Operator Generic Fundamentals Nuclear Physics – Atomic Structure

Terminal Learning 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): Describe atoms, including components, structure, and nomenclature. Use the Chart of the Nuclides to obtain information on specific nuclides. Describe Mass Defect and Binding Energy and their relationship to one another. Describe the processes by which unstable nuclides achieve stability. Describe how radiation emitted by an unstable nuclide interacts with matter, and materials typically used to shield against this radiation. Describe radioactive decay terms and Calculate activity levels, half-lives, decay constants and radioactive equilibrium. TLOs

Atoms – Introduction TLO 1 – Describe atoms, including components, structure, and nomenclature. 1.1 Using Bohr's model of an atom, describe the characteristics of the following atomic particles, including mass, charge, and location within the atom: Proton Neutron Electron 1.2 Define the following terms and given the standard notation for a given nuclide identify its nucleus and electron makeup: Nuclide Isotope Atomic number Mass number 1.3 Describe the three forces that act on particles within the nucleus and how they affect the stability of the nucleus. TLO 1

Atoms – Introduction John Dalton – determined that elements are made up of distinctly unique atoms in 1803 First modern proof for the atomic nature of matter Atoms are the smallest component of matter defining an element 100 years to prove Dalton’s theories Chemical experiments indicated the atom is indivisible Electrical and radioactivity experimentation indicated that particles of matter smaller than the atom do exist In 1906, J. J. Thompson won the Nobel Prize for establishing the existence of electrons TLO 1

Figure: Composition and Components of Atoms Atoms – Introduction 1920: Earnest Rutherford named the proton 1932: James Chadwick confirmed the existence of the neutron 1970’s: the application of the standard model of particle physics proved the existence of quarks Figure: Composition and Components of Atoms TLO 1

Atomic Structure ELO 1.1 – Using Bohr's model of an atom, describe the characteristics of the following atomic particles, including mass, charge, and location within the atom: proton, neutron, and electron. Physicist Ernest Rutherford postulated: Positive charge in an atom is at center of the atom Electrons orbit around it Niels Bohr, from Rutherford's theory and Max Planck’s quantum theory, proposed orbiting electrons in discrete fixed shells An electron in one of these orbits has a specific quantity of energy (quantum) Electron movement between shells results in a photon either emitted or absorbed ELO 1.1

Figure: Simple Carbon Atom Neutrons and Protons Protons and neutrons are located in the center of atom – called nucleus  Each element made up of atoms having a unique number of protons that defines chemical properties Neutrons are electrically neutral – no electrical charge Protons are electrically positive - electrical charge of +1 Gives nucleus positive charge One proton has a +1, two protons have +2 Neutrons and protons are essentially equal in mass > 1800 times size of an electron Figure: Simple Carbon Atom ELO 1.1

Electrons Electrons orbit the nucleus Orbit in concentric orbits referred to as orbitals or shells Mass of 1/1835 the mass of a proton or neutron Each electron has -1 electrical charge equal in magnitude to one proton For the atom to be electrically neutral, number of electrons must equal protons Electrons are bound to the nucleus by electrostatic attraction (opposite charges attract) Atom remains neutral unless some force causes a change in the number of electrons ELO 1.1

Bohr Model of Atom Bohr’s model is shown on the next slide An electron is shown to have dropped from the third shell to the first shell releasing energy Energy is released as a photon = hv h = Planck's constant - 6.63 x 10-34 J-s (joule-seconds) v = frequency of the photon Accounts for the quantum energy levels Bohr's atomic model is designed specifically to explain the hydrogen atom – has applicability as first generation model to all atoms ELO 1.1

Figure: Bohr's Model of the Hydrogen Atom Bohr Model of Atom Electron dropped from third shell to first shell causing emission of photon 𝑃ℎ𝑜𝑡𝑜𝑛 𝑒𝑛𝑒𝑟𝑔𝑦=ℎ𝑣 Where: h = Planck's constant = 6.63 x 10-34 J-s v = frequency of photon Figure: Bohr's Model of the Hydrogen Atom ELO 1.1

Atomic Measuring Units Size and mass of atoms are very small Use of normal measuring units is inconvenient Atomic Mass Unit (amu) Unit of measure for mass One amu = 1.66 x 10-24 grams Electron Volt (eV) Unit for energy eV is amount of energy acquired by single electron when it falls through potential difference of one volt 1 eV equivalent to 1.602 x 10-19 joules or 1.18 x 10-19 foot-pounds Protons eV value is + 1 and Electrons are -1 eV ELO 1.1

Atomic Measuring Units Particle Location Charge Mass Neutron Nucleus none 1.008665 amu Proton +1 eV 1.007277 amu Electron Shells around nucleus -1 eV 0.0005486 amu Figure: Nucleus and Orbital Electrons ELO 1.1

Atomic Measuring Units Knowledge Check Identify the particles included in the make-up of an atom. (More than one answer may apply.) neutron electron gamma amu Correct answers are A and B. Correct answer is A and B. ELO 1.1

Atomic Terms ELO 1.2 – Define the following terms and given the standard notation for a given nuclide, identify its nucleus and electron makeup: nuclide, isotope, atomic number, and mass number. Atomic number (Z) – # of protons Mass number (A) – Total number of neutrons & protons 𝐴=𝑍+𝑁 Nuclide – Atoms containing a unique combination of protons and neutrons 2500 specific nuclides Isotope – Atoms of the same element with the same number of protons (Z) but different number of neutrons ELO 1.2

Figure: Nomenclature for Identifying Nuclides Atomic Notation Convention for identifying elements on atomic scale Standard notation: Element symbol (X) Atomic number as subscript to lower left Z = # of protons Atomic mass number as superscript to upper left A = # of protons + neutrons Figure: Nomenclature for Identifying Nuclides ELO 1.2

Figure: Nomenclature for Identifying Nuclides Atomic Notation Atomic Number – Z Total number of protons in nucleus of atom Identifies particular element Each chemical element has unique atomic number. Helium consists of atoms with only two protons in nucleus Figure: Nomenclature for Identifying Nuclides ELO 1.2

Figure: Nomenclature for Identifying Nuclides Atomic Notation Mass Number – A Total number of nucleons (protons and neutrons) in nucleus Atoms of same element may not always contain same number of neutrons and may have different atomic mass numbers Isotopes Figure: Nomenclature for Identifying Nuclides ELO 1.2

Figure: Nomenclature for Identifying Nuclides Atomic Notation Neutron Number – N Number of neutrons in nucleus Does not appear in standard atomic notation Found by: 𝑍–𝐴 =𝑁 Figure: Nomenclature for Identifying Nuclides ELO 1.2

Isotopes Isotopes – nuclides that have same atomic number and are same element, but differ in number of neutrons Most elements have a few stable isotopes and several unstable, radioactive isotopes Oxygen has three stable isotopes that can be found in nature (oxygen-16, oxygen-17, and oxygen-18) and eight radioactive isotopes Hydrogen has two stable isotopes (hydrogen-1 and hydrogen-2) and single radioactive isotope (hydrogen-3) ELO 1.2

Figure: Isotopes of Hydrogen Isotopes of hydrogen are unique in that they are each commonly referred to by unique name instead of common chemical element name Hydrogen-1 – almost always referred to as hydrogen, but also called protium Hydrogen-2 – commonly called deuterium 1 2 𝐷 Hydrogen-3 – commonly called tritium 1 3 𝑇 Figure: Isotopes of Hydrogen ELO 1.2

Nuclides Knowledge Check State name of element and number of protons, electrons, and neutrons in nuclides listed below: Nuclide Element Protons Electrons Neutrons 1 1 𝐻 Hydrogen 1 5 10 𝐵 Boron 5 7 14 𝑁 Nitrogen 7 48 114 𝐶𝑑 Cadmium 48 66 94 239 𝑃𝑢 Plutonium 94 145 Nuclide Element Protons Electrons Neutrons 1 1 𝐻 5 10 𝐵 7 14 𝑁 48 114 𝐶𝑑 94 239 𝑃𝑢 Nuclide Element Protons Electrons Neutrons 1 1 𝐻 Hydrogen 5 10 𝐵 Boron 7 14 𝑁 Nitrogen 48 114 𝐶𝑑 Cadmium 94 239 𝑃𝑢 Plutonium 1 1 5 5 5 7 7 7 48 48 66 94 94 145 ELO 1.2

Forces Acting in the Nucleus ELO 1.3 – Describe the three forces that act on particles within the nucleus and how they affect the stability of the nucleus. Both protons and neutrons exist in an atom’s nucleus Some attractive force must exist to oppose the repulsive force between protons Forces present in the nucleus are: Gravitational forces between any two objects that have mass Electrostatic forces between charged particles Nuclear forces between nucleons (protons and neutrons) The magnitude of gravitational and electrostatic forces can be calculated based upon principles from classical physics ELO 1.3

Gravitational Force Gravitational force between two bodies is directly proportional to masses of two bodies and inversely proportional to square of distance between bodies (Newton) 𝐹 𝑔 = 𝐺 𝑚 1 𝑚 2 𝑟 2 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) ELO 1.3

Gravitational Force Greater gravitational force is because of either Larger masses Smaller distance between objects Distance between nucleons is extremely short Makes gravitational force significant Gravitational force between two protons separated by 10-20 meters is about 10-24 newtons ELO 1.3

Electrostatic Force Coulomb's Law used to calculate electrostatic force between two protons Electrostatic force is directly proportional to electrical charges of two particles and inversely proportional to square of distance between particles 𝐹 𝑒 = 𝐾 𝑄 1 𝑄 2 𝑟 2 Where: Fe = electrostatic force (newtons) K = electrostatic constant (9.0 x 109 N-m2/C2) Q1 = charge of first particle (coulombs) Q2 = charge of second particle (coulombs) r = distance between particles (meters) ELO 1.3

Electrostatic Force Electrostatic force between two protons separated by distance of 10- 20 meters is ≈ 1012 newtons Results: Electrostatic force – 1012 newtons Gravitational force – 10-24 newtons Gravitational force is so small that it can be neglected ELO 1.3

Nuclear Force If only electrostatic and gravitational forces exist stable nuclei composed of protons and neutrons can’t exist Must be some other force at work, specifically nuclear force Nuclear Force – strong attractive force that is independent of charge Acts equally only between pairs of neutrons, pairs of protons, or a neutron and a proton Has very short range - approximately equal to diameter of nucleus (10-13 cm) Attractive nuclear force drops off with distance much more quickly than electrostatic force ELO 1.3

Forces Acting in the Nucleus   Force Interaction Range 1. Gravitational Weak attractive force between all nucleons Relatively long 2. Electrostatic Strong repulsive force between like charged particles (protons) 3. Nuclear Force Strong attractive force between all nucleons Extremely short ELO 1.3

Nuclear Force In stable atoms, attractive and repulsive forces in nucleus balance If forces do not balance, the atom is unstable Nucleus will emit radiation in an attempt to achieve more stable configuration ELO 1.3

Forces Acting in the Nucleus Knowledge Check Very weak attractive force between all nucleons describes which of the forces listed below? Electrostatic Nuclear Gravitational Atomic Correct answer is C. Correct answer is C. ELO 1.3

Chart of the Nuclides TLO 2 – Use the Chart of the Nuclides to obtain information on specific nuclides. 2.1 Describe the information for stable and radioactive isotopes found on the Chart of the Nuclides. 2.2 Describe how an element’s neutron to proton ratio affects its stability. 2.3 Explain the difference between Atom percent, Atomic weight and Weight percent; and given the atom percent and atomic masses for isotopes of a particular element, calculate the atomic weight of the element. 2.4 Describe the following terms: Enriched uranium Depleted uranium Ensure students have copies of the Chart of the Nuclides TLO 2

Chart of the Nuclides ELO 2.1 – List information found on the Chart of the Nuclides for isotopes and describe how stable and radioactive isotopes are identified on the Chart of the Nuclides. Chart of the Nuclides is a two-dimensional graph plotting Number of neutrons on one axis Number of protons on the other axis Each point plotted on the graph represents a nuclide Provides a map of the radioactive behavior of isotopes of the elements Contrasts with a periodic table – maps only chemical behavior ELO 2.1

Figure: Nuclide Chart for Atomic Numbers Chart of the Nuclides The Chart of the Nuclides lists stable and unstable nuclides in addition to pertinent information about each one Chart plots box for each nuclide, with number of protons (Z) on vertical axis and number of neutrons (N = A - Z) on horizontal axis Use the complete Chart of the Nuclides and have student explore the chart for available information. Quiz class for nuclides of interest. Figure: Nuclide Chart for Atomic Numbers ELO 2.1

Figure: Stable Nuclide Chart of the Nuclides Only 287 isotopes are stable or naturally occurring Stable isotopes are listed in gray boxes and includes: Chemical Symbol Number of Nucleons % abundance in nature Isotopic mass Capture cross sections in barns Indication if it is a fission product Figure: Stable Nuclide ELO 2.1

Chart of the Nuclides Unstable nuclides are white or color boxes outside of the line of stability, these boxes contain: Chemical symbol Number of nucleons Half-life of the nuclide Mode and energy of decay (in MeV) (β-,α) Beta disintegration energy in MeV. Mass in amu when available Isomeric states Indication if it is a fission product Figure: Unstable Nuclide Figure: Line of Stability ELO 2.1

Chart of the Nuclides Knowledge Check On the Chart of the Nuclides, a stable isotope is indicated by a . white square gray square red square black square Correct answer is B. Correct answer is B. Gray boxes denote naturally occurring and stable nuclides ELO 2.1

Neutron-Proton Ratio ELO 2.2 – Describe how an element’s neutron to proton ratio affects its stability. Neutron-Proton ratio (N/Z ratio or nuclear ratio) – ratio of neutrons to protons making up the nucleus As mass numbers increase ratio of neutrons to protons increases Light elements up to calcium (Z=20), have stable isotopes with a neutron/proton ratio of one except: 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 N/Z ratio under one Uranium-238 has the highest N/Z ratio of any natural isotope at 1.59 (92 protons and 146 neutrons) Lead-208 the highest N/Z ratio of any known stable isotope at 1.54 ELO 2.2

Figure: Neutron-Proton Plot of the Stable Nuclides Neutron-Proton Ratio A nuclide existing outside of the band of stability can undergo: Alpha decay Positron emission Electron capture, or Beta emission to gain stability. Figure: Neutron-Proton Plot of the Stable Nuclides ELO 2.2

Figure: Neutron-Proton Plot of the Stable Nuclides Neutron-Proton Ratio Fission fragments have approximately the same neutron-to-proton ratio as heavy nucleus Places fragments below and to right of stability curve Beta-minus decays convert a neutron to a proton create a more stable neutron-to-proton ratio Figure: Neutron-Proton Plot of the Stable Nuclides ELO 2.2

Neutron-Proton Ratio Knowledge Check Which of the following nuclides has the higher neutron-proton ratio? Cobalt-60 Selenium-79 Silver-108 Cesium-137 Correct answer is D. Students may use Chart of the Nuclides for reference numbers, for this problem. Correct answer is D. ELO 2.2

Atomic Quantities ELO 2.3 – Explain the difference between atom percent, atomic weight and weight percent; given the atom percentages and atomic masses for isotopes of a particular element, calculate the atomic weight of the element. Isotopic calculations – determine relative amounts of isotopes in a given quantity of an element using: Atom percent Atomic weight Weight percent Relative abundance of an isotope in nature compared to other isotopes of same element is relatively constant Shown on chart of the nuclides for naturally occurring isotopes ELO 2.3

Atomic Quantities Atom Percent (a/o) Percentage of atoms of an element that are of a particular isotope Example: If a cup of water contains 8.23 x 1024 atoms of oxygen Isotopic abundance of oxygen-18 is 0.20% There are 1.65 x 1024 atoms of oxygen-18 in the cup Atomic Weight Average atomic weight of all isotopes of the element Calculated by summing products of isotopic abundance with atomic mass of isotope ELO 2.3

Atomic Weight Calculations Step Action 1. Determine the abundance of each isotope present (chart of the nuclides) 2. For each isotope determine the atomic mass (chart of the nuclides) 3. For each isotope multiply its abundance times its atomic mass 4. Sum the products of each isotope calculation ELO 2.3

Calculating Atomic Weight of Lithium Step Action  Calculation 1. Determine the abundance of each isotope present (chart of the nuclides) Lithium-6, 7.5%; Lithium-7, 92.5%  2. For each isotope determine the atomic mass (chart of the nuclides) Lithium-6, 6.015122 amu ; Lithium-7, 7.016003 amu 3. For each isotope, multiply abundance times atomic mass Li6, (.075)(6.015)= .4511 amu ; Li7, (.925)(7.016)= 6.4898 amu 4. Sum the products of each isotope calculation .4511 amu + 6.4898 amu = 6.9409 amu ELO 2.3

Weight Percent (w/o) Weight Percent (w/o) – percent weight of an element that is a particular isotope Example If a sample of material contains 100 kg of uranium that was 28 w/o uranium-235, then 28 kg of uranium-235 is in the sample ELO 2.3

Atomic Quantities Knowledge Check Calculate the atomic weight for the element silver with the following stable isotopes: Ag-107, abundance 51.84%, Mass 106.905097 amu Ag-109, abundance 48.16%, Mass 108.904752 amu 107.8681 amu 1078.6813 amu 2907.8109 amu 2.9507 amu Correct answer is A. Have students individually solve this calculation and compare results. Correct answer is A. ELO 2.3

Enrichment and Depletion ELO 2.4 – Describe the following terms: enriched uranium and depleted uranium. Natural uranium contains isotopes of: 99.2745% of uranium-238 0.72% uranium-235 0.0055% uranium-234 All isotopes have similar chemical properties But, each isotope has significantly different nuclear properties Uranium-235 is the desired material for use in reactors ELO 2.4

Enrichment and Depletion Enrichment (separating isotopes from natural quantities) is complex and expensive. For PWRs the Uranium-235 isotope must be “enriched” from natural uranium. This results in: Enriched uranium – uranium having uranium-235 isotope concentration greater than its natural value Depleted uranium – uranium having uranium-235 isotope concentration less than its natural value (0.72%) Depleted uranium is a by-product of the uranium enrichment process It has uses in both commercial and defense industries ELO 2.4

Enrichment and Depletion Knowledge Check Depleted uranium will have ___________ atomic weight than natural uranium. less the same amount greater much less Correct answer is C. Correct answer is C. ELO 2.4

Mass Defect and Binding Energy TLO 3 – Describe mass defect and binding energy and their relationship to one another. 3.1 Define mass defect and binding energy. 3.2 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. 3.3 Explain the difference between an x-ray and a gamma ray and their effects to the atom. Include an explanation for ionization, ionization energy, nucleus energy and application of the nuclear energy level diagram. Ensure students have copies of the Chart of the Nuclides and calculators 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 TLO 3

Mass Defect and Binding Energy ELO 3.1 – Define mass defect and binding energy. Laws of conservation of mass and conservation of energy continue to hold true on a nuclear level However, conversion between mass and energy occurs (E=MC2) A mass decrease results in an corresponding energy increase and vice-versa Mass does not magically appear and disappear at random The total mass and corresponding energy remains constant ELO 3.1

Mass Defect and Binding Energy Based on measurement, mass of a particular atom is always slightly less than sum of masses of individual neutrons, protons, and electrons Mass Defect (∆m) – difference between mass of atom and sum of masses of its parts Binding Energy Loss in mass, or mass defect, is due to conversion of mass to binding energy when nucleus is formed Binding Energy (BE) – amount of energy that must be supplied to nucleus to completely separate its nuclear particles (nucleons) Also defined as amount of energy that would be released if nucleus was formed from separate particles ELO 3.1

Mass Defect and Binding Energy Knowledge Check _______________ is the amount of energy that must be supplied to a nucleus to completely separate its nuclear particles. Nuclear energy Binding energy Mass defect Separation energy Correct answer is B. Correct answer is B. ELO 3.1

Calculating Mass Defect and Binding Energy ELO 3.2 – 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. Mass defect and binding energy are calculated from the following equations Always use the full accuracy of mass measurements because these differences are very small compared to the atom’s mass Rounding off to three or four significant digits prior results in a calculated mass defect of zero ELO 3.2

Mass Defect Calculations ∆𝑚= 𝑍 𝑚 𝑝 + 𝑚 𝑒 + 𝐴−𝑍 𝑚 𝑛 − 𝑚 𝑎𝑡𝑜𝑚 Where: ∆m = 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 (amu) Z = atomic number (number of protons) A = mass number (number of nucleons) ELO 3.2

Mass Defect Calculations 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( m p + m e ) 3. Determine the mass of neutrons. Subtract the atomic number (Z) from the atomic mass (A) then multiply by mass of a neutron: 𝐴−𝑍 m n 4. Add the mass of the protons, electrons and neutrons. Add the products determined in the previous two steps: 𝑍 m p + m e + 𝐴−𝑍 m n 5. Determine the difference between the mass of the nuclide and the mass of individual components. Subtract the mass of the atom of the nuclide: 𝑍 m p + m e + 𝐴 − 𝑍 m n − m atom ELO 3.2

Mass Defect Calculations Calculate the mass defect for lithium-7 given lithium-7 = 7.016003 amu. 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 (1.007276467𝑎𝑚𝑢 +.000548597 𝑎𝑚𝑢) =3.023475192 𝑎𝑚𝑢 3. Determine the mass of the neutrons. (7−3)(1.008665)=4.03466 𝑎𝑚𝑢 4. Add mass of protons, electrons, and neutrons. 3.023475192 𝑎𝑚𝑢+4.03466 𝑎𝑚𝑢 =7.0581352 𝑎𝑚𝑢 5. Determine the difference between the atomic mass of the nuclide and the mass of the individual components. 7.0581352 𝑎𝑚𝑢−7.016003 𝑎𝑚𝑢 =.042132 𝑎𝑚𝑢 ELO 3.2

Binding Energy Calculation Binding energy is energy equivalent of the mass defect Calculated using a conversion factor derived from Einstein's Theory of Relativity Einstein's Theory of Relativity is the famous equation relating mass and energy 𝐸=𝑚𝑐2 Where: c = velocity of light (𝑐=2.998× 10 8 𝑚 sec ) m = mass in amu ELO 3.2

Binding Energy 𝐸=𝑚 𝑐 2 =1 𝑎𝑚𝑢 1.6605× 10 −27 𝑘𝑔 1 𝑎𝑚𝑢 2.998× 10 8 𝑚 𝑠𝑒𝑐 1𝑁 1 𝑘𝑔–𝑚 𝑠𝑒 𝑐 2 1𝐽 1𝑁−𝑚 =1.4924× 10 −10 𝐽 1 𝑀𝑒𝑉 1.6022× 10 −13 𝐽 =931.5 𝑀𝑒𝑉 Conversion Factors: 1 𝑎𝑚𝑢=1,6605× 10 −27 𝑘𝑔 1 𝑛𝑒𝑤𝑡𝑜𝑛=1 𝑘𝑔–𝑚/sec2 1 𝑗𝑜𝑢𝑙𝑒=1 𝑛𝑒𝑤𝑡𝑜𝑛–𝑚𝑒𝑡𝑒𝑟 1 𝑀𝑒𝑉=1.6022× 10 −13 𝑗𝑜𝑢𝑙𝑒𝑠 ELO 3.2

Binding Energy Since 1 amu is equivalent to 931.5 MeV of energy, binding energy can be calculated using: 𝐵.𝐸.=∆𝑚 931.5 𝑀𝑒𝑉 1 𝑎𝑚𝑢 ELO 3.2

Binding Energy Calculation Step Description Action 1. Determine the mass defect of the nuclide  Use the equation:  ∆𝑚= 𝑍 𝑚 𝑝 + 𝑚 𝑒 + 𝐴−𝑍 𝑚 𝑛 − 𝑚 𝑎𝑡𝑜𝑚 2. Use the binding energy equation to calculate the binding energy. Use the Equation: 𝐵.𝐸.=∆𝑚 931.5 𝑀𝑒𝑉 1 𝑎𝑚𝑢 3. Calculate the binding energy Multiply delta mass from step 1 by the energy conversion to amu ELO 3.2

Binding Energy Calculation Step Description Action 1. Determine the mass defect of the nuclide  From previous calculation:  .04135 amu. 2. Use the binding energy equation to calculate the binding energy. 𝐵.𝐸.=∆𝑚 931.5 𝑀𝑒𝑉 1 𝑎𝑚𝑢 3. Calculate the binding energy B.E. =0.04135 amu 931.5 MeV 1 amu =38.5175 𝑀𝑒𝑉 ELO 3.2

Binding Energy Example Calculate the mass defect and binding energy for uranium-235. One uranium-235 atom has a mass of 235.043924 amu. Step 1: ∆𝑚= 𝑍 𝑚 𝑝 + 𝑚 𝑒 + 𝐴−𝑍 𝑚 𝑛 − 𝑚 𝑎𝑡𝑜𝑚 ∆𝑚= 92 1.007826 𝑎𝑚𝑢)+ 235−92 1.008665 𝑎𝑚𝑢 − 235.043924 𝑎𝑚𝑢 ∆𝑚=1.91517 𝑎𝑚𝑢 Step 2: 𝐵.𝐸.=1.91517 𝑎𝑚𝑢 931.5 𝑀𝑒𝑉 1 𝑎𝑚𝑢 𝐵.𝐸. =1784 𝑀𝑒𝑉 Have students individually work these calculations and compare answers. ELO 3.2

Gamma Rays and X-Rays ELO 3.3 – Explain the difference between an x-ray and a gamma ray and their effects to the atom. Include an explanation for ionization, ionization energy, nucleus energy, and application of the nuclear energy level diagram. Defined by their sources – identified by wavelength X-rays are emitted by electrons gamma rays are emitted from the nucleus – shorter wavelength They are both photons and undergo similar interactions Electrons that circle the nucleus move in fairly well-defined orbits Some of these electrons are more tightly bound in atom than others. ELO 3.3

Gamma Rays and X-Rays Electrons are attracted to the positive charge of the protons in the nucleus As they orbit further from the nucleus this attraction weakens Therefore, less energy is required to remove it Removing an electron from its orbit is called ionization The energy required for ionization is called ionization energy Only 7.38 eV is required to remove outermost electron from lead atom 88,000 eV is required to remove innermost electron ELO 3.3

Energy Levels of Atoms Ground State In a neutral atom (number of electrons = Z) it is possible for electrons to be in a variety of different orbits, each with a different energy level Ground State – state of lowest energy, the atom is normally found in Excited State Atom with more energy than ground state, is in an excited state An atom cannot stay in excited state for an indefinite period of time X-Ray Production X-Ray – a discrete bundle of electromagnetic energy emitted when an excited atom transitions to a lower-energy excited or ground state X-ray energy is equal to difference between energy levels of atom Range from several eV to 100,000 eV ELO 3.3

Energy Levels of the Nucleus Nucleons in nucleus of atom, like electrons that circle nucleus, exist in shells that correspond to energy states Energy shells of nucleus are less defined and less understood than those of electrons There is a state of lowest energy (ground state) and discrete possible excited states for a nucleus Discrete energy states for the electrons of atom are measured in eV or keV Energy levels within the nucleus are considerably greater, measured in MeV ELO 3.3

Energy Levels of the Nucleus Gamma Ray Production Nucleus in excited state will not remain for an indefinite period Nucleons in an excited nucleus transition towards their lowest energy configuration and emit a discrete bundle of electromagnetic radiation called a gamma ray (ˠ) Differences between x-rays and ˠ-rays are Energy levels Where they originate from X-rays – electron shell ˠ-rays – nucleus Nuclear Energy Level Diagram Nuclear energy-level diagram depicts ground and excited states Consists of stack of horizontal bars, one bar for each of excited states of nucleus ELO 3.3

Energy Levels of the Nucleus Vertical distance between bars represents an excited state with the bottom bar representing ground state Difference in energy between ground state and excited state is called excitation energy of excited state Ground state of nuclide has zero excitation energy Bars for excited states are labeled with their respective energy levels Figure: Energy Level Diagram – Nickel-60 ELO 3.3

Gamma Rays and X-Rays Knowledge Check Uranium-238 would be "stable" with _____ electrons orbiting the nucleus. 238 146 235 92 Correct answer is D. Correct Answer is D. 92 (same as # of protons) ELO 3.3

Nuclear Stability TLO 4 – Describe the processes by which unstable nuclides achieve stability. 4.1 Describe the conservation principles that must be observed during radioactive decay. Include an explanation of neutrinos. 4.2 Describe the following radioactive decay processes: Alpha decay Beta-minus decay Beta-plus decay Electron capture Gamma ray emission Internal conversions Isomeric transitions Neutron emission 4.3 Given the stability curve on the Chart of the Nuclides, determine the type of radioactive decay that the nuclides in each region of the chart will typically undergo. 4.4 Given a Chart of the Nuclides, describe the radioactive decay chain for a nuclide. Most naturally occurring atoms are stable and do not emit particles or energy or change state Some atoms are not stable These atoms emit radiation to achieve more stability Important to understand nuclear stability because of effects that unstable nuclei undergo to become stable Unstable nuclides achieve stability by emitting: High energy photons High energy particles TLO 4

Conservation Principles ELO 4.1 – Describe the conservation principles that must be observed during radioactive decay. Include an explanation of neutrinos. For stable nuclides, as the mass number increases the ratio of neutrons to protons increases Non-stable nuclei with an excess of neutrons undergo a transformation process known as beta (β) decay A deficiency of neutrons undergo other processes such as electron capture or positron emission Final nucleus is more stable as a result of the decay processes Ensure students have copies of the Chart of the Nuclides Some naturally occurring heavy elements, such as uranium or thorium, and their unstable decay chain elements emit radiation Uranium and thorium, have an extremely slow rate of decay All naturally occurring nuclides with atomic numbers greater than 82 are radioactive ELO 4.1

Conservation Principles Description Conservation of Electric Charge Conservation of electric charge implies that charges are neither created nor destroyed. Single positive and negative charges may neutralize each other. Possible for a neutral particle to produce one charge of each sign. Conservation of Mass Number Conservation of mass number does not allow a net change in the number of nucleons. However, the conversion of one type of nucleon to another type (proton to a neutron and vice versa) is allowed. ELO 4.1

Conservation Principles Description Conservation of Mass and Energy Conservation of mass and energy implies that the total of the kinetic energy and the energy equivalent of the mass in a system must be conserved in all decays and reactions. Mass can be converted to energy and energy can be converted to mass, but the sum of mass and mass-equivalent energy must be constant. Conservation of Momentum Conservation of momentum is responsible for the distribution of the available kinetic energy among product nuclei, particles, and/or radiation. The total amount is the same before and after the reaction even though it might be distributed differently among entirely different nuclides and/or particles. ELO 4.1

Conservation Principles Example – Xenon-135 It’s a radioactive isotope at an excited state It decays by emission of beta particle, electron or positron resulting in a neutron converting to a proton Illustrates the conservation of mass and energy The beta ejected from the nucleus no longer contributes to the atomic mass of the resultant isotope Although no longer in the nucleus, the beta particle accounts for any mass difference between the proton and neutron ELO 4.1

Conservation Principles Example – Xenon-135 Xenon-135 Decay Resultant Isotope and Energy 134.90720 amu 54 protons 81 neutrons Cesium-135 134.905977 amu 55 protons 80 neutrons ∆Mass = .001235 amu Mass is accounted for in the beta particle and energy of the gammas emitted. ELO 4.1

Conservation Principles Knowledge Check "Charges are neither created nor destroyed." Describes which of the following conservation principle? of mass of electrical charge of momentum of thermal energy Correct answer is B. Correct answer is B. ELO 4.1

Decay Processes ELO 4.2 – Describe the following radioactive decay processes: alpha decay, beta-minus decay, beta-plus decay, electron capture, gamma ray emission, internal conversions, isomeric transitions, and neutron emission. To attain stability nuclei emit radiation by a spontaneous disintegration process known as radioactive decay or nuclear decay This radiation may be electromagnetic radiation, particles, or both These are explained in this section ELO 4.2

Alpha Decay (α) Emission of alpha particles (helium nuclei) which may be represented as: 2 4 𝐻𝑒 or 2 4 𝛼 When an unstable nucleus ejects an alpha particle, atomic number is reduced by 2 and mass number decreased by 4 Uranium-234 decays by ejection of an alpha particle accompanied by emission of 0.068 MeV gamma 92 234 𝑈 → 90 230 𝑇ℎ+ 2 4 𝛼+𝛾+𝐾𝐸 Figure: Alpha Decay ELO 4.2

Alpha Decay (α) Combined KE of daughter nucleus (Thorium-230) and α particle is designated as “KE” in equation Sum of KE and gamma energy is equal to difference in mass between original nucleus and final particles Equivalent to the binding energy released, since Δm = BE Alpha particle will carry off as much as 98% of KE 92 234 𝑈 → 90 230 𝑇ℎ+ 2 4 𝛼+𝛾+𝐾𝐸 ELO 4.2

Beta Decay (β) Emission of electrons of nuclear rather than orbital origin These particles are electrons that have been expelled by excited nuclei May have a charge of either sign β- β+ ELO 4.2

Figure: Beta Minus Decay Negative electron emission, converts neutron to proton, increasing atomic number by one and leaving mass number unchanged Common mode of decay for nuclei with excess of neutrons, such as fission fragments below and to right of neutron-proton stability curve 93 239 𝑁𝑝 → 94 239 𝑃𝑢 + −1 0 𝛽 + 0 0 𝑣 The symbol on the end represents an anti-neutrino More information on neutrinos is provided later. Figure: Beta Minus Decay ELO 4.2

Beta Plus Decay Positively charged electrons (beta-plus) are known as positrons +1 0 𝑒 𝑜𝑟 +1 0 𝛽 𝑒 + 𝑜𝑟 𝛽 + Except for the charge they are nearly identical to their negative Betas. Positron emission decreases the atomic number by one but leaves the mass number unchanged by changing a proton to a neutron 7 13 𝑁 → 6 13 𝐶 + +1 0 𝛽 + 0 0 𝑣 More information on neutrinos is provided later. ELO 4.2

Electron Capture (EC, K-capture) Nuclei having excess protons may capture an inner orbit electron that immediately combines with a proton to form a neutron The electron is normally captured from the innermost orbit (K-shell), therefore, this process is also called K-capture 4 7 𝐵𝑒 + −1 0 𝑒 → 3 7 𝐿𝑖 + 0 0 𝑣 As with beta decays a neutrino is formed, its energy conserving momentum, photons are given off: Atomic mass of the product being appreciably less than the parent – gamma energy Characteristic x-rays are given off when an electron from another shell fills the vacancy in the K-shell ELO 4.2

Figure: Electron Capture or K-Capture ELO 4.2

Electron Capture vs. Positron Emission Electron capture and positron emission exist as competing processes They produce the same daughter product For positron emission to occur, the mass of the daughter product must be at least two electrons less than the mass of the parent This accounts for the ejected positron and that the daughter has one less electron than the parent If these requirements are not met, then electron capture occurs and positron emission does not ELO 4.2

Gamma Emission High-energy electromagnetic radiation originating in the nucleus It is emitted in the form of photons: Discrete bundles of energy having both wave and particle properties A daughter nuclide from decay often remains in an excited state Resolved by the nucleus dropping to the ground state by the emission of gamma radiation Gamma rays are very penetrating, often requiring several inches of metal or a couple of feet of concrete to stop (shield) ELO 4.2

Internal Conversion Normally an excited nucleus goes from the excited state to the ground state by emission of a gamma ray In some cases the gamma ray released interacts with one of the innermost orbital electrons Transfers the gamma’s energy to the electron When this occurs the atom is said to be undergoing internal conversion This energized electron is ejected from the atom with KE equal to the gamma energy minus the BE of the electron An orbital electron then drops to a lower energy state to fill the vacancy with the emission of x-rays ELO 4.2

Isomeric Transition Nuclear isomer – a nucleus in an excited state, differs in energy and behavior from other nuclei with identical atomic and mass numbers Isomeric transition – when the excited nuclear isomer drops to a lower energy level Commonly occurs immediately after particle emission May remain in an excited state for a measurable period of time before dropping to ground state It is also possible for the excited isomer to decay by alternate means An example of delayed gamma emission accompanying Beta emission is illustrated by the decay of nitrogen-16. 7 16 𝑁 → 8 16 𝑂 + 1 0 𝛽 + 0 0 𝛾 8 16 𝑂 → 8 16 𝑂 + 0 0 𝛾 ELO 4.2

Neutron Emission Non-stable nuclei may also emit neutrons (n) in order to become more stable 35 87 𝐵𝑟 𝛽 − 55.9 𝑠𝑒𝑐 → 36 87 𝐾𝑟 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑛 → 36 86 𝐾 𝑟 𝑠𝑡𝑎𝑏𝑙𝑒 Neutrons emitted from the nucleus of a radioactive atom possess a great deal of KE Capable of penetrating many materials Neutron production and interaction with matter is of great importance in nuclear physics and will be discussed in greater detail later ELO 4.2

Neutrinos Neutrino (𝝂) – emitted with positive electron emission Antineutrino ( 𝝂 ) – emitted with negative electron (Beta decay) Pass through all materials with so few interactions that energy they possess cannot be recovered Neutrinos and antineutrinos carry portion of KE that would otherwise belong to beta particle Considered for energy and momentum to be conserved Not significant in context of nuclear reactor applications ELO 4.2

Radioactivity Decay Knowledge Check Which of the following statements accurately describes alpha decay? A neutron is converted to a proton and an electron. The electron is ejected from the nucleus. A neutron is converted to a proton and a positron. The positron is ejected from the nucleus. A particle is emitted from a nucleus containing 2 neutrons and 2 protons. A particle is emitted from a nucleus containing 2 electrons and 2 protons. Correct answer is C. Correct answer is C. A particle is emitted from a nucleus containing 2 neutrons and 2 protons. ELO 4.2

Figure: Stability Curve ELO 4.3 – Given the stability curve on the Chart of the Nuclides, determine the type of radioactive decay that the nuclides in each region of the chart will typically undergo. Radioactive nuclides decay in a way that results in a daughter nuclide with a neutron-proton ratio closer to the line of stability on the Chart of the Nuclides Helps to predict the type of decay that a nuclide undergoes based on its location relative to the line of stability Figure: Stability Curve ELO 4.3

Predicting Type of Decay The line of stability illustrates the method of decay nuclides in different regions of the chart of nuclides are likely to undergo Nuclides below and to the right of the line of stability usually undergo β- decay Nuclides above and to the left of the line of stability usually undergo either β+ decay or electron capture. Nuclides likely to undergo alpha (α) decay are found in the upper right hand region Some exceptions apply especially in the region of heavy nuclides Stable isotopes are isotopes that are not radioactive They do not decay spontaneously Stable isotopes are found on the line of stability ELO 4.3

Predicting Type of Decay Figure: Types of Radioactive Decay Relative to Line of Stability ELO 4.3

Predicting Type of Decay – Examples Of the known elements 80 have at least one stable nuclide These are the first 82 elements from hydrogen to lead Exceptions are technetium-43 and promethium-61 There are a known total of 254 stable nuclides Stable, in this case, means a nuclide that has not been observed to decay against the natural background These elements have half-lives too long to be measured ELO 4.3

Stability Curve Knowledge Check Correct answers: C – Beta + D – beta - Match the 4 areas on the curve with the correct description: Alpha (α) decay Line of Stability β+ decay or electron capture β- decay Correct answers: C – Beta + D – beta - A – alpha B – line of stability 4 Correct Answers: C – Beta + D – beta - A – alpha B – line of stability ELO 4.3

Decay Chains ELO 4.4 – Given a Chart of the Nuclides, describe the radioactive decay chain for a nuclide. When an unstable nucleus decays the resulting daughter may not be stable Nucleus resulting from decay of parent is often itself unstable, and will undergo an additional decay(s) Common among the larger nuclides Steps of an unstable atom can be traced as it goes through multiple decays trying to achieve stability List of original unstable nuclide, nuclides involved as intermediate steps in decay, and final stable nuclide is known as decay chain Ensure students have their copies of the Chart of the Nuclides. ELO 4.4

Decay Chains Common method for stating decay chain is to state each nuclide involved in standard format Arrows used between nuclides to indicate where decays occur, with type of decay indicated above arrow and half-life below arrow Example: decay chain of U-238: U-238 decays, through alpha-emission, with a half-life of 4.5 billion years to thorium-234 Decays through beta-emission, with a half-life of 24 days to protactinium-234 Decays through beta-emission, with a half-life of 1.2 minutes to uranium-234 Decays through alpha-emission, with a half-life of 240 thousand years to thorium-230 ELO 4.4

Decay Chains Example Decay Chain of U-238 Continued: Thorium-230 decays, through alpha-emission, with a half-life of 77 thousand years to radium-226 Decays through alpha-emission, with a half-life of 1.6 thousand years to radon-222 Decays through alpha-emission, with a half-life of 3.8 days to polonium-218 Decays through alpha-emission, with a half-life of 3.1 minutes to lead-214 Decays through beta-emission, with a half-life of 27 minutes to bismuth-214 ELO 4.4

Decay Chains Example Decay Chain of U-238 Continued: Bismuth-214 decays through beta-emission, with a half-life of 20 minutes to polonium-214 Decays through alpha-emission, with a half-life of 160 microseconds to lead-210 Decays through beta-emission, with a half-life of 22 years to bismuth- 210 Decays through beta-emission, with a half-life of 5 days to polonium- 210 Decays through alpha-emission, with a half-life of 140 days to lead- 206, which is a stable nuclide. ELO 4.4

Decay Chains – Example Use chart of the nuclides and write decay chains for rubidium-91 and actinium-215 Continue chains until stable nuclide or nuclide with half-life greater than 1 x 106 years is reached 91 37 𝑅𝑏 𝛽 → 58.0 𝑠 91 38 𝑆𝑟 𝛽 → 9.5 ℎ𝑟𝑠 91 39 𝑌 𝛽 → 58.5 𝑑 91 40 𝑍𝑟 215 85 𝐴𝑡 𝛼 → 0.10 𝑚𝑠 211 83 𝐵𝑖 𝛼 → 2.14 𝑚𝑖𝑛 207 81 𝑇𝑙 𝛽 → 4.77 𝑚𝑖𝑛 207 82 𝑃𝑡 ELO 4.4

Emitted Radiation TLO 5 – Describe how radiation emitted by an unstable nuclide interacts with matter and materials typically used to shield against this radiation 5.1 Describe the difference between charged and uncharged particle interaction with matter. Include an explanation of specific ionization. 5.2 Describe radioactive interactions of the following types with matter: Alpha Particle Beta Particle Gamma Positron Neutron 5.3 Describe the type of material that can be used to stop (shield) the following types of radiation: Alpha particle Beta-particle Gamma ray Radiation is comprised of photons (energy waves) and particles that originate in either the nucleus or electron shells of atoms Photons and radiation particles have energy and interact with matter, transferring their energy Radiation reactions with matter are dependent on photon and particle: Mass Energy It is important to understand the physical properties of the radiation to understand the potential hazards and methods for protection TLO 5

Charged Versus Uncharged Particles ELO 5.1 – Describe the difference between charged and uncharged particle interaction with matter. Include an explanation of specific ionization. Interactions with matter vary with the different types of radiation Large, massive, charged alpha particles have very limited penetration capabilities Neutrinos, the other extreme, have a very low probability of interacting with matter, a large penetrating capability Charged vs. Uncharged Particles Radiation is classified into two groups, charged and uncharged Charged particles directly ionize the media through which they pass Uncharged particles and photons only cause ionization indirectly or by secondary radiation ELO 5.1

Charged Versus Uncharged Particles Charged Particle Interaction Charged particles have surrounding electrical fields that interact with the atomic structure of the material they are traveling through Slows the particle and accelerates electrons in the medium’s atoms The accelerated electrons acquire enough energy to escape from their parent atoms causing ionization of the affected atom Uncharged Particle Interaction Uncharged moving particles have no electrical field Only lose energy and cause ionization by collisions or scattering Photon can lose energy by: Photoelectric effect Compton Scattering Pair production ELO 5.1

Specific Ionization Ionizing radiation creates ion-pairs (+ and – charged) Specific ionization is defined as: Number of ion-pairs formed per centimeter travel in given material Specific ionization, the measure of radiation’s ionization power is: Roughly proportional to the particle's mass Square of its charge 𝐼= 𝑚 𝑧 2 𝐾.𝐸. Where: I = ionizing power m = mass of particle z = number of unit charges particle carries K.E. = kinetic energy of particle ELO 5.1

Specific Ionization Since mass for α particle is ≈ 7300 times as large as m for β particle, and z is twice as great (+2) an α will produce much more ionization per unit path length than β of same energy Larger alpha particle moves slower for given energy and thus acts on given electron for a longer time 𝐼= 𝑚 𝑧 2 𝐾.𝐸. Where: I = ionizing power m = mass of particle z = number of unit charges particle carries K.E. = kinetic energy of particle ELO 5.1

Charged Versus Uncharged Particles Knowledge Check Charged particles ionize the media they pass through. directly indirectly never Sometimes Correct answer is A. Correct Answer is A. ELO 5.1

Radioactive Interaction ELO 5.2 – Describe radioactive interactions of the following types with matter: alpha particle, beta particle, gamma, positron, and neutron. How radiation reacts with matter is dependent on the type of radiation The following types of radiation interact with matter in a specific predictable manner: Alpha particle Beta particle Gamma Positron Neutron ELO 5.2

Alpha Radiation Origin Interaction Produced from the radioactive decay of heavy nuclides and certain nuclear reactions Consists of 2 neutrons and 2 protons, same as a helium atom With no electrons, the alpha particle has a charge of +2 This strong positive charge strips electrons from the orbits of atoms Has a high specific ionization Removes electrons from the atoms it passes near Removal of electrons requires energy Alpha’s energy is reduced by each reaction Ultimately, the alpha particle expends its KE, gains 2 electrons, and becomes a helium atom Low penetration power ELO 5.2

Beta Minus Interaction Origin Interaction A beta-minus particle is an electron that has been ejected at a high velocity from an unstable nucleus Electrons have a small mass and an electrical charge of -1 Beta particles cause ionization by displacing electrons from atomic orbits Beta-minus ionization occurs from collisions with orbiting electrons. Each collision removes KE from the beta particle, causing it to slow down After a few collisions the beta particle is slowed enough to allow it to be captured as an orbiting electron in an atom ELO 5.2

Positron Radiation Origin Interaction Positively charged electrons Identical to beta- minus particles and interact with matter similarly Positrons are very short-lived and quickly annihilate via interactions with negatively charged electrons Produces two gammas with energy equal to the rest mass of the electrons (1.02 MeV) These gammas interact with matter via photoelectric effect, Compton scattering or pair-production, 2 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 0.000549 𝑎𝑚𝑢 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 931.5 𝑀𝑒𝑉 𝑎𝑚𝑢 =1.02 𝑀𝑒𝑉 Photoelectric effect, Compton scattering or pair-production covered later. ELO 5.2

Neutron Interactions Origin Interaction No electrical charge Same mass as a proton 1835 times more mass than electron 1/4 mass of an alpha particle Come primarily from nuclear reactions, such as fission, but also from decay of radioactive nuclides With no charge, the neutron has a high penetrating power Neutrons are attenuated (reduced in energy and numbers) by three major interactions: Elastic scatter Inelastic scatter Absorption Photoelectric effect, Compton scattering or pair-production covered later. ELO 5.2

Neutron Interactions Elastic Scatter Neutron collides with a nucleus and bounces off Some of the neutron’s KE is transferred to the nucleus Results in the neutron slowed and atom gaining KE Often referred to as the billiard ball effect Inelastic Scatter Same neutron/nucleus collision occurs as in elastic scatter Nucleus receives some internal energy as well as KE Slows the neutron and leaves the nucleus in an excited state Nucleus decays to original energy level and usually emits a gamma ray The gamma ray goes on to interact with matter via photoelectric effect, compton scattering or pair-production, as described later ELO 5.2

Neutron Interactions Absorption The neutron is absorbed and captured into the nucleus of an atom Atom is left in an excited state Called radiative capture if the nucleus emits one or more gamma rays to reach a stable level More probable at lower energy levels Gammas rays interact via the photoelectric effect, compton scattering or pair-production May also result in nuclear fission splitting the atom into two smaller atoms, a couple of neutrons and gamma rays Fission fragments may create additional neutrons or gamma radiation as they decay to stability ELO 5.2

Gamma Interactions Gamma Radiation Electromagnetic radiation - similar to x-ray Produced by decay of excited nuclei and by nuclear reactions Has no mass and no charge Difficult to stop and has very high penetrating power Requires several feet of concrete, several meters of water or a few inches of lead to shield Three methods of attenuating gamma rays: Photo-electric effect Compton scattering Pair-production ELO 5.2

Figure: Photoelectric Effect Gamma Interactions Photo-Electric Effect Occurs when a low energy gamma strikes an orbital electron Total energy of the gamma is expended in ejecting the electron from its orbit Atom is ionized and a high energy electron is ejected Rarely occurs with gammas having energy above 1 MeV Energy in excess of BE of electron is carried off by electron in form of KE Figure: Photoelectric Effect ELO 5.2

Figure: Compton Scattering Gamma Interactions Compton Scattering An elastic collision between an electron and a photon.  Photon has more energy than required to eject the electron from orbit, or is unable to give up all of its energy with an electron.  Since all gamma energy is not transferred, the photon must scatter. Scattered photon has less energy.  Result is ionization of the atom, a high energy beta, and a reduced energy gamma. Most predominant with gamma energy level of 1.0 to 2.0 MeV. Figure: Compton Scattering ELO 5.2

Figure: Pair Production Gamma Interactions Pair Production At higher energy levels the most likely gamma interaction A high energy gamma passes close to a heavy nucleus and disappears into an electron and positron This transformation must take place near a particle, such as a nucleus, to conserve momentum KE of the recoiling nucleus is very small; all of the photon’s energy in excess 1.02 MeV appears as KE of the pair produced The original gamma must have at least 1.02 MeV energy Electron and positron may collide and annihilate each other Figure: Pair Production ELO 5.2

Radioactive Interactions Knowledge Check Which of the following is NOT a method by which neutrons interact with matter? Inelastic scattering Elastic scattering Ionization Absorption Correct answer is C. Correct Answer is C. ELO 5.2

Shielding ELO 5.3 – Describe the type of material that can be used to stop (shield) the following types of radiation: alpha particle, beta-particle, neutron, and gamma ray. Shielding describes material placed around a radiation source used to attenuate the radiation level Effectiveness is dependent on the material used and the type of radiation Attenuation is the gradual loss in intensity of any kind of radiation flux through a medium Examples: Sunlight is attenuated by dark glasses X-rays are attenuated by lead Neutrons are attenuated by water ELO 5.3

Shielding Properties Type of Radiation Shielding Material Alpha With a strong positive charge and large mass, the alpha particle deposits a large amount of energy in a short distance. Loses energy quickly and therefore has very limited penetrating power. Alpha particles are stopped in a few centimeters of air or a sheet of paper. Beta Particle More penetrating than alpha but still relatively easy to stop Low power of penetration. The most energetic beta radiation is stopped by thin metal. ELO 5.3

Shielding Properties Type of Radiation Shielding Material Neutron With no charge is difficult to stop – has high penetrating power. Attenuated by: Elastic scattering  Inelastic scattering Absorption Most effective shield for neutrons is a material with similar mass for elastic scattering. Hydrogenous material such as water attenuates neutrons effectively. 12 inches of water is an effective shield. ELO 5.3

Shielding Properties Type of Radiation Shielding Material Gamma Ray With no mass and no charge, difficult to stop – very high penetrating power. Heavy nuclei (i.e. lead) provide large targets for gamma interactions (3 types). Although dependent on gamma energies, several meters of concrete or water or a few inches of lead are effective shielding materials. ELO 5.3

Figure: Effects Various Materials Have on Types of Radiation Shielding Properties Alpha, Beta, and Gamma Shielding Shielding thickness are referred to as 1/2 thickness or 1/10th thickness. These thicknesses are the amount of material required to reduce the original radiation field strength to 1/2 or 1/10th respectively For example: The 1/2 thickness of lead for gammas is 0.4" ˠ Figure: Effects Various Materials Have on Types of Radiation ELO 5.3

Shielding Properties Knowledge Check Which of the following materials would provide the best shielding against neutrons? Water Lead Paper Thin sheet of steel Correct answer is A. Correct Answer is A. ELO 5.3

Radioactive Decay TLO 6 – Describe radioactive decay terms and calculate activity levels, half-lives, decay constants and radioactive equilibrium. 6.1 Describe the following radioactive terms: Radioactivity Radioactive decay constant Activity Curie Becquerel Radioactive half-life 6.2 Convert between the half-life and decay constant for a nuclide. 6.3 Given the nuclide, number of atoms, half-life or decay constant, Determine current and future activity levels. 6.4 Describe the following: Radioactive equilibrium Transient radioactive equilibrium Secular radioactive equilibrium A decay rate of sample of radioactive material is not constant As individual atoms of the material decay, there are fewer of them remaining Rate of decay is directly proportional to the number of atoms The rate of decay decreases as the number of atoms decreases Knowledge of radioactive decay is important for calculating reactivity poisons in the reactor as well as understanding personnel hazards TLO 6

Radioactive Decay Terms ELO 6.1 – Describe the following radioactive terms: radioactivity, radioactive decay constant, activity, Curie, Becquerel, and radioactive half-life. To understand radioactive decay, knowledge of the terms used to describing decay rate relationships is important The following terms are described: Radioactivity Radioactive decay constant Activity Curie Becquerel Radioactive half-life ELO 6.1

Radioactive Decay Terms Radioactivity The process of certain nuclides spontaneously emitting particles or gamma radiation Occurs randomly – cannot be predicted However, the average behavior of a large sample can be accurately determined using statistical methods Radioactive Decay Constant (λ) In a given time interval a specific fraction of a given nuclei in a sample will decay Probability per unit time that an atom of a specific nuclide will decay is - the radioactive decay constant, λ (lambda) Units are inverse times such as 1/second, 1/minute, 1/hour, or 1/year Expressed as second-1, minute-1, hour-1, and year-1 ELO 6.1

Radioactive Decay Terms Activity Activity (A) is the rate of decay of a sample Measured by the number of disintegrations occurring per second A sample containing millions of atoms, activity is the product of the decay constant (λ) and number of atoms present in the sample (N) 𝐴=𝜆𝑁 Where: A = Activity of the nuclide (disintegrations/second) λ = Decay constant of the nuclide (second-1) N = Number of atoms of the nuclide in the sample ELO 6.1

Radioactive Decay Terms Radioactive Half-Life Commonly used term Estimates how quickly a nuclide is decaying Defined as the amount of time required for activity to decrease to one-half of its original value ELO 6.1

Units of Measurement for Radioactivity Two common units to measure activity are: Curie Ci, US Measurement Measures the rate of radioactive decay - activity Equal to 3.7 x 1010 disintegrations per second Approximately equivalent to the number of disintegrations that one gram of radium-226 will undergo in one second Becquerel Bq, Metric System A Becquerel is equal to one (1) disintegration per second. 1 𝐶𝑢𝑟𝑖𝑒=3.7 × 10 10 𝐵𝑒𝑐𝑞𝑢𝑒𝑟𝑒𝑙𝑠 ELO 6.1

Radioactive Decay Terms Knowledge Check Correct answers: 1-B, 2-A, 3-C, 4-D. Match the following: The decay of unstable atoms by the emission of particles and electromagnetic radiation. Unit of radioactivity equal to 3.7 x 1010 disintegrations per second. Unit of radioactivity equal to 1 disintegration per second. Probability per unit time that an atom will decay. Curie Radioactivity Becquerel Radioactive Decay Constant ELO 6.1

Convert Between Half-life and Decay Constant ELO 6.2 – Convert between the half-life and decay constant for a nuclide. With the decay constant or half-life known, calculations can be performed to determine such things as number of atoms and activity level. The relationship between half-life and the decay constant is developed from the equation: 𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡 Half-life is calculated by solving the equation for time, t, when the current activity, A, equals one-half the initial activity Ao. ELO 6.2

Convert Between Half-life and Decay Constant A relationship between half-life and decay constant can be developed: 𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡 𝐴 𝐴 𝑜 = 𝑒 −𝜆𝑡 ln 𝐴 𝐴 𝑜 =−𝜆𝑡 𝑡= − ln 𝐴 𝐴 𝑜 𝜆 If A is equal to one-half of Ao, then A/Ao is equal to one-half Substituting this in the equation yields an expression for t1/2 𝑡 1 2 = − ln 1 2 𝜆 𝑡 1 2 = ln 2 𝜆 𝑡 1 2 = 0.693 𝜆 ELO 6.2

Convert Between Half-life and Decay Constant Step Action Solution 1. Determine the half-life if decay constant is known. Use equation: 𝑡 1 2 = 0.693 𝜆 2. Determine the decay constant if half- life is known 𝜆= 0.693 𝑡 1 2 ELO 6.2

Example: Determine ג of Cesium-136 – t1/2 = 13.16 days Step Action Method Calculation 1. Determine the half- life if decay constant is known. Use equation: 𝑡 1 2 = 0.693 𝜆 Half life given as 13.16 days. 2. Determine the decay constant if half-life is known. 𝜆= 0.693 𝑡 1 2  𝜆= 0.693 13.16 𝑑𝑎𝑦𝑠 𝜆= 0.0527 −1 𝑑𝑎𝑦𝑠 ELO 6.2

Example: Determine half-life of Potassium-44 ג = .0313 -min Step Action Method Calculation 1. Determine the half-life if decay constant is known. Use equation: 𝑡 1 2 = 0.693 𝜆   𝑡 1 2 = 0.693 0.03131 −𝑚𝑖𝑛 𝑡 1 2 =22.13 𝑚𝑖𝑛 2. Determine the decay constant if half-life is known. 𝜆= 0.693 𝑡 1 2  Given ELO 6.2

Radioactive Half-Life Initial number of atoms No, population Activity decrease by one-half per unit time (half-life) Additional half of decreases occur whenever one half-life time elapses Figure: Radioactive Decay as a Function of Time in Units of Half-Life ELO 6.2

Radioactive Half-Life After five half-lives, only 1/32, or 3.1%, of original number of atoms (or activity) remains After seven half-lives, only 1/128, or 0.78%, remains Number of atoms existing after 5 to 7 half-lives is usually assumed to be negligible Figure: Radioactive Decay as a Function of Time in Units of Half-Life ELO 6.2

Half-life and Decay Constants Knowledge Check What is the decay constant for Plutonium-239, which has a half-life of 24,110 years? 2.874 x 10-5 years 2.874 x 105 years 1.67 x 10-4 years 1.67 x 104 years Correct answer is A. Correct answer is A. ELO 6.2

Calculating Activity Over Time ELO 6.3 – Given the nuclide, number of atoms, half-life or decay constant, determine current and future activity levels. The relationship between activity (A), the number of atoms present (N) and the decay constant (𝜆) is necessary in order to understanding the behavior of radioactive decay This section explains how the activity of a sample of material varies with time. Activity for a given material and a given amount of material is determined by the following equation: 𝐴=𝜆𝑁 ELO 6.3

Calculating Activity Over Time The following expressions (derived) are used to calculate the change in number of atoms present or activity over a period of time: For the number of atoms present: 𝑁= 𝑁 𝑜 𝑒 −𝜆𝑡 Where: N = number of atoms present at time t N0 = number of atoms initially present 𝜆 = decay constant (time-1) t = time Since activity and number of atoms are always proportional, they may be used interchangeably to describe any given radionuclide population: 𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡 ELO 6.3

Calculating Activity Step Action Solution 1. Determine the number of atoms present in the mass of the isotope Use the following equation: 𝑁=𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒 𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 𝑁 𝐴 1 𝑚𝑜𝑙𝑒 2. Determine the decay constant 𝜆= 0.693 𝑡 1 2 3. Determine the activity 𝐴=𝜆𝑁 ELO 6.3

Activity Calculation Example A sample of material contains 20 micrograms of californium-252. Half- life of 2.638 years. Calculate: The number of californium-252 atoms initially present The activity of the californium-252 in curies Step Action Method Calculation 1. Determine the number of atoms present in the isotope’s mass Use the equation: 𝑁=𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒 𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 𝑁 𝐴 1 𝑚𝑜𝑙𝑒 𝑁 𝛼−252 =𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒 𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 𝑁 𝐴 1 𝑚𝑜𝑙𝑒 = 20× 10 −6 𝑔 1 𝑚𝑜𝑙𝑒 252.08 𝑔 6.022× 10 23 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒 =4.78× 10 16 𝑎𝑡𝑜𝑚𝑠 ELO 6.3

Activity Calculation Example Step Action Method Calculation 2. Determine the decay constant Use the following equation: 𝑡 1 2 = 0.693 𝜆   𝑡 1 2 = 0.693 𝜆 𝜆= 0.693 2.638 𝑦𝑒𝑎𝑟𝑠 𝜆=0.263 𝑦𝑒𝑎 𝑟 −1 3. Determine the activity   Use the equation: 𝐴=𝜆𝑁  𝐴=𝜆𝑁 = 0.263 𝑦𝑒𝑎 𝑟 −1 4.78× 10 16 𝑎𝑡𝑜𝑚𝑠 1 𝑦𝑒𝑎𝑟 365.25 𝑑𝑎𝑦𝑠 1 𝑑𝑎𝑦 24 ℎ𝑜𝑢𝑟𝑠 1 ℎ𝑜𝑢𝑟 3,600 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 = 3.98× 10 8 𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 1 𝑐𝑢𝑟𝑖𝑒 3.7× 10 10 𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 =0.0108 𝑐𝑢𝑟𝑖𝑒𝑠 A sample of material contains 20 micrograms of californium-252. Half-life of 2.638 years. Calculate: The number of californium-252 atoms initially present The activity of the californium-252 in curies ELO 6.3

Variation of Radioactivity Over Time Use the following formula (or derivations) to predict the activity level of a quantity of an isotope: 𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡   Where: A = Activity at time t Ao = Activity initially present λ = decay constant t = time ELO 6.3

Variation of Radioactivity Over Time Step Action Equation 1. If the initial activity is not known determine the number of atoms present in the mass of the isotope Use this equation: 𝑁=𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒 𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 𝑁 𝐴 1 𝑚𝑜𝑙𝑒 2. Determine the decay constant if necessary 𝜆= 0.693 𝑡 1 2 3. Determine the initial activity Use the following equation: 𝐴=𝜆𝑁 4. Determine the new activity Use the following equation:   𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡 Determine the new activity Use the following equation:   A = A0e-λ t ELO 6.3

Variation of Activity Over Time Calculation Example A sample of material contains 20 micrograms of californium-252 with an activity of 0.0108 curies. Half-life of 2.638 years. Calculate: The number of californium-252 atoms that will remain in 12 years. Step Action Equation 1. If the initial activity is unknown, determine the number of atoms present in the mass of the isotope. Use the following equation: 𝑁=𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒 𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 𝑁 𝐴 1 𝑚𝑜𝑙𝑒 ELO 6.3

Variation of Activity Over Time Calculation Step Action Equation Solution 2. Determine the decay constant if necessary Use this equation: 𝜆= 0.693 𝑡 1 2 = 0.693 2.638 𝑦𝑒𝑎𝑟𝑠 =0.263 𝑦𝑒𝑎 𝑟 −1 3. Determine the initial activity Use the following equation: 𝐴=𝜆𝑁 Given: 0.0108 curies 4. Determine the new activity Use the following equation:  𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡 𝐴=0.0108 𝑒 − 0.263 𝑦𝑟 12 𝑦𝑟 =0.00046 𝑐𝑢𝑟𝑖𝑒𝑠 Determine the new activity Use the following equation:   A = A0e-λ t ELO 6.3

Plotting Radioactive Decay Step Action Method 1. Calculate the decay constant of the isotope Use the equation: 𝑡 1 2 = 0.693 𝜆 2. Use the decay constant to calculate the activity at various times Use the equation: 𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡 3. Develop a table of values from the calculations performed above Use above equations 4. Plot the points from the table on linear and semi log scales Using the correct graph paper, plot the points. ELO 6.3

Plotting Radioactive Decay Useful to plot activity of nuclide as it changes over time Used to determine when activity will fall below certain level Usually done showing activity on either linear or logarithmic scale Decay of activity of single nuclide on logarithmic scale will plot as straight line because decay is exponential ELO 6.3

Plotting Radioactive Decay – Example Demonstration Plot radioactive decay curve for nitrogen-16 over a period of 100 seconds Initial activity is 142 curies and half-life of nitrogen-16 is 7.13 seconds Plot curve on both linear rectangular coordinates and on semi-log scale First, calculate 𝜆 corresponding to half-life of 7.13 seconds 𝑡 1 2 = 0.693 𝜆 𝜆= 0.693 𝑡 1 2 𝜆= 0.693 7.13 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 𝜆=0.0972 𝑠𝑒𝑐𝑜𝑛 𝑑 −1 ELO 6.3

Plotting Radioactive Decay – Example Use the determined decay constant (𝜆=0.0972 𝑠𝑒𝑐𝑜𝑛 𝑑 −1 ) calculate the activity at various times using: 𝐴= 𝐴 𝑜 𝑒 −𝜆𝑡 The results are shown below and are then plotted on the decay charts. Time Activity 0 seconds 142.0 Ci 20 seconds 20.3 Ci 40 seconds 2.91 Ci 60 seconds 0.416 Ci 80 seconds 0.0596 Ci 100 seconds 0.00853 Ci ELO 6.3

Plotting Radioactive Decay – Example Figure: Linear and Semi-Log Plots of Nitrogen-16 Decay ELO 6.3

Plotting Radioactive Decay If a substance contains more than one radioactive nuclide, total activity is sum of individual activities of each nuclide Consider sample of material that contains: 1 x 106 atoms of iron-59 that has a half-life of 44.51 days (𝜆 = 1.80 x 10-7 sec-1) 1 x 106 atoms of manganese-54 that has a half-life of 312.2 days (𝜆 = 2.57 x 10-8 sec-1) 1 x 106 atoms of cobalt-60 that has a half-life of 1925 days (𝜆 = 4.17 x 10-9 sec-1) ELO 6.3

Plotting Radioactive Decay Initial activity of each of the nuclides is the product of number of atoms and decay constant. 𝐴 𝐹𝑒–59 = 𝑁 𝐹𝑒–59 𝜆 𝐹𝑒–59 𝐴 𝐹𝑒–59 = 1× 10 6 𝑎𝑡𝑜𝑚𝑠 1.80× 10 −7 𝑠𝑒 𝑐 −1 𝐴 𝐹𝑒–59 =0.180 𝐶𝑖 𝐴 𝑀𝑛–54 = 𝑁 𝑀𝑛–54 𝜆 𝑀𝑛–54 𝐴 𝑀𝑛–54 = 1× 10 6 𝑎𝑡𝑜𝑚𝑠 2.57× 10 −8 𝑠𝑒 𝑐 −1 𝐴 𝑀𝑛–54 =0.0257 𝐶𝑖 𝐴 𝐶𝑜–60 = 𝑁 𝐶𝑜–60 𝜆 𝐶𝑜–60 𝐴 𝐶𝑜–60 =(1× 10 6 𝑎𝑡𝑜𝑚𝑠)(4.17× 10 −9 𝑠𝑒 𝑐 −1 ) 𝐴 𝐶𝑜–60 =0.00417 𝐶𝑖 ELO 6.3

Plotting Radioactive Decay Plotting decay activities for each nuclide illustrates the relative activities of the nuclides in the sample and combined total over time Initially the activity of the shortest-lived nuclide (iron-59) dominates the total activity, then manganese-54 After most of the iron and manganese have decayed away, the only contributor to activity is cobalt-60 Figure: Combined Decay of Iron-59, Manganese-54, and Cobalt-60 ELO 6.3

Calculating Activities Knowledge Check A sample contains 100 grams of Xenon-135. The Half-life of Xenon-135 is 9.14 hours and an atomic mass 134.907 amu. Calculate the decay constant of Xenon-135 and sample activity. 0.0758 hour-1 (hr); 2.54 x 10-8 Curies 0.0758 hr-1; 9.15 x 10-11 Curies 6.334 hr-1; 2.54 x 10-8 Curies 6.334 hr-1; 9.15 x 10-11 Curies Correct answer is A. Correct answer is A. ELO 6.3

Half-Life and Decay Constants Knowledge Check A sample of Cobalt-60 contains 10 curies of activity. It has a half-life of 5.274 years. What will the activity be in 7.5 years? 3.73 Curies 5 Curies 1.999 Curies 2.68 Curies Correct answer is A. Correct answer is A. ELO 6.3

Radioactive Equilibrium ELO 6.4 – Describe the following: radioactive equilibrium and secular radioactive equilibrium. Describes the combined characteristics of parent and daughter nuclides as they reach stability Important for predicting effects of important nuclides such as Iodine and Xenon on reactor operation There are two terms that describe equilibrium: Radioactive equilibrium Secular equilibrium ELO 6.4

Radioactive Equilibrium When radioactive nuclide decay and production rates are equal. End result equilibrium # atoms Secular equilibrium Parent has an extremely long half-life Equilibrium activities are set by the half-life of the original parent Only exception is the final stable element at the end of the chain Its number of atoms are constantly increasing ELO 6.4

Radioactive Equilibrium Example Concentration of sodium-24 circulating through a sodium-cooled reactor Assume sodium-24 produced at rate of 1 x 106 atoms per second If sodium-24 did not decay, amount of sodium-24 present after some period of time could be calculated by multiplying production rate by amount of time Figure: Cumulative Production of Sodium-24 Over Time ELO 6.4

Radioactive Equilibrium Example However, sodium-24 is not stable, and it decays at a half-life of 14.96 hours Assume no sodium-24 is present initially and production starts at a rate of 1 x 106 atoms per second, the decay rate initially starts at zero because there is no sodium-24 present to decay The rate of decay will increase as the amount of sodium-24 increases Amount of sodium-24 present initially increases rapidly, then increases at continually decreasing rate until rate of decay is equal to rate of production ELO 6.4

Radioactive Equilibrium Example The amount of sodium-24 present at equilibrium is calculated by setting the production rate (R) equal to the decay rate (λ N). 𝑅=𝜆 𝑁 𝑁= 𝑅 𝜆 Where: R = production rate (atoms/second) λ = decay constant (sec-1) N = number of atoms 𝜆= 0.693 𝑡 1 2 = 0.693 14.96 ℎ𝑜𝑢𝑟𝑠 1 ℎ𝑜𝑢𝑟 3,600 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 =1.287× 10 −5 𝑠𝑒𝑐𝑜𝑛 𝑑 −1 𝑁= 𝑅 𝜆 = 1× 10 6 𝑎𝑡𝑜𝑚𝑠 𝑠𝑒𝑐 1.287× 10 5 𝑠𝑒𝑐𝑜𝑛 𝑑 −1 =7.77× 10 10 𝑎𝑡𝑜𝑚𝑠 ELO 6.4

Radioactive Equilibrium This equation is used to calculate the values of the amount of sodium-24 present at different times 𝑁= 𝑅 𝜆 1− 𝑒 −𝜆𝑡 As the time increases, the exponential term approaches zero, and the number of atoms present approaches R/λ ELO 6.4

Radioactive Equilibrium Example Figure: Approach of Sodium-24 to Equilibrium ELO 6.4

Secular Equilibrium Secular Radioactive Equilibrium Occurs when parent has extremely long half-life In long decay chain for a naturally radioactive element, such as thorium-232, each descendant builds to an equilibrium amount (constant # of atoms) all decay at a rate set by original parent. Only exception is final stable element on end of chain Number of atoms is constantly increasing because it is not decaying ELO 6.4

Radioactive Equilibrium Knowledge Check A parent nuclide that has an extremely long half-life is a description of _____________ . transient equilibrium secular equilibrium stable equilibrium unstable equilibrium Correct answer is B. Correct answer is B ELO 6.4