Kittel charles introduction to solid state physics 8th edition solution manual 9142

CHAPTER 1 The vectors xˆ + yˆ + zˆ and − xˆ − yˆ + zˆ are in the directions of two body diagonals of a cube If θ is the angle between them, their scalar product gives cos θ = –1/3, whence θ = cos −1 1/ = 90° + 19° 28' = 109° 28' The plane (100) is normal to the x axis It intercepts the a' axis at 2a' and the c' axis at 2c' ; therefore the indices referred to the primitive axes are (101) Similarly, the plane (001) will have indices (011) when referred to primitive axes The central dot of the four is at distance a cos 60° a = a ctn 60° = cos 30° from each of the other three dots, as projected onto the basal plane If the (unprojected) dots are at the center of spheres in contact, then 2 ⎛ a ⎞ ⎛c⎞ a =⎜ ⎟ +⎜ ⎟ , ⎝ ⎠ ⎝2⎠ or 2 a = c ; c = 1.633 a 1-1 CHAPTER hkA is a plane defined by the points a1/h, a2/k, and a3 / A (a) Two vectors that lie in the plane may be taken as a1/h – a2/k and a1 / h − a3 / A But each of these vectors gives zero as its scalar product with G = ha1 + ka + Aa3 , so that G must be perpendicular to the plane hkA (b) If nˆ is the unit normal to the plane, the interplanar spacing is nˆ ⋅ a1/h But nˆ = G / | G | , whence d(hkA) = G ⋅ a1 / h|G| = 2π / | G| (c) For a simple cubic lattice G = (2π / a)(hxˆ + kyˆ + Azˆ ) , The crystal plane with Miller indices whence G h + k + A2 = = d 4π a2 3a 2 (a) Cell volume a1 ⋅ a × a3 = − 3a = a a c a c xˆ (b) b1 = 2π = yˆ 4π a × a3 = − 3a | a1 ⋅ a × a3 | 3a c zˆ a c 2π ( xˆ + yˆ ), and similarly for b , b3 a (c) Six vectors in the reciprocal lattice are shown as solid lines The broken lines are the perpendicular bisectors at the midpoints The inscribed hexagon forms the first Brillouin Zone By definition of the primitive reciprocal lattice vectors VBZ = (2π)3 (a × a ) ⋅ (a × a1 ) × (a1 × a ) = (2π)3 / | (a1 ⋅ a × a ) | | (a1 ⋅ a × a ) | = (2π)3 / VC For the vector identity, see G A Korn and T M Korn, Mathematical handbook for scientists and engineers, McGraw-Hill, 1961, p 147 (a) This follows by forming 2-1 |F|2 = − exp[−iM(a ⋅ ∆k)] − exp[iM(a ⋅ ∆k)] ⋅ − exp[−i(a ⋅ ∆k)] − exp[i(a ⋅ ∆k)] − cos M(a ⋅ ∆k) sin 12 M(a ⋅ ∆k) = = − cos(a ⋅ ∆k) sin 12 (a ⋅ ∆k) (b) The first zero in sin Mε occurs for ε = 2π/M That this is the correct consideration follows from 1 sin M(πh + ε) = sin πMh cos Mε + cos sin Mε πMh       2 ± zero, as Mh is an integer S (v1 v v ) = f Σ e −2πi(x j v1 +y j v +z j v3 ) j Referred to an fcc lattice, the basis of diamond is 000; 1 Thus in the product 4 S(v1v v3 ) = S(fcc lattice) × S (basis) , we take the lattice structure factor from (48), and for the basis S (basis) = + e −i π (v1 + v + v3 ) Now S(fcc) = only if all indices are even or all indices are odd If all indices are even the structure factor of the basis vanishes unless v1 + v2 + v3 = 4n, where n is an integer For example, for the reflection (222) we have S(basis) = + e–i3π = 0, and this reflection is forbidden ∞ f G = ∫ 4πr (πa Gr)−1 sin Gr exp ( −2r a ) dr = (4 G 3a ) ∫ dx x sin x exp ( −2x Ga ) = (4 G 3a ) (4 Ga ) (1 + r G a ) 16 (4 + G a ) The integral is not difficult; it is given as Dwight 860.81 Observe that f = for G = and f ∝ 1/G4 for Ga >> (a) The basis has one atom A at the origin and one atom B at a The single Laue equation a ⋅ ∆k = 2π × (integer) defines a set of parallel planes in Fourier space Intersections with a sphere are a set of circles, so that the diffracted beams lie on a set of cones (b) S(n) = fA + fB e–iπn For n odd, S = fA – 2-2 fB; for n even, S = fA + fB (c) If fA = fB the atoms diffract identically, as if the primitive translation vector were 1 a and the diffraction condition ( a ⋅ ∆k ) = 2π × (integer) 2 2-3 CHAPTER E = (h/ 2M) (2π λ ) = (h/ 2M) (π L) , with λ = 2L U(R) = 2Nε[9.114( σ R )12 − 12.253(σ R)6 ] At equilibrium R = 1.488σ6 , and U(R ) = 2Nε( − 2.816) bcc: U(R) = 2Nε[12.132( σ R )12 − 14.454(σ R)6 ] At equilibrium R = 1.679σ6 , and U(R ) = 2Nε( − 4.305) Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is fcc: more stable than the bcc | U | = 8.60 Nε = (8.60) (6.02 × 1023 ) (50 × 10−16 ) = 25.9 × 109 erg mol = 2.59 kJ mol This will be decreased significantly by quantum corrections, so that it is quite reasonable to find the same melting points for H2 and Ne We have Na → Na+ + e – 5.14 eV; Na + e → Na– + 0.78 eV The Madelung energy in the NaCl structure, with Na+ at the Na+ sites and Na– at the Cl– sites, is αe2 (1.75) (4.80 × 10−10 ) = = 11.0 ×10−12 erg, R 3.66 × 10−8 or 6.89 eV Here R is taken as the value for metallic Na The total cohesive energy of a Na+ Na– pair in the hypothetical crystal is 2.52 eV referred to two separated Na atoms, or 1.26 eV per atom This is larger than the observed cohesive energy 1.13 eV of the metal We have neglected the repulsive energy of the Na+ Na– structure, and this must be significant in reducing the cohesion of the hypothetical crystal 5a ⎛ A αq ⎞ U(R) = N ⎜ n − ⎟ ; α = log = Madelung const R ⎠ ⎝R In equilibrium ⎛ nA αq ⎞ ∂U nA n = N ⎜ − n +1 + ⎟ = ; R = , ∂R αq R0 ⎠ ⎝ R0 and U(R ) = − Nαq (1 − ) R0 n 3-1 b U(R -R δ) = U ( R ) + ∂2U R ( R 0δ ) + , 2 ∂R bearing in mind that in equilibrium (∂U ∂R) R = ⎛ n(n + 1)A 2αq ⎞ ⎛ (n + 1) αq 2αq ⎞ ⎛ ∂2U ⎞ − = N⎜ − ⎜ ⎟ = N⎜ n+2 ⎟ 3 ⎟ R0 ⎠ R0 R0 ⎠ ⎝ ∂R ⎠R ⎝ R0 ⎝ For a unit length 2NR0 = 1, whence ⎛ ∂2U ⎞ αq (n − 1) q log 2 ∂ U − = = (n 1) ; C R ⎜ 2⎟ = ∂R R 2R R ⎝ ∂R ⎠R 0 For KCl, λ = 0.34 × 10–8 ergs and ρ = 0.326 × 10–8Å For the imagined modification of KCl with the ZnS structure, z = and α = 1.638 Then from Eq (23) with x ≡ R0/ρ we have x e− x = 8.53 × 10−3 By trial and error we find x  9.2, or R0 = 3.00 Å The actual KCl structure has R0 (exp) = 3.15 Å For the imagined structure the cohesive energy is U= -αq ⎛ p ⎞ U ⎜1⎟ , or =-0.489 R0 ⎝ R0 ⎠ q in units with R0 in Å For the actual KCl structure, using the data of Table 7, we calculate U = −0.495, q2 units as above This is about 0.1% lower than calculated for the cubic ZnS structure It is noteworthy that the difference is so slight The Madelung energy of Ba+ O– is –αe2/R0 per ion pair, or –14.61 × 10–12 erg = –9.12 eV, as compared with –4(9.12) = –36.48 eV for Ba++ O To form Ba+ and O– from Ba and O requires 5.19 – 1.5 = 3.7 eV; to form Ba++ and O requires 5.19 + 9.96 – 1.5 + 9.0 = 22.65 eV Thus at the specified value of R0 the binding of Ba+ O– is 5.42 eV and the binding of Ba++ O is 13.83 eV; the latter is indeed the stable form From (37) we have eXX = S11XX, because all other stress components are zero By (51), 3S11 = (C11 − C12 ) + (C11 + C12 ) 2 Thus Y = (C11 + C12 C11 − 2C12 ) (C11 + C12 ); further, also from (37), eyy = S21Xx, whence σ = e yy e xx = S21 S11 = − C12 (C11 + C12 ) For a longitudinal phonon with K || [111], u = v = w 3-2 ω2ρ = [C11 + 2C44 + 2(C12 + C 44 )]K 3, or v = ω K = [(C11 + 2C12 + 4C44 3ρ )]1 This dispersion relation follows from (57a) 10 We take u = – w; v = This displacement is ⊥ to the [111] direction Shear waves are degenerate in this direction Use (57a) 11 Let e xx = −e yy = e in (43) Then U = C11 ( e2 + e2 ) − C12 e2 = [ (C11 − C12 )]e ⎛ n(n + 1)A 2αq ⎞ ⎛ (n + 1) αq 2αq ⎞ ⎛ ∂2U ⎞ = N⎜ − = N⎜ − is the effective shear so that ⎜ n+2 ⎟ 3 ⎟ ⎟ R0 ⎠ R0 R0 ⎠ ⎝ ∂R ⎠R ⎝ R0 ⎝ constant 12a We rewrite the element aij = p – δij(λ + p – q) as aij = p – λ′ δij, where λ′ = λ + p – q, and δij is the Kronecker delta function With λ′ the matrix is in the “standard” form The root λ′ = Rp gives λ = (R – 1)p + q, and the R – roots λ′ = give λ = q – p b Set u (r, t) = u ei[(K 3) (x + y + z) −ωt] ; v(r, t) = v0 ei[ .] ; w(r, t) = w ei[ .] , as the displacements for waves in the [111] direction On substitution in (57) we obtain the desired equation Then, by (a), one root is ω2ρ = 2p + q = K (C11 + 2C12 + 4C 44 ) / 3, and the other two roots (shear waves) are ω2ρ = K (C11 − C12 + C44 ) / 13 Set u(r,t) = u0ei(K·r – t) and similarly for v and w Then (57a) becomes 2 ω2ρu = [C11K y + C44 (K y + K z )]u + (C12 + C44 ) (K x K y v + K x K z w ) and similarly for (57b), (57c) The elements of the determinantal equation are 3-3 M11 = C11K x + C 44 (K y + K z ) − ω ρ; 2 M12 = (C12 + C44 )K x K y ; M13 = (C12 + C44 )K x K z and so on with appropriate permutations of the axes The sum of the three roots of sum of the diagonal elements of the matrix, which is ω2ρ is equal to the (C11 + 2C44)K2, where 2 K = K x + K y + K z , whence v1 + v + v3 = (C11 + 2C44 ) ρ , 2 for the sum of the (velocities)2 of the elastic modes in any direction of K 14 The criterion for stability of a cubic crystal is that all the principal minors of the quadratic form be positive The matrix is: C11 C12 C12 C12 C11 C12 C12 C12 C11 C44 C44 C44 The principal minors are the minors along the diagonal The first three minors from the bottom are C44, C442, C443; thus one criterion of stability is C44 > The next minor is C11 C44 3, or C11 > Next: C443 (C112 – C122), whence |C12| < C11 Finally, (C11 + 2C12) (C11 – C12)2 > 0, so that C11 + 2C12 > for stability 3-4 CHAPTER 1a The kinetic energy is the sum of the individual kinetic energies each of the form Mu S The force between atoms s and s+1 is –C(us – us+1); the potential energy associated with the stretching of this bond is C(u s − u s+1 ) , and we sum over all bonds to obtain the total potential energy b The time average of 1 Mu S is Mω2 u In the potential energy we have u s +1 = u cos[ωt − (s + 1)Ka] = u{cos(ωt − sKa) ⋅ cos Ka + sin (ωt − sKa) ⋅ sin Ka} Then u s − u s +1 = u {cos(ωt − sKa) ⋅ (1 − cos Ka) − sin (ωt − sKa) ⋅ sin Ka} We square and use the mean values over time: < cos > = < sin > = ; < cos sin > = Thus the square of u{} above is u [1 − 2cos Ka + cos Ka + sin Ka] = u (1 − cos Ka) The potential energy per bond is cos Ka) this is equal to Cu (1 − cos Ka), and by the dispersion relation ω2 = (2C/M) (1 – Mω2 u Just as for a simple harmonic oscillator, the time average potential energy is equal to the time-average kinetic energy We expand in a Taylor series ⎛ ∂2u ⎞ ⎛ ∂u ⎞ u(s + p) = u(s) + pa ⎜ ⎟ + p a ⎜ ⎟ + " ; ⎝ ∂x ⎠s ⎝ ∂x ⎠s On substitution in the equation of motion (16a) we have M ∂2u ∂2u 2 = ( Σ p a C ) , p p>0 ∂t ∂x which is of the form of the continuum elastic wave equation with 4-1 v = M −1 Σ p >0 p 2a 2Cp From Eq (20) evaluated at K = π/a, the zone boundary, we have −ω2 M1u = −2Cu ; −ω2 M v = −2Cv Thus the two lattices are decoupled from one another; each moves independently At ω2 = 2C/M2 the motion is in the lattice described by the displacement v; at ω2 = 2C/M1 the u lattice moves ω2 = sin pk a A Σ (1 − cos pKa) ; p >0 M pa ∂ω2 2A = Σ sin pk 0a sin pKa ∂K M p >0 (cos (k − K) pa − cos (k + K) pa) When K = k0, ∂ω2 A = Σ (1 − cos 2k pa) , ∂K M p>0 which in general will diverge because Σ → ∞ p By analogy with Eq (18), Md u s dt = C1 (vs − u s ) + C2 (vs −1 − u s ); Md vs dt = C1 (u s − vs ) + C2 (u s +1 − vs ), whence −ω2 Mu = C1 (v − u) + C2 (ve − iKa − u); −ω2 Mv = C1 (u − v) + C2 (ueiKa − v) , and (C1 + C2 ) − Mω2 −(C1 + C2 e − iKa ) −(C1 + C2 eiKa ) (C1 + C2 ) − Mω2 =0 For Ka = 0, ω2 = and 2(C1 + C2 ) M For Ka = π, ω2 = 2C1 M and 2C M (a) The Coulomb force on an ion displaced a distance r from the center of a sphere of static or rigid conduction electron sea is – e2 n(r)/r2, where the number of electrons within a sphere of radius r is (3/4 πR3) (4πr3/3) Thus the force is –e2r/R2, and the 4-2 CHAPTER 15 1a The displacement under this force is ∞ α ( ω) e − iωt dω ∫ −∞ 2π x (t) = ∫ α ( ω) e With ω = ωR + iωI, the integral is − iωR t ωI t e dω This integral is zero for t < because we may then complete a contour with a semicircle in the upper half-plane, over which semicircle the integral vanishes The integral over the entire contour is zero because α(ω) is analytic in the upper half-plane Therefore x(t) = for t < 1b We want ∞ e − i ω t dω x (t) = ∫ ω02 − ω2 − iωρ , 2π −∞ (A) which is called the retarded Green’s function of the problem We can complete a contour integral by adding to x(t) the integral around an infinite semicircle in the upper halfplane The complete contour integral vanishes because the integrand is analytic everywhere within the contour But the integral over the infinite semicircle vanishes at t < 0, for then exp ⎡⎣ −i ( ωR + iωI ) ( − t ) ⎤⎦ = exp ( −ωI t ) exp ( iωR t ) , which → as |ω| → ∞ Thus the integral in (A) must also vanish For t > we can evaluate x(t) by carrying out a Cauchy integral in the lower half-plane The residues at the poles are ( ± 12 ω0 − 14 ρ2 ) ( exp ( − 12 ρt ) exp ⎡⎢ ∓ i ω0 − 14 ρ2 ⎣ ) t ⎤⎥ , ⎦ so that ( x ( t ) = ω0 − 14 ρ2 ) ( exp ( − 12 ρt ) sin ω0 − 14 ρ2 In the limit ω → ∞ we have α′ ( ω) → −∑ f j ω2 from (9), while from (11a) 15-1 ) t α′ ( ω ) → − ∞ sα′′ ( s ) ds πω2 ∫0 The reflected wave in vacuum may be written as − E y ( refl ) = Bz ( refl ) = A′ e − i ( kx + ωt ) , where the sign of Ey has been reversed relative to Bz in order that the direction of energy flux (Poynting vector) be reversed in the reflected wave from that in the incident wave For the transmitted wave in the dielectric medium we find E y ( trans ) = ck Bz ( trans ) εω = ε −1 Bz ( trans ) = A"e ( i kx −ωt ) , by use of the Maxwell equation c curl H = ε∂E/∂t and the dispersion relation εω2 = c2k2 for electromagnetic waves The boundary conditions at the interface at x = are that Ey should be continuous: Ey (inc) + Ey (refl) = Ey (trans), or A – A' = A'' Also Bz should be continuous, so that A + A' = ε1/2 A'' We solve for the ratio A'/A to obtain ε1/2 (A – A') = A + A', whence A' − ε1 = , A ε1 + and r≡ E ( refl ) E ( inc ) =− A' ε1 − n + ik − = = A ε1 + n + ik + The power reflectance is ⎛ n − ik − ⎞ ⎛ n + ik − ⎞ ( n − 1) + K R ( ω) = r ∗ r = ⎜ ⎟⎜ ⎟= ⎝ n − ik + ⎠ ⎝ n + ik + ⎠ ( n + 1) + K 2 (a) From (11) we have σ " ( ω) = − ∞ σ ' (s ) 2ω P∫ ds π s − ω2 In the limit ω → ∞ the denominator comes out of the integrand and we have 15-2 lim ω→∞ σ " ( ω) = ∞ σ ' ( s ) ds πω ∫0 (b) A superconductor has infinite conductivity at zero frequency and zero conductivity at frequencies up to ωg at the energy gap We can replace the lost portion of the integral (approximately σ'nωg) by a delta function σ'nωg δ(ω) in σ's(ω) at the origin Then the KK relation above gives σ ''s ( ω) = σ 'n ωg πω (c) At very high frequencies the drift velocity of the conduction electrons satisfies the free electron equation of motion mdv dt = − eE ; − iωmv = − eE , so that the current density is j = n ( −e ) v = − ine E mω and ωσ'' (ω) = ne2/m in this limit Then use (a) to obtain the desired result From (11a) we have ε ' ( ω) − = ∞ δ ( s − ωg ) ωp 4πne P∫ ds = 2 m s − ω ω − ω g n2 – K2 + 2inK = + 4πiσ0/ω For normal metals at room temperature σ0 ∼ 1017 – 1018 sec–1, so that in the infrared ω σ0 Thus n K , so that R − n and n ( 2πσ0 ω) , whence R 1− ( 2ω πσ0 ) (The units of σ0 are sec–1 in CGS.) The ground state of the line may be written ψ g = A1B1A B2 … A N BN Let the asterisk denote excited state; then if specific single atoms are excited the states are ∗ ∗ ϕ j = A1B1A B2 … A j B j … A N BN ; θ j = A1B1A B2 … A jB j … A N BN The hamiltonian acts thusly: H ϕ j = ε A ϕ j + T1θ j + T2θ j−1 ; H θ j = ε Bθ j + T1ϕ j + T2ϕ j + 15-3 An eigenstate for a single excitation will be of the form ψ k = ∑ eijka ( αϕ j + βθ j ) We j form H ψ k = ∑ eijka [αε A ϕ j + αT1θ j + αT2 θ j−1 j θ + βε B j + β T1ϕ j + β T2ϕ j + 1] ( ) = ∑ eijka [ αε A + β T1 + e − ika β T2 ϕ j j ( ) + αT1 + βε B + eika αT2 θ j ] = Eψ k = ∑ eijka [αEϕ j + β Eθ j ] This is satisfied if ( εA − Ε ) α + ( T1 + e−ika T2 ) β = 0; ( T1 + eika T2 ) α + ( εB − E ) β = The eigenvalues are the roots of εA − E T1 + e − ika T2 T1 + eika T2 ε B − E 15-4 = CHAPTER 16 e2 x ⋅ = eE; ex = r E = p; α = p E = r = a H r r 4π 4π P = inside a conducting sphere Thus p = a P = a 3E , and 3 α = p E0 = a E i = E − Because the normal component of D is continuous across a boundary, Eair = εEdiel, where Eair = 4πQ/A, with Q the charge on the boundary The potential drop between the 1⎞ ⎛ two plates is E air qd + E diel d = E air d ⎜ q + ⎟ For a plate of area A, the capacitance is ε⎠ ⎝ C= A 1⎞ ⎛ 4πd ⎜ q + ⎟ ε⎠ ⎝ It is useful to define an effective dielectric constant by 1 = +q εeff ε If ε = ∞, then εeff = 1/q We cannot have a higher effective dielectric constant than 1/q For q = 10–3, εeff = 103 The potential drop between the plates is E1 d + E2 qd The charge density Q D1 εE1 iσ = = = E2 , A 4π 4π ω by comparison of the way σ and ε enter the Maxwell equation for curl H Thus E1 + Q= 4πiσ ⎛ 4πiσ ⎞ E ; V = E 2d ⎜ + q⎟ ; εω ⎝ εω ⎠ σAi Q E ; and thus C ≡ = ω V and εeff = (1 + q ) A , ⎛ iω q ⎞ 4πd ⎜ − ⎟ ⎝ ε 4πσ ⎠ ε − ( iωεq 4πσ ) 16-1 (CGS) 5a E int = E − E int = 4π 4π E P = E0 − χ int 3 E0 4π χ 1+ b P = χ E int = χ E0 4π χ 1+ E = 2P1/a3 P2 = αE = 2αP1/a3 This has solution p1 = p2 if 2α = a ; α = a (a) One condition is, from (43), γ ( TC − T0 ) − g Ps + g Ps = The other condition is 1 γ ( Tc − T0 ) Ps − g Ps + g Ps = Thus 1 g Ps + g Ps ; 3 g4 2 g Ps = g ; Ps = g6 − g Ps + g Ps = − (b) From the first line of part (a), 2 g4 g4 g4 γ ( Tc − T0 ) = − = g6 16 g 16 g In an electric field the equilibrium condition becomes − E + γ ( T − Tc ) P + g P3 = , where the term in g6 is neglected for a second-order transition Now let P = Ps + ∆P If we retain only linear terms in ∆P , then − E + γ ( T − Tc ) ∆P + g 3Ps ∆P = , with use of (40) Further, we can eliminate Ps ∆P E = γ ( Tc − T ) 16-2 because Ps = ( γ g )( Tc − T ) Thus a ← a → i→ b cos ←i i→ ←i i→ ←i π ( na ) a ← 2a → i i i i i i Deforms to new stable structure of dimers, with lattice constant × (former constant) c 10 The induced dipole moment on the atom at the origin is p = αE, where the electric −3 field is that of all other dipoles: E = a ∑ p n = 4p a ∑ n ; the sum is over ( ) ( )( ) positive integers We assume all dipole moments equal to p The self-consistency condition is that p = α(4p/a3) (Σn–3), which has the solution p = unless α ≥ (a3/4) (1/Σn–3) The value of the summation is 1.202; it is the zeta function ζ(3) 16-3 CHAPTER 17 (a) The interference condition for a linear lattice is a cos θ = nλ The values of θ that satisfy this condition each define a cone with axis parallel to the fiber axis and to the axis of the cylindrical film Each cone intersects the film in a circle When the film is flattened out, parallel lines result (b) The intersection of a cone and a plane defines a conic section, here a hyperbola (c) Let a, b be the primitive axes of a square lattice The Laue equations (2.25) give a • ∆k = πq; b • ∆k = πr, where q, r are integers Each equation defines a set of planes The intersections of these planes gives a set of parallel lines, which play in diffraction from a two-dimensional structure the role played by reciprocal lattice points in diffraction from a three-dimensional structure In the Ewald construction these lines intersect a sphere of radius k = π/λ in a set of points In two dimensions any wavelength (below some maximum) will give points; in three dimensions only special values of λ give points of intersection because one more Laue equation must be satisfied The points correspond to the directions k' of the diffraction maxima If the photographic plate is flat the diffraction pattern (2 dim.) will appear distorted Points near the direction of the incident beam are shown (d) The lattice of surface atoms in the (110) surface of an fcc crystal is simple rectangular The long side of the rectangle in crystal (real) space is a short side in the reciprocal lattice This explains the 90° rotation between (21a) and (21b) ∫ ∞ With the trial function x exp (–ax), the normalization integral is dx x exp ( −2ax ) = 4a The kinetic energy operator applied to the trial function gives ( ) ( )( ) − h/ 2m d u dx = − h/ 2m a x − 2a exp ( −ax ) while Vu = eEx2 exp (–ax) The definite integrals that are needed have the form ∫ ∞ ( ) dx x n exp( − ax) = n! a n+1 The expectation value of the energy is < ε > = h/ 2m a + ( 3eE 2a ) , which has an extremum with respect to the range parameter a when ( ) d < ε > da = h/ 2m 2a − 3eE 2a = 0, or a = 3eEm 2h/ The value of < ε > is a minimum at this value of a, so that ( = ( h/ )( ) + ( 3eE ) ( 2h/ 2m ) ( 3eE ) ( + ) , < ε > = h/ 2m 3eEm h/ 2 13 23 23 17-1 −2 13 3eEm ) 13 where the last factor has the value 1.89 … The Airy function is treated in Sec 10.4 of the NBS Handbook of mathematical functions (a) D(ε ) = dN dk d (πk ) m m = = 2A 2 dk dε (2π / L) dk = k π= where A = L2 Note: There are two flaws in the answer m / πh 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 = ⋅ πk F2 (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 and using =k F / m = vF : kF e τ h 2π= Rs = = 2 e kF A k F vF e τ (c) Rs = 17-2 CHAPTER 18 Carbon nanotube band structure (a) b i ⋅ a j = 2πδ ij => b1 = (− 2aπ , 2π 3a b = ( 2aπ , ), 2π 3a ) (b) The angle between K and b1is 30o ; A right triangle is formed in the first BZ with two sides of length K and b1/2 Now b1 = 43πa , so: K = (b1/2)/cos(30o)= 4π/3a (c) Quantization of k along x: kx(na)=2πj= kx =2π j/na Assume n = 3i, where i is an integer Then: kx = K(j/2i) For j = 2i, kx=K Then ∆K = k y ˆj and there is a massless subband (d) For n = 10, kx =2π j/10a =K(3j/20) The closest k comes to K is for j = 7, where ∆kx = K/20 Then: ε 11 = 2=vF (4π / 3a) / 10 = 1.8 eV The next closest is for j = 6, where ∆kx = K/10, twice the previous one Therefore: ε22 = 2ε11 (e) For the lowest subband: ∆k = ( K / 20) + k y2 , so: ε = [(=K / 20vF )v F2 ]2 + (=k y v F ) This is of the desired form, with m * = =K / 20v F m * / m = =K / 20mvF = 0.12 Filling subbands ε (nx , n y ) = = 2π (nx2 + n y2 ) 2mW => States are filled up to ε (2,2) = = 2π (8) 2mW = k12,1 = 2π = (8 − 2) => 2m 2mW k1,1 = 6π 2 => n1,1 = k1,1 = W π W = k 22,1 = 2π = (8 − 5) => (2,1) subband: 2m 2mW k 2,1 = 3π 2 => n2,1 = k 2,1 = π W W (1,1) subband: 18-1 (2,1) subband: same n= + W W = 5.9 x 108 /m Breit-Wigner form of a transmission resonance (a) cos(δϕ ) ≅ − δϕ / ; | ri |= 1− | ti |2 ≅ − 12 | ti |2 − 18 | ti |4 The denominator of (29) is then: + (1− | t1 |2 )(1− | t |2 ) − 2(1 − 12 | t1 |2 − 18 | t1 |4 )(1 − 12 | t |2 − 18 | t |4 )(1 − 12 δϕ ) = 14 (| t1 | + | t |4 ) + 12 | t1 | | t |2 +δϕ = 14 (| t1 |2 + | t |2 ) + δϕ | t1 |2 | t |2 ℑ= (| t1 |2 + | t |2 ) + 4δϕ (b) δϕ = Lδk and δk / δε = ∆k / ∆ε = (π / L) / ∆ε Combining: δϕ = (2 L)(π / L)δε / ∆ε => δϕ / 2π = δε / ∆ε (c) Combining: | t1 |2 | t |2 (∆ε / 2π ) ℑ= (∆ε / 2π ) (| t1 |2 + | t |2 ) + 4δε which is (33) Barriers in series and Ohm’s law (a) 1− | r1 |2 | r2 |2 1− | r1 |2 | r2 |2 − | t1 |2 | t |2 − (1− | t1 |2 ) | r2 |2 − (1− | r1 |2 ) | t |2 = = + = + 1 ℑ | t1 |2 | t |2 | t1 |2 | t |2 | t1 |2 | t |2 = 1+ (b) σ 1D = − (| r2 |2 + | t |2 )+ | t1 |2 | r2 |2 + | r1 |2 | t |2 | r2 |2 | r1 |2 = + + which gives (36) | t1 |2 | t |2 | t |2 | t1 |2 n1D e 2τ 2k F e 2τ = , m πm and =k F = vF m But: A B = vFτ B = 2vFτ => => σ 1D = σ 1D = 2vF e 2τ 2e (2vFτ ) = h =π 2e A B h Energies of a spherical quantum dot (a) ∫ E ⋅ da = Qencl / εε o R+d V= ∫ R qdr 4πεε o r E = q / 4πεε o r Integrating from inner to outer shell: => A = ⎞ q d ⎛1 ⎟= ⎜ − 4πεε o ⎝ R R + d ⎠ 4πεε o R( R + d ) q 18-2 C= q R( R + d ) = 4πεε o V d U= (c) For d >> R , U ε 0,0 = e2 Also ε 0, = 4πεε o R e2 e2 e2 d = C 4πεε o R( R + d ) R2 A = εε o d d C ≅ 4πεε o (b) For d U ε 0,0 = e2 2m* R 4πεε o R = 2π ⋅ e2 2m* R 2m* R R = ⋅ = 4πεε o R = 2π 4πεε o = 2π π aB* ⋅ Thermal properties in 1D (a) D(ω ) = 2K L = 2π / L v πv ωD ∞ dωD(ω )=ω =L =L ⎛ k BT ⎞ ωdω U tot = ∫ ≅ = ⎟ ⎜ ∫ exp(=ω / k BT ) − πv exp(=ω / k BT ) − πv ⎝ = ⎠ Obtaining value from table of integrals: Lk B2T π π Lk B2T U tot = = 3hv =πv 2π Lk B2T CV = ∂U tot / ∂T V == 3hv (b) The heat flow to the right out of reservoir is given by: ∞ 2∞ xdx ∫ exp( x) − ∞ DR (ω ) dω=ω =ℑ =ℑ ⎛ k BT1 ⎞ π π k B2T12 ωdω JR = ∫ = ℑ ⋅v⋅ ℑ= = ⎟ ⎜ L exp(=ω / k BT1 ) − 2π ∫0 exp(=ω / k BT1 ) − 2π ⎝ = ⎠ 6h and similarly for JL The difference is: JR − JL = Let T1 = T + ∆T , T2 = T π k B2 ℑ (T − T ) => (T − T ) ≈ 2T∆T 6h 2 => 2 for small ∆T 2 JR − JL = 18-3 π k ℑ 2 B 3h ∆T which gives (78) CHAPTER 20 U = nEI The number of ways to pick n from N is N! / (N − n) !n! The number of ways to put n into N' = n'! / (N' − n) !n! ⎛ ⎞ N! N′ ! Entropy S = k B ⎜ log + log ⎟ ⎜ ⎟ ′ N − n !n! N − n !n! ( ) ( ) ⎝ ⎠ N!  N log N − ( N − n ) log ( N − n ) − n log n log ( N − n )!n! log N′ !  N′ log N′ − ( N′ − n ) log ( N′ − n ) − n log n ( N′ − n )!n! ∂U ∂S ⎛ ∂F ⎞ −T = in equilibrium; thus ⎜ ⎟ = ∂n ⎝ ∂n ⎠T ∂n N−n N′ − n ⎞ ⎛ E I = k BT ⎜ log + log ⎟ n n ⎠ ⎝ ( N − n )( N′ − n ) For n [...]... point and a corner, the electrons will spill over into the second zone in preference to filling up the corner states in the first zone Divalent elements under these conditions will be metals and not insulators 2 ε = h/ k 2m , where the free electron wavevector k may be written as the sum of a vector K in the reduced zone and of a reciprocal lattice vector G We are interested in K along the [111] direction:... this equation will have a solution only when Ka  1 Expand the sine and cosine to obtain in lowest order 1 2 The energy is ε= ( Ka ) 2 h/ 2 K 2 2m  h/ 2 P ma 2 (b) At k = π/a the determinantal equation is (P/Ka) sin Ka + cos Ka = –1 In the same limit this equation has solutions Ka = π + δ, where δ  1 We expand to obtain 1 ( P π )( −δ ) + ⎛⎜ −1 + δ2 ⎞⎟ = −1, which has the solution δ = 0 and δ = 2P/π... matrix element of H1 is equal to 8 Write (17) as H = H 0 + H1 , where (h/ / m) ∫ dV u 0 (r) k ⋅ p U 0 (r) In a cubic crystal U 0 (r) will be even or odd with respect to the      inversion operation r → − r , but p is an odd operator It follows that the diagonal matrix element    vanishes, and there is no first-order correction to the energy The function U k (r) to first order in H1 is   > 1, we have σ yx ≅ σ0 ωc τ = ne τ m 2 ωc τ 1 + ( ωc τ ) 2 σ0 E x ) ( mc eBτ ) = neB c 10 For a monatomic metal sheet one atom in thickness, n ≈ 1/d3, so that R sq ≈ mv F nd 2 e 2 ≈ mv F d e 2 If the electron wavelength... 0 Also J = L + S = 7/2 Hence the ground state is 8S7/2 b Yb+++ has 13 electrons in the f shell, leaving one hole in the otherwise filled shell Thus L = 3, S = 1/2, J = 7/2 we add S to L if the shell is more than half-filled The ground state symbol is 2F7/2 c Tb+++ has 8 f electrons, or one more than Eu++ Thus L = 3; S = 7/2 – 1/2 = 3; and J = 6 The ground state is 7F6 3a The relative occupancy probabilities... …⎟ ⎜1 + kT ⎠ ⎝ kT ⎠ → µ ⋅ ⎝ 4 2 2 Nµ µB ; χ→ = 2kT 2kT c The energy levels as a function of field are: If the field is applied to take the system from a to b we increase the entropy of the spin system from ≈ 0 to ≈ N log 2 If the magnetization is carried out constant total entropy, it is necessary that the lattice entropy be reduced, which means the temperature ↓ 4a Z = 1 + e −∆ T ; k B ∆e −∆ T... and those from Prob 5 we have E tot = E 0 (1 + ζ ) ( ) 1 − VN 2 1 + ζ 2 2 − NζµB Thus (for ζ = 9 2 9 N2 3 N 7a The Boltzmann factor gives directly, with τ = kBT... relation and R sq ≈ h/ e 2 = 137 c in Gaussian units Now 6-3 R sq ( ohms ) = 10−9 c 2 R sq ( gaussian ) ≈ ( 30 )(137 ) ohms ≈ 4.1kΩ 6-4 CHAPTER 7 1a The wavevector at the corner is longer than the wavevector at the midpoint of a side by the factor √2 As ε ∝ k2 for a free electron, the energy is higher by (√ 2)2 = 2 b In three dimensions the energy at a corner is higher by (√ 3)2 than at the midpoint