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 Blackbody Radiation is EM 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 αλ is 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 ελ is 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
σ = 5.670×10-8 W/m2-K4
To get the emission power, multiply Intensity Itotal by 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:
b = 2.898×10-3 m/K is a constant
For, T ~ 6000 K,

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:
Or per unit frequency ν:

14. Blackbody Radiation Laws: Classical Physics View
In classical physics, the energy of an oscillator is continuous, so the average is calculated as:
is the Boltzmann distribution

15. Blackbody Radiation: Classical Physics View
This gives the Rayleigh-Jeans Law
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. Max Planck postulated that
Planck’s Hypothesis
Max Planck postulated that
A system undergoing simple harmonic motion with frequency ν can only have energies
where n = 1, 2, 3,…
and h is Planck’s constant
h = 6.63×10-34 J-s

17. E is a quantum of energy
Planck’s Theory
E is a quantum of energy
For  = 3kHz

18. Planck’s Theory As before: Now energy levels are discrete, So
Sum to obtain average energy:

19. Blackbody Radiation
c is the speed of light, kB is Boltzmann’s constant, h is Planck’s constant, and T is the temperature

21. Planck’s Theory

22. Planck’s Theory

23. High Frequency - h >> kT
At room temperature, 300 K, kT= 1/40 eV
At  = 1 m:
At 300 K:

25. Blackbody Radiation from the Sun
Plank’s curve
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
peak T = 2.898×10-3 m K
At T = 5778 K:
peak = 5.015×10-7 m = 5,015 A
λmax

26. Best test of law – it is experiment agreement is better than 1/100,000

28. Electromagnetic Radiation
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
Glass
Prism
White light is made of a range of wave lengths

29. 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.

30. Assume Earth characterized by single T
Result depends only weakly on albedo
What is wrong – we have neglected the influence of the atm.

31. 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? 31

32. Albedo and Atmosphere Affect Planet Temperature32

33. Venus has 90 atm. of pressure

34. Consider a spherical cow
Sequel - Consider a cylindrical cow
Two layers- two levels of atm. Lower level emits and absorbed in second layer
Putting in the empirical formulas. (IR gets out by reradiating into free windows as well as by diffusion)
Get 16C

35. Einstein’s Photon Interpretation of Blackbody Radiation
EM Modes:
Two sine waves traveling in opposite directions create a standing wave
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:
For 3D a standing wave is proportional to:

36. Density of EM Modes, 1
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:
where, n is the distance in abstract number space from the origin (0,0,0)
To the point (n1,n2 n3)

37. 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

38. Density of EM Modes, 3
The density of modes is the number of modes per unit frequency:
This may be expressed in terms of f once n and dn/df are so expressed
This is density of modes in a volume a3
For a unit volume, the density of states is:

39. 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 = 10-6 m?
f = c, f = c/  = 3x108/10-6 = 3x1014

40. 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:
, where is the average energy of a mode of EM radiation
at frequency f and temperature T
The intensity is given by:
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

41. Blackbody Radiation
But hf = e
and , as before So