 Solution

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.