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

CHAPTER 1
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/ 3 = 90° + 19° 28' = 109° 28' .

2. 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.
3. The central dot of the four is at distance

a

cos 60°
a
= a ctn 60° =
cos 30°
3

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⎠
2

or

2 2 1 2
a = c ;
3
4

c 8
= 1.633.
a 3

1-1


CHAPTER 2

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 2 + 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 ) ,
1. The crystal plane with Miller indices

whence

1
G 2 h 2 + k 2 + A2
=
=
.
d 2 4π 2
a2

1
3a
2
1
2. (a) Cell volume a1 ⋅ a 2 × a3 = −
3a
2
0

=

1
a 0
2
1
a 0
2
0 c

1
3 a 2 c.
2

xˆ
(b) b1 = 2π

=

yˆ

4π
1
a 2 × a3
=
−
3a
2
| a1 ⋅ a 2 × a3 |
3a c 2
0

zˆ

1
a 0
2
0 c

2π 1
(
xˆ + yˆ ), and similarly for b 2 , b3 .
a
3
(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

VBZ = (2π)3

(a 2 × a 3 ) ⋅ (a 3 × a1 ) × (a1 × a 2 )
= (2π)3 / | (a1 ⋅ a 2 × a 3 ) |
3
| (a1 ⋅ a 2 × a 3 ) |

= (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.
4. (a) This follows by forming

2-1


|F|2 =

1 − exp[−iM(a ⋅ ∆k)] 1 − exp[iM(a ⋅ ∆k)]
⋅
1 − exp[−i(a ⋅ ∆k)] 1 − exp[i(a ⋅ ∆k)]

2
1 − cos M(a ⋅ ∆k) sin 12 M(a ⋅ ∆k)
.
=
=
1 − cos(a ⋅ ∆k)
sin 2 12 (a ⋅ ∆k)

(b) The first zero in sin

1

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

1
1
1
sin M(πh + ε) = sin πMh cos
Mε + cos
sin
Mε.
πMh










2
2
2
±
1
zero,
as Mh is
an integer

5. S (v1 v 2 v 3 ) = f Σ e

−2πi(x j v1 +y j v 2 +z j v3 )

j

Referred to an fcc lattice, the basis of diamond is 000;

1 1 1
. Thus in the product
4 4 4

S(v1v 2 v3 ) = S(fcc lattice) × S (basis) ,
we take the lattice structure factor from (48), and for the basis

S (basis) = 1 + e

−i

1
π (v1 + v 2 + v3 ).
2

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.

6.

∞

f G = ∫ 4πr 2 (πa 0 Gr)−1 sin Gr exp ( −2r a 0 ) dr
3

0

= (4 G 3a 0 ) ∫ dx x sin x exp ( −2x Ga 0 )
3

= (4 G 3a 0 ) (4 Ga 0 ) (1 + r G 2 a 0 ) 2
3

2

16 (4 + G 2 a 0 ) 2 .
2

The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for

Ga 0 >> 1.

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

1
a. The single Laue equation
2

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
1
a and the diffraction condition ( a ⋅ ∆k ) = 2π × (integer).
2
2

2-3


CHAPTER 3
1. E = (h/

2

2M) (2π λ ) 2 = (h/ 2 2M) (π L) 2 , with λ = 2L.
6

U(R) = 2Nε[9.114( σ R )12 − 12.253(σ R)6 ]. At equilibrium R 0 = 1.488σ6 , and
U(R 0 ) = 2Nε( − 2.816).

2. bcc:

6

U(R) = 2Nε[12.132( σ R )12 − 14.454(σ R)6 ]. At equilibrium R 0 = 1.679σ6 , and
U(R 0 ) = 2Nε( − 4.305). Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is

fcc:

more stable than the bcc.

3.

| 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.
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

αe2 (1.75) (4.80 × 10−10 ) 2
=
= 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 2 ⎞

U(R) = N ⎜ n −
⎟ ; α = 2 log 2 = Madelung const.
R ⎠
⎝R
In equilibrium

⎛ nA αq 2 ⎞
∂U
nA
n
= N ⎜ − n +1 + 2 ⎟ = 0 ; R 0 =
,
∂R
αq 2
R0 ⎠
⎝ R0
and

U(R 0 ) = −

Nαq 2
1
(1 − ).
R0
n

3-1


b.

U(R 0 -R 0 δ) = U ( R 0 ) +

1 ∂2U
2
R 0 ( R 0δ ) + . . . ,
2
2 ∂R

bearing in mind that in equilibrium (∂U ∂R) R

0

= 0.

⎛ n(n + 1)A 2αq 2 ⎞
⎛ (n + 1) αq 2 2αq 2 ⎞
⎛ ∂2U ⎞
−
= N⎜
−
⎜ 2 ⎟ = N⎜
n+2
3 ⎟
3
3 ⎟
R0 ⎠
R0
R0 ⎠
⎝ ∂R ⎠R 0

⎝ R0
⎝
For a unit length 2NR0 = 1, whence
2
⎛ ∂2U ⎞
αq 2
(n − 1) q 2 log 2
2 ∂ U
.
−
=
=
(n
1)
;
C
R
⎜ 2⎟ =
0
4
2
∂R 2 R
2R 0
R
⎝ ∂R ⎠R
0
0
0

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

x 2 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 2 ⎛ p ⎞
U
⎜1⎟ , or 2 =-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.
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),

3S11 = 2 (C11 − C12 ) + 1 (C11 + C12 ).
2

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 ).

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

3-2


ω2ρ = [C11 + 2C44 + 2(C12 + C 44 )]K 2 3,
or v = ω K = [(C11 + 2C12 + 4C44 3ρ )]1 2
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 e xx = −e yy =

1

2

e in (43). Then

U = 1 2 C11 ( 1 4 e2 + 1 4 e2 ) − 1 4 C12 e2
= 1 2 [ 1 2 (C11 − C12 )]e 2
⎛ n(n + 1)A 2αq 2 ⎞

⎛ (n + 1) αq 2 2αq 2 ⎞
⎛ ∂2U ⎞
= N⎜
−
= N⎜
−
is the effective shear
so that ⎜
n+2
3 ⎟
3
3 ⎟
2 ⎟
R0 ⎠
R0
R0 ⎠
⎝ ∂R ⎠R 0
⎝ 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 – 1 roots λ′ = 0 give λ = q – p.
b. Set

u (r, t) = u 0 ei[(K

3) (x + y + z) −ωt]

;

v(r, t) = v0 ei[. . . . .] ;
w(r, t) = w 0 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 2 (C11 + 2C12 + 4C 44 ) / 3,
and the other two roots (shear waves) are

ω2ρ = K 2 (C11 − C12 + C44 ) / 3.
13. Set u(r,t) = u0ei(K·r – t) and similarly for v and w. Then (57a) becomes
2

2

2

ω2ρu 0 = [C11K y + C44 (K y + K z )]u 0
+ (C12 + C44 ) (K x K y v 0 + K x K z w 0 )
and similarly for (57b), (57c). The elements of the determinantal equation are

3-3


M11 = C11K x + C 44 (K y + K z ) − ω 2 ρ;
2

2

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

2

2

K 2 = K x + K y + K z , whence
v1 + v 2 + v3 = (C11 + 2C44 ) ρ ,
2

2

2

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

1
2
Mu S . The force

2

between atoms s and s+1 is –C(us – us+1); the potential energy associated with the stretching of this bond is

1
C(u s − u s+1 ) 2 , and we sum over all bonds to obtain the total potential energy.
2
b. The time average of

1
1
2
Mu S is Mω2 u 2 . In the potential energy we have
2
4

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 2 > = < sin 2 > =

1
; < cos sin > = 0.
2

Thus the square of u{} above is

1 2
u [1 − 2cos Ka + cos 2 Ka + sin 2 Ka] = u 2 (1 − cos Ka).
2
The potential energy per bond is
cos Ka) this is equal to

1 2
Cu (1 − cos Ka), and by the dispersion relation ω2 = (2C/M) (1 –
2

1
Mω2 u 2 . Just as for a simple harmonic oscillator, the time average potential
4

energy is equal to the time-average kinetic energy.
2. We expand in a Taylor series

⎛ ∂2u ⎞
⎛ ∂u ⎞ 1
u(s + p) = u(s) + pa ⎜ ⎟ + p 2 a 2 ⎜ 2 ⎟ + " ;
⎝ ∂x ⎠s 2
⎝ ∂x ⎠s
On substitution in the equation of motion (16a) we have

M

∂2u

∂2u
2 2
=
(
Σ
p
a
C
)
,
p
p>0
∂t 2
∂x 2

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

4-1


v 2 = M −1 Σ

p >0

p 2a 2Cp .

3. From Eq. (20) evaluated at K = π/a, the zone boundary, we have

−ω2 M1u = −2Cu ;
−ω2 M 2 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.

4.

ω2 =

sin pk 0 a
2
A Σ
(1 − cos pKa) ;
p >0
M
pa

∂ω2 2A
=
Σ sin pk 0a sin pKa
∂K
M p >0
1
(cos (k 0 − K) pa − cos (k 0 + K) pa)
2
When K = k0,

∂ω2 A
=
Σ (1 − cos 2k 0 pa) ,
∂K M p>0
which in general will diverge because Σ 1 → ∞.

p
5. By analogy with Eq. (18),

Md 2 u s dt 2 = C1 (vs − u s ) + C2 (vs −1 − u s );
Md 2 vs dt 2 = 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 = 0 and 2(C1 + C2 ) M.
For Ka = π, ω2 = 2C1 M and 2C 2 M.

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

M  4 ×10−23 g and R  2 × 10−8 cm; thus ωD  (5 × 10−10 ) (3 × 10−46 )1 2

 3 × 1013 s −1 (c) The maximum phonon wavevector is of the order of 108 cm–1. If we suppose that ω0 is

sodium

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 ω = (2 / M)[γ (1 − cos Ka) + Σ (−1) (2e / p a )(1 − cos pKa)] .
2

P

2

3 3

p>0

At the zone boundary ω2 = 0 if

1 + σ Σ (−1) P [1 − (−1) P ]p −3 = 0 ,
p>0

or if σ Σ[1 − ( −1) ]p

−3

= 1 . The summation is 2(1 + 3–3 + 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 << 1 is
p −3 2

given by the sign of 1 = 2σ Σ ( −1) p p , which is zero when 1 – 2–1 + 3–1 – 4–1 + … = 1/2σ. The series
p

p>0

is just that for log 2, whence the root is σ = 1/(2 log 2) = 0.7213.

4-3


CHAPTER 5
1.

1
ω = ωm | sin Ka|. We solve this for K to obtain
2
2
−1
K = (2/a) sin (ω / ωm ) , whence dK/dω = (2 / a)(ωm − ω2 ) −1/ 2 and, from (15), D(ω)

(a) The dispersion relation is

= (2L/πa)(ωm − ω2 ) −1/ 2 . This is singular at ω = ωm. (b) The volume of a sphere of radius K in
3
3/2
Fourier space is Ω = 4πK / 3 = (4π / 3)[(ω0 − ω) / A] , and the density of orbitals near ω0 is
2

D(ω)= (L/2π)3 | dΩ/dω |= (L/2π)3 (2π / A 3/2 )(ω0 − ω)1/ 2 , provided ω < ω0. It is apparent that
D(ω) vanishes for ω above the minimum ω0.

2. The potential energy associated with the dilation is

1
1
1
B(∆V/V) 2 a 3 ≈ k BT . This is k BT and not
2
2
2

3
k BT , because the other degrees of freedom are to be associated with shear distortions of the lattice cell.
2
2
−47
−24
3
Thus < ( ∆V) > = 1.5 × 10 ;(∆V) rms = 4.7 × 10 cm ; and ( ∆V) rms / V = 0.125 . Now
3∆a/a ≈ ∆V/V , whence (∆a) rms / a = 0.04 .
3.

(a)

/ ρV) Σω−1 ,
< R 2 > = (h/2

where

from

(20)

for

a

Debye

spectrum

Σω−1

= ∫ dω D(ω)ω−1 = 3VωD / 4π3 v3 , whence < R 2 > = 3h/ ωD / 8π2 ρv3 . (b) In one dimension from
−1
(15) we have D(ω) = L/πv , whence ∫ dω D(ω) ω diverges at the lower limit. The mean square
1
2
/
strain
in
one
dimension
is
< (∂R/∂x) 2 > = ΣK 2 u 0 = (h/2MNv)
ΣK
2
2
2
/
= (h/2MNv)

(K D / 2) = h/ ωD / 4MNv3 .
2

2

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,

N = (A/4π2 ) (πK 2 ) = Aω2 / 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,

U = 2∫

ωD

0

where ωD is defined by N =

ωD Aω
=ω
dω ,
2
0
2πv exp(hω/τ) − 1

D(ω) n(ω,τ) =ω dω = 2∫

∫

ωD

D

D(ω) dω . In the regime =ωD >> τ , we have

U≅

2Aτ3
2πv 2 = 2

∫

∞

0

5-1

x2
dx.
ex − 1


Thus the heat capacity C = k B ∂U/∂τ ∝ T .
2

(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 ∝ T . But this only
holds at extremely low temperatures such that τ << =ωD ≈ =vN layer / L , where Nlayer/L is the number of
layers per unit length.

1 1 x
1
= (e + 1) /(e x − 1) = coth (x/2) , where
2 2
2
− x/2
− sx
− x/2
−x
x = h/ ω/k BT . The partition function Z = e Σ e = e /(1 − e ) = [2sinh (x/2)]−1 and the

5. (a) From the Planck distribution

<n>+

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

1
h/ ω coth (h/ ω/2k BT) on direct differentiation. The energy
2
< 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 = −γδV/V , or d log ω = −δ d log V . But θ ∝ ωD , whence
d log θ = −γ d log V .
∂F/∂∆ = 0 becomes B∆ = γΣ

5-2


CHAPTER 6

h/ 2 2
1. The energy eigenvalues are ε k =
k . The mean value over the volume of a sphere in k space is
2m

<ε> =

h/ 2 ∫ k 2 dk ⋅ k 2 3 h/ 2 2 3
= ⋅
k F = εF .
2m ∫ k 2 dk
5 2m
5

The total energy of N electrons is

3
U0 = N ⋅ εF .
5
2a. In general p = –∂U/∂V at constant entropy. At absolute zero all processes are at constant entropy (the
23

3 h/ 2 ⎛ 3π2 N ⎞
= N
⎜

⎟ , whence
5 2m ⎝ V ⎠

3
Third Law), so that p = −dU 0 dV, where U 0 = Nε F
5
p=

2 U0
⋅
. (b) Bulk modulus
3 V
2

dp
2 dU 0 ⎞ 2 U 0 ⎛ 2 ⎞ U 0 10 U 0
⎛ 2 U0
B = −V
.
= V⎜−
+
+⎜ ⎟
=
⎟= ⋅
2
dV
9 V
⎝ 3 V 3V dV ⎠ 3 V ⎝ 3 ⎠ V
(c) For Li,

U0 3
= (4.7 × 1022 cm −3 ) (4.7 eV) (1.6 × 10−12 erg eV)
V 5
= 2.1 × 1011 erg cm −3 = 2.1 ×1011 dyne cm −2 ,
whence B = 2.3 × 1011 dyne cm–2. By experiment (Table 3.3), B = 1.2 × 1011 dyne cm–2.
3. The number of electrons is, per unit volume, n =

∫

∞

0

dε D(ε) ⋅

1
e

(ε−µ ) τ

+1

, where D(ε) is the density

of orbitals. In two dimensions

m ∞
1
dε (ε−µ) τ
2 ∫0

πh/
+1
e
m
= 2 (µ + τ log (1 + e −µ τ )),
πh/

n=

where the definite integral is evaluated with the help of Dwight [569.1].

2 × 1033
 1057 nucleons, and roughly an equal number of electrons. In a
4a. In the sun there are
−24
1.7 × 10
white dwarf star of volume

6-1


4π
(2 × 109 )3 ≈ 3 × 1028 cm3
3
the electron concentration is ≈

1057
≈ 3 × 1028 cm −3 . Thus
28
3 × 10

h/ 2
1
1
(3π2 n)2 3 ≈ 10−27 ⋅1020 ≈ 10−7 ergs, or ≈ 3.104 eV. (b) The value of kF is not
2m
2
2
1/3
/ F  hc
/ 3 √n. (c) A
affected by relativity and is ≈ n , where n is the electron concentration. Thus ε F  hck
εF =

change of radius to 10 km = 106 cm makes the volume ≈ 4 × 1018 cm3 and the concentration ≈ 3 × 1038 cm –
. Thus ε F ≈ 10

3

−27

(3.1010 ) (1013 ) ≈ 2.10 −4 erg ≈ 108 eV. (The energy is relativistic.)

5. The number of moles per cm3 is 81 × 10–3/3 = 27 × 10–3, so that the concentration is 16 × 1021 atoms cm–
. The mass of an atom of He3 is (3.017) (1.661) × 10–24 = 5.01 × 10–24 g. Thus

3

ε F  [(1.1 × 10−54 ) 10−23 ][(30)(16) × 1021 ]2 3 ≈ 7 × 10−16 erg, or TF ≈ 5K.
6. Let E, v vary as e–iwt. Then

v=−

eE m
eτE 1 + iωτ
=−
⋅
,
−iω + (1 τ)
m 1+ (ωτ)2

and the electric current density is

j = n( − e)v =

ne 2 τ 1 + iωτ
⋅
E.
m 1+ (ωτ)2

7. (a) From the drift velocity equation

iωv x = (e m)E x + ωc v y ; iωv y = (e m)E y − ωc v x .
We solve for vx, vy to find

(ωc − ω2 )v x = iω( e m )E x + ωc ( e m )E y ;
2

(ωc − ω2 )v y = iω( e m )E y + ωc ( e m )E x .
2

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

c 2∇ 2 E = ε∂ 2 E ∂t 2 ,
we have, for solutions of the form ei(kz – ωt), the determinantal equation

ε xx ω2 − c 2 k 2 ε xy ω2
= 0.
ε yx ω2
ε yy ω2 − c 2 k 2

6-2


Here ε xx = ε yy = 1 − ωP

ω2 and ε xy = −ε yx = i ωcωp ω3 . The determinantal equation gives the

2

2

dispersion relation.
8. The energy of interaction with the ion is

e∫

r0

0

( ρ r ) 4πr 2dr = −3e2

2r0 ,

where the electron charge density is –e(3/4πr03). (b) The electron self-energy is

ρ2 ∫

r0

0

(

)(

)

dr 4πr 3 3 4πr 2 r −1 = 3e 2 5r0 .

The average Fermi energy per electron is 3εF/5, from Problem 6.1; because N V = 3 4πr0 , the average
3

is 3 ( 9π 4 )

23

h/ 2 10mr0 . The sum of the Coulomb and kinetic contributions is

2

U=−

1.80 2.21
+ 2
rs
rs

which is a minimum at

1.80 4.42
= 3 , or rs = 4.42 1.80 = 2.45.
2
rs
rs
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

jy = σ yx E x =

(

For ωcτ >> 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 is d, then mv F d ≈ h/ by the de Broglie 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 of a face. c. Unless the band gap at the midpoint of a face is
larger than the kinetic energy difference between this 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: from
2

2

Chap. 2, K = (2 π / a) (1,1,1) u, with 0 < u <

1
, will lie in the reduced zone.
2

The

lattice

G´s

of

the

reciprocal

G = ( 2 π a ) [( h − k + A ) xˆ + ( h + k − A ) yˆ +

h, k, A

are

any

integers.

are

given

( −h + k + A ) zˆ ],

(

Then ε = h/

( 2π a ) [( u + h − k + A ) + ( u + h + k − A ) + ( u − h + k + A )
2

2

2

2

by
where

2

2m

)

]. We now

have to consider all combinations of indices h, k, A for which the term in brackets is smaller than
6[3(1/2)2] or 9/2. These indices are (000);

(

) (

)

(

)

( 1 1 1 ) ; ( 100 ) , ( 0 10 ) ,

and

( 00 1 ) ;

(100), (010), and

(001); (111); 1 10 , 10 1 , and 0 1 1 ; (110), (101), and (011).
3. (a) At k = 0 the determinantal equation is (P/Ka) sin Ka + cos Ka = 1. In the limit of small positive P 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. The energy gap is
2 ⎠
⎝
E g = h/ 2 2ma 2 ( 2πδ )  h/ 2 2ma 2 ( 4P ) .
P

(

)

(

)

4. (a) There are two atoms in the basis, and we label them a and b. Then the crystal potential may be

⎛
⎝

1
1
1 ⎞
a, y + a, z + a ⎟ and the Fourier transform has

4
4
4 ⎠
1
i( G x + G y + G z ) a ⎞
⎛
4
ˆ then the exponential is
= U1G ⎜ 1 + e
⎟ . If G = 2Ax,



⎝
⎠

written as U = U1 + U 2 = U1 ( r ) + U1 ⎜ x +
components U G = U1G + U 2G


1
i Aa

e2





= eiπ = −1, and U G = 2A = 0, so that this Fourier component vanishes. Note that the quantity in


parentheses above is just the structure factor of the basis. (b) This follows directly from (44) with U set
equal to zero. In a higher order approximation we would go back to Eq. (31) where any non-vanishing
U G enters.


7-1


5. Let k = K + iH ; λ ±1 =

2
⎤
h/ 2 ⎡⎛ 1 ⎞
2
G
⎢⎜
⎟ ± iGH − H ⎥ .
2m ⎣⎢⎝ 2 ⎠
⎦⎥

The secular equation (46) is now

λ1 − ε

U

U

λ −1 − ε

= 0,

h/ 2 ⎛ 1 ⎞
σ =ε−
⎜ G⎟ ,
2m ⎝ 2 ⎠
2

and for H << G we have, with

⎛
h/ 2 ⎞ ⎛
h/ 2 ⎞
2
iG
iG
σ
+
H
⋅
σ
−
H

⎜
⎟⎜
⎟ = U1 ;
2m ⎠ ⎝
2m ⎠
⎝
2

⎛ h/ 2
⎞
2
GH ⎟ = U1 ;
σ −⎜
2m
⎝
⎠
2

h/ 2 2 U1 − σ
..
.
H = 2
h/
2m
2
G
2m
2

.

6. U(x,y) = – U[ei (2π/a) (x+y) + other sign combinations of ± x ± y]. The potential energy contains the four
reciprocal lattice vectors (2 π/a) (±1; ±1). At the zone corner the wave function ei(π/a) (x+y) is mixed with e–i
(π/a) (x+y)
. The central equations are

π π⎤
π⎤
⎡ π
; ⎥ − UC ⎢ − ; − ⎥ = 0 ;
a⎦
⎣a a⎦
⎣ a
π
π
π π
( λ − ε ) C ⎡⎢− ; − ⎤⎥ − UC ⎡⎢ ; ⎤⎥ = 0 ,
a⎦
⎣ a
⎣a a ⎦

( λ − ε ) C ⎡⎢

(

where λ = 2 h/

2

)

2m ( π a ) . The gap is 2U.
2

7-2


CHAPTER 8

1a.

E d = 13.60 eV ×

b.

r = aH × ε ×

m* 1
×  6.3 × 10−4 eV
m ε2

m
 6 × 10−6 cm
m*

c. Overlap will be significant at a concentration

N=

1

≈ 1015 atoms cm −3
3
r

4π
3

2a. From Eq. (53), n  (n 0 N d )

1/ 2

e − Ed / 2k BT , in an approximation not too good for the present example.

⎛ m*k B T ⎞
n0 ≡ 2 ⎜
2 ⎟
⎝ 2πh/ ⎠

3/ 2

≈ 4 ×1013 cm −3 ;

Ed
 1.45 ; e −1.45  0.23 .
2k BT
n  0.46 ×1013 electrons cm −3 .
1
b.
RH = −
 −1.3 × 10−14 CGS units

nec
3. The electron contribution to the transverse current is

⎛µ B
⎞
jy (e)  neµ e ⎜ e E x + E y ⎟ ;
⎝ c
⎠
⎛ −µ n B
⎞
Ex + Ey ⎟ .
⎝ c
⎠

for the holes jy (h)  neµ h ⎜
Here we have used

ωce τe =

µe B
for electrons;
c

ωch τh =

µh B
for holes.
c

The total transverse (y-direction) current is

0 = (neµ e − peµ h )(B/c)E x + (neµ e + peµ h )E y ,
2

2

(*)

and to the same order the total current in the x-direction is

jx = (peµ h + neµ e )E x .
Because (*) gives

8-1


pµ h − nµe 1
⋅ ,
pµ h + nµe c
2

Ey = Ex B

2

we have for the Hall constant

RH =

1 pµ h − nµ e

⋅
.
ec (pµ h + nµe ) 2
2

Ey
jx B

=

2

/ x / m t ; v y = hk
/ y / m t ; v z = hk
/ z /m A . The equation of motion
4. The velocity components are v x = hk

h/ dk/dt = − (e/c) v × B . Let B lie parallel to the kx axis; then
dk x / dt = 0; dk y / dt = −ωA k z ; ωA ≡ eB/mA c; dk z / dt = ωt k y ; ωt ≡ eB/m t c . We differentiate

in

k

space

is

with respect to time to obtain d k y / dt = −ωA dk z / dt ; on substitution for dkz/dt we have
2

2

d 2 k y / dt 2 + ωA ωt k y = 0 , the equation of motion of a simple harmonic oscillator of natural frequency

ω0 = (ωA ωt )1/ 2 = eB/(m A m t )1/ 2 c.
5.

Define

Qe ≡ eBτe / m e c; Q h = eBτh / m h c . In the strong field limit

Q >> 1

the

magnetoconductivity tensor (6.64) reduces to

⎛ Qe−2
ne τe ⎜ −1
Qe
σ
≈
m e ⎜⎜
⎝ 0
2

−Qe−1 0 ⎞
⎟ pe 2 τh
Qe−2 0 ⎟ +

mh
0
1 ⎟⎠

⎛ Q −h2 Q −h1
⎜ −1
−2
⎜ −Q h Q h
⎜ 0
0
⎝

0⎞
⎟
0⎟ .
1 ⎟⎠

We can write nec Qe/B for ne τe / m e and pec Qh/B for pe τh / m h . The strong field limit for σyx
2

2

follows directly. The Hall field is obtained when we set

jy = 0 =

⎛ n
ec ⎡
p ⎞ ⎤
+

⎢(n − p) E x + ⎜
⎟ Ey ⎥ .
H⎣
⎝ Qe Q h ⎠ ⎦

The current density in the x direction is

jx =

⎤
ec ⎡⎛ n
p ⎞
+
⎢⎜
⎟ E x − (n − p) E y ⎥ ;
B ⎣⎝ Q e Q h ⎠
⎦

using the Hall field for the standard geometry, we have

ec ⎡⎛ n
p ⎞
(n − p) 2
jx = ⎢⎜
+
⎟+
H ⎢⎝ Q e Q h ⎠ ⎛ n
p
+
⎜

⎢⎣
⎝ Qe Q h

8-2

⎤
⎥ Ex .
⎞⎥
⎟⎥
⎠⎦


CHAPTER 9
1.

2a.
−

−

π
= 0.78 ×108 cm −1
a

π
b

= 0.78 × 108 cm -1
−

π
= 1.57 ×108 cm −1
a

9-1

π
= 0.7 8 ×1 08 cm −1
a


b.

N = 2×

πk 2F

( 2π / k )

2

n = N/L2 = k F2 / 2π
k F = 2πn
1
n = × 1016 els/cm 2
8
k F = 0.89 ×108 cm −1
c.

3a. In the hcp structure there is one atom whose z coordinate is 0 and one at

the basis for G c =



2π
zˆ is
c 

1
c . The structure factor of
2

SGc (basis) = 1 + e− iπ = 1 − 1 = 0,


so that by the same argument as in Problem 9.4 the corresponding component UG c of the crystal potential



is zero.
b. But for U 2G c the structure factor is


S2Gc (basis) = 1 + e−i2 π = 2.


c. The two valence electrons can just fill the first BZ. All we need is an adequate energy gap at the zone
boundary and for simple hex. there is no reason against a gap.
d. In hcp there will be no gap (at least in lowest order) on the top and bottom faces of the BZ, by the
argument of part a.

9-2


4.

dk
e
5a. h/  = − v × B;
dt
c 
/
hGc
T=
evB
10−27 erg sec) (2 × 108 cm −1 ) (3 × 1010 cm s −1 )

(5 × 10−10 esu) (108 cm d −1 ) (103 gauss)
 1.2 ×10−10 sec.

9-3


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:

⎛ h/ 2 d 2
⎞
− U 0 ⎟ ψ = εψ
⎜−
2
⎝ 2m dx
⎠
h/ 2 k 2
ψ = A cos kx ; ε =
− U 0 (*)
2m
Region II:

h/ 2 d 2
ψ = εψ
−
2m dx 2

ψ = B e

Boundary condition

− qx

h/ 2 q 2
; ε=−
2m

(*)

1 dψ
continuous.
ψ dx
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
2πe
, where S = πkF2, with kF = 0.75 × 108 cm–1 from Table 6.1, for potassium. Thus
)=
/ S
H
hc

1
2
∆( ) 
 0.55 × 10−8 G −1.
2
H
137 k F e

9-4