1 / 67

Lecture #8: Blackbody Radiation, Einstein Coefficients, and Homogeneous Broadening

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!

rod
Download Presentation

Lecture #8: Blackbody Radiation, Einstein Coefficients, and Homogeneous Broadening

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture #8: Blackbody Radiation, Einstein Coefficients, and Homogeneous Broadening Substitute Lecturer: Jason Readle Thurs, Sept 17th, 2009

  2. Topic #1: Blackbody Radiation

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

  4. Blackbody Energy Density • The photon energy density for a blackbody radiator in the ν → ν + dν spectral interval is

  5. Blackbody Intensity • The intensity emitted by a blackbody surface is (Units are or J/s-cm2 or W/cm2)

  6. Blackbody Peak Wavelength • The peak wavelength for emission by a blackbody is where 1 Å = 10–8 cm

  7. 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

  8. kT (300 K) eV Example – The Sun • Also 570 nm → 17,544 cm–1

  9. Example – The Sun or Dn = 9.23 · 1012 s–1 = 9.23 THz

  10. Example – The Sun

  11. Example – The Sun Since hν = 2.18 eV = 3.49 · 10–19 J →ρ(ν) d ν / hν = 1.58 · 1010

  12. Example – The Sun Remember, Intensity = Photon Density · c or  = 4.7 · 1020 photons-cm–2-s–1 = 164 W-cm–2

  13. Example – The Sun

  14. Topic #2: Einstein Coefficients

  15. 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

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

  17. Stimulated Emission • Same phase as “stimulating” optical field • Same polarization • Same direction of propagation

  18. Putting it all together… • Assume that we have a two state system in equilibrium with a blackbody radiation field.

  19. 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

  20. 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

  21. Solving for Relative State Populations • Solving for N2/N1:

  22. Solving for Relative State Populations But… we already know that, for a blackbody,

  23. Einstein Coefficients • In order for these two expressions for ρ(ν) to be equal, Einstein said: and

  24. 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?

  25. Example – Blackbody Source

  26. Example – Blackbody Source hν = 3.2 eV l = 387.5 nm n = 7.7 · 1014 s–1

  27. Example – Blackbody Source • Blackbody emission at the surface of the emitter is

  28. 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

  29. Example – Blackbody Source And or ρ(ν)dν = 3.46 · 10–17 J-cm–3

  30. Example – Blackbody Source • The stimulated emission coefficient B21 is = 3.5 · 1024 cm3-J–1-s–2

  31. Example – Blackbody Source • Finally, the stimulated emission rate is given by = – 3.5 · 1024 cm3-J–1-s–2

  32. To reiterate… This is negligible compared to the spontaneous emission rate of A21 = 106 s–1 !

  33. 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).

  34. 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

  35. Example – Laser Source = 4.24 · 10–17 J-cm–3-Hz–1 = 83.3 photons-cm–3-Hz–1

  36. Example – Laser Source 3.5 · 1024 cm3-J–1-s–2 · 4.24 · 10–17 J-cm–3-s = 1.48 · 108 s–1

  37. Example – Laser Source • Remember, in the case of the blackbody optical source: • What made the difference?

  38. Source Comparison Total power radiated by 5000 K blackbody with R = 0.5 cm is 11.1 kW

  39. 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

  40. Ex. Photodissociation C3F7I + hν → I*

  41. 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.

  42. Topic #3: Homogeneous Line Broadening

  43. Semi-Classical Conclusion This diagram: suggests that the atom absorbs only (exactly) at

  44. The Shocking Truth!

  45. 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)

  46. 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

  47. 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.

  48. Homogeneous Broadening • In the homogenous case, we observe a Lorentzian Lineshape where ν0 ≡ line center

  49. Homogeneous Broadening Δν = FWHM Bottom line: Homogeneous → Lorentzian

  50. 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

More Related