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Learn about blackbody radiation and Einstein coefficients in this lecture by substitute lecturer Jason Readle. Understand the concept of blackbodies and their energy density and intensity, as well as Einstein coefficients for absorption, spontaneous emission, and stimulated emission. Dive into calculations, examples related to the Sun, and the implications of these concepts on different systems.
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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