Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5

In the Asakura-Oosawa model, nonadsorbing polymers are treated as ideal, mutually penetrable particles, but their centers cannot enter a layer around a hard colloid. The polymer osmotic pressure is $\Pi =\rho k_{\mathrm{B}}T$ in the ideal dilute limit, with the appropriate number density in the space dimension considered.

For fixed polymer number, reducing the inaccessible volume increases the available volume and lowers the polymer Helmholtz free energy. Equivalently, at fixed polymer chemical potential, the polymer contribution to the effective colloid potential isWhen the excluded layers overlap, their union occupies less space than two separate layers. More polymer configurations become available, so the resulting depletion force attracts the colloids. It is an entropy-driven effective interaction, even though each microscopic colloid–polymer interaction is purely repulsive.

The unrotated squares have a gap $d>a$, so their facing exclusion layers do not overlap. Rotation can bring them into contact even though their centers remain fixed. The strongest depletion force comes from parallel faces in contact over a long shared length, rather than a corner-to-corner contact.

Let $r=\ell +d$. Rotate both squares through the same angle $\theta$ relative to their original orientation. Their center displacement has a face-normal component $r\cos \theta$ and a tangential component $r\sin \theta$. Face contact requiresThe contacting faces overlap for lengthThese two mirror-related orientations, together with equivalent square rotations, maximize the leading overlap area when $a<d\ll \ell$. A small polymer gap sacrifices overlap width without a compensating leading increase in contact length.

Each face excludes polymer centers out to distance $a/2$. At face contact, the combined overlap strip has width $a$ and length $L$. Ignoring end corrections of order $a^2$, the two-dimensional Asakura-Oosawa model therefore givesThe final expression retains only the leading term in $d/\ell$. Its dimensions are correct because $\rho$ is a number per area. The zero of this effective potential energy is the well-separated, nonoverlapping-exclusion-layer configuration.

There is no universal ranking from the information supplied. The depletion force scales with polymer osmotic pressure, hence with $\rho$ in the ideal model. London dispersion forces instead depend on the material-dependent Hamaker constant, the intervening solvent, the closest approach and the contact geometry.

Close face contact can give a strong dispersion contribution; the ideal continuum expression even becomes singular without a molecular-distance cutoff. Nevertheless, refractive-index matching can greatly reduce that contribution, and increasing polymer concentration can strengthen depletion force. A quantitative comparison requires those missing material and concentration scales. The two attractive contributions may act together.

Write $q=P_{BI}$ and $p=P_{AI}=1-q$. A path with $n$ visits to $I$ makes $n-1$ unsuccessful excursions back to $B$, followed by one successful step to $A$. Every $B\to I$ step has probability one, soThe visit count has a geometric distribution, with success meaning the final $I\to A$ step.

For $p>0$, equivalently $0\le q<1$, the geometric series givesThus absorption probability is one. The qualification matters: if $p=0$, the process alternates between $B$ and $I$ forever and never reaches $A$.

Each visit to $I$ corresponds to two steps, one out of $B$ and one out of $I$. Hence a successful path with $n$ visits contains $2n$ steps. Taking its expectation over the geometric distribution givesThe last pair is $B\to I\to A$; it contributes two steps just like each previous $B\to I\to B$ excursion.

Using the differentiated geometric series,This is finite for $p>0$, equals two when every visit succeeds, and diverges as the chance of success tends to zero.

A path with $n$ visits makes $n$ departures from $B$ and $n$ departures from $I$. Its conditional mean duration is therefore $n({\tau }_B+{\tau }_I)$. Averaging over the geometric distribution,Only the mean waiting times are needed here: linearity of expectation adds their contributions without requiring each residence time to equal its mean.

The same geometric series identity givesA direct renewal check gives the identical result: $T_B={\tau }_B+T_I$ and $T_I={\tau }_I+q{T}_B$, so $p{T}_B={\tau }_B+{\tau }_I$. This also checks the units: mean first-passage time has units of time, unlike the dimensionless step count.