ch05 solid state physics kittel solutions

CHAPTER 5

1 We sol solve ve this for K to obta obtaiin 2 2 2 −1/ 2 K = (2/a) sin−1(ω / ωm) , whence whence dK /dω = (2/ a)(ωm − ω ) and, and, from (15), D(ω)

1. (a) (a) The The dispersion relation is is

ω = ωm |si |sin Ka| Ka|.

2 (b) The The volume volume o off a sph sphe ere of radius K in = (2L/πa) a)(ωm − ω2)−1/ 2 . This is singular at ω = ωm. (b) 3 3/2 Fourie ourier space is Ω = 4πK / 3 = (4π / 3)[(ω0 − ω) / A] , and the density of orbitals orbitals near ω0 is D(ω)=(L /2π)3 | dΩ/dω |= (L/2π)3(2 (2π / A 3/2)(ω0 − ω)1/2 , provided ω < ω0. I t is is appa apparent rent that that

D(ω) vanishes for ω above the minimum nimumω0. 2. The potential energy associated with the dilation is

1 1 1 B(∆V /V)2a3 ≈ kBT . This is kBT and not 2 2 2

3 kBT , because the other degrees of freedom are to be associated with shear distortions of the lattice cell. 2 2 3 −47 −24  Thu  Thus < (∆V ) > = 1.5×10 ; (∆V )rms = 4.7 ×10 cm ; and (∆V)rms / V = 0.125 . Now 3∆a/a ≈ ∆V /V , whence (∆a)rms / a = 0.04. −1 where from (20) for a Debye spectrum Σω < R 2 > = (h/ /2ρV) V ) Σω−1 , where (b) In I n one dimensi nsion from from = ∫ dω D(ω)ω−1 = 3VωD2 / 4π3v3 , whence < R 2 > = 3h/ ωD2 / 8π2 ρv3 . (b) −1 (15) we have D(ω) = L/πv , whence ∫ dω D(ω) ω diverges at the lower limit. The mean square

3.

(a)

strain

in

one

dimension

1 2

2 / /2MNv) ΣK   < (∂R/ R/∂x) x)2 > = ΣK 2u0 = (h

is

= (h/ /2MNv) (K D2 / 2) = h/ ωD2 / 4MNv3. 4. (a) The motion is constrained to each layer and is therefore essentially two-dimensional. Consider one plane of area A. A. The T here is one allowed all owed value of K per area (2π/L)2 in K spa space, or (L/2 (L /2π)2 =A/4π2 allowed values of K per unit unit area area of K space. The The total number of modes odes with with wavevector les less than K is, with wi th ω = vK,

N = (A/4π2) (πK 2) = Aω2 / 4π v2 .  The  The density ity of modes of each polari lariz zation ion type is D(ω) =dN/dω =A ω/2πv2. The thermal average phonon energy for the two polarization types is, for each layer,

∫

U=2

where ωD is defined by

ω

D

0

N=

ω Aω =ω ∫0 2πv2 exp(hω/τ) − 1dω ,

D(ω) n(ω,τ) =ω dω = 2

∫

ω

D

D

D

D(ω) dω . I n the regime =ωD >> τ , we have 2Aτ3 U≅ 2πv2= 2

∫

∞

0

5-1

x2 dx. ex − 1

 Thus the heat capacity

C = kB ∂U/∂τ ∝ T 2 .

(b) If the layers are weakly bound together, the system behaves as a linear structure with each plane as a vibrating unit. By induction from the results for 2 and 3 dimensions, we expect C ∝ T . But this only holds at extremely low temperatures such that τ < < = ωD ≈ =vNlayer / L , where Nlayer/L is the number of  layers per unit length.

1 1 1 coth(x/2) , where 2 2 2 / ω/kB T . The partition function Z = e− x/2 Σ e− sx = e− x/2 /(1− e− x ) = [2sinh(x/2)]−1 and the x= h

5. (a) From the Planck distribution

< n > + = (ex + 1)/(ex − 1) =

free energy is F = kB T log Z = kB T log[2 sinh(x/2)]. (b) With ω(∆) = ω(0) (1 – γ∆), the condition

1 / ω coth(h/ ω/2kBT) on direct differentiation. The energy h 2 < n > h/ ω is just the term to the right of the summation symbol, so that B∆ = γ U(T) . (c) By definition of  γ, we have δω / ω = −γδV /V , or d log ω = −δ d logV . But θ ∝ ωD , whence d log θ = −γ d log V .

∂F/∂∆ = 0

becomes

B∆ = γΣ

5-2