\begin{problem*}{6.57}
In diatomic molecules, the consituent atoms exert attractive forces on
each other at large distances, and repulsive forces at short
distances.  For many molecules, the Lennard-Jones law is a good
approximation to the magnitude of these forces:
\begin{equation}
 F = F_0 \left[2\left(\frac{\sigma}{r}\right)^{13} -
 \left(\frac{\sigma}{r}\right)^7 \right]
\end{equation}
Where $r$ is the center-to-center distance between the atoms in the
molecule, $\sigma$ is a length parameter, and $F_0$ is the force when
$r = \sigma$.  For an oxygen molecule, $F_0 = 9.60\E{-11}\U{N}$ and
$\sigma = 3.50\E{-10}\U{m}$.  Determine the work done by this force as
the atoms are pulled apart from $r_0 = 4.00\E{-10}\U{m}$ to $r_1 =
9.00\E{-10}\U{m}$.
\end{problem*} % problem 6.57

\begin{solution}
The work done by the force is given by
\begin{align}
 W &= \int_{r_0}^{r_1} \vect{F} \cdot d\vect{r} \\
   &= \int_{r_0}^{r_1} F \cdot dr \\
   &= \int_{r_0}^{r_1} \left\{ F_0 \left[ 2 \left(\frac{\sigma}{r}\right)^{13} -
      \left(\frac{\sigma}{r}\right)^7 \right] \right\} \cdot dr \\
   &= 2 F_0 \int_{r_0}^{r_1} \left(\frac{\sigma}{r}\right)^{13} dr
      - F_0 \int_{r_0}^{r_1} \left(\frac{\sigma}{r}\right)^7 dr \\
\end{align}
Then we note that
\begin{align}
 \int \left(\frac{a}{x}\right)^n dx
    &= a^n \int x^{-n} dx
     = a^n \frac{x^{-n+1}}{-n+1} \\
 \int_{r_0}^{r_1} \left(\frac{a}{x}\right)^n dx
    &= \frac{a^n}{1-n} \left(r_1^{1-n} - r_0^{1-n}\right)
\end{align}
And plug this into our equation for $W$
\begin{align}
 W &= 2 F_0 \frac{\sigma^{13}}{-12} \left(r_1^{-12} - r_0^{-12}\right)
      - F_0 \frac{\sigma^7}{-6} \left(r_1^{-6} - r_0 ^{-6}\right) \\
   &= \frac{-F_0 \sigma}{6} \left[ \left(\frac{\sigma}{r_1}\right)^{12}
             - \left(\frac{\sigma}{r_1}\right)^6 \right]
      + \frac{F_0 \sigma}{6} \left[ \left(\frac{\sigma}{r_0}\right)^{12}
             - \left(\frac{\sigma}{r_0}\right)^6 \right] \\
   &= \frac{-F_0 \sigma}{6} \left[ \sigma^{12} \left( r_1^{-12} - r_0^{-12} \right) - \sigma^6 \left( r_1^{-6} - r_0^{-6} \right) \right] \\
   &= \frac{ -9.50\E{-11}\U{N} \cdot 3.50\E{-10}\U{m}}{6}
      \left\{ (3.50\U{\AA})^{12} \left[ (9.00\U{\AA})^{-12} - (4.00\U{\AA})^{-12} \right]
            - (3.50\U{\AA})^{6} \left[ (9.00\U{\AA})^{-6} - (4.00\U{\AA})^{-6} \right] \right\} \\
   &= \ans{1.35\E{-25}\U{J}}
\end{align}
\end{solution}
