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C HAPTER
9
The Atomic Theory
Chapter Outline
194
9.1
T HE ATOMIC T HEORY
9.2
F URTHER U NDERSTANDING OF THE ATOM
9.3
ATOMIC S TRUCTURE
9.4
R EFERENCES
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Chapter 9. The Atomic Theory
9.1 The Atomic Theory
Lesson Objectives
The student will:
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•
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give a short history of how the concept of the atom developed.
describe the contributions of Democritus and Dalton to the atomic theory.
summarize Dalton’s atomic theory and explain its historical development.
state the law of definite proportions.
state the law of multiple proportions.
Vocabulary
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•
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atomos
Dalton’s atomic theory
law of definite proportions
law of multiple proportions
Introduction
You learned earlier in the chapter “Matter and Energy” that all matter in the universe is made up of tiny building
blocks called atoms. All modern scientists accept the concept of the atom, but when the concept of the atom was first
proposed about 2,500 years ago, ancient philosophers laughed at the idea. After all, it is difficult to be convinced
that something too small to be seen really exists. We will spend some time considering the evidence (observations)
that convinced scientists of the existence of atoms.
Democritus and the Greek Philosophers
Before we discuss the experiments and evidence that have convinced scientists matter is made up of atoms, it is only
fair to credit the man who proposed the concept of the atom in the first place. About 2,500 years ago, early Greek
philosophers believed the entire universe was a single, huge entity. In other words, “everything was one.” They
believed that all objects, all matter, and all substances were connected as a single, big, unchangeable “thing.”
One of the first people to propose the existence of atoms was a man known as Democritus, pictured in Figure
9.1. He suggested an alternative theory where atomos – tiny, indivisible, solid objects – made up all matter in the
universe. Democritus then reasoned that changes occur when the many atomos in an object were reconnected or
recombined in different ways. Democritus even extended his theory to suggest that there were different varieties
of atomos with different shapes, sizes, and masses. He thought, however, that shape, size, and mass were the only
properties differentiating the types of atomos. According to Democritus, other characteristics, like color and taste,
did not reflect properties of the atomos themselves but from the different ways in which the atomos were combined
and connected to one another.
So how could the Greek philosophers have known that Democritus had a good idea with his theory of atomos?
The best way would have been to take some careful observation and conduct a few experiments. Recall, however,
that the early Greek philosophers tried to understand the nature of the world through reason and logic, not through
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FIGURE 9.1
Democritus was known as
experimentation and observation. The Greek philosophers truly believed that, above all else, our understanding
of the world should rely on logic. In fact, they argued that the world couldn’t be understood using our senses at
all because our senses could deceive us. Therefore, instead of relying on observation, Greek philosophers tried to
understand the world using their minds and, more specifically, the power of reason (see Figure 9.2).
FIGURE 9.2
This sculpture (named
As a result, the early Greek philosophers developed some very interesting ideas, but they felt no need to justify their
ideas. You may recall from the “Introduction to Chemistry” chapter that Aristotle concluded men had more teeth than
women did. He concluded this without ever checking in anyone’s mouth because his conclusion was the “logical”
one. As a result, the Greek philosophers missed or rejected a lot of discoveries that could have made otherwise
because they never performed any experiments. Democritus’s theory would be one of these rejected theories. It
would take over two millennia before the theory of atomos (or atoms, as they’re known today) was fully appreciated.
Dalton’s Atomic Theory
Let’s consider a simple but important experiment that suggested matter might be made up of atoms. In the late 1700s
and early 1800s, scientists began noticing that when certain substances, like hydrogen and oxygen, were combined
to produce a new substance, the reactants (hydrogen and oxygen) always reacted in the same proportions by mass.
In other words, if 1 gram of hydrogen reacted with 8 grams of oxygen, then 2 grams of hydrogen would react with
16 grams of oxygen, and 3 grams of hydrogen would react with 24 grams of oxygen.
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Chapter 9. The Atomic Theory
Strangely, the observation that hydrogen and oxygen always reacted in the “same proportions by mass” wasn’t
unique to hydrogen and oxygen. In fact, it turned out that the reactants in every chemical reaction for a given
compound react in the same proportions by mass. Take, for example, nitrogen and hydrogen, which can react to
produce ammonia (NH3 ). In chemical reactions, 1 gram of hydrogen will react with 4.7 grams of nitrogen, and 2
grams of hydrogen will react with 9.4 grams of nitrogen. Can you guess how much nitrogen would react with 3
grams of hydrogen?
Scientists studied reaction after reaction, but every time the result was the same. The reactants always reacted in
the same proportions by mass or in what we call “definite proportions,” as illustrated in Figure 9.3. As a result,
scientists proposed the law of definite proportions. This law states that:
In a given type of chemical substance, the elements always combine in the same proportions by mass.
This version of the law is a more modern version. Earlier, you learned that an element is a substance made up of
only one type of atom, but when the law of definite proportions was first discovered, scientists did not know about
atoms or elements and stated the law slightly differently. We’ll stick with this modern version, though, since it is the
easiest version to understand.
FIGURE 9.3
If 1 gram of A reacts with 8 grams of B, then by the law of definite
proportions, 2 grams of A must react with 16 grams of B. If 1 gram of
A reacts with 8 grams of B, then by the law of conservation of mass, they
must produce 9 grams of C.
The law of definite proportions applies when the elements reacting together form the same product. Therefore, the
law of definite proportions can be used to compare two experiments in which hydrogen and oxygen react to form
water. The law, however, cannot be used to compare one experiment in which hydrogen and oxygen react to form
water with another experiment in which hydrogen and oxygen react to form hydrogen peroxide (peroxide is another
material that can be made from hydrogen and oxygen).
FIGURE 9.4
Unlike the early Greek philosophers, John Dalton was both a thinker and
an experimenter. He would help develop the modern conception of an
atom based on his experimental results.
A man named John Dalton (Figure 9.4) discovered this limitation in the law of definite proportions in some of
his experiments. Dalton was experimenting with several reactions in which the reactant elements formed different
products, depending on the experimental conditions he used. One common reaction that he studied was the reaction
between carbon and oxygen. When carbon and oxygen react, they produce two different substances – we’ll call
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these substances A and B. It turned out that, given the same amount of carbon, forming B always required exactly
twice as much oxygen as forming A. In other words, if you could make A with 3 grams of carbon and 4 grams of
oxygen, B could be made with the same 3 grams of carbon but with 8 grams of oxygen instead. Dalton asked himself
– why does B require twice as much oxygen as A does? Why not 1.21 times as much oxygen, or 0.95 times as much
oxygen? Why a whole number like 2?
The situation became even stranger when Dalton tried similar experiments with different substances. For example,
when he reacted nitrogen and oxygen, Dalton discovered that he could make three different substances – we’ll call
them C, D, and E. As it turned out, for the same amount of nitrogen, D always required twice as much oxygen as C
does. Similarly, E always required exactly four times as much oxygen as C does. Once again, Dalton noticed that
small whole numbers (2 and 4) seemed to be the rule. Dalton used his experimental results to propose the law of
multiple proportions:
When two elements react to form more than one substance and the same amount of one element (like oxygen)
is used in each substance, then the ratio of the masses used of the other element (like nitrogen) will be in small
whole numbers.
This law summarized Dalton’s findings, but it did not explain why the ratio was a small whole number. Dalton
thought about his law of multiple proportions and tried to develop a theory that would explain it. Dalton also knew
about the law of definite proportions and the law of conservation of mass, so what he really wanted was a theory
that explained all three laws with a simple, plausible model. One way to explain the relationships that Dalton
and others had observed was to suggest that materials like nitrogen, carbon, and oxygen were composed of small,
indivisible quantities, which Dalton called “atoms” (in reference to Democritus’s original idea). Dalton used this
idea to generate what is now known as Dalton’s atomic theory.
Dalton’s atomic theory:
1. Matter is made of tiny particles called atoms.
2. Atoms are indivisible. During a chemical reaction, atoms are rearranged, but they do not break apart, nor are
they created or destroyed.
3. All atoms of a given element are identical in mass and other properties.
4. The atoms of different elements differ in mass and other properties.
5. Atoms of one element can combine with atoms of another element to form compounds. In a given compound,
however, the different types of atoms are always present in the same relative numbers.
Historical note: Some people think that Dalton developed his atomic theory before stating the law of multiple
proportions, while others argue that the law of multiple proportions, though not formally stated, was actually
discovered first. In reality, Dalton was probably contemplating both concepts at the same time, although it is hard
to say conclusively from looking at the laboratory notes he left behind.
Lesson Summary
• 2,500 years ago, Democritus suggested that all matter in the universe was made up of tiny, indivisible, solid
objects he called atomos.
• Other Greek philosophers disliked Democritus’s atomos theory because they felt it was illogical.
• The law of definite proportions states that in a given chemical substance, the elements are always combined
in the same proportions by mass.
• The law of multiple proportions states that when two elements react to form more than one substance and the
same amount of one element is used in each substance, then the ratio of the masses used of the other element
will be in small whole numbers.
• Dalton used the law of definite proportions, the law of multiple proportions, and the law of conservation of
mass to propose his atomic theory.
• Dalton’s atomic theory states:
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1. Matter is made of tiny particles called atoms.
2. Atoms are indivisible. During a chemical reaction, atoms are rearranged, but they do not break apart,
nor are they created or destroyed.
3. All atoms of a given element are identical in mass and other properties.
4. The atoms of different elements differ in mass and other properties.
5. Atoms of one element can combine with atoms of another element to form compounds. In a given
compound, however, the different types of atoms are always present in the same relative numbers.
Review Questions
1. It turns out that a few of the ideas in Dalton’s atomic theory aren’t entirely correct. Are inaccurate theories an
indication that science is a waste of time?
2. Suppose you are trying to decide whether to wear a sweater or a T-shirt. To make your decision, you phone
two friends. The first friend says, “Wear a sweater, because I’ve already been outside today, and it’s cold.” The
second friend, however, says, “Wear a T-shirt. It isn’t logical to wear a sweater in July.” Would you decide to
go with your first friend and wear a sweater or with your second friend and wear a T-shirt? Why?
3. Decide whether each of the following statements is true or false.
(a)
(b)
(c)
(d)
Democritus believed that all matter was made of atomos.
Democritus also believed that there was only one kind of atomos.
Most early Greek scholars thought that the world was “ever-changing.”
If the early Greek philosophers hadn’t been so interested in making gold, they probably would have liked
the idea of the atomos.
4. Match the person, or group of people, with their role in the development of chemistry in the table below.
Person/Group of People
(a) early Greek philosophers
Role in Chemistry
(i) first suggested that all matter was made up of tiny,
indivisible, solid objects
(b) alchemists
(ii) tried to apply logic to the world around them
(c) John Dalton
(iii) proposed the first scientific theory relating chemical
changes to the structure, properties, and behavior of atoms
(d) Democritus
(iv) were primarily concerned with finding ways to turn
common metals into gold
5. Early Greek philosophers felt that Democritus’s atomos theory was illogical because:
(a)
(b)
(c)
(d)
no matter how hard they tried, they could never break matter into smaller pieces.
it didn’t help them to make gold.
sulfur is yellow and carbon is black, so clearly atomos must be colored.
empty space is illogical because it implies that nothing is actually something.
6. Identify the law that explains the following observation: Carbon monoxide can be formed by reacting 12 grams
of carbon with 16 grams of oxygen. To form carbon dioxide, however, 12 grams of carbon must react with
32 grams of oxygen.
7. Identify the law that explains the following observation: Carbon monoxide can be formed by reacting 12 grams
of carbon with 16 grams of oxygen. It can also be formed by reacting 24 grams of carbon with 32 grams of
oxygen.
8. Identify the law that explains the following observation: 28 grams of carbon monoxide are formed when
12 grams of carbon reacts with 16 grams of oxygen.
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9. Identify the law that explains the following observations: When 12 grams of carbon react with 4 grams of
hydrogen, they produce methane, and there is no carbon or hydrogen left over at the end of the reaction. If,
however, 11 grams of carbon react with 4 grams of hydrogen, there is hydrogen left over at the end of the
reaction.
10. Which of the following is not part of Dalton’s atomic theory?
(a)
(b)
(c)
(d)
Matter is made of tiny particles called atoms.
During a chemical reaction, atoms are rearranged.
During a nuclear reaction, atoms are split apart.
All atoms of a specific element are the same.
11. Consider the following data: 3.6 grams of boron react with 1.0 grams of hydrogen to give 4.6 grams of BH3 .
How many grams of boron would react with 2.0 grams of hydrogen?
12. Consider the following data: 12 grams of carbon and 4 grams of hydrogen react to give 16 grams of compound
A. 24 grams of carbon and 6 grams of hydrogen react to give 30 grams of compound B. Are compound A and
compound B the same? Why or why not?
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Chapter 9. The Atomic Theory
9.2 Further Understanding of the Atom
Lesson Objectives
The student will:
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explain the observations that led to Thomson’s discovery of the electron.
describe Thomson’s plum-pudding model of the atom.
draw a diagram of Thomson’s plum-pudding model of the atom and explain why it has this name.
describe Rutherford’s gold foil experiment and explain how this experiment disproved the plum-pudding
model.
• draw a diagram of the Rutherford model of the atom and label the nucleus and the electron cloud.
Vocabulary
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•
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cathode ray tube
electron
nucleus
proton
subatomic particle
Introduction
Dalton’s atomic theory held up well in a lot of the different chemical experiments that scientists performed to test
it. For almost 100 years, it seemed as if Dalton’s atomic theory was the whole truth. It wasn’t until 1897 when
a scientist named J. J. Thomson conducted some research that suggested Dalton’s atomic theory wasn’t the entire
story. Dalton had gotten a lot right - he was right in saying matter is made up of atoms; he was right in saying there
are different kinds of atoms with different mass and other properties; he was almost right in saying atoms of a given
element are identical; he was right in saying that atoms are merely rearranged during a chemical reaction; and he
was right in saying a given compound always has atoms present in the same relative numbers. But he was wrong in
saying atoms were indivisible or indestructible. As it turns out, atoms are divisible. In fact, atoms are composed of
even smaller, more fundamental particles. These particles, called subatomic particles, are particles that are smaller
than the atom. The discoveries of these subatomic particles are the focus of this chapter.
Thomson’s Plum-Pudding Model
In the mid-1800s, scientists were beginning to realize that the study of chemistry and the study of electricity were
actually related. First, a man named Michael Faraday showed how passing electricity through mixtures of different
chemicals could cause chemical reactions. Shortly after that, scientists found that by forcing electricity through
a tube filled with gas, the electricity made the gas glow. Scientists didn’t, however, understand the relationship
between chemicals and electricity until a British physicist named J. J. Thomson began experimenting with what is
known as a cathode ray tube (Figure 9.5).
The figure below shows a basic diagram of a cathode ray tube like the one Thomson would have used. A cathode
ray tube is a small glass tube with a cathode (a negatively charged metal plate) and an anode (a positively charged
metal plate) at opposite ends. By separating the cathode and anode a short distance, the cathode ray tube can generate
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FIGURE 9.5
A portrait of J. J. Thomson.
what are known as cathode rays – rays of electricity that flow from the cathode to the anode. Thomson wanted to
know what cathode rays were, where cathode rays came from, and whether cathode rays had any mass or charge.
The techniques that he used to answer these questions were very clever and earned him a Nobel Prize in physics.
First, by cutting a small hole in the anode, Thomson found that he could get some of the cathode rays to flow through
the hole in the anode and into the other end of the glass cathode ray tube. Next, he figured out that if he painted a
substance known as phosphor onto the far end of the cathode ray tube, he could see exactly where the cathode rays
hit because the cathode rays made the phosphor glow.
Thomson must have suspected that cathode rays were charged, because his next step was to place a positively
charged metal plate on one side of the cathode ray tube and a negatively charged metal plate on the other side,
as shown below. The metal plates didn’t actually touch the cathode ray tube, but they were close enough that a
remarkable thing happened. The flow of the cathode rays passing through the hole in the anode was bent upwards
towards the positive metal plate and away from the negative metal plate. In other words, instead of glowing directly
across from the hole in the anode, the phosphor now glowed at a spot quite a bit higher in the tube.
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Thomson thought about his results for a long time. It was almost as if the cathode rays were attracted to the positively
charged metal plate and repelled from the negatively charged metal plate. Thomson knew that charged objects are
attracted to and repelled from other charged objects according to the rule: opposite charges attract, like charges
repel. This means that a positive charge is attracted to a negative charge but repelled from another positive charge.
Similarly, a negative charge is attracted to a positive charge but repelled from another negative charge. Using the
“opposite charges attract, like charges repel” rule, Thomson argued that if the cathode rays were attracted to the
positively charged metal plate and repelled from the negatively charged metal plate, the rays themselves must have
a negative charge.
Thomson then did some rather complex experiments with magnets and used the results to prove that cathode rays
not only were negatively charged, but they also had mass. Remember that anything with mass is part of what we call
matter. In other words, these cathode rays must be the result of negatively charged matter flowing from the cathode
to the anode. It was here that Thomson encountered a problem. According to his measurements, these cathode rays
either had a ridiculously high charge or very, very little mass – much less mass than the smallest known atom. How
was this possible? How could the matter making up cathode rays be smaller than an atom if atoms were indivisible?
Thomson made a radical proposal: maybe atoms are divisible. He suggested that the small, negatively charged
particles making up the cathode ray were actually pieces of atoms. He called these pieces “corpuscles,” although
today we know them as electrons. Thanks to his clever experiments and careful reasoning, Thomson is credited
with the discovery of the electron.
For a demonstration of cathode ray tubes (1h), see http://www.youtube.com/watch?v=XU8nMKkzbT8 (1:09).
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Now imagine what would happen if atoms were made entirely of electrons. First of all, electrons are very, very
small; in fact, electrons are about 2,000 times smaller than the smallest known atom, so every atom would have
to contain a lot of electrons. But there’s another, bigger problem: electrons are negatively charged. Therefore, if
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atoms were made entirely out of electrons, the atoms themselves would be negatively charged, which would mean
all matter was negatively charged as well.
Of course, matter isn’t negatively charged. Most matter is what we call neutral – it has no charge at all. How can
matter be neutral if matter is composed of atoms and atoms are composed of negative electrons? The only possible
explanation is that atoms must consist of more than just electrons. Atoms must also contain some type of positively
charged material that balances the negative charge of the electrons. Negative and positive charges of equal size
cancel each other out, just like negative and positive numbers of equal size. If an atom contains an electron with a
−1 charge and some form of material with a +1 charge, overall the atom must have a (+1) + (−1) = 0 charge. In
other words, the atom would be neutral, or have no overall charge.
Based on the fact that atoms are neutral and based on Thomson’s discovery that atoms contain negative subatomic
particles called electrons, scientists assumed that atoms must also contain a positive substance. It turned out that
this positive substance was another kind of subatomic particle known as the proton. Although scientists knew that
atoms had to contain positive material, protons weren’t actually discovered, or understood, until quite a bit later.
When Thomson discovered the negative electron, he also realized that atoms had to contain positive material as
well. As a result, Thomson formulated what’s known as the plum-pudding model for the atom. According to the
plum-pudding model, the negative electrons were like pieces of fruit and the positive material was like the batter or
the pudding. In the figure below, an illustration of a plum pudding is on the left and an illustration of Thomson’s
plum-pudding model is on the right. (Instead of a plum pudding, you can also think of a chocolate chip cookie.
In that case, the positive material in the atom would be the batter in the chocolate chip cookie, while the negative
electrons would be scattered through the batter like chocolate chips.)
This made a lot of sense given Thomson’s experiments and observations. Thomson had been able to isolate electrons
using a cathode ray tube; however, he had never managed to isolate positive particles. Notice in the image above
how easy it would be to pick the pieces of fruit out of a plum pudding. On the other hand, it would be a lot harder to
pick the batter out of the plum pudding because the batter is everywhere. If an atom were similar to a plum pudding
in which the electrons are scattered throughout the “batter” of positive material, then you would expect it to be easy
to pick out the electrons and a lot harder to pick out the positive material.
Everything about Thomson’s experiments suggested the plum-pudding model was correct. According to the scien204
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Chapter 9. The Atomic Theory
tific method, however, any new theory or model should be tested by further experimentation and observation. In the
case of the plum-pudding model, it would take a man named Ernest Rutherford to prove it wrong. Rutherford and
his experiments will be the topic of the next section.
There was one thing that Thomson was unable to determine. He had measured the charge-to-mass ratio of the
electron, but he had been unable to measure accurately the charge on the electron. Instead, a different scientist
named Robert Millikan would determine the charge of the electron with his oil drop experiment. When combined
with Thomson’s charge-to-mass ratio, Millikan was able to calculate the mass of the electron. Millikan’s experiment
involved putting charges on tiny droplets of oil suspended between charged metal plates and measuring their rate of
descent. By varying the charge on different drops, he noticed that the electric charges on the drops were all multiples
of 1.6 × 10−19 C (coulomb), the charge of a single electron.
Rutherford’s Nuclear Model
Disproving Thomson’s plum-pudding model began with the discovery that an element known as uranium emits
positively charged particles called alpha particles as it undergoes radioactive decay. Radioactive decay occurs when
one element decomposes into another element. It only happens with a few very unstable elements. Alpha particles
themselves didn’t prove anything about the structure of the atom, but they were used to conduct some very interesting
experiments.
FIGURE 9.6
A portrait of Ernest Rutherford.
Ernest Rutherford (pictured in Figure 9.6) was fascinated by all aspects of alpha particles and used them as tiny
bullets that could be fired at all kinds of different materials. The results of one experiment in particular surprised
Rutherford and everyone else.
Rutherford found that when he fired alpha particles at a very thin piece of gold foil, an interesting phenomenon
happened. The diagram below helps illustrate Rutherford’s findings. Almost all of the alpha particles went straight
through the foil as if they had hit nothing at all. Every so often, though, one of the alpha particles would be deflected
slightly as if it had bounced off something hard. Even less often, Rutherford observed alpha particles bouncing
straight back at the “gun” from which they had been fired. It was as if these alpha particles had hit a wall head-on
and had ricocheted right back in the direction that they had come from.
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Rutherford thought that these experimental results were rather odd. He expected firing alpha particles at gold foil to
be like shooting a high-powered rifle at tissue paper. The bullets would break through the tissue paper and keep on
going, almost as if they had hit nothing at all. That was what Rutherford had expected to see when he fired alpha
particles at the gold foil. The fact that most alpha particles passed through did not shock him, but how could he
explain the alpha particles that were deflected? Furthermore, how could he explain the alpha particles that bounced
right back as if they had hit a wall?
Rutherford decided that the only way to explain his results was to assume that the positive matter forming the
gold atoms was not distributed like the batter in plum pudding. Instead, he proposed that the positive matter was
concentrated in one spot, forming a small, positively charged particle somewhere in the center of the gold atom. We
now call this clump of positively charged mass the nucleus. According to Rutherford, the presence of a nucleus
explained his experiments because it implied that most of the positively charged alpha particles would pass through
the gold foil without hitting anything at all. Occasionally, though, the alpha particles would actually collide with a
gold nucleus, causing the alpha particles to be deflected or even bounced back in the direction they came from.
While Rutherford’s discovery of the positively charged atomic nucleus offered insight into the structure of the atom,
it also led to some questions. According to the plum-pudding model, electrons were like plums embedded in the
positive batter of the atom. Rutherford’s model, though, suggested that the positive charge was concentrated into a
tiny particle at the center of the atom, while most of the rest of the atom was empty space. What did that mean for
the electrons? If they weren’t embedded in the positive material, exactly what were they doing? How were they held
in the atom? Rutherford suggested that the electrons might be circling or orbiting the positively charged nucleus as
some type of negatively charged cloud, like in the image below. At the time, however, there wasn’t much evidence
to suggest exactly how the electrons were held in the atom.
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A short animation of Rutherford’s experiment (1h) can be found at http://www.youtube.com/watch?v=5pZj0u_XM
bc (0:47).
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For another video discussing J.J. Thomson’s use of a cathode ray tube in his discovery of the electron (1h), see http
://www.youtube.com/watch?v=IdTxGJjA4Jw (2:54).
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Despite the problems and questions associated with Rutherford’s experiments, his work with alpha particles seemed
to point to the existence of an atomic nucleus. Between Thomson, who discovered the electron, and Rutherford,
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who suggested that the positive charges were concentrated at the atom’s center, the 1890s and early 1900s saw
huge steps in understanding the atom at the subatomic (smaller than the size of an atom) level. Although there was
still some uncertainty with respect to exactly how subatomic particles were organized in the atom, it was becoming
more and more obvious that atoms were indeed divisible. Moreover, it was clear that an atom contained negatively
charged electrons and a positively charged nucleus. In the next lesson, we’ll look more carefully at the structure of
the nucleus. We’ll learn that while the atom is made up of positive and negative particles, it also contains neutral
particles that neither Thomson nor Rutherford were able to detect with their experiments.
Lesson Summary
• Dalton’s atomic theory wasn’t entirely correct, as it was found that atoms can be divided into smaller subatomic particles.
• According to Thomson’s plum-pudding model, the negatively charged electrons in an atom are like the pieces
of fruit in a plum pudding, while the positively charged material is like the batter.
• When Ernest Rutherford fired alpha particles at a thin gold foil, most alpha particles went straight through;
however, a few were scattered at different angles, and some even bounced straight back.
• In order to explain the results of his gold foil experiment, Rutherford suggested that the positive matter in the
gold atoms was concentrated at the center of the gold atom in what we now call the nucleus of the atom.
Further Reading / Supplemental Links
The learner.org website allows users to view the Annenberg series of chemistry videos. You are required to register
before you can watch the videos, but there is no charge to register. The video called “The Atom” explores the
structure of the atom.
• http://learner.org/resources/series61.html
Review Questions
1. Decide whether each of the following statements is true or false.
(a)
(b)
(c)
(d)
(e)
Cathode rays are positively charged.
Cathode rays are rays of light, thus they have no mass.
Cathode rays can be repelled by a negatively charged metal plate.
J.J. Thomson is credited with the discovery of the electron.
Phosphor is a material that glows when struck by cathode rays.
2. Match each observation with the correct conclusion.
(a) Cathode rays are attracted to a positively charged metal plate.
i. Cathode rays are positively charged.
ii. Cathode rays are negatively charged.
iii. Cathode rays have no charge.
(b) Electrons have a negative charge.
i. Atoms must be negatively charged.
ii. Atoms must be positively charged.
iii. Atoms must also contain positive subatomic material.
(c) Alpha particles fired at a thin gold foil are occasionally scattered back in the direction that they came
from.
i. The positive material in an atom is spread throughout like the batter in pudding.
ii. Atoms contain neutrons.
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Chapter 9. The Atomic Theory
iii. The positive charge in an atom is concentrated in a small area at the center of the atom.
3. What is the name given to the tiny clump of positive material at the center of an atom?
4. Choose the correct statement.
(a)
(b)
(c)
(d)
(e)
Ernest Rutherford discovered the atomic nucleus by performing experiments with aluminum foil.
Ernest Rutherford discovered the atomic nucleus using a cathode ray tube.
When alpha particles are fired at a thin gold foil, they never go through.
Ernest Rutherford proved that the plum-pudding model was incorrect.
Ernest Rutherford experimented by firing cathode rays at gold foil.
5. Answer the following questions:
(a)
(b)
(c)
(d)
(e)
Will the charges +2 and −2 cancel each other out?
Will the charges +2 and −1 cancel each other out?
Will the charges +1 and +1 cancel each other out?
Will the charges −1 and +3 cancel each other out?
Will the charges +9 and −9 cancel each other out?
6. Electrons are ______ negatively charged metals plates and ______ positively charged metal plates.
7. What was J. J. Thomson’s name for electrons?
8. A “sodium cation” is a sodium atom that has lost one of its electrons. Would the charge on a sodium cation be
positive, negative, or neutral? Would sodium cations be attracted to a negative metal plate or a positive metal
plate? Would electrons be attracted to or repelled from sodium cations?
9. Suppose you have a cathode ray tube coated with phosphor so that you can see where on the tube the cathode
ray hits by looking for the glowing spot. What will happen to the position of this glowing spot if:
(a)
(b)
(c)
(d)
a negatively charged metal plate is placed above the cathode ray tube?
a negatively charged metal plate is placed to the right of the cathode ray tube?
a positively charged metal plate is placed to the right of the cathode ray tube?
a negatively charged metal plate is placed above the cathode ray tube, and a positively charged metal
plate is placed to the left of the cathode ray tube?
(e) a positively charged metal plate is placed below the cathode ray tube, and a positively charged metal
plate is also placed to the left of the cathode ray tube?
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9.3 Atomic Structure
Lesson Objectives
The student will:
•
•
•
•
•
•
•
•
•
identify the three major subatomic particles and their charges, masses, and location in the atom.
briefly describe the discovery of the neutron.
define atomic number.
describe the size of the nucleus in relation to the size of the atom.
explain what is meant by the atomic mass of an element and describe how atomic masses are related to carbon12.
define mass number.
explain what isotopes are and how isotopes affect an element’s atomic mass.
determine the number of protons, neutrons, and electrons in an atom.
calculate the atomic mass of an element from the masses and relative percentages of the isotopes of the
element.
Vocabulary
•
•
•
•
•
•
•
•
atomic mass
atomic mass unit
atomic number
dalton
isotopes
mass number
neutron
strong nuclear force
Introduction
Dalton’s atomic theory explained a lot about matter, chemicals, and chemical reactions. Nevertheless, it wasn’t
entirely accurate because, contrary to what Dalton believed, atoms can in fact be broken apart into smaller subunits
or subatomic particles. The first type of subatomic particle to be found in an atom was the negatively charged
electron. Since atoms are neutral, though, they must also contain positive material. In his gold foil experiment,
Rutherford proved that the positive substance in an atom was concentrated in a small area at the center of the atom,
leaving most the rest of the atom as empty space. In this lesson, we’ll examine the subatomic particles making up
the atom a little more closely.
This video gives basic information about the nucleus of atoms including comparative sizes of atom vs nucleus (1e),
see http://www.youtube.com/watch?v=Tfy0sIVfVOY (2:03).
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Chapter 9. The Atomic Theory
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Electrons, Protons, and Neutrons
Electrons have a negative charge. As a result, they are attracted to positive objects and repelled from negative
objects, including other electrons (illustrated below). To minimize repulsion, each electron is capable of staking out
a “territory” and “defending” itself from other electrons.
Protons are another type of subatomic particle found in atoms. They have a positive charge, so they are attracted
to negative objects and repelled from positive objects. Again, this means that protons repel each other (illustrated
below). However, unlike electrons, protons are forced to group together into one big clump, even though they repel
each other. Protons are bound together by what are termed strong nuclear forces. These forces are responsible for
binding the atomic nuclei together, allowing the protons to form a dense, positively charged center.
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There is a third subatomic particle known as a neutron. Rutherford proposed the existence of a neutral particle
along with the approximate mass of a proton, but it wasn’t until years later that someone proved the existence of the
neutron. A physicist named James Chadwick observed that when beryllium was bombarded with alpha particles, it
emitted an unknown radiation that had approximately the same mass as a proton, but the radiation had no electrical
charge. Chadwick was able to prove that these beryllium emissions contained a neutral particle – Rutherford’s
neutron.
As you might have already guessed from its name, the neutron is neutral. In other words, it has no charge and is
therefore neither attracted to nor repelled from other objects. Neutrons are in every atom (with one exception), and
they’re bound together with other neutrons and protons in the atomic nucleus. Again, the binding forces that help to
keep neutrons fastened into the nucleus are known as strong nuclear forces.
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Chapter 9. The Atomic Theory
Since neutrons are neither attracted to nor repelled from objects, they don’t really interact with protons or electrons
beyond being bound into the nucleus with the protons. Protons and electrons, however, do interact. Using what you
know about protons and electrons, what do you think will happen when an electron approaches a proton? Will the
two subatomic particles be attracted to each other or repelled from each other? Here’s a hint: “opposites attract, likes
repel.” Since electrons and protons have opposite charges (one negative, the other positive), you’d expect them to be
attracted to each other, as illustrated below.
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Even though electrons, protons, and neutrons are all types of subatomic particles, they are not all the same size.
When comparing the masses of electrons, protons, and neutrons, you will find that electrons have an extremely
small mass compared to the masses of either protons or neutrons (see Figure 9.7). On the other hand, the masses of
protons and neutrons are fairly similar, with the mass of a neutron being slightly greater than the mass of a proton.
Because protons and neutrons are so much more massive than electrons, almost all of the atomic mass in any atom
comes from the nucleus, which is where all of the neutrons and protons are located.
FIGURE 9.7
Electrons are much smaller than protons or neutrons. How much smaller?
If an electron was the size of a penny, a proton or a neutron would have
the mass of a large bowling ball!
Table 9.1 gives the properties and locations of electrons, protons, and neutrons. The third column shows the masses
of the three subatomic particles in grams. The second column, however, shows the masses of the three subatomic
particles in amu, or atomic mass units. An atomic mass unit (amu) is defined as one-twelfth the mass of a carbon12 atom (a carbon that has 6 protons and 6 neutrons). Atomic mass units are useful because, as you can see, the
mass of a proton and the mass of a neutron are almost exactly 1.0 in this unit system. The dalton is equivalent to
the atomic mass unit, with the two terms being different names for the same measure. The two terms are often used
interchangeably, and both will be used throughout this text.
TABLE 9.1: Subatomic Particles, Properties, and Location
Particle
Relative
(amu)
electron
proton
neutron
1
1840
1
1
Mass
Mass in Grams (g)
Electric Charge
Location
9.109383 × 10−28
1.6726217 × 10−24
1.6749273 × 10−24
−1
+1
0
outside nucleus
nucleus
nucleus
In addition to mass, another important property of subatomic particles is the charge. The fourth column in Table
9.1 shows the charges of the three subatomic particles. You already know that neutrons are neutral and thus have no
charge at all. Therefore, we say that neutrons have a charge of zero. What about electrons and protons? Electrons
are negatively charged and protons are positively charged, but the positive charge on a proton is exactly equal in
magnitude (magnitude means “absolute value”) to the negative charge on an electron. You may recall that Millikan
discovered that the magnitude of the charge on a single electron is 1.6 × 10−19 C (coulomb), which means that the
magnitude of the charge on a proton is also 1.6 × 10−19 C. In other words, a neutral atom must have exactly one
electron for every proton. If a neutral atom has 1 proton, it must have 1 electron. If a neutral atom has 2 protons, it
must have 2 electrons. If a neutral atom has 10 protons, it must have 10 electrons. Do you get the idea?
For a short animation demonstrating the properties of the electron using a cathode ray tube (1h), see http://www.y
outube.com/watch?v=4QAzu6fe8rE video (3:46).
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Chapter 9. The Atomic Theory
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Atomic Number and Mass Number
Scientists can distinguish between different elements by counting the number of protons. If an atom has only one
proton, we know it’s an atom of the element hydrogen. An atom with two protons is always an atom of the element
helium. When scientists count four protons in an atom, they know it’s a beryllium atom. An atom with three protons
is a lithium atom, an atom with five protons is a boron atom, an atom with six protons is a carbon atom. . . the list
goes on (see Figure 9.8 for more examples).
FIGURE 9.8
How would you distinguish these three
elements? You can
Since an atom of one element can be distinguished from an atom of another element by the number of protons in
the nucleus, scientists are always interested in this number and how this number differs between different elements.
Therefore, scientists give this number a special name and a special symbol. An element’s atomic number (Z) is
equal to the number of protons in the nuclei of any of its atoms. The periodic table gives the atomic number of each
element. The atomic number is a whole number usually written above the chemical symbol of each element in the
table. The atomic number for hydrogen is Z = 1 because every hydrogen atom has 1 proton. The atomic number for
helium is Z = 2 because every helium atom has 2 protons. What is the atomic number of carbon? (Answer: Carbon
has 6 protons, so the atomic number for carbon is Z = 6.)
Since neutral atoms have to have one electron for every proton, an element’s atomic number also tells you how many
electrons are in a neutral atom of that element. For example, hydrogen has atomic number Z = 1. This means that
an atom of hydrogen has one proton and, if it’s neutral, one electron. Gold, on the other hand, has atomic number Z
= 79, which means that a neutral atom of gold has 79 protons and 79 electrons.
The mass number (A) of an atom is the total number of protons and neutrons in its nucleus. Why do you think
that the mass number includes protons and neutrons, but not electrons? You know that most of the mass of an atom
is concentrated in its nucleus and that the mass of an electron is very, very small compared to the mass of either a
proton or a neutron (like the mass of a penny compared to the mass of a bowling ball). By counting the number of
protons and neutrons, scientists will have a very close approximation of the total mass of an atom.
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mass number A = (number of protons) + (number of neutrons)
An atom’s mass number is very easy to calculate once you know the number of protons and neutrons in the atom.
Notice that the mass number is not the same as the mass of the atom. You can easily relate the mass number to the
mass by recalling that both protons and neutrons have a relative mass of 1 amu.
Example:
What is the mass number of an atom that contains 3 protons and 4 neutrons?
(number of protons) = 3
(number of neutrons) = 4
mass number A = (number of protons) + (number of neutrons)
mass number A = (3) + (4) = 7
Example:
What is the mass number of an atom of helium that contains 2 neutrons?
(number of protons) = 2 (Remember that an atom of helium always has 2 protons.)
(number of neutrons) = 2
mass number A = (number of protons) + (number of neutrons)
mass number A = (2) + (2) = 4
This video summarizes the concept of the atom and to the organization of the periodic table (1a, 1e): http://www.y
outube.com/watch?v=1xSQlwWGT8M (21:05).
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Isotopes and Atomic Mass
Unlike the number of protons, which is always the same for all atoms of the same element, the number of neutrons
can be different. Atoms of the same element with different numbers of neutrons are known as isotopes. Since the
isotopes of any given element all contain the same number of protons, they have the same atomic number. However,
since the isotopes of a given element contain different numbers of neutrons, different isotopes have different mass
numbers. The following two examples should help to clarify this point.
Example:
What is the atomic number (Z) and the mass number (A) of an isotope of lithium containing 3 neutrons? A lithium
atom contains 3 protons in its nucleus.
atomic number Z = number of protons = 3
number of neutrons = 3
mass number A = (number of protons) + (number of neutrons)
mass number A = (3) + (3) = 6
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Chapter 9. The Atomic Theory
Example:
What is the atomic number (Z) and the mass number (A) of an isotope of lithium containing 4 neutrons? A lithium
atom contains 3 protons in its nucleus.
atomic number Z = number of protons = 3
number of neutrons = 4
mass number A = (number of protons) + (number of neutrons)
mass number A = (3) + (4) = 7
Notice that because the lithium atom always has 3 protons, the atomic number for lithium is always Z = 3. The mass
number, however, is A = 6 for the isotope with 3 neutrons, and A = 7 for the isotope with 4 neutrons. In nature, only
certain isotopes exist. For instance, lithium exists as an isotope with 3 neutrons and as an isotope with 4 neutrons,
but it doesn’t exists as an isotope with 2 neutrons or as an isotope with 5 neutrons.
This whole discussion of isotopes brings us back to Dalton’s atomic theory. According to Dalton, atoms of a given
element are identical. But if atoms of a given element can have different numbers of neutrons, then they can have
different masses as well. How did Dalton miss this? It turns out that elements found in nature exist as uniform
mixtures with a constant ratio of their naturally occurring isotopes. In other words, a piece of lithium always
contains both types of naturally occurring lithium (the type with 3 neutrons and the type with 4 neutrons). Moreover,
it always contains the two in the same relative amounts (or “relative abundances”). In a chunk of lithium, 93% will
always be lithium with 4 neutrons, while the remaining 7% will always be lithium with 3 neutrons.
Unfortunately, Dalton always experimented with large chunks of an element – chunks that contained all of the
naturally occurring isotopes of that element. As a result, when he performed his measurements, he was actually
observing the averaged properties of all the different isotopes in the sample. Luckily, aside from having different
masses, most other properties of different isotopes are similar.
Knowing about the different isotopes is important when it comes to calculating atomic mass. The atomic mass
(sometimes referred to as atomic weight) of an element is the weighted average mass of the atoms in a naturally
occurring sample of the element. Atomic mass is typically reported in atomic mass units. You can calculate the
atomic mass of an element, provided you know the relative abundances the element’s naturally occurring isotopes
and the masses of those different isotopes. The examples below show how this calculation is done.
Example:
Boron has two naturally occurring isotopes. In a sample of boron, 20% of the atoms are B-10, which is an isotope
of boron with 5 neutrons and a mass of 10 amu. The other 80% of the atoms are B-11, which is an isotope of boron
with 6 neutrons and a mass of 11 amu. What is the atomic mass of boron?
Solution:
To do this problem, we will calculate 20% of the mass of B-10, which is how much the B-10 isotope contributes
to the “average boron atom.” We will also calculate 80% of the mass of B-11, which is how much the B-11 isotope
contributes to the “average boron atom.”
Step One: Convert the percentages given in the question into their decimal forms by dividing each percentage by
100%:
Decimal form of 20% = 0.20
Decimal form of 80% = 0.80
Step Two: Multiply the mass of each isotope by its relative abundance (percentage) in decimal form:
20% of the mass of B-10 = 0.20 × 10 amu = 2.0 amu
80% of the mass of B-11 = 0.80 × 11 amu = 8.8 amu
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Step Three: Find the total mass of the “average atom” by adding together the contributions from the different
isotopes:
Total mass of average atom = 2.0 amu + 8.8 amu = 10.8 amu
The mass of an average boron atom, and thus boron’s atomic mass, is 10.8 amu.
Example:
Neon has three naturally occurring isotopes. In a sample of neon, 90.48% of the atoms are Ne-20, which is an
isotope of neon with 10 neutrons and a mass of 19.99 amu. Another 0.27% of the atoms are Ne-21, which is an
isotope of neon with 11 neutrons and a mass of 20.99 amu. The final 9.25% of the atoms are Ne-22, which is an
isotope of neon with 12 neutrons and a mass of 21.99 amu. What is the atomic mass of neon?
Solution:
To do this problem, we will calculate 90.48% of the mass of Ne-20, which is how much Ne-20 contributes to the
“average neon atom.” We will also calculate 0.27% of the mass of Ne-21 and 9.25% of the mass of Ne-22, which
are how much the Ne-21 and the Ne-22 isotopes contribute to the “average neon atom” respectively.
Step One: Convert the percentages given in the question into their decimal forms by dividing each percentage by
100%:
Decimal form of 90.48% = 0.9048
Decimal form of 0.27% = 0.0027
Decimal form of 9.25% = 0.0925
Step Two: Multiply the mass of each isotope by its relative abundance (percentage) in decimal form:
90.48% of the mass of Ne-20 = 0.9048 × 20.00 amu = 18.10 amu
0.27% of the mass of Ne-21 = 0.0027 × 21.00 amu = 0.057 amu
9.25% of the mass of Ne-22 = 0.0885 × 22.00 amu = 2.04 amu
Step Three: Find the total mass of the “average atom” by adding together the contributions from the different
isotopes:
Total mass of average atom = 18.10 amu + 0.057 amu + 2.04 amu = 20.20 amu
The mass of an average neon atom, and thus neon’s atomic mass, is 20.20 amu.
The periodic table gives the atomic mass of each element. The atomic mass is a number that usually appears below
the element’s symbol in each square. Notice that atomic mass of boron (symbol B) is 10.8 and the atomic mass of
neon (symbol Ne) is 20.18, both which are very close to what we calculated in our examples. Take time to notice
that not all periodic tables have the atomic number above the element’s symbol and the atomic mass below it. If you
are ever confused, remember that the atomic number should always be the smaller of the two and will be a whole
number, while the atomic mass should always be the larger of the two. (The atomic mass must include both the
number of protons and the average number of neutrons.)
Lesson Summary
• Electrons are a type of subatomic particle with a negative charge, so electrons repel each other but are attracted
to protons.
• Protons are a type of subatomic particle with a positive charge, so protons repel each other but are attracted to
electrons. Protons are bound together in an atom’s nucleus as a result of strong nuclear forces.
• Neutrons are a type of subatomic particle with no charge (they’re neutral). Like protons, neutrons are bound
into the atom’s nucleus as a result of strong nuclear forces.
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Chapter 9. The Atomic Theory
• Protons and neutrons have approximately the same mass and are both much more massive than electrons
(approximately 2,000 times as massive as an electron).
• The positive charge on a proton is equal in magnitude to the negative charge on an electron. As a result, a
neutral atom must have an equal number of protons and electrons.
• Each element has a unique number of protons. An element’s atomic number (Z) is equal to the number of
protons in the nuclei of any of its atoms.
• The mass number (A) of an atom is the sum of the protons and neutrons in the atom
• Isotopes are atoms of the same element (same number of protons) that have different numbers of neutrons in
their atomic nuclei.
• An element’s atomic mass is the average mass of one atom of that element. An element’s atomic mass can be
calculated provided the relative abundances of the element’s naturally occurring isotopes and the masses of
those isotopes are known.
• The periodic table is a convenient way to summarize information about the different elements. In addition to
the element’s symbol, most periodic tables will also contain the element’s atomic number and the element’s
atomic mass.
Further Reading / Supplemental Links
This website has a video called “Atomic Structure: The Nucleus” available.
• http://videos.howstuffworks.com/hsw/5806-atomic-structure-the-nucleus-video.htm
Review Questions
1. Decide whether each of the following statements is true or false.
(a)
(b)
(c)
(d)
(e)
The nucleus of an atom contains all of the protons in the atom.
The nucleus of an atom contains all of the neutrons in the atom.
The nucleus of an atom contains all of the electrons in the atom.
Neutral atoms of a given element must contain the same number of neutrons.
Neutral atoms of a given element must contain the same number of electrons.
2. Match the subatomic property with its description in Table 9.2.
TABLE 9.2: Table for Problem 2
Subatomic Particle
i. electron
ii. neutron
iii. proton
Characteristics
a. has an atomic charge of +1 e
b. has a mass of 9.109383 × 10−28 grams
c. is neither attracted to nor repelled from charged
objects
3. Arrange the electron, proton, and neutron in order of decreasing mass.
4. Indicate which of the following statements is true or false.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
An element’s atomic number is equal to the number of protons in the nuclei of any of its atoms.
The symbol for an element’s atomic number is A.
A neutral atom with Z = 4 must have 4 electrons.
A neutral atom with A = 4 must have 4 electrons.
An atom with 7 protons and 7 neutrons will have A = 14.
An atom with 7 protons and 7 neutrons will have Z = 14.
A neutral atom with 7 electrons and 7 neutrons will have A = 14.
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5. Use the periodic table to find the symbol for the element with:
(a)
(b)
(c)
(d)
44 electrons in a neutral atom.
30 protons.
Z = 36.
an atomic mass of 14.007 amu.
6. When will the mass number (A) of an atom be:
(a) bigger than the atomic number (Z) of the atom?
(b) smaller than the atomic number (Z) of the atom?
(c) equal to the atomic number (Z) of the atom?
7. In Table 9.3, Column 1 contains data for five different elements. Column 2 contains data for the same five
elements but with different isotopes of those elements. Match the columns by connecting isotopes of the same
element.
TABLE 9.3: Table for Problem 7
Column 1
a. an atom with 2 protons and 1 neutron
b. a Be (beryllium) atom with 5 neutrons
c. an atom with Z = 6 and A = 13
d. an atom with 1 proton and A = 1
e. an atom with Z = 7 and 7 neutrons
Column 2
i. a C (carbon) atom with 6 neutrons
ii. an atom with 2 protons and 2 neutrons
iii. an atom with Z = 7 and A = 15
iv. an atom with A = 2 and 1 neutron
v. an atom with Z = 4 and 6 neutrons
8. Match the following isotopes with their respective mass numbers in Table 25.5.
TABLE 9.4: Table for Problem 8
Column 1
(a) an atom with Z = 17 and 18 neutrons
(b) an H atom with no neutrons
(c) A He atom with 2 neutrons
(d) an atom with Z = 11 and 11 neutrons
(e) an atom with 11 neutrons and 12 protons
Column 2
i. 35
ii. 4
iii. 1
iv. 23
v. 22
9. Match the following isotopes with their respective atomic numbers in Table 9.5.
TABLE 9.5: Table for Problem 9
Column 1
(a) a B (boron) atom with A = 10
(b) an atom with A = 10 and 6 neutrons
(c) an atom with 3 protons and 3 neutrons
(d) an oxygen atom
(e) an atom with A = 4 and 2 neutrons
Column 2
i. 8
ii. 2
iii. 3
iv. 4
v. 5
10. Answer the following questions:
(a) What’s the mass number of an atom that contains 13 protons and 13 neutrons?
(b) What’s the mass number of an atom that contains 24 protons and 30 neutrons?
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Chapter 9. The Atomic Theory
11. Answer the following questions:
(a) What’s the mass number of the isotope of manganese (Mn) containing 28 neutrons?
(b) What’s the mass number of the isotope of calcium (Ca) containing 20 neutrons?
12. Answer the following questions:
(a) What’s the atomic number of an atom that has 30 neutrons, and a mass number of A = 70?
(b) What’s the atomic number of an atom with 14 neutrons, if the mass number of the atom is A = 28?
13. Answer the following questions:
(a)
(b)
(c)
(d)
What’s the mass number of a neutral atom that contains 7 protons and 7 neutrons?
What’s the mass number of a neutral atom that contains 7 electrons and 7 neutrons?
What’s the mass number of a neutral atom that contains 5 protons, 5 electrons, and 6 neutrons?
What’s the mass number of a neutral atom that contains 3 electrons and 4 neutrons?
14. Answer the following questions:
(a) What element has 32 neutrons in an atom with mass number A = 58?
(b) What element has 10 neutrons in an atom with mass number A = 19?
15. Copper has two naturally occurring isotopes. 69.15% of copper atoms are Cu-63 and have a mass of 62.93 amu.
The other 30.85% of copper atoms are Cu-65 and have a mass of 64.93 amu. What is the atomic mass of
copper?
All images, unless otherwise stated, are created by the CK-12 Foundation and are under the Creative Commons
license CC-BY-NC-SA.
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9.4. References
9.4 References
1.
2.
3.
4.
5.
6.
7.
8.
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.
.
.
.
.
.
.
.
Democritus.
Thinker.
ABC.
JohnDalton.
JJThomson.
Rutherford.
HS04-0305.
HS04-0307.
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Chapter 10. The Bohr Model of the Atom
C HAPTER
10The Bohr Model of the Atom
Chapter Outline
10.1
T HE N ATURE OF L IGHT
10.2
ATOMS AND E LECTROMAGNETIC S PECTRA
10.3
T HE B OHR M ODEL OF THE ATOM
10.4
R EFERENCES
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10.1 The Nature of Light
Lesson Objectives
The student will:
• perform calculations involving the relationship between the wavelength and frequency of electromagnetic
radiation, v = λ f .
• perform calculations involving the relationship between the energy and the frequency of electromagnetic
radiation, E = h f .
• state the velocity of electromagnetic radiation in a vacuum.
• name at least three different areas of the electromagnetic spectrum.
• when given two comparative colors or areas in the electromagnetic spectrum, identify which area has the
higher wavelength, the higher frequency, and the higher energy.
Vocabulary
•
•
•
•
•
•
•
•
amplitude
crest
electromagnetic spectrum
frequency ( f )
hertz (Hz)
trough
velocity (v)
wavelength (λ)
Introduction
In order to understand how Rutherfod’s model of the atom evolved to the current atomic model, we need to
understand some basic properties of light. During the 1600s, there was a debate about how light travels. Isaac
Newton, the English physicist, hypothesized that light consisted of tiny particles and that a beam of light would
therefore be a stream of particles. Around the same time, Christian Huygens, a Dutch physicist, suggested that light
traveled as a waveform in the same way energy travels in water.
Neither hypothesis became the dominant idea until 200 years later, when the Scottish physicist James Clerk Maxwell
proposed a wave model of light in 1864 that gained widespread support. Maxwell’s equations related electricity,
magnetism, and light so comprehensively that several physicists suggested students should major in other sciences
because everything in physics had been discovered. Scientists thought that Maxwell’s work permanently settled the
“wave versus particle” debate over the nature of light. Fortunately, quite a few students did not take their advice.
Sixty years later, German physicist Max Planck would raise the issue again and renew the debate over the nature of
light.
The Wave Form of Energy
The wave model of electromagnetic radiation is somewhat similar to waves in a rope. Suppose we tie one end of a
rope to a tree and hold the other end at a distance from the tree so that the rope is fully extended. If we then jerk
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the end of the rope up and down in a rhythmic way, the rope will go up and down. When the end of the rope we are
holding goes up and down, it pulls on the neighboring part of the rope, which will also go up and down. The up and
down motion will be passed along to each succeeding part of the rope, and after a short time, the entire rope will
contain a wave like the one shown in the image below.
The red line in the diagram shows the undisturbed position of the rope before the wave motion was initiated. The
crest is the highest point of the wave above the undisturbed position, while the trough is the lowest point of a wave
below the undisturbed position. It is important for you to recognize that the individual particles of rope do not move
horizontally. Each point on the rope only moves up and down. If the wave is allowed to dissipate, every point of the
rope will be in the exact same position it was in before the wave started. The wave in the rope moves horizontally
from the person to the tree, but no parts of rope actually move horizontally. The notion that parts of the rope are
moving horizontally is a visual illusion. Like the wave, the energy that is put into the rope by jerking it up and down
also moves horizontally from the person to the tree.
If we jerk the rope up and down with a different rhythm, the wave in the rope will change its appearance in terms of
crest height, distance between crests, and so forth, but the general shape of the wave will remain the same. We can
characterize the wave in the rope with a few measurements. The image below shows an instantaneous snapshot of
the rope so that we can indicate some characteristic values.
The distance from one crest to the next crest is called the wavelength of the wave. You could also determine
the wavelength by measuring the distance from one trough to the next or between any two identical positions on
successive waves. The symbol used for wavelength is the Greek letter lambda, λ. The distance from the maximum
height of a crest to the undisturbed position is called the amplitude of the wave. The velocity of a wave is the
distance traveled by the wave in one second. We can obtain the velocity of the wave by measuring how far a crest
travels horizontally in a unit of time. The SI unit for velocity is meters/second.
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Another important characteristic of waves is called frequency. The frequency of a wave is the number of cycles that
pass a given point per unit of time. If we choose an exact position along the path of the wave and count how many
crests pass the position per unit time, we would get a value for frequency. Based on this description, the unit for
frequency would be cycles per second or waves per second. In science, however, frequency is often denoted by 1/s
or s−1 , with “cycles” being implied rather than explicitly written out. This unit is called a hertz (abbreviated Hz),
but it means cycles per second and is written out in calculations as 1/s or s−1 . The symbol used for frequency is
the Greek letter nu, ν. Unfortunately, this Greek letter looks a very great deal like the italicized letter v. You must
be very careful when reading equations to see whether the symbol is representing velocity (v) or frequency (ν). To
avoid this problem, this text will use a lower case letter f as the symbol for frequency.
The velocity, wavelength, and frequency of a wave are all related, as indicated by the formula: v = λ f . If the
wavelength is expressed in meters and the frequency is expressed in 1/second (s−1 ), then multiplying the wavelength
times the frequency will yield meters/second, which is the unit for velocity.
Example:
What is the velocity of a rope wave if its wavelength is 0.50 m and its frequency is 12 s−1 ?
v = λ f = (0.50 m) · (12 s−1 ) = 6.0 m/s
Example:
What is the wavelength of a water wave if its velocity is 5.0 m/s and its frequency is 2.0 s−1 ?
m/s
λ = vf = 5.0
= 2.5 meters
2.0 s−1
Electromagnetic Waves
Electromagnetic radiation is a form of energy that consists of electric and magnetic fields traveling at the speed
of light. Electromagnetic waves carry this energy from one place to another and are somewhat like waves in a
rope. Unlike the wave in a rope, however, electromagnetic waves are not required to travel through a medium. For
example, light waves are electromagnetic waves capable of traveling from the sun to Earth through outer space,
which is considered a vacuum.
The energy of an electromagnetic wave travels in a straight line along the path of the wave, just like the energy in the
rope wave did. The moving light wave has associated with it an oscillating electric field and an oscillating magnetic
field. Scientists often represent the electromagnetic wave with the image below. Along the straight-line path of the
wave, there exists a positive electric field that will reach a maximum positive charge, slowly collapse to zero charge,
and then expand to a maximum negative charge. Along the path of the electromagnetic wave, this changing electric
field repeats its oscillating charge over and over again. There is also a changing magnetic field that oscillates from
maximum north pole field to maximum south pole field. Do not confuse the oscillating electric and magnetic fields
with the way light travels. Light does not travel in this weaving wave pattern. The light travels along the black line
that represents the undisturbed position. For an electromagnetic wave, the crests and troughs represent the oscillating
fields, not the path of the light.
Although light waves look different from the wave in the rope, we still characterize light waves by their wavelength,
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frequency, and velocity. We can measure along the path of the wave the distance the wave travels between one
crest and the succeeding crest. This will be the wavelength of the electromagnetic radiation. The frequency of
electromagnetic waves is still the number of full cycles of waves that pass a point in a unit of time, just like how
frequency is defined for rope waves. The velocity for all electromagnetic waves traveling through a vacuum is
the same. Although technically the velocity of electromagnetic waves traveling through air is slightly less than
the velocity in a vacuum, the two velocities are so close that we will use the same value for the velocity. In a
vacuum, every electromagnetic wave has a velocity of 3.00 × 108 m/s, which is symbolized by the lower case c. The
relationship, then, for the velocity, wavelength, and frequency of electromagnetic waves is: c = λ f .
Example:
What is the wavelength of an electromagnetic wave traveling in air whose frequency is 1.00 × 1014 s−1 ?
8 m/s
= 3.00 × 10−6 m
λ = cf = 3.00×10
1.00×1014 s−1
The Electromagnetic Spectrum
In rope waves and water waves, the amount of energy possessed by the wave is related to the amplitude of the wave;
there is more energy in the rope if the end of the rope is jerked higher and lower. In electromagnetic radiation,
however, the amount of energy possessed by the wave is only related to the frequency of the wave. In fact, the
frequency of an electromagnetic wave can be converted directly to energy (measured in joules) by multiplying
the frequency with a conversion factor. The conversion factor is called Planck’s constant and is equal to 6.6 ×
10−34 joule · seconds. Sometimes, Planck’s constant is given in units of joules/hertz, but you can show that the units
are the same. The equation for the conversion of frequency to energy is E = h f , where E is the energy in joules
(symbolized by J), h is Planck’s constant in joules·second, and f is the frequency in s−1 .
Electromagnetic waves have an extremely wide range of wavelengths, frequencies, and energies. The electromagnetic spectrum is the range of all possible frequencies of electromagnetic radiation. The highest energy form of
electromagnetic waves is gamma rays and the lowest energy form (that we have named) is radio waves.
In the image below, the electromagnetic waves on the far left have the highest energy. These waves are called gamma
rays, and they can cause significant damage to living systems. The next lowest energy form of electromagnetic waves
is called X-rays. Most of you are familiar with the penetration abilities of these waves. Although they can be helpful
in imaging bones, they can also be quite dangerous to humans. For this reason, humans are advised to try to limit as
much as possible the number of medical X-rays they have per year. After X-rays, ultraviolet rays are the next lowest
in energy. These rays are a part of sunlight, and rays on the upper end of the ultraviolet range can cause sunburn
and eventually skin cancer. The next tiny section in the spectrum is the visible range of light. The band referred to
as visible light has been expanded and extended below the full spectrum. These are the frequencies (energies) of
the electromagnetic spectrum to which the human eye responds. Lower in the spectrum are infrared rays and radio
waves.
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The light energies that are in the visible range are electromagnetic waves that cause the human eye to respond when
they enter the eye. The eye then sends signals to the brain, and the individual “sees” various colors. The waves in the
visible region with the highest energy are interpreted by the brain as violet. As the energy of the waves decreases,
the colors change to blue, green, yellow, orange, and red. When the energy of the wave is above or below the visible
range, the eye does not respond to them. When the eye receives several different frequencies at the same time, the
colors are “blended” by the brain. If all frequencies of visible light enter the eye together, the brain sees white, and
if no visible light enters the eye, the brain sees black.
All the objects that you see around you are light absorbers – that is, the chemicals on the surface of the objects
absorb certain frequencies and not others. Your eyes will then detect the frequencies that strike them. Therefore,
if your friend is wearing a red shirt, it means that the dye in that shirt reflected the red and absorbed all the other
frequencies. When the red frequency from the shirt arrives at your eye, your visual system sees red, and you would
say the shirt is red. If your only light source was one exact frequency of blue light and you shined it on a shirt that
absorbed every frequency of light except for one frequency of red, then the shirt would look black to you because no
light would be reflected to your eye.
Lesson Summary
• The wave form of energy is characterized by velocity, wavelength, and frequency.
• The velocity, wavelength, and frequency of a wave are related by the expression: v = λ f .
• Electromagnetic radiation comes in a wide spectrum that includes low energy radio waves and very high
energy gamma rays.
• The frequency and energy of electromagnetic radiation are related by the expression: E = h f .
Further Reading / Supplemental Links
This website provides more information about the properties of electromagetic waves and includes an animation
showing the relationship between wavelength and color.
• http://micro.magnet.fsu.edu/primer/java/wavebasics/index.html
Review Questions
1. Name at least three different areas in the spectrum of electromagnetic radiation.
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2.
3.
4.
5.
Which color of visible light has the longer wavelength, red or blue?
What is the velocity of all forms of electromagnetic radiation traveling in a vacuum?
How can you determine the frequency of a wave when the wavelength is known?
If the velocity of a water wave is 9.0 m/s and the wave has a wavelength of 3.0 m, what is the frequency of the
wave?
6. If a sound wave has a frequency of 256 Hz and a wavelength of 1.34 m, what is its velocity?
7. What is the relationship between the energy of electromagnetic radiation and the frequency of that radiation?
8. What is the energy, in joules, of a light wave whose frequency is 5.66 × 108 Hz?
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10.2 Atoms and Electromagnetic Spectra
Lesson Objectives
The student will:
• describe the appearance of an atomic emission spectrum.
• explain why an element can be identified by its emission spectrum.
Vocabulary
• emission spectrum
Introduction
Electric light bulbs contain a very thin wire that emits light upon heating. The wire is called a filament. The particular
wire used in light bulbs is made of tungsten. A wire made of any metal would emit light under these circumstances,
but one of the reasons that tungsten is used is because the light it emits contains virtually every frequency, making the
emitted light appear white. Every element emits light when energized, either by heating the element or by passing
electric current through it. Elements in solid form begin to glow when they are sufficiently heated, while elements
in gaseous form emit light when electricity passes through them. This is the source of light emitted by neon signs
(see Figure 10.1) and is also the source of light in a fire. You may have seen special logs created for fireplaces that
give off bright red and green colors as they burn. These logs were created by introducing certain elements into them
in order to produce those colors when heated.
FIGURE 10.1
The light emitted by the sign containing
neon gas (on the left) is different from the
light emitted by the sign containing argon
gas (on the right).
Each Element Has a Unique Spectrum
Several physicists, including Anders J. Angstrom in 1868 and Johann J. Balmer in 1875, passed the light from
energized atoms through glass prisms in such a way that the light was spread out and the individual frequencies
making up the light could be seen.
In Figure 10.2, we see the emission spectrum for hydrogen gas. The emission spectrum of a chemical element is
the pattern of frequencies obtained when the element is subjected to a specific excitation. When hydrogen gas is
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FIGURE 10.2
This is the unique emission spectrum for
hydrogen.
placed into a tube and electric current passed through it, the color of emitted light is pink. But when the light is
separated into individual colors, we see that the hydrogen spectrum is composed of four individual frequencies. The
pink color of the tube is the result of our eyes blending the four colors.
Every atom has its own characteristic spectrum; no two atomic spectra are alike. Because each element has a unique
emission spectrum, elements can be identified by using them. Figure 10.3 shows the emission spectrum of iron.
FIGURE 10.3
This is the unique emission spectrum of
iron.
You may have heard or read about scientists discussing what elements are present in the sun or some more distant
star. How could scientists know what elements are present if they have never been to these faraway places? Scientists
determine what elements are present in distant stars by analyzing the light that comes from those stars and using the
atomic spectrum to identify the elements emitting that light.
Lesson Summary
• Atoms have the ability to absorb and emit electromagnetic radiation.
• Each element has a unique emission spectrum.
Further Reading / Supplemental Links
This website “Spectral Lines” has a short discussion of atomic spectra. It also has the emission spectra of several
elements.
• http://www.colorado.edu/physics/2000/quantumzone/index.html
Review Questions
1. The emission spectrum for an element shows bright lines for the light frequencies that are emitted. The absorption spectrum of that same element shows dark lines within the complete spectrum for the light frequencies
that are absorbed. How can you explain that the bright lines in the emission spectrum of an element exactly
correspond to the dark lines in the absorption spectrum for that same element?
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10.3 The Bohr Model of the Atom
Lesson Objectives
The student will:
• describe an electron cloud containing Bohr’s energy levels.
• describe how the Bohr model of the atom explains the existence of atomic spectra.
• explain the limitations of the Bohr model and why it had to be replaced.
Vocabulary
• energy level
Introduction
By 1913, our concept of the atom had evolved from Dalton’s idea of indivisible spheres to Thomson’s plum-pudding
model and then to Rutherford’s nuclear atom theory.
Rutherford, in addition to carrying out the experiment that demonstrated the presence of the atomic nucleus, proposed that the electrons circled the nucleus in a planetary-like motion. The planetary model of the atom was attractive
to scientists because it was similar to something with which they were already familiar, namely the solar system.
Unfortunately, there was a serious flaw in the planetary model. At that time, it was already known that when a
charged particle moves in a curved path, the particle emits some form of light or radio waves and loses energy in
doing so. If the electron circling the nucleus in an atom loses energy, it would necessarily have to move closer to the
nucleus (because of the loss of potential energy) and would eventually crash into the nucleus. Scientists, however,
saw no evidence that electrons were constantly emitting energy or crashing into the nucleus. These difficulties cast
a shadow on the planetary model and indicated that it would eventually be replaced.
The replacement model came in 1913 when the Danish physicist Niels Bohr (pictured in Figure 10.4) proposed an
electron cloud model where the electrons orbit the nucleus but did not have to lose energy.
Bohr’s Energy Levels
The key idea in the Bohr model of the atom is that electrons occupy definite orbits that require the electron to have
a specific amount of energy. In order for an electron to be in the electron cloud of an atom, it must be in one of the
allowable orbits and have the precise energy required for that orbit. Orbits closer to the nucleus would require the
electrons to have a smaller amount of energy, and orbits farther from the nucleus would require the electrons to have
a greater amount of energy. The possible orbits are known as energy levels.
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FIGURE 10.4
Niels Bohr and Albert Einstein in 1925.
Bohr hypothesized that the only way electrons could gain or lose energy would be to move from one energy level to
another, thus gaining or losing precise amounts of energy. It would be like a ladder that had rungs at certain heights
(see image below). The only way you can be on that ladder is to be on one of the rungs, and the only way you could
move up or down is to move to one of the other rungs. Other rules for the ladder are that only one person can be on
a given rung and that the ladder occupants must be on the lowest rung available. Suppose we had such a ladder with
10 rungs. If the ladder had five people on it, they would be on the lowest five rungs. In this situation, no person could
move down because all the lower rungs are full. Bohr worked out the rules for the maximum number of electrons
that could be in each energy level in his model. In its normal state (ground state), this would require the atom to have
all of its electrons in the lowest energy levels available. Under these circumstances, no electron could lose energy
because no electron could move down to a lower energy level. In this way, the Bohr model explained why electrons
circling the nucleus did not emit energy and spiral into the nucleus.
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Bohr Model and Atomic Spectra
The evidence used to support the Bohr model came from the atomic spectra. Bohr suggested that an atomic spectrum
is created when the electrons in an atom move between energy levels. The electrons typically have the lowest energy
possible, but upon absorbing energy, the electrons would jump to a higher energy level, producing an excited and
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unstable state. The electrons would then immediately fall back to a lower energy level and re-emit the absorbed
energy. The energy emitted during these electron “step downs” would be emitted as light and would correspond with
a specific line in the atomic emission spectrum. Bohr was able to mathematically produce a set of energy levels for
the hydrogen atom. In his calculations, the differences between the energy levels were the exact same energies of
the frequencies of light emitted in the hydrogen spectrum. One of the most convincing aspects of the Bohr model
was that it predicted that the hydrogen atom would emit some electromagnetic radiation outside the visible range.
When scientists looked for these emissions in the infrared region, they were able to find them at the exact frequencies
predicted by the Bohr model. Bohr’s theory was rapidly accepted and he received the Nobel Prize for physics in
1922.
Shortcomings of the Bohr Model
The development of the Bohr model is a good example of applying the scientific method. It shows how the
observations of the atomic spectra led to the creation of a hypothesis about the nature of electron clouds. The
hypothesis also made predictions about emissions that had not yet been observed (the infrared light emissions).
Predicted observations such as these provide an opportunity to test the hypothesis through experimentation. When
these predictions were found to be correct, they provided evidence in support of the theory. Of course, further
observations can also provide insupportable evidence that will cause the theory to be rejected or modified. In the
case of the Bohr model of the atom, it was determined that the energy levels in atoms with more than one electron
could not be successfully calculated. Bohr’s system was only successful for atoms that have a single electron, which
meant that the Bohr model did not accurately reflect the behaviors of most atoms.
Another problem with Bohr’s theory was that the Bohr model did not explain why certain energy levels existed. As
mentioned earlier in this lesson, at the time it was already known that charged particles emit some form of light or
radio waves when moving in a curved path. Scientists have used this principle to create radio signals since 1895. This
was the serious flaw in Rutherford’s planetary model of the atom, which Bohr attempted to deal with by suggesting
his electron cloud model. Although his calculated energy levels for the hydrogen were supported by hydrogen’s
emission spectrum, Bohr did not, however, explain why only the exact energy levels he calculated were present.
Yet another problem with the Bohr model was the predicted positions of the electrons in the electron cloud. If
Bohr’s model were correct, the electron in the hydrogen atom in ground state would always be the same distance
from the nucleus. Although the actual path that the electron followed could not be determined, scientists were able
to determine the positions of the electron at various times. If the electron circled the nucleus as suggested by the
Bohr model, the electron positions would always be the same distance from the nucleus. In reality, the electron is
found at many different distances from the nucleus. In the figure below, the left side of the image (labeled as A)
shows the random positions an electron would occupy as predicted by the Bohr model, while the right side (labeled
as B) shows some actual positions of an electron.
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The Bohr model was not, however, a complete failure. It provided valuable insights that triggered the next step in
the development of the modern concept of the atom.
Lesson Summary
• The Bohr model suggests each atom has a set of unchangeable energy levels, and electrons in the electron
cloud of that atom must be in one of those energy levels.
• The Bohr model suggests that the atomic spectra of atoms is produced by electrons gaining energy from some
source, jumping up to a higher energy level, then immediately dropping back to a lower energy level and
emitting the energy difference between the two energy levels.
• The existence of the atomic spectra is support for the Bohr model of the atom.
• The Bohr model was only successful in calculating energy levels for the hydrogen atom.
This video provides a summary of the Bohr atomic model and how the Bohr model improved upon Rutheford’s
model (1i; 1g I&E, 1k I&E): http://www.youtube.com/watch?v=bDUxygs7Za8 (9:08).
MEDIA
Click image to the left for more content.
This video describes the important contributions of many scientists to the modern model of the atom. It also explains
Rutherford’s gold foil experiment (1g I&E): http://www.youtube.com/watch?v=6773jO6fMnM (9:08).
MEDIA
Click image to the left for more content.
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Further Reading / Supplemental Links
These various videos examine the components of the Bohr model of the atom.
• http://www.youtube.com/watch?v=QI50GBUJ48s&feature=related
• http://www.youtube.com/watch?v=hpKhjKrBn9s
• http://www.youtube.com/watch?v=-YYBCNQnYNM&feature=related
Review Questions
1. What is the key concept in the Bohr model of the atom?
2. What is the general relationship between the amount of energy of an electron energy level and its distance
from the nucleus?
3. According to Bohr’s theory, how can an electron gain or lose energy?
4. What happens when an electron in an excited atom returns to its ground level?
5. What concept in Bohr’s theory makes it impossible for an electron in the ground state to give up energy?
6. Use the Bohr model to explain how an atom emits a specific set of frequencies of light when it is heated or
has electric current passed through it.
7. How do scientists know that the sun contains helium atoms when no one has even taken a sample of material
from the sun?
• Bohr and Einstein. Public domain.
• Images of neon (http://en.wikipedia.org/wiki/File:NeTube.jpg ) and argon (http://en.wikipedia.org/wiki/Fil
e:ArTube.jpg ) signs created by Pslawinski, created into a composite by Richard Parsons. Neon and argon gas
signs. CC-BY-SA 2.5.
• Iron emission spectrum. Public domain.
• Hydrogen Emission Spectrum. Public domain.
All images, unless otherwise stated, are created by the CK-12 Foundation and are under the Creative Commons
license CC-BY-NC-SA.
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10.4. References
10.4 References
1.
2.
3.
4.
238
.
.
.
.
NeandArtubes.
Hspectrum.
Fespectrum.
BohrandEinstein.
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Chapter 11. The Quantum Mechanical Model of the Atom
C HAPTER
11
The Quantum Mechanical
Model of the Atom
Chapter Outline
11.1
T HE D UAL N ATURE OF L IGHT
11.2
C HARACTERISTICS OF M ATTER
11.3
Q UANTUM N UMBERS , O RBITALS , AND P ROBABILITY PATTERNS
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11.1 The Dual Nature of Light
Lesson Objectives
The student will:
• name the model that replaced the Bohr model of the atom.
• explain the concept of wave-particle duality.
• solve problems involving the relationship between the frequency and the energy of a photon.
Vocabulary
•
•
•
•
•
•
•
•
•
black body radiation
diffraction
interference
photoelectric effect
photon
quantum
quantum mechanics
quantum theory
wave-particle duality
Introduction
Further development in our understanding of the behavior of electrons in an atom’s electron cloud required some
major changes in our ideas about both matter and energy.
The branch of physics that deals with the motions of objects under the influence of forces is called mechanics.
Classical mechanics refers to the laws of motion developed by Isaac Newton in the 1600s. When the Bohr model
of the atom could not predict the energy levels of electrons in atoms with more than one electron, it seemed a new
approach to explaining the behavior of electrons was necessary. Developed in the early 1900s, this new approach
was based on the work of many scientists. The new approach came to be known as quantum mechanics (also called
wave mechanics). Quantum mechanics is the branch of physics that deals with the behavior of matter at the atomic
and subatomic level.
Properties of Waves
The controversy over the nature of light in the 1600s was partially due to the fact that different experiments with
light gave different indications about the nature of light. Energy waveforms, such as water waves or sound waves,
were found to exhibit certain characteristics, including diffraction (the bending of waves around corners) and
interference (the adding or subtracting of energies when waves overlap).
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In the image above, the sketch on the left shows a series of ocean wave crests (the long straight blue lines) striking
a sea wall. The sea wall has a gap between the cement barriers, which allows the water waves to pass through. The
energy of the water waves passes through the gap and essentially bends around the corners so that the continuing
waves are now circular and spread out on either side of the gap. The photograph on the right shows a real example
of water waves diffracting through a gap between small rocks. This type of behavior is characteristic of energy
waveforms.
When a body of water has more than one wave, like in the image above, the different waves will overlap and create
a new wave pattern. For simplicity, we will consider a case of two water waves with the same amplitude. If the
crests of both waves line up, then the new wave will have an amplitude that is twice that of the original waves. The
superposition of two crests is represented in the image above as a light area. Similarly, if a trough superimposes
over another trough, the new trough will be twice as deep (the darker areas in the image above). In both cases, the
amplitude of the new wave is greater than the amplitudes of the individual waves.
Now imagine what would happen if the crest from one water wave is superimposed on a trough from another wave.
If both waves are of equal amplitude, the upward pull of the crest and the downward pull of the trough will cancel
out. In this case, the water in that area will be flat. The amplitude of this new wave is smaller than the amplitudes of
the individual waves.
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This process of superimposing waves that occupy the same space is called interference. When the amplitude of the
new wave pattern is greater than the amplitudes of the individual waves, it is called constructive interference. When
the amplitude of the new wave pattern is smaller than the amplitudes of the individual waves, it is called destructive
interference. Interference behavior occurs with all energy waveforms.
Light as a Wave
Light also undergoes diffraction and interference. These characteristics of light can be demonstrated with what is
called a double-slit experiment. A box is sealed on all sides so that no light can enter. On one side of the box, two
very thin slits are cut. A light source placed in front of the slits will allow light to enter the two slits and shine on the
back wall of the box.
If light behaved like particles, the light would go straight from the slits to the back of the box and appear on the
back wall as two bright spots (see the left side of the image above). If light behaved like waves, the waves would
enter the slits and diffract. On the back wall, an interference pattern would appear with bright spots showing areas
of constructive interference and dark spots showing areas of destructive interference (see the right side of the image
above). When this double-slit experiment was conducted, researchers saw an interference pattern instead of two
bright spots, providing reasonably conclusive evidence that light behaves like a wave.
Light as a Particle
Although the results of the double-slit experiment strongly suggested that light is a wave, a German physicist named
Max Planck found experimental results that suggested light behaved more like a particle when he was studying black
body radiation. A black body is a theoretical object that absorbs all light that falls on it. It reflects no radiation and
appears perfectly black. Black body radiation is the energy that would be emitted from an ideal black body. In
the year 1900, Planck published a paper on the electromagnetic radiation emitted from a black object that had been
heated. In trying to explain the black body radiation, Planck determined that the experimental results could not
be explained with the wave form of light. Instead, Planck described the radiation emission as discrete bundles of
energy, which he called quanta. A quantum (singular form of quanta) is a small unit into which certain forms of
energy are divided. These “discrete bundles of energy” once again raised the question of whether light was a wave
or a particle – a question once thought settled by Maxwell’s work. Planck’s work also pointed out that the energy
of a quantum of light was related only to its frequency. Planck’s equation for calculating the energy of a bundle of
light is E = h f , where E is the energy of the photon in joules (J), f is the frequency in hertz (s−1 ), and h is Planck’s
constant, 6.63 × 10−34 J · s. (A photon is a particle of light. The word quantum is used for energy in any form; when
the type of energy under discussion is light, the words quantum and photon become interchangeable.)
Example:
What is the frequency of a photon of light whose energy is 3.00 × 10−19 joules?
3.00×10−19 J
14 Hz
f = Eh = 6.63×10
−34 J·s = 4.52 × 10
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Chapter 11. The Quantum Mechanical Model of the Atom
Example:
What is the energy of a photon whose frequency is 2.00 × 1015 s−1 ?
E = h f = (6.63 × 10−34 J · s) · (2.00 × 1015 s−1 ) = 1.33 × 10−18 J
Planck’s work became the basis for quantum theory. Quantum theory is the theory that energy can only exist
in discrete amounts (quanta). For example, we assume that we can cause an automobile to travel any speed we
choose. Quantum theory says this is not true. The problem involved in demonstrating this theory is that the scale of
a quantum of energy is much smaller than the objects we normally deal with. Imagine having a large delivery truck
sitting on a scale. If we throw one more molecule onto the truck, can we expect to see the weight change on the
scale? We cannot, because we lack instruments that can detect such a small change. Even if we added a thousand
molecules to the truck, we still would not see a difference in the truck’s weight.
For the same reason, we cannot tell that an automobile’s speed is quantized (in discrete amounts). The addition
of one quantum of kinetic energy to an automobile might change its velocity from 30.1111111111 miles per hour
to 30.1111111122 miles per hour. Therefore, according to quantum theory, a speed of 30.1111111117 mph is not
possible. (Note that these numbers are used for illustrative purposes only.) We are not able to detect this change
because we can’t measure speeds that finely. To test this theory, we must look at objects that are very tiny in order
to detect a change in one quantum.
One place where we can measure quantum-sized energy changes is in the internal vibration of molecules, or the
stretching and contracting of bond lengths. When the internal vibration of molecules is measured in the laboratory,
it is found that the vibration motion is stair-stepped. A particular molecule may be found vibrating at 3 cycles per
second 6 cycles per second or 9 cycles per second, but those molecules are never found vibrating at 1, 2, 4, 5, 7, or 8
cycles per second. (Again, these numbers are used for illustrative purposes only.) The fact that only certain vibration
levels are available to molecules is strong support for the quantum theory.
Quantum theory can also be used to explain the result of this next experiment on light called the photoelectric
effect. The photoelectric effect is a phenomenon in which electrons are emitted from the surface of a material after
the absorption of energy. This experiment involves having light strike a metal surface with enough force to knock
electrons off the metal surface. The results of the photoelectric effect indicated that if the experimenter used low
frequency light, such as red, no electrons were knocked off the metal. No matter how many light waves were used
and no matter how long the light was shined on the metal, red light could not knock off any electrons. If a higher
frequency light was used, such as blue light, then many electrons were knocked off the metal. Albert Einstein used
Planck’s quantum theory to provide the explanation for the photoelectric effect. A certain amount of energy was
necessary for electrons to be knocked off a metal surface. If light were quantized, then only particles of higher
frequency light (and therefore higher energy) would have enough energy to remove an electron. Light particles of
lower frequency (and therefore lower energy) could never remove any electrons, regardless of how many of them
were used.
As a historical side note, many people may think that Einstein won the Nobel Prize for his theory of relativity, but in
fact Einstein’s only Nobel Prize was for his explanation of the photoelectric effect.
Wave-Particle Duality
At this point, scientists had some experimental evidence (diffraction and interference) that indicated light was a
wave and other experimental evidence (black body radiation and the photoelectric effect) that indicated light was a
particle. The solution to this problem was to develop a concept known as the wave-particle duality of light. The
point of this concept is that light travels as a wave and interacts with matter like a particle. Thus when light is
traveling through space, air, or other media, we speak of its wavelength and frequency, and when the light interacts
with matter, we switch to the characteristics of a particle (quantum).
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Lesson Summary
•
•
•
•
The work of many scientists led to an understanding of the wave-particle duality of light.
Light has properties of waves and particles.
Some characteristics of energy waveforms are that they will undergo diffraction and interference.
The energy and frequency of a light photon are related by the equation E = h f .
Further Reading / Supplemental Links
This website describes the double-slit experience and provides a simulation of the double-slit experiment.
• http://www.colorado.edu/physics/2000/schroedinger/two-slit2.html
Review Questions
1. Name a phenomenon that supports the concept that light is a wave.
2. Name a phenomenon that supports the concept that light is a packet of energy.
3. Calculate the energy in joules of a photon whose frequency is 7.55 × 1014 Hz.
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Chapter 11. The Quantum Mechanical Model of the Atom
11.2 Characteristics of Matter
Lesson Objectives
The student will:
• describe a standing wave.
• state the Heisenberg uncertainty principle.
Vocabulary
• Heisenberg uncertainty principle
Wave Character of Particles
In 1924, the Frenchman Louis de Broglie, a physics graduate student at the time, suggested that if waves can
have particle-like properties as hypothesized by Planck, then perhaps particles can have some wave-like properties.
This concept received some experimental support in 1937 when investigators demonstrated that electrons could
produce diffraction patterns. (All objects, including baseballs and automobiles could be considered to have wavelike properties, but this concept is only measurable when dealing with extremely small particles like electrons.) De
Broglie’s “matter waves” would become very useful in attempts to describe the behavior of electrons inside atoms.
Standing Waves
In the chapter “The Bohr Model of the Atom,” we considered a rope wave that was created by tying one end of
the rope to a tree and by jerking the other end up and down. When a wave travels down a rope and encounters
an immovable boundary, the wave reflects off the boundary and travels back up the rope. This causes interference
between the wave traveling toward the tree and the reflected wave traveling back toward the person. If the person
moving the rope up and down adjusts the rhythm just right, the crests and troughs of the wave moving toward the
tree will coincide exactly with the crests and troughs of the reflected wave. When this occurs, the apparent horizontal
motion of the crests and troughs along the rope will cease. This is called a standing wave. In such a case, the crests
and troughs will remain in the exact same place, while the nodes between the crests and troughs do not appear to
move at all.
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In the standing wave shown above, the positions of the crests and troughs remain in the same positions. The crests
and troughs will only appear to exchange places above and below the rope. The places where the rope does not
cross the center axis line are called nodes (positions of zero displacement). These nodal positions do not change and
appear to be frozen in place. By combining the concept of a standing wave along with de Broglie’s matter waves, it
became possible to describe an electron in an electron cloud as either a particle or a standing wave.
The Heisenberg Uncertainty Principle
In all previous attempts to describe the electron’s behavior inside an atom, including in the Bohr model, scientists
tried to describe the path the electron would follow around the nucleus. The theorists wanted to describe where the
electron was located and how it would move from that position to its next position.
In 1927, a German physicist named Werner Heisenberg, a German physicist stated what is now known as the
Heisenberg uncertainty principle. This principle states that it is impossible to know both the precise location
and the precise velocity of an electron at the same time. The reason that we can’t determine both is because the act
of determining the location changes the velocity. In the process of making a measurement, we have actually changed
the measurement.
This problem is present in all laboratory work, but it is usually negligible. Consider the act of measuring the
temperature of hot water in a beaker. When you insert the thermometer into the water, the water transfers heat to the
thermometer until the thermometer is at the same temperature as the water. You can then read the temperature of the
water from the thermometer. The temperature of the water, however, is no longer the same as before you inserted
the thermometer. The water has cooled by transferring some of its heat to the thermometer. In other words, the act
of making the measurement changes the measurement. In this example, the difference is most likely not significant.
You can imagine, however, that if the mass of water was very small and the thermometer was very large, the water
would have to transfer a greater amount of heat to the thermometer, resulting in a less accurate measurement.
Consider the method that humans use to see objects. We see an object when photons bounce off the object and into
our eye or other light-measuring instrument. Recall that photons can have various wavelengths, which correspond to
different colors. If only red photons bounce back, we say the object is red. If no photons bounce back, we don’t see
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Chapter 11. The Quantum Mechanical Model of the Atom
the object. Suppose for a moment that humans were gigantic stone creatures that use golf balls, instead of photons,
to see. In other words, we see objects when the golf balls bounce off them and enter our eyes. We would be able to
see large objects like buildings and mountains successfully, because the golf balls would bounce off and reach our
eyes. Could we see something small like a butterfly with this technique? The answer is no. A golf ball has a greater
mass than a butterfly, so when the golf ball bounces off the butterfly, the motion of the butterfly will be very different
after the collision. We will know the position of the butterfly, but we won’t know the motion of the butterfly.
In the case of electrons, the photons we use to see them with are of significant energy compared to electrons and
will change the motion of the electrons upon collision. We may be able to detect the position of the electron, but its
motion is no longer the same as before the observation. The Heisenberg uncertainty principle tells us we cannot be
sure of both the location and the motion of an electron at the same time. As a result, we must give up on the idea of
determining the path an electron follows inside an atom.
Schrödinger’s Equation
The Heisenberg uncertainty principle treated the electron as a particle. In effect, the uncertainty principle stated that
the exact motion of an electron in an atom could never be determined, which also meant that the exact structure
of the atom could not be determined. Consequently, Erwin Schrödinger, an Austrian physicist, decided to treat the
electron as a wave in accordance with de Broglie’s matter waves.
Schrödinger, in considering the electron as a wave, developed an equation to describe the electron wave behavior
in three dimensions (shown below). Unfortunately, the equation is so complex that it is actually impossible to
solve exactly for atoms and ions that contain more than one electron. High-speed computers, however, can produce
very, very close approximations, and these “solutions” have provided a great deal of information about the possible
organization of electrons within an electron cloud.
When we represent electrons inside an atom, quantum mechanics requires that the wave must “fit” inside the atom
so that the wave meets itself with no overlap. In other words, the “electron wave” inside the atom must be a standing
wave. If the wave is to be arranged in the form of a circle so that it attaches to itself, the waves can only occur if
there is a whole number of waves in the circle. Consider the image below.
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On the left is an example of a standing wave. For the wave on the right, the two ends of the wave do not quite
meet each other, so the wave fails to be a standing wave. There are only certain energies (frequencies) for which
the wavelength of the wave will fit exactly to form a standing wave. These energies turn out to be the same as the
energy levels predicted by the Bohr model, but now there is a reason why electrons may only occupy these energy
levels. (Recall that one of the problems with the Bohr model was that Bohr had no explanation for why the electrons
could only occupy certain energy levels in the electron cloud.) The equations of quantum mechanics tell us about the
existence of principal energy levels, the number of energy levels in any atom, and more detailed information about
the various energy levels.
Lesson Summary
• The Heisenberg uncertainty principle states that it is impossible to know both the precise location and the
precise motion of an electron at the same time.
• Electrons in an electron cloud can be viewed as a standing wave.
• The reason that an electron in an atom may have only certain energy levels is because only certain energies of
electrons will form standing waves in the enclosed volume.
• The solutions to Schröedinger’s equation provide a great deal of information about the organization of the
electrons in the electron cloud.
Further Reading / Supplemental Links
A question and answer session on electrons behaving as waves.
• http://www.colorado.edu/physics/2000/quantumzone/debroglie.html
Review Questions
1. Which of the following statements are true?
(a) According to the Heisenberg uncertainty principle, we will eventually be able to measure both an
electron’s exact position and its exact momentum at the same time.
(b) The problem that we have when we try to measure an electron’s exact position and its exact momentum
at the same time is that our measuring equipment is not good enough.
(c) According to the Heisenberg uncertainty principle, we cannot know both the exact location and the exact
momentum of an automobile at the same time.
(d) The Heisenberg uncertainty principle applies only to very small objects like protons and electrons.
(e) The Heisenberg uncertainty principle applies only to large objects like cars and airplanes.
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Chapter 11. The Quantum Mechanical Model of the Atom
(f) The Heisenberg Heisenberg uncertainty principle applies to very small objects like protons and electrons
and to large objects like cars and airplanes.
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11.3 Quantum Numbers, Orbitals, and Probability Patterns
Lesson Objectives
The student will:
• state the relationship between the principal quantum number (n), the number of orbitals, and the maximum
number of electrons in a principal energy level.
Vocabulary
• Pauli exclusion principle
• principal quantum number
• quantum number
Introduction
Erwin Schrödinger proposed a wave equation for electron matter waves that was similar to the known equations
for other wave motions in nature. This equation describes how a wave associated with an electron varies in space
as the electron moves under various forces. Schrödinger worked out the solutions of his equation for the hydrogen
atom, and the results agreed perfectly with the known energy levels for hydrogen. Furthermore, the equation could
be applied to more complicated atoms. It was found that Schrodinger’s equation gave a correct description of an
electron’s behavior in almost every case. In spite of the overwhelming success of the wave equation in describing
electron energies, the very meaning of the waves was vague and unclear.
There are very few scientists who can visualize the behavior of an electron as a standing wave during chemical
bonding or chemical reactions. When chemists are asked to describe the behavior of an electron in an electrochemical
cell, they do not use the mathematical equations of quantum mechanics, nor do they discuss standing waves. The
behavior of electrons in chemical reactions is best understood by considering the electrons to be particles.
A physicist named Max Born was able to attach some physical significance to the mathematics of quantum mechanics. Born used data from Schrodinger’s equation to show the probability of finding the electron (as a particle)
at the point in space for which the equation was solved. Born’s ideas allowed chemists to visualize the results of
Schrodinger’s wave equation as probability patterns for electron positions.
Probability Patterns
Suppose we had a camera with such a fast shutter speed that it could capture the position of an electron at any given
moment. We could take a thousand pictures of this electron at different times and find it at many different positions
in the atom. We could then plot all the electron positions onto one picture, as seen in the sketch below.
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One way of looking at this picture is as an indication of the probability of where you are likely to find the electron
in this atom. Keep in mind that this image represents an atom with a single electron. The dots do not represent
different electrons; the dots are positions where the single electron can be found at different times. From this image,
it is clear that the electron spends more time near the nucleus than it does far away. As you move away from the
nucleus, the probability of finding the electron becomes less and less. It is important to note that there is no boundary
in this picture. In other words, there is no distance from the nucleus where the probability of finding an electron
becomes zero. However, for much of the work we will be doing with atoms, it is convenient (even necessary) to have
a boundary for the atom. Most often, chemists arbitrarily draw in a boundary for the atom, choosing some distance
from the nucleus beyond which the probability of finding the electron becomes very low. Frequently, the boundary
is placed such that 90% of the probability of finding the electron is inside the boundary.
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The image above shows boundaries drawn in at 50%, 90%, and 95% probability of finding the electron within the
boundary. It is important to remember that the boundary is there for our convenience, and there is no actual boundary
on an atom. This probability plot is very simple because it is for the first electron in an atom. As the atoms become
more complicated (more energy levels and more electrons), the probability plots also become more complicated.
All of the scientists whose names appear in the "Atom Song" have appeared in our book. Please watch the video:
http://www.youtube.com/watch?v=vUzTQWn-wfE (3:28).
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The Principal Quantum Number
Solutions to Schrödinger’s equation involve four special numbers called quantum numbers. (Three of the numbers
come from Schrödinger’s equation, and the fourth one comes from an extension of the theory.) These four numbers
completely describe the energy of an electron. Each electron has exactly four quantum numbers, and no two electrons
have the same four numbers. The statement that no two electrons can have the same four quantum numbers is known
as the Pauli exclusion principle.
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Chapter 11. The Quantum Mechanical Model of the Atom
The principal quantum number is a positive integer (1, 2, 3, . . . n) that indicates the main energy level of an
electron within an atom. According to quantum mechanics, every principal energy level has one or more sub-levels
within it. The number of sub-levels in a given energy level is equal to the number assigned to that energy level. That
is, principal energy level 1 will have 1 sub-level, principal energy level 2 will have two sub-levels, principal energy
level 3 will have three sub-levels, and so on. In any energy level, the maximum number of electrons possible is 2n2 .
Therefore, the maximum number of electrons that can occupy the first energy level is 2 (2 · 12 ). For energy level 2,
the maximum number of electrons is 8 (2 · 22 ), and for the 3rd energy level, the maximum number of electrons is 18
(2 · 32 ). Table 11.1 lists the number of sub-levels and electrons for the first four principal quantum numbers.
TABLE 11.1: Number of Sub-levels and Electrons by Principal Quantum Number
Principal Quantum Number
1
2
3
4
Number of Sub-Levels
1
2
3
4
Total Number of Electrons
2
8
18
32
The largest known atom contains slightly more than 100 electrons. Quantum mechanics sets no limit as to how
many energy levels exist, but no more than 7 principal energy levels are needed to describe the electrons of all
known atoms. Each energy level can have as many sub-levels as the principal quantum number, as discussed above,
and each sub-level is identified by a letter. Beginning with the lowest energy sub-level, the sub-levels are identified
by the letters s, p, d, f, g, h, i, and so on. Every energy level will have an s sub-level, but only energy levels 2
and above will have p sub-levels. Similarly, d sub-levels occur in energy level 3 and above, and f sub-levels occur
in energy level 4 and above. Energy level 5 could have a fifth sub-energy level named g, but all the known atoms
can have their electrons described without ever using the g sub-level. Therefore, we often say there are only four
sub-energy levels, although theoretically there can be more than four sub-levels. The principal energy levels and
sub-levels are shown in the following diagram. The principal energy levels and sub-levels that we use to describe
electrons are in red.
Orbitals
Quantum mechanics also tells us how many orbitals are in each sub-level. In Bohr’s model, an orbit was a circular
path that the electron followed around the nucleus. In quantum mechanics, an orbital is defined as an area in the
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electron cloud where the probability of finding the electron is high. The number of orbitals in an energy level is
equal to the square of the principal quantum number. Hence, energy level 1 will have 1 orbital (12 ), energy level 2
will have 4 orbitals (22 ), energy level 3 will have 9 orbitals (32 ), and energy level 4 will have 16 orbitals (42 ).
The s sub-level has only one orbital. Each of the p sub-levels has three orbitals. The d sub-levels have five orbitals,
and the f sub-levels have seven orbitals. If we wished to assign the number of orbitals to the unused sub-levels,
g would have nine orbitals and h would have eleven. You might note that the number of orbitals in the sub-levels
increases by odd numbers (1, 3, 5, 7, 9, 11, . . .).
As a result, the single orbital in energy level 1 is the s orbital. The four orbitals in energy level 2 are a single 2s
orbital and three 2p orbitals. The nine orbitals in energy level 3 are a single 3s orbital, three 3p orbitals, and five 3d
orbitals. The sixteen orbitals in energy level 4 are a the single 4s orbital, three 4p orbitals, five 4d orbitals, and seven
4f orbitals.
The chart above shows the relationship between n (the principal quantum number), the number of orbitals, and the
maximum number of electrons in a principal energy level. Theoretically, the number of orbitals and number of
electrons continue to increase for higher values of n. However, no atom actually has more than 32 electron in any of
its principal levels.
Each orbital will also have a probability pattern that is determined by interpreting Schrödinger’s equation. Earlier,
we showed that the probability pattern for an atom with a single electron is a circle. The illustration, however, is
2-dimensional. The real 3-dimensional probability pattern for the single orbital in the s sub-level is actually a sphere.
The probability patterns for the three orbitals in the p sub-levels are shown below. The three images on the left show
the probability pattern for the three p orbitals in each of the three dimensions. On the far right is an image of all
three p orbitals together. These p orbitals are said to be shaped like dumbbells (named after the objects weight lifters
use), water wings (named after the floating balloons young children use in the swimming pool), and various other
objects.
The probability patterns for the five d orbitals are more complicated and are shown below.
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Chapter 11. The Quantum Mechanical Model of the Atom
The seven f orbitals shown below are even more complicated.
You should keep in mind that no matter how complicated the probability pattern is, each shape represents a single
orbital, and the entire probability pattern is the result of the various positions that either one or two electrons can
take.
A video discussing the relationship between spectral lines and electron transitions is available at (1j) http://www.y
outube.com/watch?v=fKYso97eJs4 (3:49).
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A short animation of s and p orbitals is available on youtube.com at http://www.youtube.com/watch?v=VfBcfYR1V
Qo (1:20).
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Another example of s, p and d electron orbitals is available also on youtube.com at http://www.youtube.com/watch
?v=K-jNgq16jEY (1:37).
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Lesson Summary
• Solutions to Schrodinger’s equation involve four special numbers called quantum numbers, which completely
describe the energy of an electron.
• Each electron has exactly four quantum numbers.
• According to the Pauli Exclusion Principle, no two electrons have the same four quantum numbers.
• The major energy levels are numbered by positive integers (1, 2, 3, . . . , n), and this number is called the
principal quantum number
• Quantum mechanics also tells us how many orbitals are in each sub-level.
• In quantum mechanics, an orbital is defined as an area in the electron cloud where the probability of finding
the electron is high.
Further Reading / Supplemental Links
The following is a video on the quantum mechanical model of the atom.
• http://www.youtube.com/watch?v=IsA_oIXdF_8&feature=related
This video is a ChemStudy film called “Hydrogen Atom and Quantum Mechanics.” The film is somewhat dated but
the information is accurate. The video also contains some data supporting quantum theory.
• http://www.youtube.com/watch?v=80ZPe80fM9U
Review Questions
1.
2.
3.
4.
How many sub-levels may be present in principal energy level 3 (n = 3)?
How many sub-levels may be present in principal energy level 6 (n = 6)?
Describe the difference in the definitions of a Bohr orbital and a quantum mechanics orbital.
What is the maximum total number of electrons that can be present in an atom having three principal energy
levels?
In the first image of this chapter, the photograph showing water waves diffracting through a gap between small rocks
is from http://www.flickr.com/photos/framesof mind/554402976/ (CC-BY-SA).
All images, unless otherwise stated, are created by the CK-12 Foundation and are under the Creative Commons
license CC-BY-NC-SA.
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