Past exam of the soft condensed matter course of the University of Cambridge / 2012 / Paper 1 / 1

Use the squared three-dimensional endpoint displacements, not their arithmetic mean. The mean squared displacement estimate and corresponding diffusion coefficient areThe factor six is for three-dimensional Brownian motion. For an isolated no-slip sphere, the Stokes–Einstein relation gives the Brownian radius estimate from endpoint displacements:Here room temperature is taken as $298\mathrm{K}$; choosing $293\mathrm{K}$ gives $55\mathrm{nm}$, a much smaller difference than the sampling uncertainty. Under independent increments with a normal distribution, $\mathrm{sd}(\hat{D})/D=\sqrt{2/(3n)}\simeq 0.37$ for $n=5$. The sampling uncertainty of an endpoint diffusivity estimate therefore makes this a rough radius estimate. It assumes negligible drift and hydrodynamic coupling between probes, the supplied bulk viscosity, and actual three-dimensional tracking. Localization noise adds to the squared displacements and would bias this inferred radius downward if uncorrected.

For the three-dimensional worm-like chain, define the bending coefficient ${\kappa }_b$ by $F_b=({\kappa }_b/2)\int |{\partial }_s\mathbf{t}{|}^{2}ds$. Its persistence length is $l_p={\kappa }_b/(k_{B}T)$. Holding that coefficient and the polymer contour length $L$ fixed givesThe temperatures must be absolute. The end-to-end distance of a worm-like chain obeysFor a long flexible chain, $L\gg l_p$, the typical size is $R_{\mathrm{rms}}\simeq \sqrt{2L{l}_p}$, so warming contracts the chain, with $R_{\mathrm{rms}}(T_2)/R_{\mathrm{rms}}(T_1)\simeq \sqrt{T_1/T_2}$. The finite-length formula also decreases as $l_p$ decreases, but the rod limit $L\ll l_p$ changes very little. A large segment count alone does not guarantee $L\gg l_p$. If the material's bending coefficient itself depends on temperature, replace the first ratio by $[{\kappa }_b(T_2)/{\kappa }_b(T_1)]T_1/T_2$; the sign then needs that additional material information. This calculation concerns a generic physical filament, with no biomolecular application.

For monovalent mobile ions, the linearized Debye–Hückel screening length isHere $n_i$ is a number density in ${\mathrm{m}}^{-3}$, not a molarity. Use the question's idealized neutral-water convention, $c_{+}=c_{-}=10^{-7}\mathrm{mol}{\mathrm{L}}^{-1}$, and $T=293.15\mathrm{K}$. Both ion species contribute:For added salt concentration $c_s$ in $\mathrm{mol}{\mathrm{L}}^{-1}$, the ionic strength in this model is $I=c_s+10^{-7}\mathrm{mol}{\mathrm{L}}^{-1}$. With the numerical molarity $I_M=I/(1\mathrm{mol}{\mathrm{L}}^{-1})$,The water contribution is negligible at that salt level. The factor $10^3$ converts molarity into $\mathrm{mol}{\mathrm{m}}^{-3}$; omitting it or counting only one ion species gives a wrong screening length. Exact water autoionization and activities vary with temperature, so the first estimate uses the simplified neutral $10^{-7}\mathrm{M}$ concentrations implied by the question. At the higher salt level, ion-size effects, nonideal activities and changes in permittivity limit the numerical accuracy of this continuum estimate.