ppt

advertisement
Lecture 25 Practice problems
Final:
May 11, SEC 117
3 hours (4-7 PM), 6 problems
(mostly Chapters 6,7)
•
•
•
•
Boltzmann Statistics, Maxwell speed distribution
Fermi-Dirac distribution, Degenerate Fermi gas
Bose-Einstein distribution, BEC
Blackbody radiation
Sun’s Mass Loss
The spectrum of the Sun radiation is close to the black body spectrum with the
maximum at a wavelength  = 0.5 m. Find the mass loss for the Sun in one second.
How long it takes for the Sun to loose 1% of its mass due to radiation? Radius of the
Sun: 7·108 m, mass - 2 ·1030 kg.
max = 0.5 m 
hc
max 
5 k BT
 T
hc
5 k B max


6.6 1034  3 108
 

K

5
,
740
K
 23
6
5

1
.
38

10

0
.
5

10


2 5 k B
W
8


5
.
7

10
15h3c 2
m2K 4
4
P power emittedby a sphere  4R2 T 4
This result is consistent with the flux of the solar radiation energy received by the Earth
(1370 W/m2) being multiplied by the area of a sphere with radius 1.5·1011 m (Sun-Earth
distance).
4
2
 hc 
W
2
4
8
8
26
P  4  RSun   

4

7

10
m

5.7

10

5,740K

3.8

10
W





2 4
5
k

m
K
 B max 
the mass loss per one second
1% of Sun’s mass will be lost in
dm P 3.8 1026 W
9
 2 

4
.
2

10
kg/s
2
8
dt c
3 10 m


0.01M
2 1028 kg
t 

 4.7 1018 s  1.51011 yr
9
dm / dt 4.2 10 kg/s
Carbon monoxide poisoning
Each Hemoglobin molecule in blood has 4 adsorption sites for carrying O2.
Let’s consider one site as a system which is independent of other sites. The
binding energy of O2 is  = -0.7 eV. Calculate the probability of a site being
occupied by O2. The partial pressure of O2 in air is 0.2 atm and T=310 K.
The system has 2 states: empty ( =0) and occupied ( = -0.7 eV). So the
grand partition function is: Z  1  e    
The system is in diffusive equilibrium with O2 in air. Using the ideal gas
approximation to calculate the chemical potential:
3


 V 
2
k
T
2

mk
T

 
B
B
  k BT ln 
  k BT ln 

 
2

Nv
P
h


Q




Plugging in numbers gives:   0.6 eV
Therefore, the probability of occupied state is:
P  occupied  
e
     
Z

e
     
1 e
     
 0.98  98%
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   di exp i   9  5e 
i
5e  
1
P

 0.1
 

9  5e
1.8e  1
e   5 T 

k B ln 5
 5,900K
Problem 2 (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   di 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)
Problem 2 (partition function, average energy)
1
2
exp 2  
2  ln 1.1 T 
(f)
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 
1 e  e
1  e    e  2 
 k BT 


 



 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 (<<):
N 2 e    4e 2   e 3 
N 2  k B T
CV 

e
2
2
2



2

k BT
k BT
1 e  e
N 2 e   4e2   e3 2 N 2
CV 

2
2



2

kBT
3 kBT 2
1 e  e




T

Problem 3 (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 4 (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
vmost
prob
2k BT

m
3k BT
2 k BT
2 m

 T 
m
m
kB 3  2


2
16104  2 1.67 1027

 383K
 23
1.3810  0.1
vmost
2k BT
2 1.381023 J / K  300K

 407m / s
prob  N 2  
 26
m
5 10 kg
vmost
2 k BT
2 1.381023 J / K  300K

 1,560m / s
prob H 2  
 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 5 (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m kBT 
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 300K
- 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 6 (photon gas)
(a)(15) The black body radiation fills a cavity of volume V. The radiation energy is
4
4 4
VT 4 , the radiation pressure is P 
T
c
3c
Consider an isentropic (quasi-static and adiabatic) process of the cavity expansion
U
(TdS  dU  PdV  0). The radiation pressure performs work during the expansion and
the temperature of radiation will drop. Find how T and V are related for this process.
(b) (5) Assume that the cosmic microwave background (CMB) radiation was decoupled
from the matter when both were at 3000 K. Currently, the temperature of CMB radiation
is 2.7 K. What was the radius of the universe at the moment of decoupling compared to now?
Consider the process of expansion as isentropic.
(a) dS  0  dU   PdV
4
4 4
16
4 4
U
VT 4  dU 
T dV 
VT 3dT  
T dV
c
c
c
3c
4 4
4 4
16
dV dT
dV
dT

T dV 
T dV 
VT 3dT  0 

0
3
0
c
3c
c
3V
T
V
T
 d  lnV   3d  ln T   0  d  ln V   d  ln T 3   0  d ln VT 3    0  VT 3  const.
3
 T
  i
 T f

1
3
 
Rf Vf 
Ti 3000
(b)
 

 1111
  
Ri  Vi 
T
2.7

f
 
Thus, at the moment of decoupling, the radius of the universe was ~ 1000 time smaller.
1
3
Problem 7 (BEC)
Consider a non-interacting gas of hydrogen atoms (bosons) with the density of 11020m-3.
a)(5) Find the temperature of Bose-Einstein condensation, TC, for this system.
b)(5) Draw aqualitative graph of the number of atoms as a function of energy
of the atoms for the cases: T >> TC and T = 0.5 TC. If the total number of atoms is 11020,
how many atoms occupy the ground state at T = 0.5 TC?
c)(5) Below TC, the pressure in a degenerate Bose gas is proportional to T5/2.
Do you expect the temperature dependence of pressure to be stronger or weaker at T > TC?
Explain and draw aqualitative graph of the temperature dependence of pressure over
the temperature range 0 <T < 2 TC.
(a) Tc 
0.53  h   N 

 
k B  2 m   V 
2
6.626  10 
34 2
23

0.53
20 2 3
10
 3.3  105  K 

23
27 
1.38  10 2  1.67  10
  T 3 2 
32
N 0  N 1      1020 1   0.5   0.65  1020


  Tc  
Problem 7 (BEC) (cont.)
(c) The atoms in the ground state do not contribute to pressure. At T < TC,
two factors contribute to the fast increase of P with temperature:
(i) an increase of the number of atoms in the excited states, and
(ii) an increase of the average speed of atoms with temperature.
Above TC, only the latter factor contributes to P(T), and the rate of the pressure increase
with temperature becomes smaller than that at T < TC.
Download