Lecture 4a Blackbody Radiation

Energy Spectrum of Blackbody Radiation

- Rayleigh-Jeans Law

- Wienrsquos Law

- Stefan-Boltzmann Law

Energy Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

The average energy of photons with frequency between and +d (per unit volume)

u(T) - the energy density per unit photon energy for a photon gas in equilibrium with

a blackbody at temperature T

Su T d h g f T d

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)

3

3

8

exp 1sB

hu T h g f T

c h k T

hThud

dTuTudTudTu

u as a function of the energy

=

average number of photons within this freq range

photon energy 3 3D

phh g

f T Su T

Classical Limit (small large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated by a black body at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures 1 exp 1B B B

h h h

k T k T k T

3 2

3 3

8 8

exp 1s B

B

hu T k T

c chk T

- purely classical result (no h) can be obtained directly from the equipartition

theorem

Rayleigh-Jeans Law (contrsquod)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High Limit Wienrsquos Displacement Law

The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures 1 exp 1 expB B B

h h h

k T k T k T

33

8 exps

B

h hu T

c k T

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The (average) photon density

The total energy of photons per unit volume (the energy density of a photon gas)

45

30

8

15Bk Tu T g f dhc

the Stefan-Boltzmann Law 23

45

15

2

ch

kB the Stefan-Boltzmann

constant 44T

cTu

3 32 2

33 3

0 0 0

8 88 24

1exp 1

B Bx

B

k T kx dxn f g d d T

c c h e hchk T

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

8 4 257 10 W K m

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Energy Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

The average energy of photons with frequency between and +d (per unit volume)

u(T) - the energy density per unit photon energy for a photon gas in equilibrium with

a blackbody at temperature T

Su T d h g f T d

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)

3

3

8

exp 1sB

hu T h g f T

c h k T

hThud

dTuTudTudTu

u as a function of the energy

=

average number of photons within this freq range

photon energy 3 3D

phh g

f T Su T

Classical Limit (small large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated by a black body at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures 1 exp 1B B B

h h h

k T k T k T

3 2

3 3

8 8

exp 1s B

B

hu T k T

c chk T

- purely classical result (no h) can be obtained directly from the equipartition

theorem

Rayleigh-Jeans Law (contrsquod)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High Limit Wienrsquos Displacement Law

The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures 1 exp 1 expB B B

h h h

k T k T k T

33

8 exps

B

h hu T

c k T

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The (average) photon density

The total energy of photons per unit volume (the energy density of a photon gas)

45

30

8

15Bk Tu T g f dhc

the Stefan-Boltzmann Law 23

45

15

2

ch

kB the Stefan-Boltzmann

constant 44T

cTu

3 32 2

33 3

0 0 0

8 88 24

1exp 1

B Bx

B

k T kx dxn f g d d T

c c h e hchk T

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

8 4 257 10 W K m

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Classical Limit (small large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated by a black body at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures 1 exp 1B B B

h h h

k T k T k T

3 2

3 3

8 8

exp 1s B

B

hu T k T

c chk T

- purely classical result (no h) can be obtained directly from the equipartition

theorem

Rayleigh-Jeans Law (contrsquod)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High Limit Wienrsquos Displacement Law

The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures 1 exp 1 expB B B

h h h

k T k T k T

33

8 exps

B

h hu T

c k T

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The (average) photon density

The total energy of photons per unit volume (the energy density of a photon gas)

45

30

8

15Bk Tu T g f dhc

the Stefan-Boltzmann Law 23

45

15

2

ch

kB the Stefan-Boltzmann

constant 44T

cTu

3 32 2

33 3

0 0 0

8 88 24

1exp 1

B Bx

B

k T kx dxn f g d d T

c c h e hchk T

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

8 4 257 10 W K m

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Rayleigh-Jeans Law (contrsquod)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High Limit Wienrsquos Displacement Law

The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures 1 exp 1 expB B B

h h h

k T k T k T

33

8 exps

B

h hu T

c k T

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The (average) photon density

The total energy of photons per unit volume (the energy density of a photon gas)

45

30

8

15Bk Tu T g f dhc

the Stefan-Boltzmann Law 23

45

15

2

ch

kB the Stefan-Boltzmann

constant 44T

cTu

3 32 2

33 3

0 0 0

8 88 24

1exp 1

B Bx

B

k T kx dxn f g d d T

c c h e hchk T

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

8 4 257 10 W K m

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

High Limit Wienrsquos Displacement Law

The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures 1 exp 1 expB B B

h h h

k T k T k T

33

8 exps

B

h hu T

c k T

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The (average) photon density

The total energy of photons per unit volume (the energy density of a photon gas)

45

30

8

15Bk Tu T g f dhc

the Stefan-Boltzmann Law 23

45

15

2

ch

kB the Stefan-Boltzmann

constant 44T

cTu

3 32 2

33 3

0 0 0

8 88 24

1exp 1

B Bx

B

k T kx dxn f g d d T

c c h e hchk T

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

8 4 257 10 W K m

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The (average) photon density

The total energy of photons per unit volume (the energy density of a photon gas)

45

30

8

15Bk Tu T g f dhc

the Stefan-Boltzmann Law 23

45

15

2

ch

kB the Stefan-Boltzmann

constant 44T

cTu

3 32 2

33 3

0 0 0

8 88 24

1exp 1

B Bx

B

k T kx dxn f g d d T

c c h e hchk T

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

8 4 257 10 W K m

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Stefan-Boltzmann Law of Radiation

The (average) photon density

The total energy of photons per unit volume (the energy density of a photon gas)

45

30

8

15Bk Tu T g f dhc

the Stefan-Boltzmann Law 23

45

15

2

ch

kB the Stefan-Boltzmann

constant 44T

cTu

3 32 2

33 3

0 0 0

8 88 24

1exp 1

B Bx

B

k T kx dxn f g d d T

c c h e hchk T

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

8 4 257 10 W K m

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions 4 4c c 4

4 4J u T T T

c

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Consider a black body at 310K The power emitted by the body 24 500 mWT

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Sunrsquos Mass LossBeiser 922 The Sunrsquos mass is 2 middot1030 kg its radius is 7middot108 m and its surface temperature is 5800K Find the mass loss for the Sun in one second How many years are needed for the Sun to lose 1 of its mass by radiation

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

5 48

3 2 2 4

2 W58 10

15 m KBk

h c

424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 57 10 5740K 38 10 W

28 m KSunB

hcP R

k

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024

kg 102

010 11189

28

dtdm

Mt

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

ProblemThe cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K

(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λT) of the cosmic background radiation

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a) meVhc

11max

(b) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

27 Bk T

(b) 26442484 103721075 mWKmKWTJ CMBR

The average energy per photon

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Problem

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 earth and the energy flux of its radiation as measured on earth is 35x10-5 Wm2

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

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Problem

a) What is the energy flux of the Sunrsquos radiation at Mercurys orbit b) What is the total power absorbed by Mercury [Hint Consider that it appears

as a flat disk to the Sun and it absorbs all of the incident radiation] c) If Mercury is in thermodynamic equilibrium it will emit the same total power as

it receives from the Sun Assuming that the temperature of the hotldquo side of Mercury is uniform find this temperature

d) What is the peak frequency of the radiation absorbed by Mercury e) What is the peak frequency of the radiation emitted by Mercury

(a)

23210

26

210469

10854

104

4mW

m

W

R

PJ

orbit

Sun

(b) WmmWRJP MercuryMercury1726232 107711044210469

(c)422 MercuryMercuryMercury TRP

KmKWm

W

R

PT

Mercury

MercuryMercury 535

10765104422

10771

2

41

24826

1741

2

- hemi-sphere

Planet Mercury revolves and rotates at the same rate so one side of the planet is always facing the Sun Mercury is a distance of 58 x 1010 m from the Sun and has a radius of 244 x 106 m The radius of the Sun is 7middot108 m and its total power output is 4 x 1026 W In this problem treat the planet as if it were a black body

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)

Problem (contrsquod)

KmKWm

W

R

PT

Sun

SunSun 7955

107651074

104

4

41

24828

2641

2

(d)

HzJs

KKJ

h

Tk SunBreceived 1434

23

max 104310626

7955103818282

(e)Hz

Js

KKJ

h

Tk MercuryBemitted 1334

23

max 101310626

535103818282

• Lecture 4a Blackbody Radiation
• Energy Spectrum of Blackbody Radiation
• Classical Limit (small large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (contrsquod)
• High Limit Wienrsquos Displacement Law
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• The Greenhouse Effect
• Problem
• Problem (2)
• Problem (3)
• Problem (contrsquod)