modern physics

1. Quantum Theory of Radiation
Introduction:
Our discussion of Einstein’s theory of relativity took us into a world far beyond that of ordinary experience -
the world of objects moving at speeds close to the speed of light.
Now you are about to explore a second world that is outside ordinary experience- the subatomic world. You
will encounter a new set of surprise that, though may sometime seem bizarre, have led physicists step by
step to a deeper view of reality.
Thermal radiation:
An object and its environment can exchange energy as heat via electromagnetic waves. Energy transferred
in this way is often called thermal radiation to distinguish it from electromagnetic signals and from nuclear
radiation. When an object is in thermal equilibrium with its surroundings, it emits and absorbs energy at the
same rate. The rate at which an object radiates energy is proportional to the surface area of the object and to
the fourth power of its absolute temperature. This result found empirically by Josef Stefan in 1879 and
derived theoretically by Ludwig Boltzmann about five years later, is called the Stefan-Boltzmann law:
][, 14
ATradP εσ=
where
radP = power radiated in watts
A= object’s surface area
σ= a universal constant called Stefan’s constant,
=5.6703x10-8
W/m2
k4
ε= emissivity of the object’s surface which has a value between 0 and 1, depending on
the composition of the surface.
A surface with the maximum emissivity of 1.0 is said to be a “blackbody radiator”, but such a surface is an
ideal one and does not occur in nature.
When radiation falls on an opaque object, part of the radiation is reflected and part is absorbed. Light-
coloured objects reflect most visible radiation whereas dark objects absorb most of it. The rate at which an
object absorbs radiation is given by
][, 24
envTabsP σε=
where Tenv is the temperature of the surroundings (in Kelvins).
If an object emits more radiation that it absorbs, it cools while the surroundings absorb radiation from the
object and warm. It the object absorbs more than it emits, the object warms and the surroundings cool. The
net power radiated by and object at temperature T in an environment at temperature Tenv is
][],[ 344
TTPPP envradabsnet −=−= σε
Pnet is positive if net energy is being absorbed via radiation, and negative if it is being lost via radiation.
When an object is in thermal equilibrium with its surroundings, envTT = and the object emits and absorbs
radiation at the same rate.
Blackbody radiation:
As discussed earlier, an object that absorbs all the radiation incident upon it has an emissivity equal to 1 and
is called a blackbody. The concept of a blackbody is important because the characteristics of the radiation
emitted by such an ideal object can be calculated theoretically. Materials such as black velvet come close to
being ideal blackbodies. The best practical approximation of an ideal black body is a small hole leading to a
cavity, such as a keyhole in a closed door (Fig.1).
Radiation incident on the hole has little chance of being reflected back out of the hole before it is absorbed
by the walls of the cavity. The radiation emitted through the hole is thus characteristic of the temperature of
the object.

2. Most of the radiation emitted by an object at temperature
below about 6000
C is concentrated at wavelengths much
longer than those of visible light (visible light is
electromagnetic radiation with wavelengths between about
400 and 700 nm).
As an object is heated, the rate of energy emission increases
and the energy radiated extends to shorter and shorter
wavelengths. Fig. 2 shows the power radiated by a blackbody
as a function of wavelength for several different
temperatures. The wave length at which the power is a
maximum varies inversely with the temperature, a result
known as Wein’s displacement law:
][,
..
max 4
8982
T
Kmm
=λ
The law is used to determine the temperatures of stars from analyses of their radiation. It can also be used to
map the variation in temperature over different regions of the surfaces of an object. Such a map is called a
thermograph. Thermographs can be used to detect cancer because cancerous materials are at a slightly
higher temperature than the surrounding tissue. Eq. [4] can also be written as
KmxT ..max
3
108982 −
=λ
It may be noted that the term “displacement” in “Wein’s
displacement law” refers to the way the peak is moved or
displaced as the temperature is varied.
The spectral distribution curves shown in Fig.2 played an
important role in the history of physics. It was the
discrepancy between theoretical calculations of what the
black body spectral distribution should be using classical
thermodynamics and the experimental measurements
that led to Max Planck’s first ideas about the quantization
of energy in 1897.
Example 1:
(a) The surface temperature of the sun is about 6000 K. If the sun is assumed to be a blackbody
radiator, at what wavelength maxλ would its spectrum peak?
(b) Calculate maxλ for a blackbody al room temperature T= 300 K.
Solution:
(a) We can find maxλ from Wein’s displacement law as:
nmmx
K
Kmx
48310483
6000
108982 9
3
=== −
−
..
maxλ
(b) nmmx
K
Kmx
966010669
300
108982 6
3
=== −
−
.
..
maxλ
So, we see that the peak wavelength from the sun is in the visible spectrum. The blackbody radiation
spectrum describes the sun’s radiation fairly well.
For T= 300 K, the blackbody spectrum peaks in the infrared wave length, much longer than the wavelengths
visible to the eye. Surfaces that are not black to our eyes may act as blackbodies for infrared radiation and
absorption. Skin of human beings of all races is black to infrared radiations hence the emissivity of skin is
1.00 for its own radiation process.
Example 2:
2
Figure 1 : A hole in a cavity approximates an
ideal black body.
Figure 2: Radiated power versus wavelength
for radiation emitted by a black body.

3. Calculate the net loss in radiated energy for a naked person in a room at 200
C assuming the person to be a
blackbody with a surface area of 1.4 m2
and a surface temperature of 330
C(=306 K). (The surface
temperature of human body is slightly less than the internal temperature of 370
C because of the thermal
resistance of the skin.
Solution:
W
KmxKmWxx
TTAnetP env
111
2933064110670351 4442428
44
=
−=
−=
−
][.]..[
][σε
This large energy loss is approximately equal to the basal
metabolic rate of about 120 W. We protect ourselves from this
great loss of energy by wearing clothes.
The ultraviolet catastrophe:
Why does the blackbody spectrum have the shape shown in
Fig. 2. This problem was examined at the end of the nineteenth
century by Lord Rayleigh and James Jeans.
Since the radiation that emerges from the hole (in Fig.1) is just
a sample of the radiation inside the box, understanding the
nature of the radiation inside the box allows one to understand
the radiation that leaves through the hole.
For this reason, Rayleigh and Jeans started by considering the
radiation inside a cavity of absolute temperature T whose walls
are perfect reflectors to be a series of standing electromagnetic
waves (Fig. 3). The condition for standing waves in such a
cavity is that the path length from wall to wall, whatever the
direction, must be a whole number of half wavelength so that a
node occurs at each reflecting surface.
The number of standing waves with wavelengths between λ and λ+dλ turned out to be
][,)( 5
8
4
λ
λ
π
λλ d
V
dN =
where V is the volume of the box.
Each individual wave contributes an energy KT to the radiation in the box. This result follows from classical
thermodynamics. The radiation in the box is in thermal equilibrium with the walls at temperature T. Radiation
is reflected from the walls because it is absorbed and quickly remitted by the atoms of the walls, which in the
process oscillate at the frequency of the radiation. At temperature T, the mean thermal kinetic energy of an
oscillating atom is KT/2. For a simple harmonic oscillator, the mean kinetic energy is equal to the mean
potential energy, so the mean total energy is KT.
Let us first calculate the energy density of radiation inside the cavity:
energy density = (number of standing waves per unit volume) x
(energy per standing wave)
or, u(λ)= ][, 6
8
4
KT
λ
π
The radiancy is related to the energy density (energy per unit volume) u(λ) at the wavelength λ according to
][),()( 7
4
λλ u
c
R =
This result follows from the classical electromagnetism by calculating the amount of radiation passing
through an element of surface area within the cavity.
Finally from Eq. [7], we get
Radiancy= ][],[)( 8
4
8
4
c
KTR
λ
π
λ =
3
Fig. 3: Electromagnetic radiation in a cavity
whose walls are perfect reflectors consists of
standing waves that have nodes at the walls.

4. This result is known as the Rayleigh-Jeans formula; based
firmly on the classical theories of electromagnetism and
thermodynamics, it represents our best attempt to apply
classical physics to understanding the problem of blackbody
radiation.
In Fig. 4, the radiancy calculated from the Rayleigh-Jeans
formula is compared with the observed radiancy from
blackbody spectra. The radiancy calculated from Eq. [8]
approaches the data at long wavelengths, but at short
wavelengths the classical theory (which predicts R
0→∞→ λas ) fails miserably. The failure of the Rayleigh-
Jeans formula at short wavelength is known as the “ultraviolet
catastrophe” and represents a serious problem for classical
physics, because the theories of thermodynamics and
electromagnetism on which Rayleigh-Jeans formula is
based, have been carefully tested in many other circumstances
and found to give extremely good agreement with experiment. It
was apparent in the case of blackbody radiation that the
classical theories would not work and that a new kind of
physical theory was needed.
Planck’s hypothesis and radiation law:
The new physics that gave the correct interpretation of thermal radiation was proposed by the German
physicist Max Planck in 1900.
Planck reasoned that the reflections at the walls of the cavity resulted from radiation being absorbed and
then quickly reemitted by the atoms of the wall; during this period the atoms would oscillate at a frequency
equal to the frequency the radiation.
Planck suggested that an oscillating atom can absorb or reemit energy only in discrete bundles (called
quanta). If the energy of the quanta were proportional to the frequency of the radiation, then as the
frequencies became large, the energy would similarly become large. Since no individual wave could contain
more than KT of energy, no standing wave could exist whose energy quantum was large than KT. This
effectively limited the high frequency (low wavelength) radiant intensity and solved the ultraviolet
catastrophe.
In Planck’s theory, each oscillator can emit or absorb energy only in quantities that are integer multiples of a
certain basic quantity of energy ε,
,.......,,, 321== nnE ε
where n is the number of quanta. Furthermore, the energy of each of the quanta is determined by the
frequency
νε h=
where h is the constant of proportionality now known as Planck’s constant.
So, the main points that establish Planck’s radiation law can be summarized as:
(1) An oscillator absorbs energy from the radiation field and delivers it back to the field in quanta
of 0,ε, 2ε, 3ε,…….etc. where ε is the quanta of energy propositional to frequency ν of the
oscillator.
(2) An oscillator cannot have arbitrary energy but must occupy one of a discrete energy sets
given by νε nh= , where n is an integer.
(3) The number of oscillators emitting particular energy is given by statistical distribution law of
Boltzmann, which is
][, 90
kT
r
eNNr
ε−
=
If N is the total number of Planck oscillators and E is the total energy, then the energy per oscillator is given
by
][, 10
N
E
E >=<
4
Fig. 4: The failure of the classical Raleigh-
Jeans formula to fit the observed radiancy. At
long wavelengths the theory approaches the
data, but at short wavelengths the classical
formula fails miserably.

5. Let N0, N1, N2, ………………Nr,……………..etc. be the number of oscillators having energies 0, ε, 2ε,
………………..rε,…………..etc respectively. Then
.......................................
............................
+++++=
+++++=
r
r
NrNNNE
and
NNNNN
εεε 210
210
20
Using Maxwell’s distribution formula [9], we can write
][,][
.............................................[
...............................
.....................................
11
1
1
1
2
01
0
2
0
0000
210
kTe
N
yN
yyyN
rxeNxeNxeNN
NNNNN
r
r
ε−
−
−−−
−
=−=
+++++=
+++++=
+++++=
where
xeyandkTx −
== ε
Similarly, we can write
][,
][
][
][
......].....................................[
..................................
...........................
)(
12
1
1
1
321
2
2
20
2
0
2
02
0
12
0
000
21
kTe
kTeN
z
kTeN
zeN
rzzzeN
kT
r
eNrkTeNkTeN
NrNNE
kT
rkT
r
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
εεε
ε
ε
−
−−
−−
−−
−−−
−
=
−
=−=
+++++=
++++=
+++++=
Using [11] & [12] in [10], we can find the following expression for the average energy of an oscillator
][, 13
11 −
=
−
=>=<
−
kT
h
e
h
kTe
N
E
E
ν
ν
ε
ε
The number of oscillator per unit volume in the frequency range υ and υ+dυ can be shown to be given by
][, 14
8
3
2
c
d
dN
νπν
=
When the average energy of the oscillator is multiplied by the number of oscillators per unit volume in the
frequency range υ and υ+dυ, then we obtain the energy density νν dE belonging to the range dυ.
Thus the energy density is given by
][,. 15
1
8
3
2
−
=
kT
h
e
h
c
d
dE
ν
ννπν
νν
The equation [15] for the energy density, using 2
λ
λ
ν
λ
ν
cd
d
c −
== & , can also be represented in terms of
wavelength as
5

6. ][,.
...
16
1
18
1
18
5
23
3
3
λ
λλ
π
λ
λ
λλ
π
νλ νλ
d
kT
hc
e
hc
cd
kT
hc
e
c
c
h
dEdE
−
=
−
=|=
Eq.[15] or [16] is known as Planck’s radiation law.
From [16], we can write
1
12
1
18
44 5
2
5
−
=
−
===
kT
hc
e
hc
kT
hc
e
hcc
densityenergyx
c
RRadiancy
λλ
π
λλ
π
λ ...)(
Case 1:
We shall now derive Rayleigh-Jeans law from Eq.[16]:
For large wavelength, λT is large.
Therefore,
]18[..........1
..........][
!2
1
1 2
kT
hc
kT
hc
kT
hc
e kT
hc
λ
λλ
λ
+≈
+++=
With the help of [18], we can write Eq.[16] as
]19[..........
8
11
1
.
hc8
1
1
.
8
4
5
5
λ
λ
π
λ
λ
λ
π
λ
λ
π
λ
λ
λ
d
kT
d
kT
hc
d
e
hc
dE
kT
hc
=
−+
=
−
|=
which is Rayleigh-Jeans law.
Case 2:
We shall now derive the Wein’s law from Planck’s law.
For small wavelengths, Tλ is small. Therefore,
1>>kT
hc
eλ .
In this case, we can neglect 1 in [
1−kT
hc
e λ ] and write [
1−kT
hc
e λ ] as kT
hc
e λ . With this
approximation, we can write
,.
1
1
.
8
5
λ
λ
π
λ
λ
λ d
e
hc
dE
kT
hc
−
|=
as
]26[,.
81
.
8
55
λ
λ
π
λ
λ
π
λ λ
λ
λ de
hc
d
e
hc
dE kThc
kT
hc
−
=|=
6

7. which is Wein’s law.
Example 3. Find the intensity of light emitted from the surface of the sun in the wavelength range 600.00 to
605.00 nm.
Solution : The exact result is found by integrating radiancy within the limit 600 to 605 nm. However, result
can also be found approximately by finding the area under the radiancy R(λ) curve between these limits. The
area within these limits can be found by calculating the height of the curve at the median wavelength λ =
602.5 nm and multiplying this by the width of the interval ∆λ = 5 nm.
First we evaluate hc/λkT at λ = 602.5 nm
X=
)5800)(/1038.1)(105.602(
)/10998.2().10626.6(
239
834
KKJm
smsJ
kT
hc
−−
−
××
××
=
λ
= 4.116
R(λ) =
( )1)105.602(
)/10998.2().10626.6(2
1
1
.
2
116.459
2834
5
2
−×
××
=
−
−
−
em
smsJ
e
hc
kT
hc
π
λ
π
λ
= 7.81 ×1013
W/m3
The intensity in the 5 nm range from 600 nm to 605 nm is than approximately
R(λ)∆λ = = (7.81× 1013
W/m2
) (5 ×10-9
m) = 0.39 MW/m2
7