Example: The laboratory spectrum of
the $^{12}$C$^{16}$O
carbon-monoxide molecule shows that the fundamental $J = 1 \rightarrow
0$ transition emits a photon at $\nu = 115.271208$ GHz. (See the
on-line
spectral-line catalog called Splatalogue
for accurate frequencies of radio spectral lines.) The distance $r_{\rm
e}$ between the
C and O nuclei can be estimated from

$$r_{\rm e} = {1 \over 2 \pi}
\biggl( { h J \over m \nu} \biggr)^{1/2}$$ where the reduced mass is

$$m = {m_{\rm C} m_{\rm O} \over m_{\rm C} + m_{\rm O}} \approx m_{\rm
H} \biggl( {12 \cdot 16 \over 12 + 16} \biggr) \approx 1.67 \times
10^{-24} {\rm ~g~} \times 6.86 \approx 1.15 \times 10^{-23} {\rm
~g}~.$$ Thus the equilibrium distance between the C and O nuclei is

$$ r_{\rm e} = {1 \over 2 \pi} \biggl( { 6.63 \times 10^{-27}
{\rm ~erg~s} \times 1 \over 115.271208 \times 10^9 {\rm ~Hz~} \times
1.15 \times 10^{-23} {\rm ~g}} \biggr)^{1/2} \approx 1.13 \times
10^{-8} {\rm ~cm}$$

The centrifugal forces acting on the
nuclei increase with
$J$, so the bond will stretch
and $r_{\rm e}$ will increase slightly with $J$. Spectral lines emitted
by more
rapidly rotating $^{12}$C$^{16}$O molecules will have frequencies
slightly lower than the harmonics $2 \nu_{1-0}$, $3 \nu_{1-0}$, $...$
of the $J = 1 \rightarrow 0$ line: $2 \nu_{1-0} = 230.542416$ GHz
and
the actual $J = 2 \rightarrow 1$ frequency is $\nu_{2-1} =
230.538000$ GHz. Chemists use these line frequencies to determine
$r_{\rm e}$, and the difference between $2 \nu_{1-0}$ and $\nu_{2-1}$
is a measure of the "stiffness" of the carbon-oxygen chemical bond.
Since the actual frequency is only slightly less than the harmonic
frequency, the stiffness is quite high and it is clear that the
fundamental vibrational frequency of the CO molecule will be
much higher than the fundamental rotational frequency. Thus
the vibrational line spectrum of CO begins at infrared
wavelengths.

Equation 7D6 can
be used to calculate
frequencies for rare isotopic molecules (e.g., $^{13}$C$^{16}$O) that
might be more difficult to measure in the lab.

$${m(^{13}{\rm C}^{16}{\rm
O}) \over m(^{12}{\rm C}^{16}{\rm O})} = { 13 \cdot 16 / (13 + 16)
\over 12 \cdot 16 / (12 + 16)} \approx 1.0460$$ so we expect

$${
\nu_{1-0} (^{13}{\rm C}^{16}{\rm O}) \over \nu_{1-0} (^{12}{\rm
C}^{16}{\rm O}) } \approx \biggl[{m(^{13}{\rm C}^{16}{\rm O}) \over
m(^{12}{\rm C}^{16}{\rm O})} \biggr]^{-1}$$ $$ \nu_{1-0} (^{13}{\rm
C}^{16}{\rm O}) \approx 115.271208 / 1.0460 \approx 110.204 {\rm
~GHz}$$

The actual $^{13}$C$^{16}$O $J = 1 \rightarrow 2$ frequency is
110.201354 GHz.

This simple analysis shows that polar
diatomic molecules emit a
harmonic series of radio spectral lines at millimeter wavelengths.
Bigger and heavier linear polyatomic molecules have ladders of lines
starting at somewhat lower frequencies.

Nonlinear molecules such as the
symmetric-top ammonia (NH$_3$) with two distinct rotational axes have
more complex spectra consisting of many parallel ladders.

Molecules are excited into $E_{\rm
rot} > 0$ states by ambient radiation and by collisions in a gas.
The minimum gas temperature $T_{\rm min}$ needed for significant
collisional excitation is roughly

$$T_{\rm min} \sim { E_{\rm rot} \over k}$$ From equations 7D4 and 7D6,

$$E_{\rm rot} = {J (J + 1) \hbar^2 \over 2 I} \qquad {\rm and} \qquad
\nu = {h J \over 4 \pi^2 I}$$ so

$$T_{\rm min} \sim { J (J+1) h^2 \over 2 \cdot 4 \pi^2 I k} = { h J
\over 4 \pi^2 I} { h (J + 1) \over 2 k}$$ Thus a minimum gas kinetic
temperature

$$\bbox[border:3px blue solid,7pt]{T_{\rm min} \approx { \nu h (J + 1)
\over 2 k}}\rlap{\quad \rm {(7D7)}}$$ is
required to excite the $J \rightarrow J -1$ transition at frequency
$\nu$.