Questions: Quantized Energy Levels and Spectroscopic Transitions

Question 1

A molecule has two energy levels separated by 0.025 eV (approximately k_BT at room temperature). At room temperature, what can you say about the populations of these two levels?

• A. The upper level is essentially unpopulated — thermal energy is never enough to populate excited states
• B. The upper level is significantly populated; with ΔE ≈ k_BT, the Boltzmann factor exp(−ΔE/k_BT) ≈ e⁻¹ ≈ 0.37, so the upper level has about 37% of the ground-state population
• C. Both levels are equally populated — when ΔE equals k_BT, the levels are exactly equal
• D. The upper level is more populated — thermal energy pushes electrons upward

Answer: (B)

The Boltzmann factor exp(−ΔE/k_BT) gives the ratio of upper to lower state population (ignoring degeneracy). When ΔE = k_BT, this factor is e⁻¹ ≈ 0.37, meaning the upper level has ~37% of the lower level's population — significantly populated. This is typical for rotational levels of small molecules at room temperature. Electronic levels (ΔE ~ eV >> k_BT) have Boltzmann factors of ~10⁻¹⁷ and are essentially unpopulated, which is why electronic absorption spectra show only ground-state transitions.

Question 2

A spectroscopist observes that two energy levels exist in a molecule — confirmed by calculations — but she cannot observe any spectral line connecting them. Which explanation is most likely correct?

• A. The levels are too close together for any spectrometer to resolve
• B. The transition violates a selection rule, making the transition dipole moment zero or near-zero and the transition essentially 'dark'
• C. The upper level is unpopulated because no photons at that frequency are present in the spectrometer
• D. The transition can only be observed at very low temperatures when Boltzmann populations shift

Answer: (B)

Selection rules arise from the quantum-mechanical requirement that the transition dipole moment integral ∫ψ*_f μ ψ_i dτ be nonzero. If the wavefunctions have the wrong symmetry or angular momentum properties, this integral evaluates to zero — the transition is 'forbidden' and produces no (or a very weak) spectral line. Common examples: Δl = ±1 for electronic transitions in atoms; ΔJ = ±1 for rotational transitions; the symmetry selection rule for IR vs. Raman activity. The levels exist; the transition just has negligible probability.

Question 3

Atomic line spectra appear as sharp, discrete lines rather than continuous bands because transitions between discrete energy levels produce photons of only specific frequencies.

• T. True
• F. False

Answer: True

Each spectral line corresponds to one transition between two specific energy levels: ΔE = hν. Since the energy levels are discrete (quantized), the allowed ΔE values are a specific set — not a continuum. Each ΔE produces photons at one frequency ν = ΔE/h. The pattern of line positions is a unique fingerprint of the atom's energy-level structure. Continuous spectra arise from different sources: blackbody radiation, bremsstrahlung, or transitions to or from a continuum of states (ionization).

Question 4

If you heat a gas to a higher temperature, most rotational spectral lines become equally more intense because more molecules have energy to undergo transitions.

• T. True
• F. False

Answer: False

Heating changes the *distribution* of population across levels, not all lines equally. At higher temperature, the Boltzmann distribution spreads population over more rotational levels (since kT is larger). Lines from low-J states weaken as population migrates to higher J levels; lines from intermediate-J transitions strengthen; and lines from very high J states appear. The envelope of line intensities shifts to higher J. For vibrational or electronic transitions (where ΔE >> kT), higher temperature still primarily populates the ground state and barely affects those line intensities.

Question 5

Explain why rotational spectra of molecules have many observable lines while electronic spectra of atoms at room temperature show only absorption from the ground state.

Model answer: Rotational energy spacings (microwave region, ~0.001 eV) are much smaller than k_BT at room temperature (~0.025 eV). The Boltzmann factor exp(−ΔE/k_BT) is close to 1 for low J levels, so many rotational levels are significantly populated, and absorption from multiple levels produces many lines. Electronic energy spacings in atoms (UV/visible, ~2–10 eV) are far larger than k_BT. The Boltzmann factor exp(−E/k_BT) is ~10⁻³⁴ to 10⁻¹⁷ for excited electronic states — essentially zero. All population is in the ground state, so only ground-state absorption is observed.

This is why rotational spectroscopy produces rich spectra with hundreds of lines from a single molecule, while atomic emission spectroscopy requires high-temperature sources (flames, electrical discharges) to populate excited electronic states before emission can occur. The Boltzmann distribution is the bridge between the quantum energy ladder and what you actually see in the spectrometer.