The rotational motion of molecules has an effect on the equilibrium separation of the nuclei, a phenomenon known as bond stretching. To model this effect, consider a diatomic molecule with reduced mass $\mu$, oscillator frequency $\omega_{0}$, and internuclear separation $R _0$ when the angular momentum is zero. The effective potential energy for nonzero values of $\ell$ is then $U_{\mathrm{eff}}=\frac{1}{2} \mu \omega_{0}^{2}\left(r-R_{0}\right)^{2}+\ell(\ell+1) \frac{\hbar^{2}}{2 \mu r^{2}}$ (a) Minimize the effective potential $U_{\mathrm{eff}}$ (r) to find an equation for the equilibrium separation of the nuclei, $R_{\ell}$, when the angular momentum is $\ell$. Solve this equation approximately, assuming $\ell

Question

The rotational motion of molecules has an effect on the equilibrium separation of the nuclei, a phenomenon known as bond stretching. To model this effect, consider a diatomic molecule with reduced mass $\mu$, oscillator frequency $\omega_{0}$, and internuclear separation $R _0$ when the angular momentum is zero. The effective potential energy for nonzero values of $\ell$ is then $U_{\mathrm{eff}}=\frac{1}{2} \mu \omega_{0}^{2}\left(r-R_{0}\right)^{2}+\ell(\ell+1) \frac{\hbar^{2}}{2 \mu r^{2}}$ (a) Minimize the effective potential $U_{\mathrm{eff}}$ (r) to find an equation for the equilibrium separation of the nuclei, $R_{\ell}$, when the angular momentum is $\ell$. Solve this equation approximately, assuming $\ell<<\mu \omega_{0} R_{0}^{2} / \hbar$. (b) Near the corrected equilibrium point, $R_{\ell}$, the effective potential again is nearly harmonic and can be written approximately as $U_{\mathrm{eff}} \approx \frac{1}{2} \mu \omega^{2}\left(r-R_{\ell}\right)^{2}+U_{0}$ Find expressions for the corrected oscillator frequency $\omega$ and the energy offset $U_0$ by matching U eff and its first two derivatives at the equilibrium point $R_{\ell}$. Show that the fractional change in frequency is given by $\frac{\Delta \omega}{\omega_{0}} \approx \frac{3 \ell(\ell+1) \hbar^{2}}{2 \mu^{2} \omega_{0}^{2} R_{0}^{4}}$.

Quizlet Admin · Accepted Answer

$	extbf{a.)}$ Consider a diatomic molecule with reduced mass $\mu$, oscillator frequency $\omega_{o}$, and intermolecular separation $R_{o}$ when the angular momentum is zero.

The effective potential energy is given by:

$$
\begin{align*}
V_{eff}&=\frac{1}{2} \mu \omega_{o}^2 (r-R_{o})^2+l(l_{1}+1) \frac{h^2}{2 \mu r^2} 	ag{For nonzero values.}\
\end{align*}
$$

When we minimize the effective potential:

$U_{eff}(r)$ , take $\dfrac{du}{dr}$ and assuming $I << \mu \omega_{o} \dfrac{R_{o}^2}{h}$ , we get :

$$
R_{e}=R_o \left[\dfrac{\dfrac{l(l+1)h^2}{(\mu \omega_{o})^2}}{R_o^3} ight].
$$

$	extbf{b.)}$ Near the corrected equilibrium point $R_2$, the effective potential again is nearly harmonic and can be written:

$$
\begin{align*}
u_{eff} \approx &\frac{1}{2} \mu \omega^2 (r-R_o)^2+\mu_o 	ag{Approximately.}\
u_{o} \approx & \frac{l(l+1)h^2}{2A lR_o^2}  	ag{$\frac{d^2 u_eff}{dv}=R_i$.}\
\end{align*}
$$

Now, we have:

$$
\begin{align*}
\omega_i^2 &=\omega_o^2 + 3[(l+1)h^2] l n^2R_o ^4\
\frac{\Delta \omega}{\omega_o}&=\frac{3l (l+1)h^2}{2 \mu^2 \omega_o ^2 R_o ^4}\
\end{align*}
$$

Related questions with answers

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Mathematical Methods in the Physical Sciences

Modern Physics for Scientists and Engineers

Fundamentals of Physics

More related questions