By Charles Kittel
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Additional resources for Introduction to Solid State Physics Solution Manual
The kinetic energy of a Fermi gas of N electrons in volume V is U = N( 3 5) ( h/ 2 2m ) ( 3π2 N V ) 2 3 . Then dU/dV = – (2/3)U/V and d2U/dV2 = (10/9)U/V2. The bulk modulus B = Vd 2 U dV 2 = (10 9) U V = (10 9) (3 5) n ( mv F 2 ) = nmv F 3. The velocity of sound v = (B ρ)1 2 , where the density ρ = n (m + M) nM, whence v ( m 3M )1 2 v F . 2 2 10. The response is given, with ρ = 1/τ, by ( ) m d 2 x dt 2 + ρdx dt + ωp x = F ( t ) . 2 The conductivity σ does not enter this equation directly, although it may be written as σ = ωp2τ/4π.
A) D(ε ) = dN dk d (πk 2 ) m m 2 = = 2A 2 2 dk dε (2π / L) dk = k π= where A = L2. Note: There are two flaws in the answer m / πh 2 quoted in the text. First, the area A is missing, meaning the quoted answer is a density per unit area. This should not be a major issue. Second, the h should be replaced by = . (b) N = 2 ⋅ πk F2 2 (2π / L) => ns = N / A = k F2 / 2π L m where ns is the 2D sheet density. For a square sample, W=L, so: W ns e 2τ 2π m Rs = 2 2 and using =k F / m = vF : kF e τ h 1 2π= Rs = = 2 2 e kF A k F vF e τ (c) Rs = 17-2 CHAPTER 18 1.
The terms in U eA + U c + U K which involve exx are 1 2 2 C11e xx + C12 e xx ( e yy + e zz ) + B1α1 e xx . 2 Take ∂/∂exx: C11e xx + C12 ( e yy + e zz ) + B1α1 = 0, for minimum. 2 Further: C11e yy + C12 (e xx + e zz ) + B1α 2 = 0 . 2 C11e zz + C12 ( e xx + e yy ) + B1α 3 = 0 . 2 Solve this set of equations for exx: C − α 2 ( C11 + 2C12 ) e xx = B1 12 . ( C11 − C12 )( C11 + 2C12 ) 2 Similarly for eyy, ezz, and by identical method for exy, etc. 12-2 5a. U ( θ ) = K sin 2 θ − Ba M s cos θ Kϕ2 − Ba M s 1 2 ϕ , for θ = π + ϕ 2 and expanding about small ϕ .
Introduction to Solid State Physics Solution Manual by Charles Kittel