Conceptual Physics

As with many scientific discoveries, this discovery raised more questions. Scientists had
long known about electric charge. Since most matter is neutral, they knew there must
be positive charges to balance the negative charge of the electrons. To put it at the
microscopic level: Atoms must contain positive charges to balance the negative
electrons. But how were these positive and negative charges arranged in an atom?
Answering that question is the topic of the next section.

Atoms

Contain a nucleus and one or more
electrons

38.2 - Rutherford’s discovery of the nucleus

Nucleus: A relatively small region in the center

of an atom where the positive charge í and
most of the mass í of an atom is located.

The modern model of the atom describes it as a small positive nucleus, surrounded by
orbiting electrons. The radii of the orbits are far larger than the size of the nucleus. How
did scientists create this model of the atom? For instance, what led them to believe that
the nucleus was small compared to the size of the orbital radii?

Scientists in the early 20th century struggled to understand the nature of atoms. They Old “plum pudding” model
could not directly see the structure of atoms, but knew that they contained negatively-
charged electrons. They reasoned that atoms must contain equal amounts of positive Positive charge distributed evenly, with
and negative charges since matter tends to be electrically neutral. embedded electrons
·Wrong!
Without other data, most scientists thought that the positive charges that make up
matter were evenly distributed throughout the atom í why not? Since electrons have so
little mass, scientists knew that the positive charges carried almost all the mass of the
atom. Their mental picture of the atom looked like a blob of positively-charged cookie
dough with small chocolate chips (electrons) embedded in it. The model actually bore
the name of a more popular dessert at the time: The plum pudding model consisted of
negatively-charged electrons (plums) scattered throughout a massive cloud of positive
charge (pudding) that was distributed uniformly throughout the volume of the atom. This
model is shown in Concept 1.

Lord Rutherford made a major breakthrough in this area, winning the 1908 Nobel Prize Rutherford’s experiment
in Chemistry for his experiments. As a good scientist, he wanted data to support (or
contradict) the plum pudding model of the atom. At the time, Rutherford was studying Fired alpha particles at gold foil
alpha particles, which are massive, positively charged particles that are emitted at high Some particles scattered by a large
speed from some radioactive substances (such as radon). He realized that a beam of amount
alpha particles might serve as a tool to probe the atomic interior. Implied compact, massive nucleus

In his experiment, he aimed a beam of these particles at a thin gold foil, and measured
the distribution of the outgoing alpha particles. This is known as a scattering experiment
í observing how the particles scatter í and it is a now-common technique to probe the
details of atomic-sized systems.

A mechanical analogy to what Rutherford did would be to probe the interior of a box by
shooting a high-powered BB gun at it. If the box were filled with sponge cake, the BBs
would pass through. If it had a metal plate inside, the BBs would rebound back. Or if
there were a small metal sphere inside, a small fraction would rebound or scatter
sideways, with a distribution of angles.

Rutherford initially assumed that the alpha particles would be passing through a
“pudding” of positive charges spread uniformly throughout the foil. Relying on this
model, Rutherford predicted that most of the alpha particles should just pass straight
through or be only minimally deflected.

This was not what he observed. Much to Rutherford’s surprise, a small fraction of the Nucleus consists of positive charge

particles were scattered by 90° or more. Occasionally, an alpha particle even

rebounded from the foil, straight back at the source. He concluded that the atom must
not have a uniform distribution of positive charge inside. Instead, the large force
necessary to cause such scattering of the positively charged alpha particle could be

Copyright 2007 Kinetic Books Co. Chapter 38 699

provided only if the atom’s positive charge (and mass) were highly concentrated within

the atom, in a region called the nucleus. Nucleus is small compared to orbits of

After analyzing the data and seeing how few of the alpha particles actually scattered, electrons

Rutherford concluded that the nuclear radius must be about 10,000 times smaller than

the atomic radius, a figure that is still accepted today. Rutherford’s groundbreaking experiment proved that the atom is mostly empty space. His

atomic model is shown in Concept 3. Note that the diagram is not even close to being drawn to scale; the atomic diameter is far too small

compared to the nuclear size. If the diagram were drawn so that the nucleus were 1 cm wide, roughly the width of your pinky, the atom would

have to be drawn about 100 meters wide, about 10% longer than the length of an American football field.

38.3 - Components of the nucleus

Rutherford proved that atoms consist of a compact, very dense positively-charged
nucleus surrounded by negatively-charged electrons. In this section, we take a deeper
look at the parts of the nucleus, and introduce some common terminology and notation
that scientists use when talking about elements and nuclei.

The positive charge of the nucleus comes from particles called protons, which are about
1800 times more massive than electrons. The simplest nucleus consists of a single
proton, and the simplest atom is hydrogen, which consists of a proton and an orbiting
electron.

What distinguishes an atom of hydrogen from an atom of gold? A chemical element is Parts of the nucleus
defined by the number of protons in its nucleus. For an atom of a particular element, the
Protons
nucleus consists of Z protons, where Z is called the atomic number of the element. ·Z = number of protons

Hydrogen has a single proton, so for hydrogen, Z = 1. Gold has 79 protons, so its Neutrons
·N = number of neutrons
atomic number is 79. Protons and electrons have equal but opposite charges. This
means that an electrically neutral atom has the same number of electrons as protons,
so for example a gold atom has 79 protons and 79 electrons.

There can be more to a nucleus than just protons; there may also be other particles
present, called neutrons. These are uncharged particles that are just a bit more massive
as the proton (about 0.1% more massive). Protons and neutrons are known as
nucleons because they make up the nucleus.

The neutron number, N, states the number of neutrons. Any atom of an element always

has the same number of protons, but it can have different numbers of neutrons. Two
forms of an element with different numbers of neutrons are known as isotopes.
Hydrogen, for instance, always has a single proton, but it can have either zero neutrons
(“common hydrogen”), one neutron (an isotope called deuterium), or two neutrons
(called tritium). Deuterium and tritium are shown in Concept 2.

Different isotopes of an element will have different atomic masses because of the Isotopes
differing numbers of neutrons. The table in Concept 3 summarizes the charge and mass
properties for protons, neutrons, and electrons. Same number of protons, different
number of neutrons
Atomic masses can be measured using instruments such as the mass spectrometer.
Masses are commonly given in terms of atomic mass units, u, defined such that the
mass of the most abundant kind of carbon atom, carbon-12, has a mass of exactly 12 u.
The value of an atomic mass unit in kilograms is given in Equation 1.

The sum of Z, the number of protons, and N, the number of neutrons, is called A, the

mass number of the atom. Since electrons have little mass compared to protons and
neutrons, the mass number is very close to the entire mass of the atom when it is
expressed in atomic mass units.

An atom with a particular combination of Z and N is called a nuclide. A particular

nuclide always has the same type of nucleus. A widely adopted notation to identify a
nuclide is to write the chemical symbol of the element, with its atomic number
subscripted to the left and its mass number superscripted to the left. This is shown in
Equation 2, as applied to the most common carbon nuclide which has six protons and
six neutrons.

Neutrons and protons have similar mass
Protons far more massive than electrons

700 Copyright 2007 Kinetic Books Co. Chapter 38

Atomic mass unit
u = 1.66054 × 10í27 kg

u = atomic mass unit (amu)

38.4 - Interactive checkpoint: using Z, N and A Mass number

A=Z+N

A = mass number
Z = atomic number
N = neutron number

A neutral atom of sodium-23 has 11
protons and a mass number of 23.
How many electrons does it have?
How many neutrons does it have?

Answer:

Number of electrons =
Number of neutrons =

Copyright 2007 Kinetic Books Co. Chapter 38 701

38.5 - The strong nuclear force

Atoms contain positively-charged nuclei that attract the negative electrons, and the
nuclei contain closely packed protons and neutrons. These conclusions bring up a
major question about the nucleus. Since protons are positively charged, and positive
charges repel, why do the protons in the nucleus not fly away from one another? In this
section, we will explore the nature of the force that holds nucleons together.

One good hypothesis would be that the force of gravity attracts them. This is a good
hypothesis, but incorrect: It turns out that gravity is far too weak. (For two protons, by
using Newton’s law of gravity and Coulomb’s law, you can calculate that the attractive
force of gravity is about 1036 times weaker than the electrostatic repulsive force.)

One hypothesis down. Since gravity cannot explain the stability of the nucleus, the only Strong force
alternative is that there must be another attractive force. This fundamental force is
called the strong force.

The strong force has several important properties. It is “strong;” it manages to hold Holds particles in nucleus together

together the protons in a nucleus despite their electrostatic repulsion. It always binds Is very strong!
particles together, even if their electric charges are the same, or if they are uncharged. Always attractive, regardless of charge

The strong force causes protons to attract protons, protons to attract neutrons, and Acts only over a very short range
neutrons to attract other neutrons. The strong force acts only over a very short range.

For example, once two protons are separated by more than about 10í15 m (roughly their

own diameter), there is hardly any attraction due to the strong force, though the electrostatic repulsion is still substantial.

Although much has been learned about the properties of the strong force from experiments, there is no simple formula to relate its strength to
distance. With the electrostatic and gravitational forces, the amount of force is inversely proportional to the square of the distance between the
particles. In contrast, no simple formula can be stated for the dependence of the strong force on distance, though there are complicated
numerical approximations.

How have physicists studied the strong force? They experiment by bombarding target nuclei with high energy particles, which are influenced by
the nuclei via the strong force or the electrostatic force. These forces can change the paths of the incoming particles. By observing the
distribution of the outgoing particles, and comparing it to the predictions of theoretical models, scientists can test these models.

38.6 - Nuclear properties

Are the nucleons rigid objects, or soft, compressible ones? In other words, do they
behave like hard marbles that are clumped together, or is there some flexibility to them,
like cotton balls being crammed into a bag? These questions can be answered if we
can determine how the size of the nucleus depends on the number of nucleons in the
nucleus.

It turns out that nucleons are nearly incompressible. This conclusion can be drawn by
looking at how the radius of the nucleus relates to the number of nucleons inside. The
same experiments that physicists perform to study the strong force, where they fire
particles at nuclei, have also allowed them to measure other properties of the nucleus,
such as its size.

They have determined that the radius of an atom’s nucleus is proportional to the cube Nuclear properties
root of its mass number. As shown in Equation 1, the radius equals the cube root of the
number of neutrons and protons, multiplied by 1.2×10í15 m. Nucleons are nearly incompressible,
tightly packed
There are some striking implications of this simple-looking formula. It holds the answer
to our question about the rigidity of nucleons, and implies other facts about the nucleus. Nuclear density is roughly the same for
Consider how the radius of a sphere relates to its volume. The volume of the spherical all atoms

nucleus is equal to 4ʌ/3 times the radius cubed. If you cube the radius, using the
equation to the right, you are cubing A1/3, which equals A. In other words, the volume is
proportional to A, the number of neutrons and protons: Each time you add a nucleon,

you are adding roughly the same amount of volume to the nucleus.

The equation also allows one to conclude that the neutrons and protons must be tightly Nuclear radius increases with
packed. If there were large spaces between nucleons, then as their number increased, mass number, A
the volume would increase at an even faster rate; for example, the volume would more

than double when you doubled A. (If this is not obvious to you, consider the change in

the volume of adding a tenth planet beyond Pluto, versus adding another marble to a
bag of marbles. Adding another planet would increase the volume of the Solar System
by more than the volume of the planet itself; but as you add hard marbles to a cluster of
marbles, the volume of the cluster increases by about the volume of a single marble
each time.)

The equation tells scientists that the density of all nuclear material is constant. How do
we know this? Density equals mass divided by volume, and since the mass and volume
increase at the ratio of 1:1, the density does not change.

702 Copyright 2007 Kinetic Books Co. Chapter 38

38.7 - Sample problem: nuclear density Dependence of radius on A

R = (1.2×10í15 m)A1/3

R = nuclear radius
A = mass number

Calculate the density of the hydrogen
nucleus (A = 1, mass § 1.0 u) and of
an aluminum-27 nucleus (A = 27,
mass § 27 u). Express the answer in
kg/m3, to two significant figures.

Variables mH
mass of hydrogen nucleus RH
radius of hydrogen nucleus VH
volume of hydrogen nucleus mAl
mass of aluminum nucleus RAl
radius of aluminum nucleus VAl
volume of aluminum nucleus

What is the strategy?

1. Convert the mass of each nucleus from atomic units to kilograms using the conversion factor stated below.
2. Find the radius of each nucleus using the relationship between radius and mass number.
3. Calculate the volume of each nucleus using the radius just calculated.
4. Divide mass by volume to find the density.

Physics principles and equations

The nuclear radius grows as the cube root of the mass number.

R = (1.2×10í15 m)A1/3

The nuclear shape may be modeled as a sphere. The volume of a sphere in terms of its radius is

The definition of an atomic mass unit is
u = 1.66 × 10í27 kg

Mass density

ȡ = m/V

Copyright 2007 Kinetic Books Co. Chapter 38 703

Step-by-step solution

We begin by converting the mass of the hydrogen nucleus into kilograms, then we calculate the radius of the nucleus. Using the radius, we
calculate the nuclear volume. Finally, we divide the mass by the volume to determine the density.

Step Reason

1. mH = 1u = 1.66×10í27 kg hydrogen nuclear mass, definition of u

2. R = (1.2×10í15 m)A1/3 apply nuclear radius equation

3. calculate volume of hydrogen nucleus

4. definition of density

Now perform the same calculations for aluminum.

Step Reason

5. mAl = 27 u = 27(1.66×10í27 kg) aluminum nuclear mass, definition of u

6. R = (1.2×10í15 m)A1/3 apply nuclear radius equation

7. calculate volume of aluminum nucleus

8. definition of density

The densities are nearly identical. This is further confirmation that nucleons are tightly packed and incompressible.
The nuclear density is far beyond the density of materials in our experience. For example, consider gold, which is 19.3 times denser than water
and almost 1.7 times as dense as lead. A nucleus is 12,000,000,000,000 times denser than gold. (In case you were wondering, no, the zero
key did not get stuck down while we typed that number.) Recall that Rutherford found that the atomic radius was on the order of 10,000 times
the nuclear radius. The volume of a sphere scales as the cube of the radius, so the ratio of atomic to nuclear volume is approximately (10,000)
3. Since the mass of an atom mostly resides in the nucleus, and so much of an atom is empty, this explains the incredible density of a nucleus.

38.8 - Stable nuclei

What are the rules for the number of allowed protons and neutrons in a nucleus? Is any combination of protons and neutrons possible? Could
there be a hydrogen atom whose nucleus has 1 proton and 7 neutrons? What about a silver atom with 47 protons and no neutrons? Could
there be an element with 150 protons and any number of neutrons?
The nuclei described above do not exist. You could not even momentarily create nuclei with such extreme imbalances of protons and neutrons,
or in the last case, a nucleus with so many protons. Only certain combinations of protons and neutrons can form nuclei, and even fewer
combinations can form stable nuclei.
By stable, we mean elements that will not spontaneously decay. Gold-197 is the only stable isotope of gold, while there are several stable
oxygen isotopes: oxygen-16, oxygen-17 and oxygen-18.

704 Copyright 2007 Kinetic Books Co. Chapter 38

Some nuclides are unstable, meaning they have a limited lifespan. An unstable nucleus
is called a radionuclide, and it will spontaneously and rapidly decay or split up into more
stable pieces. Such materials are radioactive, and the details of the decay process are
the subject of another section.

In this section, we discuss what makes for a stable nucleus. The question of stability is
of fundamental importance. If all elements were unstable, then life would as we know it
would not exist í the carbon, oxygen and other elements that make up your body would
be constantly changing into other elements. Life is complicated enough!

On the other hand, if all elements were equally stable, then the nuclear fusion process
that powers stars (including the Sun) would never happen, and the universe would be
almost all hydrogen, with none of the heavier elements that make life possible.

To explain why some atoms are stable and others are not, it helps to consider a Stable nuclei

diagram of stable and unstable nuclides where Z is plotted against N. For example, Stable only for certain combinations of
even though isotopes of silver with its 47 protons (Z = 47) have been created with neutrons and protons

mass numbers as low as 96 and as high as 124, just two of these nuclides are stable (

A = 107 and 109). In other words, there must be 60 or 62 neutrons along with the 47

protons.

Unstable nuclei

Other combinations do not exist or are
unstable

Graph of stable and unstable
nuclei

Values clustered near “band of stability”

Ratio of neutrons to protons increases
with nuclear size

Larger nuclei: more neutrons required to
dilute protons

We would like you to observe three important features of this diagram. First, note that the stable nuclides are clustered around a band running
through the diagram. The unstable nuclides exist on either side of this band. Second, you can see how the stable nuclei are distributed.

Roughly speaking, for less massive nuclei, the number of protons and neutrons is approximately equal: They cluster around the line N = Z. In
contrast, for more massive nuclei, the number of neutrons exceeds the number of protons, N > Z. Third, observe that there are no stable
nuclei beyond bismuth (Z = 83).

Since neutrons are effective at diluting the repulsive electric force between protons (by spacing them out more), and the strong force binds
neutrons effectively to protons and other neutrons, it seems like having more neutrons can only bind the nucleus more tightly. You may be
wondering why an atom cannot have an extremely high ratio of neutrons to protons. For instance, why are there no hydrogen isotopes with
eight neutrons, or even just seven neutrons? Quantum mechanical principles dictate why.

In contrast, explaining why very large nuclei are unstable, even if N is closer to Z, only requires considering the nature of the strong and

electrostatic forces, not quantum mechanics.

Copyright 2007 Kinetic Books Co. Chapter 38 705

Heavy elements like uranium have a large number of protons, which all repel one another. As the number of protons increases, the repulsive
force keeps growing and growing, making the nucleus more and more unstable.

Neutrons counteract this growing instability by increasing the distance between protons, which decreases the electrostatic forces, and by
attracting each other and the protons with the strong nuclear force.

However, there comes a point when the nucleus gets too large, and will be unstable no matter how many neutrons are present. This can be
understood by considering the relative ranges of the strong and electrostatic forces. The strong force only acts between very close neighbors,
while the repulsive electrostatic force acts between all protons regardless of their position within a nucleus. When there are lots of protons
already present, and one more proton is added, it will be subject to repulsive electrostatic forces from every proton that is already there, while
the attractive strong force will only be exerted by a few very close neutrons or protons. Eventually, it becomes impossible for the nucleus to
“hold in” an additional proton because the strong force cannot overcome the electrostatic force.

38.9 - Nuclear binding energy

Binding energy: The energy that must be added
to disassemble, or unbind, a nucleus into the
protons and neutrons that make it up.

You may have heard of radioactivity, and know that uranium atoms will spontaneously
decay into other elements, while other elements such as common iron (iron-56) are
stable. Stable nuclei such as iron all have one thing in common: Their nucleons are
tightly bound. Unstable atoms are not as tightly bound. What does it mean to be “tightly
bound”?

Note that the term “stability” in nuclear physics is not making a statement about the Binding energy
tendency of an atom to enter into chemical reactions. For example, we say that iron is
“stable” in the nuclear sense, even though it rusts. When iron combines with oxygen to Energy that must be added to
form iron oxide, it is a chemical reaction, not a nuclear reaction. The iron remains iron disassemble nucleus completely
when it becomes iron oxide; it shares electrons with oxygen but the element’s nucleus
remains unchanged. Increased energy of separate particles
reflected in increased mass

What makes the protons and neutrons in a radioactive uranium atom less tightly bound
than the nucleons in an extremely stable iron atom? Can we quantify and compare the
stability of nuclei?

As it turns out, we can. One way to measure the stability of a nucleus is to try and rip
the nucleons apart, overcoming the strong force. When physicists conduct such
experiments, they find it takes energy to break the nuclear bonds (which the strong
force is responsible for), that is, to take apart a nucleus into separate protons and
neutrons. (This makes sense, because if no energy was required to separate them, they
would fall apart on their own.)

The particles that emerge when the nucleus is forced apart can be analyzed. Careful Binding energy becomes mass
measurements show that the sum of the masses of the separate nucleons is always
greater than the mass of the nucleus when it is whole. Separate particles have more mass than
assembled nucleus
Concept 1 shows this, using an isotope of hydrogen, deuterium, as an example.

Why should the mass of the nucleus increase when it is broken up? To a classical
physicist, unacquainted with Einstein’s theory of special relativity, this would be a
surprise since mass is assumed to be conserved.

However, we know that another conservation principle applies here: the total of mass
and energy (or mass-energy) must remain the same, though the individual terms may
vary. Einstein’s principle of mass-energy equivalence, summed up by the equation

E = mc2, applies. This is shown in Concept 2. It takes energy to separate the particles,

and the energy added to the nucleus to fragment it into nucleons shows up as the
“extra” mass. (We will assume for the sake of simplicity that the kinetic energies of the
particles and of the nucleus are negligible.)

The energy that must be added to completely disassemble the nucleus is known as the Assembling the nucleus
binding energy. This works in both directions. The binding energy is released when the
protons and neutrons come together to form a bound nucleus. This is illustrated in Binding energy released when nucleus
Concept 3. is assembled

The terminology could be a little confusing. You can think of it like gravity: You must Decreased energy of nucleus reflected
“add” energy to pull apart two particles, or lift a rock farther from the surface of the in decreased mass
Earth. That is analogous to the binding energy.
Mass becomes binding energy
Because of the equivalence of energy and mass, the binding energy may also be
related to mass. When energy is added to a nucleus to disassemble it, the mass of the
parts increases. On the other hand, when a nucleus is assembled, the release of

706 Copyright 2007 Kinetic Books Co. Chapter 38

binding energy from the system shows itself as a corresponding reduction of mass in
the assembled nucleus. This is sometimes called the “missing mass”.

38.10 - Interactive checkpoint: calculating the binding energy

For a carbon-12 nucleus with 6
protons and 6 neutrons, and a mass
of 11.9967 u, how much more mass
do the individual nucleons have than
the assembled carbon nucleus? This
is called the mass excess. What is
the binding energy of carbon-12’s
nucleus?
A neutron has a mass of 1.0087 u
and a proton has a mass of 1.0073 u.
1 u = 1.66×10í27 kg.

Answer: u
J
mass excess =
binding energy =

38.11 - Binding energy curve

The binding energy is a measure of stability for a nucleus, since the binding energy is
how much energy it takes to completely disassemble a nucleus. Roughly speaking, the
higher the binding energy, the harder it is to pull all the nucleons apart.

However, determining which nuclides are stable is not as simple as calculating the
binding energy. In this section, we discuss how the stability of a nuclide can be
determined.

We will use two nuclides as examples: (iron-56), and (uranium-235).

Using a table of nuclear data, one may find that the binding energy of an iron-56 atom,
which has 26 protons and 30 neutrons, is 492 MeV. The binding energy of a uranium-

235 atom (Z = 92, N = 143) is 1784 MeV.

Does this mean that the uranium nucleus is more stable since it has a greater binding Stability and binding energy
energy? Not necessarily. To compare the stability of different nuclei, a useful number to
consider is the binding energy per nucleon. To calculate this ratio, divide the binding Stability determined by binding energy
energy of the nucleus by the mass number, the total number of protons and neutrons. per nucleon
This provides a metric to compare the binding energy per nucleon in the iron isotope
with the binding energy per nucleon in uranium.

The binding energy per nucleon in uranium-235 is 7.59 MeV and the binding energy per
nucleon in iron-56 is 8.79 MeV. The binding energy per nucleon is a good measure for
stability; the fact that the binder energy per nucleon is higher for iron than uranium
correctly predicts that iron-56 is more stable than uranium-235. (In fact, the binding
energy per nucleon for iron-56 is among the highest for all nuclides.)

The graph in Concept 2 shows the binding energy per nucleon for naturally occurring

isotopes, plotted against the mass number A. This is called the binding energy curve.

Graph of binding energy per
nucleon versus mass number

Graph peaks near iron-56

Copyright 2007 Kinetic Books Co. Chapter 38 707

What is the total binding energy
for a uranium-235 nucleus?

38.12 - Shape of the binding energy curve

The graph of binding energy per nucleon versus mass number has a distinct shape that
proves to be very important. The higher on the graph an element is (indicating more
binding energy per nucleon), the more stable it is.

The most stable elements are at the highest points, with iron-56 in this region, as you
can see in Concept 1.

Very light nuclei (on the left of iron-56 in the binding energy curve) can become more Binding energy per nucleon
stable if they combine to form larger nuclei through a process called fusion. By this versus mass number
process, the binding energy per nucleon is raised, which means that energy is released.
This is the process by which stars, like the Sun, continually transform their mass into Highest in the middle
energy. In a multistep process within the Sun, hydrogen nuclei fuse together to become
helium-4 nuclei.

As mentioned, the most stable locations on the curve represent elements such as iron
and nickel. Heavier, radioactive nuclei to their right can increase their binding energy
per nucleon and become more stable by “moving to the left and up” on the curve. For
instance, you can see in Concept 3 that uranium is less stable than iron. A heavy
nucleus could become more stable by emitting particles and becoming slightly smaller
(the process of radioactive decay) or, in extreme cases, by splitting into two medium-
sized nuclei (a process called fission). This is the principle behind radioactivity and
nuclear power.

The shape of the graph also illustrates the relative distances at which the strong and
electrostatic forces effectively operate. The argument that follows is reasonably
complex but provides a good example of how graphical data can be analyzed. The
difference between these two forces can be used to explain why the graph first shows a
rapid increase of binding energy per nucleon, then levels off, and finally declines.

Earlier, we discussed the short-range nature of the strong force. It is so short-range that

it acts only between a nucleon and its nearest neighbors. The graph supports this

hypothesis. Why? When there are only a few nucleons, they are all very close and Lighter elements can undergo
every nucleon interacts with every other. For instance, when there are two nucleons, fusion and release energy
they are next to each other, and exert a strong force on one another. As a third nucleon

is added, it has two neighbors to exert a force on, so the force increases faster than the

number of nucleons. This means the binding energy per nucleon increases, so the line

has a positive slope. (Mathematically, the binding energy of the smaller nuclei increases as the square of the number of nucleons.)

As more nucleons are added, at some point they are too far apart to all be “neighbors”. For instance, when there are 100 nucleons, and
another is added, it can only interact with its close neighbors. A nucleon on one side of the nucleus is too far away to exert a significant strong
force on one on the far side. Adding a nucleon does not increase the binding energy per nucleon. This means with larger nuclei, the additional
binding energy per nucleon becomes constant. The sharp increase in binding energy per nucleon ceases.

The strong force needs to be contrasted with the electrostatic force, which acts to push apart the protons, and which acts at a greater distance
than the strong force. When a new proton is added, it is attracted only to its nearest neighbors via the strong force, but is repelled by every
other proton that is already present because the electrostatic force acts at a greater range. This, in turn, makes it easier to disassemble the
nucleus when more of the protons want to be separated.

In sum, at first the strong force dominates, causing the increase in binding energy per nucleon. But as the nucleus grows in size, the
electrostatic force plays a larger role, causing an eventual decrease in binding energy per nucleon.

708 Copyright 2007 Kinetic Books Co. Chapter 38

Heavier elements can divide and
release energy

38.13 - Fission

Fission: A heavy nucleus breaks up into two
smaller ones, releasing energy.

You may have heard the term “splitting the atom” as something that humans first
accomplished in the 20th century. In this section, you will learn what it means to “split”
an atom. Fission is the process used both in nuclear reactors to produce electrical
power and also in the first atomic bombs.

When a nucleus breaks up into smaller, more stable pieces, this is known as fission. Fission
Some unstable nuclei, such as uranium-236, do this spontaneously. When an atom
undergoes fission, it changes identity, as the new nuclei it breaks into have different Nucleus breaks up into smaller
numbers of protons. There are many ways that the nucleus can break up. For example, elements
the uranium-236 nucleus can break into Xe-140 and Sr-94, in the process releasing two ·And releases neutrons
neutrons. To see this fission process, press the refresh button in your browser and look ·And releases energy
at Concept 1.

However, being the impatient race that we are, humans learned to induce the process
to happen at a greater rate. Induced fission, also known informally as “splitting the
atom,” was first performed by Otto Hahn and Fritz Strassmann. They bombarded
uranium with neutrons and found that lighter elements (such as barium) were produced.

Why are neutrons effective at inducing fission? Was it not stated earlier that the strong
force they supply is crucial to holding a nucleus together? Yes, but it is possible to have
too much of a good thing. For a given number of protons in a nucleus í for a certain
element í the band of stability is quite narrow. Too few neutrons or too many neutrons
make the nucleus unstable.

In fact, neutrons are a natural choice to induce fission. Since they are electrically Fission and chain reactions
neutral, they can easily approach and hit the nucleus without being repelled by
electrostatic forces. If the incoming neutron has the right speed, the nucleus captures it Neutrons can be used to cause fission
and becomes even more unstable. The nucleus has one neutron too many for the Fission can be self-sustaining
number of protons, and that immediately causes the new compound nucleus to fission Releases energy
into various elements.

Energy is released during fission; in the above process, roughly 200 MeV. You can
understand why this happens by returning to the concept of binding energy per nucleon,
which is lower for very heavy nuclei than it is for intermediate-size nuclei.

The binding energy is the amount of energy that is required to separate a nucleus into its separate nucleons, or it is the energy released when
separate nucleons are brought together to form a nucleus. The more tightly bound a nucleus is, the higher the binding energy, that is, the more
energy is released. Since the intermediate-size nuclei have the highest binding energy per nucleon, this means that heavier nuclei will release
energy as they split apart and become medium-sized.

We have discussed fissioning a single atom with the release of energy. This is a crucial scientific achievement, but in order for this to be useful
(for example, in a nuclear reactor), the process needs to be self-sustaining.

What makes fission a practical process is that in a fission reaction, one or more neutrons may be released, which can then induce more fission
reactions in nearby atoms, which produce more neutrons, and so on. Why are there “extra” neutrons? Recall that for heavier nuclei, the
number of neutrons exceeds the number of protons, while for lighter nuclei, the number of neutrons and protons tends to be nearly equal. This
means there are usually some neutrons left over after the nuclear re-arrangement.

When there are enough uranium atoms so that at least one neutron, on average, is captured by another uranium atom, the critical mass has
been reached. The process is self-sustaining, and it is called a chain reaction. Press the refresh button to see the fission process occur in
Concept 2.

Copyright 2007 Kinetic Books Co. Chapter 38 709

Since humans learned to split the atom in the 20th century, the process has been put to great use. The most well-known application was the
use of nuclear fission in the “atomic” bomb. A runaway chain reaction happens very fast, releasing a lot of energy in a burst which can be used
to devastating effect.

On the positive side of the ledger, nuclear power has been used as an energy source. A slowly progressing chain reaction produces a steady
flow of heat that can be used to boil water, which then drives steam turbines to generate electricity. The poster-child for nuclear power is
France, which supplies about three-quarters of its electrical needs with nuclear reactors.

However, nuclear power is not without its risks or costs. Some of the byproducts of the fission process are highly radioactive and remain
dangerous for tens of thousands of years. A typical way to dispose of these wastes is to bury them deep in the Earth. If they leak, they
contaminate water sources.

38.14 - Fusion

Fusion: Two light nuclei fuse into a heavier one,
releasing energy.

Fusion is a process in which nuclei join together to become a single, larger nucleus.
This process also releases energy. Because positively charged nuclei repel one
another, fusion does not occur spontaneously under normal conditions on Earth.

However, fusion is commonplace in the Sun and other stars where hydrogen atoms Fusion
fuse into helium atoms. Fusion provides the energy to keep the star going, which Earth
ultimately experiences as light and heat. Fusion occurs in the Sun because of the high Light nuclei fuse into larger, more stable
temperature within the star; its interior is at about 100 million Kelvin. pieces

At this high temperature, the atoms are in an ionized state of matter called plasma. Releases energy
While normal matter consists of distinct neutral atoms, plasma is a “soup” of positive
nuclei and negative electrons. In the interior of the Sun, nuclei are hot enough and
moving quickly enough to overcome the electrostatic repulsion of their positive nuclei.
They move close enough to be bound by the attractive strong force between them, then
fuse into a single nucleus.

Why is energy released in the process? For light elements (atoms to the left of the peak on the binding energy curve), the binding energy per
nucleon increases with atomic number. As smaller nuclei are fused into larger ones, the result is a more efficient arrangement of nucleons; one
that is harder to break apart. Since the new nucleus is more efficient í more tightly bound í than the previous ones, there is energy to spare.

It may be tempting to think that fusion and fission are opposite processes, since one combines nuclei and the other splits them. It may seem
confusing as to how two “opposite” processes can release energy. The key is that fission with energy release occurs only when very heavy
nuclei break apart into medium-size nuclei, and fusion with energy release occurs when very light nuclei fuse into heavier ones. The answer
again resides on the curve of binding energy; “mid-sized” elements have the highest binding energy per nucleon. As predicted by the curve, it
requires energy to combine two medium-sized nuclei into a single large nucleus, just as it requires energy to split a medium-size nucleus into
smaller nuclei.

Fusion is critical to the universe we observe. Without it, there would be no stars, and in fact no elements heavier than Lithium (Z = 3).

However, there is also a practical reason for scientific interest in fusion. If we could create and control fusion, we could use it as an energy
source. Using fusion as an energy source has two huge appeals: the fuel (hydrogen isotopes) is present in the oceans in essentially unlimited
quantities (compared to the relative scarcity of, say, uranium) and fusion creates no radioactive byproducts.

However, before fusion reactors become commonplace, some daunting engineering challenges will have to be solved. Recreating the
conditions inside the Sun, with its enormous temperatures, is no easy feat. Furthermore, simply getting the atoms hot is not sufficient; there is a
minimum plasma density that must also be achieved so that collisions between nuclei will occur frequently enough to release energy, which
keeps the plasma hot and keeps the fusion reaction going. The combination of high temperature and high density requirements necessarily
means that the pressure must also be very high, to hold all of the reactants together.

As in the case of fission reactions, early work on fusion was directed toward nuclear weapons. Scientists solved the problem of achieving the
ultrahigh temperature and pressure conditions necessary for fusion by using a fission (“atomic”) bomb as a trigger, to both heat up and
compress the fusion fuel. Some trigger! Using atomic bombs to power a nearby fusion reactor is not a very popular proposal.

There are two less-explosive schemes that are currently being pursued to keep the superhot plasma together for fusion to occur. Magnetic
confinement uses electromagnetic fields to hold the charged particles together. Inertial confinement fusion uses a solid pellet of deuterium and
tritium that is crushed by the light pressure of perfectly timed, short-duration, high-powered laser beams from all directions. The term comes
from the fact that the particles’ own inertia keeps them in place during the laser pulse.

710 Copyright 2007 Kinetic Books Co. Chapter 38

38.15 - Radioactivity and radiation

Radioactive decay: A nucleus spontaneously
emits particles or high-energy photons and
either changes identity or becomes less excited.

Transmutation: Changing of one element to
another after Į or ȕ radiation is emitted.

Fission is not the only way that a nucleus with an unstable combination of protons and Unstable nuclei
neutrons can change into a more stable configuration. An unstable nucleus may instead
spontaneously emit a charged particle or a high-energy photon in order to reach a more Too massive or wrong ratio of protons to
stable state. It changes via a nuclear reaction. Isotopes that change (decay) like this are neutrons
said to be radioactive. It is possible for the isotope to become a different chemical
element after the decay, a process known as transmutation.

The outgoing radiation can be classified as alpha rays, beta rays, or gamma rays. (They

are represented by the Greek letters Į, ȕ, and Ȗ.) Different decay processes result in

different forms of radiation. Alpha and beta rays consist of matter particles, while

gamma rays are photons (light particles).

In an alpha decay, the initial radioactive isotope decays into a different element by

emitting an Į particle. An Į particle is made up of two protons and two neutrons. It is a

helium-4 nucleus.

The initial isotope is known as the parent. Because it loses two protons, its atomic Radioactive decay with Į or ȕ ray
number is reduced by two (the mass number is reduced by four). Since the nucleus now
has a different number of protons, it becomes another element í it has transmuted. The Nucleus decays by emitting charged
newly-formed nucleus is known as the daughter. particle
·Radioactive element has transmuted
Alpha decay occurs most commonly in heavy nuclei whose ratio of protons to neutrons, (changed to another element)

Z/N, is too large, making them unstable. An Į particle is a very stable particle, and the

daughter nucleus that is left behind is more stable (tightly bound) than the parent. To

put it another way, the net result of the radioactive decay is a reduction of the ratio Z/N.

Beta decay is characterized by the emission of an electron or antielectron. There are

two types of beta decay, negative and positive. Negative ȕ emission is represented in

Concept 4. This occurs when a neutron inside the nucleus decays into a proton, an
electron (the beta ray), and an almost zero-mass, uncharged particle known as an
antineutrino. The antineutrino’s interaction with matter is so weak that it is very hard to
detect, and so it is customarily left out of nuclear equations.

The emitted electron did not exist in the nucleus beforehand, and is not one of the
orbital electrons in the parent nucleus. When a neutron inside the nucleus turns into a
proton, electron, and neutrino (and then emits the electron as a negative beta ray), its
number of protons increases by one. The mass number stays the same. The released
electron usually zips away, leaving behind a daughter atom with a net positive charge.

In positive ȕ emission, the nucleus emits an antielectron (also called a positron). Alpha radiation

Antielectrons are essentially the same as electrons, except they have a change of Nucleus emits alpha particle (2 protons,
2 neutrons)
positivee. The decay process is shown in Concept 5. A proton inside the atom decays
Number of protons in nucleus decreases
into a neutron, an antielectron (the beta ray), and a neutrino. Like an antineutrino, a by 2
neutrino has no charge and almost zero mass. Its interaction with matter is also weak ·Transmutation
and it too is customarily left out of nuclear equations.
Mass number decreases by 4
In emitting either type of ȕ ray, the initial isotope changes atomic number, so ȕ decay

results in transmutation.

In gamma decay, the radioactive isotope emits a high-energy photon, also known as a Ȗ

ray.

Since the atomic number stays the same, the atom is the same chemical element after the decay. In this case it is the nuclear energy that
changes in the decay. A nucleus can have different energy states. When a nucleus changes from an excited, high energy state to a lower one,
a photon is emitted.

This is similar to the case when an electron falls from one energy level to another. A notable difference is that nuclear energy levels are much
more widely spaced, on the order of millions of electron volts as opposed to say five or ten. This means the emitted photon, called a gamma
ray, is much more energetic than the photon emitted by an excited atom.

A gamma decay is represented in Concept 6. An excited nucleus is denoted by an asterisk “*” after the usual symbol. How does the nucleus

get into an excited state? This usually happens after another kind of decay. In many cases of Į and ȕ decay, the product nucleus is in an
excited state, after which it emits a Ȗ ray and transitions to a lower state or to the ground state. Because the photon is electrically neutral,

transmutation does not occur during gamma radiation.

Copyright 2007 Kinetic Books Co. Chapter 38 711

Negative beta radiation

Nucleus emits negative beta particle
(electron)
Number of protons in nucleus increases
by 1
·Transmutation
Nuclear mass is unchanged

Positive beta radiation

Nucleus emits positive beta particle
(antielectron)
Number of protons in nucleus decreases
by 1
·Transmutation
Nuclear mass is unchanged

Radioactive decay with gamma
ray

Nucleus emits gamma ray (high-energy
photon)
·Becomes less excited
·No transmutation
·Nuclear mass is unchanged

712 Copyright 2007 Kinetic Books Co. Chapter 38

38.16 - Sample problem: radioactive decay

A thorium-232 nucleus (Z = 90, A =
232) is radioactive and decays by first
emitting an alpha particle, then two
negative beta particles. What are the
atomic number and the mass number
of the daughter nucleus after each of
these three decay steps?

Variables Z
A
atomic number, number of
protons

mass number, total number of
protons and neutrons

What is the strategy?

1. The Į and ȕ particles have known charge and known mass number. Subtract these from the atomic number and the mass number of
the parent nucleus to determine the daughter nucleus’s values of Z and A.

Physics principles and equations

An Į particle consists of 2 protons and 2 neutrons. Its mass number is four.

A negative ȕ particle consists of one electron. When the parent nucleus emits it, the number of protons increases by one and its mass number

is unchanged.

Step-by-step solution

We will use the notation Z0 and A0 to denote the initial values of the proton number and the mass number, Z1 and A1 to denote their values
after the first decay step, Z2 and A2 to denote their values after the second decay step, and Z3 and A3 to denote their values after the third and

final decay step.

Step Reason

1. Z0 = 90 initial values

2. Z1 = Z0 – 2 Į emission decreases nuclear charge

3. A1 = A0 – 4 Į emission decreases nuclear mass

After the first decay process, the nucleus has 88 protons and 228 nucleons in total. This is radon-228, which then emits a negative ȕ particle.

Step Reason

4. Z1 = 88 initial values

5. Z2 = Z1 + 1 ȕ emission increases nuclear charge

6. A2 = A1 ȕ emission does not affect nuclear mass

After the second decay process, there is a nucleus with 89 protons and 228 nucleons in total. This is actinium-228, which then decays by
emitting another negative ȕ particle.

Step Reason

7. Z2 = 89 initial values

8. Z3 = Z2 + 1 ȕ emission increases nuclear charge

9. A3 = A2 ȕ emission does not affect nuclear mass

After the third decay process, the nucleus has 90 protons, and a total of 228 nucleons. The end result after the three decays is again thorium
because the number of its protons is again 90. After the three decays, the element is thorium-228. This lighter thorium isotope then continues

to decay in a series of Į and ȕ decays until it becomes the stable isotope lead-208.

Copyright 2007 Kinetic Books Co. Chapter 38 713

38.17 - Radioactive decay and half-lives

Half-life: The time it takes for half of a group of
radioactive atoms to decay.

We have discussed the methods by which radioactive isotopes decay, by emitting Į, ȕ,
or Ȗ rays. What can be said about the rate at which they decay? If you have a sample of

radioactive atoms, how many of them will decay in, say, the next minute?

The answer depends on the radioactive isotope. For any isotope, the fraction of atoms Half-life
that will decay in a minute can be determined. A quantity of particular interest is the
half-life, which is the average time it takes for one-half of the radioactive material Average time for one-half of a group of
present to decay. radioactive atoms to decay

For example, let’s consider 16.00 mg of a parent nuclide, 131I, which is a radioactive
isotope of iodine. Its half-life is 8.04 days. You would find that after 8.04 days, one-half
of the parent nuclei have decayed, and 8.00 mg of 131I remains. After another 8.04
days, one-half of the remaining iodine will have decayed, and 4.00 mg remains. After a
third 8.04 days, only 2.00 mg would remain, and after 8.04 more days, 1.00 mg would
remain, and so on. Half of the remaining iodine decays every 8.04 days.

If it is possible to know how many atoms in a sample are going to decay within a certain
time interval, is it possible to know when a particular atom will decay? Numerous
experiments have shown that the answer to that question for a particular atom is no.
This is similar to the situation of flipping one thousand coins and making a prediction of
“50% heads.” The prediction will be quite accurate, though you cannot reliably predict
the outcome of any particular coin. The process of radioactive decay provided evidence
of the statistical nature of quantum mechanics, which governs processes on a
subatomic scale.

During each half-life, one-half of
remaining radioactive atoms decay

Radioactive decay is
probabilistic

Cannot be predicted for an individual
atom
Can state the probability of any atom
decaying within a certain time
·or what fraction of the atoms will decay
within a certain time

38.18 - Interactive problem: radioactive dating

Carbon-14 is a radioactive isotope of carbon that has six protons and eight neutrons in its nucleus. It is commonly used to establish a date for
organic specimens. In the first simulation, you will observe the decay of carbon-14 (C-14), and determine the half-life of that radioactive
isotope.

You are equipped with a digital timer and a gauge that reports the number of parent atoms that are present. Before the simulation starts, there
are 32 billion parent atoms. Each of the spheres on the screen represents a billion atoms. A sphere changes color when a billion atoms have
decayed.

When you press GO, the timer starts and the carbon begins to decay to nitrogen by emitting ȕ rays, and the number of daughter nitrogen

atoms begins to grow. The daughter atoms are stable.

In the simulation, time is sped up and passes in thousands of years. The number of carbon atoms is shown both graphically with a
“thermometer”-type gauge as well as numerically. When half the initial number of parent atoms has become daughter atoms, press PAUSE
and note the elapsed time. (That is, at this moment, 16 billion of the carbon atoms remain present, and the rest have decayed.)

714 Copyright 2007 Kinetic Books Co. Chapter 38

After noting the time, restart the process by pressing GO to resume the simulation and press PAUSE when half of the remaining parent atoms
have become daughter atoms í in other words, when about eight billion carbon atoms remain. How does this second time interval compare to
the first interval of time, when the number of parent atoms changed from 32 billion to 16 billion?

Repeat this process once more (or as many times as you like), each time pausing
when the number of parent atoms has fallen in half. Do you see a pattern? You are
measuring the half-life of the material, which is the average time needed for half of
the parent atoms in a radioactive sample to decay. Enter your measurement for the
half-life of carbon-14 by selecting the appropriate amount of years. Press CHECK to
see if you are correct.

In the second simulation, you will use your newly acquired skills at measuring half-
lives to investigate a crime scene. You are an environmental investigator and a
criminal is once again dumping pure samples of a radioactive lead isotope,

, into a vacant lot. Holy ecological disaster!

You have been unable to catch the perpetrator in the act, but a security camera
filmed three suspicious-looking characters in the vacant lot at different times. If you
can determine when the radioactive waste was dumped in the lot, you will know
which of these three suspects is guilty.

The factory that creates the waste is cooperating with you. They tell you that the
isotope was pure lead-209 samples that initially contained 192 billion atoms. When
you find the waste, it is about midnight. At midnight, 24 billion lead atoms remain,
which means 168 billion of the lead atoms have decayed into bismuth atoms. In
other words, one-eighth of the original lead-209 is left.

Your mission has three parts. First, determine how many half-lives have elapsed
since the pure lead-209 sample was dumped. If you are having trouble with this
piece of your detective work, return to the first simulation and calculate how many
half-lives it takes for seven-eighths of the carbon atoms to decay.

Second, measure the half-life of lead-209 using a technique similar to what you
used in the first simulation. You have the same tools as you had before.

Third, you can use the evidence from the security camera. The camera filmed Anna
in the lot 6.51 hours before you obtained the sample. Sara was loitering in the area
about 9.76 hours before this time, and a third suspect, Katherine, was filmed there
13.0 hours before the sample was found.

To put this all together: You use the value you determined for the half life of lead,
and multiply that by the number of half-lives that have passed since the lead was dumped. That tells you how long the lead was there, so you
can nail the suspect. To confirm your conclusion, drag the handcuffs in the simulation to the dastardly dumper. The simulation (and perhaps
the suspect’s reaction) will let you know if you are correct.

38.19 - Particle physics and GUTs

With this section, we come to the end of our discussion of the atom and the nucleus. It seems appropriate to take a peek ahead to the current
state of nuclear physics, and to discuss what you and others may be learning in the decades ahead.

Particle physics, also known as high-energy physics, is one of the largest subfields of current physics. Essentially, it tries to answer the
question “What is everything made of?” Ordinary matter is composed of the particles you have encountered so far í protons, neutrons, and
electrons í and in the early 1900s, that seemed to be all that was needed to answer the big question. However, starting in the 1930s, particle
physicists began discovering new, exotic particles that were created in an energy-to-mass conversion during collisions between the known
particles. Several hundred other particles have since been discovered, most of which are unstable. Some of the particles are stable if they are
left alone, but are composed of antimatter, which annihilates ordinary matter upon contact.

Early experimenters relied on cosmic rays (high-energy particles that permeate our galaxy) to initiate reactions. (The exact source of cosmic
rays is still an open question. Stars emit them during intermittent flare-ups, but supernova explosions, when stars die, are thought to be
responsible for much of the cosmic ray output.)

Later, more controlled experiments were carried out in particle accelerators, also known as atom-smashers, where particles are made to collide
with higher and higher energies. Physicists now recognize that many of the heavier particles are made up from smaller building blocks called
quarks. They understand that the smorgasbord of particles is due to the fact that when particles collide, newly-created quarks can combine with
those already present to form systems of bound quarks.

The ultimate goal of physicists is a theory of everything (how is that for ambition?). Historically speaking, great breakthroughs in physics have
often resulted in a simplification of our view of the universe. For example, Newton’s universal law of gravity showed that the laws governing
celestial orbits were the same as those governing the motion of objects falling under earth’s gravity. Maxwell and his generation showed how
electricity and magnetism, long thought to be unrelated, were really aspects of the same force. Physicists optimistically believe that the
universe has an underlying simplicity, and that the number of fundamental forces can be reduced further.

We have discussed gravity, the electromagnetic force, and the strong force. (There is also the weak force which we have not discussed.) Albert
Einstein spent a lot of his working life trying to interpret these forces as different aspects of a single “superforce”. Historically, electricity and
magnetism were united in the 1800s, and in the latter part of the 20th century, the weak force and the electromagnetic force were also joined
theoretically. The goal of further reduction with the ultimate prize of unification still continues today.

Copyright 2007 Kinetic Books Co. Chapter 38 715

As far as high-energy physicists are concerned, the goal is a Grand Unification Theory, or GUT. The current dream is to unify gravity with the
strong, weak, and electromagnetic forces. You may have heard of attempts such as superstring theory, which interpret particles, such as
electrons, as being modes of oscillation of unimaginably small “strings”. Acceptance of superstring theory demands the idea that the universe
may not consist of four dimensions (three spatial dimensions plus time), but instead ten or more dimensions. Bizarre as these ideas may seem,
they are no less bizarre than relativity, nuclear theory, and quantum physics would have seemed to a scientist of the 19th century.

38.20 - Gotchas

All carbon atoms are the same. Not true: While carbon atoms all have six protons in their nucleus, different carbon atoms may have different
numbers of neutrons. Nuclei with six protons but different numbers of neutrons are isotopes of carbon.

The nucleus is so small that an atom is about 99.99% empty space. No, but you are on the right track if you think this. Since the atomic
diameter is about 10,000 times larger than the nucleus, the atom is more like 99.9999999999% empty space.

Protons and neutrons have different charge but approximately the same mass. Yes. A proton has a charge of 1.60×10í19 C and a neutron has
no net electrical charge. The neutron is about 0.1% more massive than the proton, and they are each about 1800 times as massive as the
electron.

The energy that is required to disassemble a nucleus into its constituent parts is the same amount of energy that is released when the same
nucleus is assembled from separated nucleons. Yes; the amount of energy is the same and is called the binding energy. It requires energy to
break apart a stable nucleus, and energy is released when the nucleus is assembled.

If half of a radioactive isotope decays during one half-life, then after two half-lives, it will all be gone. No, a half-life is the average time it takes
for one-half of the radioactive material that is present to decay. After two half-lives, there will be one-quarter of the original amount left. After
three half-lives, there will be one-eighth, and so on.

38.21 - Summary

Elements are substances that cannot be divided or changed into other substances Z = atomic (proton) number
using ordinary chemical methods. An atom is the smallest piece of an element that N = number of neutrons
still has its chemical and physical properties. An atom consists of electrons orbiting
a very small, very dense nucleus. The nucleus contains both protons and neutrons,
which collectively are called nucleons.

The number of protons in an atom is the atomic number, Z. The number of neutrons Atomic mass number
is the neutron number, N. The total is known as the atomic mass number, A. Atomic
A=Z+N
masses are measured in terms of atomic mass units. The mass of a carbon-12

atom is defined to be exactly 12 u.

Isotopes are forms of an element with the same atomic number (which makes them Atomic mass unit
the same element) but different numbers of neutrons (and hence different atomic
mass numbers). u = 1.66054×10–27 kg

A fundamental, very short-range interaction called the strong force holds the Nuclear radius
nucleons together, counteracting the electrostatic repulsion between protons.
R = (1.2×10–15 m)A1/3
The nucleons are nearly incompressible and are tightly packed in the nucleus. The
nuclear radius grows roughly as the cube root of the mass number.

Nuclei are stable only for certain combinations of protons and neutrons. On a plot of Mass-energy equivalence

neutron number versus proton number, stable nuclei are represented as a band of E = mc2
stability that passes through the center of the diagram. For small nuclei, the

numbers of protons and neutrons are roughly equal, but for large nuclei, the

neutrons outnumber the protons by about 50%. Heavy nuclei need an excess of neutrons to dilute the proton concentration.

Binding energy is the energy that must be added to disassemble, or unbind, a nucleus into the protons and neutrons that make it up. The same
amount of energy is released if the nucleus is assembled from nucleons that are initially separated.

The sum of the masses of the separate nucleons is always greater than the mass of the nucleus when it is whole. The mass difference is
related to the binding energy by Einstein’s equation for mass-energy equivalence.

When the binding energy of one nucleus is greater than that of another, this means the particles are more tightly bound. For comparing how
tightly bound two nuclei are, the binding energy per nucleon is the important figure.

The shape of the binding energy per nucleon vs. mass number curve is important. The curve is highest in the middle. This means that light
nuclei can undergo the fusion process and release energy as this will increase their binding energy per nucleon. Similarly, heavy nuclei can
undergo the fission process and release energy as this also will increase their binding energy per nucleon.

In the process of radioactive decay, an unstable nucleus spontaneously emits particles or high-energy photons. If the parent nucleus emits a

charged particle such as an Į particle, negative ȕ, or positive ȕ, then the number of protons in the nucleus changes. Transmutation is the

changing of one element to another. If the parent nucleus emits gamma rays, which are uncharged photons, then transmutation does not take

place, because the number of protons in the nucleus has not changed.

The half-life of a radioactive isotope is the average time it takes for the decay of one-half of the atoms that are present in a sample.

716 Copyright 2007 Kinetic Books Co. Chapter 38

Chapter 38 Problems

Conceptual Problems

C.1 In Rutherford's scattering experiment, positive alpha particles are directed at a thin gold foil. What was the key piece of
evidence that led him to conclude that the positive charge and mass of the atom were concentrated in the nucleus?

i. All the alpha particles bounced straight back from the gold foil.
ii. None of the alpha particles were deflected by the gold foil.
iii. A small fraction were deflected from the gold foil.
iv. The alpha particles were absorbed by the gold foil.

C.2 What is the approximate ratio between an atomic diameter and a nuclear diameter?

i. 1:1
ii. 10:1
iii. 100:1
iv. 10000:1
C.3 Consider the following atoms: 12C, 14C, 14N, 14O, 16O. The atomic numbers of carbon, nitrogen, and oxygen are 6, 7, and 8,
respectively.
(a) Which atoms are isotopes of each other? Check all that apply.
(b) Which atoms have the same number of neutrons? Check all that apply.
(c) Which atoms have the same number of protons? Check all that apply.

(a) 12C and 14C
14C and 14N and 14O
14O and 16O
none of them

(b) 12C and 14C
12C and 14O
14C and 16O
12C and 14N
none of them

(c) 12C and 14C
14C and 14N and 14O
14O and 16O
none of them

C.4 The strong force acts between which of the following types of particles? Check all that apply.

proton and neutron
proton and proton
proton and electron
neutron and neutron
neutron and electron

C.5 What force or forces are responsible for keeping electrons in their orbits around the nucleus?

i. Strong force
ii. Electric (Coulomb) attraction
iii. Strong force and Coulomb attraction
iv. None of these
C.6 Consider the following atoms: 12C, 14C, 16O. Rank these in order, from smallest to largest nuclear radii.

12C 14C 16O
12C 16O 14C
14C 12C 16O
14C 16O 12C
16O 12C 14C
16O 14C 12C

C.7 Can two iron atoms fuse to form a nucleus with greater binding energy per nucleon? Explain your answer.

Yes No

C.8 Explain in your own words why a chain reaction is possible during uranium fission.

Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems 717

C.9 In the sun, 4 hydrogen nuclei undergo a multi-step process and eventually form a helium-4 nucleus (plus some positrons,
neutrinos and gamma rays, all of negligible mass). What is this an example of?

i. Nuclear fission
ii. Nuclear fusion
iii. Alpha decay
iv. Chain reaction
C.10 A radioactive sample emits both alpha particles and negative beta particles. If a uniform external magnetic field is present in
the region, what will be observed regarding the paths of the emitted particles?

i. Neither path bends.
ii. The paths of both particles bend in the same direction.
iii. Only one particle's path bends.
iv. The particles' paths bend in opposite directions.

Section Problems

Section 3 - Components of the nucleus

3.1 A neutral atom of potassium has 19 protons in its nucleus. How many electrons does it possess?

electrons

3.2 For the neutral chlorine atom , determine the (a) mass number, (b) number of protons, (c) number of neutrons (d)

number of electrons.

(a) is the mass number
(b) protons
(c) neutrons
(d) electrons

3.3 For a doubly-ionized calcium ion, which is a atom with two electrons removed (leaving it with a net charge of +2e),

determine the (a) mass number, (b) number of protons, (c) number of neutrons (d) number of electrons.

(a) is the mass number
(b) protons
(c) neutrons
(d) electrons

3.4 What is the mass of an electron, measured in atomic mass units?

u

Section 6 - Nuclear properties

6.1 In a Rutherford-like experiment that analyzes electrons that are scattered from a sample, a scientist measures the radius of a
nucleus in the sample to be 4.8×10í15 m. What of these is likely to be the mass number of the nucleus?

i. 1
ii. 2
iii. 4
iv. 16
v. 64

6.2 Find the mass number of a nucleus whose radius is approximately one-half that of uranium-240.

Section 7 - Sample problem: nuclear density

7.1 Estimate the radius of a sphere of nuclear matter whose mass is equal to that of the Earth, 5.97×1024 kg. Take 2.3×1017
kg/m3 as the density of nuclear matter.
m

718 Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems

Section 8 - Stable nuclei

8.1 Lead is an element that has 82 protons in its nucleus. Which of the following mass numbers is most likely to correspond to a
stable isotope of lead?

i. 41
ii. 123
iii. 164
iv. 206
v. 246

Section 9 - Nuclear binding energy

9.1 What is the binding energy for a nucleus of helium-4, also known as an alpha particle? A neutron has a mass of 1.00866 u, a
proton has a mass of 1.00728 u, and an alpha particle has a mass of 4.00153 u, where u = 1.66054×10í27 kg. Express the
energy in joules to three significant figures.

J

Section 13 - Fission

13.1 Consider the fission reaction that occurs when a neutron is captured by a uranium-235 atom and fissions into xenon-140,
strontium-94, and 2 neutrons. Calculate the energy released in this reaction. The masses are: neutron = 1.0087 u, uranium-
235 = 235.0439 u, xenon-140 = 139.9216 u, strontium-94 = 93.9154 u.

J

Section 14 - Fusion

14.1 In the sun, 4 hydrogen nuclei undergo a multi-step process and eventually form a helium-4 nucleus (plus some positrons,
neutrinos and gamma rays, all of negligible mass). For each helium atom produced, about 26 MeV of energy is released. (a)
Which has more mass, the 4 hydrogen atoms or the helium-4 atom? (b) What is the magnitude of the mass difference?

(a) 4 hydrogen atoms 1 helium-4 atom
(b) kg

14.2 The sun radiates electromagnetic energy at the rate of 3.91×1026 W. (a) What is the change in the sun's mass during one
second? Express the answer as a positive mass. (b) The current mass of the sun is 1.99×1030 kg. What fraction of the sun's
mass is lost in a century? Express the answer as a decimal number and assume years of 365.25 days.

(a) kg
(b)

Section 15 - Radioactivity and radiation

15.1 An excited nucleus emits a gamma ray of energy 7.5 MeV. (a) Does the nuclear mass increase, or does it decrease? (b)
What is the magnitude of the change?

(a) Increase Decrease

(b) kg

15.2 A carbon-14 atom decays into nitrogen-14 (Z = 7). What particle(s) did it emit?

i. Positive beta particle(s)
ii. Negative beta
iii. Gamma
iv. Two alpha

15.3 A uranium-238 nucleus decays into a uranium-234 nucleus. What particle(s) did it emit?

i. Four positive beta particle(s)

ii. One alpha

iii. Two alpha and two negative beta

iv. One alpha and two negative beta

Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems 719

15.4 Lead (Z = 82, A = 214) decays in a series of steps. The nucleus emits 2 negative betas, 1 alpha, 1 negative beta, 1 alpha,
and 1 negative beta. What is the final daughter product?
i. Lead-206
ii. Lead-205
iii. Lead-222
iv. Lead-210
v. None of these

15.5 Uranium (Z = 92, A = 234) decays in a series of steps. The nucleus emits 4 alphas, 1 negative beta, 2 alphas, 3 negative
betas, and finally, 1 alpha. What is the final daughter product?
i. Lead-206
ii. Lead-208
iii. Lead-210
iv. Lead-212
v. Uranium-206

15.6 Uranium-238 (Z = 92) decays by alpha emission to thorium-234 (Z = 90), by emitting a helium-4 nucleus. Your friend claims
to have witnessed a new type of decay where the uranium-238 emitted a helium-3 nucleus and became a thorium-235 atom.
Prove that this is impossible for an isolated U-238 atom. Here is some helpful data: U-238 has an atomic mass of 238.051 u,
the mass of a helium-3 nucleus is 3.014 u, and the mass of a Th-235 atom is 235.048 u.

Section 17 - Radioactive decay and half-lives

17.1 Carbon-14 has a half-life of 5730 years. After four half-lives have elapsed, what percentage of an initially pure sample would
remain unchanged?
percent

Section 18 - Interactive problem: radioactive dating

18.1 Using the simulation in the first interactive problem in this section, determine the half-life of Carbon-14.
i. 16000 years
ii. 11460
iii. 5730

iv. 2865
18.2 (a) Use the information for the second interactive problem in this section to determine the number of half-lives that elapsed

between the time the pure lead-209 was dumped and midnight. (b) Use the second simulation to measure the half-life of lead-
209. (c) Using the information given about when each suspect was spotted, determine who is the guilty party.
(a) half-lives have elapsed
(b) hours
(c) i. Anna

ii. Sara
iii. Katherine

720 Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems

Chapter 1 Answers

Answers to selected problems

C.5 (a) Increases
(b) Decreases

2.1 100000 cm
2.3 9.966 teradollars
2.5 100000 tablets
3.1 3.75eí11
3.3 5.68e8 s
3.5 1.00e8
5.1 Dime
8.1 25.0 mi/h
8.3 1.28e6 cubits
8.5 Yes
8.7 (a) 5.79e10 m

(b) 5 910 000 000 km
8.9 2.72 m
8.11 4.9 kg
11.1 6.2 ft
11.3 8 cm
11.5 3.79 km
12.1 430 m
12.3 9.87 ft
13.1 1.88 radians

Copyright 2000-2007 Kinetic Books Co. Chapter 1 Answers 721

Chapter 2 Answers 7.1 (a) Right
(b) Zero
Answers to selected problems (c) Start the ball moving slowly to the left and then
increase its speed.
C.1 No
C.5 (a) í15 m 8.1 6.67 s

(b) 15 m 9.1 1.90e+5 m/s2
C.7 Yes 9.3 (a) 0 m/s2

Answers vary (b) 9.6 m/s2
C.9 Instantaneous velocity (c) 7.7 m/s2
C.12 Zero 9.5 (a) 0 cm/s2
C.13 No (b) í3.0 cm/s2
(c) 3.0 cm/s2
Answers vary 9.7 31v/5
0.1 Negative
1.1 1.6 meters 10.1 (a) 1.0 s to 2.0 s
2.1 1 step(s) (b) 0 to 1.0 s
2.3 (a) 5.3 km (c) 4.0 to 5.0 s

(b) 5300 m 13.1 The white rabbit
3.1 0.343 km The black rabbit
3.3 2.4e+3 s
4.1 (a) 0.063 m/s 15.1 2.8 m/s

(b) 240 m/s 15.3 (a) 15.0 m
4.3 (a) 138 km (b) 3.04 s

(b) 110 km/h 15.5 a = 2(vt + d)/t2
4.5 (a) 10 ft/s
18.1 3.19 m/s
(b) 11 km/h 18.3 76.7 m/s
4.7 21.0 s 18.5 0.682 m/s2
5.1 (a) 1440 m 18.7 1.24 s
18.9 í4.2 m/s
(b) 7200 s 18.11 (a) 1.22 s
5.3 (a) 3.0 m/s
(b) 9.10 m
(b) í3.0 m/s
(c) 0 m/s
5.5 5.40 m/s

722 Copyright 2000-2007 Kinetic Books Co. Chapter 2 Answers

Chapter 3 Answers

Answers to selected problems

C.3 6.0 knots Southwest
C.5 íD
0.1 (a) (í4.0 km, 3.0 km)

(b) (3.0 km, 2.0 km)
(c) (í3.0 km, í2.0 km)
1.1 81.2
1.3 1.41e3 kg/m3
3.1 (7.00 knots, 121°)
3.3 ( 3.00, 90.0°)
4.1 (a) (3, í25) blocks
(b) No
4.3 (a) A = (3, 2)
(b) B = (í1, í3)
(c) C = (í3, 3)
5.3 (a) Submit answer on paper.
(b) D
(c) C
(d) F
(e) B and E

6.1 (a) (18, 8)
(b) (3, 6)
(c) (10, 10, 10)
(d) (a+d,b+e,c+f)

6.3 (a) a = 5
(b) b = í4

8.1 (a) (18, í6, 48)
(b) (9, í12, 15)
(c) a2, íab, íac)
(d) (í2a+18, í10í6b, í2c+12)

9.1 (a) (8, 230°)
(b) (21, 200°)
(c) (32, 80°)

Copyright 2000-2007 Kinetic Books Co. Chapter 3 Answers 723

Chapter 4 Answers

Answers to selected problems

C.3 On the Moon
Answers vary

C.5 (a) The speed of the river
(b) North
(c) No

0.1 (a) No
(b) The speed increases

1.1 (38.8,29.2) m/s
1.3 (108, 40.3) m/s
2.1 (a) (1.0, 4.0) m/s2

(b) (0.0, 5.0) m/s
2.3 ( 180, -3.75) m/s
2.5 (0.73, 1.3) m
3.1 (a) 0.64 s

(b) 33 m/s
3.3 1.20 s
3.5 66.5 m
3.7 (a) 10.5 s

(b) 284 m
(c) í29.3 m/s
(d) 39.8 m/s
3.9 b
3.11 1.07 s
3.13 1 to 9
5.1 At the professor's glove
Answers vary
7.1 -6.06 m/s
7.3 11.0 m/s
7.5 ( 2.36, 14.5) m/s
9.1 At the professor's glove
Answers vary
10.1 24.4°
10.3 ( 9.29, 20.6) m/s
10.4 35.3 m/s
10.6 (a) 1100 m/s
(b) 49.3°
10.8 4.07 s
10.10 (a) 1.93 m
(b) 2.29 s
13.1 3.1 m/s
15.1 8.0 m/s

724 Copyright 2000-2007 Kinetic Books Co. Chapter 4 Answers

Chapter 5 Answers 11.1 (a) 1.72 N
(b) 3.34 N
Answers to selected problems
11.3 (a) 172 N
C.1 Yes (b) 53.0°
Answers vary
12.1 117 N
C.5 Greater between blocks 1 and 2
Answers vary 15.1 (a) 49 N, Down
(b) 49 N, Up
C.13 Yes (c) 78 N, Right
Answers vary (d) 23 N, Left

0.1 It can be doing any of these 17.1 2.03e3 N
2.1 0 N (direction does not matter) 17.3 260 N
4.1 10 dumbells
4.3 (a) 209 N 18.1 0 m/s2
18.3 0.13
(b) 21.3 kg 18.5 0.0950
5.1 515 kg
5.3 (a) 5.0 s 22.1 (a) 59 N, Straight down
(b) 51 N, Up and to the left
(b) 7.0 s (c) 23 N, Down the plane
(c) 6.0 s (d) 78 N, Up the plane
5.5 1.2eí3 N
5.7 22500 kg 23.1 -0.301 m
5.9 8.10eí7 N/m2 23.3 (a)1.25 m/s2
5.11 (a) 0.0788 N
(b) 0.259 N (b) Upward
5.13 (a) 13.4 m/s2 23.5 (a) 4.12e+3 lbs/in
(b) 603 N
(c) Yes (b) 7.21e+5 N/m
5.15 5.1 m/s2
9.1 (a) 6.54e4 N 24.1 1.60
(b) 7.57e4 N
10.1 í0.233 m/s2 25.1 (a) 12 N, Directly to the right
(b) Any magnitude greater than 10 N, Directly to the right
(c) 10 N, Directly to the right

Copyright 2000-2007 Kinetic Books Co. Chapter 5 Answers 725

Chapter 6 Answers
Answers to selected problems

0.1 (a) Greater
(b) To the same height as the other hill

1.1 201 J
1.3 181 J
4.1 49 times higher
4.3 1410 J
5.1 (a) 8.0eí17 J

(b) 1.3e+7 m/s
5.3 1.63e+3 N
5.5 (a) 8460 J

(b) 26.0 m/s
7.1 133 N
9.1 2.1e+3 W
9.3 3.41 m/s
9.5 2.24e+5 J
10.1 $184
11.1 -8.58 J
11.3 2.04e+10 J
11.5 801 W
14.1 42.5 kg
14.3 1.44 m/s2
16.1 63.8 m
19.1 8.0 m

726 Copyright 2000-2007 Kinetic Books Co. Chapter 6 Answers

Chapter 7 Answers 727

Answers to selected problems

C.1 No
Answers vary

C.3 (a) They have the same momentum
(b) A's kinetic energy is greater

C.5 Truck impulse equal
C.7 Center of mass

Answers vary
0.1 (a) Yes

(b) No
(c) Yes
1.1 1.29 kg·m/s
1.3 31 m/s
1.5 90 kg·m/s
2.1 2.8e2 kg·m/s
3.1 -1.73e+3 N
3.3 25800 N
3.6 (a) 1.13e3 kg·m/s
(b) 4.50e4 N
3.8 (a) 4.0 N
(b) 12 kg·m/s
(c) 80 m/s
3.10 2.79e+3 N
3.12 (a) í4.09eí23 kg·m/s
(b) 0.205 N
5.1 -1.26e+3 m/s
5.3 í3.25 m/s
5.5 (a) í0.43 m/s
(b) 0.77 m/s
8.1 32 N
8.2 118 kg
9.1 (a) 2.3 kg
(b) 3.1 m/s
9.3 (a) 8.1 kg
(b) í1.9 m/s
12.1 6.0 m/s
13.1 (a) í0.40 m/s
(b) í3.6 m/s
(c) í1.8 m/s
13.3 0 m/s
13.5 2.0 kg·m/s
13.7 (a) 3.34 m/s
(b) No
14.1 4.7e6 m
14.3 (a) 6.0 m
(b) 6.3 m
16.1 (a) Impossible
(b) Elastic
(c) Inelastic

Copyright 2000-2007 Kinetic Books Co. Chapter 7 Answers

Chapter 8 Answers
Answers to selected problems

0.1 (a) Yes
(b) Yes

2.1 1.31e4 m/s
2.3 2.28e11 m
3.1 1.15 m/s2
3.3 (a) 0.0337 m/s2

(b) 0 m/s2
3.5 21 m/s2
4.1 9.0 m/s
5.1 4.50 "G's"
5.3 1.24eí4 N
5.5 6.3 N
6.1 103 s
7.1 (a) 8.0 m/s

(b) 9.5 m/s

728 Copyright 2000-2007 Kinetic Books Co. Chapter 8 Answers

Chapter 9 Answers

Answers to selected problems

C.1 Yes
Answers vary

0.1 (a) Decrease
(b) Decrease
(c) Stay the same
(d) Increase

1.1 48 m
2.1 14 rad
2.3 7.17eí4 rad
2.5 (a) 3.54 s

(b) 0.315 rad
(c) 0.500 rad/s
3.1 2.0e+3 m
3.3 1.99eí7 rad/s
3.5 (a) 0.10 rad/s
(b) 23 rad/s
4.1 -28.0 rad/s2
4.3 í0.12 rad/s2
7.1 (a) 7.08eí3 rad/s
(b) 2.01eí4 m/s2
(c) 2.05eí5 g
7.3 17 m/s
7.5 0.619 rad/s2
8.1 0.083 m/s2

Copyright 2000-2007 Kinetic Books Co. Chapter 9 Answers 729

Chapter 10 Answers
Answers to selected problems

C.1 (a) Yes
(b) Yes

C.3 False
Answers vary

0.1 (a) Yes
(b) Farthest from the axis of rotation

1.1 376 N
1.3 0.21 N·m
2.1 1.60 rad/s2
2.3 53 rad/s
3.1 19 kg·m2
3.3 1.0e2 kg·m2
4.1 1.3 s
4.3 2.0 rad/s2
4.5 5.7eí5 kg·m2
4.7 0.21
6.1 2.85e5 N
8.1 1.93e+5 kg·m2/s
9.1 49.8 kg·m2/s
10.1 143 rad/s
10.3 119 rad/s
11.1 50 rad/s
11.3 2.15 rad/s

730 Copyright 2000-2007 Kinetic Books Co. Chapter 10 Answers

Chapter 11 Answers

Answers to selected problems

C.1 b
d

C.5 Steel,
Copper,
Titanium,
Aluminum

1.1 1.47 m
1.2 0.23 m
1.4 1.4e3 N
1.6 0.302 m
1.9 1.20 m
3.1 9.2eí2 m
3.3 (a) (1.0,0.50)

(b) (1.0,1.0)
(c) (0.63,0.38)
(d) (0.69,0)
3.5 0.827 m
3.7 0.32 m
4.1 (a) 950 N
(b) Negative y
(c) 1.32 m
6.1 (a) 1.6e5 N/m2
(b) 2.0e6 N/m2
7.1 0.0257 m
7.3 2.72eí5 m
7.5 4.0e9 N/m2
7.7 82e9 N/m2
8.1 1.5 m3
8.3 í5.6eí5

Copyright 2000-2007 Kinetic Books Co. Chapter 11 Answers 731

Chapter 12 Answers

Chapter Assumptions

Unless stated otherwise, use the following values.

rEarth = 6.38×106 m
MEarth = 5.97×1024 kg

Answers to selected problems

0.1 (a) Decrease
(b) Increase

1.1 9.04e4 N
1.3 4.11e11 m
1.5 (2.0eí9,í5.2eí10) N
2.1 9.76 m/s2
2.3 24.8 m/s2
8.1 7900 m/s
9.1 7.69e3 m/s
9.3 1.02e3 m/s
11.1 3160 m/s
12.1 (a) It increases

(b) It decreases
14.1 22.7 AU
14.3 (a) 1.00 AU

(b) 0.050
14.5 22.9 AU
16.1 (a) 3.74e8 s

(b) 11.9 years
16.3 2.40e+9 s
16.5 400 hours
16.6 1.05e4 years
17.1 (a) í6.32e11 J

(b) 3.16e11 J
(c) í3.16e11 J
17.3 1.47e+13 J
17.5 í1.14e10 J
18.1 1.04e4 m/s
18.3 (a) 3.55e4 m/s
(b) 3.76e4 m/s
18.5 (a) 1.47e4 m
(b) 2.95e12 m

732 Copyright 2000-2007 Kinetic Books Co. Chapter 12 Answers

Chapter 13 Answers

Chapter Assumptions

Unless stated otherwise, use the following values:

Atmospheric pressure at the Earth's surface: Patm = 1.013×105 Pa

Density of pure water = 1000 kg/m3
Density of seawater = 1030 kg/m3
“Standard temperature and pressure” means 0°C and the atmospheric pressure stated above.

Answers to selected problems

C.3 No. The bottle will not explode. 5.1 (a) 1.49e5 Pa
Answers vary (b) 2.50e5 Pa
(c) Ben Chapman
C.5 No
Answers vary 6.1 (a) 1.01e3 kg/m3
(b) 1.03e5 Pa
C.9 The new waterline mark is lower on the hull.
6.3 1.10e+5 Pa
C.11 The ball bearing
Answers vary 7.1 6.76e+3 m3
7.3 (a) 8.18eí2 m3
C.14 The lake's water level rises very slightly
(b) 1.05e3 kg/m3
2.1 (a) 31.3 people/km2 (c) 87.0
(b) 42.7 people/km2
(c) 1.86 people/km2 8.1 (a) 79.3 N
(d) 6.10e3 people/km2 (b) 12.7 N

2.3 (a) 12 kg 9.1 43.8
(b) 26 lb
10.1 The 3.20 kg crown
3.1 (a) 6.85e6 Pa
(b) 1.57e5 Pa 11.1 63.8 N

3.3 (a) 7.64e4 Pa 13.1 36.6 m/s
(b) 1.95e3 Pa
14.1 8.36e+4 Pa
3.5 52.5 N
16.1 (a) 29.0 u
4.1 (a) 1.01e7 Pa (b) 4.81eí26 kg
(b) 5.05e7 Pa (c) 8.90e4 Pa at 1000 m
(c) 1.01e8 Pa 5.31e4 Pa at 5000 m
2.78e4 Pa at 10,000 m
4.3 (a) 10.3 m
(b) 0.760 m
(c) 760 mm
(d) 29.9 in

4.5 9.34e5 Pa

Copyright 2000-2007 Kinetic Books Co. Chapter 13 Answers 733

Chapter 14 Answers

Chapter Assumptions

The general form of the equation of motion for an object in SHM is x(t) = A cos (Ȧt + ij).
Answers to selected problems

C.1 No
Answers vary

C.2 Remove mass
C.3 kg/s
0.1 (a) Stay the same

(b) Sinusoidal function
3.1 (a) 2.78eí4 Hz

(b) 2.31eí5 Hz
4.1 0.105 rad/s
5.1 (a) 5 m

(b) 3 m
(c) 4 m

5.3 x(t) = 3.5 cos ((ʌ/2)t + 2.0)

6.1 (a) 0.40 m
(b) 2.0 s

7.1 103 m/s
8.1 0.15 m
9.1 0.018 kg·m2
10.1 3.15 m
10.3 90.0 minutes
11.1 1.20 m

734 Copyright 2000-2007 Kinetic Books Co. Chapter 14 Answers

Chapter 15 Answers

Answers to selected problems

C.1 Longitudinal
Answers vary

C.3 No
Answers vary

C.5 AM station
Answers vary

C.7 Wave on tauter string arrives first.
0.1 (a) Yes

(b) No
(c) No
4.1 2.5 m
5.1 6.0 m
6.1 0.00174 s
7.1 1.42e9 Hz
7.3 11 meters
7.5 0.497 m
8.1 347 m/s
8.3 1470 m/s
9.1 0.080 kg/m
10.1 í3.4eí2 m
10.3 (a) Right to left
(b) 3.34eí3 m
(c) 4.00 m
(d) 15.0 Hz

Copyright 2000-2007 Kinetic Books Co. Chapter 15 Answers 735

Chapter 16 Answers

Chapter Assumptions

Unless stated otherwise, use 343 m/s for the speed of sound.

Answers to selected problems

0.1 (a) Higher
(b) Stays the same

3.1 Higher
4.1 12 m
4.3 1.72eí2 W/m2
5.1 3.2eí5 W/m2
5.3 50 dB
5.5 (a) 5.0eí17 W

(b) 1.6eí10 W
(c) 5.0eí3 W
5.7 20
7.1 787 Hz
7.3 44.6 m/s
7.5 33.1 m
9.1 3.01

736 Copyright 2000-2007 Kinetic Books Co. Chapter 16 Answers

Chapter 17 Answers

Answers to selected problems

0.1 (a) Trough
(b) 2.00 m
(c) 1.00 m

4.1 97.1 Hz
4.3 2.29e+3 Hz
4.5 415 N
5.1 (a) 1.50 m

(b) 2
7.1 1860 Hz
7.2 Destructive
8.1 5.36 Hz
8.3 4.4 Hz
8.5 1.28 Hz

Copyright 2000-2007 Kinetic Books Co. Chapter 17 Answers 737

Chapter 18 Answers

Answers to selected problems

C.1 (a) Answers vary
(b) Answers vary
(c) 99 °F

C.3 It does not change
2.1 (a) 0 °F

(b) 96 °F
3.1 (a) í173.15 °C

(b) í279.67 °F
(c) 22 °C
(d) í459.67 °F
3.3 119 °C
8.1 0.00196 m
8.3 2.062eí5 1/C°
8.5 158.3°C
9.1 4.48e8 N/m2
11.1 0.103 cm3
11.3 79.3 gallons
11.5 7.54eí3 m
11.7 í0.830 %
12.1 68.9 °C
12.3 2.57e7 J
12.5 7.6eí3 K
12.7 (a) 2.7 kg
(b) 52°C
12.9 1.68e6 J
13.1 538 J/kg·K
15.1 1.37e+4 J
15.3 8.74e4 J
18.1 (a) 2.7e2 W
(b) 47 W
18.3 (a) 8.59 in
(b) 23.7 in
18.5 2.4 W
21.1 2.70e5 J

738 Copyright 2000-2007 Kinetic Books Co. Chapter 18 Answers

Chapter 19 Answers

Chapter Assumptions

Use the following values for constants:

NA = 6.02×1023
R = 8.31 J/mol·K
k = 1.38×10–23 J/K.

In problems which require you to know the atomic weights of atoms or molecules, use the following:

12.0 u for a carbon atom (C)

4.00 u for a helium atom (He)

1.00 u for a hydrogen atom (H)

14.0 for a nitrogen atom (N)

20.2 u for a neon atom (Ne)

16.0 u for an oxygen atom (O)

44.0 u for a carbon dioxide molecule (CO2)

18.0 u for a water molecule (H2O)

Answers to selected problems

3.1 1.73e5 Pa
3.3 0.150 m3
4.1 1.0e+22 molecules
4.3 (a) 44.0 u

(b) 4.40eí2 kg/mol
(c) 1.68e23 molecules
5.1 (a) 7.2 moles
(b) 4.3e24 molecules
5.3 1.24e5 Pa
5.5 í54.5 mol
5.7 741 K
7.1 2.00eí22 m3
9.1 1.1eí20 J
9.3 531 K

Copyright 2000-2007 Kinetic Books Co. Chapter 19 Answers 739

Chapter 20 Answers
Answers to selected problems

C.1 Adiabatic
0.1 (a) The heat is equal to the change in internal energy.

(b) The work done is the negative of the change in internal energy.
(c) The net heat transferred is equal to the sum of the change in internal energy and the work done.
1.1 (a) í3.4e5 J
(b) í5.1e5 J
(c) 1.7e5 J
1.3 í1700 J
3.1 (a) 9.50e3 J
(b) Work is done by the system
3.3 (a) 5.48e5 J
(b) 7.15e5 J

740 Copyright 2000-2007 Kinetic Books Co. Chapter 20 Answers

Chapter 21 Answers

Answers to selected problems

C.1 Stays the same
C.3 The room temperature goes up

Answers vary
1.1 177 J
1.3 930 J
4.1 0.222 J/K
5.1 1.45 J/K
5.3 674 J
7.1 458°C
8.1 (a) 755 K

(b) 1.21e+5 J
8.3 8.88e+3 J
8.5 (a) 28.7 %

(b) 53.2 %
9.1 (a) 47.6 %
9.3 (a) 1.22

(b) 1.33
10.1 (a) 16

(b) 88 J
10.3 (a) 1.4e5 J

(b) 97 cents
12.1 0.0281 L

Copyright 2000-2007 Kinetic Books Co. Chapter 21 Answers 741

Chapter 22 Answers

Answers to selected problems

C.1 Richard
Answers vary

C.3 (a) 0 C
(b) 0 C

C.5 (a) Rubber is a(n) insulator.
(b) Iron is a(n) conductor.
(c) Copper is a(n) conductor.
(d) Wood is a(n) insulator.

C.7 the "10 cm" end
C.9 The force does not change.
C.11 (a) Yes

(b) Answers vary
0.1 (a) Away from each other

(b) Towards each other
(c) Decreases
1.1 +3.20eí19 C
1.3 1.60eí20 kg
2.1 (a) 0 C
(b) í1.60eí19 C
2.3 3.86e+13 electrons
2.5 1.5e20 electrons
3.1 (a) íe

(b) ʌ

4.1 0 C
6.1 í3.6eí6 C
7.1 (a) Answers vary

(b) +4.00 μC

(c) 40000 dollars
8.1 2.0eí5 C
9.1 (a) Attractive

(b) 8.22eí8 N
9.3 9.32eí7 C
11.1 0.072 N
11.3 3.02e3 N
11.5 (a) NW corner íQ

SW corner +Q
SE corner íQ
(b) NE corner íQ
NW corner íQ
SW corner +Q
SE corner +Q
(c) NE corner +Q
NW corner íQ
SW corner íQ
SE corner +Q
11.7 í167 C
13.1 34 protons

742 Copyright 2000-2007 Kinetic Books Co. Chapter 22 Answers

Chapter 23 Answers

Answers to selected problems

C.1 2
C.3 (a) The combined field is zero

(b) To the right
(c) The combined field is zero
C.5 Proton
Answers vary
0.1 Greatly increases
1.1 (a) 1.75eí3 N
(b) í2.89 N
(c) í184000 N
(d) 1.87eí12 N
1.3 1.36eí21 N
1.5 600 N
2.1 2.61 N/C
2.3 (a) 5.14e11 N/C
(b) Away from the proton
5.1 (a) Closer together
(b) The direction of the force is tangent to the field lines
6.1 (a) 2.59e11 N/C, to the right
(b) 1.12e11 N/C, to the left
(c) 9.55e10 N/C, to the right
6.3 16.2 C
7.1 18 lines going into í3q

7.3 (a) q2 743
(b) q1
q3
(c) |q2| > |q1| > |q3|
(d) 3 times stronger

8.1 89.9 s
8.3 (a) í8.69eí17 N

(b) í9.54e13 m/s2
(c) 1.00e6 m/s
(d) 3.33eí3
8.5 4.64e3 m/s
8.7 (a) 4.79e10 m/s2
(b) 7.31eí7 s
(c) 1.28eí2 m
(d) 1.02eí18 J
12.1 9.76eí9 C·m
12.3 6.59 m

Copyright 2000-2007 Kinetic Books Co. Chapter 23 Answers

Chapter 24 Answers

Chapter Assumptions

Unless stated otherwise, the reference configuration for zero electric potential and zero electric potential energy is one in which there
is infinite separation between charges.

Answers to selected problems

C.1 (a) It doubles 3.1 2.6 J
(b) Yes
(c) No 3.3 (a) 2e2 N
(b) 2eí13 J
C.4 (a) No
(b) Answers vary 3.5 3.0eí12 m
(c) No
Answers vary 4.1 (a) 1.44eí9 V
(b) í1.44eí9 V
0.1 (a) Always positive (c) Near the proton
(b) Closer to (d) Magnitudes are equal

0.3 (a) It increases 4.3 2.41eí9 C
(b) A vertical line
6.1 32 V
1.1 (a) 5.4 μJ
(b) í5.4 μJ 7.1 (a) 6.34eí20 J
(b) í6.34eí20 J
(c) Closer together (c) 0.396 V

1.3 (a) í52.3 J 7.2 (a) 1.28eí20 J
(b) 52.3 J (b) 8.83e25 ions
(c) Farther apart
7.4 (a) 1.04e4 J
2.1 (a) 5.09eí3 J (b) $ 2.6eí4
(b) í5.09eí3 J
(c) 5.09eí3 J 7.5 í2.66eí2 C

2.3 (a) 2.75eí2 J 10.1 2.53e+8 V
(b) í2.75eí2 J
(c) 2.75eí2 J 10.3 (a) Sphere
(b) The surface closer to the point charge
2.5 (a) 6.71eí4 J (c) 0.32
(b) í6.71eí4 J
(c) 6.71eí4 J 11.1 í9.6eí14 J
(d) Yes 11.3 -40.9 V

11.5 710 V

11.7 (a) Planes parallel to the yz plane
(b) x = 2.24 m
(c) x = 1.87 m

744 Copyright 2000-2007 Kinetic Books Co. Chapter 24 Answers

Chapter 25 Answers

Chapter Assumptions

For any problems in this chapter that involve a household electrical outlet, assume that the voltage supplied is fixed at 120 V. While
this is not true for actual outlets, whose voltage varies and periodically exceeds 120 V, the simplification will not affect the accuracy
of your answer.

Answers to selected problems

C.1 (a) 1
(b) 3

C.3 The current stays the same
Answers vary

C.7 Noníohmic
C.9 No

Answers vary
C.11 No

Answers vary
C.13 Electric energy
1.1 (a) 6.00eí2 C

(b) 3.75e17 electrons
(c) 0 protons
1.3 12.3 C
1.5 7.4e-7 A
1.7 4.2eí7 s
1.9 (a) 3.89eí2 C/m2
(b) Down

2.1 (a) 1.1e4 s
(b) Answers vary

3.1 16.4 ȍ

3.3 4.0eí3 A
3.5 1.29 V

4.1 100 ȍ
5.1 2.4eí7 ȍ

5.3 Aluminum

5.5 (a) 10.0 ȍ
(b) 16.0 ȍ
(c) 2.00 ȍ

6.1 360°C
6.3 (a) 0.00501°Cí1

(b) Tungsten
6.5 16.3°C

6.7 51.8 ȍ

7.1 3.1 V
7.3 20 light bulbs

7.5 14 ȍ
7.7 (a) 2.4 ȍ

(b) 1.0e2 W

8.1 $7.17e20

11.1 2.07e10 W

Copyright 2000-2007 Kinetic Books Co. Chapter 25 Answers 745

Chapter 26 Answers
Answers to selected problems

C.1 (a) It increases
(b) It stays the same

C.2 (a) Increased
(b) Same amount of charge

0.1 Charge is proportional to potential difference
1.1 3.2eí10 C
1.3 7.1eí6 F
1.5 7.2eí5 C
4.1 4.00eí9 F
4.3 4.45eí1 J
4.5 9.96eí7 F
4.7 1.54eí2 J
6.1 76 J
8.1 7.2 nF
8.3 1.60e8 V/m
8.4 66.0 nF
11.1 1.6eí7 C

746 Copyright 2000-2007 Kinetic Books Co. Chapter 26 Answers

Chapter 27 Answers

Answers to selected problems

C.4 The same
Answers vary

C.6 Circuit with two light bulbs
Answers vary

C.8 Parallel
Answers vary

C.12 (a) R1 is in series with the combination of R2 and R3
(b) R2 is in parallel with R3

C.14 It will decrease to zero
Answers vary

0.1 (a) They are equal
(b) It is the same everywhere

2.1 1.17e6 J

3.1 3.0 V
3.3 (a) 6.6 W

(b) 3.6e4 J

3.5 2.0 ȍ

3.7 1.8e4 s

5.1 0.55 A

6.1 1.0 ȍ
6.3 4200 ȍ
6.5 3.23 ȍ

6.7 2.7 A
6.9 12.0 V

7.1 25.0 ȍ
10.1 1.67 ȍ
10.3 0.175 ȍ
10.5 77 ȍ
10.7 8.75 ȍ

10.9 34.8 V
10.11 4

10.13 (a) 500 ȍ
(b) 125 ȍ

11.1 60.0 ȍ
12.1 (a) 3.75 ȍ

(b) 5.00 ȍ

Copyright 2000-2007 Kinetic Books Co. Chapter 27 Answers 747

Chapter 28 Answers

Chapter Assumptions

Elementary charge, e = 1.60×10–19 C
Mass of electron, me = 9.11×10–31 kg
Mass of proton, mp = 1.67×10–27 kg

Unless stated otherwise, use 5.00×10–5 T for the strength of the Earth's magnetic field at its surface.

Answers to selected problems

C.1 Down 13.1 2.75eí3 m
Answers vary 13.3 56.3 m/s
13.5 3.67e+2 s
C.2 (a) The field is perpendicular to the plane of the wires. 13.7 2.28eí12 T
(b) Reinforce
(c) Cancel 15.1 (a) 1.29eí2 T
Answers vary (b) Decrease

C.5 (a) No 15.3 (a) 4.65eí3 m
(b) Electron (b) 5.65eí3 m
(c) 2.00eí3 m
C.6 No
Answers vary 15.5 0.0286 m

C.9 No 16.1 (a) 8.29 m/s
Answers vary (b) 1.07 m
(c) 0.872 m/s
C.12 (a) Toward you (d) 0.708 m
(b) Left
(c) Down 17.1 (a) A circle
(d) Down (b) A straight line

C.14 (a) Negative 18.1 14.1 N
(b) Positive 18.3 2.27 A
(c) Answers vary 18.5 35.2°

C.16 Away from you 20.1 (a)
Up (d)

0.1 (a) Curve 22.1 2.00eí5 T
(b) No 22.3 1.14eí3 T
(c) Yes 22.5 896 A
22.7 (a) 1.64eí26 N
7.1 (a) 6.08eí16 N
(b) Positive y direction (b) 55.4 m
22.9 4.58eí23 N, in the same direction as the current
7.3 (a) 805 m/s 22.11 (a) Opposite directions
(b) East
(b) 75.0 A
7.5 1.78 C 22.13 8.80eí3 m
22.14 6.67 m
7.7 17.3° 22.15 Direction: +z

9.1 (a) 2 Strength: 3.66eí7 T
(b) 180q
(c) 90.0°

10.1 0.417 T

10.3 0.631 T
10.5 5.76eí3 T

11.1 2.90 T

12.1 3.00e+4 m/s
12.3 2.21eí22 J
12.5 (a) 1.0e4 m/s

(b) 1.8eí2 m/s

748 Copyright 2000-2007 Kinetic Books Co. Chapter 28 Answers