ANSWERS TO
END-OF-CHAPTER QUESTIONS
CHAPTER 7: THE FIRES OF NUCLEAR FISSION
Emphasizing Essentials
1. Give two ways in which an atom of carbon can differ from another atom of carbon. Then give
three ways in which all carbon atoms differ from all uranium atoms.
Answer:
One carbon atom can differ from another in the number of neutrons (such as C-12 and C-13)
and in the number of electrons (we won’t be studying carbon ions, but these do exist). All
carbon atoms differ from all uranium atoms in their number of p, n and e and also in their
chemical properties.
2. Representations such as 14N or 15N give more information than simply the atomic symbol N.
Explain.
Answer:
The symbol N represents the element nitrogen and stands for the naturally occurring mixture of
all isotopes. The symbols 14N and 15N represent very specific isotopes with mass numbers of
14 and of 15, respectively.
3. a. How many protons does an atom of 239
94 Pu contain?
b. What element contains one more proton than uranium? Two more?
c. How many protons does radon-222 contain?

Answer:
a. 94 protons
b. Np (neptunium), Pu (plutonium)
c. 86 protons
4. Determine the number of protons and neutrons in each of these nuclei.
a. C-14 (radioactive)
b. C-12 (stable)
c. H-3 (tritium, a radioisotope of hydrogen)
d. Tc-99 (a radioisotope used in medicine)
Answer:
a. C-14 has 6 protons and 8 neutrons
b. C-12 has 6 protons and 6 neutrons
c. H-3 has 1 proton and 2 neutrons
d. Tc-99 has 43 protons and 56 neutrons
PAGE 7-1
5. E = mc2 is one of the most famous equations of the 20th century. What do the symbols in the
equation represent?
Answer:
E represents energy, m represents mass lost in a nuclear transformation, and c represents the
speed of light.
6. Give an example of a nuclear equation and of a chemical equation. In what ways are the two
equations alike? Different?
Answer:
Nuclear equation:
Chemical equation: CH4 + 2 O2  CO2 + 2 H2O
In a chemical equation, the elements do not change their identities in the process of being
converted from reactants to products. Even though they are combined differently, the same
number of atoms of each kind must appear on both sides of the equation. Usually only the
symbols for the elements are given, not their atomic or mass numbers. In a nuclear equation,
the identity of the elements may change, or at least the mass number may change. The values
for both the atomic and mass numbers are often included.
7. This nuclear equation represents a plutonium target being hit by an alpha particle. Show that
the sum of the subscripts on the left is equal to the sum of the subscripts on the right. Then do
the same for the superscripts.
Answer:
Subscripts: 94 + 2 = 96 on the left, 96 + 0 = 96 on the right
Superscripts: 239 + 4 = 243 on the left; 242+1=243 on the right
8. For the nuclear equation shown in the previous question,
a. suggest how the 42 He was produced.
b. 01 n is a product. What does this symbol represent?
c. explain why curium-243 is written in square brackets. Hint: See equation 7.1.

Answer:

a. The alpha particle may have come from the radioactive decay of another radioisotope.
b. 01 n represents a neutron.
c. Curium-243 represents an unstable intermediate in the nuclear reaction. This isotope has an
extremely short lifetime, decomposing immediately upon formation into Cm-242 with an
accompanying neutron.
9. Californium, element number 98, was first synthesized by bombarding an element with alpha
particles. The products were californium-245 and a neutron. What was the target isotope used
in this nuclear synthesis?
PAGE 7-2
Answer:
The target isotope must have been Cm-242. Here is the nuclear equation.
10. Explain the significance of neutrons in initiating and sustaining the process of nuclear fission.
In your answer, define and use the term chain reaction.
Answer:
Neutrons are needed to initiate the process of nuclear fission of U-235.
The fission products include 2 or 3 neutrons that can initiate more fission reactions. In this
manner, a self-sustaining chain reaction can be established in which the products of one
reaction initiate another.
11. Nuclear fission occurs through many different pathways. For the fission of U-235 induced by a
neutron, write a nuclear equation to form:
a. bromine-87, lanthanum-146, and more neutrons.
b. a nucleus with 56 protons, a second with a total of 94 neutrons and protons, and 2 additional
neutrons.
Answer:
a.
b.
12. This schematic diagram represents the reactor core of a nuclear power plant.
Match each letter with one of these terms.
fuel rods
cooling water into the core
cooling water out of the core
PAGE 7-3
control rod assembly
control rods
Answer:
A = control rod assembly, B = cooling water out of the core, C = control rods, D = cooling
water into the core, E = fuel rods
13. Identify the segments of the nuclear power plant diagrammed in Figure 7.6 that contain
radioactive materials and those that do not.
Answer:
The nuclear segment shown on the left in Figure 7.6 contains the reactor core, the heart of the
reactor where energy is produced. The non-nuclear portion is everything in the center and right
of the diagram, and includes the turbine and electrical generators. The non-nuclear portion also
contains the secondary and tertiary water systems, neither of which come into direct contact
with the reactor core.
14. Explain the difference between the primary coolant and the secondary coolant. The secondary
coolant is not housed in the containment dome. Why not?
Answer:
The primary coolant is the liquid surrounding the fuel bundles and control rods, a liquid that
comes in direct contact with the nuclear reactor to carry away heat. The heat from the primary
coolant is transferred to the secondary coolant, water in the steam generators that does not
come in contact with the reactor. The steam generators are separated from the nuclear reactor,
so the secondary coolant is not housed in the containment dome.
15. Boron can absorb neutrons.
a. Write the nuclear equation in which boron-10 absorbs a neutron to produce lithium-7 and an
alpha particle.
b. Boron, like cadmium, can be used in control rods. Explain.
Answer:
a.
b. Boron can be used to make control rods because it is a good neutron absorber.
16. What is an alpha particle? How is it represented? Answer these same questions for a beta
particle and for a gamma ray.
Answer:
An alpha particle is a helium nucleus consisting of 2 protons and 2 neutrons. It carries a +2
charge and is represented by α or 42 He .
A beta particle is an electron. It carries a negative one charge and is represented by the symbol
β or 01 e .

PAGE 7-4
A gamma ray is a high energy photon emitted from the nucleus of an atom. It is represented by
the symbol γ or 00γ.
17. Plutonium-239 decays by alpha emission.
a. Write the nuclear equation.
b. Plutonium is most hazardous when inhaled in particulate form. Explain.
c. Would you expect a sample of Pu-239 to decrease to background level in hours, days, or
years? Explain.
Answer:
a.
b. As a particulate, plutonium can be inhaled and become lodged in the lungs. If so, the
ionizing radiation it produces (alpha particles) can cause damage to lung cells. The product
U-235 also is radioactive and can damage tissue in the same way.
c. The half-life of Pu-239 is 24,110 years, so the timescale for a decrease to background level
would be on the order of years (actually more like centuries).
18. Iodine-131 decays by beta emission.
a. Write the nuclear equation.
b. Iodine, radioactive or not, accumulates in the body. Where?
c. Would you expect a sample of I-131 to decay in hours, days, or years? Explain.
Answer:
131
0
a. 131
53 I  54 Xe + –1 e
b. Iodine accumulates in the thyroid gland.
c. The half life of I-131 is 8.5 days, so a sample of I-131 will decay on a timescale of days.
19. Radioactive decay is accompanied by a change in the mass number, a change in the atomic
number, a change in both, or a change in neither. For the following types of radioactive decay,
which change(s) do you expect?
a. alpha emission
b. beta emission
c. gamma emission
Answer:
a. An alpha particle (helium nucleus) consists of 2 protons and 2 neutrons. If a nucleus releases
an alpha particle, both its mass number and its atomic number change.
b. A beta particle is a high-speed electron emitted from the nucleus. This emission can be
thought of as the conversion of a neutron into a proton plus this high-speed electron. If a
nucleus releases a beta particle, the atomic number increases but no change occurs in the mass
number.
c. A gamma ray is a high-energy photon of light. Gamma emission causes no change in either
the atomic number or mass number of the emitting nucleus.
PAGE 7-5
20. In a fashion similar to U-238 (see Figure 7.13), U-235 goes through a series of alpha and beta
decays before reaching a stable isotope. For practice, write the first six, which will bring you to
an isotope of radon. In order, the steps in the full radioactive decay series are a, b, a, b, a, a, a, b,
a, b, a, ending in stable Pb-207. Some steps have accompanying γ radiation, but you may omit
this.
Answer:
U 
 23190Th + 24 He
235
92
0
Th 
 231
91 Pa + -1 e
231
90
231
91
4
Pa 
 227
89 Ac + 2 He
227
89
Ac 
 22790Th + 10e
4
Th 
 223
88 Ra + 2 He
227
90
223
88
4
Ra 
 219
86 Rn + 2 He
21. Given that the average U.S. citizen receives 3600 Sv of radiation exposure per year, use the
data in Table 7.3 to calculate the percentage of radiation exposure the average U.S. citizen
receives from each of these sources.
a. food, water, and air
b. a dental X-ray twice a year
c. the nuclear power industry
Answer:
a. 2400 Sv compared to a total of 3600 Sv = 67% from food, water, and air
b. 200 Sv compared to a total of 3600 Sv = 5.6% from a dental X-ray twice a year
c. 0.09 Sv compared to a total of 3600 Sv. This depends on proximity to a nuclear power
plant, but in any case is extremely low assuming that no nuclear accidents occur.
22. What percent of a radioactive isotope would remain after two half-lives, four half-lives, and six
half-lives? What percent would have decayed after each period?
Answer:
It may be helpful to construct a chart.
# of half-lives % remaining
0
100
1
50
2
25
3
12.5
4
6.25
5
3.12
6
1.56
% decayed
0
50
75
87.5
93.75
97.88
98.44
23. Estimate the half-life of radioisotope X from this graph.
PAGE 7-6
Answer:
The half-life is 6 hours. The mass of the radioisotope falls from 100 mg to 50 mg in 6 hrs, and
then from 50 mg to 25 mg in the next 6 hours.
Concentrating on Concepts
24. The opening lines of this chapter connect acid rain and global warming to burning fossil fuels.
Explain the connection.
Answer:
Burning fossil fuels produces CO2, SO2, and NOx. (Remember, the NOx comes from the heat of
burning, not from the fossil fuels themselves.) SO2, and NOx dissolve in water to create acid
rain. CO2 is a greenhouse gas (see Chapter 3) that contributes to global warming.
25. In Consider This 7.1, you were asked to answer several questions about nuclear power. Ask the
same questions of someone at least one generation older than you and someone younger. In
comparison with your answer, what similarities and differences do you find?
Answer:
Older generations are more apt to remember the lessons of Chernobyl and the fears of Three
Mile Island. Younger generations may know nothing of either, and may have a variety of
opinions. All may be cognizant of the need to reduce carbon dioxide emissions and thus
rethinking nuclear power as an option. All generations may confuse nuclear power with
nuclear weapons.
26. The isotopes U-235 and U-238 are alike in that they are both radioactive. However, these two
isotopes have very different abundances in nature. List their natural abundances and explain
the significance of the difference.
Answer:
The natural abundances of U-238 and U-235 are 99.3% and 0.7%, respectively. U-235 can be
induced to undergo nuclear fission and thus is suitable for nuclear power plants and nuclear
weapons. Because U-235 is so rare, it is very difficult to procure in large quantities or in pure
form. This single fact has severely limited the number of nations that have nuclear weapons.
27. Consider the uranium fuel pellets used in commercial nuclear power plants.
a. Describe one way in which U-235 and U-238 can be separated.
b. Why is it necessary to enrich the uranium for use in the fuel pellets?
PAGE 7-7
c. The fuel pellets are enriched only to a few percent, rather than to 80–90%. Give three
reasons why.
d. Explain why it is not possible to separate the isotopes of uranium by chemical means.
Answer:
a. All means of separation exploit the tiny mass difference between U-235 and U-238. For
example, it is possible to separate them by converting the uranium sample to gaseous UF6 and
then use gas diffusion. A large high-speed centrifuge also can be used to separate these gas
molecules.
b. The uranium must be enriched to provide a critical mass of U-235 to sustain the chain
reaction responsible for energy production in the reactor.
c. First, the enrichment procedure is both expensive and energy intensive, so the minimum
enrichment level capable of sustaining a chain reaction is preferred. Second, reactors using
80-90% fuels have safety concerns due to the increased possibility of an uncontrolled chain
reaction. Third, such reactors would also have significant security issues. The highly enriched
fuel can be used directly in nuclear weapons, making the reactors potential terrorist targets.
d. The difference in the isotopes of uranium is in their nuclear masses. This difference is not
enough to significantly affect the chemical reactivity of the two isotopes. For chemical
separation, the isotopes of uranium would need to behave differently in a chemical reaction of
one sort or another.
28. a. Why must the fuel rods in a reactor be replaced every couple of years?
b. What happens to the fuel rods after they are taken out of the reactor?
Answer:
a. The fuel rods have to be replaced periodically because the fission products build up in them
over time. These fission products absorb neutrons, thus slowing the chain reaction.
b. Once fuel rods are removed, they are placed in pools to cool. If the reactor is in the United
States, the fuel rods are left in storage, usually in the vicinity of the reactor as there is no central
nuclear repository to accept nuclear waste. In many other parts of the world, they are
reprocessed into secondary nuclear fuels.
29. At full capacity, each reactor in the Palo Verde power plant uses only a few pounds of uranium
to generate 1243 megawatts of power. To produce the same amount of energy would require
about 2 million gallons of oil or about 10,000 tons of coal in a conventional power plant. How
is energy produced in the Palo Verde plant, compared with conventional power plants?
Answer:
The Palo Verde power plant produces energy through the process of nuclear fission. Coal and
oil burning plants generate energy by burning fossil fuels.
30. One important distinction between the Chernobyl reactors and those in the United States is that
those in Chernobyl used graphite as a moderator to slow neutrons, whereas U.S. reactors use
water. In terms of safety, give two reasons why water is a better choice.
PAGE 7-8
Answer:
Water is a better choice as a neutron adsorbent than graphite because water does not burn and
because it has a higher specific heat than graphite, and thus can more effectively adsorb and
help dissipate excess heat in the reactor.
31. If you look at nuclear equations in sources other than this textbook, you may find that the
subscripts have been omitted. For example, you may see an equation for a fission reaction
written this way.
a. How do you know what the subscripts should be? Why can they be omitted?
b. Why are the superscripts not omitted?
Answer:
a. The subscript for each element is its atomic number, which can be found in the periodic table
or in a list of elements. The subscript for the neutron is zero, which requires knowing or finding
the charge of a neutron in a reference table.
b. The superscripts cannot be omitted because nuclear equations must specify a specific
isotope and this is something that cannot be determined by looking at the periodic table or
other reference.
32. Using the model of a neutron presented in equation 7.6, explain how a high-speed electron can
be ejected from the nucleus in beta decay.
Answer:
A model for the process of beta decay is that a neutron is (somehow) transformed into a proton
and an electron. The electron is subsequently ejected from the nucleus as the beta particle,
leaving one more proton (and one less neutron) in the nucleus. Thus, the mass number does not
change since the number of protons and neutrons remains constant.
33. Coal can contain trace amounts of uranium. Explain why thorium must be found in coal as
well.
Answer:
Thorium is found with uranium because it found in the natural decay series of uranium.
34. Suppose somebody tells you that a radioisotope is “gone” after about seven half-lives. Critique
this statement, explaining both why it could be a reasonable assumption and why it might not
be.
Answer:
See question #22. After 7 half-lives, 99% of a sample has decayed (reasonably close to being
“gone”). However, the radioactivity actually is not gone, as 0.78% of the sample remains. Thus
if you start with a large amount of a radioactive substance (for example, 2000 pounds), after 7
half-lives have passed you have close to 10 pounds left. This could be considered a sizeable
amount.
PAGE 7-9
35. A website describing an X-ray procedure reports, “Despite its negative connotations, people
are exposed to more radiation on a daily basis than they may realize. For example, infrared
radiation is released whenever there is extreme heat. The sun generates ultraviolet radiation,
and a little exposure to it will tan a lighter skinned person. In addition, the body contains
naturally radioactive elements.” Examine the three examples given in this explanation. Do
they refer to nuclear or electromagnetic radiation?
Answer:
The third example, “naturally radioactive elements” is an example of exposure to nuclear
radiaton. In contrast, the first two examples, IR and UV “radiation” are not nuclear radiation.
Rather, they are electromagnetic radiation. While it may be true that “people are exposed to
more radiation on a daily basis than they may realize,” the explanation fails to differentiate
between nuclear (ionizing) radiation and electromagnetic radiation and thus is misleading.
36. Consider this representation of a Geiger–Müller counter (also called a Geiger counter), a
device commonly used to detect ionizing radiation. The probe contains a gas under low
pressure.
a. How does radiation enter the Geiger–Müller counter?
b. Why does this device only detect radiation that is capable of ionizing the gas contained in
the probe?
c. What are other methods for detecting the presence of ionizing radiation?
Answer:
a. Radiation enters through the thin window shown at the end of the tube.
b. When the gas is ionized, the ions allow an electric current to be established between the
anode and the cathode. If the radiation cannot ionize the gas, no current is generated (and no
radiation is detected).
c. Other methods of detecting radiation include film badges and scintillation counters.
37. Rapidly dividing cells are present in several places in the adult body. These include the skin,
the hair follicles, the stomach and intestines, the lining of the mouth, and the bone marrow.
Match the symptoms listed in Table 7.4 with the type of cell that was affected by the radiation.
PAGE 7-10
Answer:
Dose (rem)
0–25
25–50
Dose (Sv)
0–0.25
0.25–0.5
50–100
0.5–1
100–200
1–2
200–500
2–5
>500
>5
Likely Effect
No observable effect
White blood cell count
decreases slightly
Significant drop in white
blood cell count, lesions
Nausea, vomiting, loss of
hair
Hemorrhaging, ulcers,
possible death
Death
Cell Type Affected
Bone marrow
Bone marrow, skin
cells
Stomach and intestines,
hair follicles.
Skin, hair follicles,
stomach and intestines,
lining of mouth
All
38. Exposure to ionizing radiation can cause cancer. A beam of ionizing radiation also can be used
to cure certain types of cancer. Explain.
Answer:
Ionizing radiation can cause cancer by damaging the DNA of a cell. If the cell lives to
reproduce, the cell may be cancerous. However, a beam of strong ionizing radiation, when
directed at rapidly dividing cancer cells, this may kill the cancer cells.
39. Fluorine only has one naturally occurring radioisotope, F-19. If fluorine also occurred in nature
as F-18, would this necessarily complicate the separation of 238UF6 and 235UF6? Explain.
Answer:
The current separation is based on the difference in mass (=3) between 238U19F6 and 235U19F6. If
F-18 existed, two more compounds would be in the mix: 238U18F6 and 235U18F6. This would
complicate things, because 238U18F6 and 235U19F6 now only differ by 2 mass units, requiring an
even more careful separation.
40. a. Is depleted uranium (DU) still radioactive? Explain.
b. Is spent nuclear fuel (SNF) still radioactive? Explain.
Answer:
a. Depleted uranium is only weakly radioactive. It is primarily composed of U-238 which has a
half-life of 4.5 x 109 years and thus decays very slowly.
b. Spent fuel from a nuclear power plant is highly radioactive. It contains the fission products
of U-235 which tend to be beta emitters with relatively short half-lives. It also contains the
unburned uranium which is not very radioactive, as noted in part a.
41. It is generally believed that terrorists would be more likely to construct a nuclear bomb using
Pu-239 reclaimed from breeder reactors than using U-235. Use your knowledge of chemistry
to offer reasons for this.
PAGE 7-11
Answer:
Although uranium ore is readily available, the separation of U-235 from U-238 requires very
sophisticated technology such as gaseous diffusion or centrifugation. In contrast, plutonium
comes from spent reactor fuel. The separation of Pu-239 from uranium and other fission
products requires only chemical separations, which are far easier to carry out.
42. Weapons-grade plutonium is almost completely Pu-239. In contrast, the plutonium produced
in the normal operation of a water-cooled power reactor (reactor-grade plutonium) generally
has a higher concentration of heavier isotopes such as Pu-240 and Pu-241. Propose an
explanation for this observation.
Answer:
In a uranium reactor, U-238 absorbs a neutron to form U-239. This subsequently decays to
Np-239 which decays to Pu-239. Recall that there are a lot of neutrons present in the core of a
reactor. Pu-239 can capture one or two of these neutrons to become Pu-240.
43.
a. What are the characteristics of high-level radioactive waste (HLW)?
b. Explain how low-level waste (LLW) differs from HLW.
Answer:
a. Besides being highly radioactive, high-level nuclear waste is often in chemical forms that
are highly acidic or basic. It may also contain toxic metals. Because it contains fissionable
plutonium that could be extracted and used to construct nuclear weapons, it is also a security
risk. This waste is a byproduct of the nuclear weapons industry and nuclear power plants.
b. Low-level radioactive waste is waste contaminated with smaller quantities of radioactive
materials than high-level waste, and specifically excludes spent nuclear fuel. Examples of
low-level waste are discarded smoke detectors, radioactive pharmaceuticals, and the waste
materials from medical tests involving radioactive isotopes.
Exploring Extensions
44. Alchemists in the Middle Ages dreamed of converting base metals, such as lead, into precious
metals—gold and silver. Why could they never succeed? Today could we convert lead to gold?
Explain.
Answer:
Alchemists were perhaps the first practical chemists, but they did not have the advantage of
knowing anything about atomic structure or nuclear reactions. No chemical reaction can
produce gold from another element; a nuclear reaction is required. Even if they had envisioned
a nuclear reaction that would produce gold from another isotope, they clearly did not have the
means to accomplish this .The situation has indeed changed, and modern day chemists could
design experiments to change lead into gold. The question now is why anyone would want to,
as the cost would be prohibitive.
PAGE 7-12
45.
Make a time line of nuclear history, putting at least a dozen dates on your line. For
example, start with Becquerel’s discovery of radioactivity in 1896. Other candidates for
inclusion are Chernobyl, Hiroshima, the opening of the first commercial reactor, the discovery
of various medical isotopes, the use of uranium glazes in Fiesta ware, and the Nuclear Test Ban
Treaty.
Answer:
The Department of Energy website http://www.mbe.doe.gov/me70/history/doe_timeline.htm
is a ready source of this information. The site is divided into decades and students will find
ample events to enter onto their timelines.
46.
The Tennessee Valley authority’s nuclear reactor, Browns Ferry 1, did not operate
between 1985 and 2007. As this book went to print, it was back on line. What are the current
news reports?
Answer:
The Browns Ferry 1 reactor was re-started on May 22, 2007. Check the Tennessee Valley
Authority (TVA) website at http://www.tva.gov/sites/brownsferry.htm.
Although the restart was hailed by many, the TVA did cite cost overruns. A news article in the
Huntsville Times on December 15, 2007 reported that “Unit 1 has been off-line five times
since it returned to operation and the Nuclear Regulatory Commission recently conducted an
enhanced inspection, trying to understand the reason for the unplanned shutdowns. The NRC
has not yet released the findings from that inspection.”
47.
Explain the term decommission, as in “decommissioning a nuclear power plant.” What
technical challenges are involved? You might want to start by learning more about the
decommissioning of the Yankee Rowe facility (see Table 7.1). The resources of the Web can
help you.
Answer:
Decommissioning a nuclear power plant means shutting it down permanently. The Nuclear
Regulatory Commission’s website http://www.nrc.gov is a good source of information on the
status of nuclear power plants, functional or decommissioned, in the United States. Four stages
are involved with decommissioning:
1. removal of spent fuel from core and the spent fuel pools.
2. moving spent fuel to interim storage or reprocessing plant.
3. waiting until part of the radioactivity has decayed
4. dismantling of the plant
48. Einstein’s equation, E = mc2 applies to chemical changes as well as to nuclear reactions. An
important chemical change studied in Chapter 4 was the combustion of methane, which
releases 50.1 kJ of energy for each gram of methane burned.
a. What mass loss corresponds to the release of 50.1 kJ of energy?
PAGE 7-13
b. To produce the same amount of energy, what is the ratio of the mass of methane burned in a
chemical reaction to the mass loss converted into energy according to the equation E = mc2?
c. Use your results in parts a. and b. to comment on why Einstein’s equation, although correct
for both chemical and nuclear changes, usually is only applied to nuclear changes.
Answer:
a. E = mc2
2
 3.00 108 m 
103 J
1 kg
-10
50.1 kJ 
  mass, g  
 
 and the mass loss = 5.57  10 g
3
1 kJ
10 g 
s

b. To produce 50.1 kJ of energy, the ratio of masses is 1.00 g of methane burned to 5.57 10-10
g methane converted to energy, or about 1.80  109 to 1.
c. The amount of energy released per gram of reactant in a chemical reaction is relatively much
smaller than that released per gram of reactant in a nuclear reaction. Therefore, very little mass
is actually converted to energy in a chemical reaction, so it is reasonable to apply the Einstein
equation only to nuclear reactions.
49. When 4.00 g of hydrogen nuclei undergoes fusion to form helium in the Sun, the change in
mass is 0.0265 g and energy is released. Use Einstein’s equation, E = mc2, to calculate the
energy equivalent of this change in mass.
Answer:
E = mc2
 3.00 108 m 
1 kg
E  0.0265 g  3  

10 g
s


12
E  2.39  10 J
2
50. Under conditions like those on the Sun, hydrogen can fuse with helium to form lithium, which
in turn can form different isotopes of helium and of hydrogen.
a. What is the mass difference between a mole of the reactant and the product isotopes? Molar
masses are given below each isotope.
b. How much energy (in joules) is released in this reaction?
Answer:
a. The mass for one mole of each isotope is given. The sum of the masses of the reactants is
5.02838 g and the sum for the products is 5.00878 g. The difference is 0.0196 g.
b. The mass difference is equivalent to energy following Einstein’s equation, E = mc2.
 3.00  108 m 
1 kg
E  0.0196 g  3  

10 g
s


12
E = 1.76  10 J
2
PAGE 7-14
51. Lise Meitner and Marie Curie were both pioneers in developing an understanding of
radioactive substances. You likely have heard of Marie Curie and her work, but may not have
heard of Lise Meitner. How are these two women related in time and in their scientific work?
Answer:
Lise Meitner was an Austrian-born nuclear physicist who lived from 1878–1968. She worked
with the radiochemist Otto Hahn, discovering the element protactinium and studying the
effects of neutron bombardment on uranium. After leaving Nazi Germany in 1938, she found a
post at the Nobel Physical Institute in Stockholm. She continued her research there and,
together with her nephew Otto Frisch, did theoretical calculations to support Hahn’s
experimental work on splitting the uranium nucleus. Meitner and Frisch called the process
“nuclear fission.” During the war, she refused to work on the atomic bomb. In 1947 the
Swedish Atomic Energy Commission established a laboratory for her where she worked on an
experimental nuclear reactor.
Marie Curie, whose original name was Marja Sklodowska, was a Polish-born physicist and
chemist who lived from 1867-1934. Together with her husband, Pierre, she studied x-rays, the
harmful energy given off by some materials. She and Pierre together received the Nobel Prize
in Physics in 1903, based on their work on the radioactivity of uranium. For her individual
work, Marie was awarded the Nobel Prize in Chemistry in 1911.
Although the two women carried out research in the same fledging area of nuclear chemistry,
there is no record that they ever personally collaborated on any investigations. In fact, Marie
Curie was said to have rejected Lise Meitner’s application to work with her in Paris. More
information about these two scientists can be found at on the Web at
http://www.users.bigpond.com/Sinclair/fission/LiseMeitner.html and in Women in Chemistry,
a joint publication of the American Chemical Society and the Chemical Heritage Foundation.
52.
Taking potassium iodide tablets can protect your thyroid from exposure to radioactive
iodine, thus reducing your risk of thyroid cancer.
a. Give the chemical formula for potassium iodide.
b. By what mechanism does potassium iodide protect you?
c. How long does the protection last?
d. Are the tablets expensive? Hint: The FDA website is a good source of information for
parts b. and c.
Answer:
a. KI
b. By taking KI, you saturate your thyroid gland with nonradioactive iodide. This minimizes
the uptake of radioactive iodine by your thyroid gland, thus reducing the risk of thyroid cancer.
Taking KI does not prevent radioiodine from entering your body, however.
c. KI protects for approximately 24 hours.
d. KI can be purchased in various forms. The cost is usually low, depending (of course) on the
supplier.
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53.
A stockpile of approximately 50 metric tons of plutonium exists in the United States as a
result of disassembling warheads from the nuclear arms race. What should the fate of this
plutonium be? Hint: A search for “plutonium disposal” on the Web will bring up references.
Try also including United States and DOE as search terms.
a. Some propose that the plutonium be sent to local nuclear power plants to “burn” as
fissionable fuel. What are the advantages and disadvantages of such a course of action?
b. Others propose that it be stored permanently in a repository. Again, list the advantages and
disadvantages.
Answer:
a. Check the DOE website for up-to-date information on this issue. The plutonium from
nuclear warheads could be combined with uranium to form a mixed-oxide (MOX) fuel for use
in nuclear power plants. Those who support research into developing MOX technologies say
that the recycled fuel will reduce the need to mine uranium (a limited resource, and
environmental hazards accompany the mining) as well as reduce the amount of nuclear waste
in the U.S. Opponents fear that the creation of MOX fuel will pose a danger to the public, cost
billions of dollars, take years to complete, and encourage international commerce in plutonium,
which may end up in the hands of U.S. enemies.
b. Proponents of underground storage point out that it is a cheaper alternative both to
continuing to store the waste in its current locations and to developing MOX fuels. They also
point out the hazards of above ground storage, both in the long term and in the short.
Opponents of the development of underground storage point out the safety risks during
transportation of the waste from sites around the country, and the likelihood that the
underground storage will be not be geologically secure. Local opposition also exists at the
proposed storage site (Yucca Mountain) in the United States.
54.
Advertisements for Swiss Army watches stress their use of tritium. One ad states that the
“...hands and numerals are illuminated by self-powered tritium gas, 10 times brighter than
ordinary luminous dials...” Another advertisement boasts that the “...tritium hands and markers
glow brightly making checking your time a breeze, even at night...” Evaluate these statements
and, after doing some Web research, discuss the chemical form of tritium and its role in these
watches.
Answer:
Tritium, H-3, is a radioisotope of hydrogen. Hydrogen is a gas at room temperature, so it is
unlikely that the gas itself is contained in the watches. Several descriptions of this watch
mention the stainless steel screw-on back, again implying that no tritium gas could be inside.
The H-3 is most likely incorporated into an organic compound in the paint used for the markers
and hands of the watches. Also in the paint is a phosphor, that is, a substance that glows (is
phosphorescent) when hit by ionizing radiation (such as the beta particles emitted by tritium).
Indeed these can glow brightly, as claimed by the advertisement.
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55.
The amount of radiation you receive from medical X-rays varies
considerably, depending on the procedure. For example, dental
X-rays may be taken as “bitewings,” full mouth, or panoramic, each
involving a different dose. Select a type of medical X-ray and research
the amount of radiation involved. Report your data in both rems and
sieverts (using mrem, mSv, or Sv as appropriate).
Answer:
Students may have difficulty finding information on the Web. In our experience, dentists do
not know this information either! The American Dental Association (ADA) reports
(http://www.ada.org/public/topics/xrays_faq.asp) that the 4 films of “bitewings” have an
exposure of 0.038 mSv and the 19 films of a full mouth series is 0.150 mSv. The source cited
by the ADA was Frederiksen NL. “X-Rays: What is the Risk?” Texas Dental Journal, 1995,
vol.112(2), pages 68-72.
Noting that 1 rem = 0.0100 Sv, and 1 mrem = 10 µSv, here is the data in the other units
specified by the problem:
bitewing X-rays
0.038 mSv or 38 µSv
0.0038 rem or 3.8 mrem
full mouth series
0.150 mSv or 150 µSv
0.015 rem or 15 mrem
Table 7.3 lists 100 mSv for dental X-rays, which is in the same ballpark.
56.
The hormesis phenomenon, defined in Section 7.7, is that toxic substances in small
amounts can increase one’s resistance to the same substance in large amounts. Analogous to
the dose-response curves of Figure 7.20, the figure here indicates the zone of hormesis where
the curve dips below the control line into a therapeutic region. Use the resources of the Web to
investigate hormesis. Prepare a summary of your findings. One starting point is an article, “Is
Radiation Good for You?” at the science website The Why? Files. Hint: As in Figure 7.20,
NOAEL means no observed adverse effect level.
Answer:
Over the years, hormesis has had both strong supporters and nay-sayers. Most recently, the U.S.
National Research Council (NRC) reviewed the topic in 2006. “The committee concludes that
the assumption that any stimulatory hormetic effects from low doses of ionizing radiation will
have a significant health benefit to humans that exceeds potential detrimental effects from the
radiation exposure is unwarranted at this time.” (Quote from Health Risks from Exposure to
Low Levels of Ionizing Radiation: BEIR VII Phase 2, NRC)
57.
According to Table 7.3, smoking 1.5 packs of cigarettes a day adds 15,000 Sv to your
annual radiation dose.
a. Which radioactive isotopes are responsible for this dose in cigarette smoke?
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b. A nonsmoker, living with a 1.5-pack-a-day smoker, may receive the equivalent of 12 chest
X-rays per year as a result of the secondhand smoke. How many microsieverts would this
add to the nonsmoker’s annual dose?
Answer:
a. Polonium is the radioactive element, primarily Po-210. Its origin is believed to be the
fertilizers applied to the tobacco plants.
b. According to Table 7.3, chest X-rays add 100 mSv each. So 12 chest X-rays is the equivalent
of 1200 Sv.
58.
MRI, or magnetic resonance imaging, is an important tool for some types of medical
diagnoses.
a. What is the scientific basis for this technique?
b. What information can an MRI give a physician that cannot be obtained through direct
examination of a patient?
c. This MRI method used to be called NMR, nuclear magnetic resonance. Why do you think
the name was changed?
Answer:
a. An MRI contains a powerful magnet, and the part of the body to be imaged goes in the center
of the magnetic field. Using radio frequency waves (and a computer), the MRI collects data
point by point and constructs an image. A chemical may be injected, swallowed, or
administered intravenously to improve the contrast of the image. The website How Stuff Works
proves an excellent discussion of MRI technology.
b. MRI can provide an image of internal tissue, not available through direct examination.
c. The word nuclear, as pointed out in this chapter, carries many connotations. To some if not
all patients, this would imply use of something radioactive. This is not the case with MRI. The
“nuclear” in NMR is referring to a change of the nucleus, not to a radioisotope.
59. Deciding where to locate a nuclear power plant requires analysis of both risks and benefits
associated with the plant. If you were to play the role of a CEO of a major electrical utility
considering whether to pursue permits for the construction of a nuclear power plant in your
area, what risks and benefits would you cite?
Answer:
Benefits would include electric power for the community, no greenhouse gas emissions, no
SO2 and NOx emissions, and possibly employment for community members at the power plant.
Cost of the plant might also be cited as a benefit, depending on the alternatives. Risks would
include anything related to the nuclear fuel used in the reactor – its production, transport, use,
storage and/or transport of waste after use, and any possible release of radioisotopes to the
community due to a plant accident.
60.
As this book went to press, final plans were being drafted for the 2010 construction of a
new containment dome for Chernobyl. Accomplishing this feat will be difficult, as radiation
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levels on site are still high, and the melted core of the reactor is still thermally hot. Use the
resources of the Web to find out the current status of the project, including the safety measures
proposed to protect the workers.
Answer:
This question may not be easy to answer. According to BBC news, September 2007, “The
French construction company Novarka will build a giant arch-shaped structure out of steel,
190 metres (623 feet) wide and 200m long. It will cover the existing containment structure
which stands over the reactor and radioactive fuel that caused the accident in 1986.” We were
not yet able to find anything about the safety measures to protect the workers as this went to
print.
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