Diamond has a Debye temperature Θ_D ≈ 2200 K; lead has Θ_D ≈ 100 K. At room temperature (≈ 300 K), which material has a heat capacity closer to the classical Dulong-Petit value of 3R?
ADiamond, because a higher Θ_D means more phonon modes are activated at a given temperature
BLead, because 300 K >> 100 K so nearly all of lead's phonon modes are thermally accessible at room temperature
CBoth have the same heat capacity because Dulong-Petit applies universally above absolute zero
DDiamond, because stiff bonds store more energy per degree of freedom than weak bonds
The Debye temperature sets the scale above which Dulong-Petit applies: when T >> Θ_D, all modes are thermally accessible and C_V ≈ 3R. For lead, 300 K >> 100 K, so nearly all phonon modes are active and lead's heat capacity is close to 3R. For diamond, 300 K << 2200 K, so most high-frequency modes are frozen out; diamond's heat capacity at room temperature is well below 3R. Option A reverses the logic: high Θ_D means modes are hard to excite (stiff bonds, high frequencies), not easy.
Question 2 Multiple Choice
Why does the Debye model predict C_V ∝ T³ at low temperatures, rather than the temperature-independent 3R of Dulong-Petit?
AAt low temperatures, atoms vibrate more slowly, reducing the number of atomic collisions that transfer heat
BAt low temperatures, only low-frequency phonon modes are thermally accessible, and their density of states g(ω) ∝ ω² produces a T³ integral for total energy
CAt low temperatures, quantum zero-point energy dominates and suppresses thermal fluctuations by a factor proportional to T³
DDulong-Petit fails at low T because atoms rearrange into a different crystal structure with fewer degrees of freedom
The T³ law has two ingredients: (1) only modes with ℏω ≲ k_BT are appreciably excited — at low T, this restricts access to low-frequency, long-wavelength phonons near the bottom of the spectrum; (2) the density of states is g(ω) ∝ ω², so the number of accessible modes grows as (k_BT/ℏ)². Combining the energy per mode (~k_BT) with the count of accessible modes (~T²) gives total energy U ∝ T³ and C_V = dU/dT ∝ T³. At high T, all modes are accessible and equipartition gives the constant 3R.
Question 3 True / False
At temperatures much higher than the Debye temperature (T >> Θ_D), the Debye model recovers the classical Dulong-Petit result C_V = 3R.
TTrue
FFalse
Answer: True
This is the high-temperature limit: when k_BT >> ℏω for all phonon modes, the quantum Planck distribution for each mode reduces to the classical result and each mode contributes k_B to the heat capacity. Summing over all 3N modes gives C_V = 3Nk_B = 3R per mole. This is a necessary consistency check for any correct quantum model of solids: it must reproduce classical thermodynamics in the high-temperature limit where quantum effects are negligible.
Question 4 True / False
The Einstein model (most atoms vibrating at a single frequency) and the Debye model both correctly predict the T³ dependence of heat capacity at low temperatures.
TTrue
FFalse
Answer: False
At low temperatures, the Einstein and Debye models diverge significantly. The Einstein model predicts C_V ∝ e^(−Θ_E/T) at low T — an exponential decay, far faster than any power law — because a single uniform frequency means all modes are equally and sharply frozen out below Θ_E. The Debye model predicts the T³ law because low-frequency modes (ω → 0) are always accessible, and there are ω² of them per frequency interval. Experimental measurements fit the T³ law; the Einstein model overestimates how rapidly heat capacity drops at low temperature.
Question 5 Short Answer
Why are only low-frequency phonon modes excited at low temperatures, and how does this produce the T³ temperature dependence of heat capacity?
Think about your answer, then reveal below.
Model answer: A phonon mode contributes to heat capacity only if thermal energy k_BT is sufficient to excite it — specifically when k_BT ≳ ℏω. At low T, k_BT is small, so only modes with very low frequency ω satisfy this condition; high-frequency modes are frozen out. The number of accessible modes is those with ω ≲ k_BT/ℏ, and since the density of states is g(ω) ∝ ω², integrating up to this cutoff gives an accessible mode count proportional to (k_BT)³/ℏ³ ∝ T³. Each accessible mode contributes ~k_B to heat capacity, so C_V ∝ T³.
The T³ law is a prediction of the density-of-states structure of 3D wave propagation, not an empirical fit. The ω² density of states reflects that in three dimensions there are quadratically more long-wavelength modes than short-wavelength ones (the same geometry that gives the Planck distribution for photons). At low T, we sample only the bottom of this ω² spectrum, and the T³ result follows inevitably from the counting.