Chapter 11: Nuclear And Particle Physics
Chapter 11Chapter 11: Nuclear and Particle PhysicsThe atomic nucleus can be represented asAZExampleX5626where X symbol for the elementZ atomic number (number of protons)A atomic mass number total number of protons and neutronsFeElement: Iron-56Proton no, Z: 26Nucleon no, A : 56Neutron: 56-26 30A-Z N11.1 Binding Energy and Mass DefectEinstein Mass-Energy Relation From the theory of relativity, it leads to the idea that mass is a form of energy. Mass and energy can be related by the following relation:E mc 2whereE : amount of energym : rest mass (must be in kg)c : speed of light in vacuumUnit conversion of mass and energyThe electron-volt (eV) Is a unit of energy Is defined as the kinetic energy gained by an electron in being accelerated by apotential difference (voltage) of 1 volt.1 eV 1.60 10 19 J1 MeV 10 6 eV 1.60 10 13 JThe atomic mass unit (u) Is a unit of mass Is defined as exactly112the mass of a neutral carbon-12 atom.mass of 126C121 u 1.66 10 27 kg1u 1 atomic mass unit (u) can be converted into unit of energy by using the massenergy relation:1
Chapter 11E mc 2 1.66 10 27 3.00 108 2 1.49 10 10 JIn joule: 1 u 1.49 10 10 JIn eV/c2 or MeV/c2:E 1.49 10 10 935.1 10 6 eV/c 2 935.1 MeV/c 21.60 10 19L.O 11.1.1 Define and use mass defectMass defect is defined as the difference between the sum of the masses of individualnucleons that form an atomic nucleus and the mass of the nucleus. m Zm p Nmn M AwherempmnMAZN: mass of proton: mass of neutron: mass of nucleus: number of proton: number of neutronThe mass of a nucleus (MA) is always less than the total mass of its constituent nucleons(Zmp Nmn):M A Zm p Nmn The reduction in mass arises because the act of combining the nucleons to form thenucleus causes some of their mass to be released as energy.Any attempt to separate the nucleons would involve them being given this sameamount of energy. The energy is called the binding energy of the nucleus.L.O 11.1.2 Define and use binding energyEnergy required to separate a nucleus into its individual protons and neutrons. OR Energyreleased when nucleus is formed from its individual nucleons.The binding energy of the nucleus is equal to the energy equivalent of the mass defect. HenceE B mc 2whereEB : amount of energyΔm : rest massc : speed of light in vacuumThere are 2 methods to determine the value of Binding Energy, EB:EB in unit jouleΔm in unit kgc 3 108 m s-12EB in unit MeVΔm in unit u931.5 MeVc2 u
Chapter 11L.O 11.1.3 Determine binding energy per nucleonBinding energy per nucleon is defined as mean (average) binding energy of a nucleus.It is a measure of the nucleus stability whereBinding energy, E BBinding energy per nucleon Nucleon number, A mc 2Binding energy per nucleon AL.O11.1.3 Sketch and describe graph of binding energy per nucleon against nucleonnumberGraph of binding energy per nucleon against nucleon number For light nuclei the value of EB/A rises rapidly from 1 MeV/nucleon to 8 MeV/nucleonwith increasing mass number A.For the nuclei with A between 50 and 80, the value of EB/A ranges between 8.0 and 8.9MeV/nucleon. The nuclei in this range are very stable.62The nuclide 28Ni has the largest binding energy per nucleon (8.7945 MeV/nucleon).For nuclei with A 62, the values of EB/A decreases slowly, indicating that the nucleonsare on average, less tightly bound.For heavy nuclei with A between 200 to 240, the binding energy is between 7.5 and 8.0MeV/nucleon. These nuclei are unstable and radioactive.3
Chapter 11ExampleQuestionCalculate the binding energy of an aluminum27nucleus 13Al in MeV. Solution (Given mass of neutron, mn 1.00867 u ; massof proton, mp 1.00782 u ; speed of light invacuum, c 3 108 m s-1 and atomic mass ofaluminum, MA 26.98154 u)Calculate the binding energy per nucleon of aboron nucleus 105 B in J/nucleon. Given atomic mass of boron, MA 10.01294 u.ExerciseQuestionCalculate the binding energy in joule of a deuterium nucleus. The mass of a deuteriumnucleus is 3.34428 10 27 kg.Answer: 2.78 10 13 JCalculate the average binding energy per nucleon of the nitrogen-14 nucleusatomic mass of nitrogen-14 atom 14.003074 uAnswer: 7.47 MeV/nucleon4 N . Given147
Chapter 1111.2 RadioactivityL.O 11.2.1 Explain 𝜶, 𝜷 , 𝜷 and 𝜸 decaysRadioactivity / Radioactive decay is disintegration of unstable nucleus to a more stabledaughter nuclide with the emission of alpha, beta particles and gamma ray.Radioactive decay is a spontaneous and random process. Random means the time of decay of each atom cannot be predicted. Spontaneous means it happens by itself without external stimuli (unplanned). The decayis unaffected by normal physical or chemical processes, such as heat, pressure andchemical reactions.3 kinds of rays are produced by naturally occurring radioactivity: alpha particle (α), betaparticle (β) and gamma rays (γ).Alpha Decay (α) An -particle is a 24 He nucleus consists of two protons and two neutrons. It is positively charged particle and its value is 2e with mass of 4.001502 u.Alpha particle has least penetrating ability, can be blocked by a sheet of paper.Its motion can be deflected by both magnetic field & electric field.When a nucleus undergoes alpha decay it loses 4 nucleons (2 protons and 2 neutrons). The reaction can be represented by general equation below:AZX ParentA 4Z 2 YDaughter42Heα-particle QEnergyreleasedBeta Decay (β) Beta particle has same mass as an electron or 0.000549 u. Beta particle can be electrically positive ( β or positron ) or negative ( β- or negatron )o Symbol for positron: 10 , or 10eo Symbol for negatron: 0 1 , or 10eBeta particle has moderate penetrating ability and blocks by a few millimetersaluminum.Its motion can be deflected by both magnetic field & electric field.2 type of beta decay: β- decay and β decay5
Chapter 11β- decay β- decay occurs when an unstable nucleus has too many neutrons compared with thenumber of protons.Through the β- decay, a neutron is converted into a proton and creates a more stabledaughter nucleus.General form for β– decay isAZX Parent AZ 1Y 0 1 v Qβ-particleDaughterAntiEnergyneutrino releasedAnother elementary particle called antineutrino ( v ) emitted in this decay.Existence of antineutrino is a must in order to obey the law of conservation of energy& angular momentum (to account the missing energy in negatron decay).Example:β decay β decay occurs when an unstable nucleus has too many protons compared with thenumber of neutrons.Through the β decay, a proton is converted into a neutron and creates a more stabledaughter nucleus.General form for β decay isAZXParent AZ 1Y 0 1 v β-particle NeutrinoDaughterQEnergyreleasedAnother elementary particle called neutrino (v) emitted in this decay.Existence of neutrino is a must in order to obey the law of conservation of energy &angular momentum (to account the missing energy in positron decay).Example:6
Chapter 11Gamma Decay (γ) Gamma ray is a high energy photon. It is uncharged (neutral) and zero mass (emission of gamma ray does not change theparent nucleus into a different nuclide). Gamma ray has the most penetrating ability a few centimeters in lead. It cannot be deflected by any magnetic field. Gamma rays are emitted when an excited nucleus jumps to lower energy level. Very often, a nucleus is left in an excited energy state after it has undergone alpha or betadecay. This nucleus can then undergo 2nd decay to lower energy level by emitting ahigh energy photon called gamma (γ) ray. The decay can be written as:AZX*Excitedenergystate AZX Lowerenergystate γ-rayComparison of the properties between alpha particle, beta particle and gamma ray.Deflection in electric fieldDeflection in magnetic field7
Chapter 11ExampleQuestionSolutionWrite an equation to represent the decay. Whatis the wavelength of the 0.186 MeV γ-rayphoton emitted by radium 22688 Ra ?dN NdtLaw of radioactive decay states that “for a radioactive source, the rate of decay ( dN/dt ) isproportional to the number of radioactive nuclei present (or not yet decayed), N.”Mathematically,L.O11.2.2 State decay law and usedN NdtNegative sign means the number ofnuclei present decreases with timeDecay constant11.2.3 Define and determine activity, 𝑨 and decay constant, 𝝀dNFrom N , we getdtdNdt N Decay constant, λ is the ratio of rate of decay to the number of radioactive atoms insample. SI unit for decay constant, λ: s-1 The decay constant is a characteristic of the radioactive nuclide. It has different valuesfor different nuclides. The larger the decay constant, the greater is the rate of decay.L.OActivity, A of a radioactive sample is defined as the number of decays (or disintegrations)per second that occur.dNA dt SI unit for activity, A: Becquerel (Bq) 1 Bq 1 decay per secondAnother SI unit usually used: Curie ( Ci ) 1 Ci 3.7 1010 Bq8
Chapter 11L.O11.2.4 Use N N 0 e t or A A0 e tFrom the basic law of radioactive decay:dN NdtRearranging the equation:dN dtNSuppose that at time t 0 s, the number of undecayed nuclei in the radioactive sample is NoAt time t t, let the number of undecayed nuclei left is NIntegrate the above equation:NtdN N N 0 dt0 ln N NNln t 0t0N tN0N e tN0N N 0 e tThis equation is called exponential law of radioactive decay. It states that the disintegrationor radioactive decays of a radioactive substance reduces exponentially with time.dNdNFrom the law of radioactive decay, N and definition of activity, A dtdtA N andA N 0 e tN N 0 e t A N 0 e tandA A0 e tActivity at time tActivity at time t 09A0 N 0
Chapter 11L.O11.2.5 Define and use half lifeGraph of N (number of remaining nucleus) versus t (decay time)The decay rate of a nuclide is commonly expressed in terms of its half-life rather than thedecay constant.Half-life T½ is defined as time required for the number of radioactive nuclei to decreaseto half of the original number of nuclei.From N N 0 e t , when t T½ and N N0:2 T 1N0 N 0e 22 T 11 e 22e T 12 2 T ln 212T 12 ln 2 The half-life of any given radioactive nuclide is constant; it does not depend on thenumber of nuclei present.The units of the half-life are second (s), minute (min), hour (hr), day and year (yr).10
Chapter 11ExampleQuestionThe half life of the radioactive nucleus Radium,226388 Ra is 1.6 10 year.a) What is the decay constant of thisnucleus ?b) If a sample contains 3.0 1016 22688 Ra nucleiat t 0 s, determine its activity at this time.c) What is the activity after sample is 2.0 103year old ?Initially, a radioactive sample contains of1.0 106 nuclei. The half-life of the sample isT1/2. Calculate the number of nuclei presentafter 0.5T1/2.80% of a radioactive substance decays in 4days. Determinea)the decay constantb)its half lifeA radioactive sample contains 3.5 µg of pureC, which has a half life of 20.4 min.a) Determine the number of nuclei in thesample at t 0 s.b) What is the activity in Becquerel of thesample initially and after 8.0 h ?c) Calculate the number of radioactive nucleiremaining after 8.0 h ?11Solution
Chapter 11ExerciseQuestionA radioactive source contains 1.0 10-6 g of Pu-239. If the source emits 2300 alpha particlesper second, calculatea. the decay constantb. the half-life(Given Avogadro constant, NA 6.02 1023 mol-1)Answer: 9.13 10-13 s-1; 7.59 1011 sThorium-234 has a half-life of 24 days. The initial activity of this isotope is 10 Ci.Calculatea. the activity of the isotope after 72 daysb. the time taken for the activity to fall to 2.5 CiAnswer: 1.25 μCi; 48 daysA uranium-238 isotope which has a half-life of 4.47 x 109 years decays by emitting alphaparticle into thorium-234 nucleus 23490Th . Calculatea. the decay constantb. the mass of uranium-238 required to decay with activity of 6.00 Cic. the number of alpha particles per second for the decay of 30.0 g uranium-238Answer: 1.55 10-10 years-1; 17.86 g; 3.73 10-10 particles/second The half-life of Radon21986 Rn is 4.0 s.a. What do the numbers 86 and 219 represent in the symbol 21986 Rn ?b. Calculate the decay constant of 21986 Rn .c. Given that 219 g of Radon contains 6.02 1023 atoms, calculate the rate ofdisintegration of 1.00 g of 21986 Rn .-121Answer: 0.173 s , 4.76 10 BqRadioactive can be used for radioactive dating, i.e. a method to determine the age of anartifact based on decay rate and half-life of carbon-14. The half-life of carbon-14 is knownto be 5600 years. If a 10 gram of carbon sample from a live tree gives the decay rate of 500per hour and a 10 gram sample from an artifact gives decay rate of 100 per hour, calculatethe age of the artifact.Answer: 1.30 104 years12
Chapter 1111.3 Introduction to Particle PhysicsL.O 11.3.1 State thermionic emissionMobile electrons in metals, also called valence electrons, areresponsible for electric current conduction. If we increase thetemperature of the metal, electrons start to move faster andsome may have enough energy to escape (evaporate) from themetal. The higher the temperature, the higher will be thecurrent of escaping electrons. This temperature inducedelectron flow is called thermionic emission. In other word,thermionic emission is a process of emission of chargeparticle (known as thermion) such as electrons from thesurface of a heated metal.L.O 11.3.2 Explain the acceleration of particle by electric and magnetic fieldA particle accelerator is a machine that uses electromagnetic field to accelerate elementarycharge particles, such as electrons or protons, to very high energies. Beams of high-energyparticles are useful for fundamental and applied research in the sciences. Particle acceleratorsuse electric fields to speed up and increase the energy of a beam of particles, which are steeredand focused by magnetic fields.Electrons accelerate when placed in an electric field (E). An electron placed between anegatively charged cathode and positively charged anode will feel a force F qE . A particlewith mass m to which a force is applied will have an acceleration a F / m , therefore a particlein an electric field will experience an acceleration a qE / m .When a charged particle (such as a proton or electron) travels through a magnetic field (B) ata speed v, it has a momentum p mv and it feels a force F qvB . Unlike the force createdby an electric field, the force created by a magnetic field is perpendicular to the direction oftravel therefore the particle changes its direction (but not its speed).L.O11.3.3 State the role of electric and magnetic field in particle accelerator (LINACand Cyclotron) and detector (general principles of ionisation and deflectiononly)There are two basic types of particle accelerators: linear accelerators (LINAC) and cyclotron.Particle accelerator: LINACCharged particles produced at the beginning of the LINAC are accelerated by a continuous lineof charged hollow tubes. The voltage between a given pair of tubes is set to draw the chargedparticle in, and once the particle arrives, the voltage between the next pair of tubes is set topush the charged particle out. In other words, voltages are applied in such a way that the tubesdeliver a series of carefully synchronized electric kicks. Modern LINAC employs radio13
Chapter 11frequency (RF) cavities that set up oscillating electromagnetic fields, which propel the particleforward like a surfer on an ocean wave.Particle accelerator: CyclotronA cyclotron uses alternating electric fields and fixed magnets to accelerate particles in a circularspiral path. A particle at the center of the cyclotron is first accelerated by an electric field in agap between two D-shaped magnets (Dees). As the particle crosses over the D-shaped magnet,the particle is bent into a circular path by a Lorentz force (the force which is exerted by amagnetic field on a moving electric charge). Since this provide centripetal force:FB FCmv 2rqBrv mThis shows that the velocity is proportional to the radius, so as the particles get faster theyspiral outwards.qBv 14
Chapter 11Particle Detector - Cloud chamber/ Bubble chamber: DeflectionIf we've created particles in a collision in an accelerator,we want to be able to look at them. And that's whereparticle detectors come in. The aim of a particledetector is to quantify the momenta and discover theidentity of the particles that pass through it after beingproduced in a collision or a decay. The principle of aparticle detector is simple. It will never “see” a particledirectly, but shows where it has travelled, what signaturetracks it leaves behind and the effect it has on thedetector when it is stopped as it flies out of the collision.Cloud and bubble chambers are usually operated with a constant magnetic field perpendicularto the path of the particles. This way we can observe the particles as they spin through a spiralpattern. This happens because the magnetic force acting on the charged particles causes acentripetal pattern. By simply measuring the radius of their path we can figure out quantitiessuch as charge to mass ratio.Both cloud and bubble chambers suffer from the drawback that they cannot detect neutralparticles, since only ions (and ionizing photons) can cause any change in the chambers. Instead,we have to look for the interactions of other particles with neutral particles. For example, if wesee a charged particle interact with something we can't “see” we can guess it was a neutralparticle.Particle Detector – Ionisation Chamber: IonisationAn ionisation chamber is defined as a device used for detection or measurement of ionizingradiation. It consists basically of a sealed chamber containing a gas and two electrodes betweenwhich a voltage is maintained by an external circuit. When ionizing radiation, e.g., a photon,enters the chamber (through a foil-covered window), it ionizes one or more gas molecules. Theions are attracted to the oppositely charged electrodes; their presence causes a momentary dropin the voltage, which is recorded by the external circuit. The observed voltage drop helpsidentify the radiation because it depends on the degree of ionization, which in turn depends onthe charge, mass, and speed of the photon.15
Chapter 11L.O11.3.4 State the need of high energies required to investigate the structure ofnucleonThe scattering of high energy electrons by nucleons (protons and neutrons) can reveal theinternal structure of these particles. It was thought that nucleons were fundamental particlesbut the results of the electron scattering experiments suggest the existence of point like particleswithin the proton. As the diffraction of particles depends on their wavelength, so the moreenergetic the particle the shorter the wavelength ( E hc / ) and the finer the detail that canbe observed. So very high energies are needed to probe the very internal structure of a protonor neutron.Apart from the fact that high energies correspond to short wavelengths, another reason is manyunstable elementary particles have large masses and, due to conservation of energy, can onlybe produced by decay from highly energetic systems ( E mc 2 ). The heaviest elementaryparticle detected so far, the 'top' quark has approximately 175 GeV, nearly 200 times the massenergy of a proton. At this point it should be mentioned that the total energy in acceleratorbeams required to create such massive particles in sufficient intensities is quite substantial.L.O11.3.5 Indicate the standard quark-lepton model particles (baryons, meson,leptons and hadrons)The standard model of particle physics attempts to explain everything in the universe in termsof fundamental particles. A fundamental particle is one which cannot be broken down intoanything else. These fundamental particles are the building blocks of matter, and the thingswhich hold matter together. The standard model is repr
Chapter 11 5 11.2 Radioactivity L.O 11.2.1 Explain , , and decays Radioactivity / Radioactive decay is disintegration of unstable nucleus to a more stable daughter nuclide with the emission of alpha, beta particles and gamma ray. Radioactive decay is a spontaneous and random process. Random
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
Nuclear Chemistry What we will learn: Nature of nuclear reactions Nuclear stability Nuclear radioactivity Nuclear transmutation Nuclear fission Nuclear fusion Uses of isotopes Biological effects of radiation. GCh23-2 Nuclear Reactions Reactions involving changes in nucleus Particle Symbol Mass Charge
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
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12. Which nuclear equation represents artificial transmutation? 13. Which particle cannot be accelerated in a magnetic field? 1. alpha particle 2. beta particle 3. neutron 4. proton 14. Given the nuclear equation: What is the identity of particle X in this equation? 15. Given the equation: When t
property tests. The best particle model had a particle coefficient of restitution of 0.6; particle static friction of 0.45 for soybean-soybean contact (0.30 for soybean-steel interaction); particle rolling friction of 0.05; normal particle size distribution with standard deviation factor of 0.4; and particle shear modulus of 1.04 MPa. Keywords.
q C to heat exchanger for test and evaluation Solex/VPE/Sandia particle/sCO2 shell - and - plate heat exchanger Heat duty 100 kW T particle,in 775 q C T particle,out 570 q C T sCO2,in 550 q C T sCO2,out 700 q C I 6 0.5 kg/s High - Temperature Particle Receiver Particle receiver testing at the National Solar Thermal Test Facility
