Exam (elaborations) TEST BANK FOR Introduction to Solid State Physics

Exam (elaborations) TEST BANK FOR Introduction to Solid State Physics The vectors xˆ ˆ + + y 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 . 1 cos 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 a ctn 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 c a , 3 2 ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or 2 1 2 2 c 8 a c ; 1.633. 3 4 a 3 = = 1-1 CHAPTER 2 1. The crystal plane with Miller indices hkA 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 / A 1/h – a2/k and 1 3 a / h − a / A . But each of these vectors gives zero as its scalar product with = h k 1 2 + + 3 G a a Aa , so that G must be perpendicular to the plane hkA . (b) If nˆ is the unit normal to the plane, the interplanar spacing is 1 n a ˆ ⋅ /h . But , whence . (c) For a simple cubic lattice n G ˆ = / |G | d(hk 1 A) G= ⋅a G / h| | = 2π/ | G| G = (2π + / a)(hxˆ ˆ ky + Azˆ), whence 2 2 2 2 2 2 2 1 G h k . d 4 a + + = = π A 1 2 3 1 1 3a a 0 2 2 1 1 2. (a) Cell volume 3a a 0 2 2 0 0 a a⋅ ×a = − c 1 2 3 a c. 2 = 2 3 1 2 1 2 3 2 3 ˆ ˆ 4 1 1 (b) 2 3a a 0 | | 3a c 2 2 0 0 2 1 ( ˆ ˆ), and similarly for , . a 3 × π = π = − ⋅ × π = + x ˆ c y z a a b 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 3 2 3 3 1 1 2 3 1 2 3 1 2 3 3 C (a a ) (a a ) (a a ) ) (2 ) / | (a a a ) | | (a a a ) | / V . V ( BZ 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 2 2 1 2 2 1 2 1 exp[ iM(a k)] 1 exp[iM(a k)] |F| 1 exp[ i(a k)] 1 exp[i(a k)] 1 cosM(a k) sin M(a k). 1 cos(a k) sin (a k) − − ⋅ ∆ − ⋅ ∆ = ⋅ − − ⋅ ∆ − ⋅ ∆ − ⋅ ∆ ⋅ ∆ = = − ⋅ ∆ ⋅ ∆ (b) The first zero in 1 sin M 2 ε occurs for ε = 2π/M. That this is the correct consideration follows from 1 zero, as Mh is an integer 1 1 sin M( h ) sin Mh cos M cos Mh sin M . 2 2 ± π + ε = π ε +  π ε   1 2 5. j 1 j 2 j 3 2 i(x v +y v +z v ) S 1 2 3 (v v v ) f e j − π = Σ Referred to an fcc lattice, the basis of diamond is 111 000; . 444 Thus in the product S 1 2 3 (v v v ) = S(fcc lattice)× S (basis), we take the lattice structure factor from (48), and for the basis 1 2 3 1 i (v v v ). 2 S (basis) 1 e − π + + = + Now S(fcc) = 0 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) = 1 + e–i3π = 0, and this reflection is forbidden. 2 1 3 G 0 0 3 3 0 0 3 2 3 2 2 0 0 0 2 2 2 0 6. f 4 r ( a Gr) sin Gr exp ( 2r a ) dr (4 G a ) dx x sin x exp ( 2x Ga ) (4 G a ) (4 Ga ) (1 r G a ) 16 (4 G a ) . ∞ − = π π − = − = + + ∫ ∫ 0 The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for Ga0 1. 7. (a) The basis has one atom A at the origin and one atom 1 B at a. 2 The single Laue equation 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) = f a k ⋅ ∆ = 2π× (integer) A + 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 2 and the diffraction condition 1 ( ) 2 (integer). 2 a k ⋅ ∆ = π × 2-3 CHAPTER 3 1. 2 2 2 2 E ( = π h/ / 2M) (2 λ) = (h 2M) (π L) , with λ = 2L. 2. bcc: 12 6 U(R) = ε 2N [9.114(σ R ) −12.253(σ R) ]. At equilibrium and 6 6 R 1 0 = .488σ , U(R0 ) = ε 2N ( − 2.816). fcc: 12 6 U(R) = ε 2N [12.132(σ R ) −14.454(σ R) ]. 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 R 1 0 = .679σ , U(R0 ) = ε 2N ( − 4.305). 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 q U(R) N ; 2 log 2 Madelung const. R R ⎛ ⎞ α = − ⎜ ⎟ α = = ⎝ ⎠ In equilibrium 2 n n 1 2 0 2 0 0 U nA q n N 0 ; R R R R + ∂ α ⎛ ⎞ = −⎜ ⎟ + = = ∂ α ⎝ ⎠ A , q and 2 0 0 N q 1 U(R ) (1 ). R n α = − − 3-1 b. ( ) ( ) 2 2 0 0 0 2 0 0 1 U U(R -R ) U R R R ... , 2 R ∂ δ = + δ + ∂ bearing in mind that in equilibrium R0 ( U∂ ∂R) = 0. 2 2 2 n 2 3 3 3 0 0 0 0 U n(n 1)A 2 q (n 1) q 2 N N R R R R R R 2 + ⎛ ⎞ ∂ + ⎛ ⎞ α ⎛ + α ⎜ ⎟ = − ⎜ ⎟ = ⎜ − ⎝ ⎠ ∂ ⎝ ⎠ ⎝ 2 0 αq ⎞ ⎟ ⎠ For a unit length 2NR0 = 1, whence 0 2 2 2 2 2 2 2 4 2 0 0 0 R 0 R U q U (n 1)q log 2 (n 1) ; C R R R 2R 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 3 x e 8.53 10 . − − = × 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 2 2 0 0 -αq p U U= 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 –αe 2 /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), 3S 12 = − (C C ) +1 (C +C ). Thus 2 2 Y (C11 C12C11 2C12 11 12 = + − ) (C + C ); further, also from (37), eyy = S21Xx, whence yy xx σ = e e = S S = −C (C +C ). 9. For a longitudinal phonon with K || [111], u = v = w. 3-2 2 2 1 2 11 12 44 [C 2C 2(C C )]K 3, or v K [(C 2C 4C 3ρ)] 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 1 e e xx = − yy = 2 e in (43). Then 2 2 1 1 1 1 2 4 11 4 4 12 2 1 1 2 2 11 12 U C ( e e ) C e [ (C C )]e = + − = − 2 so that 2 2 2 n 2 3 3 3 0 0 0 0 U n(n 1)A 2 q (n 1) q 2 N N R R R R R R 2 + ⎛ ⎞ ∂ + ⎛ ⎞ α ⎛ + α ⎜ ⎟ = − ⎜ ⎟ = ⎜ − ∂⎝ ⎠ ⎝ ⎠ ⎝ 2 0 αq ⎞ ⎟ ⎠ is the effective shear constant. 12a. We rewrite the element aij = p – δ ij(λ + p – q) as aij = p – λ′ δ ij, where λ′ = λ + p – q, and δ ij 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 2p q K 11 12 44 ω ρ = + = (C + 2C + 4C )/ 3, and the other two roots (shear waves) are 2 2 K (C11 C12 C44 ω ρ = − + )/ 3. 13. Set u(r,t) = u0e i(K· r – t) and similarly for v and w. Then (57a) becomes 2 2 2 2 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 2 2 2 11 11 x 44 y z 12 12 44 x y 13 12 44 x z M C K C (K K ) M (C C )K K ; M (C C )K K . = + + −ω ρ; = + = + and so on with appropriate permutations of the axes. The sum of the three roots of 2 ω ρ is equal to the sum of the diagonal elements of the matrix, which is (C11 + 2C44)K2 , where 2 2 2 2 x y z 2 2 2 1 2 3 11 44 K K K K , whence v v v (C 2C ) ρ , = + + + + = + for the sum of the (velocities)2 of the 3 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 0. The next minor is C11 C44 3 , or C11 0. Next: C443 (C112 – C122 ), whence |C12| C11. Finally, (C11 + 2C12) (C11 – C12) 2 0, so that C11 + 2C12 0 for stability. 3-4 CHAPTER 4 1a. The kinetic energy is the sum of the individual kinetic energies each of the form 2 S 1 Mu . 2 The force between atoms s and s+1 is –C(us – us+1); the potential energy associated with the stretching of this bond is 2 s 1 1 C(u u ) 2 − s+ , and we sum over all bonds to obtain the total potential energy. b. The time average of 2 2 2 S 1 1 Mu is M u . 2 4 ω In the potential energy we have us 1 u cos[ t (s 1)Ka] u{cos( t sKa) cos Ka sin ( t sKa) sin Ka}. + = ω − + = ω − ⋅ + ω − ⋅ Th s s 1 en u u u {cos( t sKa) (1 cos Ka) sin ( t sKa) sin Ka}. − = + ω − ⋅ − − ω − ⋅ We square and use the mean values over time: 2 2 1 cos sin ; cos sin 0. 2 = = = Thus the square of u{} above is 1 2 2 2 2 u [1 2cos Ka cos Ka sin Ka] u (1 cos Ka). 2 − + + = − The potential energy per bond is 1 2 Cu (1 cos Ka), 2 − and by the dispersion relation ω2 = (2C/M) (1 – cos Ka) 1 2 2 this is equal to M u . 4 ω Just as for a simple harmonic oscillator, the time average potential energy is equal to the time-average kinetic energy. 2. We expand in a Taylor series 2 2 2 2 s s u 1 u u(s p) u(s) pa p a ; x 2 x ⎛ ⎞ ∂ ∂⎛ ⎞ + = + ⎜ ⎟ + ⎜ ⎟ + ⎝ ⎠ ∂ ∂⎝ ⎠ " On substitution in the equation of motion (16a) we have 2 2 2 2 2 2 p p 0 u u M ( p a C ) t x ∂ ∂ = Σ ∂ ∂ , which is of the form of the continuum elastic wave equation with 4-1 2 1 2 2 p p 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 2 sin pk a 4. 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 0 p 0 A (1 cos 2k pa) , K M ∂ω = Σ − ∂ which in general will diverge because p Σ 1 . → ∞ 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 ) (4πr 3 /3). Thus the force is –e 2 r/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 23 M 4 10 g −  × 8 R 2 10 cm; −  × 10 46 1 2 D (5 10 ) (3 10 ) − − ω ×  × (c) The maximum phonon wavevector is of the order of 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 p0 ω = (2 / M)[γ(1− cosKa) + Σ (−1) (2e / p a )(1− cos pKa)] . At the zone boundary ω2 = 0 if P P 3 p0 1 ( 1) [1 ( 1) ]p− + σ Σ − − − = 0 , or if . The summation is 2(1 + 3 p 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 p0 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 K (2 m /a)sin ( / ) − = ω ω 2 2 1/ 2 dK/d m (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 4 K 0 Ω = π / 3 = (4π / 3)[(ω −ω)/ A] 3/2 1/ 2 0 is , provided ω ω 3 3 3/2 D 0 (ω π )= (L/2 ) | dΩ/dω|= (L/2π) (2π / A )(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 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 3 a ∆ ≈ /a ∆V/V , whence . 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 1 3 2 d D D ( ) 3V / 4 v − = ∫ ω ω ω = ω π 3 2 3 v 2 2 R 3h D = /ω / 8π ρ . (b) In one dimension from (15) we have D(ω =) L/πv , whence 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) hω / 4MNv . 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 2 N ( = π A/4 )(πK ) = Aω / 4πv . 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 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 where ωD is defined by dω. In the regime D D N D( ) ω = ω ∫ =ωD τ , we have 3 2 2 2 x 0 2A x U dx. 2 v e 1 τ ∞ ≅ π − ∫ = 5-1 Thus the heat capacity . 2 C k = ∂ B U/∂τ ∝ T (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 . But this only holds at extremely low temperatures such that ∝ T D vNlayer τ = = ω ≈ / L, where Nlayer