# hw5 solutions

Homework 5
Duygu Can & Neşe Aral
01 Dec 2010
homework 5 solutions
1.1
Problem 4.11
Question
Show that in the limit T θ the heat capacity of a solid goes towards the the
limit CV → 3N kB , 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
CV = 3N kB × [1 −
X
an /T n ]
(1.1)
n
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:
3π 2 h̄v
U=
·
2L
τL
πh̄v
4 Z
xD
0
x3
dx
ex − 1
(1.2)
where x = πh̄vn/Lτ .
Using xD = kB θ/τ , where the Debye temperature θ is given by θ =
(h̄v/kB )(6π 2 N/V )1/3 , Eq. (??) can be written as
U=
9N kB θ
x4D
Z
xD
0
|
x3
dx .
−1
{z
}
ex
(1.3)
I
When T θ, x is very small and we can calculate the integral I by using the
Taylor expansion
x x2
x
e −1≈x 1+ +
.
(1.4)
2
6
Inserting Eq. (??) into x3 /(ex − 1) we get
−1
x3
x x2
2
≈
x
1
+
+
,
ex − 1
2
6
which can be simplified using a second Taylor expansion which gives:
x3
x3
x4
2
≈
x
−
+
.
ex − 1
2
12
2
(1.5)
Problem 4.12
Inerting Eq. (??) into Eq. (??) and integrating we get
I
T
1
θ
U = 9N kB θ 4 = 9N kB θ
− +
x
3θ 8 60T
D
2
3θ
θ
= 3N kB T −
+
.
8
20T
The heat capacity is then calculated as:
C=
1.2
∂U
= 3N kB .
∂T
Problem 4.12
Question
Consider a dielectric solid with a Debye temperature equal to 100K and with
1022 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
∂U
(1.6)
CV =
∂T V
3
12π 4 N kB T
=
(1.7)
5
θ
≈ 322erg/Kcm3
(1.8)
Then,
CV
V
=
12π 4 nkB
5
3
T
θ
(1.9)
since n = N/V .
For photons the energy per unit volume is
U
V
=
=
π2
τ4
15h̄3 c3
4σB T 4
c
(1.10)
(1.11)
Then the heat capacity of the photon per unit volume is,
CV
V
=
∂(U/V )
∂T
(1.12)
3
1. Homework 5 Solutions
16σB T 3
c
≈ 2.99 × 10−14 T 3 erg/K 4 cm3
=
(1.13)
(1.14)
Upon equating Eq.?? to Eq.?? and solving for T , we get
T
=
≈
1.3
322
× 1014
2.99
1/3
K
(1.15)
2.2 × 105 K
(1.16)
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
3
θ
T
(1.17)
Suppose that T = 10−2 K; θ = −200K; and N ≈ 1015 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 1cm−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 >
2
τ (∂U/∂τ )V
(1.18)
(1.19)
The root mean square fractional fluctuation energy is,
F2
=
=
∆U
U2
τ 2 (∂U/∂τ )V
U2
(1.20)
(1.21)
In the low temperature limit,
U (τ ) ≈
4
3piN τ 4
5(kB θ)3
(1.22)
Problem 4.13
Hence,
F2
=
=
3
kB θ
τ
3
1 θ
0.684 ×
N T
20 1
3π 4 N
(1.23)
(1.24)
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