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

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

hkT

hd e d

c


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)


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


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


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


Example – The Sun

7710

21

10575

1

10565

1103

11c

or = 9.23 · 1012 s–1

= 9.23 THz


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)(


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)(


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

AbsorptionSpontaneous

emission


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


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


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


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


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


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



Blackbody

6 cm

Atomic Ensemble



E2 = 3.2 eV

h

E1 = 0

hν = 3.2 eV

= 387.5 nm = 7.7 · 1014 s–1



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



• 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



And

J/eV106.1103

1048.6

cd)(

d)(

1910

12

orρ(ν)dν = 3.46 · 10–17 J-cm–3



34

3716

3

2121

1067.68

)105.387(s10

h8AB

= 3.5 · 1024 cm3-J–1-s–2

• The stimulated emission coefficient B21 is



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


To reiterate…

This is negligible compared to the spontaneous emission rate of

A21 = 106 s–1 !

!s4.5dt

dNN1 12

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

11 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)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



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



)(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


• Remember, in the case of the blackbody optical source:

• What made the difference?


!1010

4.5Rate sSpontaneou

Rate Stimulated 56


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


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

12 EEhc

This diagram:

suggests that the atom absorbs only (exactly) at


The Shocking Truth!

Reality

12 EEhc


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

1d)(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.


Homogeneous Broadening

• In the homogenous case, we observe a Lorentzian Lineshape

where ν0 ≡ line center

220 )/2()(

)/2(1/)(g


Homogeneous Broadening

Δν = FWHM

Bottom line: Homogeneous → Lorentzian

10

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

2u

1l

El

E

Eu


Natural Broadening

• In the case of an atomic system:

2

1

12

11

121

121


Natural Broadening

• In general

i

1i2

1

Lifetime ofupper or lower states resulting

from all processes.


Example: Sodium (Na)

3p 2P3/2

3p 2P1/2

3s 2S1/2 (Ground)

588.9 nm589.6

nm

(Both arrows indicate “resonance” transitions)


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


Example: Mercury (Hg)

63S1

404.7 nm546.1 nm 435.8 nm

3P23P13P0

253.7 nm

1S0 (Ground)


Example: Mercury (Hg)

• Remember:

A43

A42

A41

4

321

1414243sp4 }AAA{

In general,

jij

1i A


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



• 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:

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



)2(21

collcollision Collision

perturbs both upper & lower

statesCollision

freq.

coll

collcoll

1

~ Ncoll ~ pressure


Total Homogenous Broadening

• Is calculated by summing the rates of the various homogeneous broadening processes:

i

1icoll

i

1itotal

121


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



8 1

spontaneousemission

2 10

2total

s

atmospherecm1045.2

scm102319

1310

Δνtotal = 31.8 MHz +GHz9.4

· 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

Documents

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