# Difference between revisions of "Rovibrational Transitions"

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$$ \frac{1}{2} M \omega^2 L^2 \sim \frac{\hbar^2}{2mL^2} $$ | $$ \frac{1}{2} M \omega^2 L^2 \sim \frac{\hbar^2}{2mL^2} $$ | ||

$$E_{vib} \sim \hbar \omega \sim \left(\frac{m}{M}\right)^{1/2} \frac{\hbar^2}{mL^2} \sim \left(\frac{m}{M}\right)^{1/2} E_{elec} $$ | $$E_{vib} \sim \hbar \omega \sim \left(\frac{m}{M}\right)^{1/2} \frac{\hbar^2}{mL^2} \sim \left(\frac{m}{M}\right)^{1/2} E_{elec} $$ | ||

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+ | [[File:min.pdf|400px|left|alt=|Spontaneous decay profile.]] | ||

## Revision as of 11:33, 20 November 2015

### Short Topical Videos

- Rovibrational Energy Levels (Quantum Chemistry)
- Rovibrational Spectra of Diatomic Molecules (Quantum Chemistry)
- Rotation Vibration Interaction (Quantum Chemistry)

### Reference Materials

- Rotational Vibrational Coupling (Wikipedia)
- Rovibrational Spectroscopy (UC Davis)
- Rotational Vibrational Spectroscopy (Wikipedia)

<latex> \documentclass[11pt]{article} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \section*{ Ro-Vibrational Transitions }

\subsection{Order of Magnitude Energies} The Born-Oppenheimer approximation allows us to treat the electrons in a molecule as a cloud-- they are much less massive and therefore have much higher velocities than the nuclei. If $L$ is the molecular size, typical electrons have momentum $\hbar /a$ and the electronic energy spacings can be expressed as: $$ E_{elec} \sim \frac{\hbar^2}{mL^2}$$

The nuclei feel an equivalent potential that only depends on the internuclear distance and the electronic state. The internuclear potential has a minimum (see Figure \ref{min}), and vibrations about the minimum can be roughly modeled as a harmonic oscillator. This potential is about $\frac{1}{2}M\omega^2 \xi^2$, where $\omega$ is the vibration frequency and $\xi$ is the displacement from equilibrium. If the displacement is the same order as the size of the molecule, the electronic energies should be about $\hbar^2 / 2mL^2$:

$$ \frac{1}{2} M \omega^2 L^2 \sim \frac{\hbar^2}{2mL^2} $$ $$E_{vib} \sim \hbar \omega \sim \left(\frac{m}{M}\right)^{1/2} \frac{\hbar^2}{mL^2} \sim \left(\frac{m}{M}\right)^{1/2} E_{elec} $$

The nuclei can also rotate, and have rotational energies that depend on their angular momentum. If $J$ is the quantum angular momentum number,
$$ E_{rot} \sim \frac{\hbar^2 J (J+1)}{2I} $$
$$ I \sim M L^2 $$
For small $J$,
$$ E_{rot} \sim \frac{m}{M} E_{elec} $$

So the electronic, vibrational, and rotational energy states have contributions that scale with the electron-to-nucleus mass ration: $$ E_{elec} : E_{vib} : E_{rot} \sim 1: \left(\frac{m}{M}\right)^{1/2} : \left(\frac{m}{M}\right) $$

\subsection{Rotation-Vibration Spectra}
While it is possible to have a pure rotational spectrum, a pure vibrational spectrum is very unlikely: energies required to excite vibrations are much larger than those required to excite rotation. However, a combination of rotation and vibrational modes can be excited.

Rotational energies can be described using the angular momentum number $J$: $$E_{rot} = \frac{\hbar^2}{2I} J (J+1) $$

where $J = 0,\ 1,\ 2,\ ...$, $I$ is the moment of inertia, and $\mu$ is the reduced mass: $$ I = \mu L^2 $$ $$ \mu = \frac{m_1 m_2}{m_1 + m_2} $$

The vibrations can again be modeled as a harmonic oscillator: $$ E_{vib} = \hbar \omega (n + 1/2) $$ where $n = 0,\ 1,\ 2,\ ...$ and $\omega = \sqrt{k/m}$ where $k$ is the effective spring constant.

The total energy of rovibrational transitions, then, is: $$ E_{rovibe} = \hbar \omega (n+1/2) + \frac{\hbar^2}{2I} J (J+1) $$

The selection rules for rovibrational transitions tell us that $\Delta n = +1$ and $\Delta J = \pm 1$. $\Delta J = +1$ is called the R branch, and $\Delta J = -1$ is called the P branch. We can use our rovibe energy expression to find the frequency of emitted photons for the R branch: $$ \Delta J = -1 $$ $$ E_2 - E_1 = \hbar \omega (n+1 + 1/2) - \hbar \omega (n + 1/2) + \frac{\hbar^2}{2I} (J+1)(J+2) - \frac{\hbar^2}{2I} J (J+1) $$ $$ E_2 - E_1 = \hbar \omega + \frac{\hbar^2}{I} (J+1) = \hbar \omega_{obs, R} $$ $$ \omega_{obs, R} = \omega + \frac{\hbar}{I} (J+1) $$

and the P branch: $$ \Delta J = 1 $$ $$ E_2 - E_1 = \hbar \omega (n+1 + 1/2) - \hbar \omega (n + 1/2) + \frac{\hbar^2}{2I} (J-1)(J) - \frac{\hbar^2}{2I} J (J+1) $$ $$ E_2 - E_1 = \hbar \omega - \frac{\hbar^2}{I} J = \hbar \omega_{obs, P} $$ $$ \omega_{obs, P} = \omega - \frac{\hbar}{I} J $$

However, we should note that the potential is {\it not} perfectly harmonic; the slight asymmetries we can see in Figure \ref{asymm} cause the bond length to increase as $n$ increases. So if $n$ increases, $L$ increases, causing $I$ to increase. Our rovibrational energy expression depends on $I^{-1}$, so the energy {\it decreases} as $n$ increases. This effectively pulls all of our spectral lines to the left, which decreases the separation between lines on the R branch and increases the separation between lines on the P branch. Similar effects can also be found when comparing the expressions for the observed frequencies. Putting everything together, we can plot our rovibrational spectrum in Figure \ref{spectrum}.

\end{document} <\latex>