4 3 C HAPTER 5 1 a The dispersion relation is m 1 sin Ka 2 \u03c9 \u03c9 We solve this

4 3 c hapter 5 1 a the dispersion relation is m 1 sin

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4-3
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C HAPTER 5 1. (a) The dispersion relation is m 1 | sin Ka|. 2 ω = ω We solve this for K to obtain , whence and, from (15), 1 m K (2/a)sin ( / ) = ω ω 2 2 1/ 2 m dK/d (2/ a)( ) ω = ω − ω D( ) ω . This is singular at ω = ω 2 2 1/ 2 m (2L/ a)( ) = π ω − ω m . (b) The volume of a sphere of radius K in Fourier space is , and the density of orbitals near ω 3 0 4 K /3 (4 /3)[( ) / A] Ω = π = π ω − ω 3/2 1/ 2 0 is , provided ω < ω 3 3 3/2 0 D( )= (L/2 ) | d /d | (L/2 ) (2 / A )( ) ω π ω = π π ω − ω 0 . It is apparent that D( ω ) vanishes for ω above the minimum ω 0 . 2. The potential energy associated with the dilation is 2 3 B 1 1 B( V/V) a k T 2 2 . This is B 1 k T 2 and not B 3 k T 2 , because the other degrees of freedom are to be associated with shear distortions of the lattice cell. Thus and 2 47 24 rms ( V) 1.5 10 ;( V) 4.7 10 cm ; < ∆ >= × = × 3 rms ( V) / V 0.125 = . Now , whence . 3 a/a V/V ≈ ∆ rms ( a) / a 0.04 = 3. (a) , where from (20) for a Debye spectrum 2 R (h/2 V) / < >= ρ Σω 1 1 Σω , whence 2 1 3 D d D( ) 3V / 4 v = ∫ ω ω ω = ω π 3 2 3 v 2 2 D R 3h /8 / < >= ω π ρ . (b) In one dimension from (15) we have , whence D( ) L/ v ω = π 1 d D( ) ω ω ω diverges at the lower limit. The mean square strain in one dimension is 2 2 2 0 1 ( R/ x) K u (h/2MNv) K 2 / < ∂ >= Σ = Σ 2 2 3 D D (h/2MNv)(K / 2) h / 4MNv . / / = = ω 4. (a) The motion is constrained to each layer and is therefore essentially two-dimensional. Consider one plane of area A. There is one allowed value of K per area (2 π /L) 2 in K space, or (L/2 π ) 2 = A/4 π 2 allowed values of K per unit area of K space. The total number of modes with wavevector less than K is, with ω = vK, 2 2 2 N (A/4 )( K ) A / 4 v . = π π = ω π 2 The density of modes of each polarization type is D( ω ) = dN/d ω = A ω /2 π v 2 . The thermal average phonon energy for the two polarization types is, for each layer, D D 2 0 0 A U 2 D( ) n( , ) d 2 d , 2 v exp(h / ) 1 ω ω ω ω = ω ω τ ω ω = π ω τ − = = ω d ω where ω D is defined by . In the regime D D N D( ) ω = ω D ω >> τ = , we have 3 2 2 2 x 0 2A x U dx. 2 v e 1 τ π = 5-1
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