Blackbody.ppt

ECE 455: Optical Electronics
Lecture #8:
Blackbody Radiation, Einstein Coefficients, and
Homogeneous Broadening
Substitute Lecturer: Jason Readle
Thurs, Sept 17th, 2009

Topic #1:
Blackbody Radiation

What is a Blackbody?
• Ideal blackbody: Perfect absorber
– Appears black when cold!
• Emits a temperature-dependent light spectrum

Blackbody Energy Density
• The photon energy density for a blackbody radiator in the
ν → ν + dν spectral interval is
 
 
3 1
3
3
8
( ) 1
units are J cm
h
kT
h
d e d
c

 
   


 


Blackbody Intensity
• The intensity emitted by a blackbody surface is
 
3 1
2
( ) ( )
8
1
h
kT
d d c
h
e d
c

     
 


 
 
(Units are or J/s-cm2 or W/cm2)

Blackbody Peak Wavelength
• The peak wavelength for emission by a blackbody is
k
7
MAX T
Å
10
998
.
2
kT
965
.
4
hc 



where 1 Å = 10–8 cm

Example – The Sun
• Peak emission from the sun is near 570 nm and so it
appears yellow
– What is the temperature of this blackbody?
– Calculate the emission intensity in a 10 nm region centered at 570
nm.
k
7
MAX
T
Å
10
998
.
2
nm
570
~




Tk = 5260 K

Example – The Sun
• Also
10 1
14
0 7
0
3 10
5.26 10
570 10
c cm s
Hz
cm




 
   

570 nm → 17,544 cm–1
eV
18
.
2
~
h 0 

40
1

!
K
000
,
12
eV
1 

kT (300 K) eV

Example – The Sun
























 7
7
10
2
1
10
575
1
10
565
1
10
3
1
1
c
or
 = 9.23 · 1012 s–1
= 9.23 THz

Example – The Sun
  












 1
3
3
1
e
c
h
8
~
d
)
( kT
h
  
 









3
1
8
3
14
34
s
m
10
3
10
26
.
5
10
6
.
6
8 1
12
1
s
10
23
.
9
1
435
.
0
18
.
2
exp 
















3
m
J
3
10
5
.
5
~
d
)
( 






Example – The Sun
Since hν = 2.18 eV
= 3.49 · 10–19 J
→ ρ(ν) d ν / hν = 1.58 · 1010
3
cm
photons
3
m
J
3
10
5
.
5
~
d
)
( 






Example – The Sun
Remember,
Intensity = Photon Density · c
or
 = 4.7 · 1020 photons-cm–2-s–1
= 164 W-cm–2
(ν)dν = ρ(ν)dν c
 

Example – The Sun

Topic #2:
Einstein Coefficients

Absorption
• Spontaneous event in
which an atom or
molecule absorbs a
photon from an incident
optical field
• The asborption of the
photon causes the atom or
molecule to transition to
an excited state

Spontaneous Emission
• Statistical process (random phase) – emission by an
isolated atom or molecule
• Emission into 4π steradians

Stimulated Emission
• Same phase as “stimulating” optical field
• Same polarization
• Same direction of propagation
E2
h
E1
2h

Putting it all together…
• Assume that we have a two state system in equilibrium
with a blackbody radiation field.
E2
E1
Stimulated emission
Absorption
Spontaneous
emission

• For two energy levels 1 (lower) and 2 (upper) we have
– A21 (s-1), spontaneous emission coefficient
– B21 (sr·m2·J-1·s-1), stimulated emission coefficient
– B12 (sr·m2·J-1·s-1), absorption coefficient
• Bij is the coefficient for stimulated emission or absorption
between states i and j

Two Level System In The Steady State…
• The time rate of change of N2 is given by:








stimulated
21
2
s
spontaneou
21
2
2 )
(
B
N
A
N
dt
dN




 0
)
(
B
N
absorption
12
1 








Remember, ρ(ν) has units of J-cm–3-Hz–1

Solving for Relative State Populations
• Solving for N2/N1:
kT
/
h
1
2
21
21
12
1
2
e
g
g
)
(
B
A
)
(
B
N
N

















stimulated
21
2
s
spontaneou
21
2
2 )
(
B
N
A
N
dt
dN




 0
)
(
B
N
absorption
12
1 









Solving for Relative State Populations
 
1
e
g
B
g
B
1
B
A
)
(
kT
/
h
2
21
1
12
21
21






But… we already know that, for a blackbody,
1
e
1
c
h
8
)
( T
/k
h
3
3





 

• In order for these two expressions for ρ(ν) to be equal,
Einstein said:
and
h
8
A
h
n
8
c
A
B
3
21
3
3
3
21
21 





1 2
12 21
B g =B g

Example – Blackbody Source
• Suppose that we have an ensemble of atoms in State 2
(upper state). The lifetime of State 2 is
• This ensemble is placed 10 cm from a spherical blackbody
having a “color temperature” of 5000 K and having a
diameter of 6 cm
• What is the rate of stimulated emission?
1
21
A

Blackbody
6 cm
Atomic
Ensemble

E2 = 3.2 eV
h
E1 = 0
hν = 3.2 eV
  = 387.5 nm
 = 7.7 · 1014 s–1

 










d
1
e
c
c
h
8
d
)
(
kT
/
h
3
3
0
• Blackbody emission at the surface of the emitter is





 









1
e
)
10
3
(
10
)
10
7
.
7
(
10
63
.
6
8
413
.
0
eV
2
.
3
2
10
8
3
14
34
kT : 5000 K

• Assuming dν = Δν = 100 MHz,
• At the ensemble, the photon flux from the 5000 K
blackbody is:
0(ν)dν = 3.7 · 10–5 J-cm–2-s–1
 7.2 · 1013 photons-cm–2-s–1
at 387.5 nm
2
0
cm
10
cm
3
d
)
( 








 = 6.48 · 1012 photons-cm–2-s–1

And
J/eV
10
6
.
1
10
3
10
48
.
6
c
d
)
(
d
)
(
19
10
12













or
ρ(ν)dν = 3.46 · 10–17 J-cm–3

34
3
7
1
6
3
21
21
10
67
.
6
8
)
10
5
.
387
(
s
10
h
8
A
B












= 3.5 · 1024 cm3-J–1-s–2
• The stimulated emission coefficient B21 is

 























d
c
d
)
(
B
d
d
)
(
B
)
(
B
dt
dN
N
1
21
21
21
2
2
16
1
8 3
1.56 10 J-s
5.4s
10 cm



 
= – 3.5 · 1024 cm3-J–1-s–2
• Finally, the stimulated emission rate is given by

To reiterate…
This is negligible compared to the
spontaneous emission rate of
A21 = 106 s–1 !
!
s
4
.
5
dt
dN
N
1 1
2
2



Example – Laser Source
• Let us suppose that we have the same conditions as
before, EXCEPT a laser photo-excites the two level
system:
Laser
3.2 eV 2
1
1 mm
0
A21 =
106
s-1
Let Δνlaser = 108 s–1 (100 MHz, as before).

• If the power emitted by the laser is 1 W, then
– Power flux, P 2
)
cm
05
.
0
(
W
1


= 127.3 W-cm–2
Since hν = 3.2 eV
= 5.1 · 10–19 J
→ P = 2.5 · 1020 photons-cm–2-s–1

1
10
1
8
2
laser
s
-
cm
10
3
s
10
cm
-
W
3
.
127
c
P
)
(












= 4.24 · 10–17 J-cm–3-Hz–1
= 83.3 photons-cm–3-Hz–1





 )
(
B
dt
dN
N
1
21
2
2
3.5 · 1024 cm3-J–1-s–2 · 4.24 · 10–17 J-cm–3-s
= 1.48 · 108 s–1
8 1
2
6 1
Stimulated Emission Rate 1.48 10
10 !
Spontaneous Emission Rate 10
s
s



 

• Remember, in the case of the blackbody optical source:
• What made the difference?
!
10
10
4
.
5
Rate
s
Spontaneou
Rate
Stimulated 5
6




Source Comparison
Total power radiated by 5000 K blackbody with R = 0.5 cm is 11.1 kW
Laser
5000 K
Blackbody
570 (nm)
1
2 E
E
hc
nm
5
.
387



Key Points
• Moral: Despite its lower power, the laser delivers
considerably more power into the 1 → 2 atomic transition.
• Point #2: To put the maximum intensity of the blackbody
at 387.5 nm requires T  7500 K!
• Point #3: Effective use of a blackbody requires a process
having a broad absorption width

Ex. Photodissociation
ABS. C3F7I
~280 nm (nm)
I*
1.315 µm
I
C3F7I + hν → I*

Bandwidth
• In the examples, bandwidth Δν is very important
– Δν is the spectral interval over which the atom (or
molecule) and the optical field interact.

Topic #3:
Homogeneous Line Broadening

Semi-Classical Conclusion
2
1
E2
Absorption
E1
1
2 E
E
hc



This diagram:
suggests that the atom absorbs only (exactly) at

The Shocking Truth!
Reality

1
2 E
E
hc


Line Broadening
• The fact that atoms absorb over a spectral range is due to
Line Broadening
• We introduce the “lineshape” or “lineshape function” g(ν)

 =
FWHM
0

Lineshape Function
• g(ν) dν is the probability that the atom will emit (or
absorb) a photon in the ν → ν + dν frequency interval.
• g(ν) is a probability distribution
and Δν / ν0 << 1






0
1
d
)
(
g

Types of Line Broadening
• There are two general classification of line broadening:
– Homogenous — all atoms behave the same way (i.e.,
each effectively has the same g(ν).
– Inhomogeneous — each atom or molecule has a
different g(ν) due to its environment.

• In the homogenous case, we observe a Lorentzian
Lineshape
where ν0 ≡ line center
2
2
0 )
/2
(
)
(
)
/2
(
1/
)
(
g












Δν = FWHM
Bottom line: Homogeneous → Lorentzian
1
0
2
)
(
g 









Sources of Homogeneous Broadening
• Natural Broadening — any state with a finite lifetime τ sp
(τsp ≠ ∞) must have a spread in energy:
• Collisional Broadening — phase randomizing collisions

Natural Broadening
• ΔE Δt ≥  Heisenberg’s Uncertainty Principle
2
u
1
l

El
E
Eu



Natural Broadening
• In the case of an atomic system:
2
1
1
2


1
1


 
1
2
1
1
2
1 










Natural Broadening
• In general
 





i
1
i
2
1
Lifetime of
upper or lower states resulting
from all processes.

Example: Sodium (Na)
3p 2
P3/2
3p 2
P1/2
3s 2
S1/2 (Ground)
588.9
nm
589.6
nm
(Both arrows indicate “resonance” transitions)

Example: Sodium (Na)
• Radiative lifetime of the 3p 2P3/2 state is 16 ns
}
0
10
25
.
6
{
2
1
1
1
2
1
7
lower
upper

















= 9.9 · 106 s–1 ≈ 10 MHz
0



~ 2 · 10–8!
ν0 = 5.1 · 1014 Hz 

Example: Mercury (Hg)
63
S1
404.7 nm
546.1
nm 435.8 nm
3
P2
3
P1
3
P0
253.7 nm
1
S0 (Ground)

Example: Mercury (Hg)
• Remember:
A43
A42
A41
4
3
2
1
  1
41
42
43
sp
4 }
A
A
A
{ 




In general, 


j
ij
1
i A

Collisional Broadening
• An atom that radiates a
photon can be described
as a classical oscillator
with a particular phase
t
Fourier
Spectrum
ß

)
( 0




h
E
E 1
2
0




• Suppose now that we have collisions between atom A (the
radiator) and a second atom, B…
A
B

• Such collisions alter the phase of the oscillator.
t
(Arrows indicate points at which oscillator suffers collision)

• Result? Broadening of Transition!
• The rate of phase randomizing collisions is:
COL
C
C N
k
1
RATE






collisions
where:
kC (cm3 – s–1) is known as the rate constant of
collisional quenching (deactivation
of the excited atom)
NC (cm-3) is the number density of colliding atoms

)
2
(
2
1
coll
collision 




Collision
perturbs both
upper & lower
states
Collision
freq.
coll
coll
coll
1









~ Ncoll ~ pressure

Total Homogenous Broadening
• Is calculated by summing the rates of the various
homogeneous broadening processes:
 

 









i
1
i
coll
i
1
i
total
1
2
1

Example – KrF Laser
• KrF laser (λ = 248.4 nm)
• τsp = 5 ns
• kC = 2 · 10–10 cm3-s–1
• 1 atmosphere ≡ 2.45 · 1019 cm–3
1
coll
1
sp
total
1
2
1 











8 1
spontaneous
emission
2 10
2
total
s




  
atmosphere
cm
10
45
.
2
s
cm
10
2
3
19
1
3
10


 


Δνtotal = 31.8 MHz +

GHz
9
.
4
· P(atm)

Δνtotal = 31.9 MHz + 1.6 GHz · P
spontaneous collisions
Note that these terms are equal for P = 0.02 atm!

Next Time
• Inhomogeneous broadening
• Threshold gain

Blackbody.ppt

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