Lecture Presentation Chapter 21, Inc. James F. Kirby Quinnipiac University Hamden, CT
Energy: Chemical vs. Chemical energy is associated with making and breaking chemical bonds. energy is enormous in comparison. energy is due to changes in the nucleus of atoms changing them into different atoms. 13% of worldwide energy use comes from nuclear energy.
The Nucleus Remember that the nucleus is composed of the two nucleons, protons and neutrons. The number of protons is the atomic number. The number of protons and neutrons together is the mass number.
Isotopes Not all atoms of the same element have the same mass, due to different numbers of neutrons in those atoms. There are, for example, three naturally occurring isotopes of uranium: Uranium-234 Uranium-235 Uranium-238
21.1 Radioactivity and Equations 948 Radioactivity It is not uncommon for some nuclides of an element to be unstable, or radioactive. We refer to these as radionuclides. There are several ways radionuclides can decay into a different nuclide. We use nuclear equations to show how these nuclear reactions occur.
Equations In chemical equations, atoms and charges need to balance. In nuclear equations, atomic number and mass number need to balance. This is a way of balancing charge (atomic number) and mass (mass number) on an atomic scale.
Most Common Kinds of Radiation Emitted by a Radionuclide
Types of Radioactive Decay Alpha decay Beta decay Gamma emission Positron emission Electron capture
Alpha Decay Alpha decay is the loss of an α-particle (He-4 nucleus, two protons and two neutrons): 4 He 2 238 U 92 234 4 Th He 90 + 2 Note how the equation balances: atomic number: 92 = 90 + 2 mass number: 238 = 234 + 4
Beta Decay Beta decay is the loss of a β-particle (a highspeed electron emitted by the nucleus): 0 β 0 e 1 or 1 131 I 131 Xe 53 54 0 1 + e Balancing: atomic number: 53 = 54 + ( 1) mass number: 131 = 131 + 0
Gamma Emission Gamma emission is the loss of a γ-ray, which is high-energy radiation that almost always accompanies the loss of a nuclear particle: 0 γ 0
Positron Emission Some nuclei decay by emitting a positron, a particle that has the same mass as, but an opposite charge to, that of an electron: 11 C 6 0 e 1 11 B 5 + 0 e 1 Balancing: atomic number: 6 = 5 + 1 mass number: 11 = 11 + 0
Electron Capture (K-Capture) An electron from the surrounding electron cloud is absorbed into the nucleus during electron capture. 81 Rb 0 37 + e 1 81 Kr 36 Balancing: atomic number: 37 + ( 1) = 36 mass number: 81 + 0 = 81
Sources of Some Particles Beta particles: Positrons: What happens with electron capture?
Sample Exercise 21.1 Predicting the Product of a Reaction What product is formed when radium-226 undergoes alpha decay? Solution Analyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle. Plan We can best do this by writing a balanced nuclear reaction for the process. Solve The periodic table shows that radium has an atomic number of 88. The complete chemical symbol for radium-226 is therefore An alpha particle is a helium-4 nucleus, and so its symbol is The alpha particle is a product of the nuclear reaction, and so the equation is of the form where A is the mass number of the product nucleus and Z is its atomic number. Mass numbers and atomic numbers must balance, so and 226 = A + 4 88 = Z + 2
Sample Exercise 21.1 Predicting the Product of a Reaction Continued Hence, A = 222 and Z = 86 Again, from the periodic table, the element with Z = 86 is radon (Rn). The product, therefore, is nuclear equation is and the Practice Exercise 1 The plutonium-238 that is shown in the chapter-opening photograph undergoes alpha decay. What product forms when this radionuclide decays? (a) Plutonium-234 (b) Uranium-234 (c) Uranium-238 (d) Thorium-236 (e) Neptunium-237 Practice Exercise 2 Which element undergoes alpha decay to form lead-208?
Sample Exercise 21.2 Writing Equations Write nuclear equations for (a) mercury-201 undergoing electron capture; (b) thorium-231 decaying to protactinium-231. Solution Analyze We must write balanced nuclear equations in which the masses and charges of reactants and products are equal. Plan We can begin by writing the complete chemical symbols for the nuclei and decay particles that are given in the problem. Solve (a) The information given in the question can be summarized as The mass numbers must have the same sum on both sides of the equation: 201 + 0 = A Thus, the product nucleus must have a mass number of 201. Similarly, balancing the atomic numbers gives 80 1 = Z
Sample Exercise 21.2 Writing Equations Continued Thus, the atomic number of the product nucleus must be 79, which identifies it as gold (Au): (b) In this case we must determine what type of particle is emitted in the course of the radioactive decay: From 231 = 231 + A and 90 = 91 + Z, we deduce A = 0 and Z = 1. According to Table 21.2, the particle with these characteristics is the beta particle (electron). We therefore write
Sample Exercise 21.2 Writing Equations Continued Practice Exercise 1 The radioactive decay of thorium-232 occurs in multiple steps, called a radioactive decay chain. The second product produced in this chain is actinium-228. Which of the following processes could lead to this product starting with thorium-232? (a) Alpha decay followed by beta emission (b) Beta emission followed by electron capture (c) Positron emission followed by alpha decay (d) Electron capture followed by positron emission (e) More than one of the above is consistent with the observed transformation Practice Exercise 2 Write a balanced nuclear equation for the reaction in which oxygen-15 undergoes positron emission.
21.2 Patterns of Stability 952 Stability Any atom with more than one proton (anything but H) will have repulsions between the protons in the nucleus. Strong nuclear force helps keep the nucleus together. Neutrons play a key role stabilizing the nucleus, so the ratio of neutrons to protons is an important factor.
Neutron Proton Ratios For smaller nuclei (Z 20), stable nuclei have a neutron-to-proton ratio close to 1:1. As nuclei get larger, it takes a larger number of neutrons to stabilize the nucleus. The shaded region in the figure is called the belt of stability; it shows what nuclides would be stable.
Unstable Nuclei Compare a nucleus to the belt of stability. Nuclei above this belt have too many neutrons, so they tend to decay by emitting beta particles. Nuclei below the belt have too many protons, so they tend to become more stable by positron emission or electron capture.
Alpha Emission There are no stable nuclei with an atomic number greater than 83. Nuclei with such large atomic numbers tend to decay by alpha emission.
Radioactive Decay Chain Some radioactive nuclei cannot stabilize by undergoing only one nuclear transformation. They undergo a series of decays until they form a stable nuclide (often a nuclide of lead).
Stable Nuclei Magic numbers of 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82, or 126 neutrons result in more stable nuclides. Nuclei with an even number of protons and neutrons tend to be more stable than those with odd numbers.
Sample Exercise 21.3 Predicting Modes of Decay Predict the mode of decay of (a) carbon-14, (b) xenon-118. Solution Analyze We are asked to predict the modes of decay of two nuclei. Plan To do this, we must locate the respective nuclei in Figure 21.2 and determine their positions with respect to the belt of stability in order to predict the most likely mode of decay.
Sample Exercise 21.3 Predicting Modes of Decay Continued Solve (a) Carbon is element 6. Thus, carbon-14 has 6 protons and 14 6 = 8 neutrons, giving it a neutron-to-proton ratio of 1.25. Elements with Z < 20 normally have stable nuclei that contain approximately equal numbers of neutrons and protons (n/p = 1). Thus, carbon-14 is located above the belt of stability and we expect it to decay by emitting a beta particle to decrease the n/p ratio: This is indeed the mode of decay observed for carbon-14, a reaction that lowers the n/p ratio from 1.25 to 1.0. (b) Xenon is element 54. Thus, xenon-118 has 54 protons and 118 54 = 64 neutrons, giving it an n/p ratio of 1.18. According to Figure 21.2, stable nuclei in this region of the belt of stability have higher neutron-to-proton ratios than xenon-118. The nucleus can increase this ratio by either positron emission or electron capture: In this case both modes of decay are observed. Comment Keep in mind that our guidelines do not always work. For example, thorium-233, which we might expect to undergo alpha decay, actually undergoes beta emission. Furthermore, a few radioactive nuclei lie within the belt of stability. Both and for example, are stable and lie in the belt of stability. however, which lies between them, is radioactive.
Sample Exercise 21.3 Predicting Modes of Decay Continued Practice Exercise 1 Which of the following radioactive nuclei is most likely to decay via emission of a β particle? (a) nitrogen-13 (b) magnesium-23 (c) rubidium-83 (d) iodine-131 (e) neptunium-237 Practice Exercise 2 Predict the mode of decay of (a) plutonium-239, (b) indium-120.
21.3 Transmutations 956 Transmutations transmutations can be induced by accelerating a particle to collide it with the nuclide. Particle accelerators ( atom smashers ) are enormous, having circular tracks with radii that are miles long.
Other Transmutations Use of neutrons: Most synthetic isotopes used in medicine are prepared by bombarding neutrons at a particle, which won t repel the neutral particle. Transuranium elements: Elements immediately after uranium were discovered by bombarding isotopes with neutrons. Larger elements (atomic number higher than 110) were made by colliding large atoms with nuclei of light elements with high energy.
1) or 2) Writing Equations for Transmutations equations that represent nuclear transmutations are written two ways:
Sample Exercise 21.4 Writing a Balanced Equation Write the balanced nuclear equation for the process summarized as Solution Analyze We must go from the condensed descriptive form of the reaction to the balanced nuclear equation. Plan We arrive at the balanced equation by writing n and α, each with its associated subscripts and superscripts. Solve The n is the abbreviation for a neutron and a represents an alpha particle The neutron is the bombarding particle, and the alpha particle is a product. Therefore, the nuclear equation is Practice Exercise 1 Consider the following nuclear transmutation: What is the identity of nucleus X? Practice Exercise 2 Write the condensed version of the nuclear reaction
21.4 Rates of Radioactive Decay 958 Kinetics of Radioactive Decay Radioactive decay is a first-order process. The kinetics of such a process obey this equation: ln N t N 0 = kt
Half-Life The half-life of such a process is 0.693 k = t 1/2 Half-life is the time required for half of a radionuclide sample to decay.
Radiometric Dating Applying first-order kinetics and half-life information, we can date objects using a nuclear clock. Carbon dating works: the half-life of C-14 is 5700 yr. It is limited to objects up to about 50,000 yr old; after this time there is too little radioactivity left to measure. Other isotopes can be used (U-238:Pb-206 in rock).
Measuring Radioactivity: Units Activity is the rate at which a sample decays. The units used to measure activity are as follows: Becquerel (Bq): one disintegration per second Curie (Ci): 3.7 10 10 disintegrations per second, which is the rate of decay of 1 g of radium.
Measuring Radioactivity: Some Instruments Film badges Geiger counter Phosphors (scintillation counters)
Sample Exercise 21.5 Calculation Involving Half-Lives The half-life of cobalt-60 is 5.27 yr. How much of a 1.000-mg sample of cobalt-60 is left after 15.81 yr? Solution Analyze We are given the half-life for cobalt-60 and asked to calculate the amount of cobalt-60 remaining from an initial 1.000-mg sample after 15.81 yr. Plan We will use the fact that the amount of a radioactive substance decreases by 50% for every half-life that passes. Solve Because 5.27 3 = 15.81, 15.81 yr is three half-lives for cobalt-60. At the end of one half-life, 0.500 mg of cobalt-60 remains, 0.250 mg at the end of two half-lives, and 0.125 mg at the end of three half-lives. Practice Exercise 1 A radioisotope of technetium is useful in medical imaging techniques. A sample initially contains 80.0 mg of this isotope. After 24.0 h, only 5.0 mg of the technetium isotope remains. What is the half-life of the isotope? (a) 3.0 h (b) 6.0 h (c) 12.0 h (d) 16.0 h (e) 24.0 h Practice Exercise 2 Carbon-11, used in medical imaging, has a half-life of 20.4 min. The carbon-11 nuclides are formed, and the carbon atoms are then incorporated into an appropriate compound. The resulting sample is injected into a patient, and the medical image is obtained. If the entire process takes five half-lives, what percentage of the original carbon-11 remains at this time?
Sample Exercise 21.6 Calculating the Age of Objects Using Radioactive Decay A rock contains 0.257 mg of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is 4.5 10 9 yr. How old is the rock? Solution Analyze We are told that a rock sample has a certain amount of lead-206 for every unit mass of uranium-238 and asked to estimate the age of the rock. Plan Lead-206 is the product of the radioactive decay of uranium-238. We will assume that the only source of lead-206 in the rock is from the decay of uranium-238, with a known half-life. To apply first-order kinetics expressions (Equations 21.19 and 21.20) to calculate the time elapsed since the rock was formed, we first need to calculate how much initial uranium-238 there was for every 1 mg that remains today. Solve Let s assume that the rock currently contains 1.000 mg of uranium-238 and therefore 0.257 mg of lead-206. The amount of uranium-238 in the rock when it was first formed therefore equals 1.000 mg plus the quantity that has decayed to lead-206. Because the mass of lead atoms is not the same as the mass of uranium atoms, we cannot just add 1.000 mg and 0.257 mg. We have to multiply the present mass of lead-206 (0.257 mg) by the ratio of the mass number of uranium to that of lead, into which it has decayed. Therefore, the original mass of was
Sample Exercise 21.6 Calculating the Age of Objects Using Radioactive Decay Continued Using Equation 21.20, we can calculate the decay constant for the process from its half-life: Rearranging Equation 21.19 to solve for time, t, and substituting known quantities gives Comment To check this result, you could use the fact that the decay of uranium-235 to lead-207 has a half-life of 7 10 8 yr and measure the relative amounts of uranium-235 and lead-207 in the rock. Practice Exercise 1 Cesium-137, which has a half-life of 30.2 yr, is a component of the radioactive waste from nuclear power plants. If the activity due to cesium-137 in a sample of radioactive waste has decreased to 35.2% of its initial value, how old is the sample? (a) 1.04 yr (b) 15.4 yr (c) 31.5 yr (d) 45.5 yr (e) 156 yr
Sample Exercise 21.6 Calculating the Age of Objects Using Radioactive Decay Continued Practice Exercise 2 A wooden object from an archeological site is subjected to radiocarbon dating. The activity due to 14 C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14 C is 5700 yr. What is the age of the archeological sample?
Sample Exercise 21.7 Calculations Involving Radioactive Decay and Time If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 yr. (a) What is the half-life of strontium-90? (b) How much strontium-90 will remain after 5.00 yr? (c) What is the initial activity of the sample in becquerels and curies? Solution (a) Analyze We are asked to calculate a half-life, t 1/2, based on data that tell us how much of a radioactive nucleus has decayed in a time interval t = 2.00 yr and the information N 0 = 1.000 g, N t = 0.953 g. Plan We first calculate the rate constant for the decay, k, and then use that to compute t 1/2. Solve Equation 21.19 is solved for the decay constant, k, and then Equation 21.20 is used to calculate half-life, t 1/2 :
Sample Exercise 21.7 Calculations Involving Radioactive Decay and Time Continued (b) Analyze We are asked to calculate the amount of a radionuclide remaining after a given period of time. Plan We need to calculate N t, the amount of strontium present at time t, using the initial quantity, N 0, and the rate constant for decay, k, calculated in part (a). Solve Again using Equation 21.19, with k = 0.0241 yr 1, we have N t /N 0 is calculated from ln(n t /N 0 ) = 0.120 using the e x or INV LN function of a calculator: Because N 0 = 1.000 g, we have N t = (0.887)N 0 = (0.887)(1.000 g) = 0.887 g
Sample Exercise 21.7 Calculations Involving Radioactive Decay and Time Continued (c) Analyze We are asked to calculate the activity of the sample in becquerels and curies. Plan We must calculate the number of disintegrations per atom per second and then multiply by the number of atoms in the sample. Solve The number of disintegrations per atom per second is given by the decay constant, k: To obtain the total number of disintegrations per second, we calculate the number of atoms in the sample. We multiply this quantity by k, where we express k as the number of disintegrations per atom per second, to obtain the number of disintegrations per second:
Sample Exercise 21.7 Calculations Involving Radioactive Decay and Time Continued Because 1 Bq is one disintegration per second, the activity is 5.1 10 12 Bq. The activity in curies is given by We have used only two significant figures in products of these calculations because we do not know the atomic weight of 90 Sr to more than two significant figures without looking it up in a special source. Practice Exercise 1 As mentioned in the previous Practice Exercise 1, cesium-137, a component of radioactive waste, has a half-life of 30.2 yr. If a sample of waste has an initial activity of 15.0 Ci due to cesium-137, how long will it take for the activity due to cesium-137 to drop to 0.250 Ci? (a) 0.728 yr (b) 60.4 yr (c) 78.2 yr (d) 124 yr (e) 178 yr Practice Exercise 2 A sample to be used for medical imaging is labeled with 18 F, which has a half-life of 110 min. What percentage of the original activity in the sample remains after 300 min?
21.5 Detection of Radioactivity 964 Film Badges Radioactivity was first discovered by Henri Becquerel because it fogged up a photographic plate. Film has been used to detect radioactivity since more exposure to radioactivity means darker spots on the developed film. Film badges are used by people who work with radioactivity to measure their own exposure over time.
Geiger Counter A Geiger counter measures the amount of activity present in a radioactive sample. Radioactivity enters a window and creates ions in a gas; the ions result in an electric current that is measured and recorded by the instrument.
Phosphors Some substances absorb radioactivity and emit light. They are called phosphors. An instrument commonly used to measure the amount of light emitted by a phosphor is a scintillation counter. It converts the light to an electronic response for measurement.
Radiotracers Radiotracers are radioisotopes used to study a chemical reaction. An element can be followed through a reaction to determine its path and better understand the mechanism of a chemical reaction.
Medical Application of Radiotracers Radiotracers have found wide diagnostic use in medicine. Radioisotopes are administered to a patient (usually intravenously) and followed. Certain elements collect more in certain tissues, so an organ or tissue type can be studied based on where the radioactivity collects.
Positron Emission Tomography (PET Scan) A compound labeled with a positron emitter is injected into a patient. Blood flow, oxygen and glucose metabolism, and other biological functions can be studied. Labeled glucose is used to study the brain, as seen in the figure to the right.
21.6 Energy Changes in Reactions 967 Energy in Reactions There is a tremendous amount of energy stored in nuclei. Einstein s famous equation, E = mc 2, relates directly to the calculation of this energy.
Energy in Reactions To show the enormous difference in energy for nuclear reactions, the mass change associated with the α-decay of 1 mol of U-238 to Th-234 is 0.0046 g. The change in energy, ΔE, is then ΔE = (Δm)c 2 E = ( 4.6 10 6 kg)(3.00 10 8 m/s) 2 E = 4.1 10 11 J (Note: the negative sign means heat is released.)
Mass Defect Where does this energy come from? The masses of nuclei are always less than those of the individual parts. This mass difference is called the mass defect. The energy needed to separate a nucleus into its nucleons is called the nuclear binding energy.
Effects of Binding Energy on Processes Dividing the binding energy by the number of nucleons gives a value that can be compared. Heavy nuclei gain stability and give off energy when they split into two smaller nuclei. This is fission. Lighter nuclei emit great amounts of energy by being combined in fusion.
Sample Exercise 21.8 Calculating Mass Change in a Reaction How much energy is lost or gained when 1 mol of cobalt-60 undergoes beta decay, The mass of a atom is 59.933819 amu, and that of a atom is 59.930788 amu. Solution Analyze We are asked to calculate the energy change in a nuclear reaction. Plan We must first calculate the mass change in the process. We are given atomic masses, but we need the masses of the nuclei in the reaction. We calculate these by taking account of the masses of the electrons that contribute to the atomic masses. Solve A atom has 27 electrons. The mass of an electron is 5.4858 10 4 amu. (See the list of fundamental constants in the back inside cover.) We subtract the mass of the 27 electrons from the mass of the atom to find the mass of the nucleus: 59.933819 amu (27)(5.4858 10 4 amu) = 59.919007 amu (or 59.919007 g/mol) Likewise, for, the mass of the nucleus is 59.930788 amu (28)(5.4858 10 4 amu) = 59.915428 amu (or 59.915428 g/mol)
Sample Exercise 21.8 Calculating Mass Change in a Reaction Continued The mass change in the nuclear reaction is the total mass of the products minus the mass of the reactant: Thus, when a mole of cobalt-60 decays, Δm = 0.003030 g Because the mass decreases (Δm < 0), energy is released (ΔE < 0). The quantity of energy released per mole of cobalt-60 is calculated using Equation 21.22:
Sample Exercise 21.8 Calculating Mass Change in a Reaction Continued Practice Exercise 1 The nuclear reaction that powers the radioisotope thermoelectric generator shown in the chapter-opening photograph is The atomic masses of plutonium-238 and uranium-234 are 238.049554 amu and 234.040946 amu, respectively. The mass of an alpha particle is 4.001506 amu. How much energy in kj is released when 1.00 g of plutonium-238 decays to uranium-234? (a) 2.27 10 6 kj (b) 2.68 10 6 kj (c) 3.10 10 6 kj (d) 3.15 10 6 kj (e) 7.37 10 8 kj Practice Exercise 2 Positron emission from occurs with release of 2.87 10 11 J per mole of 11 C. What is the mass change per mole of 11 C in this nuclear reaction? The masses of 11 B and 11 C are 11.009305 and 11.011434 amu, respectively.
21.7 Power: Fission 970 Fission Commercial nuclear power plants use fission. Heavy nuclei can split in many ways. The equations below show two ways U-235 can split after bombardment with a neutron.
Fission Bombardment of the radioactive nuclide with a neutron starts the process. Neutrons released in the transmutation strike other nuclei, causing their decay and the production of more neutrons. This process continues in what we call a nuclear chain reaction.
Fission The minimum mass that must be present for a chain reaction to be sustained is called the critical mass. If more than critical mass is present (supercritical mass), an explosion will occur. Weapons were created by causing smaller amounts to be forced together to create this mass.
Reactors In nuclear reactors, the heat generated by the reaction is used to produce steam that turns a turbine connected to a generator. Otherwise, the plant is basically the same as any power plant.
Reactors The reactor core consists of fuel rods, control rods, moderators, and coolant. The control rods block the paths of some neutrons, keeping the system from reaching a dangerous supercritical mass.
Waste Reactors must be stopped periodically to replace or reprocess the nuclear fuel. They are stored in pools at the reactor site. The original intent was that this waste would then be transported to reprocessing or storage sites. Political opposition to storage site location and safety challenges for reprocessing have led this to be a major social problem.
21.8 Power: Fusion 975 Fusion When small atoms are combined, much energy is released. If it were possible to easily produce energy by this method, it would be a preferred source of energy. However, extremely high temperatures and pressures are needed to cause nuclei to fuse. This was achieved using an atomic bomb to initiate fusion in a hydrogen bomb. Obviously, this is not an acceptable approach to producing energy.
21.9 Radiation in the Environment and Living Systems 976 Radiation in the Environment We are constantly exposed to radiation. Ionizing radiation is more harmful to living systems than nonionizing radiation, such as radiofrequency electromagnetic radiation. Since most living tissue is ~70% water, ionizing radiation is that which causes water to ionize. This creates unstable, very reactive OH radicals, which result in much cell damage.
Damage to Cells The damage to cells depends on the type of radioactivity, the length of exposure, and whether the source is inside or outside the body. Outside the body, gamma rays are most dangerous. Inside the body, alpha radiation can cause most harm.
Exposure We are constantly exposed to radiation. What amount is safe? Setting standards for safety is difficult. Low-level, long-term exposure can cause health issues. Damage to the growth-regulation mechanism of cells results in cancer.
Radiation Dose Two units are commonly used to measure exposure to radiation: Gray (Gy): absorption of 1 J of energy per kg of tissue Rad (for radiation absorbed dose): absorption of 0.01 J of energy per kg of tissue (100 rad = 1 Gy) Not all forms of radiation harm tissue equally. A relative biological effectiveness (RBE) is used to show how much biological effect there is. The effective dose is called the rem (SI unit Sievert; 1 Sv = 100 rem) # of rem = (# of rad) (RBE)
Short-Term Exposure 600 rem is fatal to most humans. Average exposure per year is about 360 mrem.
Radon Radon-222 is a decay product of uranium-238, which is found in rock formations and soil. Most of the decay products of uranium remain in the soil, but radon is a gas. When breathed in, it can cause much harm, since it produces alpha particles, which have a high RBE. It is estimated to contribute to 10% of all lung cancer deaths in the United States.
Sample Integrative Exercise Putting Concepts Together Potassium ion is present in foods and is an essential nutrient in the human body. One of the naturally occurring isotopes of potassium, potassium-40, is radioactive. Potassium-40 has a natural abundance of 0.0117% and a half-life t 1/2 = 1.28 10 9 yr. It undergoes radioactive decay in three ways: 98.2% is by electron capture, 1.35% is by beta emission, and 0.49% is by positron emission. (a) Why should we expect 40 K to be radioactive? (b) Write the nuclear equations for the three modes by which 40 K decays. (c) How many 40 K + ions are present in 1.00 g of KCl? (d) How long does it take for 1.00% of the 40 K in a sample to undergo radioactive decay? Solution (a) The 40 K nucleus contains 19 protons and 21 neutrons. There are very few stable nuclei with odd numbers of both protons and neutrons (Section 21.2). (b) Electron capture is capture of an inner-shell electron by the nucleus: Beta emission is loss of a beta particle by the nucleus: Positron emission is loss of a positron by the nucleus:
Sample Integrative Exercise Putting Concepts Together Continued (c) The total number of K + ions in the sample is Of these, 0.0117% are 40 K + ions:
Sample Integrative Exercise Putting Concepts Together Continued (d) The decay constant (the rate constant) for the radioactive decay can be calculated from the half-life, using Equation 21.20: The rate equation, Equation 21.19, then allows us to calculate the time required: That is, it would take 18.6 million years for just 1.00% of the 40 K in a sample to decay.