= Sedimentation scale height
{title2=$H=k_BT/(\Delta\rho Vg)$}
Take identical rigid <spheres> of radius $a$, particle <density> $\rho_p$ and displaced <liquid> <density> $\rho_f$, at uniform absolute <temperature> $T$. Let $z$ increase upwards from a reference layer and write $\Delta\rho=\rho_p-\rho_f>0$. The <buoyancy> correction gives the effective gravitational <mass>
$$
V=\frac{4\pi a^3}{3},\qquad m_b=(\rho_p-\rho_f)V,\qquad U(z)-U(0)=m_bgz.
$$
The actual grain <mass> is $\rho_pV$; the displaced <liquid> subtracts $\rho_fV$ from the effective gravitational <mass>. In a dilute, noninteracting <particle suspension>, the <Boltzmann distribution> therefore gives
$$
\frac{n(z)}{n(0)}=\exp\!\left(-\frac{m_bgz}{k_BT}\right)=e^{-z/H},\qquad H=\frac{k_BT}{m_bg}=\frac{k_BT}{\Delta\rho Vg}.
$$
Here $n$ is <number density> and $k_B$ is the <Boltzmann constant>. Equivalently, the dilute <osmotic pressure> $\Pi=nk_BT$ supports the grains above it: $d\Pi/dz=-nm_bg$. Dividing by $n$ and integrating yields the same <exponential function>. No <dynamic viscosity> appears in this <equilibrium> argument. If $\rho_p<\rho_f$, the population instead increases upwards; at equal <densities> this gravitational model predicts uniform $n$.
A <microscope> counts a finite optical slab, not an infinitesimal plane. For equal viewed area $A$ and slab thickness $\delta$, an ideal uniform detection window centred at $z$ gives
$$
\overline C(z)=A\int_{z-\delta/2}^{z+\delta/2}n(s)\,ds
=2AHn(0)\sinh\!\left(\frac{\delta}{2H}\right)e^{-z/H}.
$$
Thus equal optical volumes preserve the <exponential function> count ratio exactly, even for finite $\delta$, provided the windows lie wholly inside the accessible chamber and have the same detection response. Unequal depth response or overlapping a boundary can corrupt this inference. Repeated observations estimate the mean count; closely spaced visits need not be <independent> because grains remain in view.
For an original chosen model, set $a=0.250\,\mathrm{\mu m}$, $\rho_p=1200\,\mathrm{kg\,m^{-3}}$, $\rho_f=1000\,\mathrm{kg\,m^{-3}}$, $T=298\,\mathrm K$ and $g=9.81\,\mathrm{m\,s^{-2}}$. These are illustrative inputs, not <Jean Baptiste Perrin>'s measurements. Using $k_B=1.380649\times10^{-23}\,\mathrm{J\,K^{-1}}$ gives
$$
V\simeq6.5450\times10^{-20}\,\mathrm{m^3},\qquad m_b\simeq1.3090\times10^{-17}\,\mathrm{kg},\qquad H\simeq32.04\,\mathrm{\mu m}.
$$
If equal observation effort gives expected count $400$ in the reference layer, the expected counts $20$, $40$ and $60\,\mathrm{\mu m}$ higher are approximately $214.3$, $114.8$ and $61.49$. Fractions describe <expected values> over observations, not fractional grains. Using the actual grain <mass> instead of its <buoyancy>-corrected value would make the inferred <Avogadro constant> six times too small for these chosen <densities>.
\Image[/perrin-sedimentation-height-counts.png]
{title=Counting equal optical layers in a chosen colloidal atmosphere}
{description=Original functional schematic and calculated count profile; no historical data are plotted. The reference layer is $5\,\mathrm{\mu m}$ above the chamber floor. Refocusing selects successive horizontal layers; counts use equal optical volumes and equal observation effort. The <logarithm> of the count ratio has <slope> $-1/H$.}
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