# Nanoparticle Algebra

1. Jul 26, 2007

### mbrmbrg

1. The problem statement, all variables and given/known data

$$N_{nano}=\left[1-\frac{3}{4}\left(\frac{r}{R}\right)+\frac{1}{16}\left(\frac{r}{R}\right)^3\right]N_{bulk}$$

Solve for R.

2. Relevant equations

Algebra, come to bite me in the butt...

3. The attempt at a solution

$$\frac{N_{nano}}{N_{bulk}}=1-\frac{3}{4}\left(\frac{r}{R}\right)+\frac{1}{16}\left(\frac{r}{R}\right)^3$$

$$\frac{N_{nano}}{N_{bulk}}-1=-\frac{12}{16}\left(\frac{r}{R}\right)+\frac{1}{16}\left(\frac{r}{R}\right)^3$$

$$16\left[\frac{N_{nano}}{N_{bulk}}-1\right]=\frac{r^3}{R^3}-12\left(\frac{r}{R}\right)$$

And I'm stuck...
Properly, this isn't a homework question at all; I'm trying to analyze experimental data and I'd rather not call my prof with an algebra question

2. Jul 26, 2007

### mbrmbrg

No longer *need* an answer. My professor told me to plot the data in excel and solve graphically. But I'd still like to know if it's possible to isolate R.

3. Jul 26, 2007

### daveb

It's certainly possible. Find the common denominator for the fractional parts and make cubic polynomial in R, and find the 3 roots. One is guaranteed to be real, the other two may be complex conjugates or real (can't tell just by looking at it).

4. Jul 26, 2007

### mbrmbrg

So...
$$16\left[\frac{N_{nano}}{N_{bulk}}-1\right]=\frac{r^3-12R^2}{R^3}$$

$$16R^3\left[\frac{N_{nano}}{N_{bulk}}-1\right]=r^3-12R^2$$

$$16R^3\left[\frac{N_{nano}}{N_{bulk}}-1\right]+12R^2-r^3=0$$

But how to factor this beastie?

5. Jul 26, 2007

### daveb

6. Jul 26, 2007

### mbrmbrg

:yuck: Thank you... Uhuhurgh. Maybe I'll give it a shot, though it does make me look a lot more kindly on my professor's approach of just graphing N versus R and seeing where our values of N lie on the curve.

Last edited: Jul 27, 2007