# Homework Help: Number of closely packed colloidal particles in an aggregate

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1. Mar 18, 2015

### maximus123

Hello everyone

This is sort of a geometry problem. I'm sure it has an easy answer but it just won't come to me. Here's my problem

A close packed colloidal aggregate of smaller spherical colloidal particles can be thought of as small spheres within a sphere.

I have the relationship

$N(r)\propto r^{D_f}$

The number of colloidal particles within a radius is proportional to the radius to the power of the fractal dimension. For a close packed colloidal aggregate $D_f=3$ so

$\frac{N_1(r)}{N_2(r)}=\frac{r^3}{r^2}\\ \frac{4400}{N_2(r)}=r$

So if I'm right so far then all I need is the radius of the aggregate. This is what I tried

$\frac{V_a}{V_c}=4400$

where $V_a$ is the volume of the aggregate and $V_c$ is the volume of the colloidal particle, so

$\frac{V_a}{V_c}=\frac{\frac{4}{3}\pi r_a^3}{\frac{4}{3}\pi r_c^3}=4400\\ \frac{\frac{4}{3}\pi \left(20\sigma\right)^3}{\frac{4}{3}\pi \left(\frac{1}{2}\sigma\right)^3}=4400$
where I'm using $\sigma$ to mean the diameter of the colloidal particle. This last line of calculation is clearly wrong (or at least of no use) as everything will cancel and I'll be left with some incorrect statement such as $64000=4400$.

Can anyone suggest how I might find the radius of the aggregate?

Thank you

2. Mar 23, 2015

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

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