Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S12 / 2

Let $U(x)$ be the static potential and $\gamma$ the drag coefficient, so terminal Genetic drift under force $-U^{\prime }$ is $v=-U^{\prime }/\gamma$. The probability current is $J=-D{U}^{\prime }p/(k_{B}T)$ only after imposing the desired relation; instead start independently from $J=-D{p}^{\prime }-U^{\prime }p/\gamma$. At thermal equilibrium $p\propto e^{-U/(k_{B}T)}$ and detailed balance requires $J=0$. Substitution yieldsAt 298 K with $a=0.5\mu$m, this gives approximately $4.37\times 10^{-13}$ m${}^2$ s${}^{-1}$. The argument assumes an overdamped particle in an equilibrium thermal bath, rather than arbitrary driven diffusion.

Write the displacement after $m$ independent isotropic zero-mean steps as $\mathbf{R}={\sum }_j{\mathbf{s}}_j$. Cross terms vanish, giving $\boxed{⟨R^2⟩=m\Delta }$ when $\Delta =⟨|\mathbf{s}{|}^2⟩$. Isotropy makes each Cartesian variance one third of this. Since $m\simeq kt$ and $⟨R^2⟩=6Dt$, $\boxed{D=k\Delta /6}$. If “step size” were instead defined as a single-coordinate mean square, the corresponding expression would be $k\Delta /2$; the factor of three is purely the stated three-dimensional convention.

The stationary occupation ratio is $p_1/p_2=2$. Balance of the two probability currents gives $p_1{k}_{12}=p_2{k}_{21}$, and Boltzmann occupation of the trapped macrostates gives $p_2/p_1=e^{-(G_2-G_1)/(k_{B}T)}$. ThereforeThe sketch has two wells separated by 1 micrometre, with the second free-energy level higher by $k_{B}T\log 2$ and an intervening escape barrier. Free energy includes the volume of each trapped basin; equal minimum potential heights alone would not ensure equal occupations if basin shapes differed.

In a sufficiently uniform flow the Stokes force is $F=\gamma v$ toward trap 2. Its work over separation $d$ lowers the second effective well by $Fd$. Equal effective free energies require $\gamma vd=k_{B}T\log 2$. Henceat 298 K. The temperature was not specified, so the numerical result scales as $T/298\mathrm{K}$. This approximation treats a weak uniform drag as a constant tilt and neglects changes in barrier shape and hydrodynamic interactions with boundaries.