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16.1 The Nucleus and Radioactivity
Our journey into the center of the atom begins with a brief review. You learned in Chapter 3 that the protons and neutrons in each atom are found in a tiny, central nucleus that measures about 1/100,000 the diameter of the atom itself. You also learned that the atoms of each element are not necessarily identical; they can differ with respect to the number of neutrons in their nuclei. When an element has two or more species of atoms, each with the same number of protons but a different number of neutrons, the different species are called isotopes. Different isotopes of the same element have the same atomic number, but they have a different mass number, which is the sum of the numbers of protons and neutrons in the nucleus. In the context of nuclear science, protons and neutrons are called nucleons , because they reside in the nucleus. The atom’s mass number is often called the nucleon number , and a particular type of nucleus, characterized by a specific atomic number and nucleon number, is called a nuclide. Nuclides are represented in chemical notation by a subscript atomic number (Z) and superscript nucleon number (A) on the left side of the element’s symbol (X):
Z^ X
A
Atomic number
Mass number (nucleon number)
Element symbol
For example, the most abundant nuclide of uranium has 92 protons and 146
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16.1 The Nucleus and Radioactivity 693
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neutrons, so its atomic number is 92, its nucleon number is 238 (92 + 146), and its symbol is (^23892) U. Often, the atomic number is left off of the symbol. Nuclides can also be described with the name of the element followed by the nucleon number. Therefore, (^23892) Uis commonly described as 238 U or uranium-238. Examples 16.1 and 16.2 provide
practice in writing and interpreting nuclide symbols.
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exerCise 16.1 - Nuclide Symbols
One of the nuclides used in radiation therapy for the treatment of cancer has 39 protons
and 51 neutrons. Write its nuclide symbol in the form of AZ X. Write two other ways
to represent this nuclide.
example 16.1 - Nuclide Symbols
A nuclide that has 26 protons and 33 neutrons is used to study blood chemistry. Write
its nuclide symbol in the form of AZ X. Write two other ways to represent this nuclide.
Solution
Because this nuclide has 26 protons, its atomic number, Z, is 26, identifying the element
as iron, Fe. This nuclide of iron has 59 total nucleons (26 protons + 33 neutrons), so
its nucleon number, A, is 59.
26 Fe^ or^59 Fe^ or^ iron-
59
exerCise 16.2 - Nuclide Symbols
A nuclide with the symbol 201 Tl can be used to assess a patient’s heart in a stress test. What is its atomic number and mass number? How many protons and how many neutrons are in the nucleus of each atom? Write two other ways to represent this nuclide.
example 16.2 - Nuclide Symbols
Physicians can assess a patient’s lung function with the help of krypton-81. What is this nuclide’s atomic number and mass number? How many protons and how many neutrons are in the nucleus of each atom? Write two other ways to represent this nuclide.
Solution
The periodic table shows us that the atomic number for krypton is 36, so each krypton atom has 36 protons. The number following the element name in krypton-81 is this nuclide’s mass number. The difference between the mass number (the sum of the numbers of protons and neutrons) and the atomic number (the number of protons) is equal to the number of neutrons, so krypton-81 has 45 neutrons (81 - 36).
atomic number = 36; mass number = 81; 36 protons and 45 neutrons 36 Kr^81 Kr
81
16.1 The Nucleus and Radioactivity 695
Figure 16. The Band of Stability
There are 264 stable nuclides found in nature. The graph in Figure 16.1 shows the neutron-to-proton ratios of these stable nuclides. Collectively, these nuclides fall within what is known as the band of stability.
40
Number of neutrons
Number of protons
One to one neutron to proton ratio
Band of stability
Island of stability
A nuclide containing numbers of protons and neutrons that place it outside this
band of stability will be unstable until it undergoes one or more nuclear reactions that
take it into the band of stability. We call these unstable atoms radioactive nuclides ,
and the changes they undergo to reach stability are called radioactive decay. Note
that the band of stability stops at 83 protons. All of the known nuclides with more
than 83 protons are radioactive, but scientists have postulated that there should be a
small island of stability around the point representing 114 protons and 184 neutrons.
The relative stability of the heaviest atoms that have so far been synthesized in the
laboratory suggests that this is true. (See Special Topic 3.1: Why Create New Elements .).
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Types of Radioactive Emissions One of the ways that nuclides with more than 83 protons change to reach the band of stability is to release two protons and two neutrons in the form of a helium nucleus, which in this context is called an alpha particle. Natural uranium, which is found in many rock formations on earth, has three isotopes that all experience alpha emission , the release of alpha particles. The isotope composition of natural uranium is 99.27% uranium-238, 0.72% uranium-235, and a trace of uranium-234. The nuclear equation for the alpha emission of uranium-238, the most abundant isotope, is
Two protons and two neutrons lost
The protons and neutrons leave as an alpha particle.
92U
238 90 Th
234 2 He
+^4
In nuclear equations for alpha emission, the alpha particle is written as either a or 2He
(^4). Note that in alpha emission, the radioactive nuclide changes into a different element, with an atomic number that is lower by 2 and a mass number that is lower by 4. Some radioactive nuclides have a neutron-to-proton ratio that is too high, placing them above the band of stability. To reach a more stable state they undergo beta emission (b-). In this process, a neutron becomes a proton and an electron. The proton stays in the nucleus, and the electron, which is called a beta particle in this context, is ejected from the atom. n → p + e- In nuclear equations for beta emission, the electron is written as either b, b-, or −1e
(^0). Iodine-131, which has several medical uses, including the measurement of iodine uptake by the thyroid, is a beta emitter:
e−
A neutron becomes a proton (which stays in the nucleus) and an electron (which is ejected from the atom).
53I
131 54 Xe
131 − 1 e
+^0
Energy
Note that in beta emission, the radioactive nuclide changes into a different element, with an atomic number that is higher by 1 but the same mass number.
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698 Chapter 16 Nuclear Chemistry
Because radioactive decay leads to more stable products, it always releases energy. Some of this energy is released in the form of kinetic energy, adding to the motion of the product particles, but often some of it is given off as the form of radiant energy called gamma rays. Gamma rays can be viewed as streams of high energy photons. For example, cobalt-60 is a beta emitter that also releases gamma radiation. The energy released in the beta emission leaves the product element, nickel-60, in an excited state. When the nickel-60 descends to its ground state, it gives off photons in the gamma ray region of the radiant energy spectrum. (See Section 4.1 for a review of the different forms of radiant energy.)
Excited state e-
(^6027) Co 6028 Ni* + (^) -1 (^0) e Ni + γ-photon
γ
Beta emission
Gamma photon
60 28
Nuclear Reactions and Nuclear Equations Now that we have seen some examples of nuclear reactions, let’s look more closely at how they differ from the chemical reactions we have studied in the rest of this text. Nuclear reactions involve changes in the nucleus, whereas chemical reactions involve the loss, gain, and sharing of electrons. Different isotopes of the same element may undergo very different nuclear reactions, even though an element’s isotopes all share the same chemical characteristics. Unlike chemical reactions, the rates of nuclear reactions are unaffected by temperature, pressure, and the presence of other atoms to which the radioactive atom may be bonded. Nuclear reactions, in general, give off much more energy than chemical reactions. The equations that describe nuclear reactions are different from those that describe chemical reactions because in nuclear equations charge is disregarded. If you study the nuclear changes for alpha, beta, and positron emission already described in this section, you will see that the products must be charged. For example, when an alpha particle is released from a uranium-238 nucleus, two positively charged protons are lost. Assuming that the uranium atom was uncharged initially, the thorium atom formed would have a - 2 charge. Because the alpha particle is composed of two positively charged protons and two uncharged neutrons (and no electrons), it has a +2 overall charge. (^23892) U → 23490 Th 2 − (^) + 42 He 2 +
The ions lose their charges quickly by exchanging electrons with other particles. Because we are usually not concerned about charges for nuclear reactions, and because these charges do not last very long, they are not usually mentioned in nuclear equations.
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16.1 The Nucleus and Radioactivity 699
Scientists may not be interested in the charges on the products of nuclear reactions, but they are very interested in the changes that take place in the nuclei of the initial and final particles. Therefore, nuclear equations must clearly show the changes in the atomic numbers of the nuclides (the number of protons) and the changes in their mass numbers (the sum of the numbers of protons and neutrons). Note that in each of the following equations, the sum of the superscripts (mass numbers, A) for the reactants is equal to the sum of the superscripts for the products. Likewise, the sum of the subscripts (atomic numbers, Z) for the reactants is equal to the sum of the subscripts for the products. To show this to be true, beta particles are described as (^) −^0 1e^ , and positrons are described as (^) +1e^0.
(^23892) U 23490 Th + 42 He
(^4019) K 4018 Ar + (^) + (^01) e
(^13153) I 13154 Xe + (^) − (^01) e
−^01 e^ +^12553 I^12552 Te
mass number 238 234 + 4 = 238
Alpha emission
atomic number 92 90 + 2 = 92
mass number 131 131 + 0 = 131
Beta emission
atomic number 53 54 + (−1) = 53
mass number 40 40 + 0 = 40
Positron emission
atomic number 19 18 + 1 = 19
mass number 0 + 125 = 125 125
Electron capture
atomic number − 1 + 53 = 52 52
The following general equations describe these nuclear changes:
Alpha emission AZX AZ^ −−^42 Y + 42 He
Beta emission AZX (^) Z + A 1 Y + (^) −^01 e
Positron emission AZX^ Z −A 1 Y^ +^ +^01 e
Electron capture (^) −^01 e + AZX (^) Z −A 1 Y
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Table 16.2 on the next page summarizes the nuclear changes described in this section.
16.1 The Nucleus and Radioactivity 701
c. The symbol for nitrogen-13 is (^13) 7N^ , and the symbol for a positron is (^) + 10 e^. Therefore, the beginning of our equation is 7N^ →^ _____^ +^ + 1 e
13 0 To make the subscripts balance, the subscript for the missing nuclide must be 6, so the symbol for the product nuclide is C, for carbon. The mass number stays the same in positron emission, so we write 13. (^137) N → (^136) C + (^) + (^01) e
d. The symbol for gallium-67 is (^67) 31Ga^ , and the symbol for an electron is (^) − 10 e. Therefore, the beginning of our equation is 31 Ga^ +^ −1e^ →^ _____
67 0 To balance the subscripts, the atomic number for our missing nuclide must be 30, so the symbol for the product nuclide is Zn, for zinc. The mass number stays the same in electron capture, so we write 67. 31 Ga^ +^ − 1e^ →^ 30 Zn
67 0 67
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exerCise 16.3 - Nuclear Equations
Write nuclear equations for (a) alpha emission by plutonium-239, one of the substances formed in nuclear power plants, (b) beta emission by sodium-24, used to detect blood clots, (c) positron emission by oxygen-15, used to assess the efficiency of the lungs, and (d) electron capture by copper-64, used to diagnose lung disease.
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example 16.4 - Nuclear Equations
Glenn Seaborg and his team of scientists at the Lawrence Laboratory at the University of California, Berkeley, created a number of new elements, some of which—berkelium, californium, lawrencium—have been named in honor of their work. Complete the following nuclear equations that describe the processes used to create these elements.
a. 24496 Cm^ + (^) 2He^4 → _____ + (^1) 1H + (^2) 0n^1 b. (^238) 92U^ + _____ → (^246) 98Cf + 40n^1 c. (^) _____ + (^10) 5B^ → (^) 103Lr^257 + (^5) 0n^1
Solution
First, determine the subscript for the missing formula by asking what number would make the sum of the subscripts on the left of the arrow equal the sum of the subscripts on the right. That number is the atomic number of the missing nuclide and leads us to the element symbol for that nuclide. Next, determine the superscript for the missing formula by asking what number would make the sum of the superscripts on the left of the arrow equal to the sum of the superscripts on the right.
a. 24496 Cm^ + (^4) 2He^ →^24597 Bk + (^1) 1H + (^2) 0n^1 b. (^238) 92U^ +^126 C^ → (^246) 98Cf + 410 n c.^25298 Cf^ + (^10) 5B^ → (^) 103Lr^257 + 510 n
Example 16.4 shows how you can complete a nuclear equation when one of the
symbols for a particle is missing.
702 Chapter 16 Nuclear Chemistry
exerCise 16.4 - Nuclear Equations
Complete the following nuclear equations.
a. (^14) 7N^ + (^4) 2He → _____ + (^) 1H^1
b. (^238) 92U^ + _____ → (^247) 99Es + 50n^1
c. (^) _____ + (^) 1H^2 → (^239) 93Np + (^) 0n^1
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Rates of Radioactive Decay
Because the different radioactive nuclides have different stabilities, the rates at which they decay differ as well. These rates are described in terms of a nuclide’s half-life , the time it takes for one-half of a sample to disappear. For example, radioactive carbon-14, which decays to form nitrogen-14 by emitting a beta particle, has a half- life of 5730 years. After 5730 years, one-half of a sample remains, and one-half has become nitrogen-14. After 11,460 years (two half-lives), half of that remainder will have decayed to form nitrogen-14, bringing the sample down to one-fourth of its original amount. After 17,190 years (three half-lives), half of what remained after 11,460 years will have decayed to form nitrogen-14, so one-eighth of the original sample will remain. This continues, with one-half of the sample decaying each half-life. Imagine having a pie and being told that you are only allowed to eat one- half of whatever amount is on the plate each day. The first day you eat one-half of the pie. The next day you eat half of what is there, but that’s only one-fourth of a pie (^1 2 × 1 2 ). The next day you can only eat one-eighth of the original pie (^1 2 × 1 4 or (^1 2) × 1 2 × 1 2 ), and on the next day one-sixteenth (^1 2 × 1 8 or (^1 2) × 1 2 × 1 2 × (^1 2) ). On the fifth day (after five half-lives), the piece you eat is only 1 32 of the original pie (^1 2 × 1 16 or (^1 2) × 1 2 × 1 2 × 1 2 × 1 2 ). The process continues until there is not enough pie to bother eating any. It’s a similar situation with radioactive nuclides. One-half of their amount disappears each half-life until there’s no significant amount left. The length of time necessary for a radioactive sample to dwindle to insignificance depends on its half-life and the amount that was present to begin with (Figure 16.2). In subsequent chemistry or physics courses, you might learn a general technique for using a nuclide’s half-life to predict the length of time required for any given percentage of a sample to decay. Example 16.5 gives you a glimpse of this procedure by showing how to predict the length of time required for a specific radioactive nuclide (with a given half-life) to decay to 1 2 , 1 4 , 1 8 , 1 16 , or 1 32 of its original amount. Example 16.6 shows how you can predict what fraction of a sample will remain after one, two, three, four, or five half-lives.
704 Chapter 16 Nuclear Chemistry
example 16.5 - Half-Life Radon-222, which is found in the air inside houses built over soil containing uranium, has a half-life of 3.82 days. How long before a sample decreases to 1 32 of the original amount? Solution In each half-life of a radioactive nuclide, the amount diminishes by one-half. The fraction 1 32 is (^1 2) × 1 2 × 1 2 × 1 2 × 1 2 , so five half-lives are needed to reduce the sample to that extent. For radon-222, five half-lives are 19.1 days (5 × 3.82 days).
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exerCise 16.5 - Half-Life
One of the radioactive nuclides formed in nuclear power plants is hydrogen-3, called tritium, which has a half-life of 12.26 years. How long before a sample decreases to (^1 ) of its original amount?
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example 16.6 - Half-Life One of the problems associated with the storage of radioactive wastes from nuclear power plants is that some of the nuclides remain radioactive for a very long time. An example is plutonium-239, which has a half-life of 2.44 × 10 4 years. What fraction of plutonium-239 is left after 9.76 × 10 4 years? Solution The length of time divided by the half-life yields the number of half-lives: 9.76 × 104 years 2.44 × 10 4 years
= 4 half-lives
In each half-life of a radioactive nuclide, the amount diminishes by one-half, so the fraction remaining would be^1 16 (^1 2 × 1 2 × 1 2 × 1 2 ).
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exerCise 16.6 - Half-Life
Uranium-238 is one of the radioactive nuclides sometimes found in soil. It has a half-life of 4.51 × 109 years. What fraction of a sample is left after 9.02 × 109 years?
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Radioactive Decay Series Many of the naturally-occurring radioactive nuclides have relatively short half-lives. Radon-222, which according to U.S. Environmental Protection Agency estimates causes between 5000 and 20,000 deaths per year from lung cancer, has a half-life of only 3.82 days. With such a short half-life, why are this and other short-lived nuclides still around? The answer is that although they disappear relatively quickly once they form, these nuclides are constantly being replenished because they are products of other radioactive decays. Three relatively abundant and long-lived radioactive nuclides are responsible for producing many of the other natural radioactive isotopes on earth. One of them is
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uranium-238, with a half-life of 4.51 billion years, which changes to lead-206 in a series of eight alpha decays and six beta decays (Figure 16.3). Chemists call such a sequence a nuclear decay series. Because this sequence of decays is happening constantly in soil and rocks containing uranium, all of the radioactive intermediates between uranium-238 and lead-206 are constantly being formed and are therefore still found in nature.
Figure 16. Uranium-238 Decay Series
206
210
214
218
222
226
230
234
238
Mass number
Atomic number
81 82 83 84 85 86 87 88 89 90 91 92
(^23490) Th
23892 U
(^23491) Pa (^23492) U
(^23090) Th
(^22688) Ra
(^22286) Rn
(^21884) Po
(^21482) Pb (^21483) Bi (^21484) Po
(^21082) Pb (^21083) Bi (^21084) Po
(^20682) Pb
You can see in Figure 16.3 that one of the products of the uranium-238 decay series is radium-226. This nuclide, with a half-life of 1622 years, is thought to be the second leading cause of lung cancer, after smoking. The next step in this same decay series forms radon-222. Radon-222 is also thought to cause cancer, but it does not do so directly. Radon-222 is a gas, and enters our lungs through the air. Then, because of its fairly short half-life, a significant amount of it decays to form polonium-218 while still in our lungs. Polonium and all of the radioactive nuclides that follow it in the decay series are solids that stay in the lining of the lungs, emitting alpha particles, beta particles, and gamma rays. Houses built above earth that contains uranium can harbor significant concentrations of radon, so commercial test kits have been developed to detect it. Radon is a bigger problem in colder climates and at colder times of the year because it accumulates inside houses that are sealed up tight to trap warm air. The other two important decay series are the one in which uranium-235 (with a half-life of 7.13 × 10 8 years) decays in eleven steps to lead-207 and the one in which thorium-232 (which has a half-life of 1.39 × 1010 years) decays in ten steps to lead-208.
Radon Test Kit Image provided and used with the permission of PRO-LAB, Inc.
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16.2 Uses for Radioactive Substances 707
Chernobyl accident released large quantities of it. To reduce the likelihood of thyroid damage, people were directed to take large quantities of salt containing non-radioactive iodine-127. This flooding of the thyroid glands with the non-damaging form of iodine made it less likely that the iodine-131 would be absorbed. Because alpha particles are relatively large and slow moving compared to other emissions from radioactive atoms, it is harder for them to slip between the atoms in the matter through which they pass. Alpha particles are blocked by 0.02 mm to 0.04 mm of water or about 0.05 mm of human tissue. Therefore, alpha particles that strike the outside of the body enter no further than the top layer of skin. Because beta particles are much smaller, and can move up to 90% the speed of light, they are about 100 times as penetrating as alpha particles. Thus beta particles are stopped by 2 mm to 4 mm of water or by 5 mm to 10 mm of human tissue. Beta particles from a source outside the body may penetrate to the lower layers of skin, but they will be stopped before they reach the vital organs. Gamma photons are much more penetrating, so gamma radiation from outside the body can do damage to internal organs. Although alpha and beta radiation are less damaging to us than gamma rays when emitted from external sources, both forms of radiation can do significant damage when emitted from within the body (by a source that has been eaten or inhaled). Because they lose all of their energy over a very short distance, alpha or beta particles can do more damage to localized areas in the body than the same number of gamma photons would.
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