1.1
Problem 4.11
Question
Show that in the limit T θ the heat capacity of a solid goes towards the the limit C
V
→ 3 N k
B
, in conventional units. To obtain higher accuracy when T is moderately larger than θ , the heat capacity can be expanded as a power series in 1 /T , of the form
C
V
= 3 N k
B
× [1 −
X a n
/T n
] n
(1.1)
Determine the first non-vanishing term in the sum.
Check your result by inserting T = θ and comparing with Table 4 .
2.
Solution
The internal energy of a solid is given by Eq. (4 .
41) in the book:
U =
3 π 2
2 L
·
τ L
π ¯
4
Z x
D
0 e x x 3
− 1 dx (1.2) where x = π ¯ .
Using x
D hv/k
B
=
)(6 π 2 k
N/V
B
)
θ/τ , where the Debye temperature θ is given by θ =
1 / 3 , Eq. ( ??
) can be written as
U =
9 N k
B
θ x 4
D
Z x
D
0
| x
3 e x − 1 dx .
{z
I
}
(1.3)
When T θ , x is very small and we can calculate the integral I by using the
Taylor expansion e x − 1 ≈ x 1 + x
2
+ x 2
6
.
(1.4)
Inserting Eq. ( ??
) into x
3
/ ( e x − 1) we get x 3 e x − 1
≈ x
2
1 + x
2
+ x 2
6
− 1
, which can be simplified using a second Taylor expansion which gives: x 3 e x − 1
≈ x
2 − x 3
2
+ x 4
12
.
2
(1.5)

Problem 4.12
Inerting Eq. ( ??
) into Eq. ( ??
) and integrating we get
I
U = 9 N k
B
θ x 4
D
= 3 N k
B
= 9 N k
B
θ
T −
3 θ
8
θ 2
+
20 T
T
3 θ
−
1
8
θ
+
60 T
.
The heat capacity is then calculated as:
C =
∂U
∂T
= 3 N k
B
.
1.2
Problem 4.12
Question
Consider a dielectric solid with a Debye temperature equal to 100K and with
10 22 atoms cm
− 3 . Estimate the temperature at which the photon contribution to the heat capacity would be equal to the phonon contribution evaluated at
1K.
Solution
The heat capacity of the phonons in the low temperature limit ( T θ ) is given in Eq.4.47b as
C
V
∂U
=
=
∂T
V
12 π 4 N k
B
5
≈ 322 erg/Kcm
3
T
θ
3
(1.6)
(1.7)
(1.8)
Then,
C
V
V
=
12 π 4 nk
B
5
T
θ
3 since n = N/V .
For photons the energy per unit volume is
U
V
=
=
π 2 h
3 c 3
τ
4
4 σ
B
T 4 c
Then the heat capacity of the photon per unit volume is,
C
V
V
=
∂ ( U/V )
∂T
(1.9)
(1.10)
(1.11)
(1.12)
3

1.
Homework 5 Solutions
16 σ
B
T
3
= c
≈ 2 .
99 × 10
− 14
T
3 erg/K
4 cm
3
Upon equating Eq.
??
to Eq.
??
and solving for T , we get
T =
322
2 .
99
× 10
14
≈ 2 .
2 × 10
5
K
1 / 3
K
(1.13)
(1.14)
(1.15)
(1.16)
1.3
Problem 4.13
Question
Consider a solid of N atoms in the temperature region in which the Debye
T
3 law is valid. The solid is in thermal contact with a heat reservoir. Use the results on energy fluctuations from Chapter 3 to show that the root mean square fractional energy is F is given by
F 2
= < ( ε − < ε > )
2
> / < ε >
2 ≈
0 .
07
N
θ
T
3
(1.17)
Suppose that T = 10
− 2
K; θ = − 200K; and N ≈ 10
15 for a particle 0.01 cm on a side; then F ≈ 0 .
02. At 10 − 5 K the fractional fluctuation in energy is order of unity for a dielectric particle of volume 1 cm − 3 .
Solution
The energy fluctuations in a system of N particles which are in thermal contact with a reservoir is given in Eq.3.89 as follows:
∆ U = < ( ε − < ε > )
2
>
= τ
2
( ∂U/∂τ )
V
The root mean square fractional fluctuation energy is,
F 2
=
=
∆ U
U 2
τ
2
( ∂U/∂τ )
V
U 2
In the low temperature limit,
U ( τ ) ≈
3 piN τ
4
5( k
B
θ ) 3
(1.18)
(1.19)
(1.20)
(1.21)
(1.22)
4

Hence,
F
2
=
20
3 π 4
1
N
1
= 0 .
684 ×
N k
B
θ
τ
θ
T
3
3
Problem 4.13
(1.23)
(1.24)
5