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Unformatted text preview: Introduction to Chemistry Calculations ***** 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 Half-Life 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 “hydrogen-1”); • 1 proton + 1 neutron or as 2H (termed hydrogen-2 or deuterium); and • 1 proton + 2 neutrons or as 3H (called hydrogen-3 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 silver-109 and its symbol is 109Ag . Nuclide symbols may also be written with the nuclear charge below the mass number. This is called A-Z 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 sub-atomic particles, the two with the highest mass are _________________ and _________________. 5. Write the nuclide (isotope) symbol for a single proton using A-Z 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 He-4 nucleus, and it is given the same isotopic symbol as an He-4 nucleus. Symbol for alpha particle = Example: α particle 4 = 2 He The isotope U-238 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 U-238, 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 radium-226? ***** A key to balancing nuclear reactions is to write the isotope symbols in A-Z notation. Start there for Ra-226. ***** © 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 radon-219. 2. How many protons and how many neutrons are in Rn-219? 3. How many protons and how many neutrons are in the nucleus left behind after the alpha decay of Rn-219? How many protons and neutrons are lost in the decay? 4. Lead-206 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 lead-210 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 carbon-14 undergoes beta decay. Write the isotope symbols for the two nuclear particles formed in this reaction. ***** β 14 ⎯ 6C ⎯ → Begin by converting to A-Z 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 carbon-14 is nitrogen-14. 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 iodine-131 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 I-131 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 I-131. © 2011 www.ChemReview.Net v.1g Page 557 Module 25 – Nuclear Chemistry 4. Lead-206 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 uranium-235. 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 U-235 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 U-235 and U-238 can be split, but in practice only U-235 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 U-235 must be increased. In mined uranium ore, 99.3% of nuclei are U-238 and 0.7% are U-235. For nuclear weapons, uranium must be enriched to 20-80% U-235, and over 50 kilograms of this “weapons grade” uranium must be collected. Since all isotopes, including U-235 and U-238, 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 U-235 are lighter than atoms with U-238, in gaseous substances containing uranium the particles containing U-235 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 uranium-containing 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 long-lived 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 A-Z 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, helium-4 can fuse with beryllium-8 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 math-inclined, 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 half-life 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 lower-case 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 half-life, 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 TI-type calculator might use: 1 ex . • On an RPN scientific calculator, try: 1 enter ex . Use your calculator-key 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 base-10 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 in-memory log-rule 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 odd-numbered problems first. Complete the even numbered problems for additional practice or for pre-test 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 Half-Life Calculations Half-Lives: Simple Multiples The half-life 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 half-life: 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 half-life does not change significantly at temperatures up to several thousand kelvins. As a result, each radioactive nucleus has a characteristic half-life. Some radioactive nuclides have a half-life of a few seconds; others have a half-life 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 half-life, 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 Half-Life Calculations For Simple Multiples In calculations involve radioactive half-life, 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 half-life 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 half-life, 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 half-life. • After two half-lives (double the time of the half-life), the number of nuclei remaining is half of the half that remained after the first half-life: half of 1/2 = 1/4 (25%) of the original nuclei remain and 75% have decayed. • At triple the half-life, 1/2 of 1/4 = 1/8 (12.5%) of the original nuclei remain. Apply the rules above to the following problem. Q. Fluorine-18, a radioactive isotope used in nuclear medicine, has a 1.8 hour half-life. How long will it take for 87.5% of the F-18 nuclei in a sample to decay? * ** * * A. If 87.5% has decayed, 12.5% remains. How many half-lives are required? How much time would this be? * ** * * 12.5% remains at 3 half-lives; 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 half-life 1/2 50% 50% Two half-lives 1/4 25% 75% Three half-lives 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 Pu-239 undergoes first-order radioactive decay with a half-life of 24,400 years. In a sample of constant volume containing Pu-239, a. After how many years will 25% of the original Pu-239 nuclei remain? b. After how many half-lives will the [Pu-239] be 1/16th of its original concentration? c. What percentage of the Pu-239 has decayed after exactly 3 half-lives? Rate Constants for Radioactive Decay In half-life 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 half-life, 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 half-lives, 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 half-life 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? Half-Life Calculations For Non-Simple-Multiples For a radioactive nucleus, after a time equal to one half-life ( 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 half-life, Solve this equation in symbols for half-life. * * ** * One way of several to write the equation is t1/2 ≡ ─ ln(1/2) / k This equation is one way to define radioactive half-life. In the above equations, the rate constant (k) and half-life ( 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 half-life 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 half-life. From this form, it is clear that if you know the half-life, you can find the rate constant, and if you know the rate constant you can find the half-life. To solve decay calculations, we need equations that relate the fraction remaining, half-life, 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 half-life 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 half-life. Commit the radioactive decay prompt to memory, then apply it to solve this problem. Q. Iodine-131, a radioactive isotope used to treat thyroid disorders, has a half-life of 8.1 days. What percentage of an initial [I-131] remains after 48 hours? ***** WANT: Percent [I-131]48 hrs. = % remaining DATA: t1/2 = 8.1 days = radioactive half-life 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 half-life 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 half-life, 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 [I-131] = 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 % I-131 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 half-life of tritium. 2. Strontium-90 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 Sr-90. Similar to calcium, Sr-90 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. Strontium-90 undergoes beta decay with a half-life of 28.8 years. What percentage of an original [Sr-90] 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 Po-210 is found in significant concentrations in tobacco. If 20.0% of Po-210 remains in a sample after 321 days of alpha decay, a. Estimate the half-life of Po-210. b. Calculate a precise half-life of Po-210. Compare it to your Part A estimate. 4. If 10.0% of a sample of Rn-222 remains after 12.6 days, a. estimate the half-life for Rn-222. b. Calculate a precise half-life of Rn-222. Compare it to your Part A estimate. 5. If the half-life of carbon-14 is 5,730 years, what fraction of the original carbon-14 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. First-order half-life is constant. Half remains after one half-life, half of that half (25%) remains after two half-lives. Two half-lives = 2 x 24,400 years = 48,800 years. 1b. Half remains after one half-life, 1/4th after two, 1/8th after three, 1/16th after four half-lives. 1c. Half remains after one half-life, 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 ) % [Sr-90]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 half-life, 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 % Sr-90 remains after 40 years 3a. Estimate 20% remains after 321 days. 50% remains after one half-life, and 25% after 2 half -lives. At about 300 days, 25% would remain, which is 2 half-lives, so one half-life is about ….. * * * ** About 150 days. 3b. WANT: DATA: t1/2 20.0% Po-210 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. Half-life 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 = half-life of Po-210 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 half-lives is about 12 days, then one half life would be…. * * * ** About 4 days. 4b. WANT: DATA: t1/2 90.0% Rn-222 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. Half-life 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 = half-life of Rn-222 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 [C-14] 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 half-life, 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 C-14 remaining If 0.819 = fraction remaining, 1.000 ─ 0.819 = 0.181 = fraction C-14 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 A-Z 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 half-life is constant. • At the half-life, 1/2 of the original number of nuclei remain; • At double the time of the half-life, 1/4 of the original nuclei remain; • At triple the half-life, 1/8 of the original number of nuclei remain. 5. Radioactive Decay Prompt If radioactive decay calculation includes half-life 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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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.

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