Difference between revisions of "Rovibrational Transitions"
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\usepackage{eufrak} | \usepackage{eufrak} | ||
\begin{document} | \begin{document} | ||
| − | \ | + | \section*{ Ro-Vibrational Transitions } |
| + | \subsection{Vibrational Transitions in Molecules} | ||
| + | |||
| + | Our general technique with vibrational transitions is to model them as | ||
| + | harmonic oscillators. Thus, they should have the characteristic harmonic | ||
| + | energy series: | ||
| + | $$E_n=(n+\frac12)\hbar\omega_0$$ | ||
| + | For a harmonic oscillator, $\omega_0=\sqrt{k\over m}$. We estimate that since | ||
| + | the force for a spring is $k\cdot x$, and that force should be about the | ||
| + | Coulomb force on $e^-$'s. If we say that atoms stretch with respect to | ||
| + | each other about a Bohr radius: | ||
| + | $$ka_0\sim{e^2\over a_0^2}$$ | ||
| + | $$\Delta E\big|_{vib\atop trans}\sim Ryd\cdot\sqrt{m_e\over A\cdot m_p}$$ | ||
| + | where A is the atomic mass \# of our atoms. | ||
| + | |||
| + | \subsection{Rotational Transitions in Molecules} | ||
| + | |||
| + | The thing to remember is that angular momentum comes in units of $\hbar$. | ||
\end{document} | \end{document} | ||
<\latex> | <\latex> | ||
Revision as of 16:16, 9 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{Vibrational Transitions in Molecules}
Our general technique with vibrational transitions is to model them as harmonic oscillators. Thus, they should have the characteristic harmonic energy series: $$E_n=(n+\frac12)\hbar\omega_0$$ For a harmonic oscillator, $\omega_0=\sqrt{k\over m}$. We estimate that since the force for a spring is $k\cdot x$, and that force should be about the Coulomb force on $e^-$'s. If we say that atoms stretch with respect to each other about a Bohr radius: $$ka_0\sim{e^2\over a_0^2}$$ $$\Delta E\big|_{vib\atop trans}\sim Ryd\cdot\sqrt{m_e\over A\cdot m_p}$$ where A is the atomic mass \# of our atoms.
\subsection{Rotational Transitions in Molecules}
The thing to remember is that angular momentum comes in units of $\hbar$.
\end{document} <\latex>
