A Hint of Quantum Theory — Question 3

Section 19-9 · Physics pages 3196–3197

Question 3 A two-level mode has vanishing heat capacity at both temperature extremes One mole of independent nondegenerate two-level modes has internal energy U = RΘ/(eΘ/T + 1), with constant gap parameter Θ > 0 and ground energy zero. Define its molar contribution C = dU/dT. T/Θ C/R Axes and labels specify the model; the sketch is schematic. (a) Derive C and calculate C/R at T = Θ/2. (b) Find the low- and high-temperature limits of C. Explain why the high-temperature limit differs from a classical harmonic mode with heat capacity R.
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Question 3 - Solution A two-level mode has vanishing heat capacity at both temperature extremes T/Θ C/R (a) Method and calculation x = Θ/T, dx dT = −x T , C = R x2ex (1 + ex)2 . T = Θ/2 : C/R = 4e2 (1 + e2)2 ≈.419974. (b) Interpretation and further calculation x →∞: C/R ∼x2e−x →0. x →0 : C/R ∼x2/4 →0, U →RΘ/2. Only two levels exist in this stipulated model, so the energy saturates. A harmonic oscillator has an unlimited ladder and can continue storing energy as temperature rises. Physical check A mode can be populated but contribute little additional heat capacity when its populations barely change with temperature.

Question 3

A two-level mode has vanishing heat capacity at both temperature extremes

One mole of independent nondegenerate two-level modes has internal energy U=RΘ/(eΘ/T+1)U=R\Theta/(e^{\Theta/T}+1), with constant gap parameter Θ>0\Theta>0 and ground energy zero. Define its molar contribution C=dU/dTC=dU/dT.

Diagram shown in the original SVG worksheet.

  1. Derive CC and calculate C/RC/R at T=Θ/2T=\Theta/2.

  2. Find the low- and high-temperature limits of CC. Explain why the high-temperature limit differs from a classical harmonic mode with heat capacity RR.

Question 3 - Solution

A two-level mode has vanishing heat capacity at both temperature extremes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

x=Θ/T,dxdT=−xT,C=Rx2ex(1+ex)2.x=\Theta/T,\quad\frac{dx}{dT}=-\frac xT,\quad C=R\frac{x^2e^x}{(1+e^x)^2}.T=Θ/2:C/R=4e2(1+e2)2≈.419974.T=\Theta/2:\quad C/R=\frac{4e^2}{(1+e^2)^2}\approx.419974.

(b) Interpretation and further calculation

x→∞:C/R∼x2e−x→0.x\to\infty:\quad C/R\sim x^2e^{-x}\to0.x→0:C/R∼x2/4→0,U→RΘ/2.x\to0:\quad C/R\sim x^2/4\to0,\quad U\to R\Theta/2.Only two levels exist in this stipulated model, so the energy saturates. A harmonic oscillator has an unlimited ladder and can continue storing energy as temperature rises.

Physical check

A mode can be populated but contribute little additional heat capacity when its populations barely change with temperature.