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Lecture #8: Blackbody Radiation, Einstein Coefficients, and Homogeneous Broadening. Substitute Lecturer: Jason Readle Thurs, Sept 17 th , 2009. Topic #1: Blackbody Radiation. What is a Blackbody?. Ideal blackbody: Perfect absorber Appears black when cold!
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Lecture #8: Blackbody Radiation, Einstein Coefficients, and Homogeneous Broadening Substitute Lecturer: Jason Readle Thurs, Sept 17th, 2009
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
Blackbody Intensity • The intensity emitted by a blackbody surface is (Units are or J/s-cm2 or W/cm2)
Blackbody Peak Wavelength • The peak wavelength for emission by a blackbody is 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. • Tk = 5260 K
kT (300 K) eV Example – The Sun • Also 570 nm → 17,544 cm–1
Example – The Sun or Dn = 9.23 · 1012 s–1 = 9.23 THz
Example – The Sun Since hν = 2.18 eV = 3.49 · 10–19 J →ρ(ν) d ν / hν = 1.58 · 1010
Example – The Sun Remember, Intensity = Photon Density · c or = 4.7 · 1020 photons-cm–2-s–1 = 164 W-cm–2
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
Putting it all together… • Assume that we have a two state system in equilibrium with a blackbody radiation field.
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: Remember, ρ(ν) has units of J-cm–3-Hz–1
Solving for Relative State Populations • Solving for N2/N1:
Solving for Relative State Populations But… we already know that, for a blackbody,
Einstein Coefficients • In order for these two expressions for ρ(ν) to be equal, Einstein said: and
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?
Example – Blackbody Source hν = 3.2 eV l = 387.5 nm n = 7.7 · 1014 s–1
Example – Blackbody Source • Blackbody emission at the surface of the emitter is
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 = 6.48 · 1012 photons-cm–2-s–1
Example – Blackbody Source And or ρ(ν)dν = 3.46 · 10–17 J-cm–3
Example – Blackbody Source • The stimulated emission coefficient B21 is = 3.5 · 1024 cm3-J–1-s–2
Example – Blackbody Source • Finally, the stimulated emission rate is given by = – 3.5 · 1024 cm3-J–1-s–2
To reiterate… This is negligible compared to the spontaneous emission rate of A21 = 106 s–1 !
Example – Laser Source • Let us suppose that we have the same conditions as before, EXCEPT a laser photo-excites the two level system: Let Δνlaser = 108 s–1 (100 MHz, as before).
Example – Laser Source • If the power emitted by the laser is 1 W, then • Power flux, P = 127.3 W-cm–2 Since hν = 3.2 eV = 5.1 · 10–19 J → P = 2.5 · 1020 photons-cm–2-s–1
Example – Laser Source = 4.24 · 10–17 J-cm–3-Hz–1 = 83.3 photons-cm–3-Hz–1
Example – Laser Source 3.5 · 1024 cm3-J–1-s–2 · 4.24 · 10–17 J-cm–3-s = 1.48 · 108 s–1
Example – Laser Source • Remember, in the case of the blackbody optical source: • What made the difference?
Source Comparison Total power radiated by 5000 K blackbody with R = 0.5 cm is 11.1 kW
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 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.
Semi-Classical Conclusion This diagram: suggests that the atom absorbs only (exactly) at
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)
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
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
Homogeneous Broadening Δν = FWHM Bottom line: Homogeneous → Lorentzian
Sources of Homogeneous Broadening • Natural Broadening — any state with a finite lifetime τsp (τsp ≠ ∞) must have a spread in energy: • Collisional Broadening — phase randomizing collisions