Download Chapter 27 Quantum And Relativistic Physics

Physics Including Human Applications
592 Chapter 27
Quantum And Relativistic Physics
GOALS
When you have mastered the contents of this chapter you should be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
Planck's constant
length contraction
Planck's radiation law
time dilation
Wien's law
mass-energy equivalence
photon
Compton scattering
photoelectric effect
complementarity principle
work function
deBroglie wave
relativistic mass
uncertainty principle
Quantum and Relativistic Problems
Solve problems involving Wien's law, photoelectric effect, Compton scattering,
deBroglie waves, the uncertainty principle, and relativistic physics formulations.
Tunnel Effect
Define and explain the physical significance of the tunnel effect.
PREREQUISITES
You should have mastered the goals of Chapter 4, Forces and Newton's Laws, Chapter
5, Energy, and Chapter 21, Electrical Properties of Matter, before starting this
chapter.
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
593 Chapter 27
Quantum And Relativistic Physics
27.1 Introduction
The topics we have presented in the proceeding chapters have been in the realm of
classical physics. The concepts of mass, velocity, momentum, and energy were applied
to macroscopic systems, and we expected the variables of these systems to have
continuous ranges of values. It was assumed that classical physics based on Newton's
mechanics, Maxwell's electromagnetic theory, and thermodynamics held the answers to
all the problems of the physical universe. Some physicists at the end of the nineteenth
century believed that the development of physics was complete. On the contrary, the
new models developed in physics during the first half of the twentieth century may
well signify one of the most exciting periods of human intellectual history. Quantum
physics was born and showed that classical physics failed to describe physical
phenomena of particles of 10-10 m dimensions. When velocities approach the speed of
light, Einstein's special theory of relativity replaced classical physics as the working
theory. In this chapter you will study the foundations of quantum and relativistic
physics and some consequences of the new physics.
27.2 Planck's Radiation Law and Its Derivations
Perhaps you have noticed the change of color that occurs when you dim the
incandescent lights of a room using a dimmer switch. When the switch is completely
on, the lights are bright and give off white light. As you dim the lights, you can see that
the light given off is not only less bright but is also of a different color. The dim light is
red-orange in color. We can conclude that the color, or frequency, of light emitted from
an incandescent bulb changes as we change the temperature of the filament in the bulb.
This is one phenomenon that classical physics failed to explain, that is, the frequency
distribution of the electromagnetic radiation emitted from heated objects. This radiation
is emitted in a continuous range of frequencies in a characteristic way that depends on
the temperature of the radiation source (Figure 27.1). Classical physics as applied by
Lord Rayleigh failed to describe this distribution. In fact, Rayleigh's work predicted the
"ultraviolet catastrophe," unlimited energy output at the wavelengths approaching zero.
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
594 Prior to Rayleigh's results Max Planck reported his solution to the problem. He
combined classical equations that described each end of the distribution curve with the
result being an empirical equation that fit the entire distribution curve. In order to
explain his empirical formula, Planck had to postulate that vibrating objects, or
oscillators, were responsible for the emission and absorption of thermal radiation.
Furthermore, these oscillators could only have discrete energies. They could exist only
in energy states that were integral multiples of a fundamental oscillator energy or
quantum, when the quantum energy is given by hf , where h is Planck's constant, and f is
the frequency of the oscillator.
The value of h necessary to make Planck's equation fit the radiation curves shown in
Figure 27.1 is h = 6.63 x 10-34 joule-seconds. According to Planck's postulate, the various
oscillators in a radiation source could only have energies given by
En = nhf or En = nhc/λ
(27.1)
since fλ = c (see Chapter 16) where n = 1, 2, 3, ..., c is the speed of light in meters per
second, f is the frequency in cycles per second, and λ is the wavelength in meters.
The equation that Planck found to describe the intensity or radiation from a perfect
radiator as a function of the temperature of the radiator and the wavelength of the
radiation is
E = [8πhc]/[ λ5 (exp(hc/λkT) - 1)]
(27.2)
λ
where c is the speed of light, λ is the wavelength of the radiation, k is Boltzmann's
constant, T is the absolute temperature of the radiator and E is the energy intensity
(watts/m2) emitted per unit wavelength (m).
λ
Planck's theory was not an immediate success as its postulates were quite a break
from some of the ideas of classical physics. This was much to the dismay of Planck who
had hoped to explain his quantum postulate in classical terms. After all, he had used
classical electromagnetic waves coupled to some special oscillators within the radiator.
Two important results of Planck's theory were the derivation of the Stefan-Boltzmann
law (Equation 11.8) and Wien's law (Equation 27.5). The Stefan- Boltzmann law, which
was discussed in Chapter 11, describes the rate at which energy is radiated from a body
as a function of the absolute temperature of a body. The mathematical equation is
I = P/A = σT4 watts/m2
(11.8)
where σ is Stefan's constant = 5.67 x 10-8 watts/m2-°K4, T is the absolute temperature in
degrees Kelvin, and I is the total energy per unit time per unit area emitted by a perfect
thermal radiator, or the ratio of the emitted power P divided by area A. For radiators
other than these perfect blackbody radiators we can write
I = εσT4
(27.3)
where ε is the emissivity of the object (0 < ε <1).
If an object of emissivity ε is maintained at a temperature T1 in a chamber at a
temperature of T2, the net radiation from the object is equal to the difference between
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
595 the radiation emitted from the object and absorbed from the chamber,
I = εσ (T14 - T24)
(27.4)
Wien's law is an empirical equation for the relationship between the temperature of a
perfect radiator and the wavelength of the radiation of maximum intensity. It can be
derived from Planck's law by finding the wavelength for maximum radiation intensity
from Equation 27.2. The temperature of a radiating object and the wavelength of the
maximum intensity of the emitted radiation have a constant product,
λmaxT = 2.88 x 10-3 m-°K
(27.5)
EXAMPLES
1. Given that the sun has its intensity peak output at 500 nm, find its effective blackbody
temperature.
Using Wien's law we can write:
λmaxT = 2.88 x 10-3 m-°K
T = 2.88 x 10-3 (m°K)/5.00 x 10-7 m, T = 5760°K.
This is the effective surface temperature of the sun.
2. The emissivity of the human skin is close to one. Compare the rate of energy loss of a
person (skin area = 1.5 m2) inside when the temperature is 27°C, and outside when
the temperature is -23°C. (Assume skin temperature is 32°C.)
From Equation 27.4 above you can write,
∆E/∆t = IA = εσA(T14 - T24)
where T1 is body temperature = 305°K and T2 is either the inside temperature
(300°K) or the outside temperature (250°K),
(∆E1/∆t)/(∆E2/∆t) = (σA (3054 - 3004))/( σA (3054 - 2504)) = 0.116
The rate of energy lost from the body by radiation is greatly reduced in a warm
environment.
27.3 Photoelectric Effect
While Heinrich Hertz was investigating the properties of electromagnetic radiation in
1887, he observed that the metallic electrodes of his system were more easily discharged
when the electrodes were illuminated by ultraviolet light. This interaction between light
and metal surfaces was subsequently studied in more detail and became called the
photoelectric effect. The photoelectric effect is explained by a model of light energy
interacting with a metal surface to cause electrons to be emitted from the surface. The
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
596 light energy is partially absorbed by the metal and partially transformed into the kinetic
energy of the emitted electrons.
The photoelectric effect is another example of quantitative experimental results as
shown in Figure 27.2 that could not be explained by the models of classical physics. On
the basis of the wave model of light, we found in Chapter 19 that the energy in a light
beam is proportional to its intensity, and thus an intense red light should be able to
provide enough energy to free an electron from the metal surface. The facts did not
support this; the observations showed that below some threshold frequency, no matter
how intense the light, electrons were not emitted from the surface (Fig. 27.2a). If the
intensity of the light is increased, the photoelectric current is increased if the frequency
of the light is above the threshold value (Figure 27.2b). The kinetic energy of the emitted
photoelectrons (as measured by the voltage it takes to stop them from reaching the
anode) is proportional to the frequency of the incident electromagnetic radiation (Figure
27.2a).
In 1905 Einstein offered his theory to explain the photoelectric effect. He applied
Planck's quantum postulate to the incident light. He argued that the incident light was
made up of quanta of light, which are called photons, each of energy hf. The intensity of
the light beam is a measure of the number of photons per unit area per second. Einstein
reasoned that the quantum interaction is an all or none interaction. If the photons do
not have an energy equal to the energy with which electrons are bound to the metal
(this binding energy is called the work function), there is no electron emission no matter
how intense the light may be. There is no way for the electrons in the metal to store
subthreshold photons until the threshold energy is reached. To support this,
experiments show no observable time lag between time when light is turned on and
when electrons appear.
If the photon has energy greater than the work function, the excess energy goes into
the kinetic energy of the photoelectron. This is expressed by the following equation:
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
597 energy of the incident photon of light = kinetic energy of the photoelectron + the work
function of the metal
E = hf = KE + W
(27.6)
where E is the photon energy, hf =hc/λ, W is the work function of the metal, and KE is
the kinetic energy of the photoelectron. The electron volt (1 eV = 1.60 x 10-19 J) is a
convenient energy unit for microscopic physics in this and the remaining chapters and
is the energy increase of an electron when it is accelerated through a potential of one
volt.
EXAMPLE
Light of wavelength λ = 400 nm strikes a metal and produces photoelectrons that are
brought to rest by 1.30 V. Find the work function of the metal.
The energy of the photons is E = hf = hc/λ = (6.63 x 10-34 J-sec) x (3.00 x 108 m/sec)/4.00 x
10-7 m
E = 4.97 x 10-19 J = 3.11 eV
where 1 eV = 1.60 x 10-19 J. The retarding potential of 1.30 V represents the stopping
potential for the photoelectrons. Thus the kinetic energy of the photoelectrons is e x Vs
= 1.30 eV. From Equation 27.6, we find the work function,
W = hf - KE = 3.11 eV - 1.30 eV
= 1.81 eV = 2.90 x 10-19 J
27.4 Relativistic Physics
Before we continue our discussion of the quantum properties of light, let us digress to
consider some other work that Albert Einstein did during 1905 when he was a 26-yearold patent examiner in the patent office in Berne, Switzerland. Einstein was studying
the electrodynamics of moving objects. He wrote a paper in which he proposed the two
postulates that form the basis for the special theory of relativity. Einstein's special
theory of relativity has some important consequences for the realm of quantum physics.
These consequences are derived from Einstein's two fundamental postulates.
Postulate 1: The laws of physics are the same in all reference systems at rest or moving at a
constant velocity.
Postulate 2: The speed of light is always measured to be the same value in a vacuum, and it is
independent of the state of motion of the source or observer.
We will now give some of the results of Einstein's special theory of relativity along
with the physical significance of these results.
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
598 Length Contraction: All moving lengths appear to shrink in the direction of motion.
This length contraction is frequently called Lorentz contraction. There is no contraction of
lengths perpendicular to the velocity v. This phenomenon seems to contradict everyday
experience, but it has been verified many times in modern physics experiments. The
equation for the apparent length L in the direction of motion for an object moving with
speed v is given by:
L = Lo SQR RT(1 - v2/c2)
(27.7)
where Lo is the length of the object in a system in which it is at rest, v is the speed of the
moving object and c is the speed of light.
Time Dilation: Moving clocks run slow.
Again this result seems to contradict common experience, but it too has been verified
many times in modern experimental physics. The equation for a time interval from the
point of view of a stationary observer Δto is:
∆to = ∆t SQR RT(1 - v2/c2)
(27.8)
where ∆t is the time period shown by the clock moving with the apparatus, v is the
speed of the motion of the apparatus, and c is the speed of light.
Consider the following experiment which illustrates Lorentz contraction and time
dilation. This experiment has been repeated in several ways and in each case there is
complete agreement with the consequences of Einstein's postulates.
EXAMPLE
Muons are a common component of cosmic rays that strike the top of the earth's
atmosphere. They are unstable particles that decay with an average lifetime of 2.00 x 10-6
sec when at rest. An experiment was carried out in which the number of muons
incident at the top of a mountain 3000 m above sea level was counted. These particles
were moving with a speed of 0.998c. In an average lifetime a muon would travel a
distance of only about 600 m (~c x 2 x 10-6 sec), and thus no muons would be expected to
reach sea level. When the experiment was carried out it was found that most of the
muons do reach sea level. The explanation for these observations comes from Einstein's
equation as follows: From the point of view of a person in a lab on the mountain, the
actual decay time of the muon is slowed down because the muon is a moving clock: its
average lifetime moving at speed 0.998 c was
∆to = ∆t / SQR RT(1 - v2/c2) = 2.00 x 10-6 sec / SQR RT(1 - 0.9982) = 31.6 x 10-6 sec
and the distance the muons traveled in this time was 0.998c x 31.6 x 10-6 = 9500 m, which
is greater than the mountain height, thus explaining why so many muons reached sea
level. From the point of view of the moving muon the physical result must be the same,
but here we see that the mountain (which is moving at speed 0.998c with respect to the
muon) contracts to a height of H = 3000 SQR RT(1 - 0.9982) or H = 190 m and thus the
600 m traveled is enough to reach sea level. Here we see an example how equations of
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
599 special relativity give the correct answer in each reference system.
Einstein's special theory of relativity has several important modifications for the
dynamics of particles. In order to preserve the form of Newton's second law, namely
that impulse is equal to the change in momentum (Equation 6.2), the relativistic
momentum is given the following definition
p = {m0/SQR RT(1 - v2/c2)} v
(27.9)
where m0 is the mass of the particle when it is at rest, v is the velocity of the particle, v is
the speed, or magnitude of the velocity, of the particle, and c is the speed of light. You
may notice that we can write Equation 27.9 in exactly the same form as the usual
momentum equation if we invent a quantity m that is equal to the rest mass m0 divided
by SQR RT(1 - v2/c2). We call this quantity the relativistic mass m,
m = m0/SQR RT(1 - v2/c2)
(27.10)
Mass is a measure of inertia. This equation shows us that the inertia of a particle
becomes indefinitely large as its speed approaches the speed of light. This means that it
becomes more and more difficult to accelerate a relativistic object and that it is
physically impossible to accelerate an object to the speed of light.
Einstein showed that as a result of his postulates the total energy of a relativistic
particle could be expressed in the now famous equation,
E = mc2
E = m0c2/SQR RT(1 - v2/c2)
(27.11)
where E is the total energy, m is the relativistic mass, and c is the speed of light. If the
rest energy of the particle is defined as m0c2, then the kinetic energy of a relativistic
particle can be written as
KE = E - m0c2 = m0c2 [(1/ SQR RT(1 - v2/c2)) - 1]
(27.12)
The mass-energy equivalence formula as expressed in Equation 27.11 shows that in a
fundamental way mass must be considered in a new formulation of the conservation of
energy. The difference in masses of objects due to changes in energy in classical physics
are too small to be measured, but this mass-energy equivalence is the basis of the
nuclear binding energy processes which we will discuss in Chapter 31.
EXAMPLE
Consider the mass equivalent to a potential energy increase of 40,000 J. (This is about
the energy required to lift a 10 kg mass from sea level to the top of the Empire State
Building.) The equivalent mass would be given by
m = (4.00 x 104 J)/(9.00 x 1016 (m/sec))2 = 0.444 x 10-12 kg = 4.44 x 10-10 g
This is a mass much too small to be detected with our best analytical balances.
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
600 In many atomic and nuclear processes the relativistic form of the kinetic energy of a
particle should be used. In problem 23 at the end of the chapter you are asked to show
that in the classical region of physics (where v is much smaller than c), the kinetic
energy reduces to (½) m0v2.
27.5 Compton Effect
The photon model of light gives light particlelike characteristics instead of the wave
characteristics we discussed in Chapter 19. In 1922 A. H. Compton reported that he had
observed elastic collisions between x-ray photons and electrons. He was able to explain
his results by assigning a momentum to the photons given by the equation:
p = h/λ = hf/c
(27.13)
where p is the momentum of a photon, h is Planck's constant, λ is the wavelength of the
x rays, c is the speed of light, and f is the x- ray frequency.
An experimental set up to illustrate the Compton effect is shown in Figure 27.3a. The
intensity of x rays scattered from carbon block is measured as a function of the
scattering angle and scattered x-ray wavelength. Loosely bound electrons may be freed
from the carbon by the incident x-ray photons. The freed electrons are called
photoelectrons.
The scattered photons have less energy than the incident photons. Thus, in addition
to the unmodified x rays, scattered x rays with a slightly greater wavelength are
detected at all angles except the zero scattering angle.
Compton observed collisions like the ones shown in Figure 27.3, and he could
explain the results by using the conservation of momentum applied to the photon-
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
601 electron collision. The following equations are derived from the application of the
conservation of energy and momentum to this collision (Figure 27.3b).
MOMENTUM CONSERVATION
Resolving the momentum vector into its components,
x-components:
p x = p 'x + pex
λ
λ
hf/c = (hf'/c) cosφ + pe cosθ
y-components:
p y = p 'y - pey
λ
λ
0 = (hf'/c)sinφ - pe sinθ
ENERGY CONSERVATION
hf = hf' + KE
where KE is the kinetic energy of the electron.
By the time of Compton, it was known that the classical equations for the
interchange momentum and kinetic energy of a moving object had to be modified for
objects traveling at high velocities. The modified equations for the kinetic energy and
the momentum for relativistic electrons are given by Equations 27.9 and 27.12. These
equations can be solved to obtain the following equation for the wavelength of the
scattered x-ray photon in terms of scattering angle and incident photon wavelength:
λ' -λ =( h/m0c) (1 - cosφ)
(27.14)
where λ' is the scattered photon wavelength, λ, the incident photon wavelength, and
h/m0c, is a constant called the Compton wavelength, which equals 2.43 x 10-12 m.
EXAMPLE
Given that the incident photon had a wavelength of 12.4 x 10-12 m, and the scattered
photon recoiled at 180°, find the energy transferred to the electron.
(Note: when φ = π, then the energy transferred is maximum.)
First calculate the energy lost by the photon because of the scattering:
λ' - λ =( h/m0c) (1 - cos 180 °) = 2h/m0c
λ' = λ + 2h/m0c = (12.4 + 4.86) x 10-12 m = 17.3 x 10-12 m
kinetic energy of the electron = hc/λ - hc/λ' = energy lost by the photon
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
602 energy lost by the photon = (1012/12.4 - 1012/17.3)hc
= (6.63 x 10-34 x 3.00 x 108 ) x ((4.9/(12.4 x 17.3)) x 1012 )
= 45 x 10-16 J = 28.1 keV where 1 keV = 1.60 x 10-16 J.
27.6 Complementarity Principle
The Compton effect and the photoelectric effect evoke a particle model for light. The
particle model of light is used to describe the interaction of electromagnetic radiation
with matter; the classical wave model is used to explain propagation phenomena. We
use two models for light, one model is particlelike and the other is wavelike. The aspect
of light that we observe in a particular experiment depends on the nature of the
experiment. Niels Bohr would cite this as an example of the complementarity principle.
The wave and particle models of light complement each other and give us a theory that
encompasses the many potentialities of both particle and wave behavior. Bohr applied
the complementarity principle to areas outside of physics. Bohr's goal was to discover
the interrelationships among all areas of knowledge. He viewed the complementarity
principle as an important step toward his goal. In biology, Bohr presented the molecular
model of living systems and the teleological approach to the causality model of living
systems as complementing components of the organization of the living system. In
psychology, the complements may be behaviorism and self-conscious freedom of
choice. (What complementary models would you suggest for a political system or
organization?)
27.7 The deBroglie Wave
In 1924 Louis deBroglie proposed that material particles such as electrons might have a
dual nature. He suggested that the wavelength of a particle should be defined by the
same equation that applies to the photon:
λ = h/p
(27.15)
where h is Planck's constant and p is the magnitude of the momentum of the particle.
The waves associated with matter are not electromagnetic waves. These waves are
probability waves. The intensity of the probability wave at a point is a measure of the
probability of finding the particle at the point. The test of this idea would be to do a
wave experiment with particles. One experiment that depends entirely on the unique
features of waves is a diffraction experiment. Such an experiment was done by Clinton
J. Davisson and L. H. Germer in 1924. They found that electrons were diffracted by
crystals. The results were explained by using Equation 27.15 for the electron wave
diffracted by a crystal lattice with spacing the same order of magnitude as the electron
wavelength. The wave model of electrons can be used to explain the operation of the
electron microscope. Small electron wavelengths give high resolving power for the
electron microscopes.
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
603 EXAMPLES
1. Davisson and Germer used a beam of 54.0-eV electrons in their electron diffraction
experiment. Find the wavelength of these electrons.
The wavelength l and the momentum p are given by
λ = h/p p2/2m0 = KE = eV
where e is the charge of the electron, V is the voltage, h is Planck's constant, and m0 is
the electron rest mass of 9.11 = 10-31 kg.
p= (2 m0eV)1/2
λ = h/(2 m0eV)1/2
λ = 6.63 x 10-34 J-sec / [2 x (9.11 x 10-31 kg) x 1.60 x 10-19 C x (54.0 V)]1/2
= 6.63 x 10-34 J-sec / 3.97 x 10-24 kg-m/sec
= 1.67 x 10-10 m = 0.167 nm
2. Find the speed v of a neutron that has a deBroglie wavelength of 0.100 nm.
mv = h/λ = 6.63 x 10-34 J-sec / 1.00 x 10-10 m
v = 6.63 x 10-24 / 1.67 x 10-27 = 3.97 x 103 m/sec
This is a slow neutron with a kinetic energy of 0.08 eV. This is an energy comparable to
thermal energies at room temperature and thus such neutrons are called thermal
neutrons. Thermal neutrons are used in neutron diffraction studies to measure the
distances between lattice planes in solids. Neutron diffraction is especially useful in
studying organic crystals containing hydrogen. (Can you explain why neutrons are
useful for such studies while x rays and electrons are not very useful?)
27.8 Uncertainty Principle
We have now arrived at an understanding of a property of a system called a particle
(e.g., an electron) that can be thought of as a particle or as a wave. At one instant we can
think of the electron as a small chunk of matter with an electrical charge e located at
some point in space. The next instant we can think of the electron as a cloud of
probability waves that extends throughout space. What sense then can we make of the
idea of the position of a particle?
Werner Karl Heisenberg formulated his now famous uncertainty principle in 1927 in
order to clarify the use of the terms position andvelocity when referring to quantum
phenomena. The various forms of the uncertainty principle express limits to the
preciseness of our knowledge of the magnitudes of pairs of physical variables.
Algebraically stated the Heisenberg uncertainty principle can be written as
The product of the uncertainty in momentum and position is greater than Planck's constant
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
∆px ∆x ≥ h
604 (27.16)
where ∆px is the uncertainty in the x-component of the momentum, ∆x is the uncertainty
in the position of the object in the x- direction and h is Planck's constant. Another
equivalent form is:
∆E∆t ≥ h
(27.17)
Here ∆E is the uncertainty in the energy, and ∆t is the uncertainty in the time. These
uncertainty relationships must be understood as a part of our models to explain nature;
they are not based on limitations of measuring instruments. As a result of these
relationships, the certainties of classical physics are impossible in quantum physics. In
fact, quantum theory is a theory based on a probabilistic understanding of events and
physical variables. The uncertainty principle is a consequence of this fundamental
model of physical reality as it is now formulated by physicists.
EXAMPLES
1. Find the uncertainty in velocity of a 1.0-gm particle if its position is measured to
within 0.1 mm.
m∆v = ∆p ≈ h/∆x
∆v ≈ h/m∆x = 6.63 x 10-34 J-sec / 1.0 x 10-3 kg x 10-4 m = 6.63 x 10-27 m/sec
This is an uncertainty far below any measurement capability.
2. Find the uncertainty in the energy of photons emitted in the time of 10-8 sec. (This is a
characteristic time for excitation of many atomic systems.)
∆E ∆t ≈ h
∆E ≈ 6.63 x 10-34 J-sec / 10-8 sec = 6.6 x 10-26 J = 4.1 x 10-7 eV
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
605 27.9 Tunnel Effect
One of the results of quantum theory is the prediction of events that are forbidden by
classical physics. A potential-energy barrier exists wherever a repelling force acts to
restrict a particle from a region of space. Figure 27.4 shows an energy well formed by a
finite barrier on one side and an infinite barrier on the other side. (An infinite barrier
designates a forbidden region of space.)
In this case let us assume we can confine electrons to a region of 1.00 nm using a 2.00-eV
energy barrier. This uncertainty in x gives a corresponding uncertainty in momentum of
h/∆x = 6.63 x 10-34 J-sec/1.00 x 10-9 m = 6.63 x 10-25 kg-m/sec
The corresponding uncertainty in energy is given by the equation,
∆E = (∆p)2/2m = [6.63 x 10-25 kg-(m/sec)2]2 / 2 x 9.11 x 10-31 kg = 2.41 x 10-19 J
∆E = 1.51 eV
This uncertainty means that an electron with energy of 0.5 eV or more has an
uncertainty in energy sufficient to allow it to get over the top of the 2.00-eV barrier.
Since the uncertainty in x is equal to the width of the barrier we cannot say with
certainty on which side of the barrier the electron is. In fact, a rigorous quantum
mechanical treatment of this problem shows that the electron has a probability of
tunneling through the barrier. As the barrier height and/or barrier width is increased,
the probability for tunneling through it decreases. This quantum tunneling phenomena
is helpful in understanding alpha decay in radioactive nuclei and electron behavior in
semiconductors.
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
606 SUMMARY
Use these questions to evaluate how well you have achieved the goals of this chapter.
The answers to these questions are given at the end of the summary with the number of
the section where you can find related content material.
Definitions
1. Planck's radiation law provides a correct equation for the graph of radiation ______ as
a function of ______.
2. Wien's law could be used with the spectral distribution of a radiation source to find
a. Planck's constant
b. velocity of light
c. temperature of source
d. deBroglie wavelength
e. source size
3. The photon is defined as the smallest energy packet for a given frequency and is
equal to
a. h/λ
b. hf
c. hc/λ
d. c/λ
e. hc
4. Compton scattering gives experimental evidence that the photon has particle like
a. wavelength
b. frequency
c. velocity
d. momentum
e. mass
5. The photoelectric effect shows that the electron emitted from a metal has energy that
depends on
a. metal material
b. photoelectric current
c. photon energy
d. light intensity
e. none of these
6. The uncertainty principle states that it is impossible to know exactly both a particle's
momentum and
a. energy
b. position
c. velocity
d. acceleration
e. mass
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
607 7. If a proton, which has an electric charge of +e and a rest mass equal to about 1900
times the rest mass of the electron, and an electron are accelerated from rest through
the same potential difference, the ratio of proton de Broglie wavelength to electron
de Broglie wavelength will be
a. one
b. zero
c. greater than 1
d. less than 1
e. none of these
8. An example of the complementarity principle is
a. particle- wave nature of matter
b. particle- wave nature of radiation
c. momentum-energy of particles
d. time-position of particles
e. none of these
9. Einstein's mass-energy equivalence formulation predicts a mass change associated
with energy change ∆E to be
a. ∆E / hƒ
b. ∆E / ∆p
c. hƒ / ∆E
d. ∆E / c2
e. ∆p / ∆e
10. A long stick is observed to have a length L as it moves past a speed v; its length in its
own rest frame is
a. L /SQR RT(1 -(v2/c2))
b. L SQR RT(1 -(v2/c2))
c. L
d. L(V/c)
e. L(c/v)
11. A clock moving past an observer at speed v appears to her to have a period t. The
period of a clock in a rest frame is
a. t /SQR RT(1 -(v2/c2))
b. t SQR RT(1 -(v2/c2))
c. t
d. t(v/c)
e. t(c/v)
Quantum and Relativistic Problems
12. The peak wavelength for a 10,000°K source will be
a. 2.9 x 10-7 m
b. 2.88 x 10-1 m
c. 1.2 x 10-12 m
d. 2.4 x 10-12 m
e. 4.8 x 10-12 m
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
608 13. The maximum increase of a photon wavelength undergoing Compton scattering is
(in meters)
a. 0.24 x 10-12
b. 0
c. 1.2 x 10-12
d. 2.4 x 10-12
e. 4.8 x 10-12
14. If a photon has an energy equal to the rest energy of an electron, then its wavelength
is
a. 2moc2 / hƒ
b. h / moc
c. hƒ / 2moc2
d. hƒ / 2 moc
e. 0
15. If the uncertainty in position of a microbe (10-9 kg) is 10-6 m, the uncertainty in its
speed will be (in m/sec)
a. 6.63 x 10-34
b. 6.63 x 10-28
c. 6.63 x 10-19
d. 6.63 x 10-25
e. zero
16. The length of a meter stick moving with speed 0.8c in the direction of its length will
appear how long to an observer at rest.
a. 80 cm
b. 64 cm
c. 36 cm
d. 125 cm
e. 60 cm
17. The period of a clock in its rest frame is one second. When moving at a speed of 0.6c
the period will appear to a laboratory observer to be
a. 1.67 sec
b. 0.6 sec
c. 0.64 sec
d. 1.25 sec
e. 0.8 sec
Tunnel Effect
18. The physical significance of the tunnel effect of quantum physics is that particles are
found to tunnel through regions where
a. PE
b. KE
c. KE + PE
d. KE > 0
e. none of these
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
609 Answers
1.
intensity, wavelength
(Section 27.2)
2. c. (Section 27.2)
3. b, c (Section 27.2)
4. d (Section 27.5)
5. a, c (Section 27.3)
6. b (Section 27.8)
7. d (Section 27.7)
8. a, b (Section 27.6)
9. d (Section 27.4)
(or
frequency) 10. a (Section 27.4)
11. b (Section 27.4)
12. a (Section 27.2)
13. e (Section 27.5)
13. e (Section 27.5)
14 b (Sections 27.2, 27.4, 27.7)
15. c (Section 27.8)
16 e (Section 27.4)
17. d (Section 27.4)
18. b (Section 27.9)
ALGORITHMIC PROBLEMS
Listed below are the important equations from this chapter. The problems following the
equations will help you learn to translate words into equations and to solve singleconcept problems.
Equations
En = nhf or En = nhc/λ; n = 1, 2, 3,. . .
(27.1)
E = [8πhc]/[ λ5 (exp(hc/λkT) - 1)]
(27.2)
I = εσT4
(27.3)
λmaxT = 2.88 x 10-3 m-°K
(27.5)
E = hf = KE + W
(27.6)
L = Lo SQR RT(1 - v2/c2)
(27.7)
∆to = ∆t SQR RT(1 - v2/c2)
(27.8)
p = {m0/SQR RT(1 - v2/c2)} v
(27.9)
m = m0/SQR RT(1 - v2/c2)
(27.10)
E = m0c2/SQR RT(1 - v2/c2)
(27.11)
p = h/λ = hf/c
(27.13)
λ' -λ =( h/m0c) (1 - cosφ)
(27.14)
λ = h/p
(27.15)
∆px ∆x ≥ h
(27.16)
∆E∆t ≥ h
(27.17)
λ
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
610 Problems
1. The frequency of a given source of green light is 5.5 x 1014 Hz. What is the energy of
this radiation, i.e., the energy of each photon?
2. If the absolute temperature of a filament in a light bulb is 2500°K, what is the
wavelength for the maximum intensity of the radiation from the hot filament? What
is its corresponding energy per photon of radiation? Is this bulb usable as a source of
light for reading?
3. Two bodies are exactly alike except in temperature, one is at absolute temperature of
T and the other is at absolute temperature of 1.5 T What is the ratio of the rates of
energy emission from the two bodies?
4. The value of the work function of potassium is 2.25 eV. What is the longest
wavelength that will produce a photoelectron from a potassium surface?
5. What is the momentum of a photon that has a frequency of 5.50 x 1014 Hz?
6. What is the wavelength that is associated with you when you are walking at 5
km/hour? What do you think of the prospects of detecting this wavelength?
7. Suppose you know your momentum (as for example in problem 6) to within 0.1
percent. What is the limit upon knowing your exact location at that time? Comment
upon this result.
Answers
1. 3.6 x 10-19 J = 2.3 eV
2. 1.15 x 10-6 m, 1.73 x 10- 19 J = 1.08 eV
3. 1:5.1
4. 553 nm
5. 1.22 x 10-27 kg-m/sec
6. ~10-36 m
7. ~10-32 m
EXERCISES
These exercises are designed to help you apply the ideas of a section to physical
situations. When appropriate the numerical answer is given in brackets at the end of
each exercise.
Section 27.2
1. A prominent line emitted by mercury street lights is green and has a wavelength of
5460 Å or 546 nm. What is its frequency and its energy in joules and in electron
volts? [5.49 x 1014 sec-1, 3.64 x 10-19 J, 2.27 eV]
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
611 2. One can express the energy of any photon radiation in terms of electron volts. A very
useful rule of thumb relationship is that the product of energy in terms of electron
volts and wavelength in nanometers is equal to 1240 eV-nm. Can you verify this?
3. The wavelength of maximum energy of radiation from the sun is 480 nm. What is the
frequency of radiation of this wavelength and what is the energy of this photon in
electron volts? [6.25 x 1014 sec-1, 2.59 eV]
4. Find the temperature of a perfect thermal radiator that has its peak output at 400 nm.
How much would this temperature have to be raised to increase the radiated power
per unit area by 10 percent? [7200°K, 174°K]
Section 27.3
5. In studying wave motion, you found that the intensity of a wave depends upon the
square of its amplitude. What does the intensity of light mean in terms of quantum
concepts?
6. If 5 watts of light of the yellow line of a sodium emission spectrum (589 nm) fall upon
a surface, how many photons strike the surface per second? [1.48 x 1019 photons/sec]
7. In quantum physics we say energy = h x frequency. What are the dimensions of h?
8. If sodium has a work function of 2.28 eV, what is the longest wavelength that will
produce a photoelectron from sodium? If one doubles the intensity of light of 560
nm, will a photoelectron be emitted from sodium? [545 nm]
9. Potassium has a photoelectric work function of 2.24 eV. If it is illuminated by the 436nm mercury line, what will be the maximum kinetic energy of the photoelectrons?
[0.60 eV]
10. It is noted that the threshold wavelength for a certain metal is 500 nm. Find the
energy of the photoelectrons emitted when this metal is radiated with 450 nm light.
[0.28 eV]
11. The work function of cesium is 2.0 eV. The stopping potential in an experiment is
found to be 1.1 V.
a. Find the threshold wavelength for cesium.
b. Find the wavelength of light used in the experiment. [a. 6.2 x 10-7 m; b. 4.0 x 10-7 m]
Section 27.4
12. Show that when v/c «1, the relativistic kinetic energy reduces to the classical form,
(½) m0v2
13. A particle with an average lifetime of 10-6 sec at rest moves with a speed of 0.9c.
Show that an observer in the lab would measure a lifetime of 2.3 x 10-6 sec.
14. Find the distance the particle in problem 24 will move in the lab. [6.4 x 102 m]
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
612 15. By solving Equation 27.9 for the magnitude of the velocity, v, eliminate v from
Equation 27.11 to obtain a relationship between total energy, E, momentum, p, and
rest energy, m0c 2. Show that the following statement is correct: E2 = p2 c2 + m02c4.
Section 27.5
16. Calculate the increase in wavelength of the scattered photon for each of the
following scattering angles:
a. 0°
b. 37°
c. 53°
d. 90°
e. 150°
f. 180° [a. 0; b. 4.86 x 10-4 nm; c. 9.72 x 10-4 nm; d. 2.43 x 10-3 nm; e. 4.53 x 10-3 nm; f.
4.86 x 10-3 nm]
17. If a photon behaves as a particle and obeys particle mechanics, what happens if a
photon and a particle collide head on? Set up equations obeying conservation of
momentum and conservation of kinetic energy for a collision between a photon and
an electron. Explain in words what happens.
PROBLEMS
Each of the following problems may involve more than one physical concept. When
appropriate, the answer is given in brackets at the end of the problem.
18. A photon of wavelength 7.2 x 10-12 m is scattered at π/2 radians.
a. Find the energy of the scattered photons.
b. Find the energy of the electrons after collision. [a. 129 keV; b. 43.5 keV]
19. Derive the equation for an electron deBroglie wavelength after the electron has been
accelerated from the rest throughV volts. (Assume nonrelativistic speeds.)
20. Find the energy of an electron with a wavelength equal to the circumference of the
Bohr orbit in the hydrogen atom. (Hint: circumference = 2π x 5.3 x 10-11 m) [13.8 eV]
21. Find the ratio of the de Broglie wavelength of an electron to that of a proton with the
same energy. [SQR RT(mp/me]
22. The resolving power of a microscope is proportional to the wavelength divided by
the numerical aperture. Calculate the ratio of the resolving power of an electron
microscope using 10,000-V electrons to that of a microscope using 500-nm light with
the same numerical aperture for both. [4.07 x 104]
23. Find the uncertainty in the momentum and energy of an electron within the nucleus
(Δx = 10-14 m). What does your answer suggest regarding electrons in the nucleus
[6.63 x 10-20 kg - m/sec, 1.51 x 1010 eV]
Chapter 27 Quantum and Relativistic Physics
Physics Including Human Applications
613 24. Assume the uncertainty in the lifetime of a particle is the time it takes light to travel
across the nucleus. Find the uncertainty in the energy of this particle. (This was a
calculation used to estimate the rest mass energy of the π meson, the nuclear force
field particle.) [1.99 x 10-11 J, 1.24 x 108 eV]
25. Suppose that a 40-g bullet has its position known to within the diameter of the rifle
(1 cm). Find its uncertainty in momentum. If it is aimed at a target at a distance that
it takes 3 sec to reach, by how much will it miss its target due to the uncertainty
principle? [6.63 x 10-32 kg m/sec, 4.98 x 10-30 m]
26. If the energy of a particle of mass m is fV where V is the barrier height energy
(Figure 24.4), and f is a fraction less than one, find the relation between V and the
barrier width (Δx) that will just allow tunneling using the uncertainty principle. [Δx
≤h/SQR RT(2mfV)]
27. Show that if an electron's deBroglie wavelength equals the width of a barrier 10-10 m
wide, the electron can just pass over a 145-eV barrier.
28. Use the uncertainty principle to find the energy uncertainty for an electron confined
to a space corresponding to a virus D nm in diameter. What physical significance
can you give this energy value?
[∆E = (h/D)2/2m = (hc)2/2mc2D2 = {(1240 eV nm)2/2 x 0.51 x 106 eV}(1 /D2)]
29. Calculate the ratio of photon energy to electron energy if both the photon and
electron have the same wavelength. [Ep/Ee = v/c = SQR RT(p2/(p2 + m02c2))]
30. The mass of a body is a function of velocity in accordance with Equation 27.10. Plot
the mass of an electron for velocities of 0.1c to 0.99c. (Rule of thumb: We neglect
change in mass for velocities below 0.1c.)
31. What is the mass of an electron accelerated through a potential of
a. 100 kV?
b. 2 x 106 V? [a. 1.19m0; b 4.8 m0]
32. What is the velocity of an electron accelerated through
a. 100 kV
b. 2 x 106 V [a. v ≈ 0.54 c; b. v ≈ 0.98c]
Chapter 27 Quantum and Relativistic Physics