the beginning of the quantum physics blackbody radiation and planck s hypothesis
TRANSCRIPT
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
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dEEP
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E B
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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
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c
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cTI B
B2
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33
22),(,
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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:
•
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Blackbody Radiation
c is the speed of light, kB is Boltzmann’s constant, h is
Planck’s constant, and T is the temperature
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2)( 2
2
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cI
B
Planck’s Theory
1/exp12
)( 2
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Planck’s Theory
1/exp12
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IB
High Frequency - h >> kT
At room temperature, 300 K, kT= 1/40 eV At = 1 m:
eVe
h
Jc
hh
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1099.1
102~1099.110
1031063.6
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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
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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
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2
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• 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
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Blackbody Radiation
Edf
dNfu )(
Edf
dNcfu
cfI
4)(
22
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• 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
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• But
• and , as before
• So 1/exp
12)(
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