Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2023 / Paper 2 / Section B / 23

At terminal speed the downward buoyant weight balances the Stokes drag:Positive ${\rho }_p-{\rho }_f$ gives downward settling; a lighter particle would rise. With $a=10^{-7}\mathrm{m}$, ${\rho }_p-{\rho }_f=200\mathrm{kg}{\mathrm{m}}^{-3}$, $\eta =10^{-3}\mathrm{Pa}\mathrm{s}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,about $4.4\mathrm{nm}{\mathrm{s}}^{-1}$, or $0.38\mathrm{mm}$ per day.

The very small Reynolds number makes inertial corrections to Stokes flow negligible for an isolated particle. On short particle-scale distances, Brownian motion is much faster than this systematic drift. Nevertheless the nonzero settling speed can produce concentration gradients over sufficiently long times and large sample heights; Brownian motion does not cancel gravity.

The sedimentation scale height simplifies because the same drag coefficient controls both drift and diffusion:Applying the given criterion $a\le z_0$ givesEquivalently, the gravitational Péclet number on the particle-radius scale is $a/z_0=m_{b}ga/(k_{B}T)$ and must be at most one.

At $300\mathrm{K}$ the limiting radius is $a_{\max }\simeq 8.42\times 10^{-7}\mathrm{m}=0.84\mu \mathrm{m}$, corresponding to diameter approximately $1.68\mu \mathrm{m}$. Viscosity cancels from this equilibrium criterion, although it affects the time to approach equilibrium.

For the $100\mathrm{nm}$ radius particle in part (i), $D\simeq 2.20\times 10^{-12}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $z_0\simeq 0.50\mathrm{mm}$, about $5.0\times 10^3$ particle radii. It is therefore firmly diffusion-dominated on its own size scale. This does not mean uniform concentration over an arbitrarily tall vessel: the equilibrium density still decays on the finite scale $z_0$. The supplied size criterion is an approximate gravity-versus-diffusion criterion, not a universal definition of every stable colloid.

For ideal hard spheres, potential energy is zero for nonoverlapping configurations and infinite for overlaps. At fixed temperature and volume, the accessible fluid and crystal configurations therefore have no energetic preference of the ordinary bonding kind: the phase with larger entropy has lower Helmholtz free energy.

Ordering particle centers costs some configurational freedom, but at a high packing fraction it can give each particle more usable cage/free volume than a disordered crowded arrangement. The total number of allowed vibrational and positional configurations can then be greater in a crystal despite its visible long-range order. This is entropy-driven hard-sphere crystallization.

The crystal need not be at close packing. At the stated volume fraction near $0.55$, substantial free space remains for thermal motion. Close packing near $0.74$ is the geometric upper bound for equal spheres, not the fraction at which the crystal first becomes thermodynamically favorable. Treating crystallization as an inevitable decrease in total entropy would confuse visible positional order with the full phase-space count.

Let $A$ be projected area. Close-packed circular footprints cover $\varphi A$ and the exposed flat substrate covers $(1-\varphi )A$. A hemisphere's curved area is twice its circular footprint, so the actual wetted solid area isThe Wenzel roughness factor is therefore $r=A_{\mathrm{wet}}/A=1+\varphi =1.9069$, not $\varphi$ or $2$.

Using the Wenzel wetting relation,The nanostructure makes this surface more wettable, because planar PMMA already has $\theta <90^{\circ }$. Under complete Wenzel wetting, roughening amplifies the underlying hydrophilic character; it does not turn this material hydrophobic.

The calculation assumes identical chemistry on spheres and substrate, complete liquid penetration, ideal hemispheres, a macroscopic contact angle and negligible pinning/line-tension effects. Actual dried particles may have different contact geometry, deformation, contamination or trapped air. A hydrophobic coating with planar angle above $90^{\circ }$ would make roughness increase the angle instead. Designing a texture that retains air in a Cassie–Baxter state is another route, but its stability against liquid penetration must be checked.

Define ${\gamma }_{SV}$, ${\gamma }_{SL}$ and ${\gamma }_{LV}$ as the solid-vapor, solid-liquid and liquid-vapor interfacial free energies per actual area. For the corresponding planar material, Young's contact-angle equation isConsider advancing the macroscopic liquid footprint over projected area $A$, while retaining the specified wetting only to each sphere's equator. The newly wetted upper hemispheres have actual area $2\varphi A$. The gaps occupy projected area $(1-\varphi )A$, across which a liquid-vapor interface remains instead of solid-liquid contact.

The microscopic free-energy change, apart from the outer macroscopic droplet surface, isEquating this with $-{\gamma }_{LV}\cos {\theta }^{\ast }$, the macroscopic wetting contribution, yieldsThe factor $2$ comes from the curved hemisphere area; the negative gap term is the trapped liquid-vapor interface. This combines Wenzel roughness on the wetted solid fraction with Cassie-type air gaps. It differs from both homogeneous Wenzel wetting and a Cassie formula that treats the solid contacts as flat.

For $\varphi =0.9069$ and $\theta =68^{\circ }$, this gives $\cos {\theta }^{\ast }\simeq 0.5864$ and ${\theta }^{\ast }\simeq 54.1^{\circ }$. The hybrid state is less wettable than the fully penetrated hemispherical model, but still not hydrophobic. The result is conditional on the assumed equatorial meniscus and retained air: pinning or metastability may be required, and allowing further penetration can change the equilibrium geometry.