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Revision 1.1 Sept 2016 Neutrons Student Guide GENERAL DISTRIBUTION GENERAL DISTRIBUTION: Copyright © 2016 by the National Academy for Nuclear Training. Not for sale or for commercial use. This document may be used or reproduced by Academy members and participants. Not for public distribution, delivery to, or reproduction by any third party without the prior agreement of the Academy. All other rights reserved. NOTICE: This information was prepared in connection with work sponsored by the Institute of Nuclear Power Operations (INPO). Neither INPO, INPO members, INPO participants, nor any person acting on behalf of them (a) makes any warranty or representation, expressed or implied, with respect to the accuracy, completeness, or usefulness of the information contained in this document, or that the use of any information, apparatus, method, or process disclosed in this document may not infringe on privately owned rights, or (b) assumes any liabilities with respect to the use of, or for damages resulting from the use of any information, apparatus, method, or process disclosed in this document. ii Table of Contents INTRODUCTION ..................................................................................................................... 2 TLO 1 ATOMS ...................................................................................................................... 3 Overview .......................................................................................................................... 3 ELO 1.1 Atomic Structure ............................................................................................... 4 ELO 1.2 Atomic Terms.................................................................................................... 7 ELO 1.3 Atomic Forces ................................................................................................. 10 TLO 1 Summary ............................................................................................................ 14 TLO 2 MASS DEFECT AND BINDING ENERGY ..................................................................... 15 Overview ........................................................................................................................ 15 ELO 2.1 Mass Defect and Binding Energy ................................................................... 16 ELO 2.2 Determining Mass Defect and Binding Energy .............................................. 17 ELO 2.3 Binding Energy per Nucleon ........................................................................... 22 ELO 2.4 Binding Energy Per Nucleon Curve ................................................................ 23 TLO 2 Summary ............................................................................................................ 24 TLO 3 NEUTRON INTERACTIONS WITH MATTER ................................................................. 25 Overview ........................................................................................................................ 25 ELO 3.1 Neutron Scattering Interactions....................................................................... 26 ELO 3.2 Neutron Absorption Reactions ........................................................................ 28 TLO 3 Summary ............................................................................................................ 29 TLO 4 FISSION PROCESS ..................................................................................................... 30 Overview ........................................................................................................................ 30 ELO 4.1 Excitation and Critical Energy ........................................................................ 30 ELO 4.2 Fissile and Fissionable Material ...................................................................... 31 ELO 4.3 Fission Process - Liquid Drop Model ............................................................. 34 TLO 4 Summary ............................................................................................................ 36 TLO 5 PRODUCTION OF HEAT FROM FISSION ...................................................................... 37 Overview ........................................................................................................................ 37 ELO 5.1 Energy Release Per Fission ............................................................................. 37 ELO 5.2 Fission Fragment Yield ................................................................................... 39 ELO 5.3 Fission Heat Production .................................................................................. 41 TLO 5 Summary ............................................................................................................ 42 TLO 6 PROMPT AND DELAYED NEUTRONS ......................................................................... 43 Overview ........................................................................................................................ 43 ELO 6.1 Production of Prompt and Delayed Neutrons ................................................. 43 ELO 6.2 Delayed Neutron Fraction ............................................................................... 46 ELO 6.3 Neutron Lifetimes and Generation Times ....................................................... 47 TLO 6 Summary ............................................................................................................ 49 TLO 7 NEUTRON MODERATION .......................................................................................... 50 Overview ........................................................................................................................ 50 ELO 7.1 Neutron Moderation ........................................................................................ 51 ELO 7.2 Moderator Characteristics ............................................................................... 54 TLO 7 Summary ............................................................................................................ 56 iii TLO 8 NEUTRON REACTION RATES/PROBABILITIES .......................................................... 56 Overview ....................................................................................................................... 56 ELO 8.1 Neutron Reaction Terms ................................................................................. 58 ELO 8.2 Neutron Energy Terms ................................................................................... 64 ELO 8.3 Neutron Energies versus Cross-Sections ........................................................ 66 ELO 8.4 Reaction Rate .................................................................................................. 68 ELO 8.5 Neutron Flux and Reactor Power ................................................................... 71 TLO 8 Summary ............................................................................................................ 74 KNOWLEDGE CHECK ANSWER KEY...................................................................................... 1 ELO 1.1 Atomic Structure ............................................................................................... 1 ELO 1.2 Atomic Terms ................................................................................................... 1 ELO 1.3 Atomic Forces ................................................................................................... 2 ELO 2.1 Mass Defect and Binding Energy ..................................................................... 3 ELO 2.2 Determining Mass Defect and Binding Energy ................................................ 4 ELO 2.3 Binding Energy per Nucleon ............................................................................ 4 ELO 2.4 Binding Energy Per Nucleon Curve ................................................................. 5 ELO 3.1 Neutron Scattering Interactions ........................................................................ 5 ELO 3.2 Neutron Absorption Reactions ......................................................................... 5 ELO 4.1 Excitation and Critical Energy.......................................................................... 6 ELO 4.2 Fissile and Fissionable Material ....................................................................... 6 ELO 4.3 Fission Process - Liquid Drop Model ............................................................... 7 ELO 5.1 Energy Release Per Fission .............................................................................. 7 ELO 5.2 Fission Fragment Yield..................................................................................... 8 ELO 5.3 Fission Heat Production .................................................................................... 8 ELO 6.1 Production of Prompt and Delayed Neutrons ................................................... 9 ELO 6.2 Delayed Neutron Fraction................................................................................. 9 ELO 6.3 Neutron Lifetimes and Generation Times ...................................................... 10 ELO 7.1 Neutron Moderation........................................................................................ 11 ELO 7.2 Moderator Characteristics ............................................................................... 12 ELO 8.1 Nuclear Reaction Terms ................................................................................. 13 ELO 8.2 Neutron Energy Terms ................................................................................... 13 ELO 8.3 Neutron Energies versus Cross-Sections ........................................................ 14 ELO 8.4 Reaction Rate .................................................................................................. 14 ELO 8.5 Neutron Flux and Reactor Power ................................................................... 14 iv Neutrons Revision History Revision Date 9-29-2016 Rev 0.1 Version Number Purpose for Revision Performed By 0 New Module OGF Team 0.1 Incorporated additional reviewer comments OGF Team 1 Introduction Atoms are the building blocks of matter; they are comprised of a nucleus, an orbiting field and a large volume of empty space. Atoms are the smallest components of matter that retain the identifying properties of an element. Each element is made up of atoms identified by a unique combination of subatomic particles making up their nuclei and orbiting fields. When an atom’s subatomic particle configuration is changed, the atom’s elemental identification is changed. Understanding these subatomic interactions is important to understanding the fission process that occurs in a nuclear reactor. In thermal reactors, the neutrons that cause fission are born during the fission process at a much higher energy level than required. To make fission more probable, these neutrons must be slowed down to what is known as thermal energy. To reduce high-energy neutrons to thermal energy levels a process known as moderation must take place. Pressurized water reactors (PWRs) utilize water as a moderator for thermalizing neutrons. To provide complete knowledge of fission, it is important to understand all of the various types and probabilities of neutron interactions that exist. This module will cover neutron scattering and absorption reactions, materials used for nuclear fuel, the production of heat from fission, neutron sources for shutdown and startup conditions, and the various terms used for describing and measuring neutron intensity, reaction probabilities, and reaction rates. You will then apply this background knowledge to understand the calculations for determining reactor thermal power output. Objectives At the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of 80 percent or higher on the following Terminal Learning Objectives (TLOs): 1. Describe atoms, including components, structure, and nomenclature. 2. Describe mass defect and binding energy and their relationship to one another. 3. Describe neutron interactions with matter. 4. Describe the process of nuclear fission and the types of material that can undergo fission. 5. Explain the production of heat from fission. 6. Describe the production of prompt and delayed neutrons from fission as well as the differences between them. 7. Describe the process of neutron moderation in a nuclear reactor and the characteristics of desirable moderators. 8. Explain the relationship between neutron flux, microscopic and macroscopic cross-sections, and their effect on neutron reaction rates. Rev 0.1 2 TLO 1 Atoms Overview Chemist John Dalton first proposed the modern proof for the atomic nature of matter in 1803. He theorized that unique atoms characterize each element, that the unique atoms distinguish each element from all others, and that the physical difference between different types of atoms is their weight. Subatomic Particles Because of technology limitations, it took almost 100 years to prove Dalton’s theories. Initially, chemical experiments indicated that the atom was indivisible. Later, electrical and radioactivity experimentation indicated that particles of matter smaller than the atom do exist. In 1906, Joseph John Thompson won the Nobel Prize in physics for establishing the existence of electrons. In 1920, Earnest Rutherford named the hydrogen nucleus a proton and in 1932, Sir James Chadwick confirmed the existence of the neutron. In the 1970s, the application of the standard model of particle physics proved the existence of quarks demonstrating the complexity of the atom. Many questions about the atom still exist. Experiments now in progress and in the future will provide a greater understanding of the atom. Atomic Properties Determine Nuclear Fuel Good Points Knowledge of atoms is important because their properties determine whether they would be a good nuclear fuel. Understanding the characteristics of atoms helps the student understand the nature of atomic power and the forces that control it. Objectives Upon completion of this lesson, you will be able to do the following: 1. Using Bohr's model of an atom, describe the characteristics of the following atomic particles, including mass, charge, and location within the atom: a. Proton b. Neutron c. Electron 2. Describe the nomenclature for identifying nuclides and isotopes of a given nuclide. 3. Describe the three forces that act on particles within the nucleus, and how they affect the stability of the nucleus. Rev 0.1 3 ELO 1.1 Atomic Structure Introduction Physicist Ernest Rutherford postulated that the positive charge in an atom is concentrated at the center of the atom with electrons orbiting around it. Later, Niels Bohr, combining Rutherford's theory and the quantum theory of Max Planck, proposed that orbiting electrons in discrete fixed distances from the nucleus surround the atom’s center positive charge of protons. An electron in one of these orbits, called shells, has a specific or discrete quantity of energy (quantum). Electron movement between shells results in an energy difference either emitted or absorbed in the form of a single quantum of radiant energy called a photon. Neutrons and Protons Protons and neutrons are located in a tight cluster in the center of the atom called the nucleus. Atoms comprise each element, having a unique number of protons in their nuclei. Neutrons are electrically neutral and have no electrical charge. Protons are electrically positive and exhibit an electrical charge of +1. Protons give the nucleus its positive charge. A nucleus with one proton has a +1, with two protons a charge of +2, and so on with increasing numbers of protons. Neutrons and protons are each essentially equal in mass; together they make up the mass of the nucleus. Figure: Simple Carbon-12 Atom Rev 0.1 4 Electrons Electrons are the particles that orbit the nucleus. They orbit the nucleus in concentric orbits referred to as orbitals or shells. Electrons are small and light, with a mass of only 1/1835 the mass of a proton or neutron. Each electron exhibits an electrical charge of minus one (-1) and equals in magnitude the charge of one (+1) proton. For the atom to be electrically neutral, its normal state, the number of electrons orbiting the nucleus is exactly equal to the number of protons in the nucleus. Electrons are bound to the nucleus by electrostatic attraction because opposite electrical charges attract. The atom remains neutral unless some external force causes a change in the number of electrons. Bohr's Model The figure below shows Bohr’s model using a hydrogen atom. The figure shows an electron that has dropped from the third shell to the first shell releasing energy in the process. The energy is released as a photon emission equal to h (h = Planck's constant - 6.63 x 10-34 J-s (Jouleseconds) and = frequency of the photon). Bohr's theory accounts for the quantum energy levels as measured in the laboratory. Although Bohr's atomic model specifically explains the hydrogen atom, it applies as first generation model to all atoms. Figure: Bohr's Model of the Hydrogen Atom Rev 0.1 5 Atomic Measuring Units Atoms are so small that normal measuring units are difficult to apply. Mass and energy use universally accepted units of measure on the atomic scale to standardize measurement units and calculations. It is possible to convert values expressed in these atomic scale units to non-atomic scale units if desired. Atomic Mass Unit (amu): The unit of measure for mass is the amu. One amu equals 1.66 x 10-24 grams. 6.02 x 1023 amu = 1 gram. Neutrons and protons each have a mass close to one (1) amu. Electron Volt (eV): The unit for energy is the electron volt (eV) or Megaelectron Volt (MeV). The electron volt is the amount of energy gained (or lost) by a single electron moved across a potential difference of one volt. One electron volt is equals 1.602 x 10-19 Joules (J) or 1.18 x 10-19 footpounds (lbf). A proton’s eV value is positive one (+ 1) and an electron’s eV is negative one (-1). The table below shows properties of the three subatomic particles: Properties of Three Subatomic Particles Particle Location Charge Mass (Rest) Neutron Nucleus None 1.008665 amu Proton Nucleus +1 eV 1.007277 amu Electron Shells Around Nucleus -1 eV 0.0005486 amu Knowledge Check (Answer Key) Identify the particles included in the make-up of an atom. (More than one answer may apply.) Rev 0.1 A. neutron B. electron C. gamma D. amu 6 ELO 1.2 Atomic Terms Introduction Atoms have characteristics describing their behavior. This section defines and outlines those terms. Nuclear Nomenclature The following terms describe characteristics of atoms: Atomic number Neutron number Mass number Nuclide Isotope Atomic Number The atomic number describes the number of protons in the atom’s nucleus and identifies the element. It is the Z number in the Chart of the Nuclides (atomic notation). The number of protons identifies the particular element; therefore, the atomic number identifies a particular element. For example, any atom having two protons in its nucleus has an atomic number of two (2) and is identified as the element helium. Because the number of electrons in an atom matches the number of protons (electrically neutral), the atomic number equals the number of electrons in the atom. Neutron Number The symbol N denotes the number of neutrons in a nucleus, which is the neutron number. Atomic notation does not include N; however, N is determined by subtracting the atomic number (Z) from the atomic mass number (A). Mass Number The mass number of the atom is equal to the total number of protons and neutrons in the nucleus. In atomic notation, the mass number is the A number. We calculate A as follows: 𝐴=𝑍+𝑁 Where: • Z = Atomic number • N = Neutron number Rev 0.1 7 Nuclides Nuclides are atoms that contain a unique combination of protons and neutrons. Not all proton and neutron combinations can exist in nature or as man-made or fabricated combinations. However, scientists have identified about 2,500 specific nuclides. The figure below shows the atomic notation of a nuclide with the chemical symbol (X letter) of the element, the atomic number written as a subscript, and the mass number written as a superscript. Isotope Atoms of the same element always contain the same number of protons (Z), but not always the same number of neutrons (N). This results in some atoms of an element with different atomic mass numbers (A). These atoms are isotopes. Isotopes of a particular element have different atomic mass numbers (A); however, they have the same chemical characteristics with different numbers of neutrons. This affects their radioactivity stability. Most elements have a few stable isotopes and several unstable radioactive isotopes. For example, oxygen has three stable isotopes found in nature (oxygen-16, oxygen-17, and oxygen-18) and eight unstable radioactive isotopes. Another example is hydrogen, which has two stable isotopes (hydrogen-1 and hydrogen-2) and a single radioactive isotope (hydrogen-3). Atomic Notation A convention, known as atomic or standard notation, identifies elements using the nuclear nomenclature previously described. The figure below shows standard notation: Figure: Nomenclature for Identifying Nuclides Rev 0.1 8 Identifying a Nuclide Each element has a unique chemical name, symbol, and atomic number. Any one of the three identifies the element. The chemical name or symbol followed by the mass number (for example, U-235 or uranium-235) identifies an element. Another frequently used identification method is the chemical symbol with a left superscript (for example, 235U). The nuclide concept (referring to individual nuclear species) emphasizes nuclear properties over chemical properties, while the isotope concept (grouping all atoms of each element) emphasizes chemical over nuclear. The neutron number has large effects on nuclear properties, but its effect on chemical properties is negligible for most elements. Knowledge Check (Answer Key) What is the element and number of neutrons for the following: 235 92𝑈 A. Uranium; 143 B. Uranium; 92 C. Plutonium; 143 D. Plutonium; 92 Knowledge Check (Answer Key) In the table below, complete the columns for element, protons, electrons, and neutrons. Nuclide Element Protons Electrons Neutrons 1 1𝐻 10 5𝐵 16 8𝑂 235 92𝑈 239 94𝑃𝑢 Rev 0.1 9 ELO 1.3 Atomic Forces Introduction Electrical forces in the nucleus determine the way the atomic forces behave and their respective electrical charge. Atomic Forces Acting in the Nucleus The nucleus consists of positively charged protons and electrically neutral neutrons in the Bohr model of the atom. The protons and neutrons are termed nucleons. Electrostatic and Gravitational Forces Two forces present in the nucleus are: (1) electrostatic forces between charged particles, and (2) gravitational forces between any two objects that have mass. It is For More possible to calculate the magnitude of the gravitational Information force and electrostatic force based on principles from classical physics. Gravitational Force Newton’s law of universal gravitation states that gravitational force between two bodies is directly proportional to the masses of the two bodies and inversely proportional to the square of the distance between the bodies. The equation below shows this relationship: 𝐹𝑔 = 𝐺𝑚2 𝑚2 𝑟2 Where: • Fg = gravitational force (Newtons) • m1 = mass of first body (kilograms) • m2 = mass of second body (kilograms) • G = gravitational constant (6.67 x 10-11 N-m2/kg2) • r = distance between particles (meters) Within the nucleus the nucleon mass is small, but the distance is extremely short. Calculating the gravitational force for two protons separated by a distance of 10-13 meters is about 10-17 Newtons. Rev 0.1 10 Electrostatic Force We use Coulomb's Law to calculate the electrostatic force between two protons. The electrostatic force is directly proportional to the electrical charges of the two particles and inversely proportional to the square of the distance between the particles. The equation below shows the Coulomb's Law relationship: 𝐹𝑒 = 𝐾𝑄1 𝑄2 𝑟2 Where: • Fe = electrostatic force (Newtons) • K = electrostatic constant (9.0 x 109 N-m2/C2 [Coulombs squared]) • Q1 = charge of first particle (Coulombs [C]) • Q2 = charge of second particle (Coulombs) • r = distance between particles (meters [m]) Using this equation, the electrostatic force between two protons separated by a distance of 10-13 meters is about 105 Newtons. Because the electrostatic force (105 Newtons) is much greater than the gravitational force (10-17 Newtons), the gravitational force can be neglected. Nuclear Force Without another explanation, it is impossible to have a stable nuclei composed of protons and neutrons if only the electrostatic and gravitational forces existed in the nucleus. The gravitational forces are much too weak to hold the nucleons together. Another force, called the nuclear force, is a strong attractive force independent of charge. It acts equally between pairs of neutrons, pairs of protons, or a neutron and a proton. Nuclear force acts over a short range limited to distances approximately equal to the diameter of the nucleus (10-13 meters). The attractive nuclear force between nucleons decreases with distance much quicker than the repulsive electrostatic force between protons. Rev 0.1 11 Atomic Forces Force Interaction Range 1. Gravitational Weak attractive force between all nucleons Relatively long 2. Electrostatic Strong repulsive force between like charged particles (protons) Relatively long 3. Nuclear Force Strong attractive force between all nucleons Extremely short Attractive and repulsive forces in the nucleus balance in stable atoms. If unbalanced, the nucleus emits radiation in an attempt to achieve a stable configuration. This phenomenon is discussed later in this module. Knowledge Check (Answer Key) Very weak attractive force between all nucleons describes which of the forces listed below? A. Electrostatic B. Nuclear C. Gravitational D. Atomic Neutron Proton Ratio The neutron-proton ratio (N/Z ratio or nuclear ratio) is the ratio of the number of neutrons to protons that make up the nucleus. Light elements up to calcium (Z = 20) have stable isotopes with a neutron-proton ratio of one, except for beryllium and every element with odd proton numbers from fluorine (Z = 9) to potassium (Z = 19). Helium-3 is the only stable isotope with a neutron-proton ratio under one. Uranium-238 has a high N/Z ratio of any natural isotope at 1.59; lead-208 has the highest N/Z ratio of any known stable isotope at 1.54. Rev 0.1 12 The figure below shows the distribution of the stable nuclides plotted on the same axes as the Chart of the Nuclides. The ratio of neutrons to protons in the nucleus becomes larger as the mass numbers become higher. For helium-4 (2 protons and 2 neutrons) and oxygen-16 (8 protons and 8 neutrons), this ratio is unity. For indium-115 (49 protons and 66 neutrons), the ratio of neutrons to protons has increased to 1.35, and for uranium-238 (92 protons and 146 neutrons) the neutron to proton ratio is 1.59. Figure: Neutron-Proton Plot of the Stable Nuclides A nuclide existing outside of the band of stability can undergo alpha decay, positron emission, electron capture, or beta emission to gain stability. For example, following fission, the two resulting fragments have nuclei with approximately the same high neutron-to-proton ratio as the original heavy nucleus. This high neutron-to-proton ratio with lower proton numbers places the fragments below and to the right of the stability line. Successive beta emissions, each converting a neutron to a proton, create a more stable neutron-to-proton ratio. Knowledge Check (Answer Key) Which of the following nuclides has the higher neutronproton ratio? Rev 0.1 A. Cobalt-60 B. Selenium-79 C. Silver-108 D. Cesium-137 13 TLO 1 Summary Atoms consist of three basic subatomic particles: — Protons: particles that have a positive charge and exist in the nucleus. A proton has a mass of 1 amu. — Neutrons: particles that have no electrical charge also exist in the nucleus. A neutron has approximately the same mass as a proton, about 1 amu. — Electrons: particles that have a negative charge, orbit in shells around the nucleus and have a mass about 1/1,800 the mass of a proton. Bohr model of an atom: a dense nucleus of protons and neutrons surrounded by orbiting electrons traveling in discrete orbits at fixed distances. Nuclides are atoms containing certain numbers of protons and neutrons. Isotopes are nuclides having same atomic number but with differing numbers of neutrons. Isotopes have the same chemical properties. Atomic number of an atom is the number of protons in the nucleus. Mass number of an atom is the total number of nucleons (protons and neutrons) in the nucleus. Atomic (standard) notation identifies a specific nuclide, shown below in the graphic: — Z represents the atomic number, which equals the number of protons — A represents the mass number, which equals the number of nucleons — X represents the chemical symbol of the element — Number of protons = Z — Number of neutrons = A - Z The different forces interacting within the nucleus determine the nucleus’ stability. — Gravitational force: a long-range, relatively weak attraction between masses, and negligible compared to other forces. — Electrostatic force: a relatively long-range, strong, and repulsive force that acts between positively charged protons. — Nuclear force: a short-range, attractive force between all nucleons that is able to balance the repulsive electrostatic force in a stable nucleus. Rev 0.1 14 A high neutron-to-proton ratio places nuclides below and to the right of the stability curve. Instability caused by excess neutrons is often fixed by successive beta emissions, with each converting a neutron to a proton. Atom percent (a/o) is the percentage of the atoms of an element that are of a particular isotope. Summary Now that you have completed this lesson, you should be able to do the following: 1. Using Bohr's model of an atom, describe the characteristics of the following atomic particles, including mass, charge, and location within the atom: a. Proton b. Neutron c. Electron 2. Describe the nomenclature for identifying nuclides and isotopes of a given nuclide. 3. Describe the three forces that act on particles within the nucleus and how they affect the stability of the nucleus. TLO 2 Mass Defect and Binding Energy Overview Binding energy and mass defect describe the energy associated with nuclear reactions. Understanding mass defect and binding energy and their relationship is important for understanding energies associated with atomic reactions, including fission. Objectives Upon completion of this lesson, you will be able to do the following: 1. Define the relationship between mass defect and binding energy. 2. Given the atomic mass for a nuclide and the atomic masses of a neutron, proton, and electron, calculate the mass defect and binding energy of the nuclide. 3. Explain binding energy per nucleon. 4. Explain the shape of the binding energy per nucleon versus mass number curve including its significance to fission energy. Rev 0.1 15 ELO 2.1 Mass Defect and Binding Energy Introduction Although the laws of conservation of mass and conservation of energy hold true, conversion between mass and energy occurs on a nuclear level. Instead of two separate conservation laws, a single conservation law states that the sum of mass and energy is conserved. In Einstein’s theory of relativity, energy and mass are equivalent. In fact, the rest energy and the mass are related by E = mc2, where c is the speed of light in a vacuum. Therefore, a change in the rest energy of the system is equivalent to a change in the mass of the system. The binding energy used to disassemble the nucleus appears as extra mass of the separated and stationary nucleons. In other words, the sum of the individual masses of the separated protons and neutrons is greater by some amount than the mass of the stable nucleus. Mass Defect and Binding Energy Mass defect: experimental measurements show that the mass of a particular atom is always slightly less than the sum of the atom’s individual neutrons, protons, and electron masses. This difference is the mass defect (∆m). Binding energy: a change of mass occurs from the conversion of mass to binding energy (BE) during formation of a nucleus. Binding energy is the amount of energy supplied to a nucleus to separate its nuclear particles completely. Conversely, it is the amount of energy released if separate particles formed the nucleus. Knowledge Check (Answer Key) _______________ is the amount of energy that must be supplied to a nucleus to completely separate its nuclear particles. Rev 0.1 A. Nuclear energy B. Binding energy C. Mass defect D. Separation energy 16 ELO 2.2 Determining Mass Defect and Binding Energy Introduction Binding energy is the expected reduction of potential energy in an attractive force field. When the nucleus is divided into separated nucleons, energy is required. The separated nucleons have a greater mass than the original nucleus. In atomic physics, mass defect is the difference in mass between the atom and the sum of the masses of that atom’s respective protons, neutrons, and electrons. Calculating Mass Defect The mass defect can be calculated using the below equation. It is important to use the full accuracy of mass measurements in calculating the mass defect because the difference in mass is small compared to the mass of the atom. Rounding off the masses of atoms and particles to three or four significant digits prior to the calculation results in a calculated mass defect of zero (0). NOTE: The mass of the Hydrogen atom is the same as the mass of the proton and electron used below in the equation. ∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚 Where: • Δm = mass defect (amu) • mp = mass of a proton (1.007277 amu) • mn = mass of a neutron (1.008665 amu) • me = mass of an electron (0.000548597 amu) • matom = mass of nuclide 𝐴𝑍𝑋 (amu) • Z = atomic number (number of protons) • A = mass number (number of nucleons) Rev 0.1 17 Steps for using the formula for mass defect: ∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚 Step Description Action 1. Determine the Z (atomic number) and A (atomic mass) of the nuclide. Look up information in the Chart of the Nuclides. 2. Determine the mass of the protons and electrons of the nuclide. Multiply Z times the mass of a proton and the mass of an electron: 𝑍(mp + me ). Determine the mass of the neutrons. Subtract the atomic number (Z) from the atomic mass (A) then multiply by mass of a neutron: (𝐴 − 𝑍)mn . 4. Add the mass of the protons, electrons, and neutrons. Add the products determined in the previous two steps: 𝑍(mp + me ) + (𝐴 − 𝑍)mn . 5. Determine the difference between the atomic mass of the nuclide and the mass determined above. Subtract the mass of the atom of the nuclide: [𝑍( mp + me ) + (𝐴 − 𝑍)mn ] − matom . 3. Calculating Binding Energy Binding energy is the energy equivalent to the mass defect. Calculate binding energy using a conversion factor derived from Einstein's Theory of Relativity. When the nucleus forms from its separate particles, mass defect converts to binding energy. Einstein's famous equation relating mass and energy is 𝐸 = 𝑚𝑐 2 , where c is the velocity of light (c = 2.998 x 108 meters per second [m/sec]). The energy equivalent of 1 amu is calculated by inserting this quantity of mass into Einstein's equation and applying conversion factors. Rev 0.1 18 𝐸 = 𝑚𝑐 2 1.6606 × 10−27 𝑘𝑔 𝑚 1𝑁 1𝐽 = 1 𝑎𝑚𝑢 ( ) (2.998 × 108 )( )( ) 𝑘𝑔– 𝑚 1𝑁 − 𝑚 1 𝑎𝑚𝑢 𝑠𝑒𝑐 1 𝑠𝑒𝑐 2 1 𝑀𝑒𝑉 = 1.4924 × 10−10 𝐽 ( ) 1.6022 × 10−13 𝐽 = 931.5 𝑀𝑒𝑉 Conversion Factors: 1 𝑎𝑚𝑢 = 1.6606 × 10−27 𝑘𝑔 1 N𝑒𝑤𝑡𝑜𝑛 = 1 𝑘𝑔– 𝑚 𝑠𝑒𝑐 2 1 𝐽𝑜𝑢𝑙𝑒 = 1 𝑁𝑒𝑤𝑡𝑜𝑛– 𝑚𝑒𝑡𝑒𝑟 1 𝑀𝑒𝑉 = 1.6022 × 10−13 𝐽𝑜𝑢𝑙𝑒𝑠 Since 1 amu is equivalent to 931.5 MeV of energy, the binding energy can be calculated from the following: 𝐵. 𝐸. = ∆𝑚 ( 931.5 𝑀𝑒𝑉 ) 1 𝑎𝑚𝑢 Steps to using the formula for binding energy: 𝐵. 𝐸. = ∆𝑚 ( 931.5 𝑀𝑒𝑉 ) 1 𝑎𝑚𝑢 Description Action 1. Determine the mass defect of the nuclide. Use the equation: ∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚 2. Use the binding energy equation to calculate the binding energy. Use the equation: 931.5 𝑀𝑒𝑉 𝐵. 𝐸. = ∆𝑚 ( ) 1 𝑎𝑚𝑢 Calculate the binding energy. Multiply the change in mass from Step 1 by the energy conversion for an amu. 3. Rev 0.1 19 Calculating Mass Defect Example Calculate the mass defect for lithium-7 given the mass of lithium-7 = 7.016003 amu. ∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚 Step Description Action 1. Determine the Z (atomic number) and A (atomic mass number) of the nuclide. Z = 3, A = 7 2. Determine the mass of the protons and electrons of the nuclide. 3 (1.007826 amu + 0.000548597 amu) = 3.02347979 amu 3. Determine the mass of the neutrons. (7-3) (1.008665) = 4.03466 amu 4. Add the mass of the protons, electrons and neutrons. 3.02347979 amu + 4.03466 amu = 7.058140 amu 5. Determine the difference between the atomic mass of the nuclide and the mass determined above. 7.058140 amu - 7.016003 amu = 0.042137 amu Rev 0.1 20 Example: Calculating Binding Energy of Lithium Calculate the binding energy for lithium-7: Step Description Action 1. Determine the mass defect of the nuclide. From above calculation: 0.042137 amu. 2. Use the binding energy equation to calculate the binding energy. 3. Calculate the binding energy 931.5 𝑀𝑒𝑉 𝐵. 𝐸. = ∆𝑚 ( ) 1 𝑎𝑚𝑢 BE = 0.042137 amu ( 931.5 MeV ) 1 amu = 39.2506 MeV Knowledge Check (Answer Key) Calculate the mass defect for uranium-235. One uranium-235 atom has a mass of 235.043924 amu. mp = mass of a proton (1.007277 amu) mn = mass of a neutron (1.008665 amu) me = mass of an electron (0.000548597 amu) Rev 0.1 A. 1.86471 amu B. 1.91517 amu C. 0.191517 amu D. 0.186471 amu 21 Knowledge Check (Answer Key) Calculate the binding energy for uranium-235. One uranium-235 atom has a mass defect of 1.9157 amu. A. 1784 MeV B. 178.4 MeV C. 1783 MeV D. 178.3MeV ELO 2.3 Binding Energy per Nucleon Introduction Binding energy per nucleon (BE/A) is a measure of the stability of a nucleus. The higher the binding energy per nucleon, the more stable the nucleus. The lower the binding energy per nucleon the less stable the nucleus. The next two sections go into more detail about BE/A. Binding Energy per Nucleon Binding energy per nucleon (BE/A) is equal to the average energy required to remove a single nucleon from a specific nucleus. It is determined by dividing the total binding energy of a nuclide by the total number of nucleons in its nucleus. The example below illustrates this calculation. Example: Given that the total binding energy (BE) for a U-238 nucleus is 1,804.3 MeV, calculate the binding energy per nucleon (BE/A) for U-238. Solution: 𝐵𝐸 1,804.3 𝑀𝑒𝑉 = = 7.6 𝑀𝑒𝑉 𝐴 238 Knowledge Check (Answer Key) True or False: Binding energy per nucleon is independent of the specific nuclide. Rev 0.1 A. True B. False 22 ELO 2.4 Binding Energy Per Nucleon Curve Introduction This lesson describes the relationship of binding energy per nucleon as the mass number changes. As the number of nucleons in a nucleus increases, the total binding energy also increases. Many scientific experiments have determined that the rate of increase is not linear. This non-linear relationship results in a variation of the binding energy per nucleon for nuclides of different mass numbers. Binding Energy per Nucleon Curve Example Plotting the average BE/A against the atomic mass numbers (A) shows the variation in the binding energy per nucleon, as shown in the figure below. Figure: Binding Energy per Nucleon versus Mass Number The figure above shows that as the atomic mass number (A) increases, BE/A increases, until A reaches about 60, and BE/A decreases as A increases above 60. The BE/A curve reaches a maximum value of 8.79 MeV at A = 56 and decreases to about 7.6 MeV for A = 238. No stable nuclei exist with A greater than 209. The general properties of nuclear forces determine the general shape of the BE/A curve. Very short-range attractive forces existing between nucleons hold the nucleus together and long range repulsive electrostatic (coulomb) forces existing between positively charged protons in the nucleus are forcing nucleons apart. Rev 0.1 23 As the atomic number (number of protons) and the atomic mass number of a nucleus increase, the repulsive electrostatic forces within the nucleus increase due to the greater number of protons. Increasing the proportion of neutrons in the nucleus overcomes this increased repulsion. The slope of the nuclide stability curve illustrates the effect of increasing repulsive forces. The increase in the neutron-to-proton ratio only partially compensates for the increasing proton-proton repulsive force in heavier naturally occurring elements. Less energy is required, on average, to remove a nucleon from the nucleus because repulsive forces are increasing. Therefore, the BE/A decreases. In the case of fissile materials (heaviest nuclei), only a small distortion from a spherical shape (a small energy addition) is required for the relatively large electrostatic forces attempting to force the two halves of the nucleus apart to overcome the attractive nuclear forces holding the two halves together. Consequently, the heaviest nuclei are more easily fissionable than lighter nuclei. The BE/A of a nucleus is also an indication of its stability. Generally, the higher the BE/A, the more stable the nuclide is. The increase in the BE/A as the atomic mass number decreases from 260 to 60 is the main reason for energy liberation in the fission process as the fuel isotope is split into two fragments of lower A values. The fission arrow in the figure above denotes this area. A later section will discuss delta binding energy. However, the increase in the BE/A as the atomic mass number increases from 1 to 60 is the reason that energy release is possible in a fusion event that combines rather than splits atoms. The fusion arrow in the figure above denotes this area. Knowledge Check (Answer Key) True or False. Generally, less stable nuclides have a higher BE/A than the more stable ones. A. True B. False TLO 2 Summary Binding energy and mass defect describe the energy associated with nuclear reactions. The separated nucleons have a greater mass than the original nucleus. Mass defect is the difference in mass between the atom and the sum of the masses of that atom’s respective protons, neutrons, and electrons. Rev 0.1 24 The binding energy used to disassemble the nucleus appears as extra mass of the separated and stationary nucleons. In other words, the sum of the individual masses of the separated protons and neutrons is greater by some amount than the mass of the stable nucleus. Binding energy per nucleon (BE/A) is equal to the average energy required to remove a single nucleon from a specific nucleus. It is determined by dividing the total binding energy of a nuclide by the total number of nucleons in its nucleus. BE/A can be graphed resulting in a curve of binding energy per nucleon that increases quickly through the light nuclides and reaches a maximum at a mass number of about 56. The curve decreases slowly for mass numbers greater than 60. The BE/A of a nucleus is also an indication of its stability. Generally, the higher the BE/A, the more stable the nuclide is. The increase in the BE/A as the atomic mass number decreases from 260 to 60 is the main reason for energy liberation in the fission process as the fuel isotope is split into two fragments of lower A values. TLO 3 Neutron Interactions with Matter Overview Neutrons can cause many different types of interactions. The neutron may simply scatter off the nucleus, which can result in a reduction of the neutron's energy, or be absorbed within the nucleus and removed from the system. When neutrons are absorbed into the nucleus, the nucleus becomes excited. This results in emission of gamma rays and/or particles, or it can result in fission. Objectives Upon completion of this lesson, you will be able to do the following: 1. Describe the following neutron scattering interactions, including conservation principles: a. Elastic scattering b. Inelastic scattering 2. Describe the following reactions where a neutron is absorbed in a nucleus: a. Radiative capture b. Particle ejection c. Fission Rev 0.1 25 ELO 3.1 Neutron Scattering Interactions Introduction A neutron scattering reaction occurs when a neutron strikes a nucleus and emits a single neutron. The net effect of the reaction is that the free neutron has merely bounced off, or scattered as it interacts with the nucleus. In some cases, the initial and final neutrons are not the same. There are two categories of scattering reactions, elastic, and inelastic scattering. These collisions are of great importance for the process of making thermal or low energy neutrons available for a thermal reactor. Neutron Scattering Interactions Elastic Scattering Elastic scattering, also referred to as the billiard ball effect, is most probable with light nuclides. In an elastic scattering reaction, the neutron does not contribute to nuclear excitation of the target nucleus because of a conservation of momentum. Energy completely transfers from the neutron to the target nucleus, conserving the kinetic energy of the system. The target nucleus gains an amount of kinetic energy equal to the kinetic energy the neutron loses. This kinetic energy transfer is dependent on target size and angle. The figure below illustrates an elastic scattering reaction. Figure: Elastic Scattering The following is a mathematical illustration of the conservation of momentum and kinetic energy during an elastic scattering reaction: Conservation of momentum (𝑚𝑣) (𝑚𝑛 𝑣𝑛,𝑖 ) + (𝑚 𝑇 𝑣𝑇,𝑖 ) = (𝑚𝑛 𝑣𝑛,𝑓 ) + (𝑚 𝑇 𝑣𝑇,𝑓 ) 1 Conservation of kinetic energy (2 𝑚𝑣 2 ) 1 1 1 1 2 2 2 2 ( 𝑚𝑛 𝑣𝑛,𝑖 ) + ( 𝑚 𝑇 𝑣𝑇,𝑖 ) = ( 𝑚𝑛 𝑣𝑛,𝑓 ) + ( 𝑚 𝑇 𝑣𝑇,𝑓 ) 2 2 2 2 Rev 0.1 26 Where: mn = mass of the neutron mT = mass of the target nucleus vn,i = initial neutron velocity vn,f = final neutron velocity vT,i = initial target velocity vT,f = final target velocity Inelastic Scattering In inelastic scattering, the target nucleus absorbs the incident neutron by forming a compound nucleus and transfers the kinetic energy into nuclear excitation. The compound nucleus emits a neutron of lower kinetic energy than the incident neutron, leaving the nucleus in an excited state. A gamma emission occurs if the excitation state is not at a preferred level, dropping excess energy to reach ground state. Inelastic scattering is most probable with heavy nuclides and neutrons above 1 MeV in energy. The figure below illustrates inelastic scattering. Figure: Inelastic Scattering Some amount of kinetic energy transfers into excitation energy of the target nucleus. The total kinetic energy of the outgoing neutron and nucleus is less than the kinetic energy of the incoming neutron. Therefore, there is no conservation of kinetic energy; however, momentum is conserved. Rev 0.1 27 Knowledge Check (Answer Key) True/False: The difference between elastic and inelastic scattering where neutrons are concerned is elastic scattering involves no energy being transferred into excitation energy of the target nucleus. Inelastic scattering involves a transfer of kinetic energy into excitation energy of the target nucleus. A. True B. False ELO 3.2 Neutron Absorption Reactions Introduction Most absorption reactions result in the loss of a neutron, the production of charged particles, gamma ray(s), or fission fragments and the release of neutrons and energy. Three types of absorption reaction are: Radiative capture Particle ejection Fission Radiative Capture In radiative capture, the incident neutron interacts with the target nucleus forming a compound nucleus. The compound nucleus, with the additional neutron, then decays to its ground state via gamma emission. The equation below shows an example of radiative capture with uranium-238. 1 238 239 ∗ 239 0 𝑛+ 𝑈→( 𝑈) → 𝑈+ 𝛾 0 92 92 92 0 Particle Ejection In particle ejection, an incident neutron interacts with the target nucleus, forming a compound nucleus. The new compound nucleus excites to a high enough energy level for it to eject a new particle with the incident neutron remaining in the nucleus. The nucleus may or may not exist in an excited state depending upon the mass-energy balance of the reaction after ejection of the new particle. The equation below shows a common example of this, specifically the boron-10 neutron reaction resulting in a lithium nucleus and alpha particle. 1 10 11 ∗ 7 4 𝑛 + 𝐵 → ( 𝐵) → 𝐿𝑖 + 𝛼 0 5 5 3 2 Rev 0.1 28 Fission Fission is a very important absorption reaction. An incident neutron interacts with the target nucleus, the nucleus absorbs the incident neutron, and the nucleus splits into two smaller nuclei, called fission fragments. This reaction releases multiple neutrons and a considerable amount of energy in addition to the fission fragments. A fission reaction typically produces two fission fragments, 2 to 3 neutrons, and energy (in the form of kinetic energy and gamma rays). The following shows the fission of a uranium-235 atom. 1 236 ∗ 140 93 1 235 𝑛+ 𝑈→( 𝑈) → 𝐶𝑠 + 𝑅𝑏 + 3 ( 𝑛) 0 92 55 37 0 92 Knowledge Check (Answer Key) What type of neutron interaction has occurred when a nucleus absorbs a neutron and ejects proton? A. Fission B. Fusion C. Particle ejection D. Radiative capture TLO 3 Summary Neutron Interactions Interactions where a neutron scatters off a target nucleus are either elastic or inelastic. Elastic scattering - there is conservation of both kinetic energy and momentum, and no energy is transferred into excitation energy of the target nucleus. This is most probable with light nuclides. Inelastic scattering - some amount of kinetic energy transfers into excitation energy of the target nucleus. The total kinetic energy of the outgoing neutron and nucleus is less than the kinetic energy of the incoming neutron. Therefore, there is no conservation of kinetic energy; however, there is conservation of momentum. Radiative capture is the absorption of a neutron by the target nucleus, resulting in an excited nucleus that subsequently (typically within a small fraction of a second) releases its excitation energy in the form of a gamma ray. Particle ejection occurs when a target nucleus, absorbs a neutron, resulting in an excited compound nucleus. The compound nucleus immediately ejects a particle (for example, alpha, or proton). Rev 0.1 29 Fission - an incident neutron adds sufficient energy to the target nucleus that the target nucleus splits apart, releasing two fission fragments, several neutrons, and energy. Now that you have completed this lesson, you should be able to: 1. Describe the following neutron scattering interactions, including conservation principles: a. Elastic scattering b. Inelastic scattering 3. Describe the following reactions where a neutron is absorbed in a nucleus: a. Radiative capture b. Particle ejection c. Fission TLO 4 Fission Process Overview Nuclear fission is the splitting apart of a heavy nuclide into two fission products with the release of energy and additional neutrons. The release of neutrons causes additional fissions to occur, causing a self-sustaining fission rate capable of producing sufficient heat for power production. Objectives Upon completion of this lesson, you will be able to do the following: 1. Define the following terms: a. Excitation energy (Eexc) b. Critical energy (Ecrit) 2. Define the following terms: a. Fissile material b. Fissionable material 3. Explain the fission process using the liquid drop model of a nucleus. ELO 4.1 Excitation and Critical Energy Introduction Excitation energy is the energy above ground state of a nucleus. Critical energy is the required excitation energy for fission to occur. Rev 0.1 30 Excitation Energy Excitation energy (Eexc) is the measure of how far the energy level of a nucleus is above its ground state. The last topic discussed the many neutron reactions that can cause an increase in the excitation energy of a nucleus. Critical Energy The excitation energy must be above a specific minimum value for that nuclide for fission to occur. The critical energy (Ecrit) is the minimum excitation energy required for fission to occur for a particular nuclide. The reaction excites the target nucleus by an amount equal to binding energy of the neutron plus the neutron's kinetic energy when an incident neutron strikes a target nucleus. If the binding energy is less than the required critical energy for the nucleus, additional energy is required to cause the nucleus to fission. This energy could be in the form of kinetic energy from the incident neutron. For this reason, neutrons of low kinetic energy cannot cause fission with some types of isotopes used in nuclear reactor fuels. The discussion of binding energy was covered in TLO 2. Knowledge Check (Answer Key) True or False: Excitation must be at least equal to critical energy for fission to occur. A. True B. False ELO 4.2 Fissile and Fissionable Material Introduction There are two categories of nuclear fuel materials: those that can fission with thermal neutrons and those that require neutrons of higher energies for fission to be possible. It is possible to convert some non-fissionable materials into materials capable of fission. This section will explain the following terms: Fissile material Fissionable material Rev 0.1 31 Fissile Material A fissile material consists of nuclides that will fission with incident neutrons of any energy level. They are the desired nuclides for use in thermal nuclear reactors. The advantage of fissile materials is that they can fission with neutrons possessing zero kinetic energy (thermal neutrons). Thermal neutrons have very low kinetic energy levels because they are in approximate equilibrium with the thermal energy of surrounding materials. They add essentially no kinetic energy to the reaction, however add enough BE to reach the critical energy, Ecrit. Fissile materials will fission after absorbing a thermal neutron. It is possible for fissile materials to fission with thermal neutrons because the change in binding energy supplied by the neutron addition alone is sufficient to exceed the critical energy. Examples of fissile materials are uranium-235, uranium-233, and plutonium-239. The table below lists the critical energy (Ecrit) and the binding energy change for an added neutron (BEn) to target nuclei of interest. The change in binding energy plus the kinetic energy must equal or exceed the critical energy (ΔBE + KE ≥ Ecrit) for fission to be possible. Target Nucleus Critical Energy Ecrit Binding Energy of Last Neutron BEn Required Kinetic Energy 233 92U 6.0 MeV 7.0 MeV 0 MeV 235 92U 6.5 MeV 6.8 MeV 0 MeV 239 94Pu 6.0 MeV 6.4 MeV 0 MeV Fissionable Material A fissionable material is composed of nuclides for which fission is possible. All fissile nuclides fall into this category as well as those nuclides that fission only from high-energy neutrons. The change in binding energy that occurs from neutron absorption causes an excitation energy level insufficient to reach critical energy for fission in fissionable materials. Therefore, the incident neutron must supply additional excitation energy by adding kinetic energy to the reaction. Experimental observation shows the reason for this difference between fissile and fissionable materials. A nucleus with an even number of neutrons or protons is more stable than a nucleus with an odd number. Adding a neutron to an odd numbered nucleus (changing it to an even numbered nucleus) produces higher binding energy than adding a neutron to an even numbered nucleus. Rev 0.1 32 The table below lists the critical energy (Ecrit) and the binding energy change for an added neutron (BEn) to target nuclei of interest. The change in binding energy plus the kinetic energy must equal or exceed the critical energy (ΔBE + KE ≥ Ecrit) for fission to be possible. Target Nucleus Critical Energy Ecrit Binding Energy of Last Neutron BEn Required Kinetic Energy 238 92U 7.0 MeV 5.2 MeV 1.8 MeV 240 94Pu 6.2 Mev 5.2 MeV 1.0 MeV 242 94Pu 6.2 MeV 6.5 MeV 0.3 MeV As seen in the tables above, uranium-235 can fission with thermal neutrons because its BEn is greater than its critical energy. This makes U-235 a fissile material. Uranium-238, on the other hand, has a critical energy of 7.0 MeV, which is greater than the 5.2 MeV BEn added by the neutron. Therefore, a fission neutron must have at least 1.8 MeV of kinetic energy to cause fission of U238. This makes uranium-238 a fissionable material. Uranium-238 is sometimes referred to as a “fertile” isotope. Fertile material is a material that, although not itself fissionable by thermal neutrons, can be converted into a fissile material (Pu-239) by neutron absorption and subsequent nuclei conversions. 238 U 1n 92 0 Rev 0.1 239 U 239 Np 239 Pu 92 93 94 33 Knowledge Check (Answer Key) What is the difference between a fissionable material and a fissile material? A. There is no difference between a fissionable and a fissile material, except for the number of protons and neutrons located in the nuclei of the particular materials. B. A fissionable material can become fissile by capturing a neutron with zero kinetic energy, whereas a fissile material can become fissionable by absorbing a neutron that has some kinetic energy. C. Fissile materials require a neutron with some kinetic energy in order to fission, whereas fissionable materials will fission with a neutron that has zero kinetic energy. D. Fissionable materials require a neutron with some kinetic energy in order to fission, whereas fissile materials will fission with a neutron that has zero kinetic energy. ELO 4.3 Fission Process - Liquid Drop Model Introduction In a fission reaction, the incident neutron interacts with the target nucleus, and the target nucleus absorbs the incident neutron. This creates a compound nucleus that is excited at such a high energy level (Eexc > Ecrit) that the compound nucleus splits (fissions) into two large fragments plus some neutrons. In addition, the fission process releases a large amount of energy in the form of radiation and fragment kinetic energy. Fission Definition As previously discussed, the attractive nuclear force between nucleons holds the nucleus together. In doing so, it is resisting the opposing electrostatic forces within the nucleus. Characteristics of the attractive nuclear force are: Very short range attractive force, with essentially no strength beyond nuclear scale dimensions (≈10-13 cm) Stronger than the repulsive electrostatic forces within the nucleus The force is independent of nucleon pairing. Attractive forces between pairs of neutrons, pairs of protons or neutron proton pairs are identical. Saturable, that is, a nucleon can attract only a few of its nearest neighbors. Rev 0.1 34 Fission Process Example One theory of fission considers the fission of a nucleus similar to the splitting of a liquid drop. Molecular forces hold a liquid drop together, making the drop spherical in shape and resist deformation. Nuclear forces hold the nucleus together in the same manner. The next figure illustrates the liquid drop model. Figure: Liquid Drop Model Liquid Drop Model Steps The steps illustrated above for the liquid drop model are as follows: The undisturbed nucleus in the ground state is undistorted, and its attractive nuclear forces are greater than the repulsive electrostatic forces between the protons within the nucleus. The nucleus becomes an excited compound nucleus when the incident neutron strikes the target nucleus and is absorbed. This compound nucleus temporarily contains all the charge and mass involved in the reaction, and exists in an excited state. The excitation energy added to the compound nucleus is equal to the binding energy contributed by the incident neutron plus its kinetic energy. The excitation energy may cause the nucleus to oscillate and distort in shape. If the excitation energy is greater than an isotope's specific critical energy, the oscillations may cause the compound nucleus to become dumbbellshaped as it starts to come apart. As this occurs, the attractive nuclear forces (short-range) in the neck area reduce due to saturation, while the repulsive electrostatic forces (long-range) remain almost as strong. Less force holds the nucleus together. When the repulsive electrostatic forces exceed the attractive nuclear forces, nuclear fission occurs - the nucleus breaks apart into two fission fragments and releases neutrons and energy. Rev 0.1 35 Knowledge Check (Answer Key) In the Liquid Drop Model of fission, the nucleus absorbs a neutron, becomes distorted into a dumbbell shape, and splits into two nuclei. Which of the following forces is responsible for the nucleus splitting? A. Liquid drop force B. Gravitational force C. Electrostatic force D. Fission force TLO 4 Summary Excitation energy is the energy above ground state of a nucleus. Critical energy is the required excitation energy for fission to occur. The excitation energy must be above a specific minimum value for that nuclide for fission to occur. The critical energy (Ecrit) is the minimum excitation energy required for fission to occur for a particular nuclide. There are two categories of nuclear fuel materials: those that can fission with thermal neutrons (fissile materials) and those that require neutrons of higher energies for fission to be possible (fissionable materials). Fissionable materials include fissile nuclides as well as those nuclides that fission only from high-energy neutrons. Fissile materials are the desired nuclides for use in thermal nuclear reactors because they can fission with neutrons possessing zero kinetic energy (thermal neutrons). In a fission reaction, the incident neutron interacts with the target nucleus, and the target nucleus absorbs the incident neutron. This creates a compound nucleus that is excited at such a high energy level (Eexc > Ecrit) that the compound nucleus splits (fissions) into two large fragments plus some neutrons. In addition, the fission process releases a large amount of energy in the form of radiation and fragment kinetic energy. Rev 0.1 36 One theory, the Liquid Drop Model, considers the fission of a nucleus similar to the splitting of a liquid drop. Molecular forces hold a liquid drop together, making the drop spherical in shape and resist deformation. Nuclear forces hold the nucleus together in the same manner. If enough excitation energy is added, the nucleus becomes distorted and takes on a dumbbell shape. As this occurs the attractive nuclear forces (short-range) in the neck area reduce due to saturation, while the repulsive electrostatic forces (long-range) remain almost as strong. When the repulsive electrostatic forces exceed the attractive nuclear forces, nuclear fission occurs. TLO 5 Production of Heat from Fission Overview Fission of heavy nuclides converts a small amount of mass into a large amount of energy. There are two ways to calculate the energy released by fission: you can base computations on the change in mass that occurs during the fission or by the difference in binding energy per nucleon between the fissile nuclide and the fission products. This TLO discusses the process of fission and heat production. Objectives Upon completion of this lesson, you will be able to do the following: 1. Describe the average total amount of energy released per fission event including: a. Energy released immediately from fission b. Delayed fission energy 2. Describe which fission product nuclides are most likely to result from fission. 3. Describe how heat is produced as a result of fission. ELO 5.1 Energy Release Per Fission Introduction Total energy released per fission varies from one fission event to the next, depending on what fission products the reaction forms, but the average total energy released from uranium-235 fission with a thermal neutron is approximately 200 MeV. The majority of this energy (approximately 83 percent) is from the kinetic energy of the fission fragments. Rev 0.1 37 Energy Release per Fission Details The table below shows the average energy distribution for the energy released in a U-235 thermal fission, approximately 200 MeV. Instantaneous Energy Value Kinetic Energy of Fission Fragments 165 MeV Kinetic Energy of Fission Neutrons 5 MeV Instantaneous Gamma Rays 7 MeV Capture Gamma Ray Energy 10 MeV Delayed Energy Value Kinetic Energy of Beta Particles 7 MeV Decay Gamma Rays 6 MeV Neutrinos** 10 MeV** Total Energy Released 200 MeV **Not included in total Some fission neutrons undergo radiative capture reactions, producing gamma ray emissions as their compound nuclei drop to ground state. This provides the additional 10 MeV of instantaneous energy from Capture Gamma Ray energy shown in the table above. The total energy released per fission does not include the delayed energy from neutrinos, because neutrinos rarely interact with materials inside the reactor. Delayed Fission Energy (Decay Heat) Of the 200 MeV released per fission, about seven percent (13 MeV) is released sometime after the instant of fission. Fissions mostly cease with the reactor shut down, but heat energy, referred to as decay heat, continues to be released because of the decay of fission products. After reactor shutdown, decay heat production tapers off but remains a significant source of heat for a very long time. Systems exist to remove this decay heat after reactor shutdown in order to prevent damage to the reactor core. The Three Mile Island event is an example of what can happen if this cooling source is unavailable following a shutdown. The reactor operations module discusses decay heat in detail. Rev 0.1 38 Isotope specifics, including the particular isotopes that are undergoing fission, and their classification as fissionable or fissile material, have a small effect on the amount of energy released. For instance, the fission of U-235 by a slow neutron yields nearly identical energy to that of a U-238 fission by a fast neutron. Knowledge Check (Answer Key) Which of the following statements correctly describes the amount of energy released from a single fission event? A. Approximately 200 MeV are released during fission. 13 MeV are released instantaneously, and 187 MeV are released later (delayed). B. Approximately 200 MeV are released during fission. 187 MeV are released instantaneously, and 13 MeV are released later (delayed). C. Approximately 200 eV are released during fission. 13 eV are released instantaneously, and 187 eV are released later (delayed). D. Approximately 200 eV are released during fission. 187 eV are released instantaneously, and 13 eV are released later (delayed). ELO 5.2 Fission Fragment Yield Introduction Fissions do not produce identical results on each occurrence. In fact, both the number of neutrons and the resultant fission fragments vary. Scientific experiments have developed a yield curve of fission product probabilities. Most Probable Fission Fragments Resultant fission fragments have masses that vary widely. The figure below shows the percent yield of various atomic mass numbers. The most probable pair of fission fragments for a thermal neutron fission of uranium235 have masses of about 95 and 140. Note Note Rev 0.1 The vertical axis of the fission yield curve is a logarithmic scale further amplifying the higher probability for mass numbers of 95 and 140. Rubidium-93 and cesium-140 are very likely to result from fission. 39 Figure: Uranium-235 Fission Yield for Fast and Thermal Neutrons versus Mass Number As shown in the above figure, the fission fragment yield varies considerably. An example of one fission fragment yield is: 235 𝑈+𝑛 → 236 𝑈∗ → 140 𝑋𝑒 + 94 𝑆𝑟 + 2𝑛 Notice in the example above that two neutrons result from this fission. Normally, one fission event releases two or three fast neutrons. The figure below shows average number of neutrons released per fission event for various fuels. Isotope Average Neutrons Released per Fission (v) U-233 2.492 U-235 2.418 Pu-239 2.871 Pu-241 2.927 Rev 0.1 40 Knowledge Check (Answer Key) During full power operation, which of the following combinations of isotopes are fission fragments likely to be present in a fuel assembly? A. Oxygen-18, iron-59, and zirconium-95 B. Hydrogen-2, iodine-131, and xenon-135 C. Krypton-85, strontium-90, and iodine-135 D. Hydrogen-2, hydrogen-3, and nitrogen-16 ELO 5.3 Fission Heat Production Introduction The majority of the energy liberated in the fission process releases immediately after the fission occurs. This energy appears as kinetic energy of the fission fragments and neutrons, and instantaneous gamma rays. The remaining energy releases over time after the fission occurs and appears as kinetic energy of the decay products. Fission Heat Production All of the energy released in fission, with the exception of the neutrino energy, transforms into heat via ionization and scattering. Fission fragments, with a positive charge and kinetic energy, cause ionization directly as they remove orbital electrons from the surrounding atoms through this ionization process. Kinetic energy transfers to the surrounding atoms of the fuel material, increasing temperature. Beta particles and gamma rays also give up energy through ionization, and fission neutrons interact and lose their energy through scattering. Rev 0.1 41 Knowledge Check (Answer Key) Which of the following is NOT a method by which heat is produced from fission? A. Fission fragments causing direct ionizations resulting in an increase in temperature. B. Beta particles and gamma rays causing ionizations resulting in increased temperature. C. Neutrons interacting and losing their energy through scattering, resulting in increased temperature. D. Neutrinos interacting and losing their energy through scattering and ionizations resulting in increased temperatures. TLO 5 Summary Total energy released per fission varies from one fission event to the next, depending on what fission products the reaction forms, but the average total energy released from uranium-235 fission with a thermal neutron is approximately 200 MeV. The majority of this energy (approximately 83 percent) is from the kinetic energy of the fission fragments. Fissions do not produce identical results on each occurrence. In fact, both the number of neutrons and the resultant fission fragments vary. Scientific experiments have developed a yield curve of fission product probabilities. Resultant fission fragments have masses that vary widely. The figure below shows the percent yield of various atomic mass numbers. The most probable pair of fission fragments for a thermal neutron fission of uranium235 have masses of about 95 and 140. The majority of the energy liberated in the fission process releases immediately after the fission occurs. This energy appears as kinetic energy of the fission fragments and neutrons, and instantaneous and capture gamma rays. The remaining energy releases over time after the fission occurs and appears as kinetic energy of the decay products. Rev 0.1 42 TLO 6 Prompt and Delayed Neutrons Overview Not all neutrons are born immediately following fission. Fission releases most neutrons virtually instantaneously; these are referred to as prompt neutrons. The remaining neutrons (a very small fraction), are born after the decay of certain fission products and are referred to as delayed neutrons. Although delayed neutrons are a very small fraction of the total number of neutrons, they play an extremely important role in controlling the reactor. Objectives Upon completion of this lesson, you will be able to do the following: 1. Describe the origin and production of prompt and delayed neutrons. 2. State the approximate fraction of neutrons that are born as delayed neutrons from the fission of the following fuels: a. Uranium-235 b. Plutonium-239 3. Define prompt and delayed neutron lifetimes and their generation times. ELO 6.1 Production of Prompt and Delayed Neutrons Introduction There are two ways to classify neutrons. One is to categorize according to energy, such as fast or thermal. Another is to classify according to birth time relative to a fission event. Those neutrons born immediately after a fission event are prompt neutrons, those born later from the decay of certain fission products are delayed neutrons. Rev 0.1 43 Glossary Prompt Neutrons Within about 10-14 seconds of a fission event, a majority (≈ 99.36 percent) of the neutrons are released (born). These are prompt neutrons. The number of prompt neutrons emitted during a fission event depends on the type of fuel used (U-235 averages 2.4 per fission event). The most probable energy for a prompt neutron is approximately 1 MeV, and the average energy is approximately 2 MeV. Delayed Neutrons A small portion of the neutrons born of fission, are born delayed from delayed neutron precursors. Delayed neutrons are neutrons that are born significantly after the fission process has taken place. On average, delayed neutrons are born approximately 12.7 seconds after the fission event. Delayed neutrons are born fast but at a lower energy than prompt neutrons (≈ 0.5 MeV). Delayed Neutron Precursors A delayed neutron precursor refers to delayed neutrons emitted immediately following the first beta decay of a fission fragment. An example of a delayed neutron precursor is bromine-87 (Br), shown below. Br-87 is the fission product; Kr-87 (Krypton) is the delayed neutron precursor, with Kr86 (Krypton) resulting from the delayed neutron birth. 86 𝐵𝑟 35 𝑛 𝛽− 87 86 → 𝐾𝑟 𝐾𝑟 → 36 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 36 𝑠𝑡𝑎𝑏𝑙𝑒 55.9 𝑠𝑒𝑐 Rev 0.1 44 Example It is convenient to combine the known delayed neutron precursors into groups with appropriately averaged half-life properties. These groups will vary somewhat depending on the fuel or mixture of fuel in the reactor. The table below lists the characteristics for the six delayed neutron precursor groups resulting from the thermal fission of uranium-235. Group Half Life (Seconds) Delayed Neutron Fraction Average Energy (MeV) 1 55.7 0.00021 0.25 2 22.7 0.00142 0.46 3 6.2 0.00127 0.41 4 2.3 0.00257 0.45 5 0.61 0.00075 0.41 6 0.23 0.00027 N/A Total N/A 0.0065 N/A Knowledge Check (Answer Key) Which one of the following types of neutrons has an average neutron generation lifetime of 12.7 seconds? Rev 0.1 A. Prompt B. Delayed C. Fast D. Thermal 45 Knowledge Check (Answer Key) Delayed neutrons are the neutrons that... A. have reached thermal equilibrium with the surrounding medium. B. are expelled within 10-14 seconds of the fission event. C. are produced from the radioactive decay of certain fission fragments. D. are responsible for the majority of U-235 fissions. ELO 6.2 Delayed Neutron Fraction Introduction The fraction of all neutrons produced by each delayed neutron precursor is the delayed neutron fraction for that precursor. The total fraction of all neutrons born as delayed neutrons is the delayed neutron fraction (β). Each nuclear fuel has different delayed neutron fraction (β). Delayed Neutron Fraction (β) The fraction of delayed neutrons (β) varies depending on the predominant fissile nuclide in use. The delayed neutron fractions (β) for the nuclides of most interest are as follows: Uranium-233 β = 0.0026 Uranium-235 β = 0.0064 Uranium-238 β = 0.0148 Plutonium-239 β = 0.0021 It is significant to note that uranium-235 and plutonium-239 are the two major fuels in use in PWRs. Although the β values are small, β for uranium-235 is considerably larger than plutonium-239. Over core life, uranium-235 concentration will decrease, while plutonium239 increases. This will result in lower delayed neutron fraction over core life. Rev 0.1 46 Knowledge Check (Answer Key) What is the delayed neutron fraction of uranium-235? A. 0.0064 B. 0.0148 C. 0.0021 D. 0.0026 ELO 6.3 Neutron Lifetimes and Generation Times Introduction Neutron lifetime is the time that a free neutron exists, from its birth until its loss either from leakage or by absorption. Neutron generation time is the time for a neutron from one generation to cause fission that produces the next generation of neutrons. Prompt Neutron Lifetime Prompt neutron lifetime is the average time span from prompt neutron birth until its loss from either leakage or absorption in another nucleus. This time is the sum of thermalization time and diffusion time. Diffusion time relates to absorption time of the thermal neutron and is a function of the absorption mean free path, λa, divided by the average velocity of the thermal neutron. Thermalization time is small (microseconds) compared to diffusion time (milliseconds) and is usually ignored in calculations. Simply stated, prompt neutron lifetime is the time from birth to loss by either leakage or absorption. Delayed Neutron Lifetime Delayed neutron lifetime begins at its birth from a delayed neutron precursor and ends at loss from leakage or absorption in another nucleus. We calculate delayed neutron lifetime similarly to prompt neutron lifetime. The key difference is the time of birth, which is not at the time of fission but at the time of birth from decay of one of the delayed neutron precursors. Prompt Neutron Generation Time The generation time for prompt neutrons (ℓ* — pronounced ell-star) is the total time from birth of a fast neutron in one generation to birth in the next generation. Prompt neutron generation time is equal to the prompt neutron lifetime added to the time required for a fissionable nucleus to emit a fast neutron after absorption. Rev 0.1 47 In water-moderated reactors, thermal neutrons exist for about 10-4 seconds before absorption. Taking into account losses due to leakage, the prompt neutron lifetime is equal to about 10-4 to 10-5 seconds (leakage occurs quicker). Following absorption, fission, and the birth of fast neutron(s) occurs in about 10-13 seconds — quickly. Therefore, in water-moderated thermal reactors, ℓ* is about 10-4 seconds to 10-5 seconds. Delayed Neutron Generation Time Similar to prompt neutron generation time, delayed neutron generation time equals the time of birth of a delayed neutron from a delayed neutron precursor to the time of birth of a neutron(s) in the next generation. The significant difference between prompt and delayed neutron generation time is the delay in birth from their delayed neutron precursors. As previously mentioned, the average time for decay of the delayed neutron precursors from fission of uranium-235 is 12.7 seconds. With this relatively large time interval, the delayed neutron lifetime is insignificant. Therefore, the average delayed neutron generation time is equal to approximately 12.7 seconds. The significance of delayed neutrons in relation to average generation time and reactor control is discussed in the next module – Neutron Life Cycle. Knowledge Check (Answer Key) __________ begins when it is released from a precursor and ends when it is absorbed in another nucleus. Rev 0.1 A. Delayed neutron lifetime B. Prompt neutron lifetime C. Fast neutron fraction D. Thermal neutron fraction 48 Knowledge Check (Answer Key) Neutron generation time describes... A. time from one generation to the next generation of neutrons. B. time that it takes for a neutron to become thermalized. C. time it takes for neutron precursors to emit neutrons. D. time that neutrons are born after a fission event. Knowledge Check (Answer Key) What effect on the average neutron generation time would a smaller β value produce? A. The result would be a longer average generation time. B. The result would be a shorter average generation time. C. More information is needed to determine the effect on average generation time. D. β does not have any effect on average generation time. TLO 6 Summary There are two ways to classify neutrons. One is to categorize according to energy, such as fast or thermal. Another is to classify according to birth time relative to a fission event. Those neutrons born immediately after a fission event are prompt neutrons, those born later from the decay of certain fission products are delayed neutrons. Prompt Neutrons - Within about 10-14 seconds of a fission event, a majority (≈ 99.36 percent) of the neutrons are released (born). These are prompt neutrons. The number of prompt neutrons emitted during a fission event depends on the type of fuel used (U-235 averages 2.4 per fission event). The most probable energy for a prompt neutron is approximately 1 MeV, and the average energy is approximately 2 MeV. Rev 0.1 49 Delayed Neutrons - A small portion of the neutrons born of fission, are born delayed from delayed neutron precursors. Delayed neutrons are neutrons that are born significantly after the fission process has taken place. On average, delayed neutrons are born approximately 12.7 seconds after the fission event. Delayed neutrons are born fast but at a lower energy than prompt neutrons (≈ 0.5 MeV). The significant difference between prompt and delayed neutron generation time is the delay in birth from their delayed neutron precursors. Due to the relatively large time interval for the decay of delayed neutron precursors (12.7 seconds for U-235), the delayed neutron lifetime is insignificant. Therefore, the average delayed neutron generation time is equal to approximately 12.7 seconds. TLO 7 Neutron Moderation Overview Fission neutrons are born at an average energy of 2 mega electron volt (MeV) or fast neutrons and at high temperature. Fission neutrons immediately begin to reduce their energy levels as they undergo numerous scattering reactions with nuclei in the nuclear reactor core. A number of collisions with nuclei reduce the neutron’s energy to approximately the same average kinetic energy as its surrounding atoms or molecules. This energy reduction occurs in a medium known as the moderator. A neutron in energy equilibrium with its surrounding atoms is a thermal neutron. Since the kinetic energy depends on temperature (molecular movement), the energy of a thermal neutron also depends on temperature. At 68° Fahrenheit (F), the energy of a thermal neutron is 0.025 electron volt (eV). Energies less than 1eV yield neutrons designated or known as slow neutrons. Objectives At the completion of this training session, you will be able to do the following: 1. Describe the following: a. Thermalization b. Moderator c. Moderating ratio 2. Describe the four desirable characteristics of a good moderator and explain how moderator density affects neutron moderation. Rev 0.1 50 ELO 7.1 Neutron Moderation Introduction Lowering the energy level of a neutron is essential because thermal fission requires a neutron to be at thermal equilibrium. In a nuclear reactor, fast energy neutrons born from a fission event must be slowed to the thermal energy region to maintain a chain reaction. Thermalization or moderation is the process of reducing neutron energy to the thermal range by elastic scattering. Thermalization Thermalization or moderation is the process of reducing neutron energy to the thermal range by elastic scattering. Moderator A moderator is the material used to thermalize neutrons. A desirable moderator is one that reduces the velocity of the fission neutrons, using a minimum number of scattering collisions with a low probability of neutron absorption. Slowing the neutrons in as few collisions as possible reduces their travel distance, thereby reducing the number of neutrons that leak out of the core. This shorter travel distance also reduces the number of resonance absorptions in non-fuel materials. Neutron leakage and resonance absorption are discussed in detail later in this module. Average Logarithmic Energy Decrement The average logarithmic energy decrement is the measure of neutron energy loss per collision. This term is the average decrease per collision of the logarithm of the change in neutron energy, denoted by the symbol ξ (Xi). 𝜉 = 𝑙𝑛 𝐸𝑖 𝐸𝑓 Where: ξ = logarithmic energy decrement Ei = initial energy level of neutron Ef = final energy level of neutron Rev 0.1 51 Macroscopic Slowing Down Power The logarithmic energy decrement is a convenient measure of the ability of a material to slow neutrons, but does not consider the probability of collisions taking place. Another measure of a moderator is the macroscopic slowing down power (MSDP), defined as the product of the logarithmic energy decrement and the macroscopic cross-section for scattering in the material. The equation below calculates the macroscopic slowing down power. 𝑀𝑆𝐷𝑃 = 𝜉𝛴𝑠 Where: ξ = logarithmic energy decrement 𝛴𝑠 = Macroscopic scattering cross-section Moderating Ratio Macroscopic slowing down power calculates how rapidly a neutron will thermalize in the chosen moderator. However, it does not fully explain the effectiveness of the material as a moderator. An element such as boron has a high logarithmic energy decrement and good MSDP, but it is a poor moderator because of its high probability of absorbing neutrons. The moderating ratio considers absorbing probability as well as slowing down power; and therefore, is a more complete measure of moderator effectiveness. Moderating ratio is the ratio of the microscopic slowing down power to the microscopic cross-section for absorption. The higher the moderating ratio, the more effectively the material performs as a moderator. This equation shows the calculation for the moderating ratio (MR): 𝑀𝑅 = 𝜉𝜎𝑠 𝜎𝑎 The moderating ratio characterizes the effectiveness of a material as a moderator. It considers the ratio between the absorption and scattering cross-sections factoring in neutron energy levels. It does not consider the density of the moderator. The table below compares moderating properties of different materials. Rev 0.1 52 Moderating properties of different materials are compared in the table. Material ξ Number of Collisions to Thermalize Microscopic Cross-Sections σa–σs Moderating Ratio Water (H2O) also known as light water 0.948 19.0 0.66 – 103.0 148.0 Deuterium Oxide (D2O) also known as heavy water 0.570 35.0 0.001 – 13.6 7,752.0 Beryllium (Be) 0.209 86.0 0.0092 – 7.0 159.0 Carbon (C) 0.158 114.0 0.003 – 4.8 253.0 A good moderator is one with a high moderating ratio. Any of the materials shown in the table above make good moderators; however, commercial nuclear power applications often use light water (H2O) because it is plentiful, easily obtained, and inexpensive. Another benefit of using water is that water can serve as both a moderator and a coolant. Heavy water (D2O) is by far the best performing moderator; however, its higher cost precludes its use in U.S. commercial PWRs. Knowledge Check (Answer Key) A _______________ is a material within a reactor which is responsible for thermalizing neutrons. Rev 0.1 A. fuel rod B. moderator C. poison D. reflector 53 Knowledge Check (Answer Key) The process of reducing the energy level of a neutron from the energy level at which it is produced to an energy level in the thermal range is known as _______________. A. moderating ratio B. resonance absorption C. thermalization D. inelastic scattering Knowledge Check (Answer Key) The average logarithmic energy decrement is important because... A. it can be used to determine if a material is a good moderator. B. it can be used to determine the amount of energy released from fission. C. it accounts for the change in binding energy change during fission. D. it accounts for the change in mass during fission. ELO 7.2 Moderator Characteristics Introduction The ideal moderator requires the following nuclear properties: Large scattering cross-section Small absorption cross-section Large energy loss per collision High atomic density Rev 0.1 54 Desirable Moderator Properties A material with a mass equal to a neutron, but with a large scattering crosssection is desirable as a moderator. A substance having a high absorption cross-section, acts as a poison, removing available neutrons for fission. A moderator with a low atomic density has few atoms available to thermalize neutrons. Carbon, hydrogen, beryllium, and water (both heavy and light) are moderators used in nuclear reactors. In the U.S., most reactors use light water as the moderator. Water is a good moderator because the hydrogen atoms in water are close in mass (good for elastic scattering) to the thermalized neutrons. A neutron loses more energy in a collision with an atom of nearly its own mass than in a collision with an atom whose mass is much greater than the incident neutron (inelastic scattering — billiard ball effect). Water moderators, rather than other materials, thermalize neutrons in fewer collisions, reducing both the time and distance required for thermalization. Moderator Density Effects Temperature changes affect moderator density in an operating reactor. These density changes affect the moderator’s ability to thermalize neutrons. This in turn affects the number of neutrons available for fission. Consider a PWR (light water moderator and coolant) as an example. If the moderator temperature increases, the density of the water decreases. Decreasing water density means there are fewer water atoms per unit volume to thermalize neutrons. The neutrons will have to travel further to thermalize, and this larger travel distance will take longer. Because of this and other factors discussed later, there is now an increased chance that these neutrons will not be available for fission. Fewer neutrons available for fission, means fewer fissions, resulting in less reactor power. Knowledge Check (Answer Key) Which of the items below is not a desirable property for a neutron moderator? Rev 0.1 A. Large absorption cross-section B. Large scattering cross-section C. Large energy loss per collision D. High atomic density 55 Knowledge Check (Answer Key) The density of a moderator is important because... A. it can affect the number of target nuclei available for collisions. B. it can affect the number of target nuclei available for absorption. C. it can affect the number of collisions. D. it can affect the energy loss per collision. TLO 7 Summary A moderator is the material used to thermalize neutrons. A desirable moderator is one that reduces the velocity of the fission neutrons, using a minimum number of scattering collisions with a low probability of neutron absorption. Slowing the neutrons in as few collisions as possible reduces their travel distance, thereby reducing the number of neutrons that leak out of the core. This shorter travel distance also reduces the number of resonance absorptions in non-fuel materials. Neutron leakage and resonance absorption are discussed in detail later in this module. The ideal moderator requires the following nuclear properties: Large scattering cross-section Small absorption cross-section Large energy loss per collision High atomic density TLO 8 Neutron Reaction Rates/Probabilities Overview Fission neutrons are born at an average energy of about 2 MeV and interact with reactor core materials in various absorption and scattering reactions. Scattering reactions are useful for thermalizing neutrons. Thermal neutrons may be absorbed by fissile nuclei to produce fission or absorbed in fertile material resulting in the production of fissionable fuel. Additionally, some neutrons are absorbed in structural components, reactor coolant, and other non-fuel materials resulting in the removal of neutrons from the fission process. Rev 0.1 56 To determine these neutron interaction rates, it is necessary to identify the number of neutrons available and the probability of interaction. To assist in quantifying neutron availability and reaction probabilities, we use terms including neutron flux, microscopic and macroscopic cross-section. The complexity of designing a reactor requires predicting the various reaction rates, both in specific portions of the core and averages throughout the core. One example is the calculation of the reactor's thermal output, knowing the fission rate and core volume. Objectives Upon completion of this lesson, you will be able to do the following: 1. Explain the following terms, including any mathematical relationships: a. Atomic density b. Microscopic cross-section c. Barn d. Macroscopic cross-section e. Mean free path f. Neutron flux g. Fast neutron flux h. Thermal neutron flux 2. Define the following neutron classes: a. Fast b. Intermediate c. Slow 3. Describe how the absorption and scattering cross-section of typical nuclides varies with neutron energies in the 1/V region, and the resonance absorption region. 4. Describe how changes in neutron flux and macroscopic cross-section affect reaction rates. 5. Describe the relationship between neutron flux and reactor power. Rev 0.1 57 ELO 8.1 Neutron Reaction Terms Introduction The following terms are important for understanding the theoretical concepts of nuclear reactors and are foundational for future lessons: Atomic density Microscopic cross-section Barn Macroscopic cross-section Mean free path Neutron flux Fast neutron flux Thermal neutron flux Atomic Density An important property of a material is its atomic density. The atomic density is the number of atoms of a given type per unit volume of the material. Use the following equation to calculate the atomic density of a substance. 𝑁= 𝜌𝑁𝐴 𝑀 Where: N = Atomic density (atoms/cm3) ρ = density (g/cm3) NA = Avogadro’s number (6.022 x 1023 atoms/mole) M = gram atomic weight or gram molecular weight Example: A block of aluminum has a density of 2.699 g/cm3. If the gram atomic weight of aluminum is 26.9815 g, calculate the atomic density of the aluminum. Solution: 𝑁= 𝜌𝑁𝐴 𝑀 𝑔 𝑎𝑡𝑜𝑚𝑠 (6.022 × 1023 ) 3 𝑚𝑜𝑙𝑒 𝑐𝑚 = 𝑔 26.9815 𝑚𝑜𝑙𝑒 𝑎𝑡𝑜𝑚𝑠 = 6.024 × 1022 𝑐𝑚3 2.699 Rev 0.1 58 Microscopic Cross-Section (σ) The probability of a particular reaction occurring between a neutron and a nucleus is the microscopic cross-section (σ) of the nucleus. This crosssection varies with the energy of the neutron, and is the effective area the nucleus presents to the neutron for the particular reaction. Barn is the unit of measure for this area. A larger effective area yields a greater probability for reaction. Total Microscopic Cross-Section (σT) A neutron interacts with an atom in two basic ways: scattering or absorption reaction. The probability of a neutron being absorbed is the microscopic cross-section for absorption (σa). The probability of a neutron scattering is the microscopic cross-section for scattering (σs). The sum of these two microscopic cross-sections is the total microscopic cross-section (σT). 𝜎𝑇 = 𝜎𝑎 + 𝜎𝑠 Microscopic Cross-Section for Scattering (σs) Both the absorption and the scattering microscopic cross-sections have two components. For instance, the scattering cross-section is the sum of the elastic scattering cross-section (σse) and the inelastic scattering cross-section (σsi). 𝜎𝑠 = 𝜎𝑠𝑒 + 𝜎𝑠𝑖 Microscopic Cross-Section for Absorption (σa) The microscopic absorption cross-section (σa) includes all reactions except scattering. However, for most purposes it has two categories, fission (σf) and capture (σc). 𝜎𝑎 = 𝜎𝑓 + 𝜎𝑐 The Chart of Nuclides uses two different conventions to represent microscopic cross-section for capture, depending on the type of particle ejected from the nucleus after the capture reaction. If a gamma ray results from the capture, σγ is used If an alpha particle results from the capture, σα is used A later section discusses fission and radiative capture of neutrons in detail. Rev 0.1 59 Barns Microscopic cross-sections are expressed in units of area - usually square centimeters. A square centimeter is very large compared to the effective area of a nucleus. The old story goes that a physicist once referred to the measure of a square centimeter as being "as big as a barn" when viewed on a nuclear level. The name has persisted, so now microscopic cross-sections have units of barns. The conversion to cm2 is below. 1 𝑏𝑎𝑟𝑛 = 10−24 𝑐𝑚2 Consider boron-10, for example. This nuclide has a relatively low mass with an absorption cross-section of 3,838 barns resulting in an ejected alpha particle. A boron-10 nucleon presents a very large effective area for neutron interaction. Lead-208 is a nuclide of relatively high mass and only has an absorption cross-section of 8 microbarns (8 x 10-6 barns) for the same reaction. Lead-208 presents a very small effective area for neutron interaction. Macroscopic Cross-Section (Σ) Whether or not a neutron interacts with a certain volume of material depends not only on its microscopic cross-section but also on the number of nuclei within that volume (atomic density). Macroscopic cross-section (Σ) is the probability of a given reaction occurring per unit travel of the neutron. Macroscopic cross-section is relates to microscopic cross-section (σ) by the following relationship. Σ = 𝑁𝜎 Where: Σ = macroscopic cross-section (cm-1) N = atomic density (atoms/cm3) σ = microscopic cross-section (cm2) The number of target nuclei per unit volume increases and the probability of interaction increases if atomic density increases. Rev 0.1 60 Macroscopic Calculation Most materials are composed of several elements. Most elements are composed of several isotopes; therefore, most materials involve many cross-sections, one for each isotope involved. Each macroscopic crosssection is determined to include all the isotopes for a material, which are added together as follows: 𝛴 = 𝑁1 𝜎1 + 𝑁2 𝜎2 + 𝑁3 𝜎3 +. . . . . . .. 𝑁𝑛 𝜎𝑛 Where: Nn = the number of nuclei per cm3 of the nth element σn = the microscopic cross-section of the nth element Example: Find the macroscopic thermal neutron absorption cross-section for iron, which has a density of 7.86 g/cm3. The iron microscopic cross-section for absorption is 2.56 barns and the gram atomic weight is 55.847 g. Solution: Step 1: Using the equation for atomic density, calculate the atomic density of iron. 𝑁= 𝜌𝑁𝐴 𝑀 7.68 = 𝑔 𝑎𝑡𝑜𝑚𝑠 (6.022 × 1023 ) 𝑚𝑜𝑙𝑒 𝑐𝑚3 𝑔 55.847 𝑚𝑜𝑙𝑒 = 8.48 × 1022 𝑎𝑡𝑜𝑚𝑠 𝑐𝑚3 Step 2: Using the atomic density and given microscopic cross-section, calculate the macroscopic cross-section. Σ𝑎 = 𝑁𝜎𝑎 = 8.48 × 1022 𝑎𝑡𝑜𝑚𝑠 1 × 10−24 𝑐𝑚2 (2.56 𝑏𝑎𝑟𝑛𝑠) ( ) 𝑐𝑚3 1 𝑏𝑎𝑟𝑛 = 0.217 𝑐𝑚−1 Rev 0.1 61 Microscopic versus Macroscopic Cross-Section The difference between the microscopic and macroscopic cross-sections is extremely important and deserves restatement. The microscopic crosssection (σ) represents the effective target area that a single nucleus presents to a bombarding particle (neutron). The units are in barns or cm2. The macroscopic cross-section (Σ) represents the total effective target area of the nuclei contained in 1 cm3 of the material. The units are 1/cm or cm-1. Mean Free Path The average distance of travel by a neutron before interaction is the mean free path. (λ). 1 𝜆= 𝛴 A neutron has a certain probability of undergoing a particular interaction in one centimeter of travel (Σ) in a material. The inverse of this probability describes how far the neutron will travel (average) before undergoing an interaction. An average neutron travels three mean free paths (considering all materials available to scatter it). Neutron Flux Macroscopic cross-sections for neutron reactions with a specific material determine the probability of one neutron undergoing a specific reaction per centimeter of its travel through that material. It is necessary to determine how many neutrons travel through the material, and the distance (in centimeters) the neutrons travel each second, to determine how many reactions actually occur in a field of neutrons. These calculations are based on the number of neutrons existing in one cubic centimeter at any one instant and the total distance they travel each second while in that cubic centimeter. The number of neutrons existing in a cm3 of material at any instant is the neutron density, represented by the symbol n with units of neutrons/cm3. The neutrons' velocities determine the total distance these neutrons travel each second. Neutron flux is the number of neutrons passing through the unit area (cm2) per unit time. Neutron flux (Φ) is the total path length covered by all neutrons in one cubic centimeter during one second. Its units are neutrons per centimeter squared per second (n/cm2/sec). The equation below shows the relationship between neutron flux, neutron density, and neutron velocity. Rev 0.1 62 Φ = 𝑛𝑣 Where: Φ = neutron flux (neutron/cm2-sec) n = neutron density (neutrons/cm3) v = neutron velocity (cm/sec) The term neutron flux refers to parallel beams of neutrons traveling in a single direction. The intensity (I) of a neutron beam is the product of the neutron density and the average neutron velocity. The directional beam intensity is equal to the number of neutrons per unit area and time (neutrons/cm2-sec) falling on a surface perpendicular to the direction of the beam. Neutron flux is modeled as many neutron beams traveling in various directions in a nuclear reactor. The neutron flux becomes the scalar sum of these directional flux intensities (added as numbers and not vectors), expressed as follows: 𝛷 = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼𝑛 . All of the directional beams contribute to the total reaction rate since the atoms in a reactor do not prefer neutrons coming from any particular direction. In reality, at any given point within a reactor, neutrons are traveling in all directions. There are two classes of neutron flux: thermal neutron flux and fast neutron flux in a nuclear reactor. The next section defines these neutron energies. Thermal Neutron Flux Thermal neutron flux (Φth) is the number of thermal neutrons crossing a unit area in the reactor in a given amount of time. The units are neutrons (thermal) per square centimeter per second (n/cm2/sec) as with total neutron flux. Remember that neutron flux is omni-directional, meaning neutrons can enter a particular square centimeter of reactor material from any direction. Therefore, Φth also equals the total distance of all thermal neutrons diffused (moved) in a particular unit volume in one second. Rev 0.1 63 Fast Neutron Flux Fast neutron flux (Φf) is number of fast neutrons crossing a unit area in a given amount of time. The units are neutrons (fast) per square centimeter per second (n/cm2/sec) as with thermal neutron flux. Fast neutron flux also equals the distance that fast neutrons diffuse (move) in a particular unit volume in one second. The next section defines fast neutrons. Knowledge Check (Answer Key) _______________ is the total path length traveled by all neutrons in one cubic centimeter of material during one second. A. Mean free path B. Neutron flux C. Gamma flux D. Atomic density ELO 8.2 Neutron Energy Terms Introduction There are three classes of neutron energy levels: fast, intermediate and slow. Neutrons that are in energy equilibrium with their surrounding are thermal neutrons. They are the most important for thermal reactors. Rev 0.1 64 Neutron Energy Terms The figure below illustrates the relative energy levels of neutrons and their flux distribution in a typical thermal reactor. Figure: Neutron Energy Definitions of the three classes of neutron energy levels are as follows: Fast neutrons - Fission neutrons are born as fast neutrons. They are categorized with energy levels greater than 0.1 MeV (105 eV) Intermediate neutrons - neutrons with energy levels between 1 eV and 0.1 MeV Slow neutrons - neutrons with energy levels less than 1 eV Thermal Neutrons – neutrons with an energy level of 0.025 eV (2.2 x 105 cm/sec. velocity) at 68° F. Velocity and energy increase with temperature. Knowledge Check (Answer Key) Thermal neutrons are classified in the intermediate neutron energy range. Rev 0.1 A. True B. False 65 Knowledge Check (Answer Key) A SLOW neutron is defined as one where its energy is: A. Less than 1 ev B. Between 1 eV and 100,000 ev C. Greater than 0.1 MeV D. At rest with relationship to other particles ELO 8.3 Neutron Energies versus Cross-Sections Introduction Neutron energies affect both neutron absorption and neutron scattering cross-sections. A higher energy neutron has a lower probability of interaction in general. Resonance peaks at certain specific energies possess very high cross-sections. Neutron Energies versus Cross-Sections Neutron absorption cross-sections have three unique regions of resonance probability related to neutron energy. These are the 1/v region, the resonance region, and the fast neutron region. Each region has a different relationship to changing neutron energy. Neutron scattering cross-sections are somewhat different. Resonance elastic scattering and inelastic scattering cross-sections do have resonance peaks, but they are smaller than absorption peaks. Neutron energy has little effect on potential elastic scattering cross-sections. Scattering reactions are good since they do not lose neutrons but cause them to thermalize for use as fission neutrons. Rev 0.1 66 Neutron Absorption Cross-Section versus Incident Neutron Energy The variation of absorption cross-sections with neutron energy is complex. The absorption cross-sections are small, ranging from a fraction of a barn to a few barns for slow and thermal neutrons for many elements. For a considerable number of nuclides of moderately high to high mass numbers the variation of absorption cross-sections with incident neutron energy reveals three distinct regions on an absorption cross-section versus neutron energy curve. These regions are the 1/v region, resonance peaks region and fast neutron region. The figure below illustrates these regions. Figure: Typical Neutron Absorption Cross-Section versus Neutron Energy 1/v Region In the 1/v region, the cross-section decreases steadily with increasing neutron energy; this low energy region includes thermal neutrons (< 1 eV). In this region the absorption cross-section, which is often high, is inversely proportional to the velocity (v). This region is frequently referred to as the "1/v region" because of the inverse relationship to velocity or energy (velocity and energy are directly proportional, but both are inversely proportional to cross-section). Rev 0.1 67 Resonance Region Beyond the 1/v region, there occurs the "resonance region" in which the cross-sections rise sharply to high value peaks. These "resonance peaks" occur with neutrons in the intermediate energy (epithermal) range. Resonance peaks result with affinity of the nucleus for neutrons with energies that match its discrete, quantum (preferred) energy levels. Peaks occur when neutron’s BE plus its KE are exactly equal to the amount of energy needed to raise a compound nucleus from ground state to a quantum level energy level. Resonance absorption occurs as these match. Fast Neutron Region The absorption cross-section steadily decreases as the energy of the neutron increases for higher neutron energies. This is the fast neutron region. The absorption cross-sections are usually less than 10 barns in this region. This explains the low probability for fast fissioning of neutrons in this region. Knowledge Check (Answer Key) At low neutron energies (<1 eV) the absorption crosssection for a material is ___________ proportional to the neutron velocity. (Fill in the blank). ELO 8.4 Reaction Rate Introduction Two factors are required in order to calculate the number of interactions taking place in a cubic centimeter in one second. These two factors are the total path length of all the neutrons in a cubic centimeter per second (neutron flux [Φ]), and the probability of an interaction per centimeter path length (macroscopic cross-section [Σ]). Multiply these two factors together to get the number of interactions taking place in that cubic centimeter in one second. The resulting value is the reaction rate, denoted by the symbol R. The type of reaction rate calculated will depend on the macroscopic crosssection used in the calculation. Normally, the reaction rate is of greatest interest is the fission reaction rate. Rev 0.1 68 Calculate the Reaction Rate Step Action 1. Obtain the average thermal neutron flux in the reactor. 2. Obtain the fuel macroscopic cross-section for the particular material and reaction of interest. Determine the microscopic cross-section and atomic density for material and calculate the macroscopic cross-section if the macroscopic cross is unknown using: Σ = 𝑁𝜎 Where: Σ = macroscopic cross-section (cm-1) N = atomic density (atoms/cm3) σ = microscopic cross-section (cm2) 3. Calculate the reaction rate using the following formula: 𝑅 = ΣΦ Where: R = reaction rate (reactions/cm3-sec) Σ = macroscopic cross-section (cm-1) Φ = neutron flux (neutrons/cm2-sec) Note Note Rev 0.1 This reaction rate is only calculating the reactions occurring in one cubic centimeter. This unit value requires multiplication by the core volume to determine the reactions in the entire core. The next session will discuss the conversion. 69 Reaction Rate Example: A reactor has a macroscopic fission cross-section of 0.1 cm-1, and thermal neutron flux of 1013 neutrons/cm2-sec, what is the fission rate in that cubic centimeter? Solution: 𝑅𝑓 = ΦΣf = (1 × 1013 = 1 × 1012 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 ) (0.1 𝑐𝑚−1 ) 𝑐𝑚2 – 𝑠𝑒𝑐 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 𝑐𝑚3 – 𝑠𝑒𝑐 Reaction Rate Over Core Life Reaction rate required to obtain 100% power does NOT change over core life, however, the amount of fuel (N) decreases over core life. Microscopic cross-section is a constant value. Therefore, flux must increase over core life to have the same reaction rate. At the beginning of core life (BOL) reaction rate is calculated as follows: R = 𝜎𝑁∅ To maintain 100% power over core life: R ↔ (required to obtain 100% power) 𝜎 ↔ (constant value) 𝑁 ↓, due to fuel depletion, therefore ∅ must ↑ Rev 0.1 70 Knowledge Check (Answer Key) Calculate the reaction rate (fission rate) in a one cubic centimeter section of a reactor that has a macroscopic fission cross-section of 0.2 cm-1, and a thermal neutron flux of 1014 neutrons/cm2-sec. A. R = 2.0 × 1014 B. R = 0.2 × 1013 neutrons cm2 –sec neutrons cm2 –sec C. R = 2.0 × 1013 neutrons D. 𝑅 = 20 × 1013 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 cm3 –sec 𝑐𝑚3 –𝑠𝑒𝑐 ELO 8.5 Neutron Flux and Reactor Power Introduction Multiplying the reaction rate per unit volume by the total volume of the core equals the total number of reactions occurring in the core per unit time. It is possible to calculate the rate of energy release (power) due to a certain reaction given the amount of energy involved in each reaction. Neutron Flux and Reactor Power Step-by-Step Tables The number of fissions to produce one watt of power requires the following conversion factors in a reactor where average energy per fission is 200 MeV: 1 fission = 200 MeV 1 MeV = 1.602 x 10-6 ergs 1 erg = 1 x 10-7 watt-sec 1 𝑒𝑟𝑔 1 𝑀𝑒𝑉 1 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 1 𝑤𝑎𝑡𝑡 ( )( )( ) −7 −6 1 × 10 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 1.602 × 10 𝑒𝑟𝑔 200 𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 = 3.12 × 1010 𝑠𝑒𝑐𝑜𝑛𝑑 This is equivalent to stating that 3.12 x 1010 fissions release 1 watt-second of energy. You can use this equation to calculate the power released in a reactor. Multiply the reaction rate by the volume of the reactor to obtain the total fission rate for the entire reactor. Divide the total fission rate by the number of fissions per watt-sec to obtain the power released by fission in the reactor in watts. Solve for reactor thermal power (watts or Megawatts) using the following steps based on this equation. Rev 0.1 71 Step 1. Action Obtain the following reactor data: Volume of core (cm3) Reaction rate Fission macroscopic cross-section and average thermal neutron flux or, 𝑅 = ΣΦ 2. Calculate reactor power using the following equation, substitute reaction rate if known: 𝑃= Φ𝑡ℎ Σ𝑓 𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 Where: P = power (watts) Φth = thermal neutron flux (neutrons/cm2 - sec) Σf = macroscopic cross-section for fission (cm-1) V = volume of core (cm3) Rev 0.1 72 Neutron Flux and Reactor Power Demonstration Example: Calculate reactor power given the following: 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑟𝑒 (𝑐𝑚3 ) = 20,000 𝑐𝑚3 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 = 1 × 1015 𝑐𝑚3 –𝑠𝑒𝑐 Solution: Step 1: 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑟𝑒 (𝑐𝑚3 ) = 20,000 𝑐𝑚3 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 = 1 × 1015 𝑐𝑚3 –𝑠𝑒𝑐 Step 2: 𝑃= Φ𝑡ℎ Σ𝑓 𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 Substitute: Reaction rate for thermal neutron flux multiplied by the macroscopic cross-section for fission 𝑃 = 1 × 1015 × 20,000 3.12 × 1010 𝑃 = 6.41 × 108 𝑤𝑎𝑡𝑡𝑠 = 641 𝑀𝑒𝑔𝑎𝑤𝑎𝑡𝑡𝑠𝑡ℎ𝑒𝑟𝑚𝑎𝑙 Knowledge Check (Answer Key) With a reaction rate of 2 x 1013 neutrons/cm3-sec what is the reactor power level? Assume the entire volume of the core is 10,000 cm3. A. 𝑃= B. Rev 0.1 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 (10,000 𝑐𝑚3 ) 𝑐𝑚3 – 𝑠𝑒𝑐 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 2 × 1013 𝑃= C. Φ𝑡ℎ Σ𝑓 𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 𝑃 = 6.41 × 106 𝑤𝑎𝑡𝑡𝑠 73 TLO 8 Summary This section reviewed important terms for understanding the theoretical concepts of nuclear reactors: Atomic density: number of atoms of a given type per unit volume of material Microscopic cross-section: probability of an interaction between a neutron and a target nucleus Barn: unit of microscopic cross-session (1 barn = 10-24 cm2) Macroscopic cross-section: the probability of a given interaction occurring per unit travel of the neutron Mean free path: average distance of travel by a neutron before interaction Neutron flux: number of neutrons passing through the unit area (cm2) per unit time Fast neutron flux: the number of fast neutrons crossing a unit area in a given amount of time Thermal neutron flux: the number of thermal neutrons crossing a unit area in the reactor in a given amount of time. Neutrons are also classified by energy levels: Fast (Prompt and Delayed neutrons are born fast) - >0.1 Mev Intermediate (Epithermal) - >1ev but < 0.1 Mev Slow - <1 ev Thermal - 0.025ev @ 68oF Absorption cross-sections have three unique regions of probability related to neutron energy 1/v region Resonance region Fast neutron region Rev 0.1 74 Scattering cross-sections are not really affected by energy levels of neutrons U-238 scattering cross section is ≈ 10 Barns Reactor power can be determined using the following equation: 𝑃= Φ𝑡ℎ Σ𝑓 𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 Where: P = power (watts) Φth = thermal neutron flux (neutrons/cm2 –sec) Σf = macroscopic cross-section for fission (cm-1) V = volume of core (cm3) Review Knowledge Check (Answer Key) In a comparison between a delayed neutron and a prompt neutron produced from the same fission event, the prompt neutron is more likely to... Rev 0.1 A. require a greater number of collisions to become a thermal neutron. B. be captured by U-238 at a resonance energy peak between 1 eV and 1,000 eV. C. be expelled with a lower kinetic energy. D. cause thermal fission of a U-235 nucleus. 75 Neutrons Knowledge Check Answer Key Knowledge Check Answer Key ELO 1.1 Atomic Structure Knowledge Check - Answer Identify the particles included in the make-up of an atom. (More than one answer may apply.) A. neutron B. electron C. gamma D. amu ELO 1.2 Atomic Terms Knowledge Check - Answer Match the term to its definition. 1 Protons + neutrons A. Deuterium 2 One neutron B. Neutron number 3 Number of neutrons C. Atomic number 4 Number of protons D. Mass number 1. 2. 3. 4. D – Mass number A – Deuterium B – Neutron number C – Atomic number Rev 0.1 1 Neutrons Knowledge Check Answer Key Knowledge Check - Answer What is the element and number of neutrons for the following: 235 92𝑈 A. Uranium; 143 B. Uranium; 92 C. Plutonium; 143 D. Plutonium; 92 Knowledge Check – Answer Nuclide Element Protons Electrons Neutrons 1 1𝐻 Hydrogen 1 1 0 10 5𝐵 Boron 5 5 5 16 8𝑂 Oxygen 8 8 8 235 92𝑈 Uranium 92 92 143 Plutonium 94 94 145 239 94𝑃𝑢 ELO 1.3 Atomic Forces Knowledge Check - Answer Very weak attractive force between all nucleons describes which of the forces listed below? Rev 0.1 A. Electrostatic B. Nuclear C. Gravitational D. Atomic 2 Neutrons Knowledge Check Answer Key Knowledge Check – Answer Which of the following nuclides has the higher neutronproton ratio? A. Cobalt-60 B. Selenium-79 C. Silver-108 D. Cesium-137 ELO 2.1 Mass Defect and Binding Energy Knowledge Check - Answer _______________ is the amount of energy that must be supplied to a nucleus to completely separate its nuclear particles. Rev 0.1 A. Nuclear energy B. Binding energy C. Mass defect D. Separation energy 3 Neutrons Knowledge Check Answer Key ELO 2.2 Determining Mass Defect and Binding Energy Knowledge Check - Answer Calculate the mass defect for uranium-235. One uranium-235 atom has a mass of 235.043924 amu. mp = mass of a proton (1.007277 amu) mn = mass of a neutron (1.008665 amu) me = mass of an electron (0.000548597 amu) A. 1.86471 amu B. 1.91517 amu C. 0.191517 amu D. 0.186471 amu Knowledge Check - Answer Calculate the binding energy for uranium-235. One uranium-235 atom has a mass defect of 1.9157 amu. A. 1784 MeV B. 178.4 MeV C. 1783 MeV D. 178.3MeV ELO 2.3 Binding Energy per Nucleon Knowledge Check - Answer True or False. Generally, less stable nuclides have a higher BE/A than the more stable ones. Rev 0.1 A. True B. False 4 Neutrons Knowledge Check Answer Key ELO 2.4 Binding Energy Per Nucleon Curve Knowledge Check - Answer True or False. Generally, less stable nuclides have a higher BE/A than the more stable ones. A. True B. False ELO 3.1 Neutron Scattering Interactions Knowledge Check - Answer True/False: The difference between elastic and inelastic scattering where neutrons are concerned is elastic scattering involves no energy being transferred into excitation energy of the target nucleus. Inelastic scattering involves a transfer of kinetic energy into excitation energy of the target nucleus. A. True B. False ELO 3.2 Neutron Absorption Reactions Knowledge Check - Answer What type of neutron interaction has occurred when a nucleus absorbs a neutron and ejects proton? Rev 0.1 A. Fission B. Fusion C. Particle ejection D. Radiative capture 5 Neutrons Knowledge Check Answer Key ELO 4.1 Excitation and Critical Energy Knowledge Check - Answer True or False: Excitation must be at least equal to critical energy for fission to occur. A. True B. False ELO 4.2 Fissile and Fissionable Material Knowledge Check - Answer What is the difference between a fissionable material and a fissile material? Rev 0.1 A. There is no difference between a fissionable and a fissile material, except for the number of protons and neutrons located in the nuclei of the particular materials. B. A fissionable material can become fissile by capturing a neutron with zero kinetic energy, whereas a fissile material can become fissionable by absorbing a neutron that has some kinetic energy. C. Fissile materials require a neutron with some kinetic energy in order to fission, whereas fissionable materials will fission with a neutron that has zero kinetic energy. D. Fissionable materials require a neutron with some kinetic energy in order to fission, whereas fissile materials will fission with a neutron that has zero kinetic energy. 6 Neutrons Knowledge Check Answer Key ELO 4.3 Fission Process - Liquid Drop Model Knowledge Check - Answer In the Liquid Drop Model of fission, the nucleus absorbs a neutron, becomes distorted into a dumbbell shape, and splits into two nuclei. Which of the following forces is responsible for the nucleus splitting? A. Liquid drop force B. Gravitational force C. Electrostatic force D. Fission force ELO 5.1 Energy Release Per Fission Knowledge Check - Answer Which of the following statements correctly describes the amount of energy released from a single fission event? Rev 0.1 A. Approximately 200 MeV are released during fission. 13 MeV are released instantaneously, and 187 MeV are released later (delayed). B. Approximately 200 MeV are released during fission. 187 MeV are released instantaneously, and 13 MeV are released later (delayed). C. Approximately 200 eV are released during fission. 13 eV are released instantaneously, and 187 eV are released later (delayed). D. Approximately 200 eV are released during fission. 187 eV are released instantaneously, and 13 eV are released later (delayed). 7 Neutrons Knowledge Check Answer Key ELO 5.2 Fission Fragment Yield Knowledge Check - Answer During full power operation, which of the following combinations of isotopes are fission fragments likely to be present in a fuel assembly? A. Oxygen-18, iron-59, and zirconium-95 B. Hydrogen-2, iodine-131, and xenon-135 C. Krypton-85, strontium-90, and iodine-135 D. Hydrogen-2, hydrogen-3, and nitrogen-16 (C is correct since they appear near the tops of the two curves of the Fission Product Yield curve) ELO 5.3 Fission Heat Production Knowledge Check - Answer Which of the following is NOT a method by which heat is produced from fission? Rev 0.1 A. Fission fragments causing direct ionizations resulting in an increase in temperature. B. Beta particles and gamma rays causing ionizations resulting in increased temperature. C. Neutrons interacting and losing their energy through scattering, resulting in increased temperature. D. Neutrinos interacting and losing their energy through scattering and ionizations resulting in increased temperatures. 8 Neutrons Knowledge Check Answer Key ELO 6.1 Production of Prompt and Delayed Neutrons Knowledge Check - Answer Which one of the following types of neutrons has an average neutron generation lifetime of 12.5 seconds? A. Prompt B. Delayed C. Fast D. Thermal Knowledge Check - Answer Delayed neutrons are the neutrons that... A. have reached thermal equilibrium with the surrounding medium. B. are expelled within 10-14 seconds of the fission event. C. are produced from the radioactive decay of certain fission fragments. D. are responsible for the majority of U-235 fissions. ELO 6.2 Delayed Neutron Fraction Knowledge Check - Answer What is the delayed neutron fraction of uranium-235? Rev 0.1 A. 0.0064 B. 0.0148 C. 0.0021 D. 0.0026 9 Neutrons Knowledge Check Answer Key ELO 6.3 Neutron Lifetimes and Generation Times Knowledge Check – Answer __________ begins when it is released from a precursor and ends when it is absorbed in another nucleus. A. Delayed neutron lifetime B. Prompt neutron lifetime C. Fast neutron fraction D. Thermal neutron fraction Knowledge Check - Answer Neutron generation time describes... A. time from one generation to the next generation of neutrons. B. time that it takes for a neutron to become thermalized. C. time it takes for neutron precursors to emit neutrons. D. time that neutrons are born after a fission event. Knowledge Check - Answer What effect on the average neutron generation time would a smaller β value produce? Rev 0.1 A. The result would be a longer average generation time. B. The result would be a shorter average generation time. C. More information is needed to determine the effect on average generation time. D. β does not have any effect on average generation time. 10 Neutrons Knowledge Check Answer Key ELO 7.1 Neutron Moderation Knowledge Check - Answer A _______________ is a material within a reactor which is responsible for thermalizing neutrons. A. fuel rod B. moderator C. poison D. reflector Knowledge Check - Answer The process of reducing the energy level of a neutron from the energy level at which it is produced to an energy level in the thermal range is known as _______________. Rev 0.1 A. moderating ratio B. resonance absorption C. thermalization D. inelastic scattering 11 Neutrons Knowledge Check Answer Key Knowledge Check – Answer The average logarithmic energy decrement is important because... A. it can be used to determine if a material is a good moderator. B. it can be used to determine the amount of energy released from fission. C. it accounts for the change in binding energy change during fission. D. it accounts for the change in mass during fission. ELO 7.2 Moderator Characteristics Knowledge Check - Answer Which of the items below is not a desirable property for a neutron moderator? A. Large absorption cross-section B. Large scattering cross-section C. Large energy loss per collision D. High atomic density Knowledge Check - Answer The density of a moderator is important because... Rev 0.1 A. it can affect the number of target nuclei available for collisions. B. it can affect the number of target nuclei available for absorption. C. it can affect the number of collisions. D. it can affect the energy loss per collision. 12 Neutrons Knowledge Check Answer Key ELO 8.1 Nuclear Reaction Terms Knowledge Check - Answer _______________ is the total path length traveled by all neutrons in one cubic centimeter of material during one second. A. Mean free path B. Neutron flux C. Gamma flux D. Atomic density ELO 8.2 Neutron Energy Terms Knowledge Check - Answer Thermal neutrons are classified in the intermediate neutron energy range. A. True B. False Knowledge Check - Answer A SLOW neutron is defined as one where its energy is: Rev 0.1 A. Less than 1 ev B. Between 1 eV and 100,000 ev C. Greater than 0.1 MeV D. At rest with relationship to other particles 13 Neutrons Knowledge Check Answer Key ELO 8.3 Neutron Energies versus Cross-Sections Knowledge Check - Answer At low neutron energies (<1 eV) the absorption crosssection for a material is ___________ proportional to the neutron velocity. (Fill in the blank). A. inversely ELO 8.4 Reaction Rate Knowledge Check - Answer Calculate the reaction rate (fission rate) in a one cubic centimeter section of a reactor that has a macroscopic fission cross-section of 0.2 cm-1, and a thermal neutron flux of 1014 neutrons/cm2-sec. E. R = 2.0 × 1014 F. R = 0.2 × 1013 neutrons cm2 –sec neutrons cm2 –sec G. R = 2.0 × 1013 neutrons H. 𝑅 = 20 × 1013 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 cm3 –sec 𝑐𝑚3 –𝑠𝑒𝑐 ELO 8.5 Neutron Flux and Reactor Power Knowledge Check - Answer With a reaction rate of 2 x 1013 neutrons/cm3-sec what is the reactor power level? Assume the entire volume of the core is 10,000 cm3. A. B. C. Rev 0.1 Φ𝑡ℎ Σ𝑓 𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 3 2 × 1013 3 – 𝑠𝑒𝑐 (10,000 𝑐𝑚 ) 𝑐𝑚 𝑃= 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 𝑃 = 6.41 × 106 𝑤𝑎𝑡𝑡𝑠 𝑃= 14 Neutrons Knowledge Check Answer Key Neutron Topic Review Question Knowledge Check - Answer In a comparison between a delayed neutron and a prompt neutron produced from the same fission event, the prompt neutron is more likely to... Rev 0.1 A. require a greater number of collisions to become a thermal neutron. B. be captured by U-238 at a resonance energy peak between 1 eV and 1,000 eV. C. be expelled with a lower kinetic energy. D. cause thermal fission of a U-235 nucleus. 15