Ch22 Nuclear Chemistry

Prévia do material em texto

949
Nuclear Chemistry
When methane burns in oxygen,
the C, H, and O atoms recombine to
yield and , but they still
remain C, H, and O. When aqueous
NaCl reacts with aqueous 
solid AgCl precipitates from solu-
tion, but the and ions
themselves remain the same. In
fact, in all the reactions
we’ve discussed up to this point,
only the bonds between atoms have
changed; the chemical identities of the
atoms themselves have remained
unchanged. But anyone who reads the paper or watches 
Cl�Ag�
AgNO3,
H2OCO2
(CH4)
22
22.7 Nuclear Transmutation
22.8 Detecting and Measuring
Radioactivity
22.9 Biological Effects of Radiation
22.10 Applications of Nuclear
Chemistry
� Interlude—The Origin of Chemical
Elements
22.1 Nuclear Reactions and Their
Characteristics
22.2 Nuclear Reactions and
Radioactivity
22.3 Radioactive Decay Rates
22.4 Nuclear Stability
22.5 Energy Changes During Nuclear
Reactions
22.6 Nuclear Fission and Fusion
C O N T E N T S
C h a p t e r
� More than 250 ships on
the world's oceans are
powered by nuclear
reactors, including this
Russian icebreaker.
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950 Chapter 22 Nuclear Chemistry
Carbon-12
 C
Mass number
Atomic number
 6 protons
 6 neutrons
12 nucleons
6
12
television knows that atoms can change, often resulting in the conversion of one ele-
ment into another. Atomic weapons, nuclear energy, and radioactive radon gas in
our homes are all topics of societal importance, and all involve nuclear chemistry—
the study of the properties and reactions of atomic nuclei.
22.1 Nuclear Reactions and Their Characteristics
Recall from Section 2.5 that an atom is characterized by its atomic number, Z,
and its mass number, A. The atomic number, written as a subscript to the left of
the element symbol, gives the number of protons in the nucleus. The mass
number, written as a superscript to the left of the element symbol, gives the
total number of nucleons, a general term for both protons (p) and neutrons (n).
The most common isotope of carbon, for example, has 12 nucleons: 6 protons
and 6 neutrons.
Atoms with identical atomic numbers but different mass numbers are called
isotopes, and the nucleus of a specific isotope is called a nuclide. There are 13
known isotopes of carbon, two of which occur commonly ( and ) and one of
which is produced in small amounts in the upper atmosphere by the action
of neutrons from cosmic rays on The remaining 10 carbon isotopes have been
produced artificially. Only the two commonly occurring ones are indefinitely sta-
ble; the other 11 undergo spontaneous nuclear reactions, which change their
nuclei. Carbon-14, for example, slowly decomposes to give nitrogen-14 plus an
electron, a process we can write as
The electron is often written as where the superscript 0 indicates that the
mass of an electron is essentially zero when compared to that of a proton or neu-
tron, and the subscript indicates that the charge is (The subscript in this
instance is not a true atomic number.)
Nuclear reactions, such as the spontaneous decay of are distinguished
from chemical reactions in several ways:
• A nuclear reaction involves a change in an atom’s nucleus, usually producing
a different element. A chemical reaction, by contrast, involves only a change in
distribution of the outer-shell electrons around the atom and never changes
the nucleus itself or produces a different element.
• Different isotopes of an element have essentially the same behavior in
chemical reactions but often have completely different behavior in nuclear
reactions.
• The rate of a nuclear reaction is unaffected by a change in temperature or pres-
sure or by the addition of a catalyst.
• The nuclear reaction of an atom is essentially the same, regardless of whether
the atom is in a chemical compound or in uncombined elemental form.
• The energy change accompanying a nuclear reaction is far greater than that
accompanying a chemical reaction. The nuclear transformation of 1.0 g of
uranium-235 releases for example, whereas the chemical com-
bustion of 1.0 g of methane releases only 56 kJ.
8.2 * 107 kJ,
14C,
-1.-1
�1
 0e,
 6
14C ¡
 7
14N + �1 0e
14N.
114C2
13C12C
Enrique A. Hughes and Anita
Zalts, “Radioactivity in the
Classroom,” J. Chem. Educ., Vol. 77,
2000, 613–614.
“Nuclear Chemistry: State of
the Art for Teachers,” a series
of articles in the October 1994 issue
of the Journal of Chemical Education.
S. G. Hutchinson and F. I.
Hutchinson, “Radioactivity: in
Everyday Life,” J. Chem. Educ., Vol.
74, 1997, 501–505.
Charles H. Atwood, “Teach-
ing Aids for Nuclear Chem-
istry,” J. Chem. Educ., Vol. 71, 1994,
845–847.
Energy changes for nuclear
reactions are on the order of 
a million times greater than
energy changes for chemical
reactions.
Carbon Isotopes
activity
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22.2 Nuclear Reactions and Radioactivity 951
+
−
β rays
γ rays
α rays
U
2 protons
2 neutrons
4 nucleons
 90 protons
144 neutrons
234 nucleons 92 protons
146 neutrons
238 nucleons
He + Th23892 2
4 234
90
� FIGURE 22.1 The effect of
an electric field on and 
radiation. The radioactive
source in the shielded box
emits radiation, which passes
between two electrodes. Alpha
radiation is deflected toward
the negative electrode, radia-
tion is strongly deflected
toward the positive electrode,
and radiation is undeflected.g
b
ga, b,
22.2 Nuclear Reactions and Radioactivity
Scientists have known since 1896 that many nuclides are radioactive—that is, they
spontaneously emit radiation. Early studies of radioactive nuclei, or radionuclides,
by the New Zealand physicist Ernest Rutherford in 1897 showed that there are three
common types of radiation with markedly different properties: 
and named after the first three letters of the Greek alphabet.
Alpha Radiation
Using the simple experiment shown in Figure 22.1, Rutherford found that radi-
ation consists of a stream of particles that are repelled by a positively charged
electrode, are attracted by a negatively charged electrode, and have a mass-to-
charge ratio identifying them as helium nuclei, Alpha particles thus consist
of two protons and two neutrons.
2
4He2�.
A
(A)
gamma 1g2 radiation, alpha 1a2, beta 1b2,
Because the emission of an particle from a nucleus results in a loss of two
protons and two neutrons, it reduces the mass number of the nucleus by 4 and
reduces the atomic number by 2. Alpha emission is particularly common for
heavy radioactive isotopes, or radioisotopes: Uranium-238, for example, sponta-
neously emits an particle and forms thorium-234.a
a
C. Ronneau, “Radioactivity: A
Natural Phenomenon,” J.
Chem. Educ., Vol. 67, 1990, 736–737.
An alpha particle
is a helium nucleus,
but the charge is not used
in writing nuclear equations.
+2
124He2�2
1a2
For a nuclear equation to be
balanced: sum of atomic
numbers of reactants
of atomic
numbers of products, and sum 
of nucleons of reactants
of nucleons
of products. Ionic charges are
ignored.
1superscripts2 = sum
1subscripts2 = sum
Note how the nuclear equation for the radioactive decay of uranium-238 is
written. The equation is not balanced in the usual chemical sense because the kinds
of nuclei are not the same on both sides of the arrow. Instead, a nuclear equation is
balanced when the sums of the nucleons are the same on both sides of the equation
and when the sums of the charges on the nuclei and any elementary particles (pro-
tons, neutrons, and electrons) are the same on both sides. In the decay of to
give and for example, there are 238 nucleons and 92 nuclear charges on
both sides of the nuclear equation.
Note also that we are concernedonly with charges on elementary particles and
on nuclei when we write nuclear equations, not with ionic charges on atoms. The 
particle is thus written as rather than as and the thorium resulting from
radioactive decay of is written as rather than as We ignore the
ionic charges both because they are irrelevant to nuclear disintegration and
because they soon disappear. The ion immediately picks up two electrons
from whatever it strikes, yielding a neutral helium atom, and the ion
immediately gives two electrons to whatever it is in electrical contact with, yielding
a neutral thorium atom.
 90
234Th2�
2
4He2�
 90
234Th2�.
 90
234Th
 92
238U
2
4He2�,2
4He
a
 90
234Th,2
4He
 92
238U
Separation of Alpha,
Beta, and Gamma
Rays movie
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952 Chapter 22 Nuclear Chemistry
Beta Radiation
Further work by Rutherford in the late 1800s showed that radiation consists of
a stream of particles that are attracted to a positive electrode (Figure 22.1), are
repelled by a negative electrode, and have a mass-to-charge ratio identifying them
as electrons, or Beta emission occurs when a neutron in the nucleus spon-
taneously decays into a proton plus an electron, which is then ejected. The product
nucleus has the same mass number as the starting nucleus because a neutron has
turned into a proton, but it has a higher atomic number because of the newly cre-
ated proton. The reaction of to give is an example:131Xe131I
b�.�1 0e
B
(B)
Writing the emitted particle as in the nuclear equation makes clear the
charge balance of the nuclear reaction: The subscript in the nucleus on the left
(53) is balanced by the sum of the two subscripts on the right 
Gamma Radiation
Gamma radiation is unaffected by electric fields (Figure 22.1), has no mass, and is
simply electromagnetic radiation of very high energy and thus of very short
wavelength In modern terms, we would say that radia-
tion consists of a stream of high-energy photons. Gamma radiation almost always
accompanies and emission as a mechanism for the release of energy, but it is
often not shown when writing nuclear equations because it changes neither the
mass number nor the atomic number of the product nucleus.
Positron Emission and Electron Capture
In addition to and radiation, two other common types of radioactive
decay processes also occur: positron emission and electron capture. Positron
emission occurs with conversion of a proton in the nucleus into a neutron plus
an ejected positron, or a particle that can be thought of as a “positive elec-
tron.” A positron has the same mass as an electron but an opposite charge. The
result of positron emission is a decrease in the atomic number of the product
nucleus but no change in the mass number. Potassium-40, for example, under-
goes positron emission to yield argon-40, a nuclear reaction important in geol-
ogy for dating rocks. Note once again that the sum of the two subscripts on the
right of the nuclear equation is equal to the subscript in the 
nucleus on the left.
19
40K118 + 1 = 192
b�,1
0e
ga, b,
ba
G1l = 10�11–10�14 m2.
(G)
154 - 1 = 532. 53
131I
�1
 0eb
Robert Suder, “Beta Decay
Diagram,” J. Chem. Educ., Vol.
66, 1989, 231.
A beta particle ( or )
is an electron.
b��1
 0e1b2
For the purpose of balancing
nuclear equations, the mass
of the electron is taken to be 0
and its “nuclear charge,” is taken to
be -1.
I13153
 54 protons
 77 neutrons
131 nucleons 0 nucleons
but −1 charge
 53 protons
 78 neutrons
131 nucleons
Xe + e054
131
−1
K
18 protons
22 neutrons
40 nucleons 0 nucleons
but +1 charge
19 protons
21 neutrons
40 nucleons
Ar + e01
40
18
40
19
Electron capture is a process in which the nucleus captures an inner-shell elec-
tron, thereby converting a proton into a neutron. The mass number of the product
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22.2 Nuclear Reactions and Radioactivity 953
nucleus is unchanged, but the atomic number decreases by 1, just as in positron
emission. The conversion of mercury-197 into gold-197 is an example:
Characteristics of the different kinds of radioactive decay processes are sum-
marized in Table 22.1.
Electron capture: 
core electron : neutron.
proton +
TABLE 22.1 A Summary of Radioactive Decay Processes
Change in Change in Change in 
Atomic Mass Neutron 
Process Symbol Number Number Number
Alpha emission or 
Beta emission or 0
Gamma emission or 0 0 0
Positron emission or 0
Electron capture E. C. 0 +1-1
+1-1b�10e
g0
0g
-1+1b��1 0e
-2-4-2a24He
 79 protons
118 neutrons
197 nucleons
Inner-shell
electron 80 protons
117 neutrons
197 nucleons
Au19779Hg + −1
0e19780
WORKED EXAMPLE 22.1
Write balanced nuclear equations for each of the following processes:
(a) Alpha emission from curium-242: 
(b) Beta emission from magnesium-28: 
(c) Positron emission from xenon-118: 
STRATEGY
The key to writing nuclear equations is to make sure that the number of nucleons is the
same on both sides of the equation and that the number of elementary and nuclear
charges is the same.
SOLUTION
(a) In emission, the mass number decreases by 4, and the atomic number decreases by
2, giving plutonium-238: 
(b) In emission, the mass number is unchanged, and the atomic number increases
by 1, giving aluminum-28: 
(c) In positron emission, the mass number is unchanged, and the atomic number
decreases by 1, giving iodine-118: 
� PROBLEM 22.1 Write balanced nuclear equations for each of the following
processes:
(a) Beta emission from ruthenium-106: 
(b) Alpha emission from bismuth-189: 
(c) Electron capture by polonium-204: 
� PROBLEM 22.2 What particle is produced by decay of thorium-214 to radium-210?
 90
214Th ¡
 88
210Ra + ?
 84
204Po + �1 0e : ?
 83
189Bi : 24He + ?
 44
106Ru : �1 0e + ?
 54
118Xe : 10e + 53118I
12
28Mg : �1 0e + 1328Al
b
 96
242Cm : 24He + 94238Pu
a
 54
118Xe : 10e + ?
12
28Mg : �1 0e + ?
 96
242Cm : 24He + ?
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954 Chapter 22 Nuclear Chemistry
22.3 Radioactive Decay Rates
The rates of different radioactive decay processes vary enormously. Some
radionuclides, such as uranium-238, decay at a barely perceptible rate over bil-
lions of years; others, such as carbon-17, decay within milliseconds.
Radioactive decay is kinetically a first-order process (Section 12.4), whose rate
is proportional to the number of radioactive nuclei N in a sample times the first-
order rate constant k, called the decay constant:
As we saw in Section 12.4, a first-order rate law can be converted into an inte-
grated rate law of the form
where is the number of radioactive nuclei originally present in a sample and N
is the number remaining at time t. The decay constant k has units of so that
the quantity kt is unitless.
Like all first-order processes, radioactive decay is characterized by a half-life,
the time required for the number of radioactive nuclei in a sample to drop to
half its initial value (Section 12.5). For example, the half-life of iodine-131, a
radioisotope used in thyroid testing, is 8.02 days. If today you have 1.000 g of 
then 8.02 days from now you will have only 0.500 g of remaining because one-
half of the sample will have decayed (by beta emission), yielding 0.500 g of 
After 8.02 more days (16.04 total), only 0.250 g of will remain; after a further
8.02 days (24.06 total), only 0.125 g will remain; and so on. Each passage of a half-
life causes the decay of one-half of whatever sample remains, as shown graphi-
cally by the curve in Figure 22.2. The half-life is the same no matter what the size
of the sample, the temperature, or any other external condition.
t1/2 = 8.02days 53131I ¡ 54131Xe + �1 0e
 53
131I
 54
131Xe.
 53
131I
 53
131I,
t1/2,
time�1
N0
ln a N
N0
b = -kt
Decay rate = k * N
Donald J. Olbris and Judith
Herzfeld, “Nucleogenesis! A
Game with Natural Rules for
Teaching Nuclear Synthesis and
Decay,” J. Chem. Educ., Vol. 76,
1999, 349–352.
The use of the integrated
first-order rate law for
nuclear decay is identical
with its use for chemical reactions,
except that numbers of nuclei (N
and ) replace molar concentra-
tions of chemical species.
N0
100
80
60
40
20
0
0 2 4
Number of half-lives
6 8
Sa
m
pl
e 
re
m
ai
ni
ng
 (%
)
FIGURE 22.2 The decay �
of a radionuclide over time. No
matter what the value of the
half-life, 50% of the sample
remains after one half-life, 25%
remains after two half-lives,
12.5% remains after three half-
lives, and so on.
Mathematically, the value of can be calculated from the integrated rate law
by setting at time 
 so t1/2 = ln 2k and k = ln 2t1/2
 ln P 12 N0N0 Q = -kt1/2 and ln 12 = -ln 2 = -kt1/2
t1/2:N = 1/2 N0
t1/2
First-Order Process
movie
Half-Life activity
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22.3 Radioactive Decay Rates 955
These equations say that if we know the value of either the decay constant k or the
half-life we can calculate the value of the other. Furthermore, if we know the
value of we can calculate the ratio of remaining and initial amounts of
radioactive sample at any time t by substituting the expression for k into the
integrated rate law:
Worked Example 22.2 shows how to calculate a half-life from a decay constant,
and Worked Example 22.4 shows how to determine the percentage of radioactive
sample remaining at time t.
The half-lives of some useful radioisotopes are listed in Table 22.2. As you
might expect, radioisotopes used internally in medical applications have fairly
short half-lives so that they decay rapidly and don’t cause long-term health
hazards.
 then lna N
N0
b = 1-ln 22a t
t1/2
b
 Since lna N
N0
b = -kt and k = ln 2
t1/2
N/N0
t1/2,
t1/2,
� Technetium-99m, a short-lived
radioisotope used for brain scans,
is obtained by neutron bombard-
ment of molybdenum-99 and then
stored in a “molybdenum cow” in
the form of Small
amounts are removed by passing
a saline solution through the
cylinder.
MoO4 2�.
TABLE 22.2 Half-Lives of Some Useful Radioisotopes
Radioisotope Symbol Radiation Half-Life Use
Tritium 12.33 years Biochemical tracer
Carbon-14 5730 years Archaeological dating
Phosphorus-32 14.26 days Leukemia therapy
Potassium-40 Geological dating
Cobalt-60 5.27 years Cancer therapy
Technetium- 6.01 hours Brain scans
Iodine-123 13.27 hours Thyroid therapy
Uranium-235 Nuclear reactors
The m in technetium-99m stands for metastable, meaning that it undergoes emission but does not change its mass
number or atomic number.
g*
7.04 * 108 yearsa, g
 92
235U
g
 53
123I
g
 43
99m Tc99m*
b�, g27
60Co
1.28 * 109 yearsb�1940K
b�15
32P
b�
 6
14C
b�1
3H
WORKED EXAMPLE 22.2
The decay constant for sodium-24, a radioisotope used medically in blood studies, is
What is the half-life of 
STRATEGY
Half-life can be calculated from the decay constant by using the equation
SOLUTION
Substituting the values and into the equation gives
The half-life of sodium-24 is 15.0 h.
t1/2 =
0.693
4.63 * 10-2 h�1
= 15.0 h
ln 2 = 0.693k = 4.63 * 10-2 h�1
t1/2 =
ln 2
k
24Na?4.63 * 10-2 h�1.
Radioactive Decay
activity
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956 Chapter 22 Nuclear Chemistry
WORKED EXAMPLE 22.3
The half-life of radon-222, a radioactive gas of concern as a health hazard in some
homes, is 3.823 days. What is the decay constant of 
STRATEGY
A decay constant can be calculated from the half-life by using the equation
SOLUTION
Substituting the values and into the equation gives
WORKED EXAMPLE 22.4
Phosphorus-32, a radioisotope used in leukemia therapy, has a half-life of 14.26 days.
What percent of a sample remains after 35.0 days?
STRATEGY
The ratio of remaining (N) and initial amounts of a radioactive sample at time t is
given by the equations
Taking as 100%, N can then be obtained.
SOLUTION
Substituting values for t and for into the equation gives
Taking the natural antilog of then gives the ratio 
Since the initial amount of was 100%, we can set and solve for N:
After 35.0 days, 18.3% of a sample remains, and 100% has
decayed.
- 18.3% = 81.7%32P
N
100%
= 0.183 so N = 0.183 * 100% = 18.3%
N0 = 100%32P
N
N0
= antiln 1- 1.702 = 0.183
N/N0:- 1.70
lna N
N0
b = - 0.693a 35.0 days
14.26 days
b = - 1.70
t1/2
N0
lna N
N0
b = 1- ln 22a t
t1/2
b and N
N0
= antiln c lna N
N0
b d
1N02
k =
0.693
3.823 days
= 0.181 day �1
ln 2 = 0.693t1/2 = 3.823 days
k =
ln 2
t1/2
222Rn?
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22.3 Radioactive Decay Rates 957
WORKED EXAMPLE 22.5
A sample of a radioisotope used to measure the flow of gases from smokestacks,
decays initially at a rate of 34,500 disintegrations/min, but the decay rate falls to
21,500 disintegrations/min after 75.0 min. What is the half-life of 
STRATEGY
The half-life of a radioactive decay process is given by finding in the equation
In the present instance, though, we are given decay rates at two different times rather
than values of N and Nevertheless, for a first-order process like radioactive decay,
in which the ratio of the decay rate at any time t to the decay rate at time
is the same as the ratio of N to 
SOLUTION
Substituting the proper values into the equation gives
The half-life of is 110 min.41Ar
so t1/2 = -52.0 min-0.473 = 110 min
lna21,500
34,500
b = -0.693a75.0 min
t1/2
b or -0.473 = -52.0 min
t1/2
Decay rate at time t
Decay rate at time t = 0
=
kN
kN0
=
N
N0
N0:t = 0
rate = kN,
N0.
lna N
N0
b = 1- ln 22a t
t1/2
b
t1/2
41Ar?
41Ar,
� The flow of gases from a
smokestack can be measured
by releasing and monitor-
ing its passage.
41Ar
� PROBLEM 22.3 The decay constant for mercury-197, a radioisotope used med-
ically in kidney scans, is What is the half-life of mercury-197?
� PROBLEM 22.4 The half-life of carbon-14 is 5730 years. What is its decay constant?
� PROBLEM 22.5 What percentage of remains in a sample
estimated to be 16,230 years old?
� PROBLEM 22.6 What is the half-life of iron-59, a radioisotope used medically in
the diagnosis of anemia, if a sample with an initial decay rate of 16,800 disintegra-
tions/min decays at a rate of 10,860 disintegrations/min after 28.0 days?
KEY CONCEPT PROBLEM 22.7 What is the half-life of the radionuclide that
shows the following decay curve?
 6
14C 1t1/2 = 5730 years2
1.08 * 10-2 h-1.
100
80
60
40
20
0
0 5 10
Time (days)
15 20
Sa
m
pl
e 
re
m
ai
ni
ng
 (%
)
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958 Chapter 22 Nuclear Chemistry
KEY CONCEPT PROBLEM 22.8 Copper-59 (red spheres) decays to (green
spheres) with a half-life of 1.35 min. Write a balanced nuclear equation for the process,
and tell how many half-lives have passed in the following sample:
59Ni
22.4 Nuclear Stability
Why do some nuclei undergo radioactive decay while others do not? Why, for
instance, does a carbon-14 nucleus, with six protons and eight neutrons, sponta-
neously emit a particle, whereas a carbon-13 nucleus, with six protons and seven
neutrons, is stable indefinitely? Before answering these questions, it’s important
to define what we mean by “stable.” In the context of nuclear chemistry, we’ll use
the word stable to refer to isotopes whose half-lives can be measured, even if that
half-life is onlya fraction of a second. We’ll call those isotopes that decay too
rapidly for their half-lives to be measured unstable, and those isotopes that do not
undergo radioactive decay nonradioactive, or stable indefinitely.
The answer to the question about why some nuclei are radioactive while
others are not has to do with the neutron/proton ratio in the nucleus and the
forces holding the nucleus together. To see the effect of the neutron/proton ratio
on nuclear stability, look at the grid pictured in Figure 22.3. Along the side of the
grid are divisions representing the number of neutrons in nuclei, a number arbi-
trarily cut off at 200. Along the bottom of the grid are divisions representing the
number of protons in nuclei—the first 92 divisions represent the naturally occur-
ring elements, the next 17 represent the artificially produced transuranium ele-
ments, and the divisions beyond 109 represent elements about which little or
nothing is known. (Actually, only 90 of the first 92 elements occur naturally. Tech-
netium and promethium do not occur naturally because all their isotopes are
radioactive and have very short half-lives. Francium and astatine occur on Earth
only in very tiny amounts.)
b
Darleane C. Hoffman and
Diana M. Lee, “Chemistry of
the Heaviest Elements—One Ele-
ment at a Time,” J. Chem. Educ., Vol.
76, 1999, 332–347.
Transuranium elements,
those with atomic numbers
higher than uranium, do not
occur naturally but are produced
by nuclear transmutation reactions,
discussed in Section 22.7.
59Ni
59Cu
1:1 neutron/proton ratio
Band of stability
200
180
160
140
120
100
80
60
40
20
0
0 20 40 60
Number of protons (Z)
80 100 12010 30 50 70 90 110
N
um
be
r 
of
 n
eu
tr
on
s
FIGURE 22.3 The band �
of nuclear stability indicating
various neutron/proton combi-
nations that give rise to observ-
able nuclei with measurable
half-lives. Combinations out-
side the band are not stable.
The “island of stability” near
114 protons and 184 neutrons
corresponds to a group of
superheavy nuclei that are pre-
dicted to be stable. The first
member of this group was
reported in 1999.
Nuclear Stability
activity
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22.4 Nuclear Stability 959
When the more than 3600 known nuclides are plotted on the neutron/proton
grid in Figure 22.3, they fall in a curved band sometimes called the “band of
nuclear stability.” Even within the band, only 264 of the nuclides are stable indef-
initely. The remainder decay spontaneously, although their rates of decay vary
enormously. On either side of the band is a “sea of instability” representing the
large number of unstable neutron/proton combinations that have never been
seen. Particularly interesting is the “island of stability” predicted to exist for a
few superheavy nuclides near 114 protons and 184 neutrons. The first members
of this group— and —were prepared in 1999 and do
indeed seem to be unusually stable. Isotope for example, has a half-life of
30.4 seconds.
Several generalizations can be made about the data in Figure 22.3:
• Every element in the periodic table has at least one radioactive isotope.
• Hydrogen is the only element whose most abundant stable isotope con-
tains more protons (1) than neutrons (0). The most abundant stable isotopes of
other elements lighter than calcium usually have either the same
number of protons and neutrons and for example) or have
only one more neutron than protons ( and for example).
• The ratio of neutrons to protons gradually increases for elements heavier than
calcium, giving a curved appearance to the band of stability. The most abun-
dant stable isotope of bismuth, for example, has 126 neutrons and 83 protons
• All isotopes heavier than bismuth-209 are radioactive, even though they may
occur naturally.
• Of the 264 nonradioactive isotopes, 207 have an even number of neutrons in
their nuclei. Most nonradioactive isotopes (156) have even numbers of both
protons and neutrons, 51 have an even number of neutrons but an odd num-
ber of protons, and only 4 have an odd number of both protons and neutrons
(Figure 22.4).
Observations such as those in Figure 22.3 lead to the suggestion that neutrons
function as a kind of nuclear “glue” that holds nuclei together by overcoming
proton–proton repulsions. The more protons there are in the nucleus, the more
glue is needed. Furthermore, there appear to be certain “magic numbers” of pro-
tons or neutrons—2, 8, 20, 28, 50, 82, 126—that give rise to particularly stable
nuclei. For instance, there are ten naturally occurring isotopes of tin, which has a
magic number of protons but there are only two naturally occurring iso-
topes of its neighbors on either side, indium and antimony 
Lead-208 is especially stable because it has a double magic number of nucleons: 
126 neutrons and 82 protons. We’ll see in the next section how the relative stability
of different nuclei can be measured quantitatively.
The correlation of nuclear stability with special numbers of nucleons is remi-
niscent of the correlation of chemical stability with special numbers of electrons—
the octet rule discussed in Section 6.12. In fact, a shell model of nuclear structure
has been proposed, analogous to the shell model of electronic structure. The magic
numbers of nucleons correspond to filled nuclear-shell configurations, although
the details are relatively complex.
A close-up look at a segment of the band of nuclear stability reveals some
more interesting trends (Figure 22.5). One trend is that elements with an even
atomic number have a larger number of nonradioactive isotopes than do ele-
ments with an odd atomic number. Tungsten has 5 nonradioactive iso-
topes and osmium has 7, for example, while their neighbor rhenium
has only 1.1Z = 752 1Z = 762
1Z = 742
1Z = 512.1Z = 4921Z = 502,
1
 83
209Bi2.
11
23Na,
 5
11B, 
 9
19F,
14
28Si,124He, 612C, 816O,
1Z = 202
111H2
289114,
292116287114, 288114, 289114,
Protons
Neutrons
Even
Even
Odd
Odd
156
53
51
4
� FIGURE 22.4 Numbers of
nonradioactive isotopes with
various even/odd combina-
tions of neutrons and protons.
The majority of nonradioactive
isotopes have both an even
number of protons and an
even number of neutrons. Only
4 nonradioactive isotopes have
both an odd number of pro-
tons and an odd number of
neutrons.
Magic numbers for nuclei are
analogous to noble-gas elec-
tron configurations for atoms.
A nucleus with 2, 8, 20, 28, 50, 82,
or 126 protons or neutrons is par-
ticularly stable, just as an atom
having a noble-gas electron config-
uration with 2, 10, 18, 36, 54, or 86
electrons tends to be stable.
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 959
960 Chapter 22 Nuclear Chemistry
Another trend is that radioactive nuclei with higher neutron/proton ratios
(top side of the band) tend to emit particles, while nuclei with lower
neutron/proton ratios (bottom side of the band) tend to undergo nuclear decay by
positron emission, electron capture, or emission. This makes sense if you think
about it: The nuclei on the top side of the band are neutron-rich and therefore
undergo a process that decreases the neutron/proton ratio. The nuclei on the bot-
tom side of the band, by contrast, are neutron-poor and therefore undergo
processes that increase the neutron/proton ratio. (Take a minute to convince your-
self that emission does, in fact, increase the neutron/proton ratio for heavy
nuclei in which )n 7 p.
a
a
b
130
125
120
115
110
105
100
95
90
85
80
65 70 75
Number of protons (Z)
80
N
um
be
r 
of
 n
eu
tr
on
s Beta emission
Nonradioactive
Positron emission or
electron capture
Alpha emission
FIGURE 22.5 A close-up �
look at the band of nuclearsta-
bility in the region from 
(dysprosium) through 
(gold) shows the types of
radioactive processes under-
gone by various nuclides.
Nuclides with lower
neutron/proton ratios tend to
undergo positron emission,
electron capture, or emission,
whereas nuclides with higher
neutron/proton ratios tend to
undergo emission.b
a
Z = 79
Z = 66
 
These processes increase
the neutron/proton ratio:
 L Positron emission:Electron capture:Alpha emmision: Proton ¡ Neutron + b
�
Proton + Electron ¡ Neutron
Z
A X ¡ Z-2A-4 Y + 24 He
 This process decreasesthe neutron/proton ratio: eBeta emission: Neutron ¡ Proton + b�
One further point about nuclear decay is that some nuclides, particularly those
of the heavy elements above bismuth, can’t reach a nonradioactive decay product
by a single emission. The product nucleus resulting from the first decay is itself
radioactive and therefore undergoes a further disintegration. In fact, such
nuclides must often undergo a whole series of nuclear disintegrations—a decay
series—before they ultimately reach a nonradioactive product. Uranium-238, for
example, undergoes a series of 14 sequential nuclear reactions, ultimately ending
at lead-206 (Figure 22.6).
Decay Series activity
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 960
22.5 Energy Changes During Nuclear Reactions 961
148
146
144
142
140
138
136
134
132
130
128
126
124
122
80 82 84 86
Atomic number (Z)
88 9081 83 85 87 89 91 92
N
um
be
r 
of
 n
eu
tr
on
s 
(n
)
210Pb
206Pb
210Po
210Bi
214Pb
214Po
234Th
234Pa
238U
218Po
222Rn
226Ra
230Th
214Bi
234U
� FIGURE 22.6 The decay
series from to Each
nuclide except for the last is
radioactive and undergoes
nuclear decay. The left-
pointing, longer arrows (red)
represent emissions, and the
right-pointing, shorter arrows
(blue) represent emissions.b
a
 82
206Pb.
 92
238U
� PROBLEM 22.9 Of the two isotopes and one decays by emission
and one decays by emission. Which is which?
KEY CONCEPT PROBLEM 22.10 The following series has two kinds of
processes: one represented by the shorter arrows pointing right and the other repre-
sented by the longer arrows pointing left. Tell what kind of nuclear decay process each
arrow corresponds to, and identify each nuclide A–E in the series:
a
b199Au,173Au
22.5 Energy Changes During Nuclear Reactions
We said in the previous section that neutrons appear to act as a kind of nuclear
glue by overcoming the proton–proton repulsions that would otherwise cause the
nucleus to fly apart. In principle, it should be possible to measure the strength of
the forces holding a nucleus together by measuring the amount of heat released on
forming the nucleus from isolated protons and neutrons. For example, the energy
change associated with combining two neutrons and two protons to yield a
E
D
C
A
B
N
um
be
r 
of
 n
eu
tr
on
s
154
152
150
148
146
144
90 92 94
Atomic number (Z)
96 98
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 961
962 Chapter 22 Nuclear Chemistry
helium-4 nucleus should be a direct measure of nuclear stability, just as the energy
change associated with formation of a chemical bond is a measure of the bond’s
stability.
Unfortunately, there is a problem with actually carrying out the measurement:
Temperatures rivaling those in the interior of the sun are necessary for the
reaction to occur! That is, the activation energy (Section 12.10) required to force the
elementary particles close enough for reaction is extremely high. Nevertheless, the
energy change for the process can be calculated by using the Einstein equation
which relates the energy change of a nuclear process to a corre-
sponding mass change (Section 5.5).
Take a helium-4 nucleus, for example. We know from Table 2.1 that the mass of
two neutrons and two protons is 4.031 88 amu:
Furthermore, we can subtract the mass of two electrons from the experimentally
measured mass of a helium-4 atom to find that the mass of a helium-4 nucleus is
4.001 50 amu:
Subtracting the mass of the helium nucleus from the combined mass of its con-
stituent neutrons and protons shows a difference of 0.030 38 amu. That is, 0.030 38
amu (or 0.030 38 g/mol) is lost when two protons and two neutrons combine to
form a helium-4 nucleus:
The loss in mass that occurs when protons and neutrons combine to form a
nucleus is called the mass defect of the nucleus. This lost mass is converted into
energy that is released during the nuclear reaction and is thus a direct measure of
the binding energy holding the nucleons together. The larger the binding energy,
the more stable the nucleus. Using the Einstein equation, we can calculate this
binding energy for a helium-4 nucleus:
The binding energy for helium-4 nuclei is In other words,
is released when helium-4 nuclei are formed, and
must be supplied to disintegrate helium-4 nuclei into isolated
protons and neutrons. This enormous amount of energy is more than 10 million
times the energy change associated with a typical chemical process!
2.73 * 109 kJ/mol
2.73 * 109 kJ/mol
2.73 * 109 kJ/mol.
 = 2.73 * 1012 J/mol = 2.73 * 109 kJ/mol
 = 2.73 * 1012 kg # m2/1mol # s22
 = 13.038 * 10-5 kg/mol2 13.00 * 108 m/s22
 ¢E = ¢mc2 where c = 3.00 * 108 m/s
 Mass difference = 0.030 38 amu 1or 0.030 38 g/mol2
 -Mass of 4He nucleus = -4.001 50 amu
 Mass of 2 n + 2 p = 4.031 88 amu
 Mass of helium-4 nucleus = 4.001 50 amu -Mass of 2 electrons = -122 15.486 * 10-4 amu2 = -0.001 10 amu
 Mass of helium-4 atom = 4.002 60 amu
 Total mass of 2 n + 2 p = 4.031 88 amu Mass of 2 protons = 122 11.007 28 amu2 = 2.014 56 amu
 Mass of 2 neutrons = 122 11.008 66 amu2 = 2.017 32 amu
¢m
1¢E2¢E = ¢mc2,
1107 K2
¢E = ?2 11H + 2 01n ¡ 24He
When a nucleus is formed
from protons and neutrons,
some mass (mass defect) is
converted to energy (binding
energy), as related by the Einstein
equation, ¢E = ¢mc2.
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 962
22.5 Energy Changes During Nuclear Reactions 963
To make comparisons among different nuclides easier, binding energies are usu-
ally expressed on a per-nucleon basis using electron volts (eV) as the energy unit,
where and 1 million electron volts 
Thus, the helium-4 binding energy is 7.08 MeV/nucleon:
11 MeV2 = 1.60 * 10-13 J.1 eV = 1.60 * 10-19 J
 = 7.08 MeV/nucleon
 Helium-4 binding energy = a 2.73 * 1012 J/mol
6.022 * 1023 nuclei/mol
b a 1 MeV
1.60 * 10-13 J
b a 1 nucleus
4 nucleons
b
A plot of binding energy per nucleon for the most stable isotope of each ele-
ment is shown in Figure 22.7. Since a higher binding energy per nucleon corre-
sponds to greater stability, the most stable nuclei are at the top of the curve.
Iron-56, with a binding energy of 8.79 MeV/nucleon, is the most stable isotope
known. If all the mass in the universe were somehow converted to its most stable
form, the universe would become a chunk of iron.
10
8
6
4
2
0
0 20 40 60
Atomic number (Z)
80 10010 30 50 70 90
B
in
d
in
g 
en
er
gy
 (M
eV
/
nu
cl
eo
n)
Stability peak
56Fe
2H
4He 235U
� FIGURE 22.7 The binding
energy per nucleon for the
most stable isotope of each nat-
urally occurring element. Bind-
ing energy reaches a maximum
of 8.79 MeV/nucleon at 
As a result, there is an increase
in stability when much lighter
elements fuse together to yield
heavier elements up to 
and when much heavier ele-
ments split apart to yield
lighter elements down to 
as indicated by the red arrows.
56Fe,
56Fe
56Fe.
The idea that mass and energy are interconvertible is a potentially disturbing
one because it seems to overthrow two of the fundamental principles on which
chemistryis based—the law of mass conservation and the law of energy conser-
vation. In fact, what the mass/energy interconversion means is that the two indi-
vidual laws must be combined. Neither mass nor energy is conserved separately;
only the combination of the two is conserved. Every time we do a reaction,
whether nuclear or chemical, mass and energy are interconverted, but the combi-
nation of the two is conserved. For the energy changes involved in typical chemi-
cal reactions, however, the effects are so small that the mass change can’t be
detected by even the best analytical balance. Worked Example 22.7 illustrates a
mass–energy calculation for a chemical reaction.
WORKED EXAMPLE 22.6
Helium-6 is a radioactive isotope with Calculate the mass defect (in
g/mol) for the formation of a nucleus, and calculate the binding energy in
MeV/nucleon. Is a nucleus more stable or less stable than a nucleus? (The
mass of a atom is 6.018 89 amu.)
STRATEGY
Find the mass defect by subtracting the mass of the nucleus from the mass of the
constituent nucleons, and then use the Einstein equation to find the binding
energy.
6He
6He
6He
4He6He
6He
t1/2 = 0.861 s.
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 963
964 Chapter 22 Nuclear Chemistry
SOLUTION
First, calculate the total mass of the nucleons 
Next, calculate the mass of a nucleus by subtracting the mass of two electrons
from the mass of a atom:
Then subtract the mass of the nucleus from the mass of the nucleons to find the
mass defect:
Now, use the Einstein equation to convert the mass defect into binding energy:
This value can be expressed in units of MeV/nucleon by dividing by Avogadro’s num-
ber, converting to MeV, and dividing by the number of nucleons:
The binding energy of a radioactive nucleus is 4.89 MeV/nucleon, making it less
stable than a nucleus, whose binding energy is 7.08 MeV/nucleon.
WORKED EXAMPLE 22.7
What is the change in mass (in grams) when 2 mol of hydrogen atoms combine to form
1 mol of hydrogen molecules?
STRATEGY
The problem asks us to calculate a mass defect when the energy change is
known. To do this, we have to rearrange the Einstein equation to solve for mass,
remembering that 
SOLUTION
 = - 4.84 * 10-12 kg = - 4.84 * 10-9 g
 ¢m =
¢E
c2
=
1- 436 kJ2a103 J
kJ
b a1 kg # m
2/s2
1 J
b
a3.00 * 108 m
s
b2
1 J = 1 kg # m2/s2.
¢E¢m
¢E = -436 kJ2 H ¡ H2
4He
6He
§ 2.83 * 1012 Jmol
6.022 * 1023 
nuclei
mol
¥ a 1 MeV
1.60 * 10-13 J
b a 1 nucleus
6 nucleons
b = 4.89 MeV/nucleon
 = 2.83 * 1012 J/mol = 2.83 * 109 kJ/mol
 ¢E = ¢mc2 = a0.031 41 g
mol
b a10�3 kg
g
b a3.00 * 108 m
s
b 2
 = 0.031 41 amu, or 0.031 41 g/mol
 = 16.049 20 amu2 - 16.017 79 amu2
 Mass defect = Mass of nucleons - Mass of nucleus
6He
 Mass of helium-6 nucleus = 6.017 79 amu
 - Mass of 2 electrons = -122 15.486 * 10-4 amu2 = - 0.001 10 amu
 Mass of helium-6 atom = 6.018 89 amu
6He
6He
 Mass of 4 n + 2 p = 6.049 20 amu Mass of 2 protons = 122 11.007 28 amu2 = 2.014 56 amu
 Mass of 4 neutrons = 142 11.008 66 amu2 = 4.034 64 amu
14 n + 2 p2:
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 964
22.6 Nuclear Fission and Fusion 965
The loss in mass accompanying formation of 1 mol of molecules from its
constituent atoms is far too small an amount to be detectable by any
balance currently available.
� PROBLEM 22.11 Calculate the mass defect (in g/mol) for the formation of an
oxygen-16 nucleus, and calculate the binding energy in MeV/nucleon. The mass of an
atom is 15.994 92 amu.
� PROBLEM 22.12 What is the mass change (in g/mol) for the thermite reaction of
aluminum with iron(III) oxide?
22.6 Nuclear Fission and Fusion
A careful look at the plot of atomic number versus binding energy per nucleon in
Figure 22.7 leads to some interesting and enormously important conclusions. The
fact that binding energy per nucleon begins at a relatively low value for reaches a
maximum at and then gradually tails off implies that both lighter and heavier
elements are less stable than midweight elements near iron-56. Very heavy elements
can therefore gain stability and release energy if they fragment to yield midweight
elements, while very light elements can gain stability and release energy if they fuse
together. The two resultant processes—fission for the fragmenting of heavy nuclei
and fusion for the joining together of light nuclei—have changed the world since
their discovery in the late 1930s and early 1940s.
Nuclear Fission
Certain nuclei—uranium-233, uranium-235, and plutonium-239, for example—do
more than undergo simple radioactive decay; they break into fragments when
struck by neutrons. As illustrated in Figure 22.8, an incoming neutron causes the
nucleus to split into two smaller pieces of roughly similar size.
26
56Fe,
1
2H,
¢E = -852 kJ2 Al1s2 + Fe2 O31s2 ¡ Al2 O31s2 + 2 Fe1s2
16O
4.84 * 10-9 g,
H2
Ruth Lewin Sime, “Lise Meit-
ner and the Discovery of Fis-
sion,” J. Chem. Educ., Vol. 66, 1989,
373–378.
Ruth Lewin Sime, “Lise Meit-
ner and the Discovery of Fis-
sion,” Scientific American, January
1998, 80–85.
Neutron
Neutrons
U23592
Ba14256
Kr9136
� FIGURE 22.8 A representa-
tion of nuclear fission. A
uranium-235 nucleus frag-
ments when struck by a neu-
tron, yielding two smaller
nuclei and releasing a large
amount of energy.
The fission of a nucleus does not occur in exactly the same way each time:
Nearly 400 different fission pathways have been identified for uranium-235, yield-
ing nearly 800 different fission products. One of the more frequently occurring
pathways generates barium-142 and krypton-91, along with two additional neu-
trons plus the one neutron that initiated the fission:
The three neutrons released by fission of a nucleus can induce three more
fissions yielding nine neutrons, which can induce nine more fissions yielding 
235U
0
1n +
 92
235U ¡
 56
142Ba + 3691Kr + 3 01n
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 965
966 Chapter 22 Nuclear Chemistry
27 neutrons, and so on indefinitely. The result is a chain reaction that continues to
occur even if the external supply of neutrons is cut off. If the sample size is small,
many of the neutrons escape before initiating additional fission events, and the chain
reaction soon stops. If there is a sufficient amount of though—an amount called
the critical mass—enough neutrons remain for the chain reaction to become self-
sustaining. Under high-pressure conditions that confine the to a small volume,
the chain reaction may even occur so rapidly that a nuclear explosion results. For 
the critical mass is about 56 kg, although the amount can be reduced to 15 kg by plac-
ing a coating of around the to reflect back some of the escaping neutrons.
The amount of energy released during nuclear fission can be calculated as in
Worked Example 22.8 by finding the accompanying mass change and then using
the Einstein mass–energy relationship discussed in the previous section. When
calculating the mass change, it’s simplest to use the masses of the atoms corre-
sponding to the relevant nuclei, rather than the masses of the nuclei themselves,
because the number of electrons is the same in both reactants and products and
thus cancels from the calculation.
WORKED EXAMPLE 22.8
How much energy (in kJ/mol) is released by the fission of uranium-235 to form
barium-142 and krypton-91? The atomic masses are (235.0439 amu), 
(141.9164 amu), (90.9234 amu), and n (1.008 66 amu).
STRATEGY
First calculate the mass change by subtracting the masses of the products from the
mass of the reactant, and then use the Einstein equation to convert mass to energy.
SOLUTION
Nuclear fission of releases 
� PROBLEM 22.13 An alternative pathway for the nuclear fission of produces
tellurium-137and zirconium-97. How much energy (in kJ/mol) is released in this fis-
sion pathway?
The masses are (235.0439 amu), (136.9254 amu), (96.9110 amu), and 
n (1.008 66 amu).
Nuclear Reactors
The same fission process that leads to a nuclear explosion under some conditions
can be used to generate electric power when carried out in a controlled manner in
a nuclear reactor (Figure 22.9). The principle behind a nuclear reactor is simple:
97Zr137Te235U
0
1n +
 92
235U ¡
 52
137Te + 4097Zr + 2 01n
235U
1.68 * 1010 kJ/mol.235U
 = 1.68 * 1013 kg # m2/1s2 # mol2 = 1.68 * 1010 kJ/mol
 = a0.1868 g
mol
b a1 * 10-3 kg
g
b a3.00 * 108 m
s
b 2
 ¢E = ¢mc2
 Mass change: = 0.1868 amu 1or 0.1868 g/mol2
 - Mass of 2 n = -12211.008 66 amu2 = - 2.0173 amu
 - Mass of 91Kr = - 90.9234 amu
 - Mass of 142Ba = - 141.9164 amu
 Mass of 235U = 235.0439 amu
235U
0
1n +
 92
235U ¡
 56
142Ba + 3691Kr + 3 01n
91Kr
142Ba235U
235U238U
235U,
235U
235U,
� An enormous amount of
energy is released in the explo-
sion that accompanies an
uncontrolled nuclear chain
reaction.
Nuclear Power Plant
Diagram activity
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 966
22.6 Nuclear Fission and Fusion 967
Uranium fuel is placed in a containment vessel surrounded by circulating coolant,
and control rods are added. Made of substances such as boron and cadmium, which
absorb and thus regulate the flow of neutrons, the control rods are raised and low-
ered as necessary to maintain the fission at a barely self-sustainable rate so that
overheating is prevented. Energy from the controlled fission heats the circulating
coolant, which in turn produces steam to drive a turbine and produce electricity.
Naturally occurring uranium is a mixture of two isotopes. The nonfissionable
isotope has a natural abundance of 99.3%, while the fissionable isotope
is present only to the extent of 0.7%. The fuel used in nuclear reactors is typically
made of compressed pellets of that have been isotopically enriched to a 3%
concentration of and then encased in zirconium rods. The rods are placed in a
pressure vessel filled with water, which acts as a moderator to slow the neutrons
so they can be captured more readily. No nuclear explosion can occur in a reactor
because the amount and concentration of fissionable fuel is too low and because
the fuel is not confined by pressure into a small volume. In a worst-case accident,
however, uncontrolled fission could lead to enormous overheating that could melt
the reactor and surrounding containment vessel, thereby releasing large amounts
of radioactivity to the environment.
Thirty countries around the world now obtain some of their electricity from
nuclear energy (Figure 22.10). Lithuania leads with 78%, followed by a number of
other European countries that have also made a substantial commitment to the tech-
nology. The United States has been more cautious, with only 20% of its power com-
ing from nuclear plants. Worldwide, 439 nuclear plants were in operation in early
2002, with an additional 32 under construction, most of them in Asia. Approxi-
mately 21% of the world’s electrical power is generated by nuclear reactors.
The primary problem holding back future development is the yet unsolved
matter of how to dispose of the radioactive wastes generated by the plants. It will
take at least 600 years for waste strontium-90 to decay to safe levels, and at least
20,000 years for plutonium-239 to decay.
Nuclear Fusion
Just as heavy nuclei such as release energy when they undergo fission, very
light nuclei such as the isotopes of hydrogen release enormous amounts of energy
when they undergo fusion. In fact, it’s just this fusion reaction of hydrogen nuclei to
235U
235U
UO2
235U238U
Michael Freemantle, “Ten
Years After Chernobyl: Conse-
quences are Still Emerging,” Chem.
Eng. News, April 29, 1996, 18–28.
Control
rods
Reactor
Fuel
elements
Containment
shell
Steam
Pump
Water
Steam
generator
Pump
Steam
turbine
Condenser
(steam from
turbine is
condensed)
38°C
27°C
Large water
source
Electrical
output
� FIGURE 22.9 A nuclear
power plant. Heat produced in
the reactor core is transferred
by coolant circulating in a
closed loop to a steam genera-
tor, and the steam then drives a
turbine to generate electricity.
Gregory R. Choppin, “Aspects
of Nuclear Waste Disposal of
Use in Teaching Basic Chemistry,”
J. Chem. Educ., Vol. 71, 1994,
826–829.
Torkil H. Jensen, “Fusion—A
Potential Power Source,” J.
Chem. Educ., Vol. 71, 1994, 820–823.
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 967
968 Chapter 22 Nuclear Chemistry
produce helium that powers our sun and other stars. Among the processes thought
to occur in the sun are those in the following sequence leading to helium-4:
The main appeal of nuclear fusion as a power source is that the hydrogen iso-
topes used as fuel are cheap and plentiful and that the fusion products are nonra-
dioactive and nonpolluting. The technical problems that must be solved before
achieving a practical and controllable fusion method are staggering, however. Not
the least of the problems is that a temperature of approximately 40 million kelvins
is needed to initiate the fusion process.
� PROBLEM 22.14 Calculate the amount of energy released (in kJ/mol) for the
fusion reaction of and atoms to yield a atom:
The atomic masses are (1.007 83 amu), (2.014 10 amu), and (3.016 03 amu).
22.7 Nuclear Transmutation
Only about 300 of the more than 3600 known isotopes occur naturally. The
remainder have been made by nuclear transmutation, the change of one element
into another. Such transmutation is often brought about by bombardment of an
atom with a high-energy particle such as a proton, neutron, or particle. In the
ensuing collision between particle and atom, an unstable nucleus is momentarily
a
3He2H1H
1
1H + 12H ¡ 23He
3He2H1H
2
3He + 11H ¡ 24He + 10e
2
3He + 23He ¡ 24He + 2 11H
1
1H + 12H ¡ 23He
1
1H + 11H ¡ 12H + 10e
The variation in slope of a
plot of the binding energy
versus mass number shows
that a typical fusion process
releases far more energy than the
fission processes employed in
nuclear reactors.
Lithuania
France
Belgium
Slovak Republic
Ukraine
Sweden
Bulgaria
South Korea
Slovenia
Hungary
Switzerland
Armenia
Japan
Germany
Finland
Spain
United Kingdom
Czech Republic
United States
Russia
Canada
Romania
Argentina
South Africa
Brazil
India
Netherlands
Mexico
Pakistan
China
0 20 40
Percentage of energy from nuclear power (2001)
60 10080
FIGURE 22.10 Percentage �
of electricity generated by
nuclear power in 2001.
McMFMCH22_FINAL.QXP 3/14/03 8:54 PM Page 968
22.7 Nuclear Transmutation 969
created, a nuclear change occurs, and a different element is produced. The first
nuclear transmutation was accomplished in 1917 by Ernest Rutherford, who
bombarded nuclei with particles and found that was produced:
Other nuclear transmutations can lead to the synthesis of entirely new ele-
ments never before seen on Earth. In fact, all the transuranium elements—those
elements with atomic numbers greater than 92—have been produced by bom-
bardment reactions. Plutonium, for example, can be made by bombarding
uranium-238 with particles:
The plutonium-241 that results from uranium-238 bombardment is itself
radioactive with a half-life of 14.4 years, decaying by emission to yield americium-
241. (If the name sounds familiar, it’s because americium is used commercially in
making smoke detectors.) Americium-241 is also radioactive, decaying by emis-
sion with a half-life of 432 years.
Still other nuclear transmutations are carried out using neutrons, protons, or
other particles for bombardment. The cobalt-60 used in radiationtherapy for can-
cer patients can be prepared by neutron bombardment of iron-58. Iron-58 first
absorbs a neutron to yield iron-59, the iron-59 undergoes decay to yield cobalt-
59, and the cobalt-59 then absorbs a second neutron to yield cobalt-60:
The overall change can be written as
26
58Fe + 2 01n ¡ 2760Co + �1 0e
27
59Co + 01n ¡ 2760Co
26
59Fe ¡ 2759Co + �1 0e
26
58Fe + 01n ¡ 2659Fe
b
 
 95
241Am ¡
 93
237Np + 24He
 
 94
241Pu ¡
 95
241Am + �1 0e
a
b
 92
238U + 24He ¡ 94241Pu + 01n
a
 7
14N + 24He ¡ 817O + 11H
17Oa14N
� The Fermi National Acceler-
ator Laboratory has a particle
accelerator 4 mi in circumfer-
ence that is able to accelerate
protons to energies of 1 trillion
eV.
WORKED EXAMPLE 22.9
The element berkelium was first prepared at the University of California at Berkeley in
1949 by bombardment of Two neutrons are also produced during the reac-
tion. What isotope of berkelium results from this transmutation? Write a balanced
nuclear equation.
 95
241Am.a
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 969
970 Chapter 22 Nuclear Chemistry
STRATEGY AND SOLUTION
According to the periodic table, berkelium has Since the sum of the reactant
mass numbers is and 2 neutrons are produced, the berkelium isotope
must have a mass number of 243.
� PROBLEM 22.15 Write a balanced nuclear equation for the reaction of argon-40
with a proton:
� PROBLEM 22.16 Write a balanced nuclear equation for the reaction of uranium-
238 with a deuteron 
22.8 Detecting and Measuring Radioactivity
Radioactive emissions are invisible. We can’t see, hear, smell, touch, or taste them,
no matter how high the dose. We can, however, detect radiation by measuring its
ionizing properties. High-energy radiation of all kinds is usually grouped under
the name ionizing radiation because interaction of the radiation with a molecule
knocks an electron from the molecule, thereby ionizing it.
Ionizing radiation includes not only particles, particles, and rays, but
also X rays and cosmic rays. X rays, like rays, are high-energy photons 
rather than particles; cosmic rays are energetic particles coming
from interstellar space. They consist primarily of protons, along with some and
particles.
The simplest device for detecting radiation is the photographic film badge
worn by people who routinely work with radioactive materials. Any radiation
striking the badge causes it to fog. Perhaps the best-known method for measuring
radiation is the Geiger counter, an argon-filled tube containing two electrodes
(Figure 22.11). The inner walls of the tube are coated with an electrically conduct-
ing material and given a negative charge, and a wire in the center of the tube is
given a positive charge. As radiation enters the tube through a thin window, it
strikes and ionizes argon atoms, releasing electrons that briefly conduct a tiny
electric current between the electrodes. The passage of the current is detected,
amplified, and used to produce a clicking sound or to register on a meter. The
more radiation that enters the tube, the more frequent the clicks.
b
a
10�8–10�11 m2 1l =g
gba
Molecule 
 Ionizing 
 radiation 
" Ion + e�
 92
238U + 12H ¡ ? + 2 01n
112H2:
18
40Ar + 11H ¡ ? + 01n
 95
241Am + 24He ¡ 97243Bk + 2 01n
241 + 4 = 245
Z = 97.
Ionizing radiation is any parti-
cle or photon with sufficient
energy to remove an electron
on collision with an atom or mole-
cule, thus creating an ion.
High-energy photons ( and
X rays) are the most penetrat-
ing forms of radiation. Partic-
ulate forms of radiation ( and 
particles) are less penetrating.
ba
g
This photographic film �
badge is a common device for
monitoring radiation exposure.
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 970
22.8 Detecting and Measuring Radioactivity 971
The most versatile method for measuring radiation in the laboratory is the
scintillation counter, in which a substance called a phosphor, often a sodium iodide
crystal containing a small amount of thallium iodide, emits a flash of light when
struck by radiation. The number of flashes is counted electronically and converted
into an electrical signal.
Radiation intensity is expressed in different ways, depending on what is being
measured. Some units measure the number of nuclear decay events; others mea-
sure the amount of exposure to radiation or the biological consequences of radia-
tion (Table 22.3).
+
−
–
–
–
+
+
+
+ +
–
–
Path of
radiation
Argon gas
Negatively
charged
Positively
charged
Amplifier and
counter
Battery
Wire
Window
� FIGURE 22.11 A Geiger
counter for measuring radia-
tion. As radiation enters the
tube through a thin window, it
ionizes argon atoms and pro-
duces electrons that conduct a
tiny electric current from the
negatively charged walls to the
positively charged center elec-
trode. The current flow then
registers on the meter.
� Radiation is conveniently
detected and measured using
this scintillation counter, which
electronically counts the flashes
produced when radiation
strikes a phosphor.
TABLE 22.3 Units for Measuring Radiation
Unit Quantity Measured Description
Becquerel (Bq) Decay events Amount of sample that undergoes 
1 disintegration/s
Curie (Ci) Decay events Amount of sample that undergoes 
Gray (Gy) Energy absorbed 
per kilogram of tissue
Rad Energy absorbed 
per kilogram of tissue
Sievert (Sv) Tissue damage
Rem Tissue damage 1 rem = 0.01 Sv
1 Sv = 1 J/kg
1 rad = 0.01 Gy
1 Gy = 1 J/kg tissue
3.7 * 1010 disintegrations/s
• The becquerel (Bq) is the SI unit for measuring the number of nuclear disinte-
grations occurring per second in a sample: The
curie (Ci) and millicurie (mCi) also measure disintegrations per unit time, but
1 Bq = 1 disintegration/s.
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 971
972 Chapter 22 Nuclear Chemistry
they are far larger units than the becquerel and are more often used, particu-
larly in medicine and biochemistry. One curie is the decay rate of 1 g of
radium, equal to 
For example, a 1.5 mCi sample of tritium is equal to meaning
that it undergoes 
• The gray (Gy) is the SI unit for measuring the amount of energy absorbed per
kilogram of tissue exposed to a radiation source: The rad
(radiation absorbed dose) also measures tissue exposure and is more often used
in medicine.
• The sievert (Sv) is the SI unit that measures the amount of tissue damage
caused by radiation. It takes into account not just the energy absorbed per
kilogram of tissue but also the different biological effects of different kinds of
radiation. For example, 1 Gy of radiation causes 20 times more tissue dam-
age than 1 Gy of rays, but 1 Sv of radiation and 1 Sv of rays cause the
same amount of damage. The rem (roentgen equivalent for man) is an analogous
non-SI unit that is more frequently used in medicine.
22.9 Biological Effects of Radiation
The effects of ionizing radiation on the human body vary with the kind and
energy of the radiation, the length of exposure, and whether the radiation is from
an external or internal source. When coming from an external source, X rays and 
radiation are more harmful than and particles because they penetrate clothing
and skin. When coming from an internal source, however, and particles are
particularly dangerous because all their radiation energy is given up to nearby tis-
sue. Alpha emitters are especially hazardous internally and are almost never used
in medical applications.
Because of their relatively large mass, particles move slowly (up to only
one-tenth the speed of light) and can be stopped by a few sheets of paper or by
the top layer of skin. Beta particles, because they are much lighter, move at up to
nine-tenths the speed of light and have about 100 times the penetratingpower of
a
ba
ba
g
1 rem = 0.01 Sv
gag
a
1 Gy = 1 J/kg 1 rad = 0.01 Gy
1 Gy = 1 J/kg.
11.5 mCi2a10�3 Ci
mCi
b a3.7 * 1010 Bq
Ci
b = 5.6 * 107 Bq
5.6 * 107 disintegrations/s.
5.6 * 107 Bq,
 1 Ci = 3.7 * 1010 Bq = 3.7 * 1010 disintegrations/s
 1 Bq = 1 disintegration/s
3.7 * 1010 Bq:
Rosalyn S. Yalow, “Develop-
ment and Proliferation of
Radioimmunoassay Technology,” J.
Chem. Educ., Vol. 76, 1999, 767–768.
Rosalyn S. Yalow, “Radioac-
tivity in the Service of Many,”
J. Chem. Educ., Vol. 59, 1982,
735–738.
Charles H. Atwood, “How
Much Radon is Too Much?” 
J. Chem. Educ., Vol. 69, 1992,
351–353.
TABLE 22.4 Some Properties of Ionizing Radiation
Type of Energy Penetrating Distance
Radiation Range in Water
3–9 MeV 0.02–0.04 mm
0–3 MeV 0–4 mm
X 100 eV–10 keV 0.01–1 cm
10 keV–10 MeV 1–20 cm
Distances at which one-half of the radiation has been stopped*
g
b
a
*
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 972
22.10 Applications of Nuclear Chemistry 973
particles. A block of wood or heavy protective clothing is necessary to stop 
radiation, which would otherwise penetrate and burn the skin. Gamma rays and
X rays move at the speed of light and have about 1000 times the penetrating
power of particles. A lead block several inches thick is needed to stop and X
radiation, which could otherwise penetrate and damage the body’s internal
organs. Some properties of different kinds of ionizing radiation are summarized
in Table 22.4.
The biological effects of different radiation doses are given in Table 22.5.
Although the effects sound fearful, the average radiation dose received annually
by most people is only about 120 mrem. About 70% of this radiation comes from
natural sources (rocks and cosmic rays); the remaining 30% comes from medical
procedures such as X rays. The amount due to emissions from nuclear power
plants and to fallout from atmospheric testing of nuclear weapons in the 1950s is
barely detectable.
ga
ba
TABLE 22.5 Biological Effects of Short-Term Radiation on Humans
Dose (rem) Biological Effects
0–25 No detectable effects
25–100 Temporary decrease in white blood cell count
100–200 Nausea, vomiting, longer-term decrease in white blood cells
200–300 Vomiting, diarrhea, loss of appetite, listlessness
300–600 Vomiting, diarrhea, hemorrhaging, eventual death in some cases
Above 600 Eventual death in nearly all cases
22.10 Applications of Nuclear Chemistry
Dating with Radioisotopes
Biblical scrolls are found in a cave near the Dead Sea. Are they authentic? A mummy
is discovered in an Egyptian tomb. How old is it? The burned bones of a man are dug
up near Lubbock, Texas. How long have humans lived in the area? These and many
other questions can be answered by archaeologists using a technique called
radiocarbon dating. (The Dead Sea Scrolls are 1900 years old and authentic, the mummy
is 3100 years old, and the human remains found in Texas are 9900 years old.)
Radiocarbon dating of archaeological artifacts depends on the slow and con-
stant production of radioactive carbon-14 in the upper atmosphere by neutron
bombardment of nitrogen atoms. (The neutrons come from the bombardment of
other atoms by cosmic rays.)
Carbon-14 atoms produced in the upper atmosphere combine with oxygen to
yield which slowly diffuses into the lower atmosphere, where it mixes with
ordinary and is taken up by plants during photosynthesis. When these
plants are eaten, carbon-14 enters the food chain and is ultimately distributed
evenly throughout all living organisms.
As long as a plant or animal is living, a dynamic equilibrium exists in which an
organism excretes or exhales the same amount of that it takes in. As a result, the
ratio of to in the living organism is the same as that in the atmosphere—
about 1 part in When the plant or animal dies, however, it no longer takes in
more and the ratio in the organism slowly decreases as undergoes
radioactive decay by emission, with 
 6
14C ¡
 7
14N + �1 0e
t1/2 = 5730 years.b
14C14C/12C14C,
1012.
12C14C
14C
12CO2
14CO2,
 7
14N + 01n ¡ 614C + 11H
� Radiocarbon dating places
the age of this Egyptian
mummy at 3100 years.
M. W. Rowe, “Radioactive
Dating: A Method for
Geochronology,” J. Chem. Educ.,
Vol. 62, 1985, 580–584.
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 973
974 Chapter 22 Nuclear Chemistry
At 5730 years (one half-life) after the death of the organism, the 
ratio has decreased by a factor of 2; at 11,460 years after death, the ratio
has decreased by a factor of 4; and so on. By measuring the present ratio
in the traces of any once-living organism, archaeologists can determine how long
ago the organism died. Human or animal hair from well-preserved remains, char-
coal or wood fragments from once-living trees, and cotton or linen from once-
living plants are all useful sources for radiocarbon dating. The technique becomes
less accurate as samples get older and the amount of they contain diminishes,
but artifacts with an age of 1000–20,000 years can be dated with reasonable accu-
racy. The outer limit of the technique is about 60,000 years.
Just as radiocarbon measurements allow dating of once-living organisms, sim-
ilar measurements on other radioisotopes make possible the dating of rocks.
Uranium-238, for example, has a half-life of years and decays through
the series of events shown previously in Figure 22.6 to yield lead-206. The age of a
uranium-containing rock can therefore be determined by measuring the
ratio. Similarly, potassium-40 has a half-life of years and
decays through electron capture and positron emission to yield argon-40. (Both
processes yield the same product.)
The age of a rock can be found by crushing a sample, measuring the amount of
gas that escapes, and comparing the amount of with the amount of 
remaining in the sample. It is through techniques such as these that the age of the
earth has been estimated at approximately 4.5 billion years.
WORKED EXAMPLE 22.10
Radiocarbon measurements made in 1988 on the Shroud of Turin, a religious artifact
thought by some to be the burial shroud of Christ, showed a decay rate of 14.2 dis-
integrations/min per gram of carbon. What age is implied by this result if currently
living organisms decay at the rate of 15.3 disintegrations/min per gram of carbon? The
half-life of is 5730 years.
STRATEGY
As we saw in Worked Example 22.5, the ratio of the decay rate at any time t to the
decay rate at time is the same as the ratio of N to 
To date the shroud, we need to calculate the time t that corresponds to the observed
decay rate. This can be done by solving for t in the equation
SOLUTION
Substituting the proper values into the equation gives
The Shroud of Turin is approximately 617 years old, indicating that it comes from
medieval times.
 so t = 1-0.0746215730 years2
-0.693
= 617 years
 ln a14.2
15.3
b = -0.693 a t
5730 years
b or -0.0746 = -0.693 a t
5730 years
b
ln a N
N0
b = 1- ln 22a t
t1/2
b
Decay rate at time t
Decay rate at time t = 0
=
k N
k N0
=
N
N0
N0:t = 0
14C
14C
40K40Ar40Ar
 19
40K ¡ 1840Ar + 10e
 19
40K + �1 0e ¡ 1840Ar
1.28 * 109238U/206Pb
4.47 * 109
14C
14C/12C
14C/12C
14C/12C14CM. W. Rowe, “Archeological
Dating,” J. Chem. Educ., Vol.
63, 1986, 16–20.
Ron DeLorenzo, “California
Earthquakes: Predicting the
Next Big One Using Radiocarbon
Dating,” J. Chem. Educ., Vol. 57,
1980, 601.
Kent J. Crippin and Robert D.
Cutright, “Modeling Nuclear
Decay: A Point of Integration
Between Chemistry and Mathemat-
ics,” J. Chem. Educ., Vol. 75, 1998,
434–436.
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 974
22.10 Applications of Nuclear Chemistry 975
� PROBLEM 22.17 Charcoal found in the Lascaux cave in France, site of manypre-
historic cave paintings, was observed to decay at a rate of 2.4 disintegrations/min per
gram of carbon. What is the age of the charcoal if currently living organisms decay at the
rate of 15.3 disintegrations/min per gram of carbon? The half-life of is 5730 years.14C
� What is the age of these
cave paintings?
Medical Uses of Radioactivity
The origins of nuclear medicine date to 1901, when the French physician Henri
Danlos first used radium in the treatment of a tuberculous skin lesion. Since that time,
uses of radioactivity have become a crucial part of modern medical care, both diag-
nostic and therapeutic. Current nuclear techniques can be grouped into three classes:
(1) in vivo procedures, (2) therapeutic procedures, and (3) imaging procedures.
In Vivo Procedures In vivo studies—those that take place inside the body—are
carried out to assess the functioning of a particular organ or body system. A radio-
pharmaceutical agent is administered, and its path in the body—whether it is
absorbed, excreted, diluted, or concentrated—is determined by analysis of blood
or urine samples.
An example of the many in vivo procedures using radioactive agents is the deter-
mination of whole-blood volume by injecting a known quantity of red blood cells
labeled with radioactive chromium-51. After a suitable interval to allow the labeled
cells to be distributed evenly throughout the body, a blood sample is taken, the
amount of dilution of the is measured, and the blood volume is calculated. Recall
from Section 3.8 that when a concentrated solution is diluted, the amount of solute
( in the present instance) remains the same and only the volume changes. That is,
or
where:
Therapeutic Procedures Therapeutic procedures—those in which radiation is
used to kill diseased tissue—can involve either external or internal sources of radi-
ation. External radiation therapy for the treatment of cancer is often carried out
with rays from a cobalt-60 source. The highly radioactive source is shielded by a
thick lead container and has a small opening directed toward the site of the tumor.
By focusing the radiation beam on the tumor and rotating the patient’s body, the
tumor receives the full exposure while the exposure of surrounding parts of the
body is minimized. Nevertheless, sufficient exposure occurs so that most patients
suffer some effects of radiation sickness.
g
 Vblood = Blood volume
 Cblood = Concentration of labeled cells in blood 1mCi/mL2
V0 = Volume of labeled cells injected (mL)
C0 = Concentration of labeled cells injected 1mCi/mL2
Vblood =
C0 V0
Cblood
Amount of 51Cr = C0 * V0 = Cblood * Vblood
51Cr
51Cr
� A person’s blood volume
can be found by injecting a
small amount of radioactive
chromium-51 and measuring
the dilution factor.
� Cancerous tumors can be
treated by irradiation with 
rays from this cobalt-60 source.
g
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 975
976 Chapter 22 Nuclear Chemistry
Internal radiation therapy is a much more selective technique than external
therapy. In the treatment of thyroid disease, for example, iodine-131, a powerful 
emitter known to localize in the target tissue, is administered internally. Because 
particles penetrate no farther than several millimeters, the localized produces a
high radiation dose that destroys only the surrounding diseased tissue.
Imaging Procedures Imaging procedures give diagnostic information about the
health of body organs by analyzing the distribution pattern of radioisotopes intro-
duced into the body. A radiopharmaceutical agent that is known to concentrate in a
specific tissue or organ is injected into the body, and its distribution pattern is mon-
itored by external radiation detectors. Depending on the disease and the organ, a
diseased organ might concentrate more of the radiopharmaceutical than a normal
organ and thus show up as a radioactive “hot” spot against a “cold” background.
Alternatively, the diseased organ might concentrate less of the radiopharmaceuti-
cal than a normal organ and thus show up as a cold spot on a hot background.
The radioisotope most widely used today is technetium-99m, whose short
half-life of 6.01 hours minimizes a patient’s exposure to harmful effects. Bone
scans using Tc-99m, such as that shown in Figure 22.12a, are an important tool in
the diagnosis of cancer and other pathological conditions.
Another kind of imaging procedure makes use of a technique called magnetic
resonance imaging (MRI). MRI uses no radioisotopes and has no known side-
effects. Instead, MRI uses radio waves to stimulate certain nuclei in the presence
of a powerful magnetic field. The stimulated nuclei (normally the hydrogen nuclei
in molecules) then give off a signal that can be measured, interpreted, and
correlated with their environment in the body. Figure 22.12b shows a brain scan
carried out by MRI and indicates the position of a tumor.
H2 O
131I
b
b
Mairin B. Brennan, “Positron
Emission Tomography Merges
Chemistry with Biological Imag-
ing,” Chem. Eng. News, February 19,
1996, 26–33.
Marcus E. Raichle, “Visualiz-
ing the Mind,” Scientific Amer-
ican., April 1994, 64.
Timothy J. McCarthy, Sally W.
Schwarz, and Michael J.
Welch, “Nuclear Medicine and
Positron Emission Tomography:
An Overview,”J. Chem. Educ., Vol.
71, 1994, 830–836.
(a) (b)
FIGURE 22.12 (a) A �
bone scan using radioactive
technetium-99m. (b) An MRI
brain scan, showing the posi-
tion of a tumor (the large white
area at left).
McMFMCH22_FINAL.QXP 3/14/03 8:55 PM Page 976
The Origin of Chemical
Elements
Cosmologists tell us that the universe began some 15 billion yearsago in an extraordinary event they call the “big bang.” Initially, thetemperature must have been inconceivably high, but after 1 sec-
ond, the temperature had dropped to about and elementary parti-
cles began to form: protons, neutrons, and electrons, as well as positrons
and neutrinos—neutral particles with a mass much less than that of an
electron. After 3 minutes, the temperature had dropped to and
protons began fusing with neutrons to form helium nuclei, Mat-
ter remained in this form for many millions of years until the expand-
ing universe had cooled to about 10,000 K. Electrons were then able to
bind to protons and to helium nuclei, forming stable hydrogen and
helium atoms.
The attractive force of gravity acting on regions of higher-than-
average density slowly produced massive local concentrations of
matter and ultimately formed billions of galaxies, each with many
billions of stars. As the gas clouds of hydrogen and helium con-
densed under gravitational attraction and stars formed, their tem-
peratures reached and their densities reached 
Protons and neutrons again fused to yield helium nuclei, generat-
ing vast amounts of heat and light—about per mole of pro-
tons undergoing fusion.
Most of these early stars probably burned out after a few billion years, but a
few were so massive that, as their nuclear fuel diminished, gravitational attraction
caused a rapid contraction leading to still higher core temperatures and higher
densities—up to and Much larger nuclei were now
formed, including carbon, oxygen, silicon, magnesium, and iron. Ultimately, the
stars underwent a gravitational collapse resulting in the synthesis of still-heavier
elements and an explosion visible throughout the universe as a supernova.
Matter from exploding supernovas was blown throughout the galaxy,
forming a new generation of stars and planets. Our own sun and solar sys-
tem formed only about 4.5 billion years ago from matter released by
former supernovas. Except for hydrogen and helium, all the atoms in our
bodies, our planet, and our solar system were created more than 5 billion
years ago in exploding stars.
� PROBLEM 22.18 How do elements heavier than iron arise?
5 * 105 g/cm3.5 * 108 K
6