Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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41. Which of these species would produce the greater number of

ions per mole when dissolved in water?

[Cr(C

)

]or

tetraamminediaquachromium(III) nitrate

42. Which of these species possesses the larger positive charge on

the complex ion?

tetraaquacopper(II) nitrate or

dichlorobis(ethylenediamine)iron(III) bromide

20.7 Color and Coordination Compounds

Skill Review

43. What is the electron conﬁguration for each of these transi-

tion metal ions?

a. Fe

b. Cr

c. Zn

44. What is the electron conﬁguration for each of these transi-

tion metal ions?

a. Pd

b. Ag

c. Mn

45. Consider the following two transition metal ions as free

gaseous ions. Which would have the greater number of un-

paired d electrons, Fe

or Cu

46. Which free ion has the greater number of unpaired d elec-

trons, Ti

or Co

47. Draw the orbital diagram for the d orbitals in an octahedral

complex containing each of these metal centers. (Assume

that P < ∆

a. Fe

b. Co

c. Ni

48. Draw the orbital diagram for the d orbitals in an octahedral

complex containing each of these metal centers. (Assume

that P > ∆

a. Mn

b. Fe

c. Cr

49. Repeat Problem 47, but assume that the metal centers are in-

volved in tetrahedral complexes. Although tetrahedral com-

plexes typically have ∆

> P, what would you draw if the

tetrahedral complex existed with P > ∆

50. Repeat Problem 48, but assume that the metal centers are in-

volved in square planar complexes. Assume that P > ∆ in this

problem.

51. Draw the orbital diagram for the metal center in each of these

complexes. Use the information in the spectrochemical series

to assist you in placing the orbitals.

a. [FeCl

]

−

b. [Co(CN)

]

3−

c. [Mn(CO)

]

52. Draw the orbital diagram for the metal center in each of these

complexes. Use the information in the spectrochemical series

to assist you in placing the orbitals.

a. [CuF

]

4−

b. [Ni(OH)

]

4−

c. [Cr(NO

)

]

4−

53. Which of the complexes in Problems 47 and 48 is(are)

paramagnetic?

54. Calculate the magnetic moment for each of the complexes in

Problems 47 and 48.

Chemical Applications and Practices

55. Which one of these complexes would you predict to absorb

blue light: [M(CN)

2−

], [M(H

], [MCl

2−

], or

[M(NH

)

56. Which of these complexes would you predict to absorb

the longest wavelength of visible light: [M(CN)

2−

[M(H

], [MCl

2−

], or [M(NH

)

57. The colors of common gemstones are due to the presence of

transition metal ions. The color is produced when the metal

ion absorbs visible light. Would you predict the common

gemstones to have different “colors” under infrared light?

58. Coordination compounds with Zn

ions typically are white

or colorless. Explain why this particular metal does not form

brightly colored compounds the way many other transition

metals do.

59. The crystal ﬁeld theory provides an explanation of color in

various coordination complexes. For example, [Cr(H

]

can be detected as a violet color when dissolved.

a. What colors would the complex be absorbing?

b. How many unpaired electrons does the central chromium

ion have?

c. The compound [Cr(NH

)

]

, when dissolved, appears

yellow. Would you expect it to absorb light at a higher or

lower frequency than Cr(H

]

? Explain.

d. Which ligand, NH

or H

O, is causing the greater value of

∆

60. Compare the two iron complexes [Fe(H

]

and

[Fe(CN)

]

3−

a. Which is more likely to be paramagnetic?

b. Which is more likely to absorb light of greater energy?

c. Which is more likely to be “high-spin”?

20.8 Chemical Reactions

Skill Review

61. Explain why the chelate effect typically provides for a very

favorable entropy change when a ligand exchange reaction

involves a complex going from a nonchelated complex to a

chelate complex.

62. If a ligand exchange reaction produced a large positive ∆G

value, would you expect the reaction to have a large or a small

equilibrium constant? Justify your choice.

63. A common chemical demonstration is to change a light blue

solution containing [Cu(H

]

quickly to a deep purple

solution by changing [Cu(H

]

into [Cu(NH

)

]

with

the addition of ammonia to the solution. Would you con-

sider the ﬁrst compound labile or inert? Is the exchange of

oxygen in hemoglobin considered to be representative of a

labile or an inert complex?

64. Except through loss of blood, the level of iron in humans is

fairly constant. One way that iron is moved throughout the

body, particularly from the liver, is within a molecule known

as ferritin. The iron(III) is held in a six-coordinate system

through bonds to oxygen and nitrogen that are part of several

amino acids. Would it be more logical for this molecule’s iron

site to be labile or inert with respect to other metals? Explain.

(Remember, the terms inert and labile refer to kinetic consid-

erations, not to equilibrium predictions.)

898 Chapter 20 Coordination Complexes

Focus Your Learning 899

Comprehensive Problems

65. In addition to the coordinationcomplexes of rhodium,cobalt,

and molybdenum described at the start of the chapter, use

other resources (the Internet or journals) to select another

transition metal that has a catalytic role in a chemical reaction.

66. Visit the pharmacy section of a grocery store and list the met-

als found in a mineral supplement. Use a reference (the In-

ternet or journals) to determine the primary biochemical

function of two of the metals from your list.

67. If a complex were assembled from a Co

ion, four NH

, and

two Cl

−

, would you expect to see a neutral compound, an

anion, or a cation? Show a formula that justiﬁes your answer.

68. The following complex contains an iron ion in the +3 state.

However, the resulting charge has been omitted from the

complex.Assign the charge for the complex, and indicate how

many counter ions (either Na

or Cl

−

) would have to be ion-

ically bonded to the complex to form a neutral compound.

[FeBr

]

69. The appearance of a visible color from a compound is associ-

ated with three fundamental phenomena. Describe a color-

producing compound’s properties associated with:

a. electron excitation

b. the energy of the photon of light being absorbed

c. the relative occupancy of lower and higher level orbitals

70. If 35.7 g of the complex Ca

[Fe(C

)

]

formed as a result of

using an oxalate-containing rust remover, how many grams

of iron would be removed?

71. Polydentate ligands have been very effective in applications

of soil chemistry. EDTA has been added to soil near citrus

trees to concentrate iron. Some polydentates have been used

to extract heavy metals from soil samples for further analysis.

Some internal digestive functions use polydentate ligands to

extract metals from foods we eat. Does this indicate that the

relative equilibrium constant for these reactions is larger or

smaller than one?

Thinking Beyond the Calculation

72. A new ligand is developed for use in a study to mimic the

electron transport reactions of copper metal.

a. Draw the octahedral complex that would result from the

use of this ligand and copper(II) ions.

b. Draw the d orbital diagram for an octahedral complex of

this ligand.

c. Compare the color of this octahedral complex to that of

[Cu(NH

)

]

. Where does this ligand most likely fall in

the spectrochemical series?

d. Under certain conditions, 1.00 g of copper(II) produces

3.96 g of the copper–ligand complex. What is the coordi-

nation number of the complex under these conditions?

e. The equilibrium constant for the ligand exchange of this

new ligand and chloride ions is 1.45

−4

. What does

this indicate about the new ligand?

Ethanedithiol

HS SHCH

–ligand [Cu(NH

)

]

Nuclear

Chemistry

Enrico Fermi used this very large cyclotron

at the University of Chicago in the 1950s.

The cyclotron can be used to prepare

radioactive isotopes for medical imaging.

The cyclotron generates fast-moving

subatomic particles that can be directed

at nonradioactive nuclei. The resulting

collision produces radioactive nuclei such

as ﬂuorine-18, which can be used to help

doctors observe a particular biological

function within a patient’s body.

900

Contents and Selected Applications

21.1 Isotopes and More Isotopes

21.2 Types of Radioactive Decay

21.3 Interaction of Radiation with Matter

Chemical Encounters: Radiation and Cancer

21.4 The Kinetics of Radioactive Decay

21.5 Mass and Binding Energy

21.6 Nuclear Stability and Human-made Radioactive Nuclides

21.7 Splitting the Atom: Nuclear Fission

Chemical Encounters: Nuclear Weapons

Chemical Encounters: Nuclear Reactors as a Vital Source of Electricity

21.8 Medical Uses of Radioisotopes

Chemical Encounters: Tracer Isotopes for Diagnosis

Go to college.hmco.com/pic/kelterMEE for online learning resources.

Mass

number

Atomic

number

The odds are that you know somebody who

has fought cancer. If you have talked with a friend

or relative who is a cancer patient, you may have heard

that radioactive substances are used in the process of pro-

ducing images of internal organs. You may know that radiation

can shrink a tumor or kill cancer cells. You may even be aware of the

dramatic procedure using full-body radiation that is given before a bone

marrow transplant. These applications illustrate how a health care team can

use

nuclear radiation for the beneﬁt of a cancer patient.

Although radiation can save lives, it can also damage and kill. Back in the early

1900s, early radiologists held ﬁlm plates in an X-ray beam to get pictures of their pa-

tients. These radiologists often developed sores on their hands that would not heal,

and some eventually lost parts of their ﬁngers. In August 1945, people in the Japanese

cities of Hiroshima and Nagasaki were exposed to huge bursts of radiation from the

only use of atomic bombs in warfare. Some died immediately, and others succumbed

a few weeks later from a then-unknown disease that we now call radiation sickness.

In the United States, those who worked in the nuclear weapons industry are now dis-

proportionately contracting lung, stomach, lymphatic, and other cancers.

How can nuclear radiation both cure and cause cancer? The answer to this seeming

paradox rests on our understanding of

radioactivity (the release of particles and en-

ergy accompanying a nuclear change) and how it is pro-

duced from nuclear processes. In this chapter we will exam-

ine types of radioactive decay, touching on the mathematics

of half-lives, the relationship between mass and energy, and

the interactions of ionizing radiation with matter. We also

will examine nuclear ﬁssion, a process that helps produce a

whole host of substances useful in nuclear medicine. The

story of radioactivity begins, however, at the tiny center of

the atom—its nucleus.

901

Nucleus

–

Proton

Neutron

Electron

21.1 Isotopes and More Isotopes

Nuclei that occur naturally on Earth can be as small as the nucleus of a hydrogen

atom (a single proton) or as large as uranium (92 protons plus 146 neutrons). All

of these naturally occurring atoms, and all those that are human-made, are rep-

resented by writing their element symbol in the manner described in Chapter 2.

The symbol is placed next to a superscripted mass number and a subscripted

atomic number. Some of these nuclei are stable—that is, they do not sponta-

neously decompose—and others are radioactive, decomposing to other nuclei.

As we will soon see, the size and makeup of the nucleus determine whether it is

stable or radioactive.

Recall from Chapter 2 that we can use nuclide notation, in which we list the

symbol for the element, accompanied by its atomic number and the mass num-

ber, to indicate the isotope of the element that we wish to describe. For example,

consider the element potassium, whose atomic number (Z) is 19. As we know

from the periodic table, this element has 19 protons. If the nucleus of a potassium

atom has 21 neutrons, we represent this in nuclide notation as

. The num-

ber 40, the mass number, is the sum of 19 protons and 21 neutrons. Because “19

protons” is always potassium, we can more simply represent

K, K-40 or

Human-made isotopes

Naturally occurring

isotopes of potassium

93.258% 0.012% 6.730%

potassium-40. In nuclear chemistry, we are often not concerned

with the number of electrons around a particular nucleus. We are

interested only in the nucleus of the atom, so we often omit

charges (even though they may exist). This means that in our

chemist’s shorthand for the discussion of changes in the nucleus,

we’ll consider

K. The nucleus is our only focus here.

Potassium atoms come in several varieties,shown in Figure 21.1,

some with fewer than 21 neutrons and some with more. These

varieties are known as isotopes, and each isotope is known as a

nuclide of that element, as we learned in Chapter 2.

902 Chapter 21 Nuclear Chemistry

FIGURE 21.1

Potassium has many isotopes, most of

which are listed here. Those colored

orange are radioactive. Natural abun-

dances are provided for those isotopes

that occur in nature. Note that 99.988%

of the potassium atoms on Earth are

not radioactive.

EXERCISE 21.1 Decoding Isotopes

Describe the differences and similarities in nuclear structure among the members of

each of these sets:

, and

Ar,

K, and

and

First Thoughts

The numbers at the bottom of each representation are simply a restatement of the

element symbol. The mass number is the sum of the number of neutrons and

protons. To ﬁnd the number of neutrons in a particular nuclide, just subtract the

atomic number from the mass number. Remember from Chapter 2 that numbers

written on the upper right-hand side of the nuclide notation indicate a deviation in

the number of electrons.

Solution

, and

are isotopes of carbon, each containing six protons. But each

has a different mass number and hence a different number of neutrons: 6, 7,

and 8, respectively.

Ar,

K, and

Ca have the same mass number but different numbers of

protons. Ar has 18p and 22n, K has 19p and 21n, and Ca has 20n and 20p.

and

Ca differ only in the number of electrons, so there is no differ-

ence in nuclear structure. The

Ca atom loses two electrons to form the

ion.

Further Insights

This exercise, while conceptually important, involves just notation and bookkeep-

ing. The nuclide notations, by themselves, say little about the stability or radio-

activity of an isotope. Far more useful and interesting are the topics of natural

abundance, radioactivity, and half-life, which we discuss below.

PRACTICE 21.1

Indicate the number of protons and neutrons in each of the following nuclides.

S c. radon-222

d. Tc-98

See Problems 7 and 8.

Potassium is considered an essential nutrient. Deﬁciencies in potassium can

result in muscle pain, angina, and even heart problems. Eating bananas is one way

in which we can ensure a healthy supply of potassium in our bodies. From Fig-

ure 21.1 we can see that three isotopes of potassium occur naturally:

, and

. However, whereas potassium-39 and potassium-41 possess stable nuclei,

is radioactive. This means that when we consume a banana, we get a measurable

amount of radioactive potassium-40. How much? The natural abundance of

potassium-40 is only 0.012%, or approximately 1 atom in 10,000. A typical ba-

nana has approximately 300 mg of potassium. Therefore, with each banana we

eat, we ingest approximately 0.036 mg of radioactive potassium-40.

Although potassium has 18 known isotopes, most do not occur in nature and

must be produced in a laboratory. Each of these isotopes behaves essentially the

same in chemical reactions, which involve the interaction of electrons with other

atoms. For example, on exposure to oxygen or moisture, potassium metal ionizes

to form K

. All isotopes of potassium behave in this manner. Because our planet

is wet and blanketed in oxygen, all naturally occurring potassium isotopes are

found in nature as K

. Remember, however, that we often omit the charge when

writing nuclides, and you are likely to see

K rather than

. The key idea here

is that we differentiate between chemical reactions (electron interactions between

atoms) and nuclear processes (changes within the nucleus of an individual atom).

Potassium-39 is a stable nuclide, but potassium metal certainly is not a stable

chemical when in the presence of water. On the other hand, argon is chemically

inert, but it has isotopes that are radioactive.

Natural isotopic abundances for potassium or

for any element can be found in many of the chem-

istry handbooks in the library. Table 21.1 offers a

brief look at what you’ll ﬁnd there. As we just saw,

the nuclei of elements present in nature are not

necessarily stable. In fact, some elements, such as

uranium and radon, exist naturally only in ra-

dioactive forms. Other elements, such as potas-

sium and carbon, have both stable and radioactive

isotopes. Still others, such as aluminum, have only

one stable naturally occurring form.

21.1 Isotopes and More Isotopes 903

Sources of potassium.

Application

Natural Isotopic Abundances for Selected Elements

Carbon

C 98.90%

C 1.10

Oxygen

O 99.762%

O 0.038

O 0.200

Magnesium

Mg 78.99%

Mg 10.00

Mg 11.01

Aluminum

Al 100%

TABLE 21.1

Potassium

K 93.2581%

K 0.0117

K 6.7302

Iron

Fe 5.8%

Fe 91.72

Fe 2.2

Fe 0.28

Silver

107

Ag 51.84%

109

Ag 48.16

21.2 Types of Radioactive Decay

There is no way for you hold a potassium atom in your hand and peer into its nu-

cleus. But if you could keep an eye on a few

K atoms for a period of time (per-

haps over a billion years), you would observe that some of the potassium atoms

had been replaced by atoms of calcium.

How do we account for this nuclear sleight-

of-hand?

Beta-Particle Emission

The answer in this case is beta-particle emission, a type of radioactive decay. In

beta emission, the nucleus of

ejects a

beta particle,

−1

, that travels at 90% the

speed of light. A new nucleus,

, is formed that has one more proton. Because

calcium-40 has an energetically stable nucleus (you can verify this in a table of

radioisotopes), no further nuclear reactions take place. For now, let’s write the

nuclear equation for the beta-minus emission, or “beta emission” for short, as

−



We note that the sums of the atomic masses, as well as the sums of the atomic

numbers, are the same on both sides of our equation. The nuclide that results

from the beta emission,

, is comparable in size to potassium. In addition, a

tiny product is ejected with mass number zero and a negative charge—an elec-

tron. In the context of nuclear decay, this electron is called a beta emission parti-

cle. The subscript –1 in

−

 may look strange as an atomic number. Interpret it as

a “negative one charge” rather than a “minus one proton.”

Beta emission, like many other of the nuclear reactions we will study, has the

following features:

■

A nuclide of one element, through the process of radioactive decay, is trans-

formed into a nuclide of another. The resulting

daughter nuclide may be stable

or radioactive.

■

The sum of the atomic numbers on one side of the nuclear equation is equal

to the sum on the other. For the beta emission of

K, the sum of the mass

numbers on each side of the equation is 40, and the sum of the atomic num-

bers on each side of the nuclear equation is 19.

■

Energy is released in radioactive decay reactions. Gamma rays of varying en-

ergy nearly always accompany the nuclear reaction.

How can an electron be emitted from a nucleus when there are no electrons in

the nucleus?

To see how this could happen, let’s venture down from the

macroworld into the nanoworld to look at the beta emission of a single neutron.

Neutrons aren’t something you can keep in a bottle in your chemistry lab. But if

you had the proper specialized equipment, you could demonstrate that a neutron

decays to form a proton, an electron, and an

antineutrino,

´ν

−

 +

´ν

The net effect of beta decay is the transformation of a neutron into a proton with

the release of an electron, an antineutrino, and energy. This is consistent with the

beta decay we wrote for

, where the product nuclide has one more proton.

–1



904 Chapter 21 Nuclear Chemistry

–1





In nucleus In nucleus  expelled Expelled



Video Lesson: The Nature

of Radioactivity

The antineutrino,

´ν

, is an example of antimatter. Each antimatter particle has

a mate in our world of “real” matter. For the antineutrino, this mate is the

neutrino, or “little neutron”—a particle similarly hidden within the neutron. With

no charge and probably very little mass, the neutrino required a sophisticated

piece of scientiﬁc detective work to prove its existence. However, because this

ghostly particle and its antimatter mate interact little if at all with matter, neutri-

nos typically are omitted from nuclear equations.

EXERCISE 21.2 Beta Emissions In and Around Us

A look at the world around us reveals a lot of naturally occurring radiation.

Elements responsible for this radiation include carbon-14, potassium-40, and

hydrogen-3 (tritium).

a. Write the nuclear equation for a beta emission by carbon-14.

b. If

He is formed via a beta emission, what radioisotope produced it?

First Thoughts

Beta emissions follow a set pattern: an increase of +1 in Z and no change in A.

Solution

C n

−

 +

H n

−

 +

Further Insights

We discussed the radioactivity of carbon-14 in Chapter 2. Knowing the rate of this

nuclear reaction is part of a process that enables us to predict the age of an archae-

ological artifact.

PRACTICE 21.2

Write the nuclear equation for the beta emission of

Co.

See Problems 14, 15b, 15c, and 20.

Alpha-Particle Emission

There are two other common forms of radioactivity: alpha-particle emission and

gamma-ray emission. To appreciate how these differ, let’s revisit potassium-40.

Imagine again that you were watching several individual atoms of the nuclide. No

matter how long you waited, you would not observe an

alpha decay, the emission

of a helium nucleus (

) from a larger nucleus. Why not? Potassium-40, like

many radioisotopes, does not exhibit this form of radioactivity. Why not? The

simple answer is that the nucleus is made more energetically stable by beta emis-

sion than by alpha decay.Why? The answer to this is a little more complex and will

be the focus of Sections 21.5 and 21.6.

Which elements decay by alpha emission? Radon is one example. The nuclear

reaction for the alpha decay of radon-222 is

222

Rn →

218

Po +

21.2 Types of Radioactive Decay 905

Energy is given off, along with an alpha particle (

), which is energetic but trav-

els more slowly than a beta particle, at only about 5–10% of the speed of light.

Technetium-99m is generated on demand in nuclear medicine imaging.

EXERCISE 21.3 Alpha Emissions In and Around Us

Radon, discussed earlier, is present in the atmosphere at a concentration of about

1 part in 10

. In some caves and basements, however, test kits like those shown in

Figure 21.3 can detect it at much higher concentrations. Write nuclear equations for

the alpha decay of radon-220, radon-222, and radon-219. A gamma ray accompa-

nies each of these processes.

You also may see

He or simply  used to denote the alpha particle. Is

alpha emission possible for a hydrogen or helium atom? No, because these atoms

are too small to emit an alpha particle and still have any protons left for the re-

maining nucleus. In fact, alpha emission typically does not occur for small nuclei.

It is far more common for elements above bismuth (Z = 83).

Gamma-Ray Emission

For both alpha and beta emissions, the nuclear equations we have written so far

are not complete, because a gamma ray is usually produced as well. When we

include this gamma ray (

) in the alpha decay of radon, the nuclear equation

becomes

222

Rn →

218

Po +

He +



A gamma ray, as indicated in Figure 21.2, is a high-energy photon—a form of

electromagnetic radiation with a short wavelength (typically less than a picome-

ter) traveling at the speed of light. Loss of energy from the system, in the form of

a gamma ray, is favorable because the products of the reaction would then have

less energy than the starting material. In essence, the free energy of the system is

lowered.

906 Chapter 21 Nuclear Chemistry

–11

–9

–7

–5

–3

–1

10 10

Gamma

Radio

Ultraviolet Infrared

X-ray Microwave

Wavelength (cm)

Visible

FIGURE 21.2

Gamma rays are at the high-energy end

of the electromagnetic spectrum. They

have very short wavelengths on the

order of ten-trillionths of a meter or less.

The energy of cosmic rays is comparable

to that of gamma rays, but cosmic rays

are particles and are not part of the

electromagnetic spectrum.

Gamma rays carry off excess energy in an amount depending on the particu-

lar nuclide. In some cases, lower-energy gamma rays are identical to X-rays;

however, we mentally distinguish between the two by noting that the former orig-

inated inside the nucleus and the latter outside of it. For medical purposes, as we

will see in the ﬁnal section of this chapter, both X-rays and gamma rays can ac-

complish the same diagnostic and therapeutic tasks.

Few nuclides are pure or almost pure gamma emitters. A notable example is

technetium-99m, where the m indicates a

metastable state. It represents a mis-

arrangement of protons and neutrons in a nucleus after a neutron has become a

proton. In beta emitters, this arrangement occurs so rapidly that it appears to be

simultaneous with the original beta emission. In some cases, however, it is slow

enough to be obvious:

99m

Tc n

Tc +



21.2 Types of Radioactive Decay 907

Solution

222

Rn →

218

Po +

He +



220

Rn →

216

Po +

He +



219

Rn →

215

Po +

He +



PRACTICE 21.3

Write nuclear equations for the alpha decay of

218

Po and

230

Th.

See Problems 15a, 16a, 16c, 17a, and 19.

Other Types of Radioactive Decay

The three types of radioactive decay that we have discussed are the most com-

mon, but two other modes of nuclear decay are also important to a comprehen-

sive picture of radioactive processes.

Electron capture (EC) is the combination of

an inner-orbital electron and a proton from the nucleus to form a neutron. The

mass of the nuclide doesn’t change during the process because a proton and a

neutron are similar in mass, but the atomic number decreases by one as the pro-

ton is changed to a neutron. Typically, this process is accompanied by the emis-

sion of X-rays from the nuclide. The radioactive decay of iodine-125, which is

used to diagnose problems with the pancreas and intestines, occurs by the process

of electron capture:

125

I +

−1

 →

125

In positron emission (

), a proton decays into

a neutron and a positron. A neutrino accompanies

this emission, and usually one or more gamma rays

as well. A

positron, or positive electron, is a particle

that has the same mass as an electron but carries a

charge of +1. Interestingly, the positron typically

doesn’t have a very long life, because when it comes

into contact with an electron, the two particles

combine to form two gamma rays. This type of ra-

dioactive decay is important in the lighter elements

such as aluminum-26.

Al →

 +

Table 21.2 lists the ﬁve types of radioactive

decay that are important in understanding nuclear

decay processes.

Decay Series

Radon, discussed in Exercise 21.3, is one example in which the products of the

nuclear reaction are still radioactive. All isotopes of all elements past bismuth

(Z = 83) are radioactive. So far we have described nuclear decay as though it were

a one-step process. However, a nuclide may decay to form a second radioactive

nuclide, which in turn may decay more than a dozen times in a stepwise progres-

sion toward stability that is called a

decay series.

Consider the element uranium. Two natural isotopes exist for this element,

238

U and

235

U, with natural abundances of 99.28% and 0.72%, respectively. These

isotopes exhibit a fairly extensive decay series. Both series consist of alpha and

beta emissions, as shown in Figure 21.4. The accompanying gamma rays are not

shown in these series. Why do the lines in the decay series zig-zag? Alpha decays

FIGURE 21.3

Radon test kits can be purchased and

used to measure the concentration of

radon in a basement.

Radioactive Decay Processes

Change in Change in

Type of Decay Emission Atomic Number Mass Number

Alpha-particle

−2 −4

emission

Beta-particle

−1

 +10

emission

Gamma-ray

 00

emission

Positron

 −10

emission

Electron capture X-ray −10

TABLE 21.2

Uranium ore (known as yellow cake) can

be reﬁned into pellets of uranium metal.