Questions: Boltzmann Distribution and Molecular Populations

Question 1

A molecule has a ground state at E=0 and an excited state at energy ε. At temperature T where k_BT = ε/4, approximately what fraction of molecules occupy the excited state?

• A. About 50%, because both states are available
• B. Very small (close to 0), because k_BT is much less than ε
• C. About 25%, equal to the ratio k_BT/ε
• D. Exactly ε/(2k_BT), from the Boltzmann formula

Answer: (B)

When k_BT = ε/4, the ratio ε/k_BT = 4, so the Boltzmann factor is e^(-4) ≈ 0.018 — very small. The excited state population is proportional to e^(-ε/k_BT), which becomes vanishingly small when thermal energy is much less than the level spacing. Option A is wrong because availability does not equal equal population; option C mistakes a linear ratio for an exponential one.

Question 2

A spectroscopist observes that at room temperature a vibrational spectral band has very low intensity, but the intensity increases dramatically when the sample is heated. What does this tell us about the energy spacing of the vibrational levels?

• A. The vibrational spacing is much smaller than k_BT at room temperature
• B. The vibrational spacing is much larger than k_BT at room temperature
• C. Heating increases the number of molecules, increasing intensity
• D. The energy spacing increases with temperature

Answer: (B)

Low intensity at room temperature means the absorbing state (excited vibrational level) has very low population — the Boltzmann factor e^(-E/k_BT) is small, meaning E ≫ k_BT. As temperature rises, k_BT approaches the level spacing and more molecules populate the excited state, increasing absorption intensity. Option A would predict high population (and high intensity) at room temperature. Heating does not create molecules (C), and energy spacings are fixed by molecular structure (D).

Question 3

At very high temperatures, all energy levels in a multi-level system approach equal population.

• T. True
• F. False

Answer: True

As T → ∞, k_BT becomes arbitrarily large compared to any finite energy spacing. Every Boltzmann factor e^(-E_i/k_BT) approaches e^0 = 1 regardless of E_i, so all levels carry equal weight. This is the high-temperature limit where thermal energy completely overwhelms energy level differences, and the distribution becomes uniform.

Question 4

In a two-level system, raising the temperature generally increases the fraction of molecules in the ground state.

• T. True
• F. False

Answer: False

Raising temperature increases the population of excited states and *decreases* the fraction in the ground state. The Boltzmann distribution spreads population across more states as temperature rises. At very low T, nearly all molecules are in the ground state; as T increases, population flows into excited states. The ground state fraction N₀/N_total = 1/(1 + e^(-ε/k_BT)) decreases monotonically as T increases.

Question 5

Why does the Boltzmann distribution predict that reaction rates increase with temperature, even when the reaction's overall energy change is unchanged?

Model answer: Reaction rates increase with temperature because the Boltzmann distribution shifts more molecules into high-energy states as T rises. The fraction of molecules with enough energy to surmount the activation barrier E_a is proportional to e^(-E_a/k_BT). As T increases, k_BT grows, the exponential factor becomes larger, and more molecules have enough energy to react. The activation energy has not changed — but the fraction of the population above it has.

This is the physical content of the Arrhenius equation k = A·e^(-E_a/RT), which is a direct consequence of the Boltzmann distribution. Temperature is the knob that controls which states are thermally accessible — it shifts the population distribution, not the energy levels themselves.