In rotational spectroscopy, it is conventional to denote the rotational quantum number by \(J\) rather than \(\ell\). This change in notation reflects the spectroscopic context and helps distinguish rotational motion from orbital angular momentum in other quantum problems.
Rotational term values
The rotational energy levels of a rigid diatomic molecule are most conveniently expressed using term values, defined as the energy divided by \(hc\). The rotational term values are written as
\[ F_J = B\,J(J+1), \qquad J = 0,1,2,\ldots \]
where \(B\) is the rotational constant, expressed in units of \(\text{cm}^{-1}\).
The rotational constant
The rotational constant is determined by the molecule’s moment of inertia:
\[ B = \frac{h}{8\pi^2 c I} = \frac{h}{8\pi^2 c\,\mu r^2}, \]
where:
- \(h\) is Planck’s constant,
- \(c\) is the speed of light,
- \(\mu\) is the reduced mass of the molecule,
- \(r\) is the bond length.
Because \(B \propto 1/r^2\), molecules with longer bonds have smaller rotational constants and more closely spaced rotational energy levels.
Why rotational spectroscopy is powerful
Rotational spectroscopy provides one of the most direct experimental methods for determining molecular structure. Measuring the spacing between rotational spectral lines allows \(B\) to be determined with high precision.
Once \(B\) is known, the bond length \(r\) can be calculated directly from \(B = h/(8\pi^2 c\,\mu r^2)\).
Worked idea: extracting \(B\) (and then \(r\)) from a pure rotational spectrum
A pure rotational (microwave) spectrum consists of transitions between rotational levels of a rigid diatomic molecule. Using spectroscopic term values, \(F_J = BJ(J+1)\), the transition wavenumber is obtained from \(\Delta F = F_{J+1}-F_J\).
Step 1: Use the selection rule
For an electric-dipole rotational transition in a heteronuclear diatomic molecule (like HCl), the selection rule is
\[ \Delta J = +1. \]
Step 2: Derive the line positions
The wavenumber of the transition \(J\rightarrow J+1\) is
\[ \tilde{\nu}_{J\rightarrow J+1} = F_{J+1}-F_J = B(J+1)(J+2)-BJ(J+1) = 2B(J+1). \]
This predicts a very important experimental feature: adjacent lines are evenly spaced by \(2B\).
\[ \tilde{\nu}_{(J+1)\rightarrow (J+2)}-\tilde{\nu}_{J\rightarrow (J+1)} = 2B. \]
Step 3: Extract \(B\) from the spectrum (HCl example)
In practice, you record several lines in the HCl rotational spectrum and measure their wavenumbers (in \(\text{cm}^{-1}\)). If the lines are equally spaced, compute the average spacing:
\[ \Delta \tilde{\nu} \approx 2B \quad\Longrightarrow\quad B \approx \frac{\Delta \tilde{\nu}}{2}. \]
Alternatively, if you can assign a specific transition (for example, the lowest line is often \(J=0\rightarrow 1\)), then
\[ \tilde{\nu}_{0\rightarrow 1} = 2B \quad\Longrightarrow\quad B = \frac{\tilde{\nu}_{0\rightarrow 1}}{2}. \]
For HCl, the rigid-rotor model works well, so the equal-spacing pattern is usually easy to recognize and makes \(B\) straightforward to determine.
Step 4: Use \(B\) to calculate the bond length
Once \(B\) is known, the bond length can be found from
\[ B = \frac{h}{8\pi^2 c\,\mu r^2} \quad\Longrightarrow\quad r = \sqrt{\frac{h}{8\pi^2 c\,\mu B}}. \]
Here \(\mu\) is the reduced mass of HCl, which can be computed from the atomic masses of H and Cl.
Worked example: HCl pure rotational lines → \(B\) → \(r\)
| Line # | Transition \(J' \leftarrow J''\) | Frequency \((\text{cm}^{-1})\) |
|---|---|---|
| 1 | \(1 \leftarrow 0\) | 21.19 |
| 2 | \(2 \leftarrow 1\) | 42.37 |
| 3 | \(3 \leftarrow 2\) | 63.56 |
| 4 | \(4 \leftarrow 3\) | 84.75 |
| 5 | \(5 \leftarrow 4\) | 105.93 |
| 6 | \(6 \leftarrow 5\) | 127.12 |
| 7 | \(7 \leftarrow 6\) | 148.31 |
| 8 | \(8 \leftarrow 7\) | 169.49 |
| 9 | \(9 \leftarrow 8\) | 190.68 |
| 10 | \(10 \leftarrow 9\) | 211.87 |
1) Check the spacing and identify \(2B\)
For a rigid rotor, \(\tilde{\nu}_{J\rightarrow J+1}=2B(J+1)\), so adjacent lines should be separated by \(2B\). Compute the successive differences:
\[ \Delta\tilde{\nu}\approx 21.18,\;21.19,\;21.19,\;21.18,\;21.19,\;21.19,\;21.18,\;21.19,\;21.19\ \text{cm}^{-1}. \]
The average spacing is therefore \(\Delta\tilde{\nu}\approx 21.187\ \text{cm}^{-1}\), so
\[ 2B \approx 21.187\ \text{cm}^{-1} \quad\Longrightarrow\quad B \approx 10.593\ \text{cm}^{-1}. \]
This is consistent with assigning the first line as \(J=0\rightarrow 1\), since then \(\tilde{\nu}_{0\rightarrow 1}=2B\approx 21.19\ \text{cm}^{-1}\).
2) Optional: verify the \(J\) assignments
If \(B=10.593\ \text{cm}^{-1}\), then the predicted line positions are \(\tilde{\nu}=2B(J+1)\):
\[ \tilde{\nu}_{0\to1}=2B,\; \tilde{\nu}_{1\to2}=4B,\; \tilde{\nu}_{2\to3}=6B,\;\ldots \]
Using \(2B\approx 21.187\ \text{cm}^{-1}\) gives a sequence \(21.19,\;42.37,\;63.56,\ldots\) in excellent agreement with the measured values.
3) Compute the bond length \(r\) from \(B\)
The rotational constant is related to bond length by
\[ B=\frac{h}{8\pi^2 c\,\mu r^2} \quad\Longrightarrow\quad r=\sqrt{\frac{h}{8\pi^2 c\,\mu B}}. \]
Using \(^1\text{H}\) and \(^{35}\text{Cl}\) (a common isotopologue), the reduced mass is
\[ \mu=\frac{m_\text{H}m_\text{Cl}}{m_\text{H}+m_\text{Cl}} \approx 1.63\times 10^{-27}\ \text{kg}. \]
With \(B=10.593\ \text{cm}^{-1}\) (and consistent SI units in the formula), the bond length is
\[ r \approx 1.27\times 10^{-10}\ \text{m} \;=\; 1.27\ \text{\AA}. \]
Big idea: the equal spacing of pure rotational lines gives \(2B\) directly. Once \(B\) is known, the moment of inertia (and thus the bond length) follows immediately.
Big idea: the spacing between lines in a pure rotational spectrum gives \(2B\), and once \(B\) is known the bond length \(r\) follows directly.