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The Beginning of the Quantum Physics

The Beginning of the Quantum Physics. Blackbody Radiation and Planck ’ s Hypothesis. Beginning of the Quantum Physics. • Some “ “ Problems ” ” with Classical Physics – Vastly different values of electrical resistance – Temperature Dependence of Resistivity of metals

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The Beginning of the Quantum Physics

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  1. The Beginning of the Quantum Physics Blackbody Radiation and Planck’s Hypothesis

  2. Beginning of the Quantum Physics •Some “ “Problems” ” with Classical Physics – Vastly different values of electrical resistance – Temperature Dependence of Resistivity of metals – Blackbody Radiation – Photoelectric effect – Discrete Emission Lines of Atoms – Constancy of speed of light

  3. Blackbody Radiation: • Solids heated to very high temperatures emit visible light (glow) – Incandescent Lamps (tungsten filament) • The color changes with temperature – At high temperatures emission color is whitish, at lower temperatures color is more reddish, and finally disappear – Radiation is still present, but “invisible” – Can be detected as heat • Heaters; Night Vision Goggles

  4. Blackbody Radiation: Observations • Experiment: – Focus the sun’s rays or direct a parabolic mirror with a heating spiral onto combustible material •the material will flare up and burn Materials absorb as well as emit radiation

  5. Blackbody Radiation • All object at finite temperatures radiate electromagnetic waves (emit radiation) • Objects emit a spectrum of radiation depending on their temperature and composition • From classical point of view, thermal radiation originates from accelerated charged particles in the atoms near surface of the object

  6. Blackbody Radiation – A blackbody is an object that absorbs all radiation incident upon it – Its emission is universal, i.e. independent of the nature of the object – Blackbodies radiate, but do not reflect and so are black Blackbody Radiation is EM radiation emitted by blackbodies

  7. Blackbody Radiation • There are no absolutely blackbodies in nature – this is idealization • But some objects closely mimic blackbodies: – Carbon black or Soot (reflection is <<1%) • The closest objects to the ideal blackbody is a cavity with small hole (and the universe shortly after the big bang) – Entering radiation has little chance of escaping, and mostly absorbed by the walls. Thus the hole does not reflect incident radiation and behaves like an ideal absorber, and “looks black”

  8. Kirchoff's Law of Thermal Radiation (1859) absorptivity αλis the ratio of the energy absorbed by the wall to the energy incident on the wall, for a particular wavelength. The emissivity of the wall ελis defined as the ratio of emitted energy to the amount that would be radiated if the wall were a perfect black body at that wavelength. At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity αλ= ελ If this equality were not obeyed, an object could never reach thermal equilibrium. It would either be heating up or cooling down. For a blackbody ελ= 1 Therefore, to keep your frank warm or your ice cream cold at a baseball game, wrap it in aluminum foil What color should integrated circuits be to keep them cool? • • • • • • •

  9. Blackbody Radiation Laws • Emission is continuous • The total emitted energy increases with temperature, and represents, the total intensity (Itotal) – the energy per unit time per unit area or power per unit area – of the blackbody emission at given temperature, T. • It is given by the Stefan-Boltzmann Law   4 Itotal T – σ = 5.670×10-8W/m2-K4 • To get the emission power, multiply Intensity Itotalby area A

  10. Blackbody Radiation • The maximum shifts to shorter wavelengths with increasing temperature – the color of heated body changes from red to orange to yellow-white with increasing temperature • 5780 K is the temperature of the Sun

  11. Blackbody Radiation

  12. Blackbody Radiation • The wavelength of maximum intensity per unit wavelength is defined by the Wien’ ’s Displacement Law:  T   b max – b = 2.898×10-3m/K is a constant • For, T ~ 6000 K,  3 10  . 2 898    483 nm max 6000

  13. Blackbody Radiation Laws: Classical Physics View • Average energy of a harmonic oscillator is <E> • Intensity of EM radiation emitted by classical harmonic oscillators at wavelength λ per unit wavelength: T I ) , (    2 E  3 c • Or per unit frequency ν:  2 2   I ( , T ) E 3 c

  14. Blackbody Radiation Laws: Classical Physics View • In classical physics, the energy of an oscillator is continuous, so the average is calculated as: E    k T EP  0 EP  0 ( E dE ) e dE B 0    E k T  B E    0 k T  0 P ( E dE ) P e dE B 0 E  is the Boltzmann distribution k T 0 P ( E ) P e B E k T  B

  15. Blackbody Radiation: Classical Physics View • This gives the Rayleigh-Jeans Law E      2 2 2 2 2 2 k T 3   ( , ) , ( , ) B I T I T E k T     B  3 3 2 c c c c – Agrees well with experiment long wavelength (low frequency) region • Predicts infinite intensity at very short wavelengths (higher frequencies) – “ “Ultraviolet Catastrophe” ” • Predicts diverging total emission by black bodies No “ “fixes” ” could be found using classical physics

  16. Planck’s Hypothesis Max Planck postulated that A system undergoing simple harmonic motion with frequency ν can only have energies  n E    nh where n = 1, 2, 3,… and h is Planck’s constant h = 6.63×10-34J-s

  17. Planck’s Theory  E ) 1 nh     ( E n h nh h      E is a quantum of energy For = 3kHz    E h    34 1 30    3000   s   E . 6 63 10 J s 2 10 J

  18. Planck’s Theory  2 2 • As before:  ( , ) I T E  3 c   nh  • Now energy levels are discrete,  n  k T P e B 0 ( ) P E  • So  n   k T   0 n P e B  n 0    k T   n n P e B  0  0 E n    • Sum to obtain average energy:   n   k T k T   n 1 P e e  B B 0 0     2 2 2 2 h   ( , ) I T    h 3 3 c c k T k T 1 1 e e   • B B

  19. Blackbody Radiation   2 2 h   I ( ) 2    c h  exp 1     k T B c is the speed of light, kBis Boltzmann’s constant, h is Planck’s constant, and T is the temperature

  20. Planck’s Theory   3 2 h 1   I ( )  h  1   2 c exp / k T B  h      for small : exp , /  1 / h k T k T h  k T    B B B   3 3 2 2 1  2 h k T  ( ) B I T    h  2 2 2 c c c 1 1   k T B

  21. Planck’s Theory   3 2 h 1   I ( )  h  1   2 c exp / k T B  h     for large : exp , / 1 h k T k T   B B   2   2  3 3 2 1 2 h h h     ( ) exp I        c c k T h   B exp   k   T B

  22. High Frequency - h >> kT At room temperature, 300 K, kT= 1/40 eV At = 1 m: c h h 10 63 . 6     8 3 10    34 19  19   . 1  99 10 ~ 2 10 J   6 10  19  . 1 99 10 h  . 1  24 eV  19  6 . 1 10 e  h At 300 K: . 1 24 40 49 6 . 1     kT

  23. Blackbody Radiation from the Sun Plank’ ’s curve λmax Stefan-Boltzmann Law IBB T4 IBB= T4 Stefan-Boltzmann constant  =5.67×10-8 J/m2K4 More generally: I = T4  is the emissivity Wien's Displacement Law peakT = 2.898×10-3m K At T = 5778 K:  peak= 5.015× ×10-7m = 5,015 A

  24. Energy Balance of Electromagnetic Radiation • 50% of energy emitted from the sun in visible range • Appears as white light above the atmosphere, peaked • Appears as yellow to red light due to Rayleigh scattering by the atmosphere • Earth radiates infrared electromagnetic (EM) radiation White light is made of a range of wave lengths Glass Prism

  25. Step 4: Calculate energy emitted by Earth Earth emits terrestrial long wave IR radiation Assume Earth emits as a blackbody. Calculate energy emission per unit time (Watts) Blackbody Radiation 29 Notice color change as turn up power on light bulb.

  26. Greenhouse Effect • Visible light passes through atmosphere and warms planet’s surface • Atmosphere absorbs infrared light from surface, trapping heat Why is it cooler on a mountain tops than in the valley?

  27. Albedo and Atmosphere Affect Planet Temperature T R T R a r direction he in which t path free mean the is   2 R       2 4 2 4 1 ( 4 ) 4 ( ), E   Sun Sun E E 2 4 L   E atm of IR emission randomized is 1  L    2 4 4 , the , optical depth atm  R T   E E  1 R  / 1 4 / 1 4 ` 1 ( ) 1 ( ) Sun r 2 T T a    E Sun E Albedo, a  , optical depth 1/4 1/4  1    1 a      Temp. Reduction due to Reflection 0.74 0.91 0.93 Greenhouse Temp. Increase Factor 2.9 = 1.19 = 1 Venus Earth Mars 0.7 0.3 0.25 = 70  1  0

  28. Einstein’s Photon Interpretation of Blackbody Radiation EM Modes: Two sine waves traveling in opposite directions create a standing wave    ( , ) sin( ) sin( ) 2 sin cos y x t A kx t A kx t A kx t      • For EM radiation reflecting off a perfect metal, the reflected amplitude equals the incident amplitude and the phases differ by  rad • E = 0 at the wall • For allowed modes between two walls separated by a: sin(kx) = 0 at x = 0, a •This can only occur when, ka = n, or k = n/a, n = 1,2,3… • In terms of the wavelength, k = 2/ = n/a, or /2 = a/n • This is for 1D, for 2D, a standing wave is proportional to: y k x k t y x y cos sin sin ~ ) , , ( 2 1    , / , / t k n a k n a   1 1 2 2 • For 3D a standing wave is proportional to:     ( , , , ) ~ sin sin sin cos , / , / , / y x y z t k x k y k z t k n a k n a k n a    1 2 2 1 1 2 2 3 3

  29. Density of EM Modes, 1      ˆ x ˆ y ˆ z ( , , ) ( , , ) ( , , ) k k k k k k k n n n n n n       1 2 3 1 2 3 1 2 3 1 2 3 a a a a • May represent allowed wave vectors k by points on a unit lattice in a 3D abstract number space •k = 2/. But f = c, so f = c/ = c/[(/2))(2)] = c/[(1/k)((2)]=ck/2 • f is proportional to k = n /a in 1D and can generalize to higher dimensions:    2 2 2 2 3 2 1 2 2 2 3 is proportion al to | | f k k k k k n n n n         1 a a 1  c , f ck n   2 2 a where, n is the distance in abstract number space from the origin (0,0,0) To the point (n1,n2n3)

  30. Density of EM Modes, 2 The number of modes between f and (f+df) is the number of points in number space with radii between n and (n+dn) in which n1, n2, n3,> 0, which is 1/8 of the total number of points in a shell with inner radius n and outer radius (n+dn), multiplied by 2, for a total factor of 1/4 – The first factor arises because modes with positive and negative n correspond to the same modes – The second factor arises because there are two modes with perpendicular polarization (directions of oscillation of E) for each value of f Since the density of points in number space is 1 (one point per unit volume), the number of modes between f and (f+df) is the number of points dN in number space in the positive octant of a shell with inner radius n and outer radius (n+dn) multiplied by 2 dN = 2 dV', where dV‘ = where dV‘ is the relevant volume in numbr space The volume of a complete shell is the area of the shell multiplied by its thickness, 4 n2dn The number of modes with associated radii in number space between n and (n+dn) is, therefore, dN = 2 dV‘ = (2)(1/8)4 n2dn =  n2dn • • • • •

  31. Density of EM Modes, 3 • The density of modes is the number of modes per unit frequency:  2 dN n dn dn  2 n   df df df • This may be expressed in terms of f once n and dn/df are so expressed c f n  2 a 2 2 a dn a , So n f and   c df c 3 2 2 dN dn a a a    2 2 2 ( ) ( ) 8 n f f    3 df df c c c  8 c dN • This is density of modes in a volume a3 • For a unit volume, the density of states is: 2 f  3 df

  32. Modes Density How many EM modes per unit frequency are there in a cubic cavity with sides a = 10 cm at a wavelength of  = 1 micron = 10-6m? • f = c, f = c/  = 3x108/10-6= 3x1014 3   f c df • dN a 2 8 3  1 3 10 (  ) 8 8  dN  14 2 ( 3 28 24 ) 1 8 3 ( 10 ) 10 4 . 8  10 84         3 3 ( 10 ) 3 df

  33. Blackbody Radiation • Einstein argued that the intensity of black body radiation I(f), reflects the state of thermal equilibrium of the radiation field • The energy density (energy per unit volume per unit frequency) within the black body is: dN (  )  u f  E  E , where is the average energy of a mode of EM radiation at frequency f and temperature T    df 1 c 2 c 4 dN • The intensity is given by: ( ) (  ) I f  u f  E    2 df Since (a) only ½ the flux is directed out of the black body and (b) the average component of the velocity of light In a direction normal to the surface is ½

  34. Blackbody Radiation c f I 4 dN ( )  E   df  , 3 , 2 , 1 , 0  ... E n n   8 c dN 2 f  • But 3 df   hf  n    k T   n n P e B  0 hf  0 E n     • and , as before   n hf   k T k T k T   n 1 1 P e e e   B B B 0 0   3 2 1 h  ( ) I   h  1   2 • So exp / c k T B

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