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***** Module 25 – Nuclear Chemistry
Module 25 – Nuclear Chemistry ..................................................................................549
Lesson 25A:
Lesson 25B:
Lesson 25C:
Lesson 25D:
Lesson 25E:
Lesson 25F: The Nucleus  Review ......................................................................................549
Radioactive Decay Reactions............................................................................554
Fission and Fusion .............................................................................................558
Fractions and Percentages.................................................................................561
Natural Logarithms ...........................................................................................566
Radioactive HalfLife Calculations..................................................................575
For additional lessons, visit www.ChemReview.Net © 2011 www.ChemReview.Net v.1g Page ii Module 25 – Nuclear Chemistry Module 25 – Nuclear Chemistry
Prerequisites: This module may be started at any point after Lessons 6A and 6B.
***** Lesson 25A: The Nucleus – Review
Pretest: If you recall the rules for the structure of the nucleus, try the problems in the
practice set at the end of this lesson. If you can do those problems, you may skip this
lesson.
***** Chemistry and Nuclear Reactions
The rules for nuclear reactions are quite different from those that govern chemistry. For
example,
• During chemical reactions, the nucleus is not changed in any of the reacting atoms.
Since the number of protons in the nucleus determines the identity of the atom, the
number and kind of atoms must remain the same during chemical reactions. In
nuclear reactions, nuclei can combine or divide to form different atoms. • In nuclear reactions, the amount of energy added or released per atom is nearly
always much higher in chemical reactions. • On earth, chemical reactions occur with relative ease. In every biological organism,
millions of chemical reactions occur each minute. Nuclear reactions occur naturally
during radioactive decay, and may occur when cosmic rays enter our atmosphere,
but on earth, compared to chemical reactions, nuclear reactions are rare. (In a star,
nuclear reactions are common.) Because the rules are so different, the detailed study of nuclear reactions is generally
assigned to physics rather than chemistry. We make an exception for three nuclear
reactions: radioactive decay, fission, and fusion. In nuclear chemistry, radioactive decay is
a powerful tool in the study of chemical reactions. Together, decay, fission, and fusion
explain how the atoms that we study in chemistry came to be formed. A Model for the Nucleus
Science has only a partial understanding of the nature of the atomic nucleus. When
understandings in science are limited, models are developed that may be simplified,
incomplete, or even speculative, but allow us to predict how systems will behave.
To explain the nuclear reactions that are important in chemistry, the model for the nucleus
that we utilize in chemistry is simplified compared to the models of physics, but this
simplified model can nearly always predict the impact of the structure of the nucleus on
processes of interest in chemistry and biology. © 2011 www.ChemReview.Net v.1g Page 549 Module 25 – Nuclear Chemistry Our chemistry model for the atom and its nucleus was introduced in Lesson 6B. To briefly
review:
1. Nuclear Structure
Atoms are composed three subatomic particles. In standard chemistry, our primary
focus is on electrons. In nuclear chemistry, our focus is on protons and neutrons in the
nucleus of the atom.
• Protons
o o The number of protons is the atomic number of a nucleus or atom. The number
of protons determines the name (and thus the symbol) of a nucleus or atom. The
number of protons determines the nuclear charge of an atom’s nucleus. o • Protons have a +1 electrical charge (1 unit of positive charge). Each proton has
an mass of approximately 1.0 amu (atomic mass units), which is equivalent to
1.0 grams per mole. The number of protons is a major factor in the atom’s behavior. The number of
protons in an atom is never changed by chemical reactions, but can change
during nuclear reactions. Neutrons
o Neutrons have an electrical charge of zero. A neutron has about the same mass
as a proton: 1.0 amu. o Neutrons are believed to act as the glue of the nucleus: the particles that keep
the repelling protons from flying apart. o Neutrons, like protons, are never gained or lost in chemical reactions, but the
number of neutrons and protons in an atom can change in a nuclear reaction. o Unlike the number of protons, the number of neutrons in an atom has little to
no influence on the types of chemical reactions that substances containing that
atom will undergo. However, nuclei with the same number of protons but
different numbers of neutrons will often undergo different nuclear reactions.
In addition, atoms with the same numbers of protons but differing numbers of
neutrons in their nucleus will have different masses. As a result, some physical
properties of the particles of a substance, such as their densities and the relative
speeds at which the particles move, may differ measurably if its atoms have
differing numbers of neutrons. 2. The Nucleus
All of the protons and neutrons in an atom are found in the nucleus at the center of the
atom. The diameter of a nucleus is roughly 100,000 times smaller than the effective
diameter of most atoms. However, the nucleus contains all of an atom’s positive
charge and nearly all of its mass. Electrons are located outside the nucleus and are
much lighter than protons and neutrons. © 2011 www.ChemReview.Net v.1g Page 550 Module 25 – Nuclear Chemistry 3. Types of Nuclei
Only certain combinations of protons and neutrons form a nucleus that is stable. In a
nuclear reaction, if a combination of protons and neutrons is formed that is unstable,
the nucleus will decay. In terms of stability, nuclei can be divided into three types.
• Stable nuclei are combinations of protons and neutrons that do not change in a
planetary environment such as Earth over many billions of years. • Radioactive nuclei are somewhat stable. Some radioactive nuclei exist for only a few
seconds, and others exist on average for several billion years, but they fall apart
(decay) at a constant and characteristic rate. • Unstable nuclei, if formed in nuclear reactions, decay within a few seconds. Nuclei that exist in the earth’s crust include all of the stable nuclei plus some
radioactive nuclei. All atoms with between one and 82 protons [except technetium (Tc)
with 43 protons] have at least one nucleus found in the earth’s crust that is stable.
Atoms with 83 to 92 protons exist in the earth’s crust but are always radioactive.
Atoms with 93 or more protons exist on earth only when they are created in manmade
nuclear reactions.
Radioactive atoms comprise a very small percentage of the matter on earth. Over
99.99% of the earth’s atoms have stable nuclei that have not changed since their atoms
came together to form the earth billions of years ago.
4. Terminology
Protons and neutrons are termed the nucleons. The combination of a certain number
of protons and neutrons is called a nuclide. The set of nuclides that have the same
number of protons (so they are the same element) but differing numbers of neutrons
are called the isotopes of the element. Isotopes
Some elements have only one stable nuclide; others have as many as 10 stable isotopes.
Examples: All atoms with 1 proton are called hydrogen. Two kinds of hydrogen nuclei
are stable: those with
• 1 proton; and • 1 proton and 1 neutron, an isotope that is often referred to as deuterium or
heavy hydrogen. Most hydrogen atoms found on earth are the isotope containing one proton and no
neutrons: only 1 H atom in about 6,400 contains a deuterium nucleus. However,
deuterium can be separated from the majority isotope, and it has many important uses
in chemistry. As a result, deuterium is often represented in substance formulas by its
own “atomic” symbol: a D. When both hydrogens in a water molecule contain a
deuterium nucleus, its formula may be written as D2O (instead of H2O). Water that
has a substantially higher percentage of deuterium than normal is termed heavy water.
An isotope of hydrogen consisting of one proton and two neutrons, called tritium, is
not found in the earth’s crust, but it can be isolated in measurable amounts from © 2011 www.ChemReview.Net v.1g Page 551 Module 25 – Nuclear Chemistry products in nuclear reactors. Unlike deuterium, tritium is radioactive. Half of the
atoms in a sample of tritium will decay in 12 years. Nuclide Symbols
Each nuclide has a mass number which is the sum of its number of protons and neutrons.
Mass Number of a nucleus = Protons + Neutrons
Example: All nuclei with 6 protons are carbon. If a carbon nucleus has 8 neutrons, the
mass number of the carbon isotope is 14.
A nuclide can be identified in two ways,
• by its number of protons and number of neutrons, or • by its nuclide symbol (also termed its isotope symbol). The nuclide symbol for an atom has two required parts: the element symbol and the mass
number. The mass number is written as a superscript in front of the atom symbol.
Example: The three isotopes of hydrogen can be represented as
• 1 proton + no neutrons or as 1H (a nuclide named “hydrogen1”); • 1 proton + 1 neutron or as 2H (termed hydrogen2 or deuterium); and • 1 proton + 2 neutrons or as 3H (called hydrogen3 or tritium). Uranium has two isotopes that are important commercially and historically.
• 238U , the most common isotope, contains 92 protons and 146 neutrons. • 235U , the isotope that is split in atomic bombs and nuclear power plants, contains 92 protons and 143 neutrons.
Using a table of atoms that includes atomic numbers, try this question.
Q. A nuclide with 47 protons and 62 neutrons has what nuclide symbol? *****
A. Atoms with 47 protons must be named silver, symbol Ag. The mass number of this
nuclide is 47 protons + 62 neutrons = 109 . This isotope is called silver109 and its
symbol is 109Ag . Nuclide symbols may also be written with the nuclear charge below the mass
number. This is called AZ notation, illustrated for tritium at the right. A is the
symbol for mass number and Z is the symbol for nuclear charge. 3
1H Any nucleus that includes protons is by definition an atom, and since the atom symbol also
identifies the number of protons in the nucleus, Z values are not required to identify a
nucleus. However, for many subatomic particles, the nuclear charge is not equal to the
number of protons, and showing the nuclear charge will be helpful in clearly identifying
these particles. In addition, for problems in which we must balance nuclear reactions,
knowing the nuclear charge (Z) is necessary, and showing Z will be helpful. © 2011 www.ChemReview.Net v.1g Page 552 Module 25 – Nuclear Chemistry Practice: First learn the rules above, then complete these problems. 1. The charge on the nucleus is determined by its number of _________________.
2. The mass number of a nucleus is determined by its number of _____________________.
3. Isotopes have the same number of ____________ but different numbers of ___________.
4. Of the three subatomic particles, the two with the highest mass are _________________
and _________________.
5. Write the nuclide (isotope) symbol for a single proton using AZ notation.
6. Consulting a table of elements or periodic table, fill in the blanks below.
Atom
Name Atom
Symbol Protons Helium Neutrons Atomic
Number Mass
Number Nuclide
Symbol 2
Au 118
82 206
242Pu ANSWERS
1. Protons 2. Protons + Neutrons
4. Protons and Neutrons. 5. 3. Same number Protons, different number of Neutrons 1
1H (A particle with one proton is always given the symbol H.) 6.
Atom
Name Atom
Symbol Protons Neutrons Atomic
Number Mass
Number Nuclide
Symbol Helium He 2 2 2 4 4He Gold Au 79 118 79 197 197Au Lead Pb 82 124 82 206 206Pb Plutonium Pu 94 148 94 242 242Pu In writing the nuclide or isotope symbols for atoms, the nuclear charge below the mass number is optional.
***** © 2011 www.ChemReview.Net v.1g Page 553 Module 25 – Nuclear Chemistry Lesson 25B: Radioactive Decay Reactions
Stable Nuclei
Protons have a positive electrical charge. If there is more than one proton in a nucleus, the
like charges of the protons repel and the nucleus will tend to fly apart.
Neutrons are neutral: they have a zero electrical charge and they do not repel other
particles or each other. Neutrons act in some way as the “glue” of the nucleus: if the right
number of neutrons is mixed with the protons, the repelling protons remain in the nucleus,
and the nucleus is stable.
For small nuclei, the neutron to proton ratio that results in a stable nucleus is about one to
one. As the number of protons in nuclei increases, the number of neutrons needed to form
a stable nucleus increases slightly faster: the n0/p+ ratio gradually increases.
For example:
• All stable helium nuclei have 2 protons and 2 neutrons. • Chlorine has two stable nuclei: both have 17 protons, while one has 18 and the
other 20 neutrons. • All lead have 82 protons. The four stable isotopes of lead have 122, 124, 125, or
126 neutrons. Radioactive Decay
An unstable nucleus does not have a stable ratio of neutrons to protons. Making a nucleus
stable is more complex than just adding more neutrons as “glue” or being in the right
range of ratios. Certain combinations of protons and neutrons are stable, but others are not.
A nucleus that is unstable will be radioactive. The unstable nucleus will have a tendency
to gradually expel particles until a stable neutron and proton combination is achieved. The
process of expelling particles from the nucleus is termed radioactive decay. Depending on
the nucleus, radioactive decay may occur on average in seconds or gradually for up to
billions of years.
Radioactive decay can be a powerful tool in the study of chemical reactions.
For example, most atoms with less than 84 protons have two types of isotopes: those
with a stable nucleus and those that are unstable. An atom with an unstable nucleus
will undergo radioactive decay that we can detect, but the radioactive and nonradioactive forms of the atom chemically behave the same. By substituting a
radioactive nucleus for a stable nucleus, we can “tag” the atoms in substances.
Because we can detect the location of radioactive nuclei as they decay, we can track
where they go and how they behave during chemical reactions, including reactions in
biological systems. The use of radioactive dyes in medical imaging is one example of the
importance of nuclear chemistry. © 2011 www.ChemReview.Net v.1g Page 554 Module 25 – Nuclear Chemistry There are many types of radioactive decay, but the two types encountered most often in
chemistry are alpha (α ) decay and beta (β ) decay. Alpha Decay
In alpha decay, a particle with 2 protons and 2 neutrons is ejected from the nucleus. Such a
particle is termed an alpha particle. Because it has 2 protons and 2 neutrons, an alpha
particle has the same structure as an He4 nucleus, and it is given the same isotopic symbol
as an He4 nucleus.
Symbol for alpha particle =
Example: α particle 4 = 2 He The isotope U238 undergoes radioactive decay by emitting an alpha
particle. This nuclear reaction is written as
238
92 U α
234
⎯ → 90Th
⎯ + 4 He
2 Alpha decay lowers the atomic number (and nuclear charge) of a nucleus by 2 and its mass
number by 4. Balancing Nuclear Reactions
Nuclear reactions balance differently than chemical reactions, but they balance relatively
easily. The rule is:
In nuclear reactions, both mass numbers and nuclear charge must be conserved.
This means that in a balanced nuclear reaction, on both sides of the arrow,
• The sum of the mass numbers (A, on top) must be the same, and
• The sum of the nuclear charges (Z, on the bottom) must be the same.
The result is that nuclear reactions can be balanced by simple addition and subtraction.
Apply the rule to this problem.
Q. For the alpha decay of U238, does the balancing rule apply? α
238
⎯
92 U ⎯ → 234
90Th + 4 He
2 *****
The mass numbers on top add up to 238 on both sides. The nuclear charges on the bottom
total 92 on both sides. The reaction is balanced.
The rule for balancing can be used to predict the products of radioactive decay. Try this
example in your problem notebook.
Q. Which isotope is produced by the alpha decay of radium226?
*****
A key to balancing nuclear reactions is to write the isotope symbols in AZ notation. Start
there for Ra226.
***** © 2011 www.ChemReview.Net v.1g Page 555 Module 25 – Nuclear Chemistry Radium by definition has a nucleus with 88 protons, so this reaction begins α
226
⎯
88 Ra ⎯ → Fill in the missing symbols. *****
In alpha decay, one product is always an alpha particle. Add its symbol on the right.
***** α
226
⎯
88 Ra ⎯ → 4
2 He + Use the balancing rule to write the isotopic formula for the remaining particle: the nucleus
left behind after the alpha particle is expelled.
***** α
226
⎯
88 Ra ⎯ → 4
2 He + 222
86 Rn After the decay, the nucleus has 86 protons, so it must be radon (Rn). For the mass
numbers to balance, the Rn nucleus must have a mass number of 222. Practice A
1. Write the balanced equation for the alpha decay of radon219.
2. How many protons and how many neutrons are in Rn219?
3. How many protons and how many neutrons are in the nucleus left behind after the
alpha decay of Rn219? How many protons and neutrons are lost in the decay?
4. Lead206 can be formed by the alpha decay of which radioactive isotope? Beta Decay
Beta decay is another type of radioactive decay. In beta decay, a neutron decays into a
proton and an electron, and the electron is expelled from the nucleus at high speed. An
electron formed in this manner is termed a beta particle.
Because an electrons has no protons and neutrons, its mass number is zero.
Because an electron has a negative charge, when it is formed in the nucleus its “nuclear
charge” is negative one.
Any nuclear particle that includes protons is by definition an atom, and the nuclear charge
is a positive number that is equal to the number of positive charges in the particle. This is
also the number of protons, because normally, in the nucleus of an atom, protons are the
only particles with a charge. However, on a subatomic particle, the charge may be zero or
negative. In the special case of an electron formed in the nucleus, briefly, before it is
expelled, it is a nuclear particle with a negative one charge.
In nuclear reaction equations, a beta particle can be represented in these ways:
Symbol for beta particle = © 2011 www.ChemReview.Net v.1g β particle 0 = −1 β or 0
−1 e Page 556 Module 25 – Nuclear Chemistry In beta decay, the number of neutrons in the nucleus decreases by one, but the number of
protons increases by one, so the mass number of the isotope stays the same.
Example: The equation for the beta decay of the radioactive isotope lead210 can be
written as β
210
⎯
82 Pb ⎯ → 0
−1 e+ 210
83 Bi Is the above equation balanced?
*****
Before and after the reaction, the mass numbers total 210 and the nuclear charges total 82,
so this is a balanced nuclear equation.
Using the rules for nuclear balancing, we can predict the structure and symbol for the
products of beta decay. Apply the rules to this question.
Q. The isotope carbon14 undergoes beta decay. Write the isotope symbols for the two
nuclear particles formed in this reaction.
***** β
14
⎯
6C ⎯ → Begin by converting to AZ notation: *****
One product must be a beta particle. For its symbol, you may use a β or an e . β
14
⎯
6C ⎯ →
***** 0
−1 e+ β
14
⎯
6C ⎯ → Complete the balancing.
0
−1 e+ 14
7N The isotope formed by the beta decay of carbon14 is nitrogen14. Practice B
1. From memory, write the symbols for an alpha particle and a beta particle.
2. Write balanced equations for these decay reactions.
a. 40 K ⎯β →
⎯ b. 239 Pu ⎯α →
⎯ 3. The isotope iodine131 is used for the treatment of hyperthyroidism: a condition in
which the thyroid gland produces too much thyroid hormone. In the body, iodine is
absorbed by the thyroid gland. If I131 is administered to a patient, its beta decay kills
cells in the thyroid. The result is a reduced level of thyroid hormone without surgery.
Write the symbol for the nucleus produced by the beta decay of I131. © 2011 www.ChemReview.Net v.1g Page 557 Module 25 – Nuclear Chemistry 4. Lead206 can be produced by the beta decay of which nucleus?
5. In a part of what is termed a radioactive decay series, nucleus A with 88 protons and
140 neutrons can beta decay to form nucleus B. B can emit a high speed electron to
form nucleus C, which can α decay to form nucleus D. Write isotopic symbols for A, B,
C, and D.
6. Radioactive element A with 82 protons and 132 neutrons emits a beta particle to
become element B. Element B emits an alpha particle to become element C, which can
emit a high speed electron from the nucleus to form Element D. Write the isotopic
formulas for A, B, C, and D. ANSWERS
Practice A
1. 219
86 Rn α
⎯→
⎯ 4
2 He + 215
84 Po 2. 86 Protons and 133 neutrons. 3. 84 protons and 131 neutrons: 2 protons and 2 neutrons are always lost in alpha decay.
4. 210
84 Po α
4
⎯ → 2 He
⎯ + 206
82 Pb Practice B ⎯
2a. 19 K ⎯ → −1 e β + 40
20 Ca 2b. 239
94 Pu α
4
⎯ → 2 He
⎯ β
0
⎯ → −1 e
⎯ + 131
54 Xe 4. 206
81Tl β
⎯→
⎯ e+ 228
89 Ac 40 3. 131
53 I 5. 228
88 Ra ⎯β →
⎯ 0 0
−1 A 6. β
214
0
⎯
82 Pb ⎯ → −1 ⎯β →
⎯ 0
−1 e+ B e+ α
214
4
⎯
83 Bi ⎯ → 2 He A B 228
90Th 0
−1 + 235
92 U e+ 206
82 Pb α
4
⎯
⎯→ 2 He + 224
88 Ra C + 210 Tl
81
C D β
0
⎯ → −1 e
⎯ + 210
82 Pb
D ***** © 2011 www.ChemReview.Net v.1g Page 558 Module 25 – Nuclear Chemistry Lesson 25C: Fission and Fusion
Fission
Nuclear fission is the reaction where a large nucleus divides into two smaller nuclei, both
of which contain more than two protons. If a fission reaction is accompanied by the
creation of free neutrons, those neutrons can collide with other nearby fissionable nuclei
and cause them to split.
If this “splitting of atoms” begins in a sample of fissionable nuclei that is large enough to
have critical mass, the result can be a chain reaction which releases large amounts of
energy. If the chain reaction is not controlled, the result is an atomic bomb.
In a nuclear power plant (a type of nuclear reactor), a chain reaction is controlled by
adding materials that absorb some of the free neutrons. The large amount of energy
produced by a chain reaction is then released gradually, and the resulting heat can be
harnessed to drive turbines that produce electricity.
The example of nuclear fission encountered most often is the splitting of uranium235. A
typical reaction is
235
92 U 1
+ 0n 236
92 U 140
54 Xe + 94
38 Sr +2 1
0 n + 2 x 1010 kJ/mol In this reaction, a U235 nucleus is struck by a free neutron. The neutron is at first
absorbed, but this unstable nucleus then splits into two smaller nuclei plus 2 free neutrons.
Those two neutrons can collide with other fissionable nuclei to create a chain reaction.
These reactions produce amounts of energy per mole that are millions of times larger than
that produced by chemical reactions such as the burning of fossil fuels.
A disadvantage of using fission for electricity generation is that the products include
highly radioactive isotopes. Exposure to the radiation released by radioactive decay can
cause cancer, and some of the waste products of fission remain significantly radioactive for
thousands of years. A major issue in nuclear power generation is how to store the waste
products so that they will not escape into the earth’s biological environment. Isotopic Separation
Both U235 and U238 can be split, but in practice only U235 is an effective fuel for chain
reactions. For use in nuclear power plants or weapons, naturally occurring uranium must
enriched: meaning that the percentage of nuclei that are U235 must be increased. In
mined uranium ore, 99.3% of nuclei are U238 and 0.7% are U235. For nuclear weapons,
uranium must be enriched to 2080% U235, and over 50 kilograms of this “weapons
grade” uranium must be collected.
Since all isotopes, including U235 and U238, have the same tendency to react chemically,
chemical reactions cannot effectively separate isotopes. However, particles with the lighter
isotopes will be less dense, and they will move faster at a given temperature. Since atoms
containing U235 are lighter than atoms with U238, in gaseous substances containing
uranium the particles containing U235 atoms will diffuse slightly faster. Gaseous
diffusion is therefore one method that is used to separate uranium isotopes. Because © 2011 www.ChemReview.Net v.1g Page 559 Module 25 – Nuclear Chemistry particles that are less dense tend to be moved toward the inside when spun in a circle at
high speed, use of a centrifuge is another way to separate uranium isotopes.
Both gaseous diffusion and centrifugation of uraniumcontaining substances are slow and
expensive processes. This makes it difficult to obtain the amount of highly enriched
uranium needed to build a nuclear weapon. Practice A
1. From memory, write the isotopic symbol for a free neutron.
2. Briefly describe the difference between fission and fusion.
3. Fill in the one missing isotopic symbol in this nuclear fission reaction.
235
92 U 1
+ 0n 87
35 Br + +6 1
0 n 4. Fill in the missing isotopic symbol that is the first reactant in this fission equation.
1
+ 0n 91
38 Sr + 146
56 Ba +3 1
0 n Fusion
Nuclear fusion is a reaction that combines two small nuclei to make a larger one. When a
small nucleus such as helium is the product of fusion, this reaction produces large amounts
of energy. An example of a fusion reaction is
2
1H + 3H
1 4
2 He + 1
0 n+ 2 x 109 kJ/mol In this reaction, two hydrogen nuclei are fused to form helium. Fusion is a reaction that
takes place in stars, including our sun, and in hydrogen bombs.
The primary reaction that causes stars to “burn” (release energy) is the conversion from
hydrogen to helium. At the extremely high temperatures and pressures found in stars,
lighter nuclei can fuse to form heavier nuclei, and those nuclei can undergo successive
fusion reactions. After long periods of making heavier nuclei, some stars become unstable
and explode. The atoms scattered into space from exploded stars can accumulate over time
due to gravitational attraction, forming new stars and planets. The elements that coalesced
to form our own planet billions of years ago are nuclei, or the decay products of nuclei, that
were originally formed by fusion in a star.
Because fusion can use hydrogen as its fuel, and the products of the hydrogen fusion are
generally stable nuclei rather than longlived radioactive isotopes, a nuclear reactor that
could slow and control the fusion of hydrogen would be a source of inexpensive and clean
energy. To produce the energy needed for our society, such fusion reactors could replace
the burning of fossil fuels and current nuclear power plants, both of which form products
that can harm our environment. However, while nuclear reactors can control the rate of
nuclear fission, currently no way has been discovered to engineer the gradual release of the
energy of nuclear fusion. © 2011 www.ChemReview.Net v.1g Page 560 Module 25 – Nuclear Chemistry Summary: To balance nuclear reactions, the rules you need in memory are α particle 1. 4 = 2 He ; β particle 0 = −1 β or 0
−1 e 1
, neutron = 0 n 2. Always use AZ notation: mass number on top, nuclear charge on the bottom.
3. The mass numbers and nuclear charges must be conserved: both must add to give the
same number on both sides of the arrows.
4. Fission splits a nucleus, fusion combines nuclei. Practice B
1. If a single nucleus is formed as the product of this reaction, write its isotope symbol.
2
1H + 2H
1 2. In stars that are red giants, helium4 can fuse with beryllium8 to form a single nucleus.
Write the equation for this reaction.
3. Assuming that one isotope symbol is missing from these equations, fill in the missing
isotope, then write the name for this type of reaction.
Type:
a. 235 U+ b. 238 c. 1 d. 234 1
0 92 n
4 U H+ 3 He Kr + +3 1
0 + 0
−1 _________
_________
_________ H Th n e+ _________ ANSWERS
Practice A
1.
3.
4. 1
0 n 235
92 U 2. Fission splits a nucleus, fusion combines nuclei. + 239
94 Pu 1
0 n 87
35 Br 1
+ 0n © 2011 www.ChemReview.Net v.1g 91
38 Sr +
+ 143
57 La +6 1
0 n 146
56 Ba +3 1
0 n Page 561 Module 25 – Nuclear Chemistry Practice B + 2H
1 1. 2
1H 3a. 235
92 U 3b. 234
90Th 2. 92
36 Kr
4
2 He + 1 3d. 1
+ 0n 238
92 U 3c. 1 H 4
2 He + + 4
2 He
141
56 Ba 0
−1 e+ 8
4 Be +3 1
0 12
6C _fission_ n _ alpha decay_ 234
90Th _ fusion_ 4
2 He 3
1H + beta decay_ 234
91 Pa ***** Lesson 25D: Fractions and Percentages
Pretest. If you earn a perfect score, you may skip this lesson.
1. 0.6% is what decimal equivalent?
2. 45/10,000 is what decimal equivalent and what percent?
***** Fractions and Decimal Equivalents
A fraction is a ratio: one quantity divided by another. In math, a fraction can be any ratio,
but in science, “fraction” often (but not always) refers to a part of a larger total, which will
be a smaller quantity over a larger quantity. In dealing with percentages and fractions, we
will call this
Rule 1. Fraction = Quantity A
Quantity B and often equals Part
Total = Smaller number
Larger number The decimal equivalent of a fraction is a number in decimal notation that results by
dividing the top number of the fraction (the numerator) by the bottom number (the
denominator).
An example of a fraction and its decimal equivalent is 1/2 = 0.50.
Rule 2. To find the decimal equivalent of a fraction, divide the top by the bottom.
Use your calculator to answer: Q. The decimal equivalent of 5/8 = ________
*****
A. 0.625
In chemistry calculations, the term “fraction” can refer to any fixed decimal number from
1.00 to 0.00 (such as 0.25) that can be obtained by dividing one quantity by another. In
terms of numeric value, a fraction and its decimal equivalent are the same.
Fraction = Decimal equivalent © 2011 www.ChemReview.Net v.1g Page 562 Module 25 – Nuclear Chemistry In chemistry, both 1/2 and 0.50 are termed fractions. Depending on the context, when a
chemistry problem asks “what is the fraction…” it may be asking for
• a fraction in an x/y format, or • a decimal equivalent number such as 0.25 . Usually, from the context and examples of related problems, it will be clear what type of
fraction is wanted. Calculating Percentages
A percent multiplies a decimal equivalent by 100%.
A familiar example is 1/2 = 0.50 x 100% = 50%.
For those who are not mathinclined, percentages provide more familiar numbers to
measure change than numbers with decimals. However, when a percent is required for an
answer, in most chemistry calculations you will need to solve for the decimal equivalent first,
then convert to percent. To convert a decimal equivalent to a percent, multiply by 100%
(moving the decimal twice to the right). Let’s call this
Rule 3. Percent = fraction x 100% = (decimal equivalent ) x 100%
To find a %, calculate the decimal equivalent first.
To find a %, write the fraction, then the decimal equivalent, then the %.
Example: 1/8 is what percent?
Write the fraction, then its decimal equivalent, then multiply by 100%.
1/8 = 0.125 x 100% = 12.5% Apply Rule 3 to this problem: Q. 25 is what percentage of 400? *****
The rule: if you WANT a percentage, first write the fraction, then its decimal equivalent.
In this problem, the question is: the smaller number is what part of the larger number?
Write the fraction definition and fill in the numbers.
*****
Fraction = Part = Smaller = 25
Total
Larger
400
*****
Fraction = Part = Smaller = 25
Total
Larger
400
***** = ____________ (fill in the decimal equivalent) = 0.0625 What will the percentage be? Percent = fraction x 100% = (decimal equivalent ) x 100% = 0.0625 x 100% = 6.25%
Similarly, if you are given a percentage to use in a calculation, you must change the
percentage to its decimal equivalent. Conversion calculations and mathematical equations © 2011 www.ChemReview.Net v.1g Page 563 Module 25 – Nuclear Chemistry nearly always require numeric values (the fraction or its decimal equivalent), not the
percentages that are “values x 100%”.
Since Percent = (decimal equivalent) x 100% , Decimal equivalent = Percent / 100% . Let’s call this
Rule 4 . Change a percentage to its decimal equivalent before use in conversions.
To change a percentage to its decimal equivalent, divide by 100% (moving
the decimal twice to the left).
Decimal equivalent of percent = Percent / 100%
Examples: In calculations, change 25% to 0.25 ; change 0.50% to 0.0050 Apply Rule 4 to this problem. Q. 3.5 percent of 12,000 is ?
*****
If fractions for you are easy math, solve in any way you wish. If you need a systematic
approach, try the following.
For percentages in calculations, the fundamental rules are,
• if you are given a percentage, as step one convert it to a decimal equivalent; • if you WANT a percentage, first find the decimal equivalent WANTED. In this problem, we were given a fraction, step one is to convert to the decimal
equivalent.
Decimal equivalent = percentage / 100% = 3.5% / 100% = 0.035 There are many ways to solve from here. Having converted to the fraction, it may be
intutitive that
3.5% of 12,000 = 0.035 x 12,000 =
If not, a more methodical way is to solve this equation using our equation method:
Fraction = decimal equivalent = Smaller
Larger Make a data table that matches the terms in the boxed equation.
*****
DATA:
Decimal equivalent = 0.035
Smaller = ? (Since the number be asked for is less than 100% of 12,000, it is smaller)
Larger = 12,000
SOLVE: (solve the boxed equation for the WANTED variable, then substitute.)
? = Smaller = (decimal equivalent) x (larger) = 0.035 x 12,000 = 420 © 2011 www.ChemReview.Net v.1g Page 564 Module 25 – Nuclear Chemistry Sanity check: since 10% of 12,000 is 1,200 , 3.4% should be about 400? Check.
There are many ways to solve fraction and percentage calculations. Use one that works for
you. Practice: First memorize the rules above, then do each of these problems. 1. 1/5 is what decimal equivalent and what percent?
2. 4.8% is what decimal equivalent?
3. 9.5/100,000 is what decimal equivalent and percent?
4. What percentage of 25 is 7?
5. What amount is 0.450% of 7,500. ?
6. Twelve is what percent of 24,000?
7. Complete the two problems on the pretest to this lesson. ANSWERS
Pretest: 1. 0.006 2. 0.0045, 0.45% Practice
1. Decimal equivalent of 1/5 = 0.20. Percent = decimal equivalent x 100% = 20.%
2. Decimal equivalent = percent / 100% = 4.8% /100% = 0.048
3. Move the decimal 5 times to divide by 100,000. Decimal equivalent = 0.000095 = 9.5 x 10―5
Percent = decimal equivalent x 100% = 0.000095 x 100% = 0.0095% = 9.5 x 10―3 %
4. To calculate percent, calculate fraction, then decimal equivalent, then percent.
Fraction = Part = 7 = 0.28
Total
25
Percent = fraction x 100% = decimal equivalent x 100% = 0.28 x 100% = 28%
What fraction of 25 is 7? 5. To calculate an amount, change % to decimal equivalent by dividing by 100. 0.450% = 0.00450
? = 7,500 x 0.00450 = 34
6. To calculate percent, write fraction, then decimal equivalent, then percent.
12 is what part of 24,000? = 12 is what fraction of 24,000?
Fraction = Part = 12 =
12
= 0.5 x 10―3 = 5.0 x 10―4
Total
24,000
24 x 103
This number in exponential notation is a decimal equivalent. A decimal equivalent is any numeric
value that has no denominator (which means 1 is the denominator) that is derived from a fraction.
Percent = fraction x 100% = decimal equivalent x 100% =
= 5.0 x 10―4 x 100% = 5.0 x 10―2 % = 0.050 %
7a. .Decimal equivalent of percent = percent / 100% = 0.6% /100% = 0.006 © 2011 www.ChemReview.Net v.1g Page 565 Module 25 – Nuclear Chemistry 7b. To find the decimal equivalent of a fraction, divide the top by the bottom.
45/10,000 = 0.0045
Percent = fraction x 100% = decimal equivalent x 100% = 0.0045 x 100% = 0.45%
***** Lesson 25E: Natural Logarithms
Prerequisites: Complete the earlier lesson on base 10 logarithms before beginning this
lesson.
Pretest: If you think you know this topic, try the last 4 calculations in the Practice at the
end of this lesson. If you can do those calculations, skip the lesson.
***** I. Review of Base 10 Logs
To solve halflife calculations, you will need to know the rules for natural log (ln)
calculations. The rules for natural logs parallel base 10 logs, and because our number
system is based on 10, it is easier to learn the logic of the base 10 rules first.
If you have not completed the lesson on Base 10 Logarithms in a prior module, do so
now. If you have completed that lesson, review the rules in the following summary,
then complete the practice set below. Summary: Log Rules to Commit to Memory
1. A logarithm is simply an exponent: the power to which a base number is raised.
2. A logarithm answers the question: if a number is written as a base to a power,
what is the power?
3. log buttons on a calculator find the power of a number written as 10 to a power. 4. To check log answers: when a number is written in scientific notation, its power of
10 must agree with its base 10 logarithm within ± 1.
5. The equation defining a log is log 10x = x ; the log of 100 is 2 .
6. Knowing the log of a number, to find the number, calculate the value for 10log.
This is called “taking the antilog” or “finding the inverse log.”
7. 10log x = x . Recite and repeat to remember: “10 to the log x equals x.”
As an easy example, remember: 10log 100 = 102 = 100 8. On a calculator, to convert a log value to a number,
• input the log value, then press INV LOG ; or 2nd LOG ; or • Input the log, then press 10x . or input 10, x^y , input the log , = . © 2011 www.ChemReview.Net v.1g Page 566 Module 25 – Nuclear Chemistry Practice A: Practice with the calculator you will use on tests. If you have problems
with any of these, review the previous lesson on base 10 logarithms.
1. 103.2 = (in scientific notation):________________________________ (expos ± 1 ?) ____ 2. 10─12.3 = _____________________________ (expos ± 1 ?) ____ 3. 10─0.2 = (number): _________( scientific notation): ______________(expos ± 1 ?) ____ 4. Log(10) = ___________________________________ 5. Log(1) = ___________________________________ 6. Log(1014 ) = ___________________________________ ( expo ± 1 ? __) 7. Log(2.0 x 1014 ) = ___________________________________ ( expo ± 1 ? __) 8. Log(2.0 x 10─14 ) = ______________________________ (expo ± 1 ? __) 9. Log 5.0 = ____________________________________ (expo ± 1 ? __) 10. Log 0.0050 = ____________________________________ (expo ± 1 ? __) 11. Log x = 4.7 , x = _____________________________ (expo ± 1 ? ___) 12. Log A = ─8.2 , A= ____________________________ (expo ± 1 ? ___) 13. Log D = ─0.50 , D = ____________________________ (expo ± 1 ? ___) 14. Log x = 6.6 , antilog = ____________________________ (expo ± 1 ? ___) 15. 10─9.5 = ____________________________________ (expos ± 1 ?) ____ 16. Log (3.0 x 10─5 ) = __________________________________ (expo ± 1 ? __) © 2011 www.ChemReview.Net v.1g Page 567 Module 25 – Nuclear Chemistry II. Base e Rules
As always, in the sections below, cover the answers below the * * * * * line and write
your answer to the questions that are above the line.
a. The Symbol e
In mathematical and scientific equations, the lowercase e is an abbreviation for a
number: 2.7182818… For calculations, the value e = 2.718 must be memorized.
The number e has many interesting mathematical properties. It is important in
science because it is found in many equations that predict natural phenomena. In
these equations, e is the base for values expressed in exponential notation, in the
form ex, and e is termed the natural exponential.
To solve calculations involving radioactive halflife, we will use both the number e
and the natural log function ln. Let’s consider calculations with e first.
b. Calculating with natural exponentials
We know that e1 = __________(what number?)
*****
2.718…. Using 1 and the ex button, write the key sequence that produces
that answer for e1 on your calculator.
*****
• A standard TItype calculator might use: 1 ex .
• On an RPN scientific calculator, try: 1 enter ex . Use your calculatorkey sequence to do these, then check your answers below.
1) e2 = 2) e2.5 = 3) e─1 = 4) e─2.5 = (Because the statistical basis for significant figures does not apply to logarithmic
calculations, we will use this general rule: during e and ln calculations, round
numbers in answers to 3 significant figures.)
*****
1) e2 = 7.39 2) e2.5 = 12.2 Recall that to enter a negative number, you usually use a +/ key.
3) e─1 ( = 1/e = 1/2.718.. ) = 0.368 4) e─2.5 ( = 1/e2.5 ) = 0.0821 c. Calculating Natural Logs
The ln function (the natural log) answers this question: if a number is written as e
to a power, what is the power?
Just as by definition, log 10x ≡ x , the natural log definition is © 2011 www.ChemReview.Net v.1g ln ex ≡ x Page 568 Module 25 – Nuclear Chemistry Use the natural log definition to do these without a calculator.
1) ln e0 = ______
*****
1) ln e0 = 0 2.) ln e1 = 1 3. ln e─4 = ________ 2) ln e1 = ______ 3) ln e─4 = ─4 By definition, ln e = _____ .
*****
ln e = ln e1 = 1.
Try this one in your head: ln(2.718) should equal about ____________ .
*****
ln(2.718) ≈ ln e ≈ ln e1 ≈ 1.
Now use your calculator for the same calculation: ln(2.718) = ___________
*****
Is the calculator answer close to the mental arithmetic answer?
Write down the key sequence that works: ln(2.718) =
The same steps should take the natural log of any positive number. Try these.
1) ln 314 = ________________
*****
1) ln 314 = 5.75
To check an answer, after writing it down, use the ex key and see if you return to
the number you were taking the ln of. Try that as a check on these:
2) ln 0.0050 = _________________ (after writing answer, use ex . Check? ___ 3) ln (6.02 x 1023) = _________________________ Check? ___ 4) ln (19.29 x 10─15) = _________________________ Check? ___ *****
2) ln 0.0050 = ─ 5.30 3) ln(6.02 x 1023) = 54.8 4) ln(19.29 x 10─15) = ─ 31.6 Note in part 4), a calculator does not require the input of scientific notation.
However, if you use the ex key to check your answer, it will likely return the
original number converted to scientific notation. Practice B: Use the calculator you will use on quizzes and tests. 1. e2.0 = 2. e─4.7 = 3. e─11 = 4. © 2011 www.ChemReview.Net v.1g ln 42 = Page 569 Module 25 – Nuclear Chemistry ln(9 x 105) = 5. ln 0.020 = 6. 7. ln(5.0 x 10─4) = 8. ln(10─4) = d. Converting ln Values to Numbers
A base 10 definition: 10log x = x A base e definition: eln x = x Note the similarities. Using the bottom equation, for some calculations involving ln and e you will not
need a calculator. Try this:
eln(─11) = ________
*****
eln(─11) = ─11
The equation eln x = x also means that if you know the ln value, to find the
corresponding number, make the ln value a power of e.
If the ln value = 1, the number (do this one in your head) is ___________
* * * **
e1 = 2.718…
Knowing that answer, do the same ln to number conversion on your calculator by
taking the antilog.
If the ln value = 1, the number obtained using the calculator is ___________
*****
Input the ln, then press INV or 2nd ln or press ex .
Write down or circle the key sequence that converted ln = 1 to the number 2.718…
Use your key sequence to convert the following ln values to numbers. Write first
write the number in terms of e, then the number, then the number in scientific
notation.
1) If ln = 6 , number = e = (number): ________ = (sci. notation):_____________ *****
1) If ln = 6 , number = e6 = (nbr): 403 = (sci. notation): 4.03 x 102
In base e calculations, unlike base 10, there is no obvious correlation between the
scientific notation exponent and the base e logarithm that helps in checking your
answer. However, you can check by taking the ln of the number answer and see if
it returns to the original ln value.
Try these.
2) If ln = ─4.5 , number = e © 2011 www.ChemReview.Net v.1g = (nbr): _________ = (sci. notation):_____________ Page 570 Module 25 – Nuclear Chemistry 3) ln = 57.2 , number =________________________
4) ln [A] = 0.0300 , [A] = e = (nbr. and unit): _________________ *****
2) If ln = ─4.5 , number = e─4.5 = 0.0111 = 1.11 x 10─2
3) ln = 57.2 , number = e57.2 = 6.94 x 1024
4) If ln [A] = 0.0300 , [A] = e0.0300 = 1.03 M e. Units and Logarithms
Note that in 4) above, the unit expected for a concentration has been added. From
a strict mathematical perspective, logarithms should not be taken of values with
units, and logarithm values should not have units. However, many of the
equations we write in chemistry are “shortcuts” which simplify more complex
relationships. We use these shortcuts because they speed and simplify problem
solving, but the rules for unit cancellation that must work for “real” scientific
relationships may not work when using shortcut equations.
To make shortcut equations work with logarithms, two of our rules will be
When taking the logarithm of a value with units, write the result as a value
without units.
If a WANTED quantity is based on a logarithm value, first (if needed) make all
of the supplied units in the problem consistent, then add the appropriate
consistent unit to the answer.
The quantity involved most often in log calculations will be concentration in moles
per liter. The rule will be: if a [x] is WANTED, add moles/liter (M) to the answer.
Apply that rule to the following problems. Write the answer as a number or in
scientific notation. When in doubt, check answers as you go.
5) ln [Z] = ─12.5 , [Z] =________________________
*****
5) [Z] = e ─12.5 = 3.73 x 10─6 M (add the unit of concentration: mol/L) 6) ln [R] = ─ 0.17 , [R] =________________________
7) [D] = e ─1.39 , [D] =________________________
8) ln(0.250 M) = _______________________
9) ln [A] = ─ 2.63 , [A] = ______________
*****
6) [R] = e ─0.17 = 0.844 M 7) [D] = 0.249 M 8) ln(0.250 M) = ─ 1.39 (drop the unit) © 2011 www.ChemReview.Net v.1g (add M) 9) [A] = e─ 2.63 = 0.0721 M Page 571 Module 25 – Nuclear Chemistry f. Notation with e and ln
• Some calculators use an E at the right side of the answer screen to show the
power of 10 for numbers in scientific notation. This is not the same as the
symbol e for the natural exponential. • Be careful to distinguish “taking the ln” from “the ln value.”
Ln(7.389) = _____________ . Try it. You should get close to 2.
But if ln = 7.389 , the number with that ln is __________ Try it. *****
If ln = 7.389, the number is e7.389 = 1,620
If you get lost on a natural log calculation, a good strategy is to do a similar and
simple base10 mental and calculator computation, and then apply the same logic
to the natural log case. Simple base 10 calculations can often be solved in your
head, and the formulas and steps for base 10 and base e calculations are parallel.
g. Converting between base 10 and natural logs
A general rule for logarithms of any base is logb(x) = ln(x)/ln(b) , where b is the
base. For base 10 logs, this equation becomes
Log10(x) = ln(x)/ln(10) = Log10(x) = ln(x)/2.303 This relationship is generally memorized as ln(x) = 2.303 log(x) The value of a natural log is always 2.303 times higher than the base 10 log. Summary: Add these rules to your inmemory logrule list. 9. The symbol e is an abbreviation for a number with special properties: e = 2.718...
10. The ln (natural log) function answers the question: if a number is written as e to a
power, what is the power?
11. Knowing the ln, to find the number, take the antilog. On a calculator,
• input the ln value, then press INV ln ; or • Input the ln value, then press ex . An ln is simply an exponent of e. 12. When you encounter log or ln calculations, it helps to write:
log 10x = x and 10log x = x . “The log of 10 to the x is x; 10 to the log x is x.” ln ex = x and eln x = x . Write the base 10 rules, then substitute e and ln. Note the logic: A log is an exponent.
13. ln(x) = 2.303 log(x) © 2011 www.ChemReview.Net v.1g Page 572 Module 25 – Nuclear Chemistry Practice C: Try the oddnumbered problems first. Complete the even numbered
problems for additional practice or for pretest review.
1. e5.2 = 2. e─1.7 = 4. ln 1066 = 5. ln 0.0050 = 7. ln (14.92 x 10─6) = 10. If ln = ─6.8 , number = e 11. If ln D = 7.4822 , D = 12. If ln = ─12.5 , antilog = 13. If log [A] = ─9 , [A] = 14. If log x = 13.7 , x = 15. Log A = ─13.7 , A = 16. 10─11.7 = 17. ln [B] = ─13.7 , [B] = 18. e ─11.7 = 19. ln(0.050 M) = 20. e ─0.693 = 21. If log(x) = 5.0 , ln(x) = 22. if ln(x) = 34.5 , log(x) = 23. If ln(x) = ─ ( 0.075 day─1)(4.0 days) ; x = ? 24. Given that Given that 6. ln(3 x 108) = 8. ln e6.2 = 9. eln(─42) = = (number in sci. notation):___________________ ln[A]t = ( ─ 0.0173 s─1) ( t ) + 0.693 a. If t = 20. s, [A] = ?
25. 3. e─20.75 = b. If [A] = 0.710 M, t = ? ln[A]t = ( ─ 0.0241 yrs.─1 ) ( t ) ─ 4.61 a. If [A] = 0.0025 M, t = ? b. If t = 28.8 years, [A] = ? ANSWERS
Practice A
2. 10─12.3 = 5.0 x 10─13 3. 10─0.2 = 0.63 = 6.3 x 10─1 1. 103.2 = 1.6 x 103 4. Log(10) = Log(101) = 1 5. Log(1) = Log(100) = 0 7. Log(2.0 x 1014 ) = 14.3 8. Log(2.0 x 10─14 ) = ─ 13.7 10. Log 0.0050 = ─ 2.3 11. Log x = 4.7 , x = 5.0 x 104 12. Log A = ─8.2 , A= 6.3 x 10─9 14. Log x = 6.6 , antilog = 4.0 x 106 16. Log (3.0 x 10─5 ) = ─ 4.5 © 2011 www.ChemReview.Net v.1g 6. Log(1014 ) = 14
9. Log 5.0 = 0.70 13. Log D = ─0.50 , D = 0.32
15. 10─9.5 = 3.2 x 10─10 Page 573 Module 25 – Nuclear Chemistry Practice B
2. e─4.7 = 9.10 x 10─3 1. e2.0 = 7.39 5. ln 0.020 = ─3.91 6.
8. ln(10─4) = 3. e─11 = 1.67 x 10─5 ln(9 x 105) = 13.7 ln(1 x 10─4) = 4. ln 42 = 3.74 7. ln(5.0 x 10─4) = ─7.60 ─9.21 Practice C
1. e5.2 = 181
4. ln 1066 = 6.97 5. ln 0.0050 = 7. ln (14.92 x 10─6) = ─11.1 3. e─20.75 = 9.74 x 10─10 2. e─1.7 = 0.183 6. ln (3 x 108) = 19.5 ─5.30 8. ln e6.2 = 6.2 10. If ln = ─6.8 , number = e─6.8 11. If ln D = 7.4822 , D = 1780 9. eln(─42) = ─42 = (number in std. notation): 1.11 x 10─3
12. If ln = ─12.5 , antilog = 3.73 x 10─6 13. If log [A] = ─9 , [A] = 10─9 M 15. Log A = ─13.7 , A= 2.00 x 10─14 17. ln [B] = ─13.7 , [B] = 1.12 x 10─6 M 19. ln(0.050 M) = ─3.00 (drop the unit) 21. If log(x) = 5.00 , ln(x) = ? ln(x) = 2.303 log(x) ; ln(x) = (2.303)(5.00) = 11.5 22. If ln(x) = 34.5 , log(x) = ? ln(x) = 2.303 log(x) ; log(x) = 34.5/2.303 = 15.0 23. If ln(x) = ─ ( 0.075 day─1)(4.0 days) ; x = ? 14. If log x = 13.7 , x = 5.01 x 1013
16. 10─11.7 = 2.00 x 10─12
18. e ─11.7 = 8.29 x 10─6
20. e ─0.693 = 0.500 ln(x) = ─ 0.300 (day─1)(days) = ─ 0.300 day0 = ─ 0.300 (1) = ─ 0.300
x = eln(x) = e─0.300 = 0.741
24. Given that ln[A]t = ( ─ 0.0173 s─1) ( t ) + 0.693 a. Strategy: to find [A]t , first solve for ln[A]t
? = ln[A]t = ( ─ 0.0173 s─1 ) ( 20.0 s ) + 0.693
= ─ 0.346 + 0.693 = 0.347
WANTED is [A] at 20 s. Known is: ln[A]20 s = 0.347
*****
[A] = eln[A] = e0.347 = 1.41 M Solve for [A]20 s. (If a [ ] is wanted, add M as unit) b. Strategy: Solve for t in symbols first.
*****
t = ln[A] ─ 0.693 = ln[0.710 M] ─ 0.693 = ─ 0.342 ─ 0.693
─ 0.0173 s─1
─ 0.0173 s─1
─ 0.0173 s─1 © 2011 www.ChemReview.Net v.1g = Page 574 Module 25 – Nuclear Chemistry =
25. Given that ─ 1.035
=
─1
─ 0.0173 s 59.9 s = t ( 1/ s─1 = (s─1) ─1 = s ) ln[A]t = ( ─ 0.0241/yr. ) ( t ) ─ 4.61 a. Strategy: Solve for t in symbols first.
*****
t = ln[A] + 4.61 = ln[0.0025 M] + 4.61
─ 0.0241 /yr.
─ 0.0241 /yr.
= ─ 1.38
=
0.0241 /yr. = ─ 5.99 + 4.61 =
─ 0.0241 /yr. 57.3 years = t b. Strategy: to find [A]t , first solve for ln[A]t
? = ln[A]t = ( ─ 0.0241 /yr. ) (28.8 yr. ) ─ 4.61
WANTED is [A]. Known is: ln[A] = ─ 5.30
*****
[A] = eln[A] = e─ 5.30 = 0.0050 M = ─ 0.694 ─ 4.61 = ─ 5.30
Solve for [A].
If a [ ] is wanted, add M. ***** Lesson 25F: Radioactive HalfLife Calculations
HalfLives: Simple Multiples
The halflife of a reactant (symbol t1/2) is the time required for half of the particles of the
reactant to be used up in a reaction.
A radioactive nucleus has a characteristic halflife: in any sample, the time in which half
of the nuclei decay is constant.
The rate of chemical reactions changes with changes in temperature, but in radioactive
decay, a nuclear reaction, the time of a halflife does not change significantly at
temperatures up to several thousand kelvins.
As a result, each radioactive nucleus has a characteristic halflife. Some radioactive
nuclides have a halflife of a few seconds; others have a halflife of billions of years.
However, in any significant sample of a given nucleus, the time in which half of the nuclei
decay is always the same. While it is not possible to predict when any one nucleus will
decay, for any sample more than a hundred of a given nuclide, if we calculate (or look up)
the halflife, we how long it will take for half of the nuclei to decay, and we can calculate
how long it will take for any percentage of those nuclei in any sample to decay. © 2011 www.ChemReview.Net v.1g Page 575 Module 25 – Nuclear Chemistry HalfLife Calculations For Simple Multiples
In calculations involve radioactive halflife, the two variables will generally be the time
that the nuclei in a sample have decayed and the percentage of the nuclei that remain. If
the time period for the decay is equal to either the halflife or a simple multiple of the halflife, answers can be calculated by mental arithmetic.
• If a nucleus has decayed for a time equal to one halflife, 1/2 of the original nuclei
have decayed and half remain.
If the nuclei are in a sample that has a constant volume (which should be assumed
unless other conditions are stated), 1/2 of the original concentration of the reactant
remains after one halflife. • After two halflives (double the time of the halflife), the number of nuclei
remaining is half of the half that remained after the first halflife: half of 1/2 = 1/4
(25%) of the original nuclei remain and 75% have decayed. • At triple the halflife, 1/2 of 1/4 = 1/8 (12.5%) of the original nuclei remain. Apply the rules above to the following problem.
Q. Fluorine18, a radioactive isotope used in nuclear medicine, has a 1.8 hour halflife.
How long will it take for 87.5% of the F18 nuclei in a sample to decay? * ** * *
A. If 87.5% has decayed, 12.5% remains. How many halflives are required? How
much time would this be?
* ** * *
12.5% remains at 3 halflives; 3 x 1.8 hours = 5.4 hours
To solve radioactive decay calculations for these “simple multiple” cases, given any one of
headings in the table below, you will need to be able to fill in the rest of the table from
memory. This should not be difficult: note that in the two middle columns, each number
is simply half of the one above.
For Radioactive Nuclei,
at time = Fraction
Remaining Percent
Remaining Percent
Decayed 1 100% 0% One halflife 1/2 50% 50% Two halflives 1/4 25% 75% Three halflives 1/8 12.5% 87.5% 0 We will also use these rules in estimates to check calculations that are not easy multiples. © 2011 www.ChemReview.Net v.1g Page 576 Module 25 – Nuclear Chemistry Practice A: Write the table above until, given the top row, you can fill in four rows
below from memory. Then complete these problems. 1. The nucleus of Pu239 undergoes firstorder radioactive decay with a halflife of
24,400 years. In a sample of constant volume containing Pu239,
a. After how many years will 25% of the original Pu239 nuclei remain?
b. After how many halflives will the [Pu239] be 1/16th of its original concentration?
c. What percentage of the Pu239 has decayed after exactly 3 halflives? Rate Constants for Radioactive Decay
In halflife calculations that do not involve simple multiples, we can solve using rate
equations and the math of natural logs.
Each radioactive nucleus has a rate constant (k) that is characteristic: a value that is
constant. Different radioactive nuclei will have different values for k .
One way to write the equation that predicts the decay rate for radioactive nuclei is
ln [A]t = ─kt which can be written ln(fraction remaining) = ─kt [A]0 In the first equation, [A]t / [A]0 is the fraction of nuclei remaining at time = t .
For example: after one halflife, half (50%) of the original sample remains and the
fraction remaining is 0.50 . After two radioactive half lives, the fraction remaining would be?
*****
After two halflives, the percentage remaining is 1/4 (25%), and the fraction remaining
is therefore 0.25 .
Since the reactant is being used up over time, the value of [A] after time = t will be less
than it was at time = 0, and the value of the fraction will be less than one. The decimal
equivalent of the fraction will always have the form 0.xxx .
The units used to calculate the fraction can be concentration or any consistent units that
are proportional to concentration, including the mass or number of particles in a sample.
Radioactive halflife calculations often involve fractions or percentages, and in those cases
the form of the equation above that includes (fraction remaining ) will be the most
convenient to use. However, to use the equation with the term (fraction remaining), you
must calculate using fractions or the decimal equivalent of fractions, rather than
percentages.
If a percentage is WANTED, you will need to solve for the fraction first. If a percentage is
given, you will need to convert to its decimal equivalent fraction to use in the equation. © 2011 www.ChemReview.Net v.1g Page 577 Module 25 – Nuclear Chemistry When using fractions and percentages, recall that
• A fraction can be expressed as x/y or as a decimal equivalent value that is x
divided by y. • Percentage = fraction x 100% • Percentage remaining = 100% ─ percentage decayed • Fraction remaining = 1.000 ─ fraction decayed • The fractions in decay calculations will have a value between 1.00 and 0 (such
as 0.25). and decimal fraction = percent/100% (For practice in converting between percentages and fractions, see Lesson 25C.) Practice B: Complete each of these before continuing. 1. If 90% of a sample has decayed, what is the fraction remaining?
2. If the fraction of a sample that has decayed is 0.40, what percent remains?
3. If the value for ln(fraction of sample remaining) is ─ 1.386 ,
a. What is the fraction of the sample remaining?
b. What percentage has decayed?
4. For the decay of a radioactive nucleus, if the rate constant of the reaction is
k = 0.04606 hours─1 , what percentage remains after 50.0 hours? HalfLife Calculations For NonSimpleMultiples
For a radioactive nucleus, after a time equal to one halflife ( t1/2 ), half of a sample has
decayed and half remains. Substituting into the equation
ln(fraction remaining) = ─kt
we can write ln(1/2) = ─k t1/2 at a time equal to one halflife,
Solve this equation in symbols for halflife. * * ** *
One way of several to write the equation is
t1/2 ≡ ─ ln(1/2) / k This equation is one way to define radioactive halflife. In the above equations, the rate constant (k) and halflife ( t1/2 ) are variables: their
numeric values will differ for different radioactive nuclei. The term ln(1/2) is a constant:
it can be converted to a number without units. Use your calculator to find its fixed
decimal value.
ln(1/2) = __________________
* * ** * © 2011 www.ChemReview.Net v.1g Page 578 Module 25 – Nuclear Chemistry ln(1/2) = ln(0.500) = ─ 0.693
Substitute this numeric value into the equation above that defines halflife and simplify.
* * ** *
t1/2 = ─ (─0.693) / k which simplifies to t1/2 = 0.693 / k This last equation above is often listed in textbooks as a definition of radioactive halflife.
From this form, it is clear that if you know the halflife, you can find the rate constant, and
if you know the rate constant you can find the halflife.
To solve decay calculations, we need equations that relate the fraction remaining, halflife,
time, and k . Several combinations of the equations above can be used, but the best
equations may be those that are easy to remember. In these lessons, we will solve using
what we will call the
Radioactive Decay Prompt
If radioactive decay calculation includes or halflife and a fraction or percentage of a
sample, and the answer cannot be calculated using simple multiples, write in the
DATA:
ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Note that the second equation is simply a special case of the first: when the fraction
remaining is 1/2, the time is equal to the halflife.
Commit the radioactive decay prompt to memory, then apply it to solve this problem.
Q. Iodine131, a radioactive isotope used to treat thyroid disorders, has a halflife of
8.1 days. What percentage of an initial [I131] remains after 48 hours?
*****
WANT:
Percent [I131]48 hrs. = % remaining
DATA: t1/2 = 8.1 days = radioactive halflife
48 hours = 2.0 days = t (choose any consistent time unit) Strategy: Write the equations that relate the symbols in the problem.
For radioactive decay calculations that include halflife and fraction or
percentage, write and use
ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Percentage remaining = fraction remaining x 100%
If needed, adjust your work and solve from here.
* * ** *
The variable that links both of the prompt equations is k .
Since we know the halflife, the second prompt equation will find k .
Knowing k and t, the first prompt equation will find ln(fraction remaining). © 2011 www.ChemReview.Net v.1g Page 579 Module 25 – Nuclear Chemistry Knowing ln(fraction remaining), the fraction remaining can be found using
Fraction = eln(fraction) and Percentage [I131] = Fraction x 100% Apply those steps and solve.
*****
ln(1/2) = ─k t1/2 , has two variables k and t1/2 , and we know t1/2 .
k = ─ ln(1/2) / t1/2 = ─ (─ 0.693) / t1/2 = + 0.693 / 8.1 days = 0.0856 day─1 = k
ln( fraction remaining) = ─kt = ─ ( 0.0856 day─1)(2.0 days) = ─ 0.171
Fraction = eln(fraction) = e─0.171 = 0.843 = 84 % I131 remains after 2 days Practice C: If you are unsure of the answer to a part, check it before doing the next part. 1. The rate constant for the decay of the tritium isotope of hydrogen is 0.0562 years─1 .
Calculate the halflife of tritium.
2. Strontium90 is a radioactive nuclide found in fallout: dust particles in the cloud
produced by the atmospheric testing of nuclear weapons. In chemical and biological
systems, strontium behaves much like calcium. If dairy cattle consume crops exposed
to dust or rain containing fallout, dairy products containing calcium will also contain
Sr90. Similar to calcium, Sr90 will be deposited in the bones of dairy product
consumers, including children. In part for this reason, most (but not all) nations
conducting nuclear tests signed a 1963 treaty which banned atmospheric testing.
Strontium90 undergoes beta decay with a halflife of 28.8 years. What percentage of
an original [Sr90] in bones will remain after 40.0 years?
3. The element Polonium was first isolated by Dr. Marie Sklodowska Curie and named for
her native Poland. Radioactive Po210 is found in significant concentrations in tobacco.
If 20.0% of Po210 remains in a sample after 321 days of alpha decay,
a. Estimate the halflife of Po210.
b. Calculate a precise halflife of Po210. Compare it to your Part A estimate.
4. If 10.0% of a sample of Rn222 remains after 12.6 days,
a. estimate the halflife for Rn222.
b. Calculate a precise halflife of Rn222. Compare it to your Part A estimate.
5. If the halflife of carbon14 is 5,730 years, what fraction of the original carbon14 in a
sample has decayed after 1650 years? Estimate, then calculate. © 2011 www.ChemReview.Net v.1g Page 580 Module 25 – Nuclear Chemistry ANSWERS
Practice A
1a. Firstorder halflife is constant. Half remains after one halflife, half of that half (25%) remains after two
halflives. Two halflives = 2 x 24,400 years = 48,800 years.
1b. Half remains after one halflife, 1/4th after two, 1/8th after three, 1/16th after four halflives.
1c. Half remains after one halflife, 1/4th after two, 1/8th after three; 1/8 = 0.125 = 12.5% remains, so
87.5% has decayed. Practice B
1. If 90% has decayed, 10% remains, and fraction remaining = 10% / 100% = 0.100
2. If fraction decayed = 0.40, percentage decayed = 0.40 x 100% = 40%, and percentage remaining =
100% ─ 40% = 60%
3a. WANT:
Strategy:
DATA: % of sample remaining
To find a percentage, find the fraction first.
ln(fraction remaining) = ─ 1.386
Knowing a value for ln(fraction remaining), to find fraction remaining, use
Fraction remaining = eln(fraction remaining) * * * **
= e─1.386 = 0.250
3b. If the fraction remaining is 0.25, the percentage remaining is 25%, and the percentage decayed is 75%.
4. WANT:
DATA: % of sample remaining . To find percentage, find fraction first.
0.04606 hours─1 = k
50.0 hours = t Strategy: The equation that relates the three equation terms is
ln(fraction remaining) = ─kt
Knowing k and t, ln(fraction) and then fraction can be found. * * * **
ln( fraction remaining ) = ─kt = ─ (0.04606 hours─1 )( 50.0 hours ) = ─ 2.303
Fraction remaining = eln(fraction remaining) = e─2.303 = 0.100
The percentage remaining is fraction x 100% = 10.0% Practice C
1. WANT: t1/2 for tritium DATA: 0.0562 years─1 = k Strategy: In radioactive decay calculations that include half life and fraction or percentage, write
ln(fraction remaining) = ─kt © 2011 www.ChemReview.Net v.1g and ln(1/2) = ─k t1/2 Page 581 Module 25 – Nuclear Chemistry In this problem, only the second equation is needed to relate the symbols in the WANTED
and DATA.
SOLVE: t1/2 = ln(1/2) / ─k = (─ 0.693) / ─ 0.0562 years─1 = 12.3 years ( 1/ years─1 = (years─1) ─1 = years )
% [Sr90]40.0 yrs. remaining = 2. WANT:
DATA: 28.8 yrs. = t1/2
40.0 yrs = t
In radioactive decay calculations that include half life and fraction or percentage, write
ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Percentage = Fraction x 100%
From halflife, k can be found. From k and t, ln(fraction) and then fraction can be found.
SOLVE: Since ln(1/2) = ─k t1/2 , k = ─ ln(1/2) / t1/2 = ─ (─ 0.693) / t1/2 = 0.693 / 28.8 yrs. = 0.02406 yrs.─1 = k
ln(fraction remaining) = ─kt = ─ ( 0.02406 yrs.─1 )( 40.0 yrs. ) = ─ 0.9624 = ln(fraction) Fraction = eln(fraction) = e─0.9624 = 0.382 = 38.2 % Sr90 remains after 40 years
3a. Estimate 20% remains after 321 days. 50% remains after one halflife, and 25% after 2 half lives.
At about 300 days, 25% would remain, which is 2 halflives, so one halflife is about …..
* * * **
About 150 days.
3b. WANT:
DATA: t1/2
20.0% Po210 remains
fraction remaining = 20% / 100% = 0.200
t = 321 days.
See radioactive decay, half life, and fraction or percentage? Write
ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Percentage = Fraction x 100%
Knowing the fraction and t, k can be found from the first prompt equation. Halflife can then
be found from the second equation. Solving for k in symbols first:
k = ─ ln(fraction) }/ t = ─ {ln(0.200) }/ (321 days) = ─ (─1.61) / (321 days) = 5.01 x 10─ 3 days─1
t1/2 = ─ln(1/2) / k = + 0.693 / 5.01 x 10─ 3 days─1 = 138 days = halflife of Po210
Is this answer close to the estimate in Part A? © 2011 www.ChemReview.Net v.1g Page 582 Module 25 – Nuclear Chemistry 4a. Estimate 10% remains after 12.6 days. 12.5% remains after three half –lives, which is close to 10%.
If three halflives is about 12 days, then one half life would be….
* * * **
About 4 days.
4b. WANT:
DATA: t1/2
90.0% Rn222 has decayed, so 10.0% remains, and fraction remaining = 0.100.
t = 12.6 days.
ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Fraction remaining = 10% / 100% = 0.100
Knowing the fraction and t, k can be found from the first equation. Halflife can then be
found from the second equation.
k = ─ ln(fraction) / t = ─ ln(0.100) / (12.6 days) = ─ (─2.30) / (12.6 days) = 0.183 days─1
t1/2 = ─ ln(1/2) / k = +0.693 / 0.183 days─1 = 3.79 days = halflife of Rn222 Close to the estimate in Part A?
5. Estimate: 0.50 is the fraction gone after about 6,000 years, so maybe 0.15 is the fraction gone in
about a third of that time? Calculate:
WANT: fraction [C14] decayed = 1.000 ─ fraction remaining DATA: t1/2 = 5,730 yrs.
t = 1,650 yrs.
In radioactive decay calculations that include half life and fraction or percentage, write
ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 From halflife, k can be found. From k and t, ln(fraction) and then fraction can be found.
SOLVE: k = ─ ln(1/2) / t1/2 = ─ (─ 0.693) / t1/2 = + 0.693 / 5730 yrs. = 1.21 x 10─ 4 years─1
ln( fraction remaining ) = ─kt = ─ (1.21 x 10─ 4 yr─1 )( 1,650 yrs. ) = ─ 0.1996
Fraction remaining = eln(fraction remaining) = e─0.1996 = 0.819 fraction C14 remaining
If 0.819 = fraction remaining, 1.000 ─ 0.819 = 0.181 = fraction C14 decayed ***** © 2011 www.ChemReview.Net v.1g Page 583 Module 25 – Nuclear Chemistry Summary: Nuclear Chemistry
1. α particle 4 = 2 He ; β particle 0 = −1 β or 0
−1 e 1
, neutron = 0 n 2. To balance nuclear reactions,
a. Always use AZ notation: mass number on top, nuclear charge on the bottom.
b. Both mass numbers and nuclear charge must be conserved; each must add to give
the same number before and after the reaction.
3. Fission splits a nucleus, fusion combines nuclei.
4. In the decay of radioactive isotopes, the halflife is constant.
• At the halflife, 1/2 of the original number of nuclei remain; • At double the time of the halflife, 1/4 of the original nuclei remain; • At triple the halflife, 1/8 of the original number of nuclei remain. 5. Radioactive Decay Prompt
If radioactive decay calculation includes halflife and a fraction or percentage of a
sample, and the answer cannot be calculated using simple multiples,
write in the DATA: ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 and use the math of natural logs (Lesson 25D) to solve.
6. Fraction = Quantity A
Quantity B and often equals Part
Total = Smaller number
Larger number A numeric fraction may be expressed in an x/y format or as a decimal equivalent.
In chemistry, both 1/2 and its decimal equivalent 0.50 are termed fractions.
7. In calculations,
• If you are given a percentage, as step one convert it to a decimal equivalent; • If you WANT a percentage, first find the decimal equivalent WANTED. • Fraction = percent/100% and Percentage = fraction x 100% 8. When using the radioactive decay prompt,
• To use the equation with (fraction remaining), you must work in fractions, not
percentages. • Fraction remaining = 1.000 ─ fraction decayed • Percentage remaining = 100% ─ percentage decayed • The fractions in decay calculations will have a value between 1.00 and 0
(such as 0.25).
##### © 2011 www.ChemReview.Net v.1g Page 584 The ELEMENTS –
The third column shows the atomic number: The
protons in the nucleus of the atom.
The fourth column is the molar mass, in
grams/mole. For radioactive atoms, ( ) is the
molar mass of most stable isotope.
Actinium
Aluminum
Americium
Antimony
Argon
Arsenic
Astatine
Barium
Berkelium
Beryllium
Bismuth
Boron
Bromine
Cadmium
Calcium
Californium
Carbon
Cerium
Cesium
Chlorine
Chromium
Cobalt
Copper
Curium
Dysprosium
Erbium
Europium
Fermium
Fluorine
Francium
Gadolinium
Gallium
Germanium
Gold
Hafnium
Helium
Holmium
Hydrogen
Indium
Iodine
Iridium
Iron
Krypton
Lanthanum
Lawrencium
Lead
Lithium Ac
Al
Am
Sb
Ar
As
At
Ba
Bk
Be
Bi
B
Br
Cd
Ca
Cf
C
Ce
Cs
Cl
Cr
Co
Cu
Cm
Dy
Er
Eu
Fm
F
Fr
Gd
Ga
Ge
Au
Hf
He
Ho
H
In
I
Ir
Fe
Kr
La
Lr
Pb
Li 89
13
95
51
18
33
84
56
97
4
83
5
35
48
20
98
6
58
55
17
24
27
29
96
66
68
63
100
9
87
64
31
32
79
72
2
67
1
49
53
77
26
36
57
103
82
3 (227)
27.0
(243)
121.8
40.0
74.9
(210)
137.3
(247)
9.01
209.0
10.8
79.9
112.4
40.1
(249)
12.0
140.1
132.9
35.5
52.0
58.9
63.5
(247)
162.5
167.3
152.0
(253)
19.0
(223)
157.3
69.7
72.6
197.0
178.5
4.00
164.9
1.008
114.8
126.9
192.2
55.8
83.8
138.9
(257)
207.2
6.94 Lutetium
Magnesium
Manganese
Mendelevium
Mercury
Molybdenum
Neodymium
Neon
Neptunium
Nickel
Niobium
Nitrogen
Nobelium
Osmium
Oxygen
Palladium
Phosphorus
Platinum
Plutonium
Polonium
Potassium
Praseodymium
Promethium
Protactinium
Radium
Radon
Rhenium
Rhodium
Rubidium
Ruthenium
Samarium
Scandium
Selenium
Silicon
Silver
Sodium
Strontium
Sulfur
Tantalum
Technetium
Tellurium
Terbium
Thallium
Thorium
Thulium
Tin
Titanium
Tungsten
Uranium
Vanadium
Xenon
Ytterbium
Yttrium
Zinc
Zirconium Lu
Mg
Mn
Md
Hg
Mo
Nd
Ne
Np
Ni
Nb
N
No
Os
O
Pd
P
Pt
Pu
Po
K
Pr
Pm
Pa
Ra
Rn
Re
Rh
Rb
Ru
Sm
Sc
Se
Si
Ag
Na
Sr
S
Ta
Tc
Te
Tb
Tl
Th
Tm
Sn
Ti
W
U
V
Xe
Yb
Y
Zn
Zr 71
12
25
101
80
42
60
10
93
28
41
7
102
76
8
46
15
78
94
84
19
59
61
91
88
86
75
45
37
44
62
21
34
14
47
11
38
16
73
43
52
65
81
90
69
50
22
74
92
23
54
70
39
30
40 175.0
24.3
54.9
(256)
200.6
95.9
144.2
20.2
(237)
58.7
92.9
14.0
(253)
190.2
16.0
106.4
31.0
195.1
(242)
(209)
39.1
140.9
(145)
(231)
(226)
(222)
186.2
102.9
85.5
101.1
150.4
45.0
79.0
28.1
107.9
23.0
87.6
32.1
180.9
(98)
127.6
158.9
204.4
232.0
168.9
118.7
47.9
183.8
238.0
50.9
131.3
173.0
88.9
65.4
91.2 ...
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