Rotational levels

Recall in problem set # 4, we considered the energy levels and wave functions of a particle of mass $m$ constrained to move on a ring of radius $R$ in the $xy$ plane. The energy levels are given by

$$E_n = {\hbar^2 \over 2mR^2}n^2\;\;\;\;\;\;\;\;\;\;n=0,1,2,.....$$

The wave functions are given by

$$\psi_n(\theta) = {1 \over \sqrt{2\pi}}e^{in\theta}$$

where $\theta$ is the angle made by the position vector of the particle and the positive $x$ axis. If we now consider a diatomic molecule with nuclear masses $m_1$ and $m_2$ and equilibrium bond length $R_e$ rotating in the $xy$ plane, as shown in the figure below:

Figure: Rigid diatomic rotating in the $xy$ plane.

The origin is placed at the molecule's center of mass

$${\bf R}= {m_1 {\bf r}_1 + m_2 {\bf r}_2 \over m_1 + m_2}$$

(In the figure, $m_2 > m_1$.) The energy levels of the molecule rotating in the plane wave exactly the same as those of a particle on a ring, namely

$$E_n = {\hbar^2 \over 2I}n^2$$

Here $I$ denotes the moment of inertia of the molecule, given by

$$I = \mu R_e^2$$

and $\mu$ is the reduced mass of the molecule

$$\mu = {m_1 m_2 \over m_1 + m_2}$$

If the molecule rotates in three dimensions, then we need two angles $\theta$ and $\phi$ to characterize it. We will take these to be the angles in the spherical polar coordinate system ($\theta$ is the polar angle and $\phi$ is the azimuthal angle). The energy levels are given by

$$E_n = {\hbar^2 \over 2I}n(n+1)\;\;\;\;\;\;\;\;\;\;n=0,1,2,....$$

In order to avoid confusion between $n$, the radial quantum number for the hydrogen atom, we use a different letter for the rotational energy levels. We use the symbol $J$ and write the rotational levels as

$$E_J = {\hbar^2 \over 2I}J(J+1)\;\;\;\;\;\;\;\;\;\;J=0,1,2,...$$

The rotational wave functions are exactly the same as the angular part of the hydrogen-atom wave functions, namely the spherical harmonics $Y_{Jm}(\theta,\phi)$. Here, the quantum number $m$ characterizes the $z$-component of the molecule's angular momentum and takes on the values $m=-J,...,J$. Hence, each rotational energy level has a degeneracy $g(E_J)=(2J+1)$.

Now, if the molecule absorbs a photon of frequency $\nu$, the molecule can undergo a transition between rotational energy levels. What is the energy difference between the energy level with $J+1$ and the energy level with $J$? This is given by

$$\Delta E_{(J+1),J} = {\hbar^2 \over 2I}(J+1)(J+2) - {\hbar^2 \over 2I}J(J+1)$$

$$= {\hbar^2 \over 2I}\left[J^2 + 3J + 2 - J^2 - J\right]$$

$$= {\hbar^2 \over 2I}(2J+2)$$

$$= {\hbar^2 \over I}(J+1)$$

Therefore, the frequencies at which transitions can occur are given by

$$h\nu = \Delta E_{(J+1),J} = {\hbar^2 \over I}(J+1)$$

$$\nu = {\hbar^2 \over Ih}(J+1)$$

$$= {\hbar^2 \over 2\pi I\hbar}(J+1)$$

$$= {\hbar \over 2\pi I}(J+1)$$

$$\equiv 2B(J+1)$$

where the rotational constant $B$ is given by

$$B = {\hbar \over 4\pi I}$$

Example: The molecule NaH is found to undergo a rotational transition from $J=0$ to $J=1$ when it absorbs a photon of frequency 2.94 $\times 10^{11}$ Hz. What is the equilibrium bond length of the molecule?

We use $J=0$ in the formula for the transition frequency

$$\nu = 2B = {\hbar \over 2\pi I} = {\hbar \over 2\pi \mu R_e^2}$$

Solving for $R_e$ gives

$$R_e = \sqrt{{\hbar \over 2\pi \mu\nu}}$$

The reduced mass is given by

$$\mu = {m_{Na}m_H \over m_{Na} + m_H} = {(22.989)(1.0078) \over 22.989 + 1.0078}$$

$$= 0.9655$$

which is in atomic mass units or relative units. In order to convert to kilograms, we need the conversion factor 1 au = 1.66$\times$10$^{-27}$ kg. Multiplying this by 0.9655 gives a reduced mass of 1.603$\times$10$^{-27}$ kg. Substituting in for $R_e$ gives

$$R_e = \sqrt{(1.055 \times 10^{-34}{\rm J}\cdot{\rm s}) \over 2\pi(1.603\times 10^{-27}{\rm kg})(2.94 \times 10^{11} {\rm Hz})}$$

$$= 1.89\times 10^{-10}\;{\rm m} = 1.89\;{\rm\AA}$$