43
C
HAPTER
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
m
K
(2/a)sin
(
/
)
−
=
ω ω
2
2
1/ 2
m
dK/d
(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
0
4 K /3
(4
/3)[(
) / A]
Ω =
π
=
π
ω − ω
3/2
1/ 2
0
is
, provided
ω
<
ω
3
3
3/2
0
D(
)= (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
ω
0
.
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
, whence
.
3 a/a
V/V
∆
≈ ∆
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
2
1
3
D
d
D(
)
3V
/ 4
v
−
= ∫ ω
ω ω
=
ω
π
3
2
3
v
2
2
D
R
3h
/8
/
<
>=
ω
π ρ
. (b) In one dimension from
(15) we have
, whence
D(
)
L/ v
ω =
π
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)
h
/ 4MNv .
/
/
=
=
ω
4. (a) The motion is constrained to each layer and is therefore essentially twodimensional. Consider one
plane of area A. There is one allowed value of K per area (2
π
/L)
2
in K space, or (L/2
π
)
2
= A/4
π
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
)( K )
A
/ 4 v .
=
π
π
=
ω
π
2
The density of modes of each polarization type is D(
ω
) = dN/d
ω
= A
ω
/2
π
v
2
. 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
d
ω
where
ω
D
is defined by
. In the regime
D
D
N
D(
)
ω
=
ω
∫
D
ω
>> τ
=
, we have
3
2
2
2
x
0
2A
x
U
dx.
2 v
e
1
∞
τ
≅
π
−
∫
=
51