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PART 1: PARTICLES AND WAVES
PREVIEW
A photon is the smallest particle of light, and has an energy which is proportional to its
frequency. The photon nature of light is the principle behind the photoelectric effect, in
which the absorption of photons of a certain frequency causes electrons to be emitted
from a metal surface. The Compton effect also verifies the photon nature of light by
showing that momentum is conserved in a collision between a photon and an electron.
The Heisenberg uncertainty principle states that both the position and momentum of a
subatomic particle cannot be measured precisely. Since light waves exhibit particle
(photon) properties, de Broglie suggested that particles, such as electrons, can exhibit
wave properties.
QUICK REFERENCE
Important Terms
blackbody radiation
the radiation emitted by a blackbody, or perfect emitter and absorber of light, due
to its temperature
Compton effect
the interaction of photons with electrons resulting in the increased wavelengths of
the photons and kinetic energy of the electrons
Compton wavelength of an electron
half the maximum wavelength change of a photon in a Compton scattering with
an electron
de Broglie wavelength
the wavelength associated with a moving particle with a momentum mv
Heisenberg uncertainty principle
the more accurately one determines the position of a subatomic particle, the less
accurately its momentum is known
photoelectric effect
the ejection of electrons from certain metals when exposed to light of a minimum
frequency
photon
the smallest particle of light
Planck’s constant
the quantity that results when the energy of a photon is divided by its frequency
quantized
a quantity that cannot be divided into smaller increments forever, for which there
exists a minimum, quantum increment
quantum mechanics
the study of the properties of matter using its wave properties
wave-particle duality
under certain circumstances, waves can behave like particles, and particles can
behave like waves
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work function
the minimum energy required to release an electron from a metal
Equations and Symbols
E  hf
c  f
where
E = energy of a photon
c = speed of light = 3 x 108 m/s
f = frequency of light
λ = wavelength of light
Wo = work function of a photoemissive
surface (denoted by  on the AP
Physics exam)
h = Planck’s constant = 6.63 x 10-34 J s =
4.14 x 10-15 eV s
fo = threshold frequency of a
photoemissive surface
KEmax = maximum kinetic energy of
electrons emitted in the
photoelectric effect
e = charge on one electron
Vstop = voltage needed to stop the
emission of electrons
  = wavelength of a photon after being
scattered by a collision with an
electron
θ = angle between the scattered photon
and electron after they collide
p = momentum of a photon
m = mass of a moving particle
v = speed or velocity
Wo  hf o
KEmax  hf  Wo  eVstop
   
p
h
1  cos 
mc
h

h
mv
E  pc

The Wave-Particle Duality, Blackbody Radiation and
Planck’s Constant and Photons and the Photoelectric Effect
In prior chapters we treated light as a wave. But there are circumstances when light
behaves more like it is made up of individual bundles of energy, separate from each
other, but sharing a wavelength, frequency, and speed. The quantum of light is called the
photon.
c
c
light wave
photon
2
In the late 19th century an effect was discovered by Heinrich Hertz which could not be
explained by the wave model of light. He shined ultraviolet light on a piece of zinc metal,
and the metal became positively charged. Although he did not know it at the time, the
light was causing the metal to emit electrons. This effect of using light to cause electrons
to be emitted from a metal is called the photoelectric effect. According to the theory of
light at the time, light was considered a wave, and should not be able to “knock”
electrons off of a metal surface. At the turn of the 20th century, Max Planck showed that
light could be treated as tiny bundles of energy called photons, and the energy of a
photon was proportional to its frequency. Thus, a graph of photon energy E vs. frequency
f looks like this:
Energy
slope 
E
h
f
ΔE
Δf
frequency
The slope of this line is a constant that occurs many times in the study of quantum
phenomena called Planck’s constant. Its symbol is h, and its value is 6.62 x 10-34 J s (or
J/Hz). The equation for the energy of a photon is
E = hf
or, since f 
E
c

,
hc

The energy of a photon is proportional to its frequency, but inversely proportional to its
wavelength. This means that a violet has a higher frequency and energy than a red
photon.
Oftentimes when dealing with small amounts of energy like that of photons or electrons,
we may prefer to use a very small unit of energy called the electron-volt (eV). The
conversion between joules and electron-volts is
1 eV  1.6 x10 19 J
Planck’s constant can be expressed in terms of electron-volts as
h  4.14 x10 15 eV s
3
In 1905, Albert Einstein used Planck’s idea of the photon to explain the photoelectric
effect: one photon of energy which is higher than the energy (work function )which
binds the electron to the metal is absorbed by one electron in the metal surface, giving the
electron enough energy be released from the metal. Any energy left over from the photon
after the work function has been met becomes the kinetic energy of the electron.
Photon E
KEmax  E photon  

KEmax  hf  hf 0
metal
e
where fo is called the threshold frequency, which KE
is the minimum frequency the incoming photon
must have to dig the electron out of the metal surface.
Example 1
The metal sodium has a threshold frequency which corresponds to yellow light. Describe
what will happen if
(a) yellow light is shined on the sodium surface,
(b) red light is shined on the metal surface,
(c) green light is shined on the metal surface,
(d) bright green light is shined on the metal surface.
Solution
(a) If yellow light is shined on a sodium surface, the yellow photons will be absorbed by
electrons in the metal, causing them to be released, but there will be no energy left over
for the electrons to have any kinetic energy.
(b) Red light has a lower frequency and energy than yellow light, therefore red photons
do not have enough energy to release the electrons from the sodium surface.
(c) Green light has a higher frequency and energy than yellow light, and therefore a green
photon will be absorbed by a sodium electron and the electron will be released from the
metal and have kinetic energy.
(d) If a brighter (more photons) green light is shined on the surface, more electrons will
be emitted, since one photon can be absorbed by one electron. If these electrons are
funneled into a circuit, we can use them as current in an electrical device.
The photoelectric effect is the principle behind any process in which light produces
electricity, such as a solar calculator or an auto-focus camera.
The graph of maximum kinetic energy of a photoelectron vs. frequency of light incident
on a sodium surface would look like this:
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Kinetic
Energy
slope 
E
h
f
frequency
R
O
Y
G
B
V
Note that the electrons have no kinetic energy up to the threshold frequency (color), and
then their kinetic energy is proportional to the frequency of the incoming light.
Example 2
photon
photoemissive
surface
e
+
Adjustable Voltage
Light is shined on a photoemissive surface of work function  = 2.0 eV and electrons are
released with a kinetic energy KEmax = 4.0 eV.
(a) What voltage, called the stopping voltage Vstop, would be necessary to stop the
emission of electrons?
(b) Determine the energy of each of the incoming photons in eV and in Joules.
(c) Determine the frequency of the incoming photons in Hz.
(d) If photons of wavelength λ = 2.5x10-7m were shined on this photoemissive surface,
would electrons be emitted from the surface? Justify your answer.
Solution
(a) Stopping the emission of electrons requires work equal to the maximum kinetic
energy of the electrons:
5
KEmax  W  qeVstop


KEmax 4.0 eV  1.6 x10 19 J / eV

 4.0Volts
qe
1.6 x10 19 C
Thus, it would take 4.0 V to stop electrons with a kinetic energy of 4.0 eV, as we might
expect.
Vstop 
(b)
E photons  KEmax    4.0 eV  2.0 eV  6.0 eV


6.0eV 1.6 x10 19 J / eV  9.6 x10 19 J
9.6 x10 19 J
 1.5 x1015 Hz
h
6.6 x10 34 J / Hz
(d) The wavelength of these incoming photons corresponds to a frequency of
c 3x108 m / s
f  
 1.2 x1015 Hz
7
 2.5 x10 m
(c) f 
E photon

Now we need to check to see if this frequency is higher than the threshold frequency f0:

2.0eV
f0  
 4.8 x1014 Hz.
h 4.14 x10 15 eV / Hz
The incoming frequency is higher than the threshold frequency, so electrons will be
emitted from the metal surface.
The Momentum of a Photon and the Compton Effect
Since a photon has energy, does it follow that it has momentum? Recall in an earlier
chapter that we defined momentum as the product of mass and velocity. But a photon has
no mass. It turns out that in quantum physics, photons do have momentum which is
inversely proportional to its wavelength. The equation for the momentum of a photon is
h
p

Photons can and do impart momentum to sub-atomic particles in collisions that follow
the law of conservation of momentum. This phenomena was experimentally verified by
Arthur Compton in 1922. Compton aimed x-rays of a certain frequency at electrons, and
when they collided and scattered, the x-rays were measured to have a lower frequency
indicating less energy and momentum. The scattering of x-ray photons from an electron
with a loss in energy of the x-ray photon is called the Compton effect. It is difficult to
understand how a photon, having only energy and no mass, can collide with a particle
like an electron and change its momentum, but this has been verified experimentally
many times.
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Example 3
Before
After
c
c
electron
electron
photon
photon
A photon is fired at an electron which is initially at rest. The photon strikes the electron
and reverses its direction, as shown in the diagram representing the photon and electron
after the collision.
(a) Determine the shift in wavelength of the photon as a result of the collision.
The photon is an x-ray with a wavelength of 6.62 x 10-11 m.
(b) Determine
i. the initial momentum of the photon before the collision with the electron.
ii. the final momentum of the photon after the collision with the electron.
(c) Find the final momentum of the electron after the collision.
Solution
(a) The Compton equation gives the shift in wavelength:
h
6.62 x10 34 J / Hz
1  cos  
1  cos180  4.84 x10 12 m
   
31
8
mc
9.11x10 kg 3x10 m / s


6.62 x10 J / Hz
 1.00 x10 23 kgm / s
11

6.62 x10 m
ii. The momentum of the photon after the collision p΄corresponds to the new
wavelength λ΄:
h
h
6.62 x10 34 J / Hz

p 


  9.32 x10 24 kgm / s
11
12
     6.62 x10 m  4.84 x10 m
(b) i. p 
h

34

The momentum of the photon is given a negative sign, since it reverses direction after the
collision.
(c) By the law of conservation of momentum, the momentum lost by the photon must
have been gained by the electron.
p  p  p   9.32 x10 24 kgm / s  1.00 x10 23 kgm / s  1.93x10 23 kgm / s

 

lost by the photon. Thus, the momentum gained by the electron is 1.93 x 10-23 kgm/s.
7
The de Broglie Wavelength and the Wave Nature of Matter
In 1924, Louis de Broglie reasoned that if a wave such as light can behave like a particle,
having momentum, then why couldn’t particles behave like waves? If the momentum of a
photon can be found by the equation p 
h

, then the wavelength can be found by
h
. De Broglie suggested that for a particle with mass m and speed v, we could write
p
h
the equation as  
, and the wavelength of a moving particle could be calculated.
mv
This hypothesis was initially met with a considerable amount of skepticism until it was
shown by Davisson and Germer in 1927 that electrons passing through a nickel crystal
were diffracted through the crystal, producing a diffraction pattern on a photographic
plate. Thus, de Broglie’s hypothesis that particles could behave like waves was
experimentally verified. Nuclear and particle physicists must take into account the wave
behavior of subatomic particles in their experiments. We typically don’t notice the wave
properties of objects moving around us because the masses are large in comparison to
subatomic particles and the value for Planck’s constant h is extremely small. But the
wavelength of any moving mass is inversely proportional to the momentum of the object.

The Heisenberg Uncertainty Principle
Since a photon is the smallest and most unobtrusive measuring device we have available
to us, and even a photon has too large of a momentum to make accurate measurements of
the speed and position of sub-atomic particles, we must admit to an uncertainty that will
always exist in quantum measurements. This limit to accuracy at this level was
formulated by Werner Heisenberg in 1928 and is called the Heisenberg uncertainty
principle. It can be stated like this:
There is a limit to the accuracy of the measurement of the speed (or momentum) and
position of any sub-atomic particle. The more accurately we measure the speed of a
particular particle, the less accurately we can measure its position, and vice-versa.
REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. The smallest, discrete value of any
quantity in physics is called the
(A) atom
(B) molecule
(C) proton
(D) electron
(E) quantum
2. The smallest discrete value of
electromagnetic energy is called the
(A) photon
(B) proton
(C) electron
(D) neutron
(E) quark
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3. Which of the following photons has
the highest energy?
(A) x-ray
(B) ultraviolet light
(C) green light
(D) microwave
(E) radio wave
6. Light is shined on a metal surface
which exhibits the photoelectric
effect according to the graph shown.
What color(s) correspond to the
threshold frequency of the metal?
(A) red only
(B) red and orange
(C) red, orange, yellow, and green
(D) blue only
(E) blue, indigo, and violet
4. The threshold frequency of zinc for
the photoelectric effect is in the
ultraviolet range. Which of the following
will occur if x-rays are shined on a zinc
metal surface?
(A) No electrons will be emitted from
the metal.
(B) Electrons will be released from the
metal but have no kinetic energy.
(C) Electrons will be released from the
metal and have kinetic energy.
(D) Electrons will be released from the
metal but then will immediately be
recaptured by the zinc atoms.
(E) Electrons will simply move from one
zinc atom in the metal to another
zinc atom in the metal.
KE
R
O
Y
G
B
V
7. Which of the following is true of the
momentum of a photon?
(A) It is proportional to the wavelength
of the photon.
(B) It is inversely proportional to the
wavelength of the photon.
(C) It is inversely proportional to the
square of the wavelength of the
photon.
(D) It is proportional to the mass of the
photon.
(E) It is equal to the energy of the
photon.
5. A metal surface has a threshold
frequency for the photoelectric effect
which corresponds to green light. If blue
light is shined on this metal,
(A) no electrons will be emitted from the
metal.
(B) the number of emitted electrons is
proportional to the brightness
(intensity) of the blue light.
(C) the electrons will have no kinetic
energy.
(D) more electrons will be emitted than
if green light were shined on the
metal.
(E) electrons will be emitted from the
metal, but since the light is not
green, only a few electrons will be
released.
8. The Heisenberg uncertainty principle
implies that
(A) Electrons are too small to be studied.
(B) Every photon is exactly the same
size.
(C) The more you know about the
momentum of an electron, the less
you can know about its position.
9
(D) The more you know about the
energy of a photon, the less you can
know about its frequency.
(E) You cannot state with accuracy the
number of electrons in an atom.
(A) It is never large enough to measure
(B) It is proportional to the speed of the
particle.
(C) It is inversely proportional to the
momentum of the particle.
(D) It is equal to Planck’s constant.
(E) It has no effect on the behavior of
electrons.
9. Which of the following statements is
true for the de Broglie wavelength of a
moving particle?
10. When a photon transfers momentum to an electron, the wavelength of the photon
(A) increases
(B) decreases
(C) remains the same
(D) is equal to the wavelength of the
electron
(E) is always in the x-ray range
Free Response Question
Directions: Show all work in working the following question. The question is worth 15 points,
and the suggested time for answering the question is about 15 minutes. The parts within a
question may not have equal weight.
1. (15 points)
photon
photoemissive
surface
e
+
Adjustable Voltage
Light of a certain wavelength is shined on a photoemissive surface, ejecting electrons as
shown above. The graph below show the maximum kinetic energy of each electron
(x 10-20 J) vs. frequency of the incoming light (x 1014 Hz).
(a) On the graph below, draw the line that is your estimate of the best straight-line fit to
the data points.
10
(b) Using your graph, find a value for Planck’s constant, and briefly explain how you
found the value.
(c) From the graph, estimate the threshold frequency of the photoemissive surface.
(d) Photons of frequency 7.0 x 1014 Hz are shined on the metal surface. Determine the
i. kinetic energy of the emitted electrons
ii. speed of the emitted electrons
iii. de Broglie wavelength of the emitted electrons.
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. E
The quantum is the smallest discrete value of any quantity, such as the electron for charge
and the photon for light.
2. A
A photon is the smallest bundle of light energy.
3. A
11
The x-ray has the highest frequency of the choices, and since energy is proportional to
frequency, has the highest energy as well.
4. C
Since the frequency of x-rays is higher than the ultraviolet threshold frequency, electrons
will be emitted from the metal and have kinetic energy left over.
5. B
After the threshold frequency is met, the number of photons (brightness) dictate how
many electrons are emitted, since one photon can release one electron. Thus, a brighter
light will release more electrons.
6. D
The electrons begin being released when blue light is shined on the metal, so blue has the
threshold (minimum) frequency for this metal.
7. B
Since the equation for the momentum of a photon is p = h/ , the momentum is inversely
proportional to the wavelength of the photon, implying that a photon with a shorter
wavelength has a higher momentum than one with a longer wavelength.
8. C
Heisenberg’s uncertainty principle states that you have to sacrifice your knowledge of the
position of any subatomic particle to know its momentum accurately, and vice-versa.
9. C
According to the equations for the de Broglie wavelength, the higher the momentum of
the particle, the shorter its wavelength.
10. A
A decrease in the photon’s momentum corresponds to an increase in the photon’s
wavelength, since momentum and wavelength are inversely proportional to each other.
Free Response Question Solution
(a) 3 points
The best-fit straight line represents the average of the data points, and therefore there
would be some data points above the line and some below the line. The best-fit line does
not necessarily connect the first and last points, and does not necessarily pass through any
particular data point. You should always use a straight-edge to draw a best-fit line that
you know to be straight.
12
(b) 4 points
Planck’s constant is equal to the slope of the graph. Let’s choose two convenient points
on the line to find the slope, (4.0 x 1014 Hz, 0) and (8.0 x 1014 Hz, 20.0 x 10-20 J). Then
the slope would be
KE
20.0 x10 20 J  0
h

 5.7 x10 34 J / Hz
14
14
f
8.0 x10 Hz  4.5 x10 Hz
This value for h is close to the actual value of h within a reasonable margin of error.
(c) 2 points
The threshold frequency can be found by marking the place where the graph crosses the
frequency axis. From the graph, f0 = 4.5 x 1014 Hz.
(d) 6 points
i. From the graph, the KE associated with the frequency of 7.0 x 1014 Hz is 14 x 10-20 J,
or 1.4 x 10-19 J.
1
KE  mv 2
2
ii.
2 KE
2 1.4 x10 19 J
v

 5.5 x10 5 m / s
31
m
9.1x10 kg


h
6.6 x10 34 J / Hz

 1.3x10 9 m
iii.  
31
5
mv 9.1x10 kg 5.5 x10 m / s 
13
PART 2: THE NATURE OF THE ATOM
PREVIEW
The atom is the smallest particle of an element that can be identified with that element.
The atom consists of a nucleus surrounded by electrons which are in quantized, or
discrete, energy levels. An electron can only change energy levels when it absorbs or
emits energy. The energy emitted as a result of a downward energy level transition is
typically in the form of a photon, the smallest particle of light, and the energy of the
emitted photon is equal to the difference between the initial and final energy of the
electron.
QUICK REFERENCE
Important Terms
atom
the smallest particle of an element that can be identified with that element; the
atom consists of protons and neutrons in the nucleus, and electrons in orbitals
around the nucleus.
electron
the smallest negatively charged particle; electrons orbit the nucleus of the atom
energy level
amount of energy an electron has while in a particular orbit around the
nucleus of an atom
excited state
the energy level of an electron in an atom after it has absorbed energy
ground state
the lowest energy level of an electron in an atom
ionization energy
the energy needed to completely remove an electron from its orbital in an atom
line spectrum
discrete lines which are emitted by a cool excited gas
principal quantum number
an integer number n which determines the total energy of an atom
quantum model of the atom
atomic model in which only the probability of locating an electron is known
x – rays
high frequency and energy electromagnetic waves which are produced when high
– energy electrons strike a metal target in an evacuated tube
14
Equations and Symbols
E  hf 
where
hc

E = energy of a photon
c = speed of light = 3 x 108 m/s
f = frequency of light
λ = wavelength of light
Ef – Ei = difference between a final
energy level of an electron in an
atom and its initial energ
c  f
E photon  E f  Ei
Line Spectra
The ancient Greeks were the first to document the concept of the atom. They believed
that all matter is made up of tiny indivisible particles. In fact, the word atom comes from
the Greek word atomos, meaning “uncuttable”. But a working model of the atom didn’t
begin to take shape until J.J. Thomson’s discovery of the electron in 1897. He found that
electrons are tiny negatively charged particles and that all atoms contain electrons. He
also recognized that atoms are naturally neutral, containing equal amounts of positive and
negative charge, although he was not correct in his theory of how the charge was
arranged.
You may remember studying Thomson’s “plum-pudding” model of the atom, with
electrons floating around in positive fluid. A significant improvement on this model of
the atom was made by Ernest Rutherford around 1911, when he decided to shoot alpha
particles (helium nuclei) at very thin gold foil to probe the inner structure of the atom. He
discovered that the atom has a dense, positively charged nucleus with electrons orbiting
around it.
electron in
orbit
Nucleus
In 1913, Niels Bohr made an important improvement to the Rutherford model of the
atom. He observed that excited hydrogen gas gave off a spectrum of colors when viewed
through a spectrosope. But the spectrum was not continuous, that is, the colors were
bright, sharp lines which were separate from each other. It had long been known that
every low pressure, excited gas emitted its own special spectrum in this way, but Bohr
was the first to associate the bright-line spectra of these gases, particularly hydrogen,
with a model of the atom. Section 30.2 in your textbook has excellent photographs of
continuous and bright-line spectra.
15
He proposed that the electrons orbiting the nucleus of an atom do not radiate energy in
the form of light while they are in a particular orbit, but only when they change orbits.
Furthermore, an electron cannot orbit at just any radius around the nucleus, but only
certain selected (quantized) orbits.
The Bohr Model of the Hydrogen Atom
The two postulates of the Bohr model of the atom are summarized below:
1. Electrons orbiting the nucleus of an atom can only orbit in certain quantized orbits,
and no others. These orbits from the nucleus outward are designated n =1, 2, 3…, and
the electron has energy in each of these orbits E1, E2, E3, and so on. The energies of
electrons are typically measured in electron-volts (eV). The lowest energy (in the
orbit nearest the nucleus) is called the ground state energy E1. (Fig. A)
n=2
n=1
E2
E1
E3
E3
E2
E2
photon
photon E = E2-E1
E1
Fig. A
Fig. B
E1
Fig. C
2. Electrons can change orbits when they absorb or emit energy.
(a) When an electron absorbs exactly enough energy to reach a higher energy level, it
jumps up to that level. If the energy offered to the electron is not exactly enough
to raise it to a higher level, the electron will ignore the energy and let it pass.
(Fig. B)
(b) When an electron is in a higher energy level, it can jump down to a lower energy
level by releasing energy in the form of a photon of light. The energy of the
emitted photon is exactly equal to the difference between the energy levels the
electron moves between.
(Fig. C)
16
Example 1
Consider the energy level diagram for a particular atom shown below:
Energy above
ground state
E4 = 7 eV
E3 = 6 eV
E2 = 4 eV
E1 = 0
An electron begins in the ground state of this atom.
(a) How much energy must be absorbed by this electron to reach the 4th energy level?
(b) How many possible photons can be emitted from this atom if the electron starts in the
4th energy level? Sketch the possible transitions on the diagram above using arrows to
indicate a transition between levels.
(c) The electron drops from E4 to E2 and emits a photon, then drops from E2 to E1 and
emits a second photon.
i. Calculate the frequency and wavelength of the photon emitted when the electron
drops from E4 to E2.
ii. Calculate the frequency and wavelength of the photon emitted when the electron
drops from E2 to E1.
(d) Are either, both, or neither of the photons emitted in part (c) above in the visible
range? How can you tell?
Solution
(a) E = E4 – E1 = 7 eV – 0 eV = 7 eV
E4 = 7 eV
(b) Six possible transitions
E3 = 6 eV
E2 = 4 eV
E1 = 0
17
(c) i. E42 = E4 – E2 = 7 eV – 4 eV = 3 eV
3 eV
E
f  
 7.2 x1014 Hz
15
h 4.14 x10 eV / Hz
c
3x10 8 m / s
 
 4.1x10 7 m
14
f 7.2 x10 Hz
ii. E21 = E2 – E1 = 4 eV – 0 eV = 4 eV
4 eV
E
f  
 9.7 x1014 Hz
15
h 4.14 x10 eV / Hz
c
3x10 8 m / s
 
 3.1x10 7 m
14
f 9.7 x10 Hz
(d) The range of visible wavelengths is about 4 x 10-7 m to 7 x 10-7 m. The photon
emitted in the transition from E4 to E2 is in this visible range, but the photon emitted in
the transition from E2 to E1 is not in this range.
REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. An emission spectrum is produced
when
(A) electrons in an excited gas jump up
to a higher energy level and release
photons.
(B) electrons in an excited gas jump
down to a lower energy level and
release photons.
(C) electrons are released from the outer
orbitals of an excited gas.
(D) an unstable nucleus releases energy.
(E) light is shined on a metal surface and
electrons are released.
18
Energy above
ground state
E5 = 5 eV
E4 = 4 eV
E3 = 3 eV
E2 = 2 eV
E5 = -0.54 eV
E4 = -0.85 eV
E1 = 0
E3 = -1.5 eV
2. Consider the electron energy level
diagram for a particular atom shown. An
electron is in the ground state energy
level. If a photon of energy 6 eV is given
to the electron, which of the following
will occur?
(A) The electron will ignore the photon
since the photon’s energy does not
match the energy levels.
(B) The electron will absorb the photon,
jump up to the 5-eV level shown,
and convert the remainder of the
photon’s energy into kinetic energy,
but will stay in the 5-eV energy
level.
(C) The electron will absorb the photon,
jump out of the atom completely,
and convert the remainder of the
photon’s energy into kinetic energy.
(D) The electron will absorb the photon,
jump up to the 5-eV level, then back
down to the 4 eV level.
(E) The electron will jump up to the 3eV level, then immediately back
down the ground state.
E2 = -3.4 eV
E1 = -13.6 eV
3. Consider the electron energy level
diagram for hydrogen shown. An
electron in the ground state of a
hydrogen atom has an energy of
- 13.6 eV. Which of the following
energies is NOT a possible energy
for a photon emitted from hydrogen?
(A) 1.9 eV
(B) 13.6 eV
(C) 0.65 eV
(D) 11.1 eV
(E) 10.2 eV
4. The reason why electrons can only
orbit at certain circumferences is
(A) some electrons are larger than others
(B) the energy of electrons gets smaller
as the circumference gets larger
(C) electrons do not radiate energy when
they are in a particular orbit
(D) the atom is mostly empty space
(E) a whole number of de Broglie
wavelengths of the electron must fit
into the orbit.
19
Free Response Question
Directions: Show all work in working the following question. The question is worth 10 points,
and the suggested time for answering the question is about 10 minutes. The parts within a
question may not have equal weight.
1. (10 points)
E=0
E4 = - 0.85 eV
E3 = -1.51 eV
E2 = -3.4 eV
E1 = -13.6 eV
The energy level diagram for hydrogen is shown above. A free electron comes close
enough to the hydrogen atom that it is captured and makes a transition to the third energy
level of the atom. Then the electron makes a transition to the first energy level.
(a) Sketch arrows on the diagram above representing the two transitions made by the
electron.
(b) Calculate the wavelength of the photon emitted as the electron makes the transition to
the third energy level.
While the electron is in the ground state it absorbs a 17-eV photon.
(c) Briefly describe what happens to the electron as a result of absorbing the 17-eV
photon.
(d) Calculate the de Broglie wavelength of the electron after absorbing the 17-eV photon.
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. B
When electrons jump back to lower energy levels, they emit energy as photons.
2. C
When an electron absorbs enough energy to completely escape the atom we say that the
atom is ionized, and the energy remaining, in this case 1 eV, is converted to kinetic
energy.
20
3. D
An electron emits a photon of energy which corresponds exactly to the difference in two
energy levels, and 11.1 eV does not correspond to any energy differences in the hydrogen
atom.
4. E
If a whole number of electron wavelengths does not fit into a particular circumference,
the electron wave would destructively interfere and could not exist in that orbit.
Free Response Question Solution
E=0
(a) 2 points
photon
(b) 3 points
hc
hc


E E3  0
E4 = - 0.85 eV
E3 = -1.51 eV
E2 = -3.4 eV
1240 eV nm
 821.2nm
1.51 eV
(c) 2 points
It takes 13.6 eV to release the
electron from the ground state,
and the remaining energy of 3.4 eV
is the kinetic energy of the freed electron.

(d) 3 points
The speed of the ejected electron is
v

2 KE

m

photon
E1 = -13.6 eV

23eV  1.6 x10 19 J / eV
 1.0 x10 6 m / s
31
9.1x10 kg
h
6.6 x10 34 J / Hz

 7.1x10 10 m
31
6
mv 9.1x10 kg 1.0 x10 m / s 
21
PART 3: NUCLEAR PHYSICS AND RADIOACTIVITY
PREVIEW
The modern view of the atom includes electrons in energy levels around the nucleus of
the atom. The nucleus contains positively-charged protons and neutral neutrons, each of
which are made up of quarks. The nucleus is held together by the strong nuclear force,
and the binding energy in the nucleus is a result of some of the mass of the particles (the
mass defect) in the nucleus being converted into energy by the relationship E = mc2. The
atomic number is the number of protons in the nucleus, and the atomic mass number is
the number of nucleons (protons and neutrons) in the nucleus. Nuclear changes can take
place, but the total amount of atomic mass in the process must remain constant.
QUICK REFERENCE
Important Terms
alpha particle
positively charged particle consisting of two protons and two neutrons
atomic mass number (A)
the number of protons and neutrons in the nucleus of an atom
atomic mass unit
the unit of mass equal to 1/12 the mass of a carbon-12 nucleus; the
atomic mass rounded to the nearest whole number is called the mass number
atomic number (Z)
the number of protons in the nucleus of an atom
beta particle
high speed electron emitted from a radioactive element when a neutron
decays into a proton
binding energy
the nuclear energy that binds protons and neutrons in the nucleus of the
atom
element
a substance made of only one kind of atom
isotope
a form of an element which has a particular number of neutrons, that is, has the
same atomic number but a different mass number than the other elements which
occupy the same place on the periodic table
mass defect
the mass equivalent of the binding energy in the nucleus of an atom by E =
mc2
neutron
an electrically neutral subatomic particle found in the nucleus of an atom
nuclear reaction
any process in the nucleus of an atom that causes the number of
22
protons and/or neutrons to change
nucleons
protons or a neutrons
strong nuclear force
the force that binds protons and neutrons together in the nucleus of
an atom
transmutation
the changing of one element into another by a loss of gain of one or more protons
Equations and Symbols
E  mc 2
where
1u  1.6726 x10 27 kg  931.5 MeV
A
Z X
ΔE = binding energy of the nucleus
Δm = mass defect of the nucleus
c = speed of light = 3 x 108 m/s
u = atomic mass unit
X = element symbol
A = atomic mass number (number of
protons and neutrons)
Z = atomic number (number of protons)
The Mass Defect of the Nucleus and Nuclear Binding Energy
The nucleus is made up of positively charged protons and neutrons, which have no
charge. The proton has exactly the same charge as an electron, but is positive. The
neutron is actually made up of a proton and an electron bound together to create the
neutral particle. A proton is about 1800 times more massive than an electron, which
makes a neutron only very slightly more massive than a proton. We say that a proton has
a mass of approximately one atomic mass unit, u. The atomic number (Z) of an element is
equal to the number of protons found in an atom of that element, and fundamentally is an
indication of the charge on the nucleus of that element. All atoms of a given element have
the same atomic number. In other words, the number of protons an atom has defines what
kind of element it is. The total number of neutrons and protons in an atom is called the
mass number (A) of that element. The symbol ZA X is used to show both the atomic
number and the mass number of an X atom, where Z is the atomic number and A is the
mass number.
Even though the number of protons must be the same for all atoms of a particular
element, the number of neutrons, and thus the mass number, can be different. Atoms of
the same element with different masses are known as isotopes of one another. For
example, carbon-12 is a carbon atom with 6 protons and 6 neutrons, while carbon-14 is a
carbon atom with 6 protons and 8 neutrons. We would write these two isotopes of carbon
as 126C and 146C .
23
The table below summarizes the basic features of protons, neutrons and electrons. Notice
that we use an H to symbolize the proton, since the proton is a hydrogen nucleus.
Particle
proton
Symbol
1
1H
neutron
electron
Relative mass
1
Charge
+1
Location
nucleus
n
1
0
nucleus
e or e-
0
-1
electron orbitals
around the
nucleus
1
0
0
1
Example 1
Find the number of protons, electrons, and neutrons in a neutral atom of iron
56
26
Fe .
Solution
This isotope of iron has an atomic number of 26 and a mass number of 56. Therefore, it
will have 26 protons, 26 electrons, and 56 – 26 = 30 neutrons.
Since positive charges repel each other, one might wonder why protons stay together in
the nucleus of the atom. There is a force holding the protons together which is greater
than the electrostatic repulsion between them called the strong nuclear force, and it is a
result of the binding energy of the nucleus.
According to Einstein’s famous equation E = mc2, mass and energy can be converted into
one another. When an nucleus is assembled, each proton and neutron gives up a little of
its mass to be converted into binding energy.
Example 2
The nitrogen atom 147 N is composed of 7 protons and 7 neutrons, which gives a total of 14
atomic mass units (u). But if these particles are combined into a
the resulting mass of the nitrogen nucleus is 14.003074 u.
(a) Find the mass defect of the nitrogen nucleus.
(b) What is the binding energy of the nitrogen nucleus?
(c) What is the binding energy per nucleon?
Solution
(a) The sum of the masses of the protons and neutrons is
7(1.007 825 u) = 7.054775 u
7(1.008 665 u) = 7.060655 u
14.115430 u
The mass defect is
14.115430 u
- 14.003074 u
0.112356 u
24
14
7
N nitrogen nucleus,
(b) The equivalence between mass in atomic mass units and energy in million electronvolts (MeV) is 1 u = 931 MeV. Then the binding energy of the nucleus is
BE = (MD)(931 MeV/u) = (0.112356 u)(931 MeV/u) = 104.60344 MeV
(c)
104.60344 MeV
BE
MeV

 7.471674
nucleon
14 nucleons
nucleon
Radioactivity
At the end of the 19th century, there were elements discovered that continuously emitted
mysterious rays. These elements were identified as being radioactive. A radioactive
element spontaneously emits particles from its nucleus because the energy of the nucleus
is unstable. Examples of naturally-occurring radioactive elements are uranium 238
92 U ,
radium
226
88
Ra , and carbon 146C .
There are four types of particles that can be emitted when an element undergoes
radioactive decay:
1. Alpha decay . Uranium, for example, undergoes alpha decay, meaning that it emits an
alpha particle from its nucleus. An alpha particle is a helium nucleus, consisting of 2
protons and 2 neutrons. When an element emits an alpha particle, its nucleus loses 2
atomic numbers and 4 mass numbers, and thus changes into another element, called
the daughter element. But what would this element be? We can write the nuclear
equation for the radioactive decay of uranium as
U  Az X  24He
238
92
where X is the daughter element and 24 He is the alpha particle. The atomic number on the
left must equal the sum of the atomic numbers on the right, since charge and mass are
conserved in this process. The same is true for the mass numbers on the left and right. So,
the daughter element has an atomic number Z = 92-2 = 90 and a mass number A = 238 –
4 = 234. Uranium decays into the daughter element 234
90Th , thorium.
2. Beta decay. A beta particle is the name given to an electron emitted from the nucleus
of a radioactive element. But what is an electron doing in the nucleus of an atom?
Remember that we discussed the neutron in the nucleus of an atom as being a proton
and an electron bound together. Beta decay is really just a neutron emitting an
electron and becoming a proton. Thus, the daughter element resulting from beta decay
is one atomic number higher than the parent nucleus, but the mass number essentially
does not change. For example, carbon 146C is a radioactive element that undergoes
beta decay. The decay equation is
25
14
6
C  ZA X  10 e
We use the same symbol for a beta particle as we do for an electron. The daughter
element must have an atomic number of 6 – (-1) = 7 and a mass number of 14 – 0 = 14.
The daughter element is 147 N , nitrogen.
3. Gamma decay. Some radioactive elements emit a gamma ray, a very high energy
electromagnetic wave which has no charge or mass, so only the energy of the nucleus
changes, and neither Z nor A change.
4. Positron decay. A positron is exactly like an electron except for the fact that it is
positively charged. A positron is not a proton, as their masses and other features are
very different. Positron decay equations are typically not included on the AP Physics
B exam.
CHAPTER REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. The neutral element
24
magnesium 12
Mg has
(A) 12 protons, 12 electrons, and 24
neutrons.
(B) 12 protons, 12 electrons, and 12
neutrons.
(C) 24 protons, 24 electrons, and 12
neutrons.
(D) 24 protons, 12 electrons, and 12
neutrons.
(E) 12 protons, 24 electrons, and 24
neutrons.
masses of the individual particles that
make up the nucleus. This missing mass,
called the mass defect, has been
(A) converted into the binding energy of
the nucleus.
(B) given off in a radioactive decay
process.
(C) converted into electrons.
(D) converted into energy to hold the
electrons in orbit.
(E) emitted as light.
2. All isotopes of uranium have
(A) the same atomic number and the
same mass number.
(B) different atomic numbers but the
same mass number.
(C) different atomic numbers and
different mass numbers.
(D) the same atomic number but
different mass numbers.
(E) no electrons.
4. The isotope of thorium 234
90Th
undergoes alpha decay according to the
equation
234
A
4
90Th  Z X  2 He . The element X is
3. Six protons and six neutrons are
brought together to form a carbon
nucleus, but the mass of the carbon
nucleus is less than the sum of the
26
(A)
238
92
(B)
230
88
Ra
(C)
236
94
Pu
(D)
238
88
Ra
(E)
230
92
U
U
60
Co undergoes
5. The isotope of cobalt 27
beta decay according to the equation
60
A
0
27 Co Z X  1 e . The element X is
60
Fe
(A) 26
(B)
56
25
Mn
(C)
60
28
Ni
60
Cu
(D) 29
(E)
61
27
Co
Free Response Question
Directions: Show all work in working the following question. The question is worth 10 points,
and the suggested time for answering the question is about 10 minutes. The parts within a
question may not have equal weight.
1. (10 points)
Particle or
nucleus
1
0n
1
1
13
7
H
N
14
7
N
Mass (u)
1.008 665 u
1.007 825 u
13.005738 u
14.003074 u
13
7
N  01n147N  energy
A neutron is bound to a nitrogen nucleus, as shown in the equation above.
(a) Find the mass defect and binding energy that holds a neutron to the nitrogen atom
(b) The U.S. uses about 1020 J of energy per year. How many nitrogen atoms would have to give
up a neutron and release its binding energy to provide the energy needed in the U.S. for a year?
ANSWERS AND EXPLANATIONS TO CHAPTER REVIEW QUESTIONS
Multiple Choice
1. B
The atomic number 12 implies both 12 protons and 12 electrons, and the mass number 24 is the
sum of protons and neutrons, giving 12 neutrons.
27
2. D
All isotopes of a particular element must have the same atomic number (number of protons),
since this number identifies the element, but can have a different mass number (number of
neutrons).
3. A
Each particle that makes up the nucleus gives up a little mass to be converted into energy by E =
mc2 to bind the nucleus together.
4. B
The atomic number Z of element X is found by 90 = Z + 2, so Z = 88, and the mass number A is
found by 234 = A + 4, so A = 230. The element X is radium.
5. C
The atomic number Z of element X is found by 27 = Z + (-1), so Z = 28, and the mass number A
is found by 60 = A + 0, so A = 60. The element X is nickel.
Free Response Question Solution
(a) 5 points
13
7
N  01n147N  energy
(13.005738 u) + (1.008665 u) = 14.003074 u + (mass defect)
Mass defect = 0.011329 u
BE = (MD)(931 MeV/u) = (0.011329 u)( 931 MeV/u) = 10.547299 MeV
(b) 5 points
Converting MeV to J:
10.547299 MeV(1.6 x 10-13 J) = 1.69 x 10-12 J
Number of nitrogen atoms =
10 20 J
 5.93 x 10 31 atoms
12
1.69 x10 J / atom
28
PART 4: IONIZING RADIATION, NUCLEAR ENERGY, AND
ELEMENTARY PARTICLES
PREVIEW
The nucleus of the atom can undergo changes, but the total amount of atomic mass in the process
must remain constant. Nuclear reactions may be induced by bombardment of one nucleus with
another, such as in the processes of fusion and fission. In general, when a nuclear change takes
place, energy is released.
The content contained in sections 2, 3, 5, and 8 (Example 9) of chapter 32 of the textbook
is included on the AP Physics B exam.
QUICK REFERENCE
Important Terms
chain reaction
nuclear process producing more neutrons which in turn can create more
nuclear processes, usually applied to fission
critical mass
the minimum amount of mass of fissionable material necessary to sustain
a nuclear chain reaction
elementary particles
the particles (quarks and leptons) of which all matter is composed
neutron
an electrically neutral subatomic particle found in the nucleus of an atom
nuclear fission
the splitting of a heavy nucleus into two smaller ones
nuclear fusion
the combining of two light nuclei into one larger one
nuclear reaction
any process in the nucleus of an atom that causes the number of
protons and/or neutrons to change
nuclear reactor
device in which nuclear fission or fusion is used to generate electricity
quark
one of the elementary particles of which all protons and neutrons are made
29
Equations and Symbols
A
Z
X
where
X = element symbol
A = atomic mass number (number of protons and neutrons)
Z = atomic number (number of protons)
DISCUSSION OF SELECTED SECTIONS
Nuclear Fission
Fission is a process in which a large nucleus splits in to smaller nuclei. Fission is usually caused
artificially by shooting a slow neutron at a large atom such as uranium which absorbs the neutron
and splits into two smaller atoms, along with the release of more neutrons and some energy.
For example, a fission reaction occurs when uranium 235
92 U absorbs a slow neutron and then splits
into xenon and strontium, releasing three neutrons and some energy. The equation for this fission
reaction is
94
1
U  01n140
54 Xe 38 Sr  2 0 n  energy
235
92
Once again, the sum of the atomic and mass numbers on the left equal the sum of the atomic and
mass numbers on the right. The three neutrons which are produced in this reaction can be used to
split three more uranium atoms, which produce three more neutrons in each reaction, each of
which can split three more uranium atoms, and so on. This is called a chain reaction, and is used
to sustain the release of energy in fission reactions. However, before a chain reaction can be
sustained, there must be a minimum
amount of fissionable material, such as uranium or plutonium, present. The minimum amount of
fissionable material that must be present to sustain a chain reaction is called the critical mass.
Nuclear Fusion
Fusion occurs when small nuclei combine into larger nuclei and energy is released. Many stars,
including our sun, power themselves by fusing four hydrogen nuclei to make one helium
nucleus. The fusion process is clean and a large amount of energy is released,
30
which is why researchers here on earth are trying to find ways to use fusion as an alternative
energy source.
For example, the element tritium 13 H is combined with 12 H , a hydrogen isotope called deuterium,
to form helium 24 He and a neutron, along with the release of energy. The equation for this fusion
reaction is
3
1
H  12H  24 He  01n
Note that the sum of the atomic numbers on the left must equal the sum of the atomic numbers
on the right. The same is true of the mass numbers.
REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. Consider the following nuclear fission equation:
which belongs in the blank is
(A) 3795 Kr
(B)
94
36
Kr
(C)
94
35
Br
(D)
94
37
Rb
(E)
94
56
1
U  01n139
56 Ba  ____30 n . The element
235
92
Ba
2. Consider the following nuclear fusion equation: 12 H 12H  ____ 01n . The element which
belongs in the blank is
(A) 24 He
(B) 23 He
(C) 23 H
(D) 14 H
(E) 34 He
3. Nitrogen is bombarded with an alpha particle, producing a nucleus of
(A) a neutron
(B) an electron
(C) a beta particle
(D) a proton
(E) an alpha particle
31
17
8
O and