Diamond has an unusually low heat capacity at room temperature compared to most solids, well below the classical Dulong-Petit value of 3R. The Einstein model explains this because:
ADiamond atoms are arranged in a rigid lattice that prevents vibration entirely
BDiamond's high bond stiffness gives a large Einstein temperature Θ_E, so room temperature is in the 'quantum frozen' regime where oscillators cannot be thermally excited
CDiamond has fewer atoms per mole than most elements, so its total heat capacity is lower
DDiamond is a semiconductor, so electrons rather than lattice vibrations dominate its heat capacity
The Einstein temperature Θ_E = ℏω_E/k sets the crossover scale between classical (T >> Θ_E, C_V ≈ 3R) and quantum-frozen (T << Θ_E, C_V → 0) regimes. Diamond's stiff C-C covalent bonds give very high vibrational frequencies ω_E, so Θ_E ≈ 1320 K — far above room temperature (~300 K). At room temperature, thermal energy kT is insufficient to excite most oscillators to their first quantum level, so they remain frozen in their ground states and contribute little to the heat capacity. This was a direct triumph of the Einstein model: explaining diamond's anomalous heat capacity from quantum mechanics.
Question 2 Multiple Choice
The Dulong-Petit law (C_V ≈ 3R for elemental solids at high temperatures) is both a result of classical equipartition AND a prediction of the Einstein quantum model. What does this consistency tell us?
AThe quantum model must be wrong at high temperatures because it agrees with the classical result
BQuantum mechanics reduces to classical mechanics in the high-temperature limit, where thermal energy greatly exceeds the quantum level spacing
CBoth models make identical predictions at all temperatures, differing only in computational complexity
DThe equipartition theorem is a quantum result, so its agreement with Einstein's model is expected
When T >> Θ_E, the thermal energy kT greatly exceeds the oscillator level spacing ℏω. In this limit, the discrete quantum energy levels are so closely spaced relative to kT that they effectively form a continuum, and the quantum calculation converges to the classical equipartition result: each of the 3N oscillators gets kT of mean energy, giving C_V = 3Nk = 3R. This is the **correspondence principle**: quantum mechanics must reproduce classical results where classical mechanics is known to work. Einstein's model satisfies this at high T — it was wrong only at low T, where quantum discreteness becomes essential.
Question 3 True / False
The Einstein model correctly predicts that the heat capacity of a solid approaches zero as temperature approaches absolute zero.
TTrue
FFalse
Answer: True
True. This was a key success of the Einstein model — it explained, for the first time, why C_V → 0 as T → 0. When T << Θ_E, thermal energy kT is insufficient to excite even the first quantum level ℏω. The oscillators are effectively frozen in their zero-point ground states, unable to absorb heat (since the next available energy level is a discrete jump away). This quantum 'freezing' has no classical explanation: classically, every oscillator always contributes kT to the energy regardless of temperature, predicting a constant C_V = 3R all the way to T = 0.
Question 4 True / False
At very low temperatures, the Einstein model's prediction of how rapidly heat capacity approaches zero matches the experimentally measured T³ behavior of real solids.
TTrue
FFalse
Answer: False
False. The Einstein model predicts exponential decay: C_V ∝ exp(−Θ_E/T), which falls *too steeply* at low temperatures. Experiments find C_V ∝ T³ (a power law). The discrepancy arises because the Einstein model assumes all atoms vibrate at the same frequency ω_E, which ignores the full spectrum of vibrational modes in a real solid. Long-wavelength acoustic modes (sound waves) have low frequencies and remain thermally active even at very low T, producing the T³ behavior. The Debye model fixes this by using a realistic distribution of frequencies.
Question 5 Short Answer
Why does quantum discreteness — the fact that harmonic oscillator energies are restricted to εₙ = (n + ½)ℏω — explain the drop in heat capacity at low temperatures, when classical mechanics predicts no such drop?
Think about your answer, then reveal below.
Model answer: A classical oscillator can absorb arbitrarily small amounts of energy — any increment of heat can excite it slightly. So at any temperature above zero, every classical oscillator contributes to the heat capacity. A quantum oscillator, by contrast, has discrete energy levels spaced ℏω apart. To absorb any energy at all, the oscillator must receive at least ℏω in one step. When kT << ℏω, the probability of the oscillator being thermally excited to the first level above the ground state is exponentially small (Boltzmann factor e^{-ℏω/kT} ≈ 0). Nearly all oscillators remain frozen in their ground states, contributing nothing to heat capacity. The drop in C_V is therefore a direct consequence of energy quantization: you cannot deposit less than one quantum, and one quantum becomes inaccessibly expensive at low temperatures.
This is why Einstein called his 1907 paper a decisive test of Planck's quantum hypothesis. If energy were continuous, heat capacity would stay at 3R down to absolute zero. The observed decline was irreconcilable with classical physics and demanded quantization. Einstein's model was the first successful application of quantum ideas beyond blackbody radiation.