The Beginning of the Quantum Physics

Blackbody Radiation and Plancks Hypothesis

Beginning of the Quantum PhysicsSome Problems with Classical PhysicsVastly different values of electrical resistance Temperature Dependence of Resistivity of metalsBlackbody RadiationPhotoelectric effectDiscrete Emission Lines of AtomsConstancy of speed of light

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 disappearRadiation is still present, but invisibleCan be detected as heatHeaters; Night Vision Goggles

Blackbody Radiation: Observations

Experiment:Focus the suns rays or direct a parabolic mirror with a heating spiral onto combustible materialthe material will flare up and burn

Materials absorb as well as emit radiation

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

Blackbody RadiationA blackbody is an object that absorbs all radiation incident upon itIts emission is universal, i.e. independent of the nature of the objectBlackbodies radiate, but do not reflect and so are blackBlackbody Radiation is EM radiation emitted by blackbodies

Blackbody RadiationThere are no absolutely blackbodies in nature this is idealizationBut some objects closely mimic blackbodies:Carbon black or Soot (reflection is

Kirchoff's Law of Thermal Radiation (1859)absorptivity 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 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 = 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 = 1Therefore, to keep your frank warm or your ice cream cold at a baseball game, wrap it in aluminum foilWhat color should integrated circuits be to keep them cool?

Blackbody Radiation LawsEmission 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

= 5.67010-8 W/m2-K4To get the emission power, multiply Intensity Itotal by area A

Blackbody RadiationThe maximum shifts to shorter wavelengths with increasing temperaturethe color of heated body changes from red to orange to yellow-white with increasing temperature

5780 K is the temperature of the Sun

Blackbody Radiation

Blackbody RadiationThe wavelength of maximum intensity per unit wavelength is defined by the Wiens Displacement Law:

b = 2.89810-3 m/K is a constantFor, T ~ 6000 K,

Average energy of a harmonic oscillator is

Intensity of EM radiation emitted by classical harmonic oscillators at wavelength per unit wavelength:

Or per unit frequency :

Blackbody Radiation Laws: Classical Physics View

Blackbody Radiation Laws: Classical Physics ViewIn classical physics, the energy of an oscillator is continuous, so the average is calculated as:

is the Boltzmann distribution

Blackbody Radiation: Classical Physics ViewThis gives the Rayleigh-Jeans Law

Agrees well with experiment long wavelength (low frequency) regionPredicts infinite intensity at very short wavelengths (higher frequencies) Ultraviolet CatastrophePredicts diverging total emission by black bodies

No fixes could be found using classical physics

Plancks HypothesisMax Planck postulated that A system undergoing simple harmonic motion with frequency can only have energies

where n = 1, 2, 3, and h is Plancks constanth = 6.6310-34 J-s

Plancks TheoryE is a quantum of energy

For = 3kHz

Plancks TheoryAs before: Now energy levels are discrete,

So

Sum to obtain average energy:

Blackbody Radiation

c is the speed of light, kB is Boltzmanns constant, h is Plancks constant, and T is the temperature

Plancks Theory

Plancks Theory

High Frequency - h >> kTAt room temperature, 300 K, kT= 1/40 eV At = 1 m:At 300 K:

Blackbody Radiation from the SunPlanks curvemaxStefan-Boltzmann LawIBB T4

IBB = T4

Stefan-Boltzmann constant =5.6710-8 J/m2K4

More generally:I = T4 is the emissivity

Wien's Displacement Lawpeak T = 2.89810-3 m K

At T = 5778 K:peak = 5.01510-7 m = 5,015 A

50% of energy emitted from the sun in visible rangeAppears as white light above the atmosphere, peakedAppears as yellow to red light due to Rayleigh scattering by the atmosphereEarth radiates infrared electromagnetic (EM) radiationEnergy Balance of Electromagnetic Radiation White light is made of a range of wave lengths

*Step 4: Calculate energy emitted by EarthEarth emits terrestrial long wave IR radiationAssume Earth emits as a blackbody.Calculate energy emission per unit time (Watts)Blackbody RadiationNotice color change as turn up power on light bulb.

Greenhouse EffectVisible light passes through atmosphere and warms planets surface

Atmosphere absorbs infrared light from surface, trapping heatWhy is it cooler on a mountain tops than in the valley?

Albedo and Atmosphere Affect Planet Temperature

Albedo, a

(, optical depth

Temp. Reduction due to Reflection

Greenhouse Temp. Increase Factor

Venus

0.7

0.74

= 70

2.9

Earth

0.3

0.91

( 1

= 1.19

Mars

0.25

0.93

( 0

= 1

_1296933091.unknown

_1297422893.unknown

Einsteins Photon Interpretation of Blackbody RadiationTwo 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:EM Modes:

Density of EM Modes, 1 May represent allowed wave vectors k by points on a unit lattice in a 3D abstract number spacek = 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)

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 modesThe second factor arises because there are two modes with perpendicular polarization (directions of oscillation of E) for each value of fSince 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 2dN = 2 dV', where dV = where dV is the relevant volume in numbr spaceThe volume of a complete shell is the area of the shell multiplied by its thickness, 4 n2dnThe number of modes with associated radii in number space between n and (n+dn) is, therefore, dN = 2 dV = (2)(1/8)4 n2dn = n2dnDensity of EM Modes, 2

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:

Modes DensityHow 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

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: The intensity is given by:Since (a) only the flux is directed out of the black bodyand (b) the average component of the velocity of light In a direction normal to the surface is , where is the average energy of a mode of EM radiationat frequency f and temperature T

Blackbody Radiation But and , as before So

*************************Best test of law it is experiment agreement is better than 1/100,000***Assume Earth characterized by single TResult depends only weakly on albedoWhat is wrong we have neglected the influence of the atm.*****Venus has 90 atm. of pressure*Consider a spherical cowSequel - Consider a cylindrical cowTwo layers- two levels of atm. Lower level emits and absorbed in second layerPutting in the empirical formulas. (IR gets out by reradiating into free windows as well as by diffusion)Get 16C ******