Difference between revisions of "Rovibrational Transitions"

From AstroBaki
Jump to navigationJump to search
Line 68: Line 68:
 
$$ \omega_{obs, P} = \omega - \frac{\hbar}{I} J $$
 
$$ \omega_{obs, P} = \omega - \frac{\hbar}{I} J $$
  
[[File:asymm.jpg|300px|left|alt=|Asymmetries in potential.]]  
+
[[File:asymm.jpg|300px|left|alt=|Asymmetries in potential: note that bond length increases with increasing n.]]  
  
 
However, we should note that the potential is {\it not} perfectly harmonic; the slight asymmetries we can see in the potential 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.
 
However, we should note that the potential is {\it not} perfectly harmonic; the slight asymmetries we can see in the potential 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.

Revision as of 13:20, 20 November 2015

Short Topical Videos

Reference Materials

<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, 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 the potential 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.


\end{document} <\latex>