CH 27 – Quantum Physics
Quantum mechanics is the branch of physics that deals with systems at the atomic level.
Some of the consequences of quantum mechanics are that energy is quantized, particles
have wave-like characteristics, and there are inherent uncertainties with which we can
determine the position, momentum, and energy of a particle.
Blackbody Radiation
Quantum mechanics was developed in the early part of the 1900’s in order to explain
some observations that could not be explained by classical physics. One of these was
Blackbody Radiation. All objects emit electromagnetic radiation by virtue of their
thermal energy. The radiation consists of a distribution of wavelengths, and the intensity
of the radiation and the wavelength distribution changes with the temperature of the
object. A blackbody is an object which absorbs all radiation that is incident upon it at all
wavelengths. One way to make a nearly perfect blackbody would be to have an object
with a cavity and a small pinhole in the cavity. Then if the pinhole was small, any
radiation impinging upon the pinhole would be absorbed and reflected inside the walls of
the cavity and would have little chance of getting back out through the pinhole. Thus, the
pinhole would be a blackbody (and would look perfectly black). It turns out that a
perfect blackbody is also a perfect radiator of electromagnetic radiation. That is, it will
emit as much or more radiation (per unit of surface area) as any other object at the same
temperature.
The figure to the right gives the intensity of radiation emitted by a blackbody as a
function of wavelength for different temperatures. Some important properties of the
distribution function are as follows.
1. There is a broad distribution of wavelengths.
2. The distribution peaks at a wavelength that
decreases with increasing temperature.
3. The total intensity of the radiation increases
rapidly with increasing temperature.
The wavelength corresponding to maximum
intensity is given by Wien’s displacement law,
which is
max T  constant  0.2898 x10 2 m  K
In this expression, the temperature T must be in
Kelvin.
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Example:
What is the peak wavelength of the electromagnetic radiation from a blackbody at 20oC?
Solution:
max 
0.2898 x10 2 m  K 0.2898x10 2 m  K

 9.89 x10  6 m  9.89  m
T
293K
This wavelength is in the infrared part of the spectrum and is not visible.
Example:
Now repeat the above calculation assuming T = 1000oC.
Solution:
0.2898x10 2 m  K 0.2898x10 2 m  K
max 

 1.27 x10  6 m  1.27  m
T
2273K
This wavelength is still in the infrared part of the spectrum. However, it is close enough
to the visible so that a good part of the distribution will be in the red region and the object
will look red hot.
The total intensity (power per unit area) of the radiation from all wavelengths is given by
Stefan’s law, which is
I
P
 e T 4
A
The quantity e is the emissivity and is a measure of how good the object radiates and
absorbs radiation. For a black body, e = 1. In general, 0 < e  1.  is the StefanBoltzman constant and has the value
  5.6704 x10 8 W  m 2  K 4
Example:
An object radiates 200 W of electromagnetic radiation at 20oC. How much does it radiate
at 200oC?
2
Solution:
I 2 T24 (473) 4


 6.79
I1 T14 (293) 4
I 2  6.79 I1  (6.79)( 200W )  1358W
To a certain extent, blackbody radiation can be described using classical physics.
Because of the temperature of an object, the charges in the object oscillate back and forth
and radiate electromagnetic radiation somewhat like tiny antennas. However, classical
physics predicts that the intensity of the radiation should continually increase as the
wavelength of the radiation decreases, whereas experimentally it is found that the
intensity of the radiation approaches zero as the wavelength decreases to zero. Planck
was able explain the observed dependence of the intensity on wavelength and
temperature by assuming that the charges in the radiator oscillated at discrete values
given by
E n  nhf
where n is an an integer and h = 6.6 x 10-34 Js. This was one of the first successful
theories of the quantum nature of matter and light.
Photoelectric Effect
The photoelectric effect is the ejection of electrons from a metal by incident light. This
phenomenon was first observed in the late 1800s by Hertz. An analysis of photoelectric
effect measurements provides strong evidence of the photon theory of light.
Light of a specific wavelength and frequency is incident upon a metal surface. The
ejected electrons are collected by a second metal surface and the resulting current is
measured. By applying a sufficiently large negative electric potential to the second
surface with respect to the first, the ejected electrons can be repelled and the current can
be reduced to zero. This stopping voltage gives the maximum kinetic energy of the
ejected electrons.
KEmax  eVs
Several results of the photoelectric experiment cannot be explained by the classical
theory of light:

No photoelectrons are produced if the frequency of the light is below a certain cutoff
value.
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
The maximum kinetic energy of the photoelectrons is independent of the intensity of
the light.

The maximum kinetic energy of the photoelectrons increases with the frequency of
the light.

The electrons are ejected from the metal surface almost instantaneously when the
light is applied.
Einstein was able to explain these results by proposing that light consisted of quanta
(photons) with energy
E  hf
where h is the constant that appeared in Planck’s theory of blackbody radiation.
Conservation of energy would require that the energy of the absorbed photon was
converted into work to extract the electron from the metal with the excess energy going
into the kinetic energy of the ejected electron. That is,
hf  KE  work .
The minimum work needed to extract an electron is called the work function, , for
which the kinetic energy of the electron is a maximum. Thus, we have
hf  KEmax   ,
or
KEmax  hf  
Example:
The work function for Zn is 4.31 ev. What maximum wavelength of light would be
required to produce photoelectrons from Zn?
Solution:
KEmax  0  hf    hc /   
  hc /   ( 6.6 x10  34 Js )( 3 x108 m / s ) /( 4.31evx1.6 x10 19 J / ev )
 2.88 x10  7 m  288 nm ( UV )
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If the wavelength of the light were 200 nm, what would be the maximum kinetic energy
of the ejected electrons?
Solution:
KEmax  hf    hc /    
 ( 6.6 x10  34 Js )( 3x108 m / s ) / 2 x10  7 m  ( 4.31evx1.6 x10 19 J / ev )
 3.00 x10 19 J  1.78 ev
X-ray Production
If very high energy electrons strike a metal, then x-rays are produced. The figure below
shows an x-ray tube in which the electrons are accelerated though a potential difference
and strike a tungsten target.
As the electrons are deflected by nuclei inside the target they loose kinetic energy by
radiating photons. The maximum photon energy would correspond to the electron being
completely stopped with all the energy being converted to that of a photon.
eV  hf max 
hc
min
,
or
min 
hc
eV
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So x-rays would be produced with a range of
wavelengths ranging from min to infinity. In addition
to this continuous range of wavelengths (called
bremsstrahlung radiation) there can be some discrete
wavelengths where the intensity is strong. These
characteristic wavelengths are a result of quantum
transitions within the atoms. The impinging electrons
may have enough kinetic energy to kick out an electron
from an inner orbit of an atom. An x-ray photon would
then be produced when an electron from an outer orbit
drops down to fill this vacancy.
Example:
What accelerating voltage would be needed to produce
x-rays whose shortest wavelength is 1 nm?
Solution:
V 
hc
emin

( 6.6 x10 34 Js )( 3 x108 m / s )
( 1.6 x10 19 C )( 1x10  9 m )
 1,238V
X-ray Diffraction
One of the important applications of x-rays is in analyzing the structure of materials. In
crystal diffraction, monochromatic x-rays are diffracted from a crystal lattice. The
diffracted intensity is a maximum
at angles that depend on the
wavelength, the types of atoms in
the crystal, and their atomic
spacing. The figure below shows
x-ray beams reflected from
adjacent atomic planes. In order
for the reflected beams to be in
phase, the additional distance
traveled by the lower beam must
be an integral number of
wavelengths. This so-called
Bragg condition is
2d sin   m ,
where m = 1, 2, 3, … and d is the distance between the atomic planes. Note that this
formula is similar, but not the same, as the formula for diffraction maximum from a
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diffraction grating. In the formula above,  is measured from the atomic planes, not the
perpendicular to the planes. Also, note the factor of 2.
Compton Effect
In the Compton Effect, high energy photons are scattered from the electrons in a solid
and the wavelength of the scattered photons is measured as a function of the scattering
angle. This effect, discovered by Arthur Compton, provides further evidence of the
quantum nature of light.
This process can be analyzed using conservation of energy and momentum. Assuming
the energy and momentum expressions for the photon and the relativistic expressions for
the energy and momentum of the electron, the difference in the wavelengths of the
scattered and incident photons can be calculated to give
  '  
h
( 1  cos  ) ,
mc
where  is the scattering angle of the photon.
Example:
If the wavelength of the incident photon is  = 0.05 nm, what is the wavelength of the
scattered photon at 45o?
Solution:
 
( 6.6 x10 34 Js )
 31
8
( 1  cos 45o )  7.1x10 13 m
( 9.11x10 kg )( 3x10 m / s )
'      0.05nm  0.0007nm  0.0507nm
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How much kinetic energy is imparted to the electron?
Solution:
KE 
hc


hc
1 1
1
1
 hc(  )  ( 6.6 x10  34 )( 3x108 )(

)

9
'
 '
0.05 x10
0.0507 x10  9
 5.47 x10 17 J  342 ev
Wave-Particle Duality of Matter
We have previously seen how light can exhibit both wave-like and particle-like
properties. The wave-like properties are demonstrated in interference and diffraction
effects. The particle-like properties are demonstrated in phenomena in which the concept
of the photon is important, such as the photoelectric effect and the Compton effect. For
the photon, the energy and momentum relationships are
E  hf and p 
h

De Broglie suggested that particles with a non-zero rest mass such as electrons and
protons should also have both particle-like and wave-like properties and postulated that
the above relationships for the photon should also hold for particles with mass.
Example:
What is the wavelength of an electron whose kinetic energy if 10 keV?
Solution:
Since KE << mc2, we can use the classical expressions to find the momentum.
KE 
1 mv 2
2
p2

2m
p  2m( KE )  2( 9.1x10  31 kg )( 1x10 4 ev )( 1.6 x10 19 J / ev )
 5.40 x10  23 kg  m / s
h
6.63 x10  34 J  s
 
 1.23 x10 11 m  0.0123 nm

23
p 5.40 x10
kg  m / s
This wavelength is of a sufficient size to use electrons to image materials on an atomic
and molecular level, as in transmission electron microscopes.
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Example:
What is the wavelength of a baseball traveling at 30 m/s?
Solution:
Mass of baseball = 0.145 kg
h
h
6.63x10  34 J  s
 

 1.52 x10  34 m
p mv ( 0.145kg )( 30m / s )
This wavelength is so small that it is impossible to detect the wavelike nature of a
baseball.
Wave Functions
Schroedinger postulated that the wave-like nature of particles with mass could be
described by a ‘wave function’, , that is a function of position and time. He postulated
a mathematical equation, now called the ‘Schroedinger equation’, which can be used to
solve for the wave function. The physical meaning of  is that 2 is a measure of the
probability of finding the particle at a certain position at a certain time.  is somewhat
like the electromagnetic field wave associated with the photon. The intensity of the
electromagnetic wave is proportional to E2 or B2.
Uncertainty Principle
One of the consequences of the wave concept of particles is that there is an uncertainty in
precisely locating the position of the particle.
wave function
probability function
In general, the wave function and the probability function have a finite extent in space, as
shown above. So, rather than having an exact location of the position of the particle, we
could only say that it has a probability of being within a certain region of space.
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The fact that the wave function is localized rather than being a sinusoidal wave extending
infinitely through space means that the wavelength, and thus the momentum, of the
particle are also not precisely defined. The more we try to localize the wave function and
the position of the particle, the more uncertain is the wavelength and momentum of the
particle. This so-called Heisenberg uncertainty relationship is given by
p x x 
h
4
where px and x are the uncertainties with which one can determine the momentum and
position of the particle.
Another way of looking at the uncertainty relationship is as follows. If we wanted to try
to precisely determine the position of a particle, we would have to shine light on it. Since
light is a wave, the best we could do is determine the position to within approximately
one wavelength of the light due to diffraction effects. So, we would have x  . In
addition, at least one photon with momentum p = h/ would collide with the particle and
transfer at least this much momentum to the particle, which would increase its
momentum uncertainty. If we try to decrease the uncertainty in determining x by
decreasing the wavelength of the light, then we increase the uncertainty in the momentum
by increasing the momentum transfer from the photon.
There also exists a similar uncertainty relationship between energy and time, given by
Et 
h
4
This says that it is impossible to precisely determine energy of a particle in a finite time
t. The shorter the time of the measurement, the larger is the uncertainty in the energy
and vice versa.
Example:
An electron is observed to have a speed of 5 x 105 m/s to within an accuracy of 1%.
What does this say about the uncertainty with which we can locate the electron?
Solution:
p  mv  ( 9.11x10 31 kg )( 5 x103 m / s )  4.56 x10 27 kg  m / s
x 
h
6.63x10 34 J  s

 1.16 x10 8 m  11.6 nm
4 p 4 ( 4.56 x10  27 kg  m / s )
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Example:
The half life of the charged pi meson is 2.6 x 10-8 s and its rest mass energy is 139.6 Mev.
What is the uncertainty in its rest mass energy?
Solution:
E 
h
( 6.63x10 34 J  s )

 2.03x10  27 J / 1.6 x10 19 J / ev  1.3x10 8 ev

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4 t
4 ( 2.6 x10 s )
Thus, the short lifetime has virtually no effect on how precisely one can determine the
rest mass energy.
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