The Boltzmann Distribution

Introduction

Of all the results in statistical thermodynamics, one equation is used more than any other: the Boltzmann distribution . It tells us what fraction of molecules occupies each energy state at thermal equilibrium. It explains why reaction rates rise steeply with temperature, why spectral lines have the intensities they do, and why the atmosphere thins with altitude. Everything that follows in this unit builds on it.

Core explanation

Statement. For a system at temperature T in thermal equilibrium, the fraction of molecules in a state i of energy εᵢ is

nᵢ/N = exp(−εᵢ/kT) / q

where q = Σⱼ exp(−εⱼ/kT) is the sum of Boltzmann factors over all states. Dividing by q ensures that the fractions add up to 1. It is common to write β = 1/kT, so each state carries a Boltzmann factor exp(−βεᵢ).

Origin. The distribution is the dominant configuration of a large system. Maximising ln W subject to fixed N and fixed total energy (using Lagrange multipliers) gives nᵢ ∝ exp(−βεᵢ). Comparing the resulting thermodynamic predictions with those of a perfect gas identifies β = 1/kT. Thus the distribution is not an extra assumption: it is simply the overwhelmingly most probable way of sharing energy.

Population ratios. The partition function cancels when comparing two states:

nⱼ/nᵢ = exp[−(εⱼ − εᵢ)/kT]

This is the most useful practical form. Only the energy difference matters, so the choice of zero of energy is irrelevant for ratios.

Key features.

- Lower-energy states are always more populated than higher-energy states at any positive temperature. - As T → 0, all molecules fall into the ground state. - As T → ∞, the exponentials all approach 1 and the states become equally populated. Populations never invert at equilibrium. - The decisive quantity is Δε/kT. If Δε ≪ kT the states are nearly equally populated; if Δε ≫ kT the upper state is essentially empty.

States versus levels. The formula above refers to individual states . If a level contains g states of equal energy (degeneracy g), its population is multiplied by g. A degenerate upper level can therefore hold more molecules than a lower one, as later pages show for molecular rotation.

Units. Energies may be in joules per molecule with k, in J mol⁻¹ with R, or in cm⁻¹ with kT/hc (about 207 cm⁻¹ at 298 K). What matters is that numerator and denominator are in the same units.

Formulae

nᵢ/N = exp(−εᵢ/kT)/q; nⱼ/nᵢ = exp(−Δε/kT); per mole, nⱼ/nᵢ = exp(−ΔE m/RT); kT/hc = 0.695 cm⁻¹ K⁻¹ × T.

Step-by-step reasoning

To find a population ratio:

1. Identify the energy difference Δε between the two states. 2. Express kT in the same units (joules, J mol⁻¹ via RT, or cm⁻¹). 3. Calculate the dimensionless ratio Δε/kT. 4. Evaluate exp(−Δε/kT); multiply by the ratio of degeneracies if comparing levels rather than states.

Visual explanation

Draw a ladder of equally spaced states and represent each population by a horizontal bar. The bar lengths fall away exponentially up the ladder. At low temperature only the bottom bar is long; at high temperature the bars shrink slowly and many rungs are occupied.

Real-world analogy

Think of people climbing a hillside to picnic, where each step up costs effort. Most settle near the bottom; fewer climb higher, and the number falls by the same factor for every extra metre. On an energetic day (higher temperature) people spread further up the slope, but the bottom is still the most crowded.

Real-world example

The Earth's atmosphere follows a Boltzmann distribution in gravitational potential energy: n(h)/n(0) = exp(−mgh/kT). For nitrogen near 288 K, the density falls by a factor of e for roughly every 8.7 km of height. This is why climbers need supplementary oxygen at the altitude of the highest mountains.

Why?

Why is the dependence exponential? Energy is shared among an enormous number of molecules. Each additional unit of energy given to one molecule must be taken from the rest, reducing the number of ways the rest can be arranged by a constant factor. Repeated multiplication by a constant factor is exactly an exponential.

Common misconception

"At high temperature, most molecules move into the upper states." At thermal equilibrium, the populations of higher states approach, but never exceed, those of lower states. More molecules in an upper state than a lower one (population inversion, as in a laser) is a non-equilibrium situation.

Worked example

Question: Two states are separated by 1000 cm⁻¹. Calculate the ratio of upper to lower populations at 298 K and at 1000 K.

Reasoning: At 298 K, kT/hc = 0.695 × 298 = 207 cm⁻¹, so Δε/kT = 1000/207 = 4.83 and the ratio is exp(−4.83) ≈ 8.0 × 10⁻³. At 1000 K, kT/hc = 695 cm⁻¹, so Δε/kT = 1.44 and the ratio is exp(−1.44) ≈ 0.24.

Answer: About 0.008 at 298 K and 0.24 at 1000 K — a thirtyfold increase for a roughly threefold rise in temperature.

Quick check

1. What is the ratio of populations of two non-degenerate states whose energies differ by exactly kT? Answer: exp(−1) ≈ 0.37, so the upper state holds about 37% as many molecules as the lower one.

Exam focus

Examiners expect confident use of nⱼ/nᵢ = exp(−Δε/kT) with consistent units. Common errors are mixing molar and molecular energies (use R with J mol⁻¹, k with J per molecule) and forgetting degeneracy factors when levels rather than states are compared.

Advanced insight

The Boltzmann distribution is the classical limit of quantum statistics. Fermi–Dirac statistics (for fermions such as electrons) and Bose–Einstein statistics (for bosons such as ⁴He) differ from it when the average occupancy of a state approaches 1. For molecular gases at ordinary temperatures, occupancies are tiny, and Boltzmann statistics is essentially exact.

Summary

At thermal equilibrium the fraction of molecules in a state of energy εᵢ is exp(−εᵢ/kT)/q. Population ratios depend only on Δε/kT. Lower states are always more populated than higher states, all molecules are in the ground state as T → 0, and states become equally populated only as T → ∞. The distribution emerges as the dominant configuration of a large system sharing a fixed energy.

Practice questions

1. Calculate the ratio of populations of two states separated by 5.0 kJ mol⁻¹ at 300 K. Answer: ΔE/RT = 5000/(8.314 × 300) = 2.00, so the ratio is exp(−2.00) ≈ 0.14. 2. The vibrational wavenumber of HCl is 2886 cm⁻¹. What fraction of the ground-state population is in v = 1 at 298 K? Answer: 2886/207 ≈ 13.9, so the ratio is exp(−13.9) ≈ 9 × 10⁻⁷ — essentially all molecules are in v = 0. 3. At what temperature is the population ratio of two states separated by 200 cm⁻¹ equal to 0.50? Answer: exp(−200/(0.695T)) = 0.50 gives 200/(0.695T) = ln 2 = 0.693, so T ≈ 415 K. 4. Explain why the choice of zero of energy does not affect population ratios. Answer: Shifting all energies by a constant multiplies every Boltzmann factor by the same factor, which cancels in any ratio and in the normalisation by q.