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CHAPTER 1
- The vectors xˆ + yˆ +zˆ and − xˆ − ˆy +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 = = ° + ° = °
- 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'
- 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
2
a c
a ,
or
2 2
2 1 c 8
a c ; 1..
3 4 a 3
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CHAPTER 2
- The crystal plane with Miller indices hk is a plane defined by the points a
1
/h, a
2
/k, and. (a)
Two vectors that lie in the plane may be taken as a
3
a /
1
/h – a
2
/k and
1 3
a / h − a /. But each of these vectors
gives zero as its scalar product with
1 2
h k
3
G = a + a + a , so that G must be perpendicular to the plane
hk. (b) If is the unit normal to the plane, the interplanar spacing is
n
1
n a⋅ /h. But ,
whence. (c) For a simple cubic lattice
n =G / | G|
1
d(hk ) = G ⋅ a / h| G| = 2 π/ | G|
(2 / a)(h ˆ k ˆ ˆ) G = π x + y + z
,
whence
2 2 2 2
2 2 2
1 G h k
d 4 a
π
1 2 3
3a a 0
- (a) Cell volume 3a a 0
a ⋅ a × a = −
c
2
3 a c.
2 3
1
2
1 2 3
2 3
(b) 2 3a a 0
3a c
( ), 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.
- By definition of the primitive reciprocal lattice vectors
3 3 2 3 3 1 1 2
1 2 3 3
1 2 3
3
C
(a a ) (a a ) (a a )
) (2 ) / | (a a a ) |
| (a a a ) |
/ V.
BZ
V (
× ⋅ × × ×
= 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.
- (a) This follows by forming
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f B
; for n even, S = f
A
B
. (c) If f
A
= f
B
the atoms diffract identically, as if the primitive translation vector
were
a
and the diffraction condition
( ) 2 (integer ).
a ⋅ ∆ k = π ×
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CHAPTER 3
2 2 2 2
E (h 2M) (2 ) (h 2M) ( L) , with 2L
= / π λ = / π λ =.
- bcc:
12 6
U(R) = 2 N [9.114(ε σ R ) − 12.253( σ R) ]. At equilibrium and
6 6
0
R = 1.488σ ,
0
U(R ) = 2 N (ε − 2.816).
fcc:
12 6
U(R) = 2 N [12.132(ε 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
0
R = 1.679σ ,
0
U(R ) = 2 N (ε − 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 H 2
and Ne.
- We have Na → 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
0 n 1 2 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
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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).
- We take u = – w; v = 0. This displacement is ⊥ to the [111] direction. Shear waves are degenerate in
this direction. Use (57a).
- Let
1
2 xx yy
e = − e = e
in (43). Then
2 2
1 1 1 1
2 4 4 4 11 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
q
α
is the effective shear
constant.
12a. We rewrite the element a ij
= p –
δ
ij
(λ + p – q) as a
ij
= 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 ,
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 44
ω ρ = 2p + q = K (C + 2C +4C ) / 3,
and the other two roots (shear waves) are
2 2
11 12 44
ω ρ = K (C − C +C ) / 3.
- 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
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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
(C 11
)K
2
, 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.
- The criterion for stability of a cubic crystal is that all the principal minors of the quadratic form be
positive. The matrix is:
C 11
C C 12 12
C
12
C C
11 12
C
12
C C
12 11
C 44
C 44
C
44
The principal minors are the minors along the diagonal. The first three minors from the bottom are C 44
,
C 44
2
, C 44
3
; thus one criterion of stability is C 44
- The next minor is
C 11
C 44 11 44 11 12 12 11
3
, or C > 0. Next: C
3
(C
2
2
), whence |C | < C. Finally, (C 11 12 11 12
2
0, so
that C 11
12
0 for stability.
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2 1 2 2
p
p 0
v M p a C
−
- 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 0
2
0
0 0
p 0
p 0
2 sin pk a
- A (1 cos pKa) ;
M pa
2A
sin pk a sin pKa
K M
(cos (k K) pa cos (k K) pa)
ω = Σ −
∂ω
When K = k 0
,
2
0
p 0
A
(1 cos 2k pa) ,
K M
∂ω
which in general will diverge because
p
- 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 )
(C C e ) (C C ) M
−
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 =
- (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
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force constant is e
2
/R
3
. The oscillation frequency ω D
is (force constant/mass)
1/
, or (e
2
/MR
3
)
1/
. (b) For
sodium and thus
23
M 4 10 g
−
×
8
R 2 10 cm;
−
×
10 46 1 2
D
− −
ω × ×
(c) The maximum phonon wavevector is of the order of 10
13 1
3 10 s
−
×
8
cm
is
associated with this maximum wavevector, the velocity defined by ω 0
/K max
≈ 3 × 10
5
cm s
, generally a
reasonable order of magnitude.
- The result (a) is the force of a dipole e p
u p 0
on a dipole e u
0
at a distance pa. Eq. (16a)
becomes
2 P 2 3 3
p>
ω = (2 / M)[ (1γ − cos Ka) + Σ −( 1) (2e / p a )(1 −cos pKa)].
At the zone boundary ω
2
= 0 if
P P 3
p>
1 ( 1) [1 ( 1) ]p
−
or if. The summation is 2(1 + 3
p 3
[1 ( 1) ]p 1
−
σ Σ − − =
- …) = 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
p 3 2
p>
1 2 ( 1) p p ,
−
= σ Σ − which is zero when 1 – 2
–1 –1 –
- 3 – 4 + … = 1/2σ. The series
is just that for log 2, whence the root is σ = 1/(2 log 2) = 0.7213.
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Thus the heat capacity.
2
B
C = k ∂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 layer
vN / L τ << ω ≈ , where N layer
/L is the number of
layers per unit length.
- (a) From the Planck distribution
x x
n (e 1) /(e 1) coth (x/2)
< > + = + − = , where
. The partition function
B
x h /k T
= /ω
x/2 sx x/2 x 1
Z e e e /(1 e ) [2sinh (x/2)]
− − − −
−
and the
free energy is F = k B
T log Z = k
B
T log[2 sinh(x/2)]. (b) With ω(∆) = ω(0) (1 – γ∆), the condition
∂F/ ∂∆ = 0 becomes
B
B h coth (h /2k T)
∆ = 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 U (T) ∆ = γ. (c) By definition
of γ, we have δω ω = −/ γδ V/V , or d log ω = −δ d log V. But , whence
.
D
θ ∝ ω
d log θ = −γd log V
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CHAPTER 6
- The energy eigenvalues are
2
2
k
h
k.
2m
ε = The mean value over the volume of a sphere in k space is
2 2 2 2
2
F F 2
h k dk k 3 h 3
k.
2m k dk 5 2m 5
< ε > = = ⋅ = ε
The total energy of N electrons is
0 F
U N
= ⋅ ε.
2a. In general p = – ∂U/∂V at constant entropy. At absolute zero all processes are at constant entropy (the
Third Law), so that
0
p = −dU dV, where
0 F
U N
= ε
2 3
2 2
3 h 3 N
N
5 2m V
/ ⎛ π ⎞
, whence
0
2 U
p
3 V
= ⋅. (b) Bulk modulus
2
0 0 0 0
2
dp 2 U 2 dU 2 U 2 U 10 U
B V V
dV 3 V 3V dV 3 V 3 V 9 V
0
(c) For Li,
22 3 12 0
11 3 11 2
U 3
(4.7 10 cm )(4.7 eV ) (1.6 10 erg eV )
V 5
2.1 10 erg cm 2.1 10 dyne cm ,
− −
− −
= × ×
= × = ×
whence B = 2.3 × 10
11
dyne cm
- . By experiment (Table 3.3), B = 1.2 × 10
11
dyne cm
.
- The number of electrons is, per unit volume,
( )
0
n d D( )
e 1
∞
ε−μ τ
= ε ε ⋅ ,
where D(ε) is the density
of orbitals. In two dimensions
2
0
2
m 1
n d
h e 1
m
( log (1 e )),
h
∞
(ε−μ) τ
−μ τ
= ε
π / +
= μ + τ +
π
∫
where the definite integral is evaluated with the help of Dwight [569.1].
4a. In the sun there are
33
57
24
−
×
×
nucleons, and roughly an equal number of electrons. In a
white dwarf star of volume
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Here
2
xx yy P xy yx c p
1 and i.
2
ε = ε = − ω ω ε = −ε = ω ω ω
2 3
The determinantal equation gives the
dispersion relation.
- The energy of interaction with the ion is
0
r
2 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
0
r
2 3 2 1 2
0
0
dr 4 r 3 4 r r 3e 5r.
−
ρ π π =
The average Fermi energy per electron is 3ε F
/5, from Problem 6.1; because
3
0
N V = 3 4 rπ , the average
is
2 3
2 2
0
3 9 4 h 10mr
π /. The sum of the Coulomb and kinetic contributions is
2
s s
U
r r
which is a minimum at
s 2 3
s s
, or r 4.42 1.80 2..
r r
The binding energy at this value of r s
is less than 1 Ry; therefore separated H atoms are more stable.
- From the magnetoconductivity matrix we have
c
y yx x 0 x 2
c
j E E
ω τ
= σ = σ
For ω c
τ >> 1, we have
2
yx 0 c
σ ≅ σ ω τ = ne τ m mc eBτ = neB c.
- For a monatomic metal sheet one atom in thickness, n ≈ 1/d
3
, so that
2 2 2
sq F F
R ≈ mv nd e ≈ mv d e.
If the electron wavelength is d, then
F
mv d ≈ h/ by the de Broglie relation and
2
sq
R h e 137 c
in Gaussian units. Now
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9 2
sq sq
R ohms 10 c R gaussian
30 137 ohms
4.1 k.
−
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- Let
2
2
1
h 1
k K i ; G iG
2m 2
2
H H
±
= + λ = ± − ⎢ ⎥
⎜ ⎟
H.
The secular equation (46) is now
1
1
U
U
−
λ − ε
λ − ε
and for H << G we have, with
2
2
h 1
G ,
2m 2
σ = ε −
2 2
2
1
2
2
2 2
1
2 2
. 1
2
2
h h
iG iG U ;
2m 2m
h
G U ;
2m
h U
2m h
G
2m
2
H H
H
H
σ + ⋅ σ − =
σ − =
/ − σ
- U(x,y) = – U[e
i (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 e
i(π/a) (x+y)
is mixed with e
–i
(π/a) (x+y)
. The central equations are
C ; UC ; 0
a a a a
C ; UC ; 0
a a a a
π π π π ⎡ ⎤ ⎡ ⎤
λ − ε − − − =
π π π π ⎡ ⎤ ⎡ ⎤
λ − ε − − − =
where
2
2
2 h 2m a.
λ = / π The gap is 2U.
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CHAPTER 8
4
d 2
m* 1
1a. E 13.60 eV 6.3 10 eV
m
−
= × × ×
ε
6
H
m
b. r a 6 10 c
m*
−
= × ε × × m
c. Overlap will be significant at a concentration
15 3
3 4
3
N 10 atoms c
r
−
π
= ≈ m
2a. From Eq. (53), , in an approximation not too good for the present example.
d B
1/ 2 E / 2k T
0 d
n (n N ) e
−
3/ 2
13 3 B
0 2
m*k T
n 2 4 10 cm
2 h
−
≡ ≈ ×
π/
⎝ ⎠
d
B
13 3
E
1.45 ; e 0..
2k T
n 0.46 10 electrons cm.
−
−
×
14
H
b. R 1.3 10 CGS units
nec
−
= − − ×
- The electron contribution to the transverse current is
e
y e x
B
j (e) ne E E ;
c
μ ⎛ ⎞
μ +
y
for the holes
n
y h x
B
j (h) ne E E.
c
−μ ⎛ ⎞
μ +
y
Here we have used
e h
ce e ch h
B B
for 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 y
0 = (neμ − peμ )(B/c)E + (neμ + peμ )E , (*)
and to the same order the total current in the x-direction is
x h e
j (pe ne )E.
x
= μ + μ
Because (*) gives
