PHY3101 Modern Physics Lecture Notes
Quanta 2
Quanta 2
Disclaimer: These lecture notes are not meant to replace the course textbook. The
content may be incomplete. Some topics may be unclear. These notes are only meant to
be a study aid and a supplement to your own notes. Please report any inaccuracies to the
professor.
Black-Body Radiation
How do night-vision goggles (or cameras) work? You might know that these
devices record infrared light, which is not visible to the naked eye. But why is there
infrared light at night, when the sun (and even the moon) is not visible? Perhaps you
know that infrared light comes from heated objects, but then why do heated objects emit
light? This was the question that 19th century physicists tried to solve using the physics
known at the time: electromagnetism and thermodynamics, cornerstones of 19th century
physics. They set out to explain the spectrum of emitted electromagnetic radiation from
heated bodies, but they failed. In the end, a new quantum, the photon, had to be invoked.
It was a subtle realization, however, because the spectrum of radiation emitted by hot
bodies does not appear quantized at all. A continuous spectrum of wavelengths is
emitted.
Black body: an idealized object which absorbs all incident radiation and emits none. It
makes the calculations simpler because the black-body radiation spectrum will be
independent of the material property.
One way to imagine a black body is not think of it as a body at all, but rather a
cavity with a small hole. Radiation can make it in through the hole, but after internal
reflections, very unlikely to escape (hence it is absorbed). When the cavity reaches
thermal equilibrium, the emitted radiation from the hole equals the absorbed radiation.
This emitted electromagnetic spectrum is what we wish to explain.
Cavity
Hole
We will skip the classical calculation of black-body radiation applying only
electromagnetism and thermodynamics to our black-body cavity example. The final
result is known as the Rayleigh-Jeans Law:
dR
2ckT
 I ,T 
d
4
b g
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PHY3101 Modern Physics Lecture Notes
Quanta 2
b g
The intensity I  , T is the total power radiated per unit area, R, per unit wavelength at a
given temperature T. The speed of light, c, arises from electromagnetism, and the
Boltzmann constant k  138
.  1023 J / K arises from thermodynamics. This calculation
assumes that the average energy of an atomic oscillator (which generates standing waves
in the cavity) is kT.
Crisis
Using this classical results, we can calculate the total power radiated per unit area
from the heated object:
R
zb g

0
I  , T d 
1
3

0

The total power is apparently infinite! This cannot be true, however, by empirical
observation. This crisis is known as the Ultraviolet Catastrophe, because the integral
diverges at the wavelength goes to zero. Classical physics yielded a nonsense result.
Empirical Results
Here is what the radiation spectrum actually looks like.
Intensity
T = 6000 K
T = 5000 K
T = 4000 K
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Wavelength (nm)
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PHY3101 Modern Physics Lecture Notes
Quanta 2
Before discussing the new physics needed, let’s understand what is known
empirically about black-body radiation.
1. The total power radiated (integral of intensity over wavelength) increases with
temperature, as illustrated in the previous figure. This is summarized in the
empirical Stefan-Boltzmann Law:
bg zb g
R T  I  , T d    T 4
  emissivity of body ( = 1 for a black body)
  5.67  108 W m-2 K -4 Stefan - Boltzmann constant
2. The black-body spectrum exhibits a peak at a certain wavelength, and this peak
shifts to smaller wavelengths as the temperature increases. In fact, it is inversely
related to temperature, as expressed in the empirical Wien’s Displacement Law:
 peak T  2.898  103 m- K
The Rayleigh-Jeans Law exhibits no peak at all, and diverges as   0
For example, if we consider the sun as a black-body, we can derive its surface
temperature. The peak emission wavelength is about 500 nm, so this corresponds
to a temperature of about 6000 K.
For ordinary temperatures of about 300 K, the peak emission is in the infrared
region. That is why night-vision devices work.
The universe itself has a temperature remnant of the Big Bang. This emission in
the microwave region (   1 mm ), so corresponds to a cold temperature of 3 K.
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Quanta 2
Derivation by Max Planck
Planck was able to derive the mathematical form of the black-body radiation
spectrum in 1900 by making a rather radical assumption. He assumed that the energy of
the atomic oscillators contained in the black-body was quantized: En  nhf . That is, the
energy depends on an integer n=1,2,3,…, the frequency f, and a new constant h. When an
oscillator emits radiation, it corresponds to the energy difference between two energy
levels:
E  En  En 1  h f
With this assumption, Planck was able to derive the intensity of the black-body
radiation (which we will skip):
Planck’s Radiation Law
2
b g
I , T 
2c h
1
5
e hc / kT  1
We can look at the limit of very large wavelengths:
hc I
hc
F
 1
G
J
HkT K kT 
2ckT
Ib
,Tg


lim  exp
lim 
4
So we recover the Rayleigh-Jeans Law in the limit of large wavelengths, but
Planck’s derivation solves the ultraviolet catastrophe. By fitting this formula to the
empirical data on black-body radiation, Planck derived a new constant of nature:
h  6.626  1034 J - s
Planck’s Constant
It is small, so atomic oscillators emit nearly continuous energy; but it is not
exactly zero. Nature is quantized at a fundamental level. The quantization is necessary
to solve the ultraviolet catastrophe.
Using Planck’s formula, it is also possible to derive the Stefan-Boltzmann Law
and Wien’s displacement Law (which we will skip).
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