- #1
maximus123
- 50
- 0
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
I have the relationship
[itex]N(r)\propto r^{D_f}[/itex]
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 [itex]D_f=3[/itex] so
[itex]\frac{N_1(r)}{N_2(r)}=\frac{r^3}{r^2}\\
\frac{4400}{N_2(r)}=r
[/itex]
So if I'm right so far then all I need is the radius of the aggregate. This is what I tried
[itex]\frac{V_a}{V_c}=4400
[/itex]
where [itex]V_a[/itex] is the volume of the aggregate and [itex]V_c[/itex] is the volume of the colloidal particle, so
[itex]\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
[/itex]
where I'm using [itex]\sigma[/itex] 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 [itex]64000=4400[/itex].
Can anyone suggest how I might find the radius of the aggregate?
Thank you
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.If a close packed colloidal aggregate contains 4400 monodisperse spherical colloidal particles (CP), how many colloidal particles would there be in a colloidal aggregate (or ‘floc’) of fractal dimension 2.00 with the same radius and made up of the same type of colloidal particles. Assume that the radius of both types of colloidal aggregate is 20 CP particle diameters.
I have the relationship
[itex]N(r)\propto r^{D_f}[/itex]
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 [itex]D_f=3[/itex] so
[itex]\frac{N_1(r)}{N_2(r)}=\frac{r^3}{r^2}\\
\frac{4400}{N_2(r)}=r
[/itex]
So if I'm right so far then all I need is the radius of the aggregate. This is what I tried
[itex]\frac{V_a}{V_c}=4400
[/itex]
where [itex]V_a[/itex] is the volume of the aggregate and [itex]V_c[/itex] is the volume of the colloidal particle, so
[itex]\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
[/itex]
Can anyone suggest how I might find the radius of the aggregate?
Thank you