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 $^{39}K^{37}Cl$}