CH 28 – 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. In this section we discuss
some of the phenomena that support the wave-particle nature of both light and particles
and quantum physics.
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.

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
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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 )
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.
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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
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?
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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
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.
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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
How much kinetic energy is imparted to the electron?
Solution:
KE 
hc


hc
1 1
1
1
 hc(  )  ( 6.6 x10  34 )( 3x108 )(

)
'
 '
0.05 x10  9 0.0507 x10  9
 5.47 x10 17 J  342 ev
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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.
Example:
What is the wavelength of a baseball traveling at 30 m/s?
Solution:
Mass of baseball = 0.145 kg
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
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.
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
2
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where px and x are the uncertainties with which one can determine the momentum and
position of the particle. (Note: Some textbooks give the uncertainty principle with h/4
on the right of the above equation. It depends on how the uncertainty is defined.)
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
2
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

 2.32 x10 8 m  23.2 nm
2 p 4 (4.56 x10  27 kg  m / s)
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?
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Solution:
E 
h
(6.63x10 34 J  s )

 4.06 x10  27 J / 1.6 x10 19 J / ev  2.6 x10 8 ev

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2 t
2 (2.6 x10 s )
Thus, the short lifetime has virtually no effect on how precisely one can determine the
rest mass energy.
Particle in a box
The wave nature of a particle can lead to quantization of energy if the particle is confined
within a finite region of space. A simple example is a particle confined to move freely in
one dimension along a line of length L. Since the particle can't exist beyond the ends of
the line, then the wave function must be zero at the ends. Thus,
Ln

2
, or  
2L
n
where n = 1, 2, .... Then the energy (kinetic) is
2
p 2 (h /  ) 2
2 h
En 

n
2m
2m
8mL2
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