Black Body radiation



Hot filament glows.
Classical physics cant
explain the observed
wavelength distribution of
EM radiation from such a
hot object.
This problem is
historically the problem
that leads to the rise of
quantum physics during
the turn of 20th century
Need for Quantum Physics



Problems remained from classical mechanics
that relativity didn’t explain
Attempts to apply the laws of classical physics
to explain the behavior of matter on the atomic
scale were consistently unsuccessful
Problems included:

blackbody radiation


The electromagnetic radiation emitted by a heated object
photoelectric effect

Emission of electrons by an illuminated metal
Quantum Mechanics
Revolution



Between 1900 and 1930, another revolution
took place in physics
A new theory called quantum mechanics was
successful in explaining the behavior of
particles of microscopic size
The first explanation using quantum
mechanics was introduced by Max Planck

Many other physicists were involved in other
subsequent developments
Blackbody Radiation

An object at any temperature is known
to emit thermal radiation


Characteristics depend on the temperature
and surface properties
The thermal radiation consists of a
continuous distribution of wavelengths from
all portions of the em spectrum
Blackbody Radiation, cont.


At room temperature, the wavelengths of the
thermal radiation are mainly in the infrared
region
As the surface temperature increases, the
wavelength changes


It will glow red and eventually white
The basic problem was in understanding the
observed distribution in the radiation emitted
by a black body

Classical physics didn’t adequately describe the
observed distribution
Blackbody Radiation, final


A black body is an ideal system that
absorbs all radiation incident on it
The electromagnetic radiation emitted
by a black body is called blackbody
radiation
Blackbody Approximation



A good approximation of a
black body is a small hole
leading to the inside of a
hollow object
The hole acts as a perfect
absorber
The nature of the radiation
leaving the cavity through
the hole depends only on
the temperature of the
cavity
Blackbody Experiment Results

The total power of the emitted radiation
increases with temperature


Stefan’s law (from Chapter 20):
P = sAeT4
The peak of the wavelength distribution shifts
to shorter wavelengths as the temperature
increases


Wien’s displacement law
lmaxT = 2.898 x 10-3 m.K
Real life blackbody


A close
approximation of
blackbody radiator
The colour of the
light emitted from
the charcoal
depends only upon
the temperature
example
This figure shows two stars in
the constellation Orion.
Betelgeuse appears to glow
red, while Rigel looks blue in
color. Which star has a higher
surface temperature?
(a) Betelgeuse
(b) Rigel
(c) They both have the same
surface temperature.
(d) Impossible to determine.
Stefan’s Law – Details

P = sAeT4


P is the power
s is the Stefan-Boltzmann constant


s = 5.670 x 10-8 W / m2 . K4
Stefan’s law can be written in terms of
intensity

I = P/A = sT4

For a blackbody, where e = 1
Wien’s Displacement Law


lmaxT = 2.898 x 10-3 m.K

lmax is the wavelength at which the curve

peaks
T is the absolute temperature
The wavelength is inversely proportional
to the absolute temperature

As the temperature increases, the peak is
“displaced” to shorter wavelengths
Intensity of Blackbody
Radiation, Summary


The intensity increases
with increasing
temperature
The amount of radiation
emitted increases with
increasing temperature


The area under the
curve
The peak wavelength
decreases with
increasing temperature
Exmple



Find the peak wavelength of the
blackbody radiation emited by
(A) the Sun (2000 K)
(B) the tungsten of a lightbulb at 3000 K
Solutions




(A) the sun (2000 K)
By Wein’s displacement law,
3
2.898

10
mK
lmax 
2000K
 1.4 m
(infrared)
(B) the tungsten of a lightbulb
at 3000 K
lmax
3
2.898

10
mK

 0.5 m

Yellow-green
5800K
Rayleigh-Jeans Law

An early classical attempt to explain
blackbody radiation was the RayleighJeans law
2πck BT
I  λ,T  
λ4

At long wavelengths, the law matched
experimental results fairly well
Rayleigh-Jeans Law, cont.


At short wavelengths,
there was a major
disagreement between
the Rayleigh-Jeans law
and experiment
This mismatch became
known as the ultraviolet
catastrophe

You would have infinite
energy as the wavelength
approaches zero
Max Planck


Introduced the
concept of “quantum
of action”
In 1918 he was
awarded the Nobel
Prize for the
discovery of the
quantized nature of
energy
Planck’s Theory of Blackbody
Radiation




In 1900 Planck developed a theory of
blackbody radiation that leads to an equation
for the intensity of the radiation
This equation is in complete agreement with
experimental observations
He assumed the cavity radiation came from
atomic oscillations in the cavity walls
Planck made two assumptions about the
nature of the oscillators in the cavity walls
Planck’s Assumption, 1

The energy of an oscillator can have only
certain discrete values En

En = nhƒ



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
n is a positive integer called the quantum
number
h is Planck’s constant
ƒ is the frequency of oscillation
This says the energy is quantized
Each discrete energy value corresponds to
a different quantum state
Planck’s Assumption, 2

The oscillators emit or absorb energy
when making a transition from one
quantum state to another


The entire energy difference between the
initial and final states in the transition is
emitted or absorbed as a single quantum
of radiation
An oscillator emits or absorbs energy only
when it changes quantum states
Energy-Level Diagram




An energy-level diagram
shows the quantized
energy levels and allowed
transitions
Energy is on the vertical
axis
Horizontal lines represent
the allowed energy levels
The double-headed
arrows indicate allowed
transitions
More About Planck’s Model


The average energy of a wave is the average
energy difference between levels of the
oscillator, weighted according to the
probability of the wave being emitted
This weighting is described by the Boltzmann
distribution law and gives the probability of a
state being occupied as being proportional to
e E kBT where E is the energy of the state
Planck’s
Model,
Graphs
Planck’s Wavelength
Distribution Function

Planck generated a theoretical
expression for the wavelength
distribution
2πhc 2
I  λ,T   5 hc λk T
B
λ e
 1


h = 6.626 x 10-34 J.s
h is a fundamental constant of nature
Planck’s Wavelength
Distribution Function, cont.


At long wavelengths, Planck’s equation
reduces to the Rayleigh-Jeans
expression
At short wavelengths, it predicts an
exponential decrease in intensity with
decreasing wavelength

This is in agreement with experimental
results
Example: quantised oscillator vs
classical oscillator


A 2.0 kg block is attached to a massless
spring that has a force constant k=25
N/m. The spring is stretched 0.40 m
from its EB position and released.
(A) Find the total energy of the system
and the frequency of oscillation
according to classical mechanics.
Solution (A)


Mechanical Energy, E = ½ kA2 = …= 2.0 J
1 k
Frequency, f 
 ...  0.56Hz
2
m

(B) Assuming that the energy is
quantised, find the quantum number n
for the system oscillating with this
amplitude.
QUICK QUIZ 40.1
(For the end of section 40.1)
In a certain experiment, you pass a current through a wire
and measure the spectrum of light that is emitted from the
glowing filament. For a current I1, you measure the wavelength
of the highest intensity (also called lmax) to be l1. You then
increase the voltage across the wire by a factor of 8 and the
current increases by a factor of 2 to 2I1. The wavelength of the
highest intensity will shift to a) 4l1, b) 2l1, c) 2l1, d) l1/ 2, e)
l1/ 2, or f) l1/ 4.
QUICK QUIZ 40.1 ANSWER
(e). Assuming that the wire behaves like a blackbody, the
wavelength with the highest intensity will be inversely
proportional to the absolute temperature according to Wien’s
displacement law, or lmax 1/T. In addition, the power
radiated from the wire is proportional to the absolute
temperature to the fourth power; from Stefan’s law, or P  T4.
Also, the power dissipated in the wire is given by P = IV.
Combining these equations, one finds that,
1
1
1
1
1
lmax   1/ 4 

 .
1/ 4
1/ 4
T P
( IV )
(2  8)
2
QUICK QUIZ 40.2
The oscillators in a blackbody may oscillate a) at only certain
frequencies, b) with only certain energies, c) at only certain
frequencies and with only certain energies, d) with only certain
energies for a certain frequency, or e) at only certain frequencies
for a certain energy.
QUICK QUIZ 40.2 ANSWER
(d). The condition that the oscillators in a blackbody can
have only discrete energy values according to Equation 40.4,
En = nhf, does not imply that the oscillator may only oscillate
at discrete frequencies. In fact, the blackbody spectrum is
continuous precisely because all oscillation frequencies are
allowed. What Equation 40.4 does imply is that if an
oscillator is oscillating at a specific frequency, then it may
only occupy certain discrete energy states, and that
transitions are only allowed between these discrete energy
states.