 Solution

For radius $a=10\mathrm{nm}$ and gold density $2.0\times 10^4\mathrm{kg}{\mathrm{m}}^{-3}$, $m=8.38\times 10^{-20}\mathrm{kg}$. The supplied Stokes model gives $\zeta =3.96\times 10^{-12}\mathrm{kg}{\mathrm{s}}^{-1}$ and $D=1.05\times 10^{-9}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ at $300\mathrm{K}$. Therefore the standard one-second diffusive rms distances are $45.7\mu \mathrm{m}$ in one coordinate and $79.2\mu \mathrm{m}$ in three dimensions.

Using the source's supplied shorthand literally with $m⟨v^2⟩=3k_{B}T$ instead gives $\sqrt{3k_{B}Tt/\zeta }=56.0\mu \mathrm{m}$. That shorthand lacks the factor two in the standard long-time mean-square displacement; it should not be identified with either of the preceding standard rms conventions. None of these rms values is a cumulative path length.

An ideal drag-free fall from rest gives $g{t}^2/2=4.91\mathrm{m}$. Actual gravitational settling in the assumed viscous model is much smaller: ${\tau }_p=m/\zeta =2.12\times 10^{-8}\mathrm{s}$, $u_t=mg/\zeta =0.208\mu \mathrm{m}{\mathrm{s}}^{-1}$, and displacement in one second is approximately $0.208\mu \mathrm{m}$. Thus diffusion overwhelms settling, even though ideal vacuum fall exceeds the rms diffusion distance. For rarefied-gas nanoparticle drag, at a 20-nm size in argon, continuum no-slip drag may require a rarefaction/slip correction; pressure and gas mean-free-path data are absent, so these numerical transport estimates are conditional on the supplied Stokes model.