Paul did a demonstration with a drop of soap on the end of a pin, which had a diameter of ~ 2 mm. When placed on a water-pepper surface, the drop spread out in a circular puddle with a radius of about 115 mm. How can we use these observations to estimate the size of atoms?
We shall assume that the 'puddle' had a thickness which is no less than the size of an atom (it might be more). So if we could just figure the thickness...
Let's assume the volume of the soap drop is the same as the volume of the soap puddle. Assuming the drop is spherical, it's volume is $$v_d = \frac{4}{3}\pi r^3$$ and since the radius $r$ is 1/2 of the diameter (2mm), we conclude that the radius is 1 mm, and so the volume of the drop is $$v_d = \frac{4}{3}\pi (1 mm)^3=4.189\ mm^3$$
The volume of the puddle is its area=$\pi r_p^2$ times it's thickness, which I'll call $t$. If diameter is 115 mm, the radius is 57.5 mm, and the volume of the puddle is: $$\pi r^2 * t = \pi (57.5mm)^2 * t = 10,387 mm^2 * t$$
Now, setting the volume of the drop equal to the volume of the puddle gives us an algebraic equation that we can solve for $t$: $$\begineq 4.189 mm^3 &=& 10,387 mm^2 *t\\ \frac{4.189 mm^3}{10,387 mm^2} &=& t \\ 0.0004033 mm = 4.03\times 10^{-4} mm &=& t\endeq$$
To get an answer in meters, we multiply this number of millimeters by the conversion factor 1 m / 1000 mm (that is multiply by $10^{-3}$.) This finally gives us our estimate of $t$: $$t = 4.03\times 10^{-7} m$$