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}

Substituting:

\begin{eqnarray}\label{8} \nu=\frac{E_{J+1}-E_{J}}{h}=\nonumber\\h(J+1)(J+2)[B_e- \alpha_e(v+1/2)]-hJ(J+1)[B_e-\alpha_e(v+1/2)]=\nonumber\\2(J+1)[B_e-\alpha_e(v+1) /2)] \end{eqnarray}

Calling $B_v=B_e-\alpha_e(v+1/2)$ equation (3) leaves us:

\begin{equation}\label{9} \nu=2(J+1)B_v \end{equation}

with J=0,1,2,3....

For rotational transitions with v=0 the average rotational constant $B_v$ becomes $B_0=B_e-1/2\alpha_e$

The wavenumber of rotational transitions is obtained by dividing the frequency by the speed of light.

\begin{equation}\label{10} \bar{\nu}=\frac{\nu}{c} \end{equation}

The pure rotation spectrum is made up of a series of lines equally spaced at $2B_0, 4B_0, 6B_0$

**Question:** If the rotational transition J=2 to 3 for a diatomic molecule occurs at $\lambda =2.00 cm$, calculate $\lambda$ for the transition J=6 to 7 of this molecule.

**Solution:** $\lambda_{67}=1/14\bar{B}_e$

**Question:** The transition $J=0\rightarrow 1, v=0\rightarrow 1$ for $^{1}H^{79}Br$ occurs at 500.7216 GHz, and for $^{1}H^ {81}Br$ at 500.5658 GHz. (a) Calculate the bond distance $R_o$ in each of the molecules. Neglect centrifugal distortion. (b) Predict the frequency of the transition $J=1\rightarrow 2$ with $v=0\rightarrow 0$ for the $^{1}H^{79}Br$. Data: atomic masses 1H:1.0078; ^{79}Br:$78.9183

**Solution:** (a) $\mu=1.6524\times 10^{-24}\;g/molec$; $R_e(^{1}H^{79}Br)=1.4242\times 10^{-10}\;m$; $R_e(^{1}H^{81}Br)=1.4242\times 10^{-10}\;m$. (b) $\nu_{12}=2\nu_{10}$

**Question:** The rotational transition $J=2\rightarrow 3$ for the ground vibrational state of $^{39}K^{37}Cl$ occurs at 22410 MHz. Neglecting centrifugal distortion, predict the frequency of the pure rotational transition $ J=0\rightarrow 1$ of $^{39}K^{37}Cl$}