Boltzmann Populations and Spectral Intensity

Point a microwave spectrometer at HCl gas and you see a comb of absorption lines with wildly unequal heights. The strongest line is not at the lowest rotational transition. It appears several lines further along. Why should a "cold" gas absorb most strongly at some intermediate energy rather than at the smallest possible jump? The answer lies in two competing effects: how many molecules occupy each starting level, and how many quantum states share that level. Together these determine the population that can actually undergo a transition, and therefore how intense the resulting spectral line appears. This article builds the Boltzmann-population argument for spectral intensity from first principles. You will see how the partition function apportions molecules among energy levels. You will see how degeneracy multiplies that population. Finally you will combine both to predict the intensity pattern of a rotational spectrum, with fully worked numerical examples.

Key takeaways

Line intensity is proportional to the population difference between the two states linked by a transition.
Population of a level depends on its Boltzmann factor $e^{-E/k_BT}$ and its degeneracy $g$ .
For rotational levels, degeneracy $g_J = 2J+1$ grows with $J$ , competing against the falling Boltzmann factor.
This competition produces a peaked intensity distribution with a maximum at a specific $J_{max}$ , not at $J=0$ .
Temperature shifts $J_{max}$ : hotter gas populates higher $J$ , shifting peak intensity to higher- $J$ lines.
Key idea: Spectral intensity tracks the product of Boltzmann probability and degeneracy, not the Boltzmann factor alone.

The core insight

The intensity of a spectral line is governed by the number of molecules poised to absorb or emit — the population of the initial state — and that population is the product of a degeneracy factor (how many states share an energy) and a Boltzmann factor (how thermally accessible that energy is), so intensity patterns rise, peak, and fall as these two factors trade off with increasing energy.

The physical picture

A spectral transition occurs between two energy levels, and the transition dipole moment (covered in Transition Intensities and the Transition Dipole Moment) sets the intrinsic probability of the jump. But the observed intensity of a line in a real sample depends on something else too: how many molecules are actually sitting in the initial state, ready to make that jump. Even a transition with a large transition dipole moment produces a weak line if almost no molecules occupy the starting level. The Boltzmann distribution tells us that the number of molecules $N_i$ in a state of energy $E_i$ , relative to the total $N$ , falls off exponentially with energy:

\frac{N_i}{N} = \frac{g_i e^{-E_i/k_BT}}{q}

Here $g_i$ is the degeneracy of level $i$ — the number of distinct quantum states sharing that one energy — and $q$ is the molecular partition function, a normalizing sum over all levels. For rotational levels of a linear molecule, $E_J = BhcJ(J+1)$ and each level carries degeneracy $g_J = 2J+1$ , arising from the $2J+1$ orientations of angular momentum ( $M_J = -J, \ldots, +J$ ) that share the same energy. This degeneracy is the crucial extra ingredient beyond a bare Boltzmann factor. As $J$ increases, the exponential term shrinks the population, but the degeneracy term $(2J+1)$ grows linearly and multiplies it back up. At low $J$ , degeneracy dominates and populations rise with $J$ . At high $J$ , the exponential wins and populations collapse. The product has an interior maximum — this is the physical origin of the classic "comb with a hump" appearance of a rotational spectrum.

Figure

Population of rotational levels versus J, showing the Boltzmann and degeneracy factors and their product

The same reasoning applies to vibrational, electronic, or any other level scheme: the intensity you record in an absorption or emission spectrum is proportional to the population difference between the two combining levels, modulated by their degeneracies.

The core relations

\frac{N_i}{N} = \frac{g_i\, e^{-E_i/k_BT}}{q}, \qquad q = \sum_j g_j\, e^{-E_j/k_BT}

where: $N_i/N$ is the fraction of molecules in level $i$ , $g_i$ its degeneracy, $E_i$ its energy, and $q$ the partition function summing over all levels $j$ .

I_{i\to f} \;\propto\; (N_i - N_f)\,\left|\mu_{fi}\right|^2 \,\nu

where: $I$ is the transition intensity, $N_i - N_f$ the population difference between initial and final states, $\mu_{fi}$ the transition dipole moment, and $\nu$ the transition frequency (from the density-of-states/Einstein-coefficient prefactors).

E_J = BhcJ(J+1), \qquad g_J = 2J+1

where: $B$ is the rotational constant (cm⁻¹), $h$ Planck's constant, $c$ the speed of light, and $J = 0,1,2,\ldots$

J_{max} \approx \sqrt{\frac{k_BT}{2hcB}} - \frac{1}{2}

where: $J_{max}$ is the rotational level with the largest population, obtained by maximizing $g_J e^{-E_J/k_BT}$ with respect to $J$ .

Derivation AFinding the most populated rotational level

Goal. Starting from the relative population of rotational level $J$ , find the value of $J$ that maximizes the population, and hence predict which absorption line in a rotational spectrum should be most intense.

Step 1. Write the relative population. For a linear rotor, the number of molecules in level $J$ relative to the total is proportional to degeneracy times the Boltzmann factor:

N_J \propto (2J+1)\, e^{-BhcJ(J+1)/k_BT}

We treat $J$ as a continuous variable for the purpose of locating the maximum, since $B$ is usually small compared with $k_BT/hc$ (many rotational levels are thermally populated).

Step 2. Differentiate with respect to J. Let $x = J(J+1)$ so that $E_J = Bhcx$ . Differentiating $N_J$ with respect to $J$ and setting the derivative to zero:

\frac{d}{dJ}\left[(2J+1)\,e^{-BhcJ(J+1)/k_BT}\right] = 0

Expanding the derivative using the product rule gives:

2\,e^{-BhcJ(J+1)/k_BT} + (2J+1)\left(-\frac{Bhc(2J+1)}{k_BT}\right)e^{-BhcJ(J+1)/k_BT} = 0

Step 3. Simplify. The exponential factor is never zero, so it cancels, leaving:

2 = \frac{Bhc(2J+1)^2}{k_BT}

Step 4. Solve for J. Rearranging for $(2J+1)^2$ :

(2J+1)^2 = \frac{2k_BT}{Bhc}

Taking the square root and isolating $J$ :

J_{max} = \sqrt{\frac{k_BT}{2Bhc}} - \frac{1}{2}

Step 5. Interpret. This result shows $J_{max}$ grows as $\sqrt{T}$ and shrinks as $\sqrt{B}$ grows (heavier or longer molecules, with smaller $B$ , populate higher $J$ at a given temperature). The corresponding spectral line, the $J_{max} \to J_{max}+1$ transition, should be the most intense line in the rotational absorption spectrum, provided the transition dipole moment does not vary much across low $J$ .

\boxed{J_{max} \approx \sqrt{\dfrac{k_BT}{2hcB}} - \dfrac{1}{2}}

Worked examples

Example 1 — Foundational: relative populations of two rotational levels of HCl. For gas-phase HCl, $B = 10.59\ \text{cm}^{-1}$ . Find the ratio of populations $N_2/N_0$ at $T = 298\ \text{K}$ . Step 1: Convert $B$ to an energy-equivalent temperature-scaled quantity. Using $E_J = BhcJ(J+1)$ :

E_2 = (10.59\ \text{cm}^{-1})(6.626\times10^{-34}\ \text{J s})(2.998\times10^{10}\ \text{cm s}^{-1})(2)(3)

First compute $hc = (6.626\times10^{-34})(2.998\times10^{10}) = 1.986\times10^{-23}\ \text{J cm}$ .

E_2 = (10.59)(1.986\times10^{-23}\ \text{J})(6) = 1.262\times10^{-21}\ \text{J}

Step 2: Compute $k_BT$ at 298 K:

k_BT = (1.381\times10^{-23}\ \text{J K}^{-1})(298\ \text{K}) = 4.115\times10^{-21}\ \text{J}

Step 3: Boltzmann factor for $J=2$ :

e^{-E_2/k_BT} = e^{-1.262\times10^{-21}/4.115\times10^{-21}} = e^{-0.3067} = 0.7359

Step 4: Include degeneracies. $g_2 = 2(2)+1 = 5$ , $g_0 = 2(0)+1 = 1$ . So:

\frac{N_2}{N_0} = \frac{g_2}{g_0}\, e^{-(E_2-E_0)/k_BT} = 5 \times 0.7359 = 3.68

Sanity check: the answer exceeds 1, confirming that $J=2$ is more populated than $J=0$ at room temperature — consistent with the hump in the population curve occurring at $J>0$ . This matches the qualitative picture in the figure above.

Example 2 — Applied: locating the most intense line in the HCl rotational spectrum and comparing to a heavier molecule. Using the boxed formula, predict $J_{max}$ for HCl ( $B = 10.59\ \text{cm}^{-1}$ ) at 298 K, and compare with CO ( $B = 1.931\ \text{cm}^{-1}$ ) at the same temperature. Step 1: Compute $k_BT$ in units matching $hcB$ (cm⁻¹). Using $k_B = 1.381\times10^{-23}\ \text{J K}^{-1}$ and $hc = 1.986\times10^{-23}\ \text{J cm}$ :

\frac{k_BT}{hc} = \frac{(1.381\times10^{-23})(298)}{1.986\times10^{-23}\ \text{J cm}} = 207.2\ \text{cm}^{-1}

Step 2: For HCl, apply the boxed formula:

J_{max} = \sqrt{\frac{207.2}{2(10.59)}} - \frac{1}{2} = \sqrt{9.786} - 0.5 = 3.128 - 0.5 = 2.63

Rounding to the nearest integer, $J_{max} = 3$ (or the transition $J=3\to4$ dominates), consistent with the labeled peak in the population figure. Step 3: For CO, repeat with $B = 1.931\ \text{cm}^{-1}$ :

J_{max} = \sqrt{\frac{207.2}{2(1.931)}} - \frac{1}{2} = \sqrt{53.65} - 0.5 = 7.325 - 0.5 = 6.83

Rounding, $J_{max} \approx 7$ for CO.

Sanity check: CO has a much smaller rotational constant (larger moment of inertia, since CO is heavier and stiffer bonds shift energy spacing less per unit $J$ ) than HCl, so its rotational levels are more closely spaced. More levels fit below $k_BT$ , so the population maximum shifts to higher $J$ — exactly the trend the formula predicts. Both results agree with textbook microwave spectra of HCl (strongest line near $J=3\to4$ ) and CO (strongest lines further along the ladder, around $J=7\to8$ ).

Limiting cases & common traps

Situation	What happens	Why
$T \to 0$	Only $J=0$ populated; $J_{max}=0$	Boltzmann factor collapses for any $E>0$
$T \to \infty$	$J_{max}\to\infty$ , spectrum spreads over many lines	Degeneracy term dominates over an ever-flatter exponential
Large $B$ (light molecule, e.g. H₂)	Few levels populated, $J_{max}$ small even near room T	Level spacing $\propto B$ is large compared with $k_BT$
Small $B$ (heavy molecule, e.g. I₂)	Many levels populated, $J_{max}$ large	Level spacing small compared with $k_BT$ , so spectrum resembles a continuum envelope
Vibrational spectra	Almost only $v=0$ populated at 298 K	Vibrational spacing (hundreds to thousands of cm⁻¹) far exceeds $k_BT \approx 207\ \text{cm}^{-1}$

The chemist behind it

Ludwig Boltzmann's statistical interpretation of entropy and his population law gave chemists the tool to connect microscopic energy-level structure to macroscopic, measurable quantities — including the intensity pattern of a spectrum. When spectroscopists in the twentieth century began recording rotational and vibrational spectra with enough resolution to see individual lines, it was Boltzmann's distribution, combined with the degeneracy counting rules from quantum mechanics, that explained why some lines in a rotational band are dramatically taller than others. This synthesis of statistical mechanics and spectroscopy became a routine diagnostic: measuring the intensity envelope of a rotational spectrum lets chemists read off the temperature of a sample, from interstellar gas clouds to combustion flames.

Origins · 1 minLudwig Boltzmann and how this was discovered

Ludwig Boltzmann

1844–1906 · Austrian physicist whose statistical distribution law underlies every population-weighted spectral intensity calculation

Ask the tutor: How does adding centrifugal distortion to the rotational energy formula shift the predicted position of the most intense line?

Why does degeneracy matter for spectral intensity at all, if the Boltzmann factor already tells us the population of a single quantum state?

The Boltzmann factor $e^{-E/k_BT}$ gives the relative probability of occupying one specific quantum state. But a spectral level often groups several degenerate states of identical energy — for rotation, the $2J+1$ values of $M_J$ . All of these states absorb at the same frequency, so their populations add. The total population of the level is the single-state Boltzmann factor multiplied by the number of degenerate states, $g_i e^{-E_i/k_BT}$ , and it is this total that sets the line intensity.

Does the intensity formula I is proportional to (N_i minus N_f) times the transition dipole squared apply to emission as well as absorption?

Yes, with the initial and final states swapped in role. In absorption, $N_i$ is the lower-energy population and $N_f$ the upper; net absorption requires $N_i > N_f$ , which is guaranteed at thermal equilibrium since lower levels are always more populated for equal degeneracy. In emission, the initially excited (upper) population decays, and intensity again tracks that population, weighted by the same transition dipole moment and an extra frequency-dependent factor from the Einstein A coefficient.

Why is J_max around 3 for HCl but closer to 7 for CO, when both are diatomic gases at the same temperature?

$J_{max}$ depends on the ratio of thermal energy to rotational constant, scaling as the square root of $k_BT$ divided by $B$ . HCl has a small reduced mass and a short, stiff bond, giving a large rotational constant $B = 10.59\ \text{cm}^{-1}$ , so its energy levels are widely spaced. CO's rotational constant is only $1.931\ \text{cm}^{-1}$ , roughly five times smaller, so many more levels fit below $k_BT$ , pushing the population maximum to higher $J$ .

How does this population argument connect to what comes after, in microwave rotational spectroscopy?

The Boltzmann-and-degeneracy argument developed here explains the overall envelope, or intensity profile, of a rotational absorption spectrum, but it treats each line as infinitely sharp. Real spectra show finite linewidths from Doppler motion, pressure (collisional) broadening, and the finite lifetime of excited states. Microwave Rotational Spectroscopy builds on this population framework and adds those broadening mechanisms to explain the observed lineshape.

Can this same reasoning be applied to vibrational or electronic spectra, not just rotational ones?

Yes, the logic is general: intensity tracks population, and population tracks degeneracy times Boltzmann factor. The difference is scale. Vibrational level spacings (hundreds to thousands of cm⁻¹) and electronic spacings (tens of thousands of cm⁻¹) are so much larger than $k_BT \approx 207\ \text{cm}^{-1}$ at room temperature that essentially only the ground vibrational and electronic state is populated. That is why vibrational "hot bands" from $v=1$ are weak at room temperature but grow noticeably at higher temperatures.