# Rotational and vibrational spectra in diatomics

- Details
- Written by: Germán Fernández
- Category: Rotational and Vibrational Spectra in Diatomics
- Hits: 317

The selection rules allow us to know the rotational and vibrational transitions allowed. The integral that determines the selection rules is:

\begin{equation}\label{1-1} \int{\psi^{\ast}\hat{\mu} \psi'}dq \end{equation}

where $\mu$ is the electric dipole moment. $\psi$ and $\psi'$ are the two states involved by the transition. Applying the Born-Oppenheimer approximation the wave function can be separated into the product of the nuclear wave function times the electronic $\psi =\psi_N \psi_{ele}$. Since the vibrational-rotary transition does not change the electronic state $\psi'=\psi'_N \psi'_{ele}$. Substituting into the integral (\ref{1-1}):

\begin{equation}\label{2} \int{\psi^{\ast}\hat{\mu} \psi'}dq=\int\int{\psi_{ele}^{\ast}\psi_{ N}^{\ast}\hat{\mu}\psi_{ele}\psi'_{N}dq_{ele}}dq_N \end{equation}

- Details
- Written by: Germán Fernández
- Category: Rotational and Vibrational Spectra in Diatomics
- Hits: 333

They are transitions with $\Delta v=0$ that take place in the microwave and far IR region. For a diatomic molecule to present a pure rotation spectrum, it must have a non-zero electric dipole moment.

Pure rotation spectra are usually absorption with $\Delta J=+1$ transitions. Due to the population of excited levels at room temperature, different transitions are observed that give rise to several lines in the spectrum.

**absorption frequencies**

Knowing that the energy difference between two consecutive levels is $h\nu$ we can write $E_{J+1}-E_J = h\nu$. Clearing the frequency:

\begin{equation}\label{6} \nu=\frac{E_{J+1}-E_{J}}{h} \end{equation}

Where $E_J=hB_e J(J+1)-h\alpha_e(v+\frac{1}{2})J(J+1)$ we do not consider the centrifugal distortion term because it is small. Taking out common factor to $hJ(J+1)$:

\begin{equation}\label{7} E_J=hJ(J+1)[B_e-\alpha_e(v+1/2)] \end{equation}

- Details
- Written by: Germán Fernández
- Category: Rotational and Vibrational Spectra in Diatomics
- Hits: 315

Infrared photons produce vibrational, though not electronic, transitions, giving rise to rotation-vibration spectra. For the spectrum of rotational vibration to be observed, it is necessary for the dipole moment to change during vibration. For this reason, homonuclear diatomic molecules do not present IR spectrum of rotational vibration.

The rotation-vibration spectrum is generally observed as an absorption spectrum with transitions $\Delta v =+1 (+2, +3, ...)$ where the transitions in parentheses, called overtones, are less likely. Simultaneously $\Delta J =+1$ producing transitions between rotational levels simultaneous to the vibrational ones.

The rotation-vibration spectrum consists of a series of bands that correspond to transitions between two vibrational levels $v''$ and $v'$ and a series of lines, corresponding to rotational transitions.

**Vibration bands**

In principle, we will calculate the frequency of a transition where v changes but J is zero in both the initial and final states. This transition gives us the position of the band: origin.

\begin{equation}\label{11} \bar{\nu}_{o}=\frac{E_{v'}-E_{v''}}{hc} \end{equation}

The vibrational energy for state v is given by:

\begin{equation}\label{12} E_{v}=h\nu_{e}\left(v+\frac{1}{2}\right)-h\nu_{e}x_{e}\left( v+\frac{1}{2}\right)^2 \end{equation}