At room temperature, which rotational quantum number J has the highest population for a typical diatomic molecule?
AJ = 0 — the lowest energy state is always the most populated
BJ = 1 — the first excited state gains population from Boltzmann factors
CAn intermediate J value (e.g., 3–8 for many small molecules) — where the Boltzmann factor and degeneracy together are maximized
DThe highest thermally accessible J — higher states are always favored at room temperature
Each J level has a degeneracy of (2J+1), meaning there are more quantum states at higher J. The population of level J is proportional to (2J+1)·exp(−E_J/kT). At low J, degeneracy is small and Boltzmann suppression is weak, so population rises with J. At high J, Boltzmann suppression dominates and population falls. The result is a peak at an intermediate J_max ≈ √(kT/2hcB) − 1/2. The common misconception (option A) ignores degeneracy entirely.
Question 2 Multiple Choice
The spacing between adjacent lines in the pure rotational microwave spectrum of molecule X is twice the spacing observed for molecule Y. What can you conclude?
AMolecule X has twice the molar mass of molecule Y
BMolecule X has twice the moment of inertia of molecule Y
CMolecule X has half the moment of inertia of molecule Y, consistent with a shorter bond or lighter atoms
DMolecule X has a stronger dipole moment than molecule Y
Line spacing in a pure rotational spectrum equals 2B, where B = ℏ²/(2I). If X has twice the spacing, then B_X = 2B_Y, which means I_X = I_Y/2 — molecule X has half the moment of inertia. Since I = μr² (μ = reduced mass, r = bond length), this could reflect a shorter bond length, lighter atoms, or both. Molar mass alone (option A) doesn't determine moment of inertia — the geometry and bond length matter. Dipole moment (option D) affects spectral intensity but not line spacing.
Question 3 True / False
The lines in a pure rotational microwave spectrum of a rigid diatomic molecule are equally spaced.
TTrue
FFalse
Answer: True
For a rigid rotor, E_J = BJ(J+1). The transition energy for ΔJ = +1 from level J to J+1 is E(J+1) − E(J) = B[(J+1)(J+2) − J(J+1)] = 2B(J+1). As J increases by 1, each successive transition energy increases by 2B — giving equally spaced lines separated by 2B. This elegant regularity is why rotational spectra serve as precise molecular fingerprints. (Note: centrifugal distortion causes slight deviations at high J in real molecules.)
Question 4 True / False
Raising the temperature of a gas sample generally increases the population of the J = 0 rotational level at the expense of higher J levels.
TTrue
FFalse
Answer: False
This is the opposite of what happens. Increasing temperature provides more thermal energy, populating higher J levels. The J = 0 state has degeneracy 1, and at elevated temperature, the Boltzmann distribution shifts the population peak to higher J values — J = 0 actually becomes relatively less populated. At absolute zero, all molecules would be in J = 0, but as temperature rises, the population redistributes toward higher J states.
Question 5 Short Answer
Why do the intensities of rotational spectral lines first increase and then decrease as J increases, rather than simply decreasing from J = 0? What physical factors compete to produce this pattern?
Think about your answer, then reveal below.
Model answer: Two competing factors control the intensity. First, degeneracy: each level J has (2J+1) degenerate substates, so higher J levels have more ways to be occupied — this factor alone would push population toward higher J. Second, Boltzmann suppression: the energy E_J = BJ(J+1) increases rapidly with J, so the thermal occupancy factor exp(−E_J/kT) decreases at high J. At low J, degeneracy wins and population rises; at high J, Boltzmann suppression wins and population falls. The peak occurs at an intermediate J where the two effects balance.
This intensity envelope is diagnostic of the temperature of the gas — the peak shifts to higher J at higher temperatures because the Boltzmann suppression weakens. Astrophysicists use this property to determine the temperature of cold interstellar molecular clouds from the relative intensities of rotational emission lines, without needing a thermometer in space.