Baixe Kittel Charles-Introduction To Solid State Physics 8Th Edition-Solution Manual e outras Manuais, Projetos, Pesquisas em PDF para Física, somente na Docsity! CHAPTER 1 1. The vectors ˆ ˆ ˆ+ +x y z and ˆ ˆ ˆ− − +x y z 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 . 1cos 1/ 3 90 19 28' 109 28'−θ = = °+ ° = ° 2. The plane (100) is normal to the x axis. It intercepts the a' axis at and the c' axis at ; therefore the indices referred to the primitive axes are (101). Similarly, the plane (001) will have indices (011) when referred to primitive axes. 2a' 2c' 3. The central dot of the four is at distance cos 60 actn 60 cos30 3 a a° = ° = ° 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 2 2 a ca , 23 ⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ or 2 22 1 c 8a c ; 1.633. 3 4 a 3 = = 1-1 CHAPTER 2 1. The crystal plane with Miller indices hk is a plane defined by the points a1/h, a2/k, and . (a) Two vectors that lie in the plane may be taken as a 3 /a 1/h – a2/k and 1 3/ h /−a a . But each of these vectors gives zero as its scalar product with 1 2h k 3= + +G a a a , so that G must be perpendicular to the plane . (b) If is the unit normal to the plane, the interplanar spacing is hk n̂ 1ˆ /h⋅n a . But , whence . (c) For a simple cubic lattice ˆ / | |=n G G 1d(hk ) G / h| | 2 / | G|= ⋅ = πa G ˆ ˆ ˆ(2 / a)(h k )= π + +G x y z , whence 2 2 2 2 2 2 2 1 G h k . d 4 a + + = = π 1 2 3 1 13a a 0 2 2 1 12. (a) Cell volume 3a a 0 2 2 0 0 ⋅ × = −a a a c 21 3 a c. 2 = 2 3 1 2 1 2 3 2 3 ˆ ˆ 4 1 1(b) 2 3a a 0 | | 2 23a c 0 0 2 1 ˆ ˆ( ), and similarly for , . a 3 × π = π = − ⋅ × π = + x ˆ c y z a ab a a a x y b b (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. 3. By definition of the primitive reciprocal lattice vectors 3 32 3 3 1 1 2 1 2 33 1 2 3 3 C (a a ) (a a ) (a a )) (2 ) / | (a a a ) | | (a a a ) | / V . BZV (2 (2 ) × ⋅ × × × = π ⋅ × ⋅ × = π = π For the vector identity, see G. A. Korn and T. M. Korn, Mathematical handbook for scientists and engineers, McGraw-Hill, 1961, p. 147. 4. (a) This follows by forming 2-1 CHAPTER 3 1. 2 2 2 2E (h 2M) (2 ) (h 2M) ( L) , with 2L/ /= π λ = π λ .= 2. bcc: 12 6U(R) 2N [9.114( R ) 12.253( R) ].= ε σ − σ At equilibrium and 6 6 0R 1.488= σ , 0U(R ) 2N ( 2.816).= ε − fcc: 12 6U(R) 2N [12.132( R ) 14.454( R) ].= ε σ − σ At equilibrium and Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is more stable than the bcc. 6 6 0R 1.679= σ , 0U(R ) 2N ( 4.305).= ε − 23 16 9 3. | U | 8.60 N (8.60) (6.02 10 ) (50 10 ) 25.9 10 erg mol 2.59 kJ mol. − = ε = × × = × = This will be decreased significantly by quantum corrections, so that it is quite reasonable to find the same melting points for H2 and Ne. 4. 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 2 10 2 12 8 e (1.75) (4.80 10 ) 11.0 10 erg, R 3.66 10 − − − α × = = × × 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. 2 n A qU(R) N ; 2 log 2 Madelung const. R R ⎛ ⎞α = − α = =⎜ ⎟ ⎝ ⎠ In equilibrium 2 n 02n 1 2 0 0 U nA q nN 0 ; R R R R+ ⎛ ⎞∂ α = − + = =⎜ ⎟ ∂ α⎝ ⎠ A , q and 2 0 0 N q 1U(R ) (1 ). R n α = − − 3-1 b. ( ) ( ) 2 2 0 0 0 0 02 1 UU(R -R ) U R R R .. . , 2 R ∂ δ = + δ + ∂ bearing in mind that in equilibrium R0 ( U R) 0.∂ ∂ = 2 2 n 2 3 3 32 0 0 00 U n(n 1)A 2 q (n 1) q 2N N R R R R RR 2 + ⎛ ⎞ ⎛⎛ ⎞∂ + α + α = − = −⎜ ⎟ ⎜⎜ ⎟∂⎝ ⎠ ⎝ ⎠ ⎝ 2 0 q ⎞α ⎟ ⎠ For a unit length 2NR0 = 1, whence 0 2 2 2 2 2 04 22 2 0 0R 0 R U q U (n 1)q log 2(n 1) ; C R R R2R R ⎛ ⎞∂ α ∂ − = − = =⎜ ⎟∂ ∂⎝ ⎠ . 6. 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 = 4 and α = 1.638. Then from Eq. (23) with x ≡ R0/ρ we have 2 x 3x e 8.53 10 .− −= × By trial and error we find or Rx 9.2, 0 = 3.00 Å. The actual KCl structure has R0 (exp) = 3.15 Å . For the imagined structure the cohesive energy is 2 2 0 0 -αq p UU= 1- , or =-0.489 R R q ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ in units with R0 in Å. For the actual KCl structure, using the data of Table 7, we calculate 2 U 0.495, q = − 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. 7. 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. 8. From (37) we have eXX = S11XX, because all other stress components are zero. By (51), 11 11 12 11 123S 2 (C C ) 1 (C C ).= − + + Thus 2 211 12 11 12 11 12Y (C C C 2C ) (C C );= + − + further, also from (37), eyy = S21Xx, whence yy 21 11 12 11 12xxe e S S C (C C )σ = = = − + . 9. For a longitudinal phonon with K || [111], u = v = w. 3-2 2 2 11 44 12 44 1 2 11 12 44 [C 2C 2(C C )]K 3, or v K [(C 2C 4C 3 )]ρ ω ρ = + + + = ω = + + This dispersion relation follows from (57a). 10. We take u = – w; v = 0. This displacement is ⊥ to the [111] direction. Shear waves are degenerate in this direction. Use (57a). 11. Let 12xx yye e= − = e in (43). Then 2 21 1 1 1 2 4 4 411 12 21 1 2 2 11 12 U C ( e e ) C e [ (C C )]e = + − = − 2 so that 2 2 n 2 3 3 32 0 0 00 U n(n 1)A 2 q (n 1) q 2N N R R R R RR 2 + ⎛ ⎞ ⎛⎛ ⎞∂ + α + α = − = −⎜ ⎟ ⎜⎜ ⎟∂⎝ ⎠ ⎝ ⎠ ⎝ 2 0 q ⎞α ⎟ ⎠ is the effective shear 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 – 1 roots λ′ = 0 give λ = q – p. b. Set i[(K 3) (x y z) t] 0 i[. . . . .] 0 i[. . . . .] 0 u (r, t) u e ; v(r, t) v e ; w(r, t) w e , + + −ω= = = as the displacements for waves in the [111] direction. On substitution in (57) we obtain the desired equation. Then, by (a), one root is 2 2 11 12 442p q K (C 2C 4C ) / 3,ω ρ = + = + + and the other two roots (shear waves) are 2 2 11 12 44K (C C C ) / 3.ω ρ = − + 13. Set u(r,t) = u0ei(K·r – t) and similarly for v and w. Then (57a) becomes 2 2 22 0 11 y 44 y z 12 44 x y 0 x z 0 u [C K C (K K )]u (C C ) (K K v K K w ) ω ρ = + + + + + 0 and similarly for (57b), (57c). The elements of the determinantal equation are 3-3 2 1 2 2 pp 0 v M p a C− > = Σ . 3. From Eq. (20) evaluated at K = π/a, the zone boundary, we have 2 1 2 2 M u 2Cu ; M v 2Cv . −ω = − −ω = − 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. 2 0 2 0 0 0 p 0 p 0 sin pk a24. A (1 cos pKa) ; M pa 2A sin pk a sin pKa K M 1 (cos (k K) pa cos (k K) pa) 2 > > ω = Σ − ∂ω = Σ ∂ − − + When K = k0, 2 0p 0 A (1 cos 2k pa) , K M > ∂ω = Σ − ∂ which in general will diverge because p 1 .Σ →∞ 5. By analogy with Eq. (18), 2 2 s 1 s s 2 s 1 s 2 2 s 1 s s 2 s 1 s 2 iKa 1 2 2 iKa 1 2 Md u dt C (v u ) C (v u ); Md v dt C (u v ) C (u v ), whence Mu C (v u) C (ve u); Mv C (u v) C (ue v) , and − + − = − + − = − + − −ω = − + − −ω = − + − 2 iK 1 2 1 2 iKa 2 1 2 1 2 (C C ) M (C C e ) 0 (C C e ) (C C ) M −+ − ω − + a = − + + − ω 2 1 2 2 1 2 For Ka 0, 0 and 2(C C ) M. For Ka , 2C M and 2C M. = ω = + = π ω = 6. (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 force constant is e2/R3. The oscillation frequency ωD is (force constant/mass)1/2, or (e2/MR3)1/2. (b) For sodium and thus 23M 4 10 g−× 8R 2 10 cm;−× 10 46 1 2D (5 10 ) (3 10 ) − −ω × × (c) The maximum phonon wavevector is of the order of 1013 13 10 s−× 8 cm–1. If we suppose that ω0 is associated with this maximum wavevector, the velocity defined by ω0/Kmax ≈ 3 × 105 cm s–1, generally a reasonable order of magnitude. 7. The result (a) is the force of a dipole ep up on a dipole e0 u0 at a distance pa. Eq. (16a) becomes 2 P 2 3 3 p>0 (2 / M)[ (1 cos Ka) ( 1) (2e / p a )(1 cos pKa)] .ω = γ − + Σ − − At the zone boundary ω2 = 0 if P P 3 p>0 1 ( 1) [1 ( 1) ]p−+ σ Σ − − − = 0 , or if . The summation is 2(1 + 3p 3[1 ( 1) ]p 1−σ Σ − − = –3 + 5–3 + …) = 2.104 and this, by the properties of the zeta function, is also 7 ζ (3)/4. The sign of the square of the speed of sound in the limit Ka is given by the sign of 1<< p 3 2 p>0 1 2 ( 1) p p ,−= σ Σ − which is zero when 1 – 2–1 + 3–1 – 4–1 + … = 1/2σ. The series is just that for log 2, whence the root is σ = 1/(2 log 2) = 0.7213. 4-3 CHAPTER 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 mK (2/a)sin ( / ) −= ω ω 2 2 1/ 2mdK/d (2 / a)( ) −ω = ω −ω D( )ω . This is singular at ω = ω2 2 1/ 2m(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 ω3 04 K / 3 (4 / 3)[( ) / A]Ω = π = π ω −ω 3/2 1/ 2 0 is , provided ω < ω3 3 3/2 0D( )= (L/2 ) | d /d | (L/2 ) (2 / A )( )ω π Ω ω = π π ω −ω 0. It is apparent that D(ω) vanishes for ω above the minimum ω0. 2. The potential energy associated with the dilation is 2 3 B 1 1B( 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 24rms( 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 2R (h/2 V) −/< >= ρ Σω 1 1−Σω , whence 21 3Dd D( ) 3V / 4 v −= ∫ ω ω ω = ω π 3 2 3v22 DR 3h / 8/< >= ω π ρ . (b) In one dimension from (15) we have , whence D( ) L/ vω = π 1d D( ) −∫ ω ω ω diverges at the lower limit. The mean square strain in one dimension is 22 2 0 1( R/ x) K u (h/2MNv) K 2 /< ∂ ∂ >= Σ = Σ 2 2 3D 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 (2π/L)2 in K space, or (L/2π)2 = A/4π2 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 2N (A/4 ) ( K ) A / 4 v .= π π = ω π 2 The density of modes of each polarization type is D(ω) = dN/dω = Aω/2πv2. The thermal average phonon energy for the two polarization types is, for each layer, D D 20 0 AU 2 D( ) n( , ) d 2 d , 2 v exp(h / ) 1 ω ω ω ω = ω ω τ ω ω = π ω τ −∫ ∫ ω dω where ωD is defined by . In the regime D D N D( ) ω = ω∫ Dω >> τ , we have 3 2 2 2 x0 2A xU dx. 2 v e 1 ∞τ ≅ π −∫ 5-1 9 3 28 34 (2 10 ) 3 10 cmπ × ≈ × 3 the electron concentration is 57 28 3 28 10 3 10 cm . 3 10 −≈ ≈ × × Thus 2 2 3 27 20 7 4 F h 1 1(3 n) 10 10 10 ergs, or 3.10 eV. 2m 2 2 2 − −/ε = π ≈ ⋅ ≈ ≈ (b) The value of kF is not affected by relativity and is ≈ n1/3, where n is the electron concentration. Thus 3F Fhck hc/ /ε √n. (c) A change of radius to 10 km = 106 cm makes the volume ≈ 4 × 1018 cm3 and the concentration ≈ 3 × 1038 cm – 3. Thus (The energy is relativistic.) 27 10 13 4 8F 10 (3.10 ) (10 ) 2.10 erg 10 eV. − −ε ≈ ≈ ≈ 5. The number of moles per cm3 is 81 × 10–3/3 = 27 × 10–3, so that the concentration is 16 × 1021 atoms cm– 3. The mass of an atom of He3 is (3.017) (1.661) × 10–24 = 5.01 × 10–24 g. Thus 54 23 21 2 3 16 F [(1.1 10 ) 10 ][(30)(16) 10 ] 7 10 − −ε × × ≈ × − erg, or TF ≈ 5K. 6. Let E, v vary as e–iwt. Then eE m e E 1 iv , i m 2 τ + ωτ = − = − ⋅ − ω+ (1 τ) 1+ (ωτ) and the electric current density is 2ne 1 ij n( e)v E. m 2 τ + ωτ = − = ⋅ 1+ (ωτ) 7. (a) From the drift velocity equation x x c y y yi v (e m)E v ; i v (e m)E v .ω = +ω ω = −ωc x We solve for vx, vy to find 2 c x x c 2 c y y c ( )v i e m E e m E ( )v i e m E e m E 2 2 ω −ω = ω( ) +ω ( ) ω −ω = ω( ) +ω ( ) y x ; . We neglect the terms in ωc2. Because j = n(–e)v = σE, the components of σ come out directly. (b) From the electromagnetic wave equation 2 2 2c E E t2 ,∇ = ε∂ ∂ we have, for solutions of the form ei(kz – ωt), the determinantal equation 2 2 xx xy 2 2 yx yy c k 0. c k 2 2 2 2 ε ω − ε ω = ε ω ε ω − 6-2 Here 2xx yy P xy yx c p1 and i . 2ε = ε = −ω ω ε = −ε = ω ω ω2 3 The determinantal equation gives the dispersion relation. 8. The energy of interaction with the ion is ( )0 r 2 2 00 e r 4 r dr 3e 2ρ π = −∫ r , where the electron charge density is –e(3/4πr03). (b) The electron self-energy is ( ) ( )0r2 3 2 1 2 00 dr 4 r 3 4 r r 3e 5r . −ρ π π =∫ The average Fermi energy per electron is 3εF/5, from Problem 6.1; because 3 0N V 3 4 r= π , the average is ( )2 3 22 03 9 4 h 10mr/π . The sum of the Coulomb and kinetic contributions is 2 s s 1.80 2.21U r r = − + which is a minimum at s2 3 s s 1.80 4.42 , or r 4.42 1.80 2.45. r r = = = The binding energy at this value of rs is less than 1 Ry; therefore separated H atoms are more stable. 9. From the magnetoconductivity matrix we have ( ) c y yx x 0 x2 c j E E 1 .ω τ= σ = σ + ω τ For ωcτ >> 1, we have ( )( )2yx 0 c ne m mc eB neB cσ ≅ σ ω τ = τ τ = . 10. For a monatomic metal sheet one atom in thickness, n ≈ 1/d3, so that 2 2 2 sq F FR mv nd e mv d e≈ ≈ . If the electron wavelength is d, then Fmv d h/≈ by the de Broglie relation and 2 sqR h e 137 c/≈ = in Gaussian units. Now 6-3 ( ) ( ) ( ) ( ) 9 2 sq sqR ohms 10 c R gaussian 30 137 ohms 4.1k . −= ≈ ≈ Ω 6-4 CHAPTER 8 4 d 2 m* 11a. E 13.60 eV 6.3 10 eV m −= × × × ε 6 H mb. r a 6 10 c m* −= ×ε× × m c. Overlap will be significant at a concentration 15 3 34 3 1N 10 atoms c r − π = ≈ m 2a. From Eq. (53), , in an approximation not too good for the present example. d BE / 2k T1/ 20 dn (n N ) e − 3/ 2 13 3B 0 2 m*k Tn 2 4 10 cm 2 h −⎛ ⎞≡ ≈ ×⎜ ⎟/π⎝ ⎠ ; 1.45d B 13 3 E 1.45 ; e 0.23 . 2k T n 0.46 10 electrons cm . − −× 14 H 1b. R 1.3 10 CGS units nec −= − − × 3. The electron contribution to the transverse current is e y e x Bj (e) ne E E ; c µ⎛ ⎞µ +⎜ ⎟ ⎝ ⎠ y for the holes ny h x Bj (h) ne E E . c −µ⎛ ⎞µ +⎜ ⎟ ⎝ ⎠ y Here we have used e h ce e ch h B Bfor electrons; for holes. c c µ µ ω τ = ω τ = The total transverse (y-direction) current is 2 2 e h x e h y0 (ne pe )(B/c)E (ne pe )E , (*)= µ − µ + µ + µ and to the same order the total current in the x-direction is x h ej (pe ne )E .x= µ + µ Because (*) gives 8-1 2 2 h e y x h e p n 1E E B p n c µ − µ ,= ⋅ µ + µ we have for the Hall constant 2 2 y h e H 2 x h E p n1R . j B ec (p n ) µ − µ = = ⋅ µ + µe t z y c. hc 1 4. The velocity components are . The equation of motion in k space is . Let B lie parallel to the k x x t y y t z zv hk / m ; v hk / m ; v hk /m/ / /= = = h dk/dt (e/c) v B/ = − × x axis; then . We differentiate with respect to time to obtain ; on substitution for dk x y z z t y tdk / dt 0; dk / dt k ; eB/m c; dk / dt k ; eB/m c= = −ω ω ≡ = ω ω ≡ 2 2 yd k / dt dk / dt= −ω z/dt we have , the equation of motion of a simple harmonic oscillator of natural frequency 2 2y td k / dt k 0+ω ω = 1/ 2 1/ 2 0 t t( ) eB/(m m )ω = ω ω = 5. Define . In the strong field limit Q the magnetoconductivity tensor (6.64) reduces to e e e h hQ eB / m c; Q eB / m≡ τ = τ >> 2 1 2 1 e e h h2 2 1 2 1 2e h e e h h e h Q Q 0 Q Q 0 ne peQ Q 0 Q Q 0 m m 0 0 1 0 0 1 − − − − − − − − ≈ ⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟τ τ σ + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . e h We can write nec Qe/B for and pec Q 2 ene / mτ h/B for . The strong field limit for σ 2 hpe / mτ yx follows directly. The Hall field is obtained when we set y x e h ec n pj 0 (n p) E E H Q Q y . ⎡ ⎤⎛ ⎞ = = − + +⎢ ⎥⎜ ⎟ ⎝ ⎠⎣ ⎦ The current density in the x direction is x x e h ec n pj E (n B Q Q y p) E ; ⎡ ⎤⎛ ⎞ = + − −⎢ ⎥⎜ ⎟ ⎝ ⎠⎣ ⎦ using the Hall field for the standard geometry, we have 2 x x e h e h ec n p (n p)j E H Q Q n p Q Q ⎡ ⎤⎛ ⎞ − = + +⎢ ⎥⎜ ⎟ ⎛ ⎞⎝ ⎠⎢ ⎥+⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ . 8-2 CHAPTER 9 1. 2a. 8 10.78 10 cm a −π− = × 8 10.78 10 cm a −π− = × 8 11.57 10 cm a −π− = × 8 -0.78 10 cm b π = × 1 9-1 b. The electron moves in a direction normal to the Fermi surface -- more or less in a straight line if the Fermi surface is close to planar in the region of interest. The magnetic field puts a wiggle on the motion, but the field does not make the electron move in a helix, contrary to the behavior of a free electron. 6a. Region I: 2 2 02 2 2 h d U 2m dx h kA cos kx ; 2m ψ ψ ψ ⎛ ⎞/ − − = ε⎜ ⎟ ⎝ ⎠ / = ε = 0U (*)− Region II: 2 2 2 2 2 qx h d 2m dx h qB e ; 2m ψ ψ ψ − / − = ε / = ε = − (*) Boundary condition 1 d dx ψ ψ continuous. k tan (ka / 2) q ,= (**) with k and q related to ε as above. b. The lazy way here is to show that the ε’s in the equations marked (*) above are equal when k and q are connected by (**), with ε = –0.45 as read off Fig. 20. This is indeed so. 7a. 1 2( ) H hc π ∆ = / e S , where S = πkF2, with kF = 0.75 × 108 cm–1 from Table 6.1, for potassium. Thus 8 1 2 F 1 2( ) 0.55 10 G . H 137 k e − −∆ × 9-4 b. F F c F 3 2 6 2 v mc hk cR v ; R e B e B 0.5 10 cm R 0.7 10 cm . ω − − / = = = × π × 8. Write (17) as , where 0H H H= + 1 1 (h / m) k pH /= ⋅ . Then (18) is an eigenfunction of with the eigenvalue . In this representation the diagonal matrix element of is equal to In a cubic crystal will be even or odd with respect to the inversion operation , but is an odd operator. It follows that the diagonal matrix element vanishes, and there is no first-order correction to the energy. The function to first order in is 0H 2 2 n (0) h k / 2m/ε + 1H 0 0(h / m) dV u (r) k p U (r)./ ⋅∫ 0U (r) r → − r p kU (r) 1H 1 k 0 n j j0 | | n0U (r) U (r) , (0) (0)j H< >′= + Σ ε − ε and the energy to second order is 2 2 2 n n n j | n0 | k p | j0 | (k) (0) (hk) / 2m (h / m) . (0) (0)j < ⋅ > ′/ /ε = ε + + Σ ε − ε The effective mass ratio is the coefficient of , or 2 2h k / 2m/ 2 * n j | n0 | p | j0 |m 21 . m m (0) (0j ) < > ′= + Σ ε − ε 9a. n m n m n m ik' r ik r1 * kk'k k' ik (r r )1 k dV w*(r r ) w (r r ) N e e dV (r) N e ψ ψ⋅ − ⋅− ⋅ −− − − = Σ Σ = Σ ∫ ∫ (r) where the summation is zero unless n = m, when it is equal to N. b. nik(x x )1 2n 0 k w(x x ) N U (x) e .−−− = Σ The summation is equal to 9-5 n dpn n n N 2 i2 p(x x ) Nai2 p(x x ) Na P N 2 i (x x ) a i (x x ) a n n n e e e e sin [ x x , i2 x x Na (x x ) Na π −π − − π − − π − Σ − π( − = = π( − ) π − ∫ a ]) whence 1 2 n n 0 n sin [ x x a ]w (x x ) N u (x) (x x ) a π( − ) − = π − . 10a. jy = σ0 (Q–1 Ex + sEy) = 0 in the Hall geometry, whence Ey = – Ex/sQ. b. We have jx = σ0 (Q–2 Ex – Q–1 Ey), and with our result for Ey it follows that 2 1 2 x 0j (Q s Q ) Eσ − − −= + x , whence 2x x 0 sE j (Q ) s 1 ρ = = σ + . 9-6 ( ) 2 2 02 B 1 B B . ⎛ ⎞∂ ∂ λ + − = −Φ δ⎜ ⎟∂ρ ρ ∂ρ⎝ ⎠ ρ This equation has the solution ( ) ( ) ( )20 0B 2 K ,ρ = Φ πλ ρ λ where K0 is a hyperbolic Bessel function* infinite at the origin and zero at infinity: ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 2 0 1 22 0 B 2 n ; B 2 2 exp ρ << λ ρ Φ πλ λ ρ ρ >> λ ρ Φ πλ πλ ρ −ρ λ . 0. The total flux is the flux quantum: ( ) ( )0 0 0 0 2 d B dx x K x ∞ ∞ π ρ ρ ρ = Φ = Φ∫ ∫ 5. It is a standard result of mechanics that 1grad c t.−= − ϕ− ∂ ∂E A If grad ϕ = 0, when we differentiate the London equation we obtain ( )22 Lj t c 4 E.∂ ∂ = πλ Now j = nqv and ( )2j t nq v t nq m E.∂ ∂ = ∂ ∂ = Compare the two equations for ∂j/∂t to find 22 2 Lc 4 nq m.πλ = *Handbook of mathematical functions, U.S. National Bureau of Standards AMS 55, sec. 9.6. 6. Let x be the coordinate in the plane of the junction and normal to B, with w 2 x w 2.− ≤ ≤ The flux through a rectangle of width 2x and thickness T is 2xTB = φ (x). The current through two elements at x and –x, each of width dx is ( ) ( ) ( ) ( )0 0dJ J w cos e x hc dx J w cos 2xTeB hc dx ,/ /= Φ =⎡ ⎤⎣ ⎦ and the total current is ( ) ( ) ( )( ) w 2 0 0 0 sin wTBe hc J J w cos xTe B hc dx J . wTBe hc / /= = /∫ 7a. For a sphere ( ) aH inside B 4 M 3;= − π for the Meissner effect whence ( )H inside 4 M,= − π aB 8 M= − π 3. b. The external field due to the sphere is that of a dipole of moment µ = MV, when V is the volume. In the equatorial plane at the surface of the sphere the field of the sphere is 3 aa 4 M 3 B−µ = − π = 2. The total field in this position is 3Ba/2. 10-3 10-4 CHAPTER 11 1. From Eq. (10), 2 2 2 eN r 6mc .χ = − < > Here 0 2r a 22 2 03 00 1r 4 r dr e a ∞ − < > = ⋅ π ⋅ = π ∫ 3a . The numerical result follows on using N = 6.02 × 1023 mol–1. 2a. Eu++ has a half-filled f shell. Thus S = 7 × 1/2 = 7/2. The orbitals mL = 3, 2, 1, 0, –1, –2, –3 have one spin orientation filled, so that L = ΣmL = 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 are ( ) ( ) ( ) B kT B kT B kT s______ e Here stand for k ______ e ______ e ______ 1 − ∆+µ −∆ − ∆−µ ∆ ∆ The average magnetic moment is ( ) ( )B kT B kTe e Z − ∆−µ − ∆+µ− < µ > = µ where ( ) ( )B kT B kTkTZ 1 e e e− ∆−µ − ∆+µ−∆= + + + . b. At high temperatures kTe 1−∆ → and 11-1 c. For B = 0 and ζ = 0. 2 2tot 0 02 2 E E20 1 40 4 FE VN 0 if V 9 2 9 N 3 N ε∂ − < > = ∂ζ 7a. The Boltzmann factor gives directly, with τ = kBT ( ) ( )2 2B B B e eU tanh e e C k dU d k k T sech k T , ∆ τ −∆ τ ∆ τ −∆ τ − = −∆ = −∆ ∆ τ + = τ = ∆ ∆ B ; because d tanh x/dx = sech2x. b. The probability P(∆) d∆ that the upper energy level lies between ∆ and ∆ + d∆, referred to the midpoint as the zero of energy, is P(∆) d∆ = (d∆) / ∆0. Thus, from (a), ( ) ( ) ( ) ( ) 0 0 0 0 0 2 2 2 B 0 0 x 2 2 B 0 0 U d tanh C k d sech k dx x sech ∆ ∆ < >= − ∆ ∆ ∆ ∆ τ < > = ∆ ∆ ∆ τ ∆ τ = τ ∆ ∫ ∫ ∫ , x , where x ≡ ∆/τ. The integrand is dominated by contributions from 0 < ∆ < τ, because sech x decreases exponentially for large values of x. Thus ( ) 2 2B 0 0 C k dx x sech x . ∞ < > τ ∆ ∫ 8. B Be e 2 sinh x 1 2 cosh x 1 2 cosh x µ τ −µ τ< µ > − = = µ + + 11-4 CHAPTER 12 1. We have Thus ikS S e ⋅δρ+δ ρ= . ( ) ( ) ( ) x yik y x y z y x x y z dS 2JS 6 e S dt h 2JS 6 2 cos k a cos k a cos k a S ; h dS 2JS 6 2 cos k a cos k a cos k a S . dt h ρ ⋅δ ρ δ ρ ρ ρ ⎛ ⎞= −∑⎜ ⎟/⎝ ⎠ ⎛ ⎞ ⎡ ⎤= − + +⎜ ⎟ ⎣ ⎦/⎝ ⎠ ⎛ ⎞ ⎡ ⎤= − − + +⎜ ⎟ ⎣ ⎦/⎝ ⎠ These equations have a solution with time-dependence ∼ exp(–iωt) if ( ) ( )x y2JS h 6 2 cos k a 2 cos k a 2 cos k a/ω = − − − z . 2. ( ) ( )k k k U n h h d n/ /= ∑ ω = ω ω ω< ω >∫ D . If ω = Ak2, then 1 2d dk 2Ak 2 A ,ω = = ω and ( ) 1 2 3 21 2 4 1 1 . 8 A 4 A2 A π ω ω ω = = π πω D 3/ 2 Then 3 2 2 3 2 h h 1U d 4 A e 1/β ω / = ω ω . π −∫ At low temps, ( ) ( ) 3 2 5 2 5 2x gamma zeta0 function function 1 x 1 5 5dx ;1 e 1 2 2h h ∞ ⎛ ⎞ ⎛ ⎞= Γ ς⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠/ /β β∫ ∫ [See Dwight 860.39] ( )5 2 2 3 2 3 2BU 0.45 k T / A h/π ( )3 2B BC dU dT 0.113 k k T hA ./= 12-1 3. ( ) ( ( ) A B A B B A M T C B M M B applied field M T C B M M = −µ − ε = = − ε −µ ) Non-trivial solution for B = 0 if ( )C T C C 0; T C C T C + ε µ = = µ − µ + ε ε Now find ( )A B CM M B at T Tχ = + > : ( ) ( ) ( ) ( )C 2CMT 2CH CM ; T C T . = − ε +µ χ = + µ + ε ∴θ = µ + ε µ − ε 4. The terms in which involve ee cU U U+ + K xx are ( )2 211 xx 12 xx yy zz 1 1 xx1 C e C e e e B e .2 + + + α Take ∂/∂exx: ( ) 211 xx 12 yy zz 1 1C e C e e B 0, for minimum.+ + + α = Further: ( ) 2 11 yy 12 xx zz 1 2 2 11 zz 12 xx yy 1 3 C e C (e e ) B 0 . C e C e e B 0 . + + + α = + + + α = Solve this set of equations for exx: ( ) ( ) ( ) 2 12 2 11 12 xx 1 11 12 11 12 C C 2C e B C C C 2C −α + = − + . Similarly for eyy, ezz, and by identical method for exy, etc. 12-2 CHAPTER 13 1. Consider a coil which when empty has resistance R0 and inductance L0. The impedance is Z0 = R0 – iωL0. When the coil is filled with material of permeability the impedance is 1 4µ = + πχ ( ) ( )0 0 0 0Z R i L 1 4 R i L 1 4 4 i′ ′′= − ω + πχ = − ω + πχ + π χ , or ( )0 0 0 R L Z R 4 L i L 1 4 .′′ ′= + πωχ − ω + πχ 2a. x x ˆdF dF dxx̂ F dt dt dt = + + . x y z R ˆdyˆ ˆdF dx dzF F F dt dt dt dt ⎛ ⎞⎛ ⎞= + + +⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . Now ( ) ( ) ( ) x ˆdyˆ ˆdx dzˆ ˆx ; y ; z . dt dt dt ˆdxF F . dt = Ω× = Ω× = Ω× + ⋅⋅⋅⋅ = Ω× ˆ b. ; R dM dMM B M M B . dt dt ⎛ ⎞= γ × +Ω× = γ ×⎜ ⎟ ⎝ ⎠ R dM M B dt ⎛ ⎞Ω⎛ ⎞ = γ × +⎜ ⎟⎜ ⎟ γ⎝ ⎠ ⎝ ⎠ . c. With we have 0 ˆB zΩ = −γ 1 R dM ˆM B x , dt ⎛ ⎞ = γ ×⎜ ⎟ ⎝ ⎠ so that precesses about with a frequency ω = γBM x̂ 1. The time t1/2 to give t1/2ω = π is t1/2 = π/γB1. d. The field rotates in the xy plane with frequency 1B 0B .Ω = γ 13-1 3a. 2 2 z i j j k aB I N ⎛ ⎞< > = ∑∑ <⎜ ⎟ ⎝ ⎠ z kI > , where for 1I 2 = we have z zj k jk 1I I 4 < > = δ . Thus 2 2 2 i j jk a 1 aB . N 4 4N ⎛ ⎞< > = ∑δ =⎜ ⎟ ⎝ ⎠ k b. 4 4 z z i j k jk m aB I I I N ⎛ ⎞< > = ∑ < >⎜ ⎟ ⎝ ⎠ z z mI . Now z z z z j k m jk k m jk m j km jm k 1I I I I [ 16 ], and < >= δ δ δ + δ +δ δ + δ δ δ 4 4 2 4 2 i a 1 a 3NB [N 3N ] N 16 N 16 ⎛ ⎞ ⎛ ⎞< > = + −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . 4. For small θ, we have . Now the magnetic energy density 2KU K− θ MU BM cos= − θ− 21BM BM 2 − + θ , so that with proper choice of the zero of energy the anisotropy energy is equivalent to a field AB 2K M= along the z axis. This is valid for 1θ << . For a sphere the demagnetizing field is parallel to M and exerts no torque on the spin system. Thus B0 + BA is the effective field. 5. We may rewrite (48) with appropriate changes in M, and with Banisotropy = 0. Thus ( ) ( ) A A A B B A B B B A A B i M i M M M M ; i M i M M M M . + + + + + + − ω = − γ λ + λ − ω = γ λ + λ The secular equation is A B A A B B B A M M 0 , M M γ λ −ω γ λ = −γ λ γ λ −ω or 13-2 ( )2 A B B AM Mω −ω γ λ − γ λ = 0 . One root is ω0 = 0; this is the uniform mode. The other root is ( )0 A B B AM Mω = λ γ − γ = 0; this is the exchange mode. 13-3 ( ) ( ) ( ) ( ) ( ) ( ) 6 11 18 1 8 6 14 F 1 2 1 22 23 20 2 p 16 1 1 10 9 10 10 s ; 1 v 1.6 10 4 10 0.4 10 s ; 4 ne m 10 10 23 10 10 1.5 10 s . − − − − − − − σ = × ρ = τ = × × × ω = π × × × × 1 7 The homogeneous equation has a solution of the form ( ) (tx t 0 Ae sin t ,−λ> = ω + φ) where ( ) 1 222 p 2⎡ω = ω + ρ⎣ ⎤ ⎦ and λ = ρ/2. To this we add the particular solution x = –e/mω and find A and φ to satisfy the initial conditions x(0) = 0 and ( )x 0 0.= 11. The Laplacian whence 2 0,∇ ϕ = 2 2 2 d f K f 0 . dz − = This has solutions ( ) ( ) Kz K z d f Ae for z 0 f Ae for z d f B cosh K z d 2 for 0 z d − − = < = > = − < .< This solution assures that ϕ will be continuous across the boundaries if B = A/cosh(Kd/2). To arrange that the normal component of D is continuous, we need ε(ω) ∂ϕ/∂z continuous, or ε(ω) = – tanh(Kd/2). 14-3 CHAPTER 15 1a. The displacement under this force is ( ) ( ) i t1x t d . 2 e ∞ − ω −∞ = α ω ω π ∫ With ω = ωR + iωI, the integral is ( ) R Ii t te e d− ω ω .α ω ω∫ This integral is zero for t < 0 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) = 0 for t < 0. 1b. We want ( ) i t 2 2 0 1 e dx t , 2 i ∞ − ω −∞ ω = π ω −ω − ωρ∫ (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 half- plane. 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 ( ) ( ) ( ) ( )R I I Rexp i i t exp t exp i t ,⎡ ⎤− ω + ω − = −ω ω⎣ ⎦ which → 0 as |ω| → ∞. Thus the integral in (A) must also vanish. For t > 0 we can evaluate x(t) by carrying out a Cauchy integral in the lower half-plane. The residues at the poles are ( ) ( ) ( ) 1 1 2 22 22 21 1 1 1 0 02 4 2 4exp t exp i t , ⎡ ⎤± ω − ρ − ρ ω − ρ⎢ ⎥⎣ ⎦ ∓ so that ( ) ( ) ( ) ( ) 1 1 2 22 22 21 1 1 0 04 2 4x t exp t sin t .= ω − ρ − ρ ω − ρ 2. In the limit ω → ∞ we have ( ) 2jf′α ω → − ω∑ from (9), while from (11a) 15-1 ( ) ( )2 0 2 s s ds ∞ ′ ′α ω → − α πω ∫ .′ 3. The reflected wave in vacuum may be written as ( ) ( ) ( )i kx + ty zE refl B refl A e ,− ω′− = = 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 ( ) ( ) ( ) ( ) y z i kx t1 2 z E trans ck B trans B trans A"e ,−ω− = εω = ε = 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 = 0 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 1 2 1 2 A' 1 , A 1 − ε = ε + and ( ) ( ) 1 2 1 2 E refl A' 1 n ik 1r . E inc A 1 n ik 1 ε − + − ≡ = − = = ε + + + The power reflectance is ( ) ( ) ( ) 2 2 2 2 n 1 Kn ik 1 n ik 1R r r n ik 1 n ik 1 n 1 K − +− − + −⎛ ⎞⎛ ⎞ω = ∗ = =⎜ ⎟⎜ ⎟− + + +⎝ ⎠⎝ ⎠ + + . 4. (a) From (11) we have ( ) ( )2 2 0 ' s2" P s ∞ σω σ ω = − π −ω∫ ds. In the limit ω → ∞ the denominator comes out of the integrand and we have 15-2 CHAPTER 16 1. 2 33 3 H e x eE; ex r E p; p E r a r r ⋅ = = = α = = = . 2. i 0 4E E P 3 π = − = 0 inside a conducting sphere. Thus 3 3 0 4p a P a 3 π = = E , and 3 0p E aα = = . 3. 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 two plates is air diel air 1E qd E d E d q⎛+ = +⎜ ⎞ ⎟ε⎝ ⎠ . For a plate of area A, the capacitance is AC . 14 d q = ⎛ ⎞π +⎜ ⎟ε⎝ ⎠ It is useful to define an effective dielectric constant by eff 1 1 q .= + ε ε If ε = ∞, then εeff = 1/q. We cannot have a higher effective dielectric constant than 1/q. For q = 10–3, εeff = 103. 4. The potential drop between the plates is E1 d + E2 qd. The charge density 1 1 2 Q D E i E , A 4 4 ε σ = = = π π ω (CGS) by comparison of the way σ and ε enter the Maxwell equation for curl H. Thus 1 2 2 4 i 4 iE E ; V E dπ σ π σ⎛ ⎞+ = ⎜ ⎟εω εω⎝ ⎠ q ;+ 2 AiQ σ= ω E ; and thus Q AC 1 i qV 4 d 4 ≡ = ω⎛ ⎞π −⎜ ⎟ε πσ⎝ ⎠ , and ( ) ( )eff 1 q 1 i q 4 ε ε = + − ωε πσ . 16-1 5a. Eintint 0 0 4 4E E P E 3 3 π π = − = − χ . 0int EE 41 3 = π + χ . b. int 0P E E41 3 χ = χ = π + χ . 6. E = 2P1/a3. P2 = αE = 2αP1/a3. This has solution p1 = p2 0 if 3 3 12 a ; a 2 α = α = . 7 (a). One condition is, from (43), ( ) 2 4C 0 4 s 6 sT T g P g P 0γ − − + = . The other condition is ( ) 2 4 6c 0 s 4 s 6 s 1 1 1T T P g P g P 0 . 2 4 6 γ − − + = Thus 2 4 2 4 s 6 s 4 s 6 s 1 1g P g P g P g P ; 2 3 − + = − + 4 2 2 4 6 s 4 s 6 g2 1 3g P g ; P . 3 2 4 g = = (b) From the first line of part (a), ( ) 2 2 2 4 4 4 c 0 6 6 g g g3 9 3T T 4 g 16 g 16 g γ − = − = 6 . 8. In an electric field the equilibrium condition becomes ( ) 3c 4E T T P g P 0− + γ − + = , where the term in g6 is neglected for a second-order transition. Now let . If we retain only linear terms in , then sP P P= + ∆ P∆ ( ) 2c 4 sE T T P g 3P P 0− + γ − ∆ + ∆ = , with use of (40). Further, we can eliminate because 2sP ( ) ( )2s 4 cP g T T= γ − . Thus ( )cP E 1 2 T T∆ = γ − . 16-2 9 a. ( )a cos na a→ ← → ← → ← π ← → i i i i i i b. 2a← → i i i i i i Deforms to new stable structure of dimers, with lattice constant 2 × (former constant). c. 10. The induced dipole moment on the atom at the origin is p = αE, where the electric field is that of all other dipoles: ( ) ( )( )33 3 nnE 2 a p 4p a −= ∑ = ∑ ; 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 = 0 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 18 1. Carbon nanotube band structure. (a) ),(),,(2 3 22 23 22 1 aaaaijji πππππδ =−==>=⋅ bbab . (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 = a3 4π , 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 and there is a massless subband. jk y ˆ=∆K (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: 10/)3/4(211 avF πε = = 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: 222 )20/( ykK +=∆k , so: 2222 )(])20/[( FyFF vkvvK +=ε This is of the desired form, with FvKm 20/* = . 12.020//* == FmvKmm . 2. Filling subbands )( 2 ),( 222 22 yxyx nnmW nn += πε => States are filled up to )8( 2 )2,2( 2 22 mW πε = (1,1) subband: )28( 22 2 222 1,1 2 −= mWm k π => W k π61,1 = => W kn 622 1,11,1 == π (2,1) subband: )58( 22 2 222 1,2 2 −= mWm k π => W k π31,2 = => W kn 322 1,21,2 == π 18-1 (2,1) subband: same. WW n 3462 += = 5.9 x 108 /m. 3. Breit-Wigner form of a transmission resonance (a) ; 2/1)cos( 2δϕδϕ −≅ 481 2 2 12 ||||1||1|| iiii tttr −−≅−= The denominator of (29) is then: )1)(||||1)(||||1(2)||1)(||1(1 221 4 28 12 22 14 18 12 12 12 2 2 1 δϕ−−−−−−−−+ tttttt = 2222 2 14 122 2 2 12 14 2 4 14 1 )|||(|||||)|||(| δϕδϕ ++=+++ tttttt 222 2 2 1 2 2 2 1 4)|||(| ||||4 δϕ++ =ℑ tt tt . (b) kLδδϕ 2= and επεδεδ ∆=∆∆= /)/(// Lkk . Combining: εδεπδϕ ∆= /)/)(2( LL => εδεπδϕ ∆= /2/ (c) Combining: 222 2 2 1 2 22 2 2 1 4)|||(|)2/( )2/(||||4 δεπε πε ++∆ ∆ =ℑ tt tt which is (33). 4. Barriers in series and Ohm’s law (a) 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 |||| ||)||1(||)||1(11 |||| ||||||||11 |||| ||||11 tt trrt tt ttrr tt rr −−−− += −− += − = ℑ 2 1 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 || || || ||1 |||| ||||||||)|||(|11 t r t r tt trrttr ++= +++− += which gives (36). (b) m ek m en FD D π ττσ 22 1 1 2 == , and FF vm k = => h veev FF D )2(22 22 1 τ π τσ == But: ττ FBFB vv 2== => h e B D 2 1 2 =σ . 5. Energies of a spherical quantum dot (a) => Integrating from inner to outer shell: oenclA Qd εε/=⋅∫ aE 24/ rqE oπεε= V = )(4 11 44 2 dRR dq dRR q r qdr oo dR R o + =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + −=∫ + πεεπεεπεε 18-2 d dRR V qC o )(4 +== πεε and therefore )(4 22 dRR de C eU o + == πεε . (b) For d << R , d A d RC oo εεπεε =≅ 2 4 . (c) For d >> R , R eU oπεε4 2 = . Also 2* 22 0,0 2 Rm πε = => 22 2*2 0,0 2 4 ππεεε Rm R eU o ⋅= *222 *2 22 2*2 0,0 22 4 2 4 Boo a RRmeRm R eU πππεεππεεε =⋅=⋅= 6. Thermal properties in 1D (a) v L vL KD ππ ω == 1 /2 2)( ∫∫∫ ∞∞ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= − ≅ − = 0 2 00 1)exp(1)/exp(1)/exp( )( x xdxTk v L Tk d v L Tk DdU B BB tot D πω ωω πω ωωωω Obtaining value from table of integrals: hv TLk v TLkU BBtot 36 222222 ππ π == hv TLkTUC B VtotV 3 2/ 22π ==∂∂= (b) The heat flow to the right out of reservoir 1 is given by: ℑ=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ℑ= − ℑ =ℑ − ⋅⋅= ∫∫ ∞∞ h TkTk Tk d Tk dv L DJ BB BB R R 6621)/exp(21)/exp( )( 21 2222 1 0 10 1 ππ πω ωω πω ωωω and similarly for JL. The difference is: ( )2221 22 6 TT h kJJ BLR − ℑ =− π Let => TTTTT =∆+= 21 , ( ) TTTT ∆≈− 22221 for small ∆T. => T h kJJ BLR ∆ ℑ =− 3 22π which gives (78). 18-3