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INSTRUCTOR’S MANUAL TO ACCOMPANY

CHARLES KITTEL

INTRODUCTION TO SOLID

STATE PHYSICS

EIGHTH EDITION

JOHN WILEY & SONS, INC.

Copyright © 2005 by John Wiley & Sons, Inc.

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TABLE OF CONTENTS

• Chapter 1 1-
• Chapter 2 2-
• Chapter 3 3-
• Chapter 4 4-
• Chapter 5 5-
• Chapter 6 6-
• Chapter 7 7-
• Chapter 8 8-
• Chapter 9 9-
• Chapter 10 10-
• Chapter 11 11-
• Chapter 12 12-
• Chapter 13 13-
• Chapter 14 14-
• Chapter 15 15-
• Chapter 16 16-
• Chapter 17 17-
• Chapter 18 18-
• Chapter 20 20-
• Chapter 21 21-
• Chapter 22 22-

CHAPTER 1

1. The vectors x ˆ + y ˆ + z ˆ and − x ˆ − y ˆ + z ˆ 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 cos 30 (^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 a^ c a ,

3 2

⎝ ⎠ ⎝^ ⎠

or

2 2 1 2 c 8 a c ; 1.633. 3 4 a 3

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 cos M(a k) sin^ M(a^ k) . 1 cos(a k) (^) sin (a k)

(b) The first zero in

sin M 2

ε occurs for ε = 2π/M. That this is the correct consideration follows from

zero,^1 as Mh is an integer

sin M( h ) sin Mh cos M cos Mh sin M. 2 2 ±

π + ε = π ε + π ε  

2 i(x v +y v +z v )j 1 j 2 j 3 S (v v v ) 1 2 3 f e j

− π = Σ

Referred to an fcc lattice, the basis of diamond is

Thus in the product

S(v v v ) 1 2 3 = 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 v 1 + v 2 + v 3 = 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 3 1 G 0 0

3 3 0 0

3 3 2 22 0 0 0

2 22 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/G

4 for

Ga 0 >>1.

7. (a) The basis has one atom A at the origin and one atom

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 = 2 π×(integer)

A + fB e

–iπn

. For n odd, S = fA –

fB; for n even, S = fA + fB. (c) If fA = fB the atoms diffract identically, as if the primitive translation vector

were

a 2

and the diffraction condition

( ) 2 (integer). 2

a ⋅ ∆ k = π ×

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 R 0

( 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 2NR 0 = 1, whence

0

2 2 2 2 2 2 4 0 2 2 (^0) R 0 0

R

U q U (n 1) q log 2 (n 1) ; C R R (^) 2R R R

⎛ ∂ ⎞ α ∂ −

⎜ ⎟ =^ −^ =^ = ⎝ ∂^ ⎠ ∂

.

6. For KCl, λ = 0.34 × 10

   ergs and ρ = 0.326 × 10 - Å. For the imagined modification of KCl with the 

ZnS structure, z = 4 and α = 1.638. Then from Eq. (23) with x ≡ R 0 /ρ we have

2 x 3 x e 8.53 10.

− − = ×

By trial and error we find x  9.2, or R 0 = 3.00 Å. The actual KCl structure has R 0 (exp) = 3.15 Å. For

the imagined structure the cohesive energy is

2

2 0 0

• αq p U U= 1- , or =-0. R R q

in units with R 0 in Å. For the actual KCl structure, using the data of Table 7, we calculate 2

U

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 /R 0 per ion pair, or –14.61 × 10

• 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 R 0 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 = S 11 XX, because all other stress components are zero. By (51),

3S 11 = 2 (C 11 − C 12 ) + 1 (C 11 +C 12 ).

Thus

2 2 Y = (C 11 + C C 12 11 − 2C 12 ) (C 11 +C 12 );

further, also from (37), eyy = S 21 Xx,

whence (^) yy 21 11 12 11 12 xx

σ = e e = S S = − C (C + C ).

9. For a longitudinal phonon with K || [111], u = v = w.

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 exx = − eyy= 12 e in (43). Then

1 1 2 1 2 1 2 11 4 4 4 12

1 1 2 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 ω ρ = 2p + q = K (C 11 + 2C 12 +4C 44 ) / 3,

and the other two roots (shear waves) are

2 2 ω ρ = K (C 11 − C 12 +C 44 ) / 3.

13. Set u(r,t) = u 0 e

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

C HAPTER 4

1a. The kinetic energy is the sum of the individual kinetic energies each of the form

2 S

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

C(u u ) 2

− (^) s + , and we sum over all bonds to obtain the total potential energy.

b. The time average of

(^2 2 ) S

Mu is M u. 2 4

ω In the potential energy we have

u s 1 u cos[ t (s 1)Ka] u{cos( t sKa) cos Ka

sin ( t sKa) sin Ka}.

• =^ ω −^ +^ =^ ω −^ ⋅

• ω − ⋅

Then u s u (^) s 1 u {cos( t sKa) (1 cos Ka)

sin ( t sKa) sin Ka}.

− (^) + = ω − ⋅ −

− ω − ⋅

We square and use the mean values over time:

cos sin ; cos sin 0. 2

Thus the square of u{} above is

u [1 2cos Ka cos Ka sin Ka] u (1 cos Ka). 2

The potential energy per bond is

Cu (1 cos Ka), 2

− and by the dispersion relation ω

2 = (2C/M) (1 –

cos Ka)

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 p 2 p 0

u u M ( p a C ) t > x

which is of the form of the continuum elastic wave equation with

2 1 2 2 p p 0

v M p a C

−

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 ω

2 = 2C/M 2 the

motion is in the lattice described by the displacement v; at ω

2 = 2C/M 1 the u lattice moves.

(^2 )

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

(cos (k K) pa cos (k K) pa) 2

ω = Σ −

∂ω = Σ ∂

When K = k 0 ,

2

0 p 0

A

(1 cos 2k pa) , K M >

∂ω = Σ − ∂

which in general will diverge because p

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 – e

2 n(r)/r

2 , where the

number of electrons within a sphere of radius r is (3/4 πR

3 ) (4πr

3 /3). Thus the force is –e

2 r/R

2 , and the

C HAPTER 5

1. (a) The dispersion relation is (^) m

| sin Ka|. 2

ω = ω We solve this for K to obtain

, whence and, from (15),

1 K (2/a) sin ( / (^) m)

− = ω ω

(^2 2) 1/ 2 dK/d (2 / a)( (^) m )

− ω = ω − ω D( ω)

. This is singular at ω = ω

(^2 2) 1/ 2 (2L/ a)( (^) m )

− = 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 Ω = 4 K / 3π = (4 π / 3)[( ω − ω 0 ) / A]

3/

1/ 2

0 is

, provided ω < ω

3 3 3/ D( ω)= (L/2 ) | d π Ω/d ω =| (L/2 ) (2π π / A )( ω − ω 0 ) 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

B( V/V) a k T 2 2

∆ ≈. This is (^) B

k T 2

and not

B

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 ( V) 1.5 10 ;( V) (^) rms 4.7 10 cm ;

− − < ∆ > = × ∆ = ×

3 ( ∆V) (^) rms / V = 0.125. Now

3 a/a∆ ≈ ∆V/V , whence ( ∆a) (^) rms / a =0.04.

3. (a) , where from (20) for a Debye spectrum

2 R (h/2 V)

− < > = / ρ Σω

1 − 1 Σω

, whence

1 2 3 d D( ) 3V (^) D / 4 v

− = ∫ ω ω ω = ω π

3 2 3 v

2 2 < R > = 3h/ ωD / 8π ρ. (b) In one dimension from

(15) we have D( ω =) L/ vπ , whence

1 d D( )

− ∫ ω ω ω diverges at the lower limit. The mean square

strain in one dimension is

2 2 2 0

( R/ x) K u (h/2MNv) K 2

(^2 2 ) = (h/2MNv) (K/ (^) D / 2) = h/ ωD / 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/4π

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 π ) ( K )π = A ω / 4 v .π

2

The density of modes of each polarization type is D(ω) = dN/dω = Aω/2πv

2

. The thermal average phonon

energy for the two polarization types is, for each layer,

D D

0 0 2

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

∫ D

=ω >> τ, we have

3 2

2 2 0 x

2A x U dx. 2 v e 1

τ^ ∞ ≅ π −

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

τ << =ω (^) D ≈=vN (^) layer/ L, where Nlayer/L is the number of

layers per unit length.

5. (a) From the Planck distribution

(^1 1) x x 1 n (e 1) /(e 1) coth (x/2) 2 2 2

< > + = + − = , where

x = h/ ω/k T (^) B. The partition function

x/2 sx x/2 x 1 Z e e e /(1 e ) [2sinh (x/2)]

− − − − = Σ = − =

− and the

free energy is F = kBT log Z = kBT log[2 sinh(x/2)]. (b) With ω(∆) = ω(0) (1 – γ∆), the condition

∂F/ ∂∆ = 0 becomes B

B h coth (h /2k T) 2

∆ = γΣ / ω /ω on direct differentiation. The energy

< n > h/ω is just the term to the right of the summation symbol, so that B∆ = γU (T ). (c) By definition

of γ, we have δω ω = −γδ/ V /V , or d log ω = −δ d log V. But , whence

.

θ ∝ ω D

d log θ = −γd log V

(2 10 ) 3 10 cm

π × ≈ × 3

the electron concentration is

57 28 3 28

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

ε = π ≈ ⋅ ≈ ≈ (b) The value of kF is not

affected by relativity and is ≈ n

1/ , where n is the electron concentration. Thus

3 εF  hck/ (^) F hc/ √n. (c) A

change of radius to 10 km = 10

6 cm makes the volume ≈ 4 × 10

18 cm

3 and the concentration ≈ 3 × 10

38 cm

3

. Thus (The energy is relativistic.)

27 10 13 4 8 F 10 (3.10^ ) (10^ )^ 2.10^ erg^10 eV.

− − ε ≈ ≈ ≈

5. The number of moles per cm

3 is 81 × 10

• /3 = 27 × 10 - , so that the concentration is 16 × 10

21 atoms cm

3

. The mass of an atom of He

3 is (3.017) (1.661) × 10

• = 5.01 × 10 - g. Thus 54 23 21 2 3 16 F [(1.1^10 ) 10^ ][(30)(16)^10 ]^7

− − ε  × × ≈ ×

− erg, or TF ≈ 5K.

6. Let E, v vary as e

–iwt

. Then

eE m e E 1 i v , i m

2

τ + ωτ = − = − ⋅ − ω + (1 τ) 1+ (ωτ)

and the electric current density is

2 ne 1 i j n( e)v E. m

2

τ + ωτ = − = ⋅ 1 + (ωτ)

7. (a) From the drift velocity equation

i ωv (^) x = (e m)E (^) x + ωc v (^) y ; i ωv (^) y = (e m)E (^) y− ωcv. 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 ωc

2

. Because j = n(–e)v = σE, the components of σ come out directly. (b) From the

electromagnetic wave equation

2 2 2 c E E t

2 ∇ = ε∂ ∂ ,

we have, for solutions of the form e

i(kz – ωt) , the determinantal equation

2 2 xx xy 2 2 yx yy

c k

c k

2 2

2 2

ε ω − ε ω

ε ω ε ω −

Here

2 xx yy 1 P and^ xy yx i^ c p.

2 ε = ε = − ω ω ε = −ε = ω ω ω

(^2 ) The determinantal equation gives the

dispersion relation.

8. The energy of interaction with the ion is

r (^0 2 )

0 0

e ρ r 4 r drπ = −3e 2

r ,

where the electron charge density is –e(3/4πr 0

3 ). (b) The electron self-energy is

2 r^03 2 1 0 0

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 N V = 3 4 rπ 0 , the average

is ( )

2 3 (^22) 3 9 π 4 h/ 10mr 0. The sum of the Coulomb and kinetic contributions is

2 s (^) s

U

r (^) r

which is a minimum at

2 3 s s s

, 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 2 0 x

c

j E E

1

ω τ = σ = σ

• ω τ

For ωcτ >> 1, we have ( )( )

2 σyx ≅ σ 0 ω τ =c ne τ m mc eB τ = neB c.

10. For a monatomic metal sheet one atom in thickness, n ≈ 1/d

3 , so that

2 2 2 R (^) sq ≈ mv (^) F nd e ≈ mv d eF.

If the electron wavelength is d, then mv dF ≈ h/ by the de Broglie relation and

2 R (^) sq≈ h e/ =137 c

in Gaussian units. Now