Transcript Slide 1

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
I total  T
4
– σ = 5.670×10-8 W/m2-K4
• To get the emission power, multiply Intensity Itotal by area A
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:
max  T  b
– b = 2.898×10-3 m/K is a constant
• For, T ~ 6000 K,
max
3
2.898 10

6000
 483nm
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:
2
I ( , T )  3 E
c
• Or per unit frequency ν:
2 2
I ( , T )  3 E
c
Blackbody Radiation Laws:
Classical Physics View
• In classical physics, the energy of an oscillator is
continuous, so the average is calculated as:


 EP( E)dE  EP e

0
E 
0

 P( E)dE
0
P( E)  P0 e

E
kBT

0

P e
0

E
k BT
E
k BT
dE
 k BT
dE
0
is the Boltzmann distribution
E  kBT
Blackbody Radiation:
Classical Physics View
• This gives the Rayleigh-Jeans Law
2 E
2 k BT
2 2
2 2
I ( , T ) 

, I ( , T )  3 E  2 k BT
3
3
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
Planck’s Hypothesis
Max Planck postulated that
A system undergoing simple harmonic motion with
frequency ν can only have energies
E  n  nh
where n = 1, 2, 3,…
and h is Planck’s constant
h = 6.63×10-34 J-s
Planck’s Theory
E  nh
E  (n  1)h  nh  h
E is a quantum of energy
For  = 3kHz
E  h
E  6.63  10 34 J  s  3000s 1  2  10 30 J
Planck’s Theory
• As before:
2 2
I ( , T )  3 E
c
• Now energy levels are discrete,

• So P ( E ) 
P0 e

n
k BT
Pe
n 0
  nh

n
k BT
0

E  n 
• Sum to obtain average energy:
 nP e
n 0

n

k BT
n 0
•


e k BT  1
2 2
 3
c
h
e
h
k BT
n
k BT
0
 P0e
2 2
I ( , T )  3
c

1



e k BT  1
Blackbody Radiation
2
I ( )  2
c
2
h
 h 
exp
 1
 k BT 
c is the speed of light, kB is Boltzmann’s
constant, h is Planck’s constant, and T
is the temperature
Planck’s Theory
2h
1
I ( ) 
2
c
exph / kBT   1
3
for small : h  k BT , exph / k BT   1  h / k BT
2h 3
1
2 3 k BT  2
I ( ) 

 2T
2
2
h
c
c
c
1
1
k BT
Planck’s Theory
2h
1
I ( ) 
2
c
exph / kBT   1
3
for large : h  k BT , exph / k BT   1
2h 3
I ( ) 
c2
 h 
1
2h 3

exp

2
c
k
T
 h 
B


exp

k
T
 B 
High Frequency - h >> kT
At room temperature, 300 K, kT= 1/40 eV
At  = 1 m:
h  h
c

34
 6.6310
3 108
19
19

1
.
99

10
~
2

10
J
6
10
h 1.991019

 1.24 eV
19
e
1.6 10
At 300 K:
h
 1.24  40  49 .6  1
kT
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
peak T = 2.898×10-3 m K
At T = 5778 K:
peak = 5.015×10-7 m = 5,015 A
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
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.
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
RE2

2
4
2
4
(1  a )4RSunTSun  4RETE (
),
2
4rE
  Latm
 is themean free pathin which the directionof IR emission is randomized
L
1
 4RE2TE4
,   atm , theopticaldepth
1

RSun `
TE  TSun (1  a )1/ 4 (1   )1/ 4
2rE
Venus
Earth
Mars
Albedo, a
1  a 
0.7
0.3
0.25
Temp. Reduction due
to Reflection
0.74
0.91
0.93
1/ 4

, optical
depth
= 70
1
0
1   
1/ 4

Greenhouse Temp.
Increase Factor
2.9
= 1.19
=1
Einstein’s Photon Interpretation of
Blackbody Radiation
EM Modes:
Two sine waves traveling in opposite directions create a standing wave
y( x, t )  A sin(kx  t )  A sin(kx  t )  2 A sin kx cos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( x, y, t ) ~ sin k1 x sin k2 y cost , k1  n1 / a, k2  n2 / a
• For 3D a standing wave is proportional to:
y( x, y, z, t ) ~ sin k1x sin k2 y sin k2 z cost , k1  n1 / a, k2  n2 / a, k3  n3 / a
Density of EM Modes, 1





ˆ
ˆ
k  k1 x  k 2 y  k3 zˆ  (k1 , k 2 , k3 )  (n1 , n2 , n3 )  (n1 , n2 , n3 )
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:



f is proport ion
al t o | k | k  k12  k 22  k32 
n12  n22  n32  n
a
a
1
c
f 
ck 
n,
2
2a
where, n is the distance in abstract number space from the origin (0,0,0)
To the point (n1,n2 n3)
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
Density of EM Modes, 3
• The density of modes is the number of modes per unit frequency:
dN n 2 dn
2 dn

 n
df
df
df
• This may be expressed in terms of f once n and dn/df are so expressed
c
f
n
2a
2a
dn 2a
So, n 
f and 
c
df
c
dN
2a 2 2a
a3 2
2 dn
 n
(
f ) ( )  8 3 f
df
df
c
c
c
• This is density of modes in a volume a3
• For a unit volume, the density of states is:
dN 8 2
 3 f
df
c
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
dN
a3 2
 8 3 f
df
c
dN
(101 )3
8 ( 3 28 24 )
14 2
1
 8
(
3

10
)

10

8
.
4

10
 84
8 3
df
(3 10 )
3
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:
u ( f ) 
dN
 E  , where E  is the average energy of a mode of EM radiation
df
at frequency f and temperature T
• The intensity is given by:
I( f ) 
1c
c dN
u ( f ) 
 E
22
4 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
Blackbody Radiation
I( f ) 
• But
E  n , n  0,1,2,3...
dN 8 2
 3 f
df
c
  hf

• and
E  n 
c dN
 E
4 df
 nP e
n 0


n
k BT
0
 P0e
n

k BT



e k BT  1
n 0
• So
2h 3
1
I ( ) 
c 2 exph / k BT   1
hf

e
hf
k BT
, as before
1