Lecture 27 Debye Model of Solids Phonon Gas

1 2 3 3N

ħ 2

32

1 3

1

3

1

22 hNnhkrrmU i

N

i

N

iii

In 1907 Einstein developed the first quantum-mechanical model of solids that was able to qualitatively describe the low-T heat capacity of the crystal lattice Although this was a crucial step in the right direction the model was too crude

In Einsteinrsquos model all oscillators are identical and their frequencies are the same

However Einsteinrsquos model ignores the fact that the atomic vibrations are coupled together the potential energy of an atom in the crystal depends on the distance from its neighbors

23

11

3

1

2

2

1

2

1

N

jiji

N

ii rrkrmU

This energy is not a sum of single-particle energies Thus the calculation of the partition function may look rather difficult But a system of N coupled three-dimensional oscillators is equivalent to a system of 3N independent one-dimensional oscillators the price to be paid is that the independent oscillators are not of the same frequency the normal modes of vibration of a solid have a wide range of frequencies These modes are not related to the motion of single atoms but to the collective motion of all atoms in the crystal ndash vibrational modes or sound waves

Einsteinrsquos Model of a SolidIn 1907 Einstein in the first application of quantum theory to a problem other than radiation modeled a solid body containing N atoms as a collection of 3N harmonic oscillators The partition function of a single oscillator

22

1

exp12

exp

exp2

exp2

1expexp

0001

h

h

h

hh

hnZn

nnnn

cosechThe oscillators are

independent of each other thus

NZZ 1

The mean energy

1exp

1

2

1

2coth

2

1ln 1

h

hh

hZE

This looks familiar the same energy would have a photon of frequency

The internal energy is not a directly measurable quantity and instead we measure the heat capacity

2

22

1exp

exp33

h

hhNkENU

Uk

dT

dUTC BBV

Limits high T (kBTgtgth) BV NkTC 3 equipartition

low T (kBTltlth) ThhNkTC BV expexp3 2

The Einstein model predicts much too low a heat capacity at low temperatures

Debyersquos Theory of the Heat Capacity of Solids

If we quantize this elastic distortion field similar to the quantization of the e-m field we arrive at the concept of phonons the quanta of this elastic field For the thermal phonons the wavelength increases with decreasing T

Nobel 1936

These low-energy modes remain active at low temperatures when the high-frequency modes are already frozen out Large values of that correspond to these modes justify the use of a continuum model

mmKTTk

hc

B

s 1010~11041

1021066~1 7

34

334

There is a close analogy between photons and phonons both are ldquounconservedrdquo bosons Distinctions (a) the speed of propagation of phonons (~ the speed of sound waves) is by a factor of 105 less than that for light (b) sound waves can be longitudinal as well as transversal thus 3 polarizations (2 for photons) and (c) because of discreteness of matter there is an upper limit on the wavelength of phonons ndash the interatomic distance

Debyersquos model (1912) starts from the opposite point of view treating the solid as a continuum ie the atomic structure is ignored A continuum has vibrational modes of arbitrary low frequencies and at sufficiently low T only these low frequency modes are excited These low frequency normal modes are simply standing sound waves

s

B

chhTk cS ndash the sound velocity

Density of States in Debye Model

Each normal mode is a quantized harmonic oscillator The mean energy of each mode

1exp

1

2

1

hhE

dh

ghUdTEgU

DD

0

0

0 1exp

is the total energy per unit volume The U0 term comes from the zero-point motion of atoms It reduces the cohesive energy of the solid (the zero point motion in helium is sufficient to prevent solidification at any T at normal pressure) but since it does not depend on T it does not contribute to C Note that we ignored this term for phonons where it is In QED this unobservable term is swept under the rug by the process known as renormalization

2

3

63k

kG

3

2

2

6

13

scG

3

212

scd

dGg

For a macroscopic crystal the spectrum of sound waves is almost continuous and we can treat is a continuous variable As in the case of photons we start with the density of states per unit frequency g() The number of modes per unit volume with the wave number lt k

- multiplied by 3 since a sound wave in a solid can have three polarizations (two transverse and one longitudinal)

and

31

0 4

33

ncndg sD

D

- this eq only holds for sufficiently low

2

kcs

cS ~3 kms a~02 nm D~1013 Hz

(= large wavelengths and the continuous approximation is valid) There is also an upper cut-off for the frequencies ( interatomic distances) the so-called Debye frequency D which depends on the density n

The Heat Capacity in Debyersquos Model

At high temperatures all the modes are excited (the number of phonons does not increase any more) and the heat capacity approaches the equipartition limit C=3NkB

At low temperatures we can choose the upper limit as (the high-frequency modes are not excited the energy is too low) How low should be T

333

45

5

16T

ch

k

dT

dUC

s

B

Thus Debyersquos model predicts that in the limit of sufficiently low T the heat capacity due to vibrations of the crystal lattice (in a metal electrons also contribute to C) must vary as T3 and not as 2exp(- hkBT) as in Einsteinrsquos model (Roughly speaking the number of phonons ~ T3 their average energy is proportional to T)

33

45

0

0

05

4

1exp s

B

ch

TkUd

h

ghUU

The low-T heat capacity

Kak

hcT

B

s 1000~1011041

10210661023

334

TNkUdAUU B

D

30

0

2

0

Debye TemperatureThe material-specific parameter is the sound speed If the temperature is properly normalized the data for different materials collapse onto a universal dependence

343

33

45

5

12

5

16

D

B

s

B TNkT

ch

kC

31

4

3

n

k

hc

B

sD

The normalization factor is called the Debye temperature

It is related to the maximum frequency D the Debye frequency

The higher the sound speed and the density of ions the higher the Debye temperature However the real phonon spectra are very complicated and D is better to treat as an experimental fitting parameter

DDBsD hTkn

c

31

4

3

Problem (blackbody radiation)The spectrum of Sun plotted as a function of energy peaks at a photon energy of 14 eV The spectrum for Sirius A plotted as a function of energy peaks at a photon energy of 24 eV The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun How does the radius of Sirius A compare to the Sunrsquos radius

2424

T

LRTRL

The temperature according to Wienrsquos law is proportional to the energy that corresponds to the peak of the photon distribution

671

4142

24

22

SunSirius

SunSirius

Sun

Sirius

TT

LL

R

R

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Einsteinrsquos Model of a SolidIn 1907 Einstein in the first application of quantum theory to a problem other than radiation modeled a solid body containing N atoms as a collection of 3N harmonic oscillators The partition function of a single oscillator

22

1

exp12

exp

exp2

exp2

1expexp

0001

h

h

h

hh

hnZn

nnnn

cosechThe oscillators are

independent of each other thus

NZZ 1

The mean energy

1exp

1

2

1

2coth

2

1ln 1

h

hh

hZE

This looks familiar the same energy would have a photon of frequency

The internal energy is not a directly measurable quantity and instead we measure the heat capacity

2

22

1exp

exp33

h

hhNkENU

Uk

dT

dUTC BBV

Limits high T (kBTgtgth) BV NkTC 3 equipartition

low T (kBTltlth) ThhNkTC BV expexp3 2

The Einstein model predicts much too low a heat capacity at low temperatures

Debyersquos Theory of the Heat Capacity of Solids

If we quantize this elastic distortion field similar to the quantization of the e-m field we arrive at the concept of phonons the quanta of this elastic field For the thermal phonons the wavelength increases with decreasing T

Nobel 1936

These low-energy modes remain active at low temperatures when the high-frequency modes are already frozen out Large values of that correspond to these modes justify the use of a continuum model

mmKTTk

hc

B

s 1010~11041

1021066~1 7

34

334

There is a close analogy between photons and phonons both are ldquounconservedrdquo bosons Distinctions (a) the speed of propagation of phonons (~ the speed of sound waves) is by a factor of 105 less than that for light (b) sound waves can be longitudinal as well as transversal thus 3 polarizations (2 for photons) and (c) because of discreteness of matter there is an upper limit on the wavelength of phonons ndash the interatomic distance

Debyersquos model (1912) starts from the opposite point of view treating the solid as a continuum ie the atomic structure is ignored A continuum has vibrational modes of arbitrary low frequencies and at sufficiently low T only these low frequency modes are excited These low frequency normal modes are simply standing sound waves

s

B

chhTk cS ndash the sound velocity

Density of States in Debye Model

Each normal mode is a quantized harmonic oscillator The mean energy of each mode

1exp

1

2

1

hhE

dh

ghUdTEgU

DD

0

0

0 1exp

is the total energy per unit volume The U0 term comes from the zero-point motion of atoms It reduces the cohesive energy of the solid (the zero point motion in helium is sufficient to prevent solidification at any T at normal pressure) but since it does not depend on T it does not contribute to C Note that we ignored this term for phonons where it is In QED this unobservable term is swept under the rug by the process known as renormalization

2

3

63k

kG

3

2

2

6

13

scG

3

212

scd

dGg

For a macroscopic crystal the spectrum of sound waves is almost continuous and we can treat is a continuous variable As in the case of photons we start with the density of states per unit frequency g() The number of modes per unit volume with the wave number lt k

- multiplied by 3 since a sound wave in a solid can have three polarizations (two transverse and one longitudinal)

and

31

0 4

33

ncndg sD

D

- this eq only holds for sufficiently low

2

kcs

cS ~3 kms a~02 nm D~1013 Hz

(= large wavelengths and the continuous approximation is valid) There is also an upper cut-off for the frequencies ( interatomic distances) the so-called Debye frequency D which depends on the density n

The Heat Capacity in Debyersquos Model

At high temperatures all the modes are excited (the number of phonons does not increase any more) and the heat capacity approaches the equipartition limit C=3NkB

At low temperatures we can choose the upper limit as (the high-frequency modes are not excited the energy is too low) How low should be T

333

45

5

16T

ch

k

dT

dUC

s

B

Thus Debyersquos model predicts that in the limit of sufficiently low T the heat capacity due to vibrations of the crystal lattice (in a metal electrons also contribute to C) must vary as T3 and not as 2exp(- hkBT) as in Einsteinrsquos model (Roughly speaking the number of phonons ~ T3 their average energy is proportional to T)

33

45

0

0

05

4

1exp s

B

ch

TkUd

h

ghUU

The low-T heat capacity

Kak

hcT

B

s 1000~1011041

10210661023

334

TNkUdAUU B

D

30

0

2

0

Debye TemperatureThe material-specific parameter is the sound speed If the temperature is properly normalized the data for different materials collapse onto a universal dependence

343

33

45

5

12

5

16

D

B

s

B TNkT

ch

kC

31

4

3

n

k

hc

B

sD

The normalization factor is called the Debye temperature

It is related to the maximum frequency D the Debye frequency

The higher the sound speed and the density of ions the higher the Debye temperature However the real phonon spectra are very complicated and D is better to treat as an experimental fitting parameter

DDBsD hTkn

c

31

4

3

Problem (blackbody radiation)The spectrum of Sun plotted as a function of energy peaks at a photon energy of 14 eV The spectrum for Sirius A plotted as a function of energy peaks at a photon energy of 24 eV The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun How does the radius of Sirius A compare to the Sunrsquos radius

2424

T

LRTRL

The temperature according to Wienrsquos law is proportional to the energy that corresponds to the peak of the photon distribution

671

4142

24

22

SunSirius

SunSirius

Sun

Sirius

TT

LL

R

R

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Debyersquos Theory of the Heat Capacity of Solids

If we quantize this elastic distortion field similar to the quantization of the e-m field we arrive at the concept of phonons the quanta of this elastic field For the thermal phonons the wavelength increases with decreasing T

Nobel 1936

These low-energy modes remain active at low temperatures when the high-frequency modes are already frozen out Large values of that correspond to these modes justify the use of a continuum model

mmKTTk

hc

B

s 1010~11041

1021066~1 7

34

334

There is a close analogy between photons and phonons both are ldquounconservedrdquo bosons Distinctions (a) the speed of propagation of phonons (~ the speed of sound waves) is by a factor of 105 less than that for light (b) sound waves can be longitudinal as well as transversal thus 3 polarizations (2 for photons) and (c) because of discreteness of matter there is an upper limit on the wavelength of phonons ndash the interatomic distance

Debyersquos model (1912) starts from the opposite point of view treating the solid as a continuum ie the atomic structure is ignored A continuum has vibrational modes of arbitrary low frequencies and at sufficiently low T only these low frequency modes are excited These low frequency normal modes are simply standing sound waves

s

B

chhTk cS ndash the sound velocity

Density of States in Debye Model

Each normal mode is a quantized harmonic oscillator The mean energy of each mode

1exp

1

2

1

hhE

dh

ghUdTEgU

DD

0

0

0 1exp

is the total energy per unit volume The U0 term comes from the zero-point motion of atoms It reduces the cohesive energy of the solid (the zero point motion in helium is sufficient to prevent solidification at any T at normal pressure) but since it does not depend on T it does not contribute to C Note that we ignored this term for phonons where it is In QED this unobservable term is swept under the rug by the process known as renormalization

2

3

63k

kG

3

2

2

6

13

scG

3

212

scd

dGg

For a macroscopic crystal the spectrum of sound waves is almost continuous and we can treat is a continuous variable As in the case of photons we start with the density of states per unit frequency g() The number of modes per unit volume with the wave number lt k

- multiplied by 3 since a sound wave in a solid can have three polarizations (two transverse and one longitudinal)

and

31

0 4

33

ncndg sD

D

- this eq only holds for sufficiently low

2

kcs

cS ~3 kms a~02 nm D~1013 Hz

(= large wavelengths and the continuous approximation is valid) There is also an upper cut-off for the frequencies ( interatomic distances) the so-called Debye frequency D which depends on the density n

The Heat Capacity in Debyersquos Model

At high temperatures all the modes are excited (the number of phonons does not increase any more) and the heat capacity approaches the equipartition limit C=3NkB

At low temperatures we can choose the upper limit as (the high-frequency modes are not excited the energy is too low) How low should be T

333

45

5

16T

ch

k

dT

dUC

s

B

Thus Debyersquos model predicts that in the limit of sufficiently low T the heat capacity due to vibrations of the crystal lattice (in a metal electrons also contribute to C) must vary as T3 and not as 2exp(- hkBT) as in Einsteinrsquos model (Roughly speaking the number of phonons ~ T3 their average energy is proportional to T)

33

45

0

0

05

4

1exp s

B

ch

TkUd

h

ghUU

The low-T heat capacity

Kak

hcT

B

s 1000~1011041

10210661023

334

TNkUdAUU B

D

30

0

2

0

Debye TemperatureThe material-specific parameter is the sound speed If the temperature is properly normalized the data for different materials collapse onto a universal dependence

343

33

45

5

12

5

16

D

B

s

B TNkT

ch

kC

31

4

3

n

k

hc

B

sD

The normalization factor is called the Debye temperature

It is related to the maximum frequency D the Debye frequency

The higher the sound speed and the density of ions the higher the Debye temperature However the real phonon spectra are very complicated and D is better to treat as an experimental fitting parameter

DDBsD hTkn

c

31

4

3

Problem (blackbody radiation)The spectrum of Sun plotted as a function of energy peaks at a photon energy of 14 eV The spectrum for Sirius A plotted as a function of energy peaks at a photon energy of 24 eV The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun How does the radius of Sirius A compare to the Sunrsquos radius

2424

T

LRTRL

The temperature according to Wienrsquos law is proportional to the energy that corresponds to the peak of the photon distribution

671

4142

24

22

SunSirius

SunSirius

Sun

Sirius

TT

LL

R

R

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Density of States in Debye Model

Each normal mode is a quantized harmonic oscillator The mean energy of each mode

1exp

1

2

1

hhE

dh

ghUdTEgU

DD

0

0

0 1exp

is the total energy per unit volume The U0 term comes from the zero-point motion of atoms It reduces the cohesive energy of the solid (the zero point motion in helium is sufficient to prevent solidification at any T at normal pressure) but since it does not depend on T it does not contribute to C Note that we ignored this term for phonons where it is In QED this unobservable term is swept under the rug by the process known as renormalization

2

3

63k

kG

3

2

2

6

13

scG

3

212

scd

dGg

For a macroscopic crystal the spectrum of sound waves is almost continuous and we can treat is a continuous variable As in the case of photons we start with the density of states per unit frequency g() The number of modes per unit volume with the wave number lt k

- multiplied by 3 since a sound wave in a solid can have three polarizations (two transverse and one longitudinal)

and

31

0 4

33

ncndg sD

D

- this eq only holds for sufficiently low

2

kcs

cS ~3 kms a~02 nm D~1013 Hz

(= large wavelengths and the continuous approximation is valid) There is also an upper cut-off for the frequencies ( interatomic distances) the so-called Debye frequency D which depends on the density n

The Heat Capacity in Debyersquos Model

At high temperatures all the modes are excited (the number of phonons does not increase any more) and the heat capacity approaches the equipartition limit C=3NkB

At low temperatures we can choose the upper limit as (the high-frequency modes are not excited the energy is too low) How low should be T

333

45

5

16T

ch

k

dT

dUC

s

B

Thus Debyersquos model predicts that in the limit of sufficiently low T the heat capacity due to vibrations of the crystal lattice (in a metal electrons also contribute to C) must vary as T3 and not as 2exp(- hkBT) as in Einsteinrsquos model (Roughly speaking the number of phonons ~ T3 their average energy is proportional to T)

33

45

0

0

05

4

1exp s

B

ch

TkUd

h

ghUU

The low-T heat capacity

Kak

hcT

B

s 1000~1011041

10210661023

334

TNkUdAUU B

D

30

0

2

0

Debye TemperatureThe material-specific parameter is the sound speed If the temperature is properly normalized the data for different materials collapse onto a universal dependence

343

33

45

5

12

5

16

D

B

s

B TNkT

ch

kC

31

4

3

n

k

hc

B

sD

The normalization factor is called the Debye temperature

It is related to the maximum frequency D the Debye frequency

The higher the sound speed and the density of ions the higher the Debye temperature However the real phonon spectra are very complicated and D is better to treat as an experimental fitting parameter

DDBsD hTkn

c

31

4

3

Problem (blackbody radiation)The spectrum of Sun plotted as a function of energy peaks at a photon energy of 14 eV The spectrum for Sirius A plotted as a function of energy peaks at a photon energy of 24 eV The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun How does the radius of Sirius A compare to the Sunrsquos radius

2424

T

LRTRL

The temperature according to Wienrsquos law is proportional to the energy that corresponds to the peak of the photon distribution

671

4142

24

22

SunSirius

SunSirius

Sun

Sirius

TT

LL

R

R

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

The Heat Capacity in Debyersquos Model

At high temperatures all the modes are excited (the number of phonons does not increase any more) and the heat capacity approaches the equipartition limit C=3NkB

At low temperatures we can choose the upper limit as (the high-frequency modes are not excited the energy is too low) How low should be T

333

45

5

16T

ch

k

dT

dUC

s

B

Thus Debyersquos model predicts that in the limit of sufficiently low T the heat capacity due to vibrations of the crystal lattice (in a metal electrons also contribute to C) must vary as T3 and not as 2exp(- hkBT) as in Einsteinrsquos model (Roughly speaking the number of phonons ~ T3 their average energy is proportional to T)

33

45

0

0

05

4

1exp s

B

ch

TkUd

h

ghUU

The low-T heat capacity

Kak

hcT

B

s 1000~1011041

10210661023

334

TNkUdAUU B

D

30

0

2

0

Debye TemperatureThe material-specific parameter is the sound speed If the temperature is properly normalized the data for different materials collapse onto a universal dependence

343

33

45

5

12

5

16

D

B

s

B TNkT

ch

kC

31

4

3

n

k

hc

B

sD

The normalization factor is called the Debye temperature

It is related to the maximum frequency D the Debye frequency

The higher the sound speed and the density of ions the higher the Debye temperature However the real phonon spectra are very complicated and D is better to treat as an experimental fitting parameter

DDBsD hTkn

c

31

4

3

Problem (blackbody radiation)The spectrum of Sun plotted as a function of energy peaks at a photon energy of 14 eV The spectrum for Sirius A plotted as a function of energy peaks at a photon energy of 24 eV The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun How does the radius of Sirius A compare to the Sunrsquos radius

2424

T

LRTRL

The temperature according to Wienrsquos law is proportional to the energy that corresponds to the peak of the photon distribution

671

4142

24

22

SunSirius

SunSirius

Sun

Sirius

TT

LL

R

R

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Debye TemperatureThe material-specific parameter is the sound speed If the temperature is properly normalized the data for different materials collapse onto a universal dependence

343

33

45

5

12

5

16

D

B

s

B TNkT

ch

kC

31

4

3

n

k

hc

B

sD

The normalization factor is called the Debye temperature

It is related to the maximum frequency D the Debye frequency

The higher the sound speed and the density of ions the higher the Debye temperature However the real phonon spectra are very complicated and D is better to treat as an experimental fitting parameter

DDBsD hTkn

c

31

4

3

Problem (blackbody radiation)The spectrum of Sun plotted as a function of energy peaks at a photon energy of 14 eV The spectrum for Sirius A plotted as a function of energy peaks at a photon energy of 24 eV The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun How does the radius of Sirius A compare to the Sunrsquos radius

2424

T

LRTRL

The temperature according to Wienrsquos law is proportional to the energy that corresponds to the peak of the photon distribution

671

4142

24

22

SunSirius

SunSirius

Sun

Sirius

TT

LL

R

R

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Problem (blackbody radiation)The spectrum of Sun plotted as a function of energy peaks at a photon energy of 14 eV The spectrum for Sirius A plotted as a function of energy peaks at a photon energy of 24 eV The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun How does the radius of Sirius A compare to the Sunrsquos radius

2424

T

LRTRL

The temperature according to Wienrsquos law is proportional to the energy that corresponds to the peak of the photon distribution

671

4142

24

22

SunSirius

SunSirius

Sun

Sirius

TT

LL

R

R

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Final 2006 (blackbody radiation)

The frequency peak in the spectral density of radiation for a certain distant star is at 17 x 1014 Hz The star is at a distance of 19 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 35x10-5 Wm2

a) (5) What is the surface temperature of the star b) (5) What is the total power emitted by 1 m2 of the surface of the starc) (5) What is the total electromagnetic power emitted by the star d) (5) What is the radius of the star

Kk

hT

B

30007210381

10711066

72 23

1434

(a)

(b)

(c)

26442484 106430001075 mWKmKWTJ

(d)

WmWmrJrpower 31252172 10611053109144

mmRJ

rJrRrJrRJR

SSSS

116

51722 1025

1064

1053109144

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If

not why(d) (15) What is approximately the number of CMBR photons hitting the earth per

second per square meter [ie photons(sm2)]

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

Jh 22max 10041

Jhc 22

max

1081

The maxima u(λT) and u(T) do not correspond to the same photon energy As increases the frequency range included in unit wavelength interval increases as 2 moving the peak to shorter wavelengths

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Problem 2006 (blackbody radiation) cont

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Problem (blackbody radiation)

The planet Jupiter is a distance of 778 x 1011 meters from the Sun and is of radius 715 x 107 meters Assume that Jupiter is a blackbody even though this is not entirely correct Recall that the solar power output of Sun is 4 x 1026 W

2

211

26

2652

107874

104

4mW

m

W

R

PJ

orbit

Sun

WmmWRJP JupiterJupiter172722 104810157652

424 JupiterJupiterJupiter TRP

KmKWm

W

R

PT

Jupiter

JupiterJupiter 123

10765101574

1048

4

41

24827

1741

2

a) What is the energy flux of the Suns radiation at the distance of Jupiters orbit

b) Recognizing that Suns energy falls on the geometric disk presented to the Sun by the planet what is the total power incident on Jupiter from the Sun

c) Jupiter rotates at a rather rapid rate (one revolution per 04 earth days) and therefore all portions of the planet absorb energy from the Sun Hence all portions of the surface of this planet radiate energy outward On the basis of this information find the surface temperature of Jupiter

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

d) Estimates of the surface temperature of Jupiter indicate that it is clearly above 500 Kelvins On the basis of your answer to c) what might you conclude about this planet

There must be some other source of energy to produce a surface temperature of 500 K

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Problem (blackbody radiation)

Tungsten has an emissivity of 03 at high temperatures Tungsten filaments operate at a temperature of 4800 K

a) At what frequency does a tungsten filament radiate the most energy b) What is the powerunit surface emitted by the tungsten filament c) If the surface area of a tungsten filament is 001 cm2 what is the power output of

the bulb d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb e) What fraction of radiation incident on a tungsten filament is reflected

[Answers a) 288 x 1014 Hz b) 903 x106 Wm2 c) 903 watts d) 0180 watts and e) 70]

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Problem (radiation pressure)

(a) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )

(b) At what temperature will the pressure of the photon gas be equal to 10-5 Pa(c) The temperature at the Sunrsquos center is 107 K What is the pressure of the

radiation Compare it to the pressure of the gas at the center of the Sun which is 1011 bar

(d) Calculate the pressure of the Sunrsquos radiation on the Earthrsquos surface given that the power of the radiation reaching earth from the Sun is 1400 Wm2

414

4

3

3

4

cP

TTc

P P=105 Pa KT 5

41

8

58

104110754

1011033

P=10-5 Pa KT 45010754

101103341

8

58

(a)

(b)

(c) barPaP 712

8

478

105210521033

1010754

- at this T the pressure of radiation is still negligible in the balance for mechanical equilibrium

(d)

Pac

IPI

cTuTuP 6

8106

1033

14004

3

44

3

1

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15

Problem (blackbody radiation)

The black body temperature is 3000K Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation

Tk

hc

B5max

d

Tkhc

hcdTu

B

1exp

18

5

dhc

Tkcd

hc

TkhccdTu

cdJ BB

15exp

152

15exp

158

4

4 4

5

5

5

near the maximum

2 Wm394834

5238

101310115exp

1

1031066

30001038151032

dJ

the power emitted by a unit area in all directions dTuc

Jd 4

• Lecture 27 Debye Model of Solids Phonon Gas
• Einsteinrsquos Model of a Solid
• Debyersquos Theory of the Heat Capacity of Solids
• Density of States in Debye Model
• The Heat Capacity in Debyersquos Model
• Debye Temperature
• Problem (blackbody radiation)
• Final 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation) cont
• Slide 11
• Slide 12
• Slide 13
• Problem (radiation pressure)
• Slide 15