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– Blackbody Radiation– Photoelectric effect– Discrete Emission Lines of Atoms– Constancy 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 disappear

– Radiation is still present, but “invisible”

– Can be detected as heat• Heaters; Night Vision Goggles

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

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

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

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”

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?

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

4TItotal

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

Blackbody Radiation

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,

bT max

nm 4836000

10898.2 3

max

• 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:

Blackbody Radiation Laws: Classical Physics View

Ec

TI 3

22),(

Ec

TI3

2),(

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

Tk

dEeP

dEeEP

dEEP

dEEEP

E B

TkE

TkE

B

B

00

00

0

0

)(

)(

TkE B

TkE

BePEP

0)(

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

Tkc

Ec

TITk

c

E

cTI B

B2

2

3

2

33

22),(,

22),(

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

nhnE

Planck’s Theory

hnhhnE

nhE

)1(

J1023000J1063.6 30134

ssE

hE

E is a quantum of energy

For = 3kHz

Planck’s Theory

• As before:

• Now energy levels are discrete,

• So

• Sum to obtain average energy:

•

Ec

TI3

22),(

10

0

00

Tk

n

Tk

nn

Tk

n

BB

B

eeP

ePnnE

1

2

1

2),(

3

2

3

2

Tk

h

Tk BB e

h

ce

cTI

nh

00

0)(

n

Tk

n

Tk

n

B

B

eP

ePEP

Blackbody Radiation

c is the speed of light, kB is Boltzmann’s constant, h is

Planck’s constant, and T is the temperature

1exp

2)( 2

2

Tkh

h

cI

B

Planck’s Theory

1/exp12

)( 2

3

Tkhch

IB

Planck’s Theory

1/exp12

)( 2

3

Tkhch

IB

High Frequency - h >> kT

At room temperature, 300 K, kT= 1/40 eV At = 1 m:

eVe

h

Jc

hh

24.1106.1

1099.1

102~1099.110

1031063.6

19

19

19196

834

16.494024.1 kT

hAt 300 K:

Blackbody Radiation from the SunBlackbody Radiation from the SunPlank’s curve

λmax

Stefan-Boltzmann LawIBB T4

IBB = T4

Stefan-Boltzmann constant =5.67×10-8 J/m2K4

More generally:I = T4

is the emissivity

Wien's Displacement Lawpeak T = 2.898×10-3 m K

At T = 5778 K:peak = 5.015×10-7 m = 5,015 A

• 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

Energy Balance of Electromagnetic 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

Notice color change as turn up power on light bulb.

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?

Albedo and Atmosphere Affect Planet Temperature

Albedo, a 1/4

1 a , optical depth

1/41

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

E

SunSunE

atmEE

atmEESunSun

E

E

r

RaTT

LTR

LTRTRa

r

R

2)1()1(

depth optical the,,1

14

randomized isemission IR ofdirection hein which tpath freemean theis

),(44)1(4

`4/14/1

42

42422

2

Einstein’s Photon Interpretation of Blackbody Radiation

ankankanktzkykxktzyxy /,/,/,cossinsinsin~),,,( 332211221

tkxAtkxAtkxAtxy cossin2)sin()sin(),( 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:

ankanktykxktyxy /,/,cossinsin~),,( 221121

EM Modes:

Density of EM Modes, 1),,(),,(),,(ˆˆˆ 321321321321 nnn

aan

an

ankkkzkykxkk

• 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:

,22

1

||toalproportionis 23

22

21

23

22

21

na

cckf

na

nnna

kkkkkf

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 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

Density of EM Modes, 2

Density of EM Modes, 3

23

322 8)

2()

2(

22,

2

fc

a

c

af

c

a

df

dnn

df

dN

c

a

df

dnandf

c

anSo

na

cf

df

dnn

df

dnn

df

dN 22

• 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

23

8f

cdf

dN • This is density of modes in a volume a3

• For a unit volume, the density of states is:

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

84104.8103

8)103(

)103(

)10(8

8

1)24283(21438

31

23

3

df

dN

fc

a

df

dN

Blackbody Radiation

Edf

dNfu )(

Edf

dNcfu

cfI

4)(

22

1)(

• 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 ½ 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 ½

, where is the average energy of a mode of EM radiationat frequency f and temperature T

E

Blackbody Radiation

110

0

00

Tk

hf

Tk

n

Tk

nn

Tk

n

BBB

B

e

hf

eeP

ePnnE

Edf

dNcfI

4)(

• But

• and , as before

• So 1/exp

12)(

2

3

Tkhc

hI

B

23

8f

cdf

dN

...3,2,1,0, nnE

hf