Download Solution - Physics for All | Physics at LUMS

Condensed Matter Physics.
PSS 1
Problem Solving Section 1
Problem 1:
Copper has a mass density ρm = 8.95 gmcm−3 and an electrical resistivity ρ = 1.55 ×
10−8 ohm − m at room temperature. Calculate,
(a). The concentration of the conduction electrons.
(b). The mean free time τ .
(c). The Fermi energy EF .
(d). The Fermi velocity vF and the mean free path at Fermi level.
Solution:
(a) We have to find the concentration of conduction electrons n =
Given that, mass density of copper Cu =
n=
mass
volume
N
.
V
= 8.95 gcm−3 .
N
·
mass/mass density
(1)
Cu has valency 1 and mass number is 64 amu. By plugging N = 1 and mass density in
Eq(1), we get
1(8.95)
64(1.67 × 10−27 )
= 8.47 × 1022 cm−3 .
n =
(b) The relation for electrical conductivity is given by,
σ=
ne2 τ
,
m
1
σ= ,
ρ
Feb. 24, 2011
(2)
(3)
1
Condensed Matter Physics.
where,
τ
PSS 1
is relaxation time and ρ is resistivity.
For copper,
given that
ρcu = 1.55 × 10−8 ohm − m. Using Eq(2) and Eq(3), relaxation time is
τ =
m
ne2 ρ
9.11 × 10−31
8.47 × 1028 (1.602 × 10−19 )2 1.55 × 10−8
= 2.677 × 10−14 s.
=
(c) The Fermi energy for 3-D is given by,
EF
2/3
h2 3N
=
8me πV
0.66
h2 3n
=
.
8me π
(4)
0.66
(6.625 × 10−34 )2 3(8.47 × 1028 )
=
.
8(9.11 × 10−31 )
3.14
= 6.7 ev
≈ 7 ev.
(d) The Fermi momentum is given by,
pF = ~kF = mvF
~kF
vF =
·
me
From Eq(4), kF is
kF = (3π 2 n)1/3
1/3
2
28
= 3(3.14) (8.47 × 10 )
= 1.326 × 1010 m−1 .
Feb. 24, 2011
2
Condensed Matter Physics.
PSS 1
Thus, Fermi velocity is
6.625 × 10−34 1.326 × 1010
=
2(3.14)
9.11 × 10−31
= 1.536 × 106 ms−1 ,
vF
and mean free path of a conduction electron at Fermi level is
lF = vF τ
= (1.536 × 106 )(2.704 × 10−14 )
= 4.153 × 10−8 m
Problem 2:
Find the Hall coefficient for germanium if for a given sample (length 1 cm, breadth 5 mm.
thickness 1 mm) a current of 5 milliamperes flown from a 1.35 volts supply develops a Hall
voltage of 20 millivolts across the specimen in a magnetic field of 0.45 Wbm−2 .
Solution:
We have to find Hall coefficient RH ,
1
(Hjx )
nec
= RH Hjx ,
Ey = −
where Ey is Hall’s field and RH is
RH =
Feb. 24, 2011
Ey
·
Hjx
3
Condensed Matter Physics.
PSS 1
Therefore, firstly we have to find Ey
Vy
d
20 × 10−3
=
1 × 10−3
= 2 voltm−1 .
Ey =
(d = thickness)
Area of cross section = (Breadth)(thickness)
= (5 × 10−3 )(1 × 10−3 )
= 5 × 10−6 m2 .
jx = current density =
5 × 10−3
=
5 × 10−6
= 1 × 103 ampm−2 .
current
area
The Hall coefficient is,
2 voltm−1
(0.45 wbm−2 )(1 × 103 ampm−2 )
= 4.44 × 10−3 voltm3 amp−1 wb−1 .
RH =
Problem 3:
Use the equation,
−
→ −
→
−
dv →
v
m
+
= −e E ,
dt
τ
Feb. 24, 2011
4
Condensed Matter Physics.
PSS 1
−
for the electron drift velocity →
v to show that the conductivity at frequency ω is
1 + iωτ
σ(ω) = σ(0)
,
1 + (ωτ )2
where σ(0) = ne2 τ /m.
Solution:
Given that, the magnitude form of equation of motion is
dv v
m
+
= −eE.
dt τ
(5)
For v = v0 e−iωt , we have
dv
= −iωv0 e−iωt
dt
= −iωv.
By plugging v and
dv
dt
in Eq(5),
v0 e−iωt
−iωt
m −iωv0 e
+
= −eE
τ
1
eE
−iω + v = −
τ
m
−eE/m
v = 1
( τ − iω)
−eEτ 1 + iωτ
=
,
m
1 + ω2τ 2
and the current density is
Feb. 24, 2011
5
Condensed Matter Physics.
PSS 1
J = nqv
−eEτ 1 + iωτ
= n(−e)
m 1 + ω2τ 2
ne2 Eτ 1 + iωτ
=
m
1 + ω2τ 2
ne2 τ 1 + iωτ
=
E·
m
1 + ω2τ 2
(6)
Using the relation for electrical conductivity,
ne2 τ
·
m
J = σE.
σ0 =
(7)
By comparing Eq(6) and Eq(7),
ne2 τ
σ(ω) =
m
1 + iωτ
1 + ω2τ 2
1 + iωτ
= σ(ω = 0)
,
1 + ω2τ 2
which is the required condition.
Problem 4:
a. Calculate the mean free energy for Mg at 0 K. The density of Mg is 1.74 gcm−3 .
b. How does EF compare to kT for Mg at room temperature? What is the value of the
Fermi temperature?
Solution:
(a) The mean free energy at T = 0 K is
E
N
3
= kB TF
5
3
= EF ,
5
<E> =
Feb. 24, 2011
(8)
6
Condensed Matter Physics.
PSS 1
where, Fermi energy is
2/3
h2 3N
EF =
.
8me πV
(9)
Given that, density of Mg = 1.74 gcm−3 .
N
V
2(1.74 × 10−3 )
=
24(1.67 × 10−27 )
= 8.68 × 1028 m−3 .
n =
Plug n, me , h in Eq(9)
EF
2/3
(6.625 × 10−34 )2 3(8.68 × 1028 )
=
8(9.11 × 10−31 )
3.14
−18
≈ 1.14 × 10
J,
and mean free energy is
3
(1.14 × 10−18 )
5
= 6.84 × 10−19 J.
<E> =
(b) In order to compare EF to KT at Troom = 300 K, take their ratio
EF
1.14 × 10−18
=
kB T
300(1.38 × 10−23 )
= 275
it means that EF = 275kB T .
The Fermi temperature is the temperature at which EF = kB TF ,
Feb. 24, 2011
7
Condensed Matter Physics.
PSS 1
EF
kB
1.14 × 10−18
=
1.38 × 10−23
= 82, 608 K.
TF =
Problem 5:
Estimate the electronic contribution of specific heat kmol of copper at 4 K and 300 K. The
Fermi energy of copper is 7.05 ev and is assumed to be temperature independent.
Solution:
Given that, Fermi energy of Cu is EF = 7.05 ev. Using the relation for electronic specific
heat Cv ,
π 2 kB T nkB
2 EF
2
π 2 kB
T
=
n.
2 EF
Cv =
Therefore, electronic contribution of specific heat kmol at T = 4 K is
(3.14)2 (1.38 × 10−23 )2 4
2 7.05(1.602 × 10−19 )
= 2.00 Jkmol−1 K−1 .
Cv =
Similarly, Cv at T = 300 K is
Cv = 150 Jkmol−1 K−1 .
Problem 6:
A uniform copper wire of length 0.5 m and diameter 0.3 mm has a resistance of
Feb. 24, 2011
8
Condensed Matter Physics.
PSS 1
0.12 Ω at 293 K. If the thermal conductivity of the specimen at the same temperature is
390 Wm−1 K−1 , calculate the Lorentz number. Compare this value with the theoretical value.
Solution
Given that,
Length of Cu wire = 0.5 m.
diameter = 0.3 × 10−3 m.
radius = 1.5 × 10−4 m.
resistance = 0.12 Ω.
thermal conductivity = 390 Wm−1 K−1 .
We have to calculate the Lorentz number L as,
L=
Ke
,
σT
(10)
where σ is electrical conductivity.
σ =
=
1
ρ
1
RA/L
A = πr2 = 7.065 × 10−8 m2
0.5
0.12(7.065 × 10−8 )
= 5.89 × 107 Ω−1 m−1 .
=
By plugging the values of σ, Ke and T = 293 K in Eq(10)
390
5.89 × 107 (293)
= 2.26 × 10−8 WΩK−2 .
Lexp =
Feb. 24, 2011
9
Condensed Matter Physics.
PSS 1
On the other hand, the theoretical value of Lorentz number can be evaluated by using the
relation
Ltheo
2
π 2 kB
=
3
e
(3.14)2 (1.38 × 10−23 )2
=
3 (1.602 × 10−19 )2
= 2.84 × 10−8 WΩK−2 ,
Ltheo > Lexp .
Comparing the above two values of Lorentz numbers, we observe that the theoretical value
is about 1.26 times higher than the experimental one.
Problem 7:
Calculate the Hall coefficient of sodium based on free electron model. Sodium has bcc
structure and the side of the cube is 4.28 A0 .
Solution
Given that, Sodium has bcc structure, therefore number of atoms per unit cell is 2 and the
side of the cube is a = 4.28 A0
Number of electrons
volume
2
= 3
a
2
=
(4.28 × 10−10 )3
= 2.55 × 1028 m−3 .
n =
Hall coefficient is given by,
Feb. 24, 2011
10
Condensed Matter Physics.
PSS 1
RH = −
=
1
nec
1
10−19 )(2.55
(1.602 ×
× 1028 )(3 × 108 )
= −8.163 × 10−19 m2 s−1 .
Feb. 24, 2011
11