Lecture 27 Overview
Final:
May 8, SEC 117
3 hours (4-7 PM), 6 problems
(mostly Chapters 6,7)
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•
•
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Boltzmann Statistics, Maxwell speed distribution
Fermi-Dirac distribution, Degenerate Fermi gas
Bose-Einstein distribution, BEC
Blackbody radiation
Problem 1 (partition function, average energy)
The neutral carbon atom has a 9-fold degenerate ground level and a 5-fold
degenerate excited level at an energy 0.82 eV above the ground level. Spectroscopic
measurements of a certain star show that 10% of the neutral carbon atoms are in the
excited level, and that the population of higher levels is negligible. Assuming thermal
equilibrium, find the temperature.
Z   d i exp  i   9  5e  
i
5e  
1
P

 0.1
 

9  5e
1.8e  1
e   5 T 

k B ln 5
 5,900 K
Problem (final 2005, partition function)
Consider a particle with five microstates with energies 0, , , , and 2 (  = 1 eV ) in
equilibrium with a reservoir at temperature T = 0.5 eV.
1. Find the partition function of the particle.
2. Find the average energy of the particle.
3. What is the average energy of 10 such particles?
  
 2 
  exp  
  1  0.406  0.018  1.424
Z1  1  3 exp  
 k BT 
 k BT 
the average energy of a
single particle:

    i P i  
i
the same
result you’d
get from this:
 2 
  2  exp  

 k BT 
 k BT   1 eV  0.406  0.036  0.310 eV
1.424
  
 2 
  exp  

1  3 exp  
 k BT 
 k BT 
  3 exp  
 
 
1 Z
3  exp        exp  2   2 

Z 
1  3 exp     exp  2 
the average energy of N = 10 such particles:
U10  N U  10  0.310 eV  3.1 eV
Problem 2006 (partition function, average energy)
Consider a system of N particles with only 3 possible energy levels separated by  (let the
ground state energy be 0). The system occupies a fixed volume V and is in
thermal equilibrium with a reservoir at temperature T. Ignore interactions between particles
and assume that Boltzmann statistics applies.
(a) (2) What is the partition function for a single particle in the system?
(b) (5) What is the average energy per particle?
(c) (5) What is probability that the 2 level is occupied in the high temperature limit, kBT >>
? Explain your answer on physical grounds.
(d) (5) What is the average energy per particle in the high temperature limit, kBT >> ?
(e) (3) At what temperature is the ground state 1.1 times as likely to be occupied as the 2
level?
(f) (25) Find the heat capacity of the system, CV, analyze the low-T (kBT<<) and high-T
(kBT >> ) limits, and sketch CV as a function of T. Explain your answer on physical grounds.
(a)
Z   d i exp  i   1  e    e 2 
i
(b)
(c)
(d)

1 Z
  e     2 e 2 
e    2e 2 
 


Z 
1  e    e 2 
1  e   e 2 
e 2 
1  2
1
P


1  e    e 2  1  1    1  2 3
e    2e 2 
1 2
 



 
 2 
1 e  e
111
all 3 levels are populated with
the same probability
(e)
(f)
Problem 2006 (partition function, average energy)
1
2
exp  2  
2  ln 1.1 T 
CV 
1.1
d 
k B ln 1.1
d  d
dU
N
N
dT
dT
d dT



1    e     2 2e  2 
e    2e  2    e     2 e  2 

 N  

2 
 
 2 
 
 2  2
k
T
1

e

e

1

e

e
 B 



 


 N 2  e    4e  2  1  e    e  2   e    2e  2  e    2e  2 

 
2
2 
1  e    e  2 
 k BT 





 e    4e  2   e  2   4e 3   e 3   4e  4   e  2   4e 3   4e  4 

2

1  e    e  2 
N 2 e    4e  2   e 3 

CV
k BT 2 1  e    e  2  2
N 2

k BT 2


Low T (>>):
high T (<<):





 
2 
3 

N e  4e
e
N
k BT

e
k BT 2 1  e    e 2  2
k BT 2
N 2 e    4e 2   e 3 2 N 2
CV 

k BT 2 1  e    e  2  2
3 k BT 2
CV 

2

2




T

Problem (Boltzmann distribution)
A solid is placed in an external magnetic field B = 3 T. The solid contains weakly
interacting paramagnetic atoms of spin ½ so that the energy of each atom is ±
 B,  =9.3·10-23 J/T.
(a) Below what temperature must one cool the solid so that more than 75 percent
of the atoms are polarized with their spins parallel to the external magnetic
field?
(b) An absorption of the radio-frequency electromagnetic waves can induce
transitions between these two energy levels if the frequency f satisfies he
condition h f = 2  B. The power absorbed is proportional to the difference in
the number of atoms in these two energy states. Assume that the solid is in
thermal equilibrium at  B << kBT. How does the absorbed power depend on
the temperature?
(a)
(b)
   
 2B 
P1 

 exp   1 2   exp  
P 2 
k
T
k
T
B


 B 
 2B 
  0.333
exp  
k
T
 B 
T
2B
 36.8 K
k B ln 3
The absorbed power is proportional to the difference in the number of atoms in
these two energy states:
 2B 
 2B  2B
  1  1 
 
Power  P1  - P 2   1  exp  
k
T
k
T
B
 B 

 k BT
The absorbed power is inversely proportional to the temperature.
Problem (Boltzmann distribution)
Consider an isothermic atmosphere at T=300K in a uniform gravitational field. Find
the ratio of the number of molecules in two layers: one is 10 cm thick at the earth’s
surface, and another one is 1 km thick at a height of 100 km. The mass of an air
molecule m= 5·10-26 kg, the acceleration of free fall g=10 m/s2.
 mgh 
k T
dh  (area )n0 B
N i  (area )  n0 exp  
mg
 k BT 
h1
h2
(area )n0
N10cm
N1km
k BT
mg
 mgh   mgh 
 
d 
 
exp

k
T
k
T
B

  B 
mgh1 / k BT
mgh2 / k BT
  mgh1 
 mgh2 


 

exp


exp
 

k
T
k
T
B
B



 
 mg  0m 
 mg  0.1m 
  exp  

exp  
k
T
k
T
B
B






 mg 1 105 m 
 mg 1.01105 m 
  exp  

exp  
k
T
k
T
B
B




 5 10  26 kg 10m / s 2  0.1m 

exp  0   exp  
 23
1
.
38

10
J
/
K

300
K



 5 10  26 kg 10m / s 2 1105 m 
 5 10  26 kg 10m / s 2 1.01 105 m 
  exp  

exp  
 23
 23
1
.
38

10
J
/
K

300
K
1
.
38

10
J
/
K

300
K




1.2 10 5
 18.5 - more air in the 10-cm-thick layer at the earth’s surface
5.69 10 6  5.04 10 6
Problem (Maxwell distr.)
Find the temperature at which the number of molecules in an ideal Boltzmann gas
with the values of speed within the range v - v+dv is a maximum.
 m 

Pv, T , m   
2

k
T
B 

3/ 2
 mv 2 

4v exp  
2
k
T
B 

2
maximum:
Pv, T 
0
v
Problem 2006 (maxwell-boltzmann)
(a) Find the temperature T at which the root mean square thermal speed of a hydrogen
molecule H2 exceeds its most probable speed by 400 m/s.
(b) The earth’s escape velocity (the velocity an object must have at the sea level to escape
the earth’s gravitational field) is 7.9x103 m/s. Compare this velocity with the root mean
square thermal velocity at 300K of (a) a nitrogen molecule N2 and (b) a hydrogen molecule
H2. Explain why the earth’s atmosphere contains nitrogen but not hydrogen.
vrms
3k BT

m
2k BT
vmost prob 
m
3k BT
2 k BT
2 m

 T 
m
m
kB 3  2


2
16 10 4  2 1.67 10 27

 383K
 23
1.38 10  0.1
2k BT
2 1.38 1023 J / K  300 K
vmost prob N 2  

 407m / s
 26
m
5 10 kg
2k BT
2 1.38 1023 J / K  300 K
vmost prob H 2  

 1,560m / s
 27
mH 2
3.4 10 kg
Significant percentage of hydrogen molecules in the “tail” of the Maxwell-Boltzmann
distribution can escape the gravitational field of the Earth.
Problem (degenerate Fermi gas)
The density of mobile electrons in copper is 8.5·1028 m-3, the effective mass = the
mass of a free electron.
(a) Estimate the magnitude of the thermal de Broglie wavelength for an electron at
room temperature. Can you apply Boltzmann statistics to this system? Explain.
h
6.6 1034
9
Q 


4
.
3

10
m
1/ 2
1/ 2
31
 23
2mkBT 
6.28  9.110 1.38 10  300


V
h3
 26
3
volume per particle   VQ 

8

10
m
N
2mkBT 3 / 2
- Fermi distribution
(b) Calculate the Fermi energy for mobile electrons in Cu. Is room temperature sufficiently
low to treat this system as degenerate electron gas? Explain.
h 2  3N

EF 
8m   V



2/3
6.6 10 

3
28 


8
.
5

10
31 
8  9.1 10  

34 2
2/3
 1.110 18 J  6.7 eV  k B 300 K
- strongly degenerate
(c) If the copper is heated to 1160K, what is the average number of electrons in the
state with energy F + 0.1 eV?
n
1
  F
exp 
 k BT

  1


1
 0.27
 0.1 eV 
exp 
 1
 0.1 eV 
Problem 2006 (electrons in Lithium)
Metallic Lithium has a Fermi temperature of 5.5x104 K and a Debye temperature of 400K.
(a) (7) Find the total density of electron gas in Lithium. Assume that the effective mass of
electrons equals the free electron mass.
(b) (10) Find (approximately) the density of electrons in Lithium that can carry current at
300K.
(c) (8) Find at what temperature the phonon and electron contributions to the heat capacity
become equal.
(a)
(b)
h2  3 
EF 
 n
8m   
g   
nE F  k B T 
2/3
  8mE 
n  2F 
3 h 
 4.7 10 28 m 3
3n
3n






E

F
3/ 2
2 EF
2 EF
E F  k BT
 n  g  d 
E F  k BT
(c)
3/ 2
3n
3n k BT 3n T
 0.5  2k BT 

 n  0.008  3.8 1026 m 3
2 EF
2 EF
2 TF
The phonon contribution
to the heat capacity:
The electron contribution
to the heat capacity:
C ph
12 4  T 


 Nk B
5  D 
Ce 
2
2
Nk B
k BT
EF
3
5 3D
T
 5K
24 2 TF