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Chapter 31 Nuclear Physics and Radioactivity Chapter 31 NUCLEAR PHYSICS AND RADIOACTIVITY PREVIEW The modern view of the atom includes electrons in energy levels around the nucleus of the atom. The nucleus contains positively-charged protons and neutral neutrons, each of which are made up of quarks. The nucleus is held together by the strong nuclear force, and the binding energy in the nucleus is a result of some of the mass of the particles (the mass defect) in the nucleus being converted into energy by the relationship E = mc2. The atomic number is the number of protons in the nucleus, and the atomic mass number is the number of nucleons (protons and neutrons) in the nucleus. Nuclear changes can take place, but the total amount of atomic mass in the process must remain constant. The content contained in sections 1 – 4, and 10 (Example 13) of chapter 31 of the textbook is included on the AP Physics B exam. QUICK REFERENCE Important Terms alpha particle positively charged particle consisting of two protons and two neutrons atomic mass number (A) the number of protons and neutrons in the nucleus of an atom atomic mass unit the unit of mass equal to 1/12 the mass of a carbon-12 nucleus; the atomic mass rounded to the nearest whole number is called the mass number atomic number (Z) the number of protons in the nucleus of an atom beta particle high speed electron emitted from a radioactive element when a neutron decays into a proton binding energy the nuclear energy that binds protons and neutrons in the nucleus of the atom element a substance made of only one kind of atom isotope a form of an element which has a particular number of neutrons, that is, has the same atomic number but a different mass number than the other elements which occupy the same place on the periodic table 346 Chapter 31 Nuclear Physics and Radioactivity mass defect the mass equivalent of the binding energy in the nucleus of an atom by E = mc2 neutron an electrically neutral subatomic particle found in the nucleus of an atom nuclear reaction any process in the nucleus of an atom that causes the number of protons and/or neutrons to change nucleons protons or a neutrons strong nuclear force the force that binds protons and neutrons together in the nucleus of an atom transmutation the changing of one element into another by a loss of gain of one or more protons Equations and Symbols E mc 2 where 1u 1.6726 x10 27 kg 931.5 MeV A Z X ΔE = binding energy of the nucleus Δm = mass defect of the nucleus c = speed of light = 3 x 108 m/s u = atomic mass unit X = element symbol A = atomic mass number (number of protons and neutrons) Z = atomic number (number of protons) Ten Homework Problems Chapter 31 Problems 1, 2, 10, 12, 14, 17, 18, 19, 22, 51 DISCUSSION OF SELECTED SECTIONS 31.3 The Mass Defect of the Nucleus and Nuclear Binding Energy The nucleus is made up of positively charged protons and neutrons, which have no charge. The proton has exactly the same charge as an electron, but is positive. The neutron is actually made up of a proton and an electron bound together to create the neutral particle. A proton is about 1800 times more massive than an electron, which makes a neutron only very slightly more massive than a proton. We say that a proton has a mass of approximately one atomic mass unit, u. The atomic number (Z) of an element is equal to the number of protons found in an atom of that element, and fundamentally is an indication of the charge on the nucleus of that element. All atoms of a given element have 347 Chapter 31 Nuclear Physics and Radioactivity the same atomic number. In other words, the number of protons an atom has defines what kind of element it is. The total number of neutrons and protons in an atom is called the mass number (A) of that element. The symbol ZA X is used to show both the atomic number and the mass number of an X atom, where Z is the atomic number and A is the mass number. Even though the number of protons must be the same for all atoms of a particular element, the number of neutrons, and thus the mass number, can be different. Atoms of the same element with different masses are known as isotopes of one another. For example, carbon-12 is a carbon atom with 6 protons and 6 neutrons, while carbon-14 is a carbon atom with 6 protons and 8 neutrons. We would write these two isotopes of carbon as 126C and 146C . The table below summarizes the basic features of protons, neutrons and electrons. Notice that we use an H to symbolize the proton, since the proton is a hydrogen nucleus. Particle proton Symbol 1 1H neutron electron Relative mass 1 Charge +1 Location nucleus n 1 0 nucleus e or e- 0 -1 electron orbitals around the nucleus 1 0 0 1 Example 1 Find the number of protons, electrons, and neutrons in a neutral atom of iron 56 26 Fe . Solution This isotope of iron has an atomic number of 26 and a mass number of 56. Therefore, it will have 26 protons, 26 electrons, and 56 – 26 = 30 neutrons. Since positive charges repel each other, one might wonder why protons stay together in the nucleus of the atom. There is a force holding the protons together which is greater than the electrostatic repulsion between them called the strong nuclear force, and it is a result of the binding energy of the nucleus. According to Einstein’s famous equation E = mc2, mass and energy can be converted into one another. When an nucleus is assembled, each proton and neutron gives up a little of its mass to be converted into binding energy. Example 2 The nitrogen atom 147 N is composed of 7 protons and 7 neutrons, which gives a total of 14 atomic mass units (u). But if these particles are combined into a the resulting mass of the nitrogen nucleus is 14.003074 u. 348 14 7 N nitrogen nucleus, Chapter 31 Nuclear Physics and Radioactivity (a) Find the mass defect of the nitrogen nucleus. (b) What is the binding energy of the nitrogen nucleus? (c) What is the binding energy per nucleon? Solution (a) The sum of the masses of the protons and neutrons is 7(1.007 825 u) = 7.054775 u 7(1.008 665 u) = 7.060655 u 14.115430 u The mass defect is 14.115430 u - 14.003074 u 0.112356 u (b) The equivalence between mass in atomic mass units and energy in million electronvolts (MeV) is 1 u = 931 MeV. Then the binding energy of the nucleus is BE = (MD)(931 MeV/u) = (0.112356 u)(931 MeV/u) = 104.60344 MeV (c) 104.60344 MeV BE MeV 7.471674 nucleon 14 nucleons nucleon 31.4 Radioactivity At the end of the 19th century, there were elements discovered that continuously emitted mysterious rays. These elements were identified as being radioactive. A radioactive element spontaneously emits particles from its nucleus because the energy of the nucleus is unstable. Examples of naturally-occurring radioactive elements are uranium 238 92 U , radium 226 88 Ra , and carbon 146C . There are four types of particles that can be emitted when an element undergoes radioactive decay: 1. Alpha decay . Uranium, for example, undergoes alpha decay, meaning that it emits an alpha particle from its nucleus. An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. When an element emits an alpha particle, its nucleus loses 2 atomic numbers and 4 mass numbers, and thus changes into another element, called the daughter element. But what would this element be? We can write the nuclear equation for the radioactive decay of uranium as U Az X 24He 238 92 349 Chapter 31 Nuclear Physics and Radioactivity where X is the daughter element and 24 He is the alpha particle. The atomic number on the left must equal the sum of the atomic numbers on the right, since charge and mass are conserved in this process. The same is true for the mass numbers on the left and right. So, the daughter element has an atomic number Z = 92-2 = 90 and a mass number A = 238 – 4 = 234. Uranium decays into the daughter element 234 90Th , thorium. 2. Beta decay. A beta particle is the name given to an electron emitted from the nucleus of a radioactive element. But what is an electron doing in the nucleus of an atom? Remember that we discussed the neutron in the nucleus of an atom as being a proton and an electron bound together. Beta decay is really just a neutron emitting an electron and becoming a proton. Thus, the daughter element resulting from beta decay is one atomic number higher than the parent nucleus, but the mass number essentially does not change. For example, carbon 146C is a radioactive element that undergoes beta decay. The decay equation is 14 6 C ZA X 10 e We use the same symbol for a beta particle as we do for an electron. The daughter element must have an atomic number of 6 – (-1) = 7 and a mass number of 14 – 0 = 14. The daughter element is 147 N , nitrogen. 3. Gamma decay. Some radioactive elements emit a gamma ray, a very high energy electromagnetic wave which has no charge or mass, so only the energy of the nucleus changes, and neither Z nor A change. 4. Positron decay. A positron is exactly like an electron except for the fact that it is positively charged. A positron is not a proton, as their masses and other features are very different. Positron decay equations are typically not included on the AP Physics B exam. 350 Chapter 31 Nuclear Physics and Radioactivity CHAPTER 31 REVIEW QUESTIONS For each of the multiple choice questions below, choose the best answer. 1. The neutral element 24 magnesium 12 Mg has (A) 12 protons, 12 electrons, and 24 neutrons. (B) 12 protons, 12 electrons, and 12 neutrons. (C) 24 protons, 24 electrons, and 12 neutrons. (D) 24 protons, 12 electrons, and 12 neutrons. (E) 12 protons, 24 electrons, and 24 neutrons. 4. The isotope of thorium 234 90Th undergoes alpha decay according to the equation 234 A 4 90Th Z X 2 He . The element X is 2. All isotopes of uranium have (A) the same atomic number and the same mass number. (B) different atomic numbers but the same mass number. (C) different atomic numbers and different mass numbers. (D) the same atomic number but different mass numbers. (E) no electrons. (A) 238 92 (B) 230 88 Ra (C) 236 94 Pu (D) 238 88 Ra (E) 230 92 U U 60 Co undergoes 5. The isotope of cobalt 27 beta decay according to the equation 60 A 0 27 Co Z X 1 e . The element X is 60 Fe (A) 26 (B) 56 25 Mn (C) 60 28 Ni 60 29 (D) Cu (E) 3. Six protons and six neutrons are brought together to form a carbon nucleus, but the mass of the carbon nucleus is less than the sum of the masses of the individual particles that make up the nucleus. This missing mass, called the mass defect, has been (A) converted into the binding energy of the nucleus. (B) given off in a radioactive decay process. (C) converted into electrons. (D) converted into energy to hold the electrons in orbit. (E) emitted as light. 351 61 27 Co Chapter 31 Nuclear Physics and Radioactivity Free Response Question Directions: Show all work in working the following question. The question is worth 10 points, and the suggested time for answering the question is about 10 minutes. The parts within a question may not have equal weight. 1. (10 points) Particle or nucleus 1 0n Mass (u) 1.008 665 u 1 1 13 7 H 1.007 825 u N 13.005738 u 14 7 N 14.003074 u 13 7 N 01n147N energy A neutron is bound to a nitrogen nucleus, as shown in the equation above. (a) Find the mass defect and binding energy that holds a neutron to the nitrogen atom (b) The U.S. uses about 1020 J of energy per year. How many nitrogen atoms would have to give up a neutron and release its binding energy to provide the energy needed in the U.S. for a year? ANSWERS AND EXPLANATIONS TO CHAPTER 31 REVIEW QUESTIONS Multiple Choice 1. B The atomic number 12 implies both 12 protons and 12 electrons, and the mass number 24 is the sum of protons and neutrons, giving 12 neutrons. 2. D All isotopes of a particular element must have the same atomic number (number of protons), since this number identifies the element, but can have a different mass number (number of neutrons). 3. A Each particle that makes up the nucleus gives up a little mass to be converted into energy by E = mc2 to bind the nucleus together. 352 Chapter 31 Nuclear Physics and Radioactivity 4. B The atomic number Z of element X is found by 90 = Z + 2, so Z = 88, and the mass number A is found by 234 = A + 4, so A = 230. The element X is radium. 5. C The atomic number Z of element X is found by 27 = Z + (-1), so Z = 28, and the mass number A is found by 60 = A + 0, so A = 60. The element X is nickel. Free Response Question Solution (a) 5 points 13 7 N 01n147N energy (13.005738 u) + (1.008665 u) = 14.003074 u + (mass defect) Mass defect = 0.011329 u BE = (MD)(931 MeV/u) = (0.011329 u)( 931 MeV/u) = 10.547299 MeV (b) 5 points Converting MeV to J: 10.547299 MeV(1.6 x 10-13 J) = 1.69 x 10-12 J Number of nitrogen atoms = 10 20 J 5.93 x 10 31 atoms 12 1.69 x10 J / atom 353