometrical capacitance for the same mass, partly offset by its slightly greater transport resistance. The mass coefficient $\alpha$ encodes the assumed plate shape; replacing it by bulk ice density would change the supplied model.

In a mixed-phase cloud, saturation vapour pressure over ice is lower than over supercooled liquid water. Air can therefore be subsaturated with respect to liquid yet supersaturated with respect to ice: droplets evaporate while ice crystals grow, transferring water through the vapour phase. This is the Wegener–Bergeron–Findeisen mechanism. Subsequent aggregation, riming and melting can produce atmospheric precipitation much faster than condensation alone.

The large difference is not a universal fixed ice/liquid time ratio. Atmospheric supersaturation evolves, latent heating increases thermal resistance, crystal habits and capacitance change, and surface attachment or ventilation may matter. Ice nucleation and the availability of freezing nuclei also precede the calculated growth. The boxed time describes the specified already-present crystals under maintained conditions, not a complete prediction of the time until rain begins.

The dew-point depression is $6\mathrm{K}$, soKeeping $T_{D,s}=276.15\mathrm{K}$, $H\simeq 158.5\mathrm{km}$ and $c_p=1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ gives approximately $747\mathrm{m}$, consistent with the rounded estimate.

The real cloud base can differ because entrainment changes the parcel's atmospheric mixing ratio and temperature, surface humidity varies, and the environmental pressure profile need not match the assumed parcel profile. Finite cloud-droplet activation requires some atmospheric supersaturation, so the first visible cloud droplets need not occur exactly at the thermodynamic lifting condensation level. A parcel must also actually be lifted that far; environmental static stability can prevent this. The constant-latent heat and first-order Taylor expansion approximations produce smaller numerical differences.

A rising parcel cools, lowering the saturation pressure. Once its water vapor supply exceeds that value, atmospheric supersaturation develops. Homogeneous nucleation of pure water is strongly inhibited at the small atmospheric supersaturations typical of clouds; existing aerosol particles provide cloud condensation nuclei. Their dry sizes and composition, together with the parcel's ascent rate, determine how many cloud droplets activate.

Köhler theory combines the Kelvin equation, which raises equilibrium vapor pressure over a curved surface, with Raoult's law, which lowers it for a chemical solution. In a dilute diameter convention, with dry diameter $D_d$ and wet diameter $D$,Here $s=e_v/E(T)-1$ is atmospheric supersaturation, ${\sigma }_w$ is surface tension, and ${\kappa }_h$ measures aerosol hygroscopicity, not the dry-adiabatic process exponent. The maximum of the Köhler curve is the critical supersaturation. Larger or more hygroscopic particles generally have lower thresholds. Cloud-droplet activation requires overcoming that maximum; total aerosol number is not the activated number.

Given an observed aerosol number size distribution $n_j(D_d)=d{N}_j/d\ln D_d$ for composition classes $j$, first assign each class a hygroscopicity and compute its critical supersaturation. If the parcel reaches maximum atmospheric supersaturation $s_{\max }$, the equilibrium-threshold estimate isFor each fitted lognormal distribution, this is its upper-tail integral; it can be evaluated analytically with an error function or numerically from measured bins.

The value $s_{\max }$ must be calculated, not guessed from the aerosol distribution alone. Prescribe parcel upward velocity $w$, initial pressure, temperature and humidity, then integrate expansion cooling, latent heat release, conservation of total water, and diffusional cloud droplet growth. Schematically,where $a_s,b_s>0$, $G$ includes water vapor diffusion and thermal conduction, and $q_l$ is liquid mass fraction. Rising motion produces atmospheric supersaturation; growing particles consume it. Their competition selects the peak and which particles cross their activation barriers. Assume a well-mixed closed parcel, liquid water, soluble spherical particles, specified mixing state, and negligible collision-coalescence and fallout during activation. Slow kinetic growth, weakly soluble particles or entrainment require a more detailed treatment than the threshold integral.

Let $1<S<S_c$ be the ambient saturation ratio. There are two equilibrium radii: $r_{-}<r_c<r_{+}$. Condensation occurs if $S>S_{\mathrm{eq}}(r)$; evaporation occurs if $S<S_{\mathrm{eq}}(r)$. Therefore droplets with $r<r_{-}$ grow towards $r_{-}$, while those with $r_{-}<r<r_c$ shrink towards $r_{-}$. The smaller equilibrium is stable.

Above $r_c$, droplets with $r_c<r<r_{+}$ still undergo evaporation and ultimately approach $r_{-}$. Droplets with $r>r_{+}$ grow, and the larger equilibrium $r_{+}$ is unstable. Merely exceeding $r_c$ is therefore not sufficient for continued growth at subcritical ambient atmospheric supersaturation. For example, if $(S-1)/(S_c-1)=0.6$, the two intersections have approximately $r_{-}/r_c=0.676$ and $r_{+}/r_c=2.349$; a droplet of radius $1.5r_c$ undergoes evaporation. If instead $S>S_c$, there is no equilibrium barrier in this approximation and droplets activate into growing cloud droplets.

Köhler theory explains activation of cloud condensation nuclei, rather than rapid production of atmospheric precipitation. Diffusion-controlled droplet growth is limited by diffusion of water vapor and removal of released latent heat; approximately $d(r^2)/dt$ is proportional to the ambient atmospheric supersaturation. It slows in radius and tends to narrow a droplet size distribution. The available water vapor, changing temperature and shared depletion of atmospheric supersaturation also invalidate unlimited growth at prescribed $S$. Formation of rain-sized drops normally requires collision-coalescence, while colder clouds can additionally use cloud ice nucleation versus ice growth.

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.

The ambient water vapor partial pressure divided by the equilibrium saturation pressure over a stated reference surface at the same temperature. Relative to flat pure water, atmospheric supersaturation is $s=S-1$. A curved solute-containing cloud droplet generally has a different equilibrium saturation ratio from unity.