Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Nanoparticle Algebra

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data

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

    Solve for R.

    2. Relevant equations

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

    3. The attempt at a solution

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

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

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


    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 :redface:
     
  2. jcsd
  3. Jul 26, 2007 #2
    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.
     
  4. Jul 26, 2007 #3
    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).
     
  5. Jul 26, 2007 #4
    So...
    [tex]16\left[\frac{N_{nano}}{N_{bulk}}-1\right]=\frac{r^3-12R^2}{R^3}[/tex]

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

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

    But how to factor this beastie?
     
  6. Jul 26, 2007 #5
  7. Jul 26, 2007 #6
    :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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook