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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 ·10 8 m, mass - 2 ·10 30 kg.

 max = 0.5  m   max 

hc

5

k B T

T

hc

5

k B

 max    6 .

6  10  34 5  1 .

38  10  23  3  10 8  0 .

5  10  6

K

 5 , 740

K

 

P

 power emitted by a sphere   4 

R

2 

T

4   2  5

k B

4 15

h

3

c

2  5 .

7  10  8 W m 2 K 4 This result is consistent with the flux of the solar radiation energy received by the Earth (1370 W/m 2 ) being multiplied by the area of a sphere with radius 1.5·10 11 m (Sun-Earth distance).

P

 4  

R Sun

 2   

hc

5

k B

 max   4  4     2   8 W 4   5,740K  4   the mass loss per one second 1% of Sun’s mass will be lost in

dm

dt P c

2  3 .

8  3  10  10 8 26 m  W 2  4 .

2  10 9 kg/s 

t

 0 .

01

M dm

/

dt

 2  10 28 kg 4 .

2  10 9 kg/s  4 .

7  10 18 s  1.5

 10 11 yr

Carbon monoxide poisoning

Each Hemoglobin molecule in blood has 4 adsorption sites for carrying O 2 .

Let’s consider one site as a system which is independent of other sites. The binding energy of O 2 is  = -0.7 eV. Calculate the probability of a site being occupied by O 2 . The partial pressure of O 2 in air is 0.2 atm and T=310 K.

The system has 2 states: empty (  grand partition function is:

Z

=0) and occupied (  1

e

   = -0.7 eV). So the The system is in diffusive equilibrium with O 2 in air. Using the ideal gas approximation to calculate the chemical potential:   

k T B

ln  

V Nv Q

   

k T B

ln  

k T B P

2 

mk T B h

2 3 2   Plugging in numbers gives:

 

0.6 eV

Therefore, the probability of occupied state is:

P

 occupied  

e

 

Z

  1

e

 

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

 

i d i

exp   

i

  9  5

e

 

P

 9 5

e

   5

e

   1 1 .

8

e

  1  0 .

1

e

  5

T

 

k B

ln 5  5 , 900

K

Problem

2

Consider a system of

N

(partition function, average energy)

particles with only 3 possible energy levels separated by ground state energy be 0). The system occupies a fixed volume

V

 (let the and is in thermal equilibrium with a reservoir at temperature and assume that Boltzmann statistics applies.

T

. Ignore interactions between particles (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,  ? Explain your answer on physical grounds.

k B T

>> (d) (5) What is the average energy per particle in the high temperature limit,

k B T

>>  ?

(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,

k B T

>>  ) limits, and sketch

C V

as a function of

C V

, analyze the low-

T

(

k B T

<<  ) and high-

T T

. Explain your answer on physical grounds.

(a) (b)

Z

   

i

1

d i

Z

exp     

Z

  

i

 1   1 

e

  

e

 

e

  

e

 2     2 

e

 2  

e

 2   

e

1    

e

  2

e

 2  

e

 2 

(c)

P

 1 

e

 2 

e

  

e

 2   1  1  1   2   1  2   1 3 all 3 levels are populated with the same probability

(d)

   1

e

   

e

  2

e

 2  

e

 2    1 1   1 2  1  

(e) (f)

Problem

2

(partition function, average energy)

exp

C V

   2

dU dT

   

N

1 1 .

1

d

dT

2  

N

d

ln 

d

 1 .

1

d

dT T

k B

2  ln 1 .

1    

N

k

  

N

B T

2 2

k

1

B T

     2        4

e

 1 2  

e

 1      2 

e

  2  2

e

  1 

e

 

e

2 2         

e

e e

   2      2  2

e

2

e

  2 1 2      

e

  

e

    

e

e

   2 2

e

    2 2     2  

e

 2   

N

 2

k B T

2

e

   4

e

 2  

e

 2   4

e

 3   1  

e

 3 

e

  

e

 4

e

 4  2   2  Low

T

high

T

 (  >>  ): (  <<  ):

N

 2

k B T

2

e

  1   

e

4

e

 2    

e

 2  

e

 3  2 

C V C V

N

 2 

k B T N

 2

k B T

2 2

e

   1   4

e e

  2  1

e

  

e

   4

e

  2

e

  2    

e

e

  2  3  

e

 2  3  2   

N

k B T

2 3 2 2

e N

k B T

 2 2

k

B T

e

 2   4

e

 3   4

e

 4 

C V T

(a)

Problem 3 (Boltzmann distribution)

A solid is placed in an external magnetic field interacting paramagnetic atoms of spin 

B

,  =9.3

·10 -23 J/T.

B

= 3 T. The solid contains weakly ½ so that the energy of each atom is ± (a) Below what temperature must one cool the solid so that more than 75 percent (b) of the atoms are polarized with their spins parallel to the external magnetic field?

An absorption of the radio-frequency electromagnetic waves can induce transitions between these two energy levels if the frequency condition

h f = 2

f

satisfies he

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 << k B T

. How does the absorbed power depend on the temperature?

P P

 

 1

 

2  exp     1   2

k B T

   exp    2 

B k B T

  exp    2 

B k B T

   0 .

333

T

 2 

B k B

ln 3  36 .

8 K

(b)

The absorbed power is proportional to the difference in the number of atoms in these two energy states: Power 

P

   

1 2  1  exp    2 

B k B T

   1    1  2 

B k B T

   2 

B k B T

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 H 2 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 7.9x10

3 m/s. Compare this velocity with the root mean square thermal velocity at 300K of (a) a nitrogen molecule N 2 and (b) a hydrogen molecule H 2 . Explain why the earth’s atmosphere contains nitrogen but not hydrogen.

v rms

 3

k B T m

3

k B T m

 2

k B T m v most prob

  

T

k B

 2

k B T m

 2

m

3  2  2  16  10 4  2  1 .

67 1 .

38  10  23   10  27 0 .

1  383

K v most prob

  2  2

k B T m

 2  1 .

38  10  23

J

/

K

5  10  26

kg

 300

K

 407

m

/

s v most prob

  2  2

k B T m H

2  2  1 .

38  10  23

J

3 .

4  10  27 /

K kg

 300

K

 1 , 560

m

/

s

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

·10 28 mass of a free electron.

m -3 , the effective mass = the (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.

Q

2 

mk h B T

1 / 2  6 .

6  6 .

28  9 .

1  10  31  10  34  1 .

38  10  23  300  1 / 2  4 .

3  10  9 m

volume per particle

V N



V Q

h

2 

mk

3

B T

3 / 2  8  10  26 m 3 - 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.

E F

h

2 8

m

  3

N

V

  2 / 3  8  6 .

6   10  34  2 9 .

1  10  31    3 8 .

5  10 28   2 / 3  1 .

1  10  18 J  6.7

eV  k B 3 00

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 exp  

k

B T F

   1  exp 1 0 .

1 eV 0 .

1 eV  1  0 .

27

Problem 6 (photon gas)

(a)(15) The black body radiation fills a cavity of volume V. The radiation energy is

U

 4 

c VT

4 , the radiation pressure is

P

 4  3

c T

4 Consider an isentropic (quasi-static and adiabatic) process of the cavi ty expansion (

TdS

dU

PdV

 0). The radiation pressure performs work during the expansion and the temperature of radiation will drop. Find how and

V

are related for this process.

(b) (5) Assume that the c osmic 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 dU

 

PdV U

  4 4 

c c

VT

4

T dV

4 

d

 ln

V

    3

d dU

4  3

c

 ln 

T

 4 

c

4

T dV

 16 

c d

 16 

c

3

VT dT

 ln

V

 

d

  4  3

c dV

 ln

T

3 3

V

 

dT T d

 ln

dV V

   3

dT T

(b)

R f R i

  

V f V i

  1 3    

T T f i

 3   1 3  

T T f i

 3000 2.7

 1111  0

VT

3  const.

Thus, at the moment of decoupling, the radius of the universe was ~ 1000 time smaller.

Problem 7 (BEC)

Consider a non-interacting gas of hydrogen atoms (bosons) with the density of 1  10 20 m -3 . a)(5) Find the temperature of Bose-Einstein condensation,

T C ,

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 >> T C

and

T

= 0.5

T C .

If the total number of atoms is 1  10 20 , how many atoms occupy the ground state at

T

= 0.5

T C

?

c)(5) Below T C , the pressure in a degenerate Bose gas is proportional to T 5/2 . Do you expect the temperature dependence of pressure to be stronger or weaker at T > T C ? Explain and draw aqualitative graph of the temperature dependence of pressure over the temperature range 0 <

T

< 2

T

C .

(a) T c  0.53

k B

 

h

2 2 

m



N V

2 3  0.53

 23

2    34 

  

27 2 3   5

N

0 

N

  1   

T T c

 3 2     10 20  1    3 2  20

Problem 7 (BEC) (cont.)

(c) The atoms in the ground state do not contribute to pressure. At

T

<

T C

, 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

T C

, only the latter factor contributes to

P(T)

, and the rate of the pressure increase with temperature becomes smaller than that at

T < T C

.