Thermal Energy at the Nanoscale Homework Solution - Week 3

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Julianna Parsons
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1 Thermal Energy at the Nanoscale Homework Solution - Week 3 Spring 3. Graphene ZA mode specific heat (a) The cutoff wavevector K Q is found by equating the number of states in k-space within a circle of radius K Q to the total number of unit cells N. πkq ( π ) L = N K Q = (4πη a ) / () where η a is the number of unit cells per unit area. The cutoff frequency ω Q is then obtained from the dispersion relation. ω Q = 4Cπη a () (b) The specific heat c v,za is given by: c v,za = ω Q hω f o BE T D Q,D(ω)dω (3) where D Q,D (ω) is the two-dimensional density of states under the quadratic dispersion model: D Q,D (ω) = dn L dω = dn dk L dk dω = ( ) d πk L dk (π/l) CK = (4) 4πC Thus the DOS of the ZA mode is a constant. Substituting the above expression into the integral in Eq. (3), substituting for the derivative of the Bose-Einstein distribution function, we obtain: c v,za = 4πC ω Q ( ) θq/t T = η a k B θ Q exp( hω/k B T ) hω hω (exp( hω/k B T ) ) k B T dω x e x dx (5) (e x )

2 where x = hω/k B T and θ Q = hω Q /k B (analogous to the Debye temperature ). For temperatures much less than θ Q, the upper limit of the integral in Eq. (5) can be taken to be and the specific heat is proportional to T. This result explains the linear dependence of the specific heat of graphene at low temperatures. At higher temperatures, the ZA mode becomes fully populated and the linear LA and TA modes contribute to specific heat. This explains the quadratic dependence at higher temperatures (see Section 3..4 of textbook).. Specific heat of metals (a) Specific heat c v is a sum of electron and phonon contributions. c v = c v,e + c v,p = π k B η e E F T + 34η ak B T 3 (6) θd 3 Hence, c v T = π kb η e E }{{ F } y intercept + 34η ak B θ 3 D } {{ } slope T (7) (b) Using the experimental data and the result from part (a),.8 mj/mol deg = J/cc deg = π k B η e E F (8) Substituting k B = J/K, η e =.34 cm 3, we obtain E F =.74 9 J =.7 ev. (c) From the slope of the given graph and the result in part (a),.57 mj/mol deg 4 = J/cc deg 4 = 34η ak B θ 3 D (9) Substituting η a =.33 atoms/cc, we obtain = 9. K. 3. Thermal conductivity from kinetic theory The following expression for heat flux q z was obtained in the lecture: In D, v z = v, Λ z = Λ. q z u v z Λ z = v u T zλ z T = c T vv z Λ z q z = c v vλ T In D, v z = v cos θ, Λ z = Λ cos θ. We average the heat flux over an angle of π radians. q z = c v vλ T = c vvλ T π cos θdθ π () () ()

3 In 3D, the heat flux is averaged over a solid angle of π steradians. q z = c v vλ T = 3 c vvλ T π π π/ cos θ sin θdθdψ From Eqs. (,, 3), thermal conductivity κ in d dimensions is given by, (3) Substituting the low temperature result for specific heat c v, we get: ( T κ = η a k B vλ κ = d c vvλ (4) ) d x d+ e x dx (e x ) (5) Note that the Debye approximation is used in the above expression for specific heat. Also we have neglected multiple phonon polarizations. At low temperatures, the thermal conductivity shows the same temperature dependence as the specific heat and scales as T d. 4. Specific heat of a diatomic chain (a) The Einstein frequency ω E is calculated by taking an average of the minimum and maximum frequencies of the optical branch. See Section.6 of the textbook for derivations of the minimum (ω + (K = π/a)) and maximum (ω + (K = )) frequencies. ω E = (ω +(K = ) + ω + (K = π/a)) = ( ) g g µ + m ( g m + m + ) m = m m (6) The Debye frequency ω D is just the product of the group velocity of the acoustic branch at the Brillouin zone center and the Debye cutoff wavevector K D. In D, K D = πη a = π/a since the unit cell density is /a. See Section.6 of the textbook for a derivation of the group velocity of the acoustic branch at the center of Brillouin zone. ω D = v g (K = )K D gµ π = a m m a g = π (m + m ) From Eqs. (6, 7), the ratio of Einstein and Debye temperatures is given by: θ E = ω E = ( m m m ) ω D π m m m (7) (8) 3

4 T c v,d η ak B c v,e η ak B c v,d c v,d +c v,e % % % Table : Acoustic and optical phonon specific heats of a diatomic chain with m /m =. (b) The specific heat of the acoustic branch is given by (see Section 3..4 in textbook): ( ) θd/t T c v,d = η a k B x e x dx (9) (e x ) For T/ =., and, c v,d /η a k B =.38,.973 and.993 respectively (the integral was evaluated numerically). Observe that the specific heat is very close to the Dulong and Petit limit of c v,d = η a k B for temperatures higher than the Debye temperature. The specific heat of the optical branch is given by (see Section 3.. in textbook): χ E eχ E c v,e = η a k B () (e χ E ) where χ E = θ E /T. For m /m =, θ E / =.6 (using the result derived in part (a) of this problem). Hence θ E /T =.6 /T. For T/ =., and, χ E =.6,.6 and.63 respectively. Thus c v,e /η a k B =.7,.884 and.969 for T/ =., and respectively. Note that the optical phonon specific heat is almost zero for T/ =.. This is because the high frequency optical mode is not populated at such low temperatures. The optical phonon specific heat also approaches the Dulong and Petit law for high temperatures. Table shows a summary of the results we have calculated. At low temperatures, the heat capacity of the acoustic branch dominates over the optical branch. For temperatures above the Debye temperature, c v,d /c v,e indicating that both the acoustic and optical modes contribute equally to the specific heat. (c) Figure shows snapshots from the CDF tool where the acoustic and optical contributions to the total specific heat are plotted as a function of temperature. Clearly the acoustic branch dominates the specific heat at low temperatures and the fractional contributions tend to.5 for very high temperatures. As the mass ratio increases, the Einstein frequency moves farther from the Debye frequency. Hence the temperature at which the acoustic and optical contributions become equal also increases with increasing mass ratio. 4

5 Figure : Acoustic and optical contributions to specific heat. 5

Phonons II - Thermal Properties (Kittel Ch. 5)

Phonons II - Thermal Properties (Kittel Ch. 5) Heat Capacity C T 3 Approaches classical limit 3 N k B T Physics 460 F 2006 Lect 10 1 Outline What are thermal properties? Fundamental law for probabilities