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{\bf Question}

The bond extension $y$ of a vibrating diatomic molecule satisfies
the differential equation

$$m \frac{d^2y}{dt^2}= -\frac{dV}{dy}$$

where m is the reduced mass and $V(y)$ is the potential energy. An
approximation to $V(y)$ is given by the Morse potential:

$$Y(y)=d(1-e^{-by})^2$$ where $d>0$ and $b>0$ are positive
constants.  Using the Morse potential $V(y)$:
\begin{description}
\item[(a)]
For what values of y does $V(y)=0$?  Find the turning points of
V(y) and calculate $lim_{y\rightarrow -\infty}V(y)$ and
$lim_{y\rightarrow -\infty}V(y)$.  Sketch a ROUGH graph of V(y).

\item[(b)]
Write down the Taylor series expansion for $e^{-by}$ about $y=0$.
Hence shoe that when $y$ is small the Morse potential is
approximately

$$V(y) \approx d(by)^2.$$
\item[(c)]
Using your approximation for V(y) when y is small, show that the
molecule vibrates with a frequency

$$f=\frac{1}{2\pi}\sqrt{\frac{2db^2}{m}}.$$
\end{description}


{\bf Answer}

\begin{description}
\item[(a)]
\begin{eqnarray*}
{\rm Morse\ potential\ } V(y) & = & d(1 - e^{-by})^2 \\ V(y) = 0 &
\Leftrightarrow & d(1 - e^{-by})^2 = 0 \\ & \Leftrightarrow & (1 -
e^{-by})^2 = 0 \\ & \Leftrightarrow &  e^{-by} = 1 \\ {\rm since\
} b > 0, V(0) & = & 0 \Leftrightarrow y = 0
\end{eqnarray*}
Differentiating gives $\frac{dV}{dy} = 2d(1 - e^{-by})be^{-by}$

So turning points occur when $2d(1 - e^{-by})be^{-by} = 0$

This only happens when $(1 - e^{-by}) = 0$, and so when $y = 0$

There is only one turning point at y = 0.

Nature of turning point:  $ \left\{ \begin{array}{rrr}
{\frac{dV}{dy}<0} & {when} & {y<0} \\ {\frac{dV}{dy}>0} & {when} &
{y>0} \end{array} \right.$

So V(y) has a local minimum at y = 0.

As $y \rightarrow +\infty$, $e^{-by} \rightarrow 0$.  So $
\displaystyle \lim_{y \to +\infty}V(y) = d$

As $y \rightarrow -\infty$, $e^{-by} \rightarrow +\infty$.  So $
\displaystyle \lim_{y \to -\infty}V(y) = + \infty$

Graph of $V(y)$ against $y$:

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Also see Figure 16.37 of (Figure 16.36 in the new edition)

P.W. Atkins, Physical Chemistry, Oxford, 1994.

(Copies at QA453 in the short loans section of the library.)

\item[(b)]
Taylor expansion about $y = 0$:

$e^{-by} = 1 + (-by) + \frac{1}{2!}(-by)^2 + \frac{1}{3!}(-by)^3
+...$

For small y, $ e^{-by} \approx 1 - by$ (neglect terms of order 2
and above)

so for small y, $V(y) \approx d(1 - [1-by])^2 = d(by)^2$

\item[(c)]
\begin{eqnarray*} {\rm Using\ approximation\ } V(y) & \approx &
d(by)^2 \\ {\rm and\ } \frac{dV}{dy} & \approx & 2db^2y
\end{eqnarray*}
So the differential equation for the bond extension y becomes $$m
\frac{d^2y}{dt^2} = -2db^2y {\rm \ or \ } m \frac{d^2y}{dt^2} +
2db^2y = 0$$ which is the equation for simple harmonic motion. The
auxiliary equation is $$\lambda^2 + \frac{2db^2}{m} = 0$$ with a
pair of complex conjugate roots $\displaystyle \lambda = \pm i
\sqrt{ \frac{2db^2}{m}}$

Hence the general solution is
$$y =   A\cos \left(t\sqrt{ \frac{2db^2}{m}}\right) + B \sin \left(t\sqrt{
\frac{2db^2}{m}}\right)$$
Which is periodic with time period
$$T = \frac{2\pi}{\sqrt{ \frac{2db^2}{m}}}$$
and frequency
$$f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{ \frac{2db^2}{m}}.$$

\end{description}


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