Boltzmann Distribution: Population Ratio of Energy Levels

By Aniket Bhardwaj · 23 October 2026 · Updated 23 October 2026 · Statistical Thermodynamics

The Boltzmann distribution answers a basic question of statistical thermodynamics: at temperature T, how many molecules sit in a higher energy level compared with a lower one? It sits behind the intensities of rotational and vibrational spectral lines, the temperature dependence of reaction rates and the idea of the partition function. In IIT-JAM, GATE and CSIR-NET physical chemistry it appears as short numericals that depend on two things only: keeping units consistent and not forgetting degeneracy.

A four bar population ratio chart shows values 0.00244, 0.1347, 0.548, and 0.818 at 100, 300, 1000, and 3000 K. — As temperature rises, the upper-to-lower population ratio approaches 1.

The formula

For two levels with energies E₁ and E₂ (E₂ > E₁) in a system at thermal equilibrium:

N₂ ÷ N₁ = (g₂ ÷ g₁) × exp(−ΔE ÷ k_BT)
Molar form: N₂ ÷ N₁ = (g₂ ÷ g₁) × exp(−ΔE_m ÷ RT)
Wavenumber form: ΔE ÷ k_BT = (hc/k_B) × ν̃ ÷ T, with hc/k_B = 1.4388 cm K

N₂, N₁ are the numbers of molecules in the two levels. g is the degeneracy, the number of distinct states at that energy. ΔE = E₂ − E₁ is the gap in joules per molecule. k_B = 1.380649 × 10⁻²³ J K⁻¹ is the Boltzmann constant. ΔE_m is the same gap per mole (J mol⁻¹), used with R = 8.314 J K⁻¹ mol⁻¹. ν̃ is the wavenumber of the gap in cm⁻¹. Use k_B with J per molecule, or R with J per mole, never one with the other.

The exponent ΔE ÷ k_BT is a pure number. It compares the gap with the thermal energy k_BT. If the gap is much larger than k_BT, the upper level is almost empty; if the gap is much smaller, the two levels are nearly equally populated (apart from g).

Worked example 1: gap equal to k_BT

Question. Two non-degenerate levels are separated by ΔE = 4.142 × 10⁻²¹ J. Find N₂ ÷ N₁ at 300 K.

Thermal energy. k_BT = 1.380649 × 10⁻²³ × 300 = 4.142 × 10⁻²¹ J.

Exponent. ΔE ÷ k_BT = 1.000.

Ratio. N₂ ÷ N₁ = e⁻¹ = 0.3679. Fraction in the upper level of a two-level system = 0.3679 ÷ (1 + 0.3679) = 0.269, and in the lower level 0.731.

Worked example 2: a gap given per mole

Question. Two non-degenerate levels differ by 5.0 kJ mol⁻¹. Find the population ratio at 100 K, 300 K, 1000 K and 3000 K, and the temperature at which the ratio reaches 0.5.

Convert to J. ΔE_m = 5000 J mol⁻¹.

At 300 K. RT = 8.314 × 300 = 2494 J mol⁻¹. Exponent = 5000 ÷ 2494 = 2.005. Ratio = e^−2.005 = 0.1347.

Other temperatures. At 100 K: exponent 6.014, ratio 0.00244. At 1000 K: exponent 0.6014, ratio 0.548. At 3000 K: exponent 0.2005, ratio 0.818.

Ratio = 0.5. ln 2 = ΔE_m ÷ RT, so T = 5000 ÷ (8.314 × 0.6931) = 867.6 K.

T (K)	ΔE_m ÷ RT	N₂ ÷ N₁
100	6.014	0.00244
300	2.005	0.1347
1000	0.6014	0.548
3000	0.2005	0.818

As T rises, the ratio climbs towards 1 for non-degenerate levels but never exceeds it. At infinite temperature the populations depend only on g.

Worked example 3: a gap in wavenumbers

Question. A vibrational level lies 2000 cm⁻¹ above the ground level (a round number chosen for practice, not a specific molecule). Find the population ratio at 300 K and at 1000 K, taking both levels as non-degenerate.

At 300 K. Exponent = 1.4388 × 2000 ÷ 300 = 9.592. Ratio = e^−9.592 = 6.83 × 10⁻⁵.

At 1000 K. Exponent = 1.4388 × 2000 ÷ 1000 = 2.878. Ratio = e^−2.878 = 0.0563.

At room temperature almost every molecule is in the lowest vibrational level. This is why a spectrum of a gas at room temperature is dominated by transitions starting from the ground state, while a hot gas shows additional lines from excited levels.

Worked example 4: degeneracy matters

Question. A non-degenerate ground level (g₁ = 1) lies below a triply degenerate level (g₂ = 3). The gap is 0.5 k_BT at the temperature of interest. Find N₂ ÷ N₁ and the fraction of molecules in the upper level.

Ratio. N₂ ÷ N₁ = 3 × e^−0.5 = 3 × 0.6065 = 1.820.

Fraction in the upper level. 1.820 ÷ (1 + 1.820) = 0.645.

The upper level holds more molecules than the lower one, even though its energy is higher, because there are three states there to occupy. The Boltzmann factor applies to each state; the degeneracy multiplies it for a whole level.

Worked example 5: finding the gap from populations

Question. At 400 K the upper level of a non-degenerate pair has one quarter of the population of the lower. Find ΔE per mole.

Solution. ln(N₂÷N₁) = −ΔE_m ÷ RT, so ΔE_m = RT ln 4 = 8.314 × 400 × 1.3863 = 4610 J mol⁻¹ = 4.61 kJ mol⁻¹.

Another case. If the ratio at 300 K were 10⁻³: ΔE_m = 8.314 × 300 × ln 1000 = 17.2 kJ mol⁻¹.

Common mistakes

Mixing k_B and R. k_B goes with joules per molecule; R goes with joules per mole. A kJ mol⁻¹ gap must be turned into J mol⁻¹ first.
Dropping the degeneracy factor. For rotational levels g = 2J + 1, so the most populated level is not the lowest.
Forgetting the minus sign. The exponent is negative; a ratio above 1 for non-degenerate levels is always an error.
Using the ratio when T is very low. The formula is fine, but check that your calculator does not round e^−x to zero too early.
Confusing cm⁻¹ with energy. A wavenumber must be multiplied by hc to become an energy, or you can use hc/k_B = 1.4388 cm K directly.
Treating the ratio as a fraction. N₂ ÷ N₁ is not N₂ ÷ N_total. Add the levels to get the total.

Exam relevance

Situation	What to remember
Gap >> k_BT	Upper level almost empty
Gap << k_BT	Ratio approaches g₂ ÷ g₁
Rising temperature	Upper-level fraction increases
Degenerate levels	Multiply the Boltzmann factor by g₂ ÷ g₁
Gap in cm⁻¹	Exponent = 1.4388 ν̃ ÷ T

Check the current official notification for your exam's syllabus. The ideas here are standard, but the depth asked (two levels, many levels, partition function) varies.

The suite has no dedicated Boltzmann-distribution tool. Use the scientific calculator for the exponential and logarithm steps, and keep the constants from the Scientific Constants panel nearby to avoid typing errors.

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