Lecture 25. Blackbody Radiation (Ch. 7)
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Transcript Lecture 25. Blackbody Radiation (Ch. 7)
Lecture 4a. Blackbody Radiation
Energy Spectrum of Blackbody Radiation
- Rayleigh-Jeans Law
- Wien’s Law
- Stefan-Boltzmann Law
Energy Spectrum of Blackbody Radiation
The average energy of photons with frequency u , T d h g f , T d
S
between and +d (per unit volume):
h g
3D
ph
f , T
3
uS , T
photon
energy
=
8 h
3
us , T h g f , T 3
c exp h / kBT 1
u as a function of the energy:
average number
of photons within
this freq. range
u , T d u , T d
- the spectral density of the
black-body radiation
(the Plank’s radiation law)
u , T u , T
8
u , T
hc3
d
u h , T h
d
3
1
exp
k BT
u(,T) - the energy density per unit photon
energy for a photon gas in equilibrium with
a blackbody at temperature T.
Classical Limit (small , large ), Rayleigh-Jeans Law
At low frequencies or high temperatures:
8 h
us , T 3
c
3
h
exp
1
kBT
Rayleigh-Jeans Law
This equation predicts the socalled ultraviolet catastrophe
– an infinite amount of energy
being radiated by a black body
at high frequencies or short
wavelengths.
8 2
3 kBT
c
h
h
h
1 exp
1
kBT
k
T
kBT
B
- purely classical result (no h), can be
obtained directly from the equipartition
theorem
Rayleigh-Jeans Law (cont’d)
u as a function of the wavelength:
3
u , T d u , T d
c
h
d
hc
8
1
hc 8 hc
u
,
T
d
2
hc3 exp hc 1 2 5 exp hc 1
k T
k T
B
B
In the classical limit of large :
u , T large
8 k BT
4
1
4
High Limit, Wien’s Displacement Law
At high frequencies/low temperatures:
h
8 h 3
us , T 3 exp
c
k
T
B
h
h
h
1 exp
1
exp
kBT
k
T
k
T
B
B
Nobel 1911
The maximum of u() shifts toward higher frequencies with increasing temperature. The
position of maximum:
3
h
3x 2
k
T
du
d
x 3e x
B
const
const x
0
2
x
d
e
1
h
h
e 1
exp
1
d
k BT
k BT
max 2.8
k BT
h
3 x e x 3
x 2.8
h max
2.8
k BT
u(,T)
Wien’s displacement law
- discovered experimentally
by Wilhelm Wien
- the “most likely” frequency of a photon in a
blackbody radiation with temperature T
Numerous applications
(e.g., non-contact radiation thermometry)
Solar Radiation
The surface temperature of the Sun - 5,800K.
max
As a function of energy, the spectrum of
sunlight peaks at a photon energy of
hc
0.5 m
5 k BT
umax h max 2.8kBT 1.4 eV
- close to the energy gap in Si, ~1.1 eV, which has been so far the best material for solar cells
Spectral sensitivity of human eye:
Stefan-Boltzmann Law of Radiation
The (average) photon density:
n
0
8
f g d 3
c
2
h
8 k BT
d
0
c3 h
exp
1
k
T
B
The total energy of photons per unit volume :
(the energy density of a photon gas)
4 4
u T
T
c
3
x 2 dx
kB 3
8
0 e x 1 hc T 2.4
- increases as T 3
u T g f d
0
2 5 k B
15h 3c 2
4
the StefanBoltzmann Law
3
8 5 k BT
15 hc
the Stefan-Boltzmann
constant
3
5.7 108 W / K 4 m 2
The average energy per photon:
u T
8 5 k BT hc
4
k BT 2.7 k BT
3
3
N
15hc 8 k BT 2.4 15 2.4
4
4
3
(just slightly less than the “most” probable energy)
Power Emitted by a Black Body
For the “uni-directional” motion, the flux of energy per unit area
c u
T
energy density u
c 1s
Integration over all angles
provides a factor of ¼:
1m2
power emitted by unit area
1
cu
4
(the hole size must be >> the wavelength)
Thus, the power emitted by a unit-area
surface at temperature T in all directions:
c
c 4 4
J u T
T T4
4
4 c
The total power emitted by a black-body sphere of radius R:
4R 2 T 4
Some
4
2
Consider a black body at 310K. The power emitted by the body: T 500W / m
numbers:
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).
Sun’s Mass Loss
Beiser 9.22. The Sun’s mass is 2 ·1030 kg, its radius is 7·108 m, and its surface
temperature is 5,800K. 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?
P power emittedby a sphere 4R T
2
2 5 k B 4
W
8
5.8
10
15h3c 2
m2 K 4
4
This result is consistent with the flux of the solar radiation energy received by the Earth
(1370 W/m2) being multiplied by the area of a sphere with radius 1.5·1011 m (Sun-Earth
distance).
4
P 4 RSun
2
2
hc
W
4
8
8
26
4
7
10
m
5.7
10
5,740K
3.8
10
W
2 4
mK
2.8 k B max
the mass loss per one second
1% of Sun’s mass will be lost in
dm P 3.8 1026 W
2
4.2 109 kg/s
2
dt c
3 108 m
0.01M
2 1028 kg
18
11
t
4
.
7
10
s
1.5
10
yr
9
dm / dt 4.2 10 kg/s
The Greenhouse Effect
Absorption:
2
Power out 4 RE TE
4
2
4 R
Power in RE TSun Sun
Rorbit
the flux of the solar radiation energy
received by the Earth ~ 1370 W/m2
Emission:
2
R
TE Sun
4 Rorbit
Rorbit = 1.5·1011 m
Transmittance of the Earth atmosphere
= 1 – TEarth = 280K
In reality
= 0.7 – 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:
2 1/ 4
TSun
RSun = 7·108 m
Problem
The cosmic microwave background radiation (CMBR) has a temperature of
approximately 2.7 K.
(a) What wavelength λmax (in m) corresponds to the maximum spectral density u(λ,T) of
the cosmic background radiation?
(b) What is approximately the number of CMBR photons hitting the earth per second
per square meter [i.e. photons/(s·m2)]?
(a)
(b)
hc
6.6 1034 3 108
3
max
1
.
1
10
m 1.1mm
23
5 k BT 5 1.3810 2.7
hc
max
1.1 m eV
J TCMBR 5.7 108 W / K 4 m2 2.7 K 4 3 106W / m2
4
The average energy per photon:
4
2.7 kBT
W
J 2
3 106
m
photons
16 photons
N
3
10
2
J
2.7 1.3810 23 2.7
s m2
sm
Problem
The frequency peak in the spectral density of radiation for a certain distant star
is at 1.7 x 1014 Hz. The star is at a distance of 1.9 x 1017 m away from earth
and the energy flux of its radiation as measured on earth is 3.5x10-5 W/m2.
a)
b)
c)
d)
(a)
What is the surface temperature of the star?
What is the total power emitted by 1 m2 of the surface of the star?
What is the total electromagnetic power emitted by the star?
What is the radius of the star?
h
6.6 1034 1.7 1014
T
3000K
23
2.7 k B
1.3810 2.7
(b)
J T 4 5.7 108 W / K 4 m2 3000 K 4 4.6 106W / m2
(c)
power 4 r 2 J r 4 1.9 1017 m2 3.5 105W / m2 1.6 1031W
(d) 4 R J RS 4 r J r
2
S
2
4
J r
3.5 10 5
17
RS r
1.9 10 m
5.2 1011 m
6
J RS
4.6 10
Problem
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 5.8 x 1010 m from the Sun, and has a radius of 2.44
x 106 m. The radius of the Sun is 7·108 m and its total power output is 4 x 1026 W. In this
problem treat the planet as if it were a black body
a)
b)
c)
d)
e)
(a)
(b)
(c)
What is the energy flux of the Sun’s radiation at Mercury's orbit?
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.]
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 "hot“ side of
Mercury is uniform, find this temperature.
What is the peak frequency of the radiation absorbed by Mercury?
What is the peak frequency of the radiation emitted by Mercury?
PSun
4 1026 W
3
2
J
9
.
46
10
W
/
m
2
2
4Rorbit
4 5.8 1010 m
PMercury J R
2
Mercury
2
4
PMercury 2RMercury
TMercury
2
6
2
- hemi-sphere
1/ 4
TMercury
9.46 10 W / m 2.44 10 m 1.77 1017 W
3
PMercury
2R 2
Mercury
1/ 4
1.77 10 W
2
6
8
4 2
2 2.4410 m 5.7610 W / K m
17
535K
Problem (cont’d)
(d)
PSun
TSun
2
4RSun
(e)
received
max
emitted
max
1/ 4
1/ 4
26
4
10
W
2
4 7 108 m 5.76108W / K 4 m2
5,795K
k BTSun
1.381023 J / K 5,795K
2.8
2.8
3.4 1014 Hz
34
h
6.62 10 Js
2.8
k BTMercury
h
1.381023 J / K 535K
13
2.8
3
.
1
10
Hz
34
6.6210 Js