ece 455: optical electronics lecture #8: blackbody radiation, einstein coefficients, and homogeneous...
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ECE 455: Optical Electronics
Lecture #8:
Blackbody Radiation, Einstein Coefficients, and Homogeneous Broadening
Substitute Lecturer: Jason Readle
Thurs, Sept 17th, 2009
ECE 455: Optical Electronics
Topic #1: Blackbody Radiation
ECE 455: Optical Electronics
What is a Blackbody?
• Ideal blackbody: Perfect absorber– Appears black when cold!
• Emits a temperature-dependent light spectrum
ECE 455: Optical Electronics
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
hkT
hd e d
c
ECE 455: Optical Electronics
Blackbody Intensity
• The intensity emitted by a blackbody surface is
3 1
2
( ) ( )
81
hkT
d d c
he d
c
(Units are or J/s-cm2 or W/cm2)
ECE 455: Optical Electronics
Blackbody Peak Wavelength
• The peak wavelength for emission by a blackbody is
k
7
MAX TÅ10998.2
kT965.4hc
where 1 Å = 10–8 cm
ECE 455: Optical Electronics
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 TÅ10998.2
nm570~
Tk = 5260 K
ECE 455: Optical Electronics
Example – The Sun
• Also10 1
140 7
0
3 105.26 10
570 10
c cm sHz
cm
570 nm → 17,544 cm–1
eV18.2~h 0 401
!K000,12eV1
kT (300 K) eV
ECE 455: Optical Electronics
Example – The Sun
7710
21
10575
1
10565
1103
11c
or = 9.23 · 1012 s–1
= 9.23 THz
ECE 455: Optical Electronics
Example – The Sun
1
3
3
1ec
h8~d)( kTh
318
31434
sm103
1026.5106.68 1121
s1023.91435.018.2
exp
3mJ3105.5~d)(
ECE 455: Optical Electronics
Example – The Sun
Since hν = 2.18 eV = 3.49 · 10–19 J
→ ρ(ν) d ν / hν = 1.58 · 1010 3cm
photons
3mJ3105.5~d)(
ECE 455: Optical Electronics
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
ECE 455: Optical Electronics
Example – The Sun
ECE 455: Optical Electronics
Topic #2: Einstein Coefficients
ECE 455: Optical Electronics
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
ECE 455: Optical Electronics
Spontaneous Emission
• Statistical process (random phase) – emission by an isolated atom or molecule
• Emission into 4π steradians
ECE 455: Optical Electronics
Stimulated Emission
• Same phase as “stimulating” optical field
• Same polarization
• Same direction of propagation
E2
h
E1
2h
ECE 455: Optical Electronics
Putting it all together…
• Assume that we have a two state system in equilibrium with a blackbody radiation field.
E2
E1
Stimulated emission
AbsorptionSpontaneous
emission
ECE 455: Optical Electronics
Einstein Coefficients
• 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
ECE 455: Optical Electronics
Two Level System In The Steady State…
• The time rate of change of N2 is given by:
stimulated
212
sspontaneou
2122 )(BNAN
dtdN 0)(BN
absorption
121
Remember, ρ(ν) has units of J-cm–3-Hz–1
ECE 455: Optical Electronics
Solving for Relative State Populations
• Solving for N2/N1:
kT/h
1
2
2121
12
1
2
egg
)(BA)(B
NN
stimulated
212
sspontaneou
2122 )(BNAN
dtdN 0)(BN
absorption
121
ECE 455: Optical Electronics
Solving for Relative State Populations
1egBgB
1BA
)(kT/h
221
11221
21
But… we already know that, for a blackbody,
1e
1
c
h8)(
T/kh3
3
ECE 455: Optical Electronics
Einstein Coefficients
• In order for these two expressions for ρ(ν) to be equal, Einstein said:
and
h8A
hn8
cAB
3
2133
3
2121
1 212 21B g =B g
ECE 455: Optical Electronics
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?
121A
ECE 455: Optical Electronics
Example – Blackbody Source
Blackbody
6 cm
Atomic Ensemble
ECE 455: Optical Electronics
Example – Blackbody Source
E2 = 3.2 eV
h
E1 = 0
hν = 3.2 eV
= 387.5 nm = 7.7 · 1014 s–1
ECE 455: Optical Electronics
Example – Blackbody Source
d1ec
ch8d)(
kT/h3
3
0
• Blackbody emission at the surface of the emitter is
1e)103(
10)107.7(1063.68
413.0eV2.3
210
831434
kT : 5000 K
ECE 455: Optical Electronics
Example – Blackbody Source
• 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 cm10cm3
d)(
= 6.48 · 1012 photons-cm–2-s–1
ECE 455: Optical Electronics
Example – Blackbody Source
And
J/eV106.1103
1048.6
cd)(
d)(
1910
12
orρ(ν)dν = 3.46 · 10–17 J-cm–3
ECE 455: Optical Electronics
Example – Blackbody Source
34
3716
3
2121
1067.68
)105.387(s10
h8AB
= 3.5 · 1024 cm3-J–1-s–2
• The stimulated emission coefficient B21 is
ECE 455: Optical Electronics
Example – Blackbody Source
dcd)(
B
dd)(
B
)(Bdt
dNN1
21
21
212
2
161
8 3
1.56 10 J-s5.4s
10 cm
= – 3.5 · 1024 cm3-J–1-s–2
• Finally, the stimulated emission rate is given by
ECE 455: Optical Electronics
To reiterate…
This is negligible compared to the spontaneous emission rate of
A21 = 106 s–1 !
!s4.5dt
dNN1 12
2
ECE 455: Optical Electronics
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
11 mm
0
A21 = 106 s-1
Let Δνlaser = 108 s–1 (100 MHz, as before).
ECE 455: Optical Electronics
Example – Laser Source
• If the power emitted by the laser is 1 W, then
– Power flux, P 2)cm05.0(
W1
= 127.3 W-cm–2
Since hν = 3.2 eV = 5.1 · 10–19 J
→ P = 2.5 · 1020 photons-cm–2-s–1
ECE 455: Optical Electronics
Example – Laser Source
11018
2
laser
s-cm103s10
cm-W3.127
cP
)(
= 4.24 · 10–17 J-cm–3-Hz–1
= 83.3 photons-cm–3-Hz–1
ECE 455: Optical Electronics
Example – Laser Source
)(Bdt
dNN1
212
2 3.5 · 1024 cm3-J–1-s–2 · 4.24 · 10–17 J-cm–3-s
= 1.48 · 108 s–1
8 12
6 1
Stimulated Emission Rate 1.48 1010 !
Spontaneous Emission Rate 10
s
s
ECE 455: Optical Electronics
• Remember, in the case of the blackbody optical source:
• What made the difference?
Example – Laser Source
!1010
4.5Rate sSpontaneou
Rate Stimulated 56
ECE 455: Optical Electronics
Source Comparison
Total power radiated by 5000 K blackbody with R = 0.5 cm is 11.1 kW
Laser
5000 KBlackbody
570 (nm)
12 EEhc
nm5.387
ECE 455: Optical Electronics
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
ECE 455: Optical Electronics
Ex. Photodissociation
ABS. C3F7I
~280 nm (nm)I*
1.315 µm
I
C3F7I + hν → I*
ECE 455: Optical Electronics
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.
ECE 455: Optical Electronics
Topic #3: Homogeneous Line Broadening
ECE 455: Optical Electronics
Semi-Classical Conclusion
2
1
E2
Absorption
E1
12 EEhc
This diagram:
suggests that the atom absorbs only (exactly) at
ECE 455: Optical Electronics
The Shocking Truth!
Reality
12 EEhc
ECE 455: Optical Electronics
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
ECE 455: Optical Electronics
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
1d)(g
ECE 455: Optical Electronics
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.
ECE 455: Optical Electronics
Homogeneous Broadening
• In the homogenous case, we observe a Lorentzian Lineshape
where ν0 ≡ line center
220 )/2()(
)/2(1/)(g
ECE 455: Optical Electronics
Homogeneous Broadening
Δν = FWHM
Bottom line: Homogeneous → Lorentzian
10
2)(g
ECE 455: Optical Electronics
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
ECE 455: Optical Electronics
Natural Broadening
• ΔE Δt ≥ Heisenberg’s Uncertainty Principle
2u
1l
El
E
Eu
ECE 455: Optical Electronics
Natural Broadening
• In the case of an atomic system:
2
1
12
11
121
121
ECE 455: Optical Electronics
Natural Broadening
• In general
i
1i2
1
Lifetime ofupper or lower states resulting
from all processes.
ECE 455: Optical Electronics
Example: Sodium (Na)
3p 2P3/2
3p 2P1/2
3s 2S1/2 (Ground)
588.9 nm589.6
nm
(Both arrows indicate “resonance” transitions)
ECE 455: Optical Electronics
Example: Sodium (Na)
• Radiative lifetime of the 3p 2P3/2 state is 16 ns
}01025.6{21
1121
7
lowerupper
= 9.9 · 106 s–1 ≈ 10 MHz
0
~ 2 · 10–8!ν0 = 5.1 · 1014 Hz
ECE 455: Optical Electronics
Example: Mercury (Hg)
63S1
404.7 nm546.1 nm 435.8 nm
3P23P13P0
253.7 nm
1S0 (Ground)
ECE 455: Optical Electronics
Example: Mercury (Hg)
• Remember:
A43
A42
A41
4
321
1414243sp4 }AAA{
In general,
jij
1i A
ECE 455: Optical Electronics
Collisional Broadening
• An atom that radiates a photon can be described as a classical oscillator with a particular phase
t
FourierSpectrum
ß
)( 0
hEE 12
0
ECE 455: Optical Electronics
Collisional Broadening
• Suppose now that we have collisions between atom A (the radiator) and a second atom, B…
A
B
ECE 455: Optical Electronics
Collisional Broadening
• Such collisions alter the phase of the oscillator.
t
(Arrows indicate points at which oscillator suffers collision)
ECE 455: Optical Electronics
Collisional Broadening
• Result? Broadening of Transition!
• The rate of phase randomizing collisions is:
COLCC Nk
1RATE
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
ECE 455: Optical Electronics
Collisional Broadening
)2(21
collcollision Collision
perturbs both upper & lower
statesCollision
freq.
coll
collcoll
1
~ Ncoll ~ pressure
ECE 455: Optical Electronics
Total Homogenous Broadening
• Is calculated by summing the rates of the various homogeneous broadening processes:
i
1icoll
i
1itotal
121
ECE 455: Optical Electronics
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
1coll
1sptotal
121
ECE 455: Optical Electronics
Example – KrF Laser
8 1
spontaneousemission
2 10
2total
s
atmospherecm1045.2
scm102319
1310
Δνtotal = 31.8 MHz +GHz9.4
· P(atm)
ECE 455: Optical Electronics
Example – KrF Laser
Δνtotal = 31.9 MHz + 1.6 GHz · P spontaneous collisions
Note that these terms are equal for P = 0.02 atm!
ECE 455: Optical Electronics
Next Time
• Inhomogeneous broadening
• Threshold gain