Chapter 10 Chapter 10: Nuclear Physics Nuclear ? Chapter 10 Nuclear Physics Chapter 10: Nuclear Physics Homework: All questions on the Multiple- ... Fundamental Forces of Nature - Review

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1Chapter 10Nuclear PhysicsChapter 10: Nuclear PhysicsHomework: All questions on the Multiple-Choice and the odd-numbered questions onExercises sections at the end of the chapter.Copyright Houghton Mifflin Company. All rights reserved. 10 | 2Nuclear Physics The characteristics of the atomic nucleus areimportant to our modern society. Diagnosis and treatment of cancer and otherdiseases Geological and archeological dating Chemical analysis Nuclear energy and nuclear diposal Formation of new elements Radiation of solar energy Household smoke detectorIntroCopyright Houghton Mifflin Company. All rights reserved. 10 | 3Early Thoughts about Elements The Greek philosophers (600 200B.C.) were the first people to speculateabout the basic substances of matter. Aristotle speculated that all matter onearth is composed of only fourelements: earth, air, fire, and water. He was wrong on all counts!Section 10.1 Copyright Houghton Mifflin Company. All rights reserved. 10 | 4Symbols of the Elements Swedishchemist, JonsJakob Berzelius(early 1800s)used one or twoletters of theLatin name todesignate eachelement.Section 10.1Copyright Houghton Mifflin Company. All rights reserved. 10 | 5Symbols of the Elements Since Berzelius time most elementshave been symbolized by the first oneor two letters of the English name. YOU are expected to know the namesand symbols of the 45 elements listedon Table 10.2.Section 10.1 Copyright Houghton Mifflin Company. All rights reserved. 10 | 6Names and Symbols of Common ElementsSection 10.12Copyright Houghton Mifflin Company. All rights reserved. 10 | 7The Atom All matter is composed of atoms. An atom is composed of three subatomicparticles: electrons (-), protons (+), andneutrons (0) The nucleus of the atom contains the protonsand the neutrons (also called nucleons.) The electrons surround (orbit) the nucleus. Electrons and protons have equal butopposite charges.Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 8Major Constituents of an AtomSection 10.2U=unified atomic mass unitsCopyright Houghton Mifflin Company. All rights reserved. 10 | 9The Atomic Nucleus Protons and neutrons have nearly thesame mass and are 2000 times moremassive than an electron. Discovery Electron (J.J. Thomson in1897), Proton (Ernest Rutherford in1918), and Neutron (James Chadwick in1932)Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 10Rutherford's Alpha-Scattering Experiment J.J. Thomsons plum pudding model predicted thealpha particles would pass through the evenlydistributed positive charges in the gold atoms.a particle = helium nucleusSection 10.2Copyright Houghton Mifflin Company. All rights reserved. 10 | 11Rutherford's Alpha-Scattering Experiment Only 1 out of 20,000 alpha particlesbounced back. Rutherford could only explain this byassuming that each gold atom had itspositive charge concentrated in a verysmall nucleus. Diameter of nucleus = about 10-14 m Electron orbit diameter = about 10-10 m Atomic Mass is concentrated in thenucleus (>99.97%)Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 12Atomic Mass is Concentrated in theNucleus! Therefore the volume (or size) of anatom is determined by the orbitingelectrons. The diameter of an atom is approximately10,000 times the diameter of the nucleus. If only nuclear material (protons andneutrons) could be closely packed into asphere the size of a ping-pong ball itwould have the incredible mass of 2.5billion metric tons!Section 10.23Copyright Houghton Mifflin Company. All rights reserved. 10 | 13VisualRepresentation ofa Nucleus Tightly PackedProtons andNeutronsSection 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 14Atomic Designations Atomic Number (Z) the # of protons in thenucleus (defines the element the # ofprotons is always the same for a givenelement) Atomic Number also designates the numberof electrons in an element. If an element either gains or loses electrons,the resulting particle is called and ion. For example, if a sodium atom (Na) loses anelectron it becomes a sodium ion (Na+.)Section 10.2Copyright Houghton Mifflin Company. All rights reserved. 10 | 15More Atomic Designations Mass Number (A) protons + neutrons, orthe total number of nucleons Isotope when the number of neutrons varyin the nucleus of a given element (alwayssame number of protons) Only 112 elements are known, but the totalnumber of isotopes is about 2000.Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 16Section 10.2Copyright Houghton Mifflin Company. All rights reserved. 10 | 17Isotopes Some elements have several isotopes(like carbon 12C, 13C, 14C) Isotopes of a single element have thesame chemical properties (due to samenumber of electrons), but they havedifferent masses (due to varying numberof neutrons.) Due to their various masses isotopesbehave slightly different during reactions.Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 18Carbon Isotopes - ExampleSymbolProtons(Z)Neutrons(N)Mass #(A)12C 6 6 1213C 6 7 1314C 6 8 14Section 10.24Copyright Houghton Mifflin Company. All rights reserved. 10 | 19Three Isotopes of HydrogenIn naturally occurring Hydrogen - 1 atom in 6000 isdeuterium and 1 in 10,000,000 is tritium. Heavy water = D2OSection 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 20CommonIsotopes ofsome of theLighterElementsSection 10.2Copyright Houghton Mifflin Company. All rights reserved. 10 | 21Determining the Composition of an Atom Atomic Number (Z) = 9 \ protons = 9 & electrons = 9 Mass Number (A) = 19 A = N + Z {N = Neutron Number} \ N = A Z = 19 9 = 10 neutrons = 10Section 10.29 Determine the number of protons, electrons,and neutrons in the fluorine atom 19FCopyright Houghton Mifflin Company. All rights reserved. 10 | 22Atomic Review Protons & Neutrons in nucleus Electrons orbit around nucleus Mass Number (A) = protons + neutrons Atomic Number (Z) = # of protons Neutron Number (N) = # of neutrons Isotope an element with different # ofneutrons (same # of protons)Section 10.2Copyright Houghton Mifflin Company. All rights reserved. 10 | 23Atomic Mass The weighted average mass of an atom ofthe element in a naturally occurring sample The Atomic Mass is measured in unifiedunifiedatomic mass units (u)atomic mass units (u) basically the weightof a proton or neutron. The 12C atom is used as the standard, and isassigned the Atomic Mass of exactlyexactly 12 u. The weighted average mass of all carbon isslightly higher than 12 (12.011) becausesome is 13C and 14C.Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 24Schematic Drawing of a Mass SpectrometerCopyright Houghton Mifflin CompanyThe ion with the greatest mass is deflected the least, the ionwith the least mass is deflected the most.Section 10.25Copyright Houghton Mifflin Company. All rights reserved. 10 | 25A Mass SpectrogramShowing the Three Isotopes of Neon and their relativeabundancesSection 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 26Calculating an Elements Atomic Mass Naturally occurring chlorine is a mixtureconsisting of 75.77% 35Cl (atomic mass =34.97 u) and 24.23% 37Cl (atomic mass =36.97 u). Calculate the atomic mass for theelement chlorine. Calculate the contribution of each Cl isotope. 0.7577 x 34.97 u = 26.50 u (35Cl) 0.2423 x 36.97 u = 8.96 u (37Cl) Total = 35.46 u = Atomic Mass for ClSection 10.2Copyright Houghton Mifflin Company. All rights reserved. 10 | 27Fundamental Forces of Nature - Review We have previously discussed twofundamental forces of nature gravitationaland electromagnetic. The electromagnetic force between a proton(+) and an electron (-) is 1039 greater thanthe gravitational forces between the twoparticles. Therefore the electromagnetic forces arethe only important forces on electrons andare responsible for the structure of atoms,molecules and all matter in general.Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 28The (Strong) Nuclear Force Remember that the nucleus of any atom isextremely small and packed with acombination of neutrons and protons (+.) According to Coulombs Law like chargesrepel each other. Therefore the repulsiveforces in a nucleus are huge and the nucleusshould fly apart. There must exist a third fundamental forcethat somehow holds the nucleus together. For a large nucleus the forces arecomplicated.Section 10.2Copyright Houghton Mifflin Company. All rights reserved. 10 | 29Large Nucleus & the Nuclear Force An individual proton is only attracted by the 6 or 7 closestnucleons, but is repelled by all the other protons. When the # of protons exceeds 83, the electrical repulsionovercomes the nuclear force, and the nucleus is unstable. Spontaneousdisintegration ordecay occurs toadjust for the neutron-proton imbalance.Section 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 30Standard Model Physicists have also identified a weaknuclear force within an atom. This is a short-range force that revealsitself principally in beta decay. Physicists have organized three of theknown atomic forces (electromagnetic,weak nuclear, and strong nuclear) into asingle unifying theory called thestandard model.Section 10.26Copyright Houghton Mifflin Company. All rights reserved. 10 | 31Atomic Review Mass Number (A) protons + neutrons,or the total number of nucleons Isotope when the number of neutronsvary in the nucleus of a given element(always same number of protons) Atomic Number (Z) number of protonsSection 10.2 Copyright Houghton Mifflin Company. All rights reserved. 10 | 32Radioactivity Radioactivity (radioactive decay) thespontaneous process of nuclei undergoing achange by emitting particles or rays Nuclide a specific type of nucleus 238U or 14C Radionuclides (radioactive isotopes orradioisotopes) nuclides whose nucleiundergo spontaneous decay (disintegration) Substances that give off such radiation aresaid to be radioactiveSection 10.3Copyright Houghton Mifflin Company. All rights reserved. 10 | 33Radioactive Decay Parent nucleus the original nucleusbefore decay Daughter nucleus (or daughter product) the resulting nucleus after decay Radioactive nuclei can decay(disintegrate) in three common ways Alpha decay Beta decay Gamma decaySection 10.3 Copyright Houghton Mifflin Company. All rights reserved. 10 | 34Radioactive Decay (disintegration) Gamma decay occurs when a nucleusemits a gamma ray (g) and becomes aless energetic form of the same nucleusSection 10.32 Alpha decay disintegration of a nucleusinto a nucleus of another element, w/ theemission of an alpha particle (a) - a heliumnucleus (4He)-1 Beta decay a neutron is transformedinto a proton, w/ the emission of a betaparticle (b) an electron ( 0e)Copyright Houghton Mifflin Company. All rights reserved. 10 | 35Three Components of Radiationfrom RadionuclidesSection 10.3Alpha(a), Beta(b), Gamma(g)Copyright Houghton Mifflin Company. All rights reserved. 10 | 36Nuclear Decay Equations - Examples In a nuclear decay equation, the sums ofthe mass numbers (A) and the sums ofthe atomic numbers (Z) will be theequivalent on each sideSection 10.3 Alpha decay = 232Th 228Ra + 4He9088 2 Beta decay = 14C 14N + 0e6 7 -1 Gamma decay = 204Pb* 204Pb + g82827Copyright Houghton Mifflin Company. All rights reserved. 10 | 37The Products of Alpha Decay Example Must determine the mass number (A), theatomic number (Z), and the chemical symbolfor the daughter productSection 10.3 238 U undergoes alpha decay. Write the equationfor the process92 238 U ?92 238 U _??_ + 4He92 2 238 U _234_ + 4He92 290 238 U _234Th_ + 4He92 290Copyright Houghton Mifflin Company. All rights reserved. 10 | 38Five Common Forms of Nuclear RadiationsSection 10.3Copyright Houghton Mifflin Company. All rights reserved. 10 | 39Decay Series of Uranium-238 to Lead-206Section 10.3 Copyright Houghton Mifflin Company. All rights reserved. 10 | 40Identifying Radionuclides Which nuclides are unstable (radioactive)and which are stable? An interesting pattern emerges: Most stable nuclides have an even number ofboth protons and neutrons (even-even nuclides) Most unstable nuclides have an odd number ofboth protons and neutrons (odd-odd nuclides) A nuclide will be radioactive if: Its atomic number (Z) is > than 83Section 10.3 n8Copyright Houghton Mifflin Company. All rights reserved. 10 | 43A Plot of Number of Neutrons (N) Versus Numberof Protons (Z) for the Nuclides Showing band ofstabilitySection 10.3 Copyright Houghton Mifflin Company. All rights reserved. 10 | 44Half-Life of a Radinuclide Half-Life the time it takes for half of thenuclei of a given sample to decay In other words after one half-life hasexpired, only one-half of the originalamount of radionuclide remainsundecayed After 2 half-lives only one-quarter ( of) of the original amount of theradionuclide remains undecayedSection 10.3Copyright Houghton Mifflin Company. All rights reserved. 10 | 45Decay of Thorium-234 over Two Half-LivesThorium-234 has a Half-Life of 24 daysSection 10.3 Copyright Houghton Mifflin Company. All rights reserved. 10 | 46Decay Curvefor AnyRadionuclideSection 10.3Copyright Houghton Mifflin Company. All rights reserved. 10 | 47Half-Lives of Some RadionuclidesSection 10.3 Copyright Houghton Mifflin Company. All rights reserved. 10 | 48Finding the Number of Half-Livesand the Final Amount What fraction and mass of a 40 mg sample ofiodine-131 (half-live = 8d) will remain in 24d? Step 2 Start with the given amount No =40mg, and half it 3 times (3 half-lives) Once No/2 = 20 mg (after 8 days) Twice No/4 = 10 mg (after 16 days) Thrice No/8 = 5 mg (after 24 days)Section 10.3 Step 1 find the number of half-lives that havepassed in 24 days: 24 days = 3 half-lives8d/half-life9Copyright Houghton Mifflin Company. All rights reserved. 10 | 49Finding the Number of Half-Lives and theFinal Amount Confidence Exercise What fraction of Stontium-90 produced in1963 (half-live = 29y) will remain in 2021? Step 2 Start with the given amount = No After 1 half-life No/2 = (after 29 years) After 2 half-lives No /4 = (after 58 years)\\ One fourth of the original Strontium-90 remainsSection 10.3 Step 1 find the number of half-lives that havepassed in 58 years: 58 years = 2 half-lives29y/half-lifeCopyright Houghton Mifflin Company. All rights reserved. 10 | 50Finding the Elapsed Time How long would it take a sample of 14C todecay to one-fourth its original activity?(half-live of 14C is 5730 years) Solution: No No/2 No/4 \14C wouldneed to decay for two half-lives in order tobe reduced to its original activity. (2 half-lives)(5730 y/half-life) = 11,460yearsSection 10.3Copyright Houghton Mifflin Company. All rights reserved. 10 | 51Determining the Half-Live of aRadioactive Isotope (Radionuclide) In order to determine the half-life of aparticular radionuclide, we must monitor theactivity of a known amount in the laboratory Activity the rate of emission of the decayparticles (usually in counts per minute, cpm) When (time) the initial activity rate has fallento one-half we have reached One Half-Life Measured with a Geiger Counter --Section 10.3 Copyright Houghton Mifflin Company. All rights reserved. 10 | 52Geiger CounterWhen a high-energy particle from a radioactivesource enters the window it ionizes an argonatom, giving off a small pulse of current, whichis counted & amplified into the familiar clicksSection 10.3Copyright Houghton Mifflin Company. All rights reserved. 10 | 53Nuclear Reactions We know that radioactive nuclei canspontaneously change into nuclei of otherelements, a process called transmutation. Scientists wondered if the reverse waspossible. Could a particle (proton or neutron) be addedto a nucleus to change it into anotherelement? The answer is yes, and this process iscalled a nuclear reaction.Section 10.4 Copyright Houghton Mifflin Company. All rights reserved. 10 | 54Nuclear Reactions The result was an artificial transmutationof a nitrogen isotope into an oxygenisotope.Section 10.4 4 He + 14 N 17 O = 1 H2 7 8 1 In 1919 Ernest Rutherford produced thefirst nuclear reaction by bombarding 14Nwith alpha (4He) particles.210Copyright Houghton Mifflin Company. All rights reserved. 10 | 55Nuclear Reaction General Form Note that the conservation of mass numberand conservation of atomic number holds innuclear reactions, just like in nuclear decay. 18 = total mass # on each side 9 = total atomic # on each side The general form for a nuclear reaction is a + A B + b a is the particle that bombards A to form nucleus Band emitted particle bSection 10.4 4 He + 14 N 17 O = 1 H2 7 8 1Copyright Houghton Mifflin Company. All rights reserved. 10 | 56Common Particles Encounteredin Nuclear ReactionsSection 10.411 1 In addition to the particles in the table below,protons (1H), deuterons (2H),and tritons (3H) arecommonly encountered in nuclear reactions.Copyright Houghton Mifflin Company. All rights reserved. 10 | 57Completing an Equation for a NuclearReaction an Example Complete the equation for the protonbombardment of lithium-7. Note, the sum of the mass #s on left = 8. The mass # on the right must also = 8,therefore the missing particle must have amass # = 7. The sum of the atomic #s on left = 4 Therefore the sum of the atomic #s on rightmust also equal 4.Section 10.4 1H + 7Li ???? + 1n1 3 0Copyright Houghton Mifflin Company. All rights reserved. 10 | 58Completing an Equation for a NuclearReaction an Example (cont.) The missing particle must have an atomicnumber = 4 Therefore the missing particle has a massnumber of 7 and an atomic number of 4. This element is 74Be (beryllium.) Completed equationSection 10.4 1H + 7Li ???? + 1n1 3 0 1H + 7Li 7Be + 1n1 3 04Copyright Houghton Mifflin Company. All rights reserved. 10 | 59Confidence Exercise Note the sum of the mass #s on left = 29 The mass # on the right must also = 29,therefore the missing particle must have amass # = 25 The sum of the atomic #s on left = 14 Therefore the sum of the atomic #s on rightmust also equal 14.Section 10.4 Complete the equation for the deuteronbombardment of aluminum-27 2H + 27Al ???? + 4He1 13 2Copyright Houghton Mifflin Company. All rights reserved. 10 | 60Confidence Exercise (cont.) The missing particle must have an atomicnumber = 12 Therefore the missing particle has a massnumber of 25 and an atomic number of 12.Section 10.4 2H + 27Al 25Mg + 4He1 13 212 2H + 27Al ???? + 4He1 13 2 Completed equation This element is 25Mg (magnesium.)1211Copyright Houghton Mifflin Company. All rights reserved. 10 | 61Nuclear Reactions Rutherfords discovery of thetransmutation of 14N into 17O wasactually an accident! But the implications of this discovery wereenormous! One element could now be changed intoanother completely different element!Section 10.4 Copyright Houghton Mifflin Company. All rights reserved. 10 | 62Nuclear Reactions The age-old dream of the alchemists hadcome true. It was now possible to actually make gold(Au) from other more common elements! Unfortunately, the above process to makegold is VERY expensive. About $1,000,000 per ounce!Section 10.4 1H + 200Hg 197Au + 4He1 80 79 2Copyright Houghton Mifflin Company. All rights reserved. 10 | 63Nuclear Reactions Neutrons produced in nuclear reactions canbe used to induce other nuclear reactions. Hence, a chain reaction is possible. Since neutrons have an electrical charge,they are particularly efficient in penetratingthe nucleus and inducing a reaction. Without an electrical charge neutrons are notinfluenced by positive and negative atomiccharges.Section 10.4 Copyright Houghton Mifflin Company. All rights reserved. 10 | 64Transuranium Elements Transuranium Elements those withatomic number greater than 92 All of these elements are synthetic. Created in the lab by bombarding a lighternucleus with alpha particles or neutrons For Example Section 10.4 Or 1n + 238U 239Np + 0e0 92 93 -1 58Fe + 209Bi 266Mt + 1n26 83 109 0Copyright Houghton Mifflin Company. All rights reserved. 10 | 65Nucleosynthesis H, He, and Li are thought to haveformed in the Big Bang. Be up to Fe were likely made in thecores of stars by fusion. Atoms of heavier then Fe are thought tohave formed during supernovaexplosions of stars, when there was anabundance of neutrons and medium-sized atoms.Section 10.4 Copyright Houghton Mifflin Company. All rights reserved. 10 | 66Uses of Radionuclides Radionuclides have many uses inmedicine, chemistry, biology, geology,agriculture, and industry. One medical use involves a radioactiveisotope of iodine, 123I, it is used in adiagnostic measurement of the thyroidgland. Americum-241, a synthetictransuranium radionuclide, is used inmost common home smoke detectors.Section 10.412Copyright Houghton Mifflin Company. All rights reserved. 10 | 67Smoke Detector A weak radioactive source ionizes the air and sets upa small current. If smoke particles enter, the currentis reduced, causing an alarm.Section 10.4 Copyright Houghton Mifflin Company. All rights reserved. 10 | 68Uses of Radionuclides In both chemistry and biology,radioactive tracers (14C & 3H) are usedto tag an atom within a molecule. In this way the reaction pathways of drugs& hormones may be determined. In geology, the predictable decay rate ofradioactive elements in rocks andminerals allow age determination. In industry, tracer radionuclides helpmanufacturers test durability.Section 10.4Copyright Houghton Mifflin Company. All rights reserved. 10 | 69Nuclear Fission Fission the process in which a large nucleussplits into two intermediate-size nuclei With the emission of neutrons and The conversion of mass into energy For example, 236U fissions into two smallernuclei, emits several neutrons, and releasesenergy.Section 10.5 Or 236U 140Xe + 94Sr + 1n92 54 38 0 236U 132Sn + 101Mo + 3 1n92 50 42 0Copyright Houghton Mifflin Company. All rights reserved. 10 | 70Completing the Equation for Fissionan Example Complete the following equation for fission. Atomic #s are balanced, 92 on both sides Therefore the atomic # for the unknown is 0. Mass #s are not balanced. 236 on the left & 232 on the right Therefore 4 additional units of mass areneeded on the right.Section 10.5 236U 88KR + 144Ba +????92 36 56Copyright Houghton Mifflin Company. All rights reserved. 10 | 71Completing the Equation for Fissionan Example (cont.) The missing particle must have: An atomic number of 0 A mass number of 4 But no single particle exists with thoseproperties Therefore the missing particle is actually 4neutrons.Section 10.5 236U 88KR + 144Ba +????92 36 56 236U 88KR + 144Ba +4 1n92 36 56 0Copyright Houghton Mifflin Company. All rights reserved. 10 | 72Completing the Equation for FissionConfidence ExerciseSection 10.5 Atomic #s are not balanced. The atomic # for the unknown must be 56. Mass #s are not balanced. The mass # for the unknown must be 146. Complete the following equation for fission 236U 88Sr + ???? + 2 1n92 36 0 The unknown must be 146Ba.56 236U 88Sr + 146Ba + 2 1n92 36 56 013Copyright Houghton Mifflin Company. All rights reserved. 10 | 73Nuclear Fission Three Important Features The products of fission are alwaysradioactive. Some of the products have half-lives ofthousands of years. Nuclear waste disposal problems Relatively large amounts of energy areproduced. Neutrons are released.Section 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 74Chain Reaction In an expanding chain reaction, one initialreaction triggers a growing number ofsubsequent reactions. In the case of 236U each fission emits twoneutrons. Each of these two neutrons can hit another 235U,resulting in the fission of two additional 235U nucleiand the release of energy and four neutrons. As the chain expands, the number ofneutrons emitted and the energy outputincreases.Section 10.5Copyright Houghton Mifflin Company. All rights reserved. 10 | 75Fission 235U absorbs a neutron, initially changes into 236U andthen immediately undergoes fission, releasing twoneutrons and energy.Section 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 76Chain Reaction If all of the emittedneutrons hit another235U nucleus, thisresults in anincreasing numberof emitted neutrons,fission reactions,and energy release. An expanding chainreaction occurs.Section 10.5Copyright Houghton Mifflin Company. All rights reserved. 10 | 77Self-Sustaining Chain Reaction A steady release of energy can be attainedwhen each fission event causes only onemore fission event - a self-sustaining chainreaction For a self-sustaining chain reaction toproceed, the correct amount andconcentration of fissionable material (235U)must be present. If there is too much fissionable materialpresent an expanding chain reaction occurs. If there is too little fissionable material presentthe chain reaction will stop .Section 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 78Critical Mass Critical Mass the minimum amount offissionable material necessary tosustain a chain reaction About 4kg (baseball size) of pure 235U Subcritical Mass no chain reactionoccurs Supercritical Mass an expandingchain reaction occursSection 10.514Copyright Houghton Mifflin Company. All rights reserved. 10 | 79Subcritical and Supercritical MassesSection 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 80Uranium Natural uranium contains 99.3% 238U and only0.7% is the fissionable 235U isotope. Therefore, the 235U must be concentrated orenriched in order to create either a self-sustaining or expanding chain reaction. In U.S. nuclear reactors the 235U has beenenriched to about 3%, enough for a self-sustaining chain reaction. In contrast, nuclear weapons require anenrichment of 90% or more 235U, enough for asudden release of energySection 10.5Copyright Houghton Mifflin Company. All rights reserved. 10 | 81Atomic Bomb In order to create a fission bomb (atomicbomb), a supercritical mass of fissionablematerial must be formed and held together. Subcritical portions of the fissionable materialare held apart until detonation. A conventional explosion brings thesubcritical segments together to create asupercritical mass capable of a explosiverelease of energy. This energy sudden energy release is due to anexpanding chain reaction of the fissionablematerial.Section 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 82Nuclear Reactors Nuclear reactors must have a controlled andcontinual release of fission energy. Within the reactor core, long fuel rods andcontrol rods are placed. The fuel rods contain the fissionable material. This is the heat source. The control rods contain neutron-absorbingmaterial, such as B or Cd. These rods control the rate of nuclear fission andthereby the amount of heat produced.Section 10.5Copyright Houghton Mifflin Company. All rights reserved. 10 | 83Nuclear Reactors Since the fission rate in the fuel rods cannotbe directly controlled The control rods are inserted or withdrawnfrom between the fuel rods to control thenumber of neutrons being absorbed The reactor core is basically a heat source The heat is continually removed by the coolant(H2O) flowing through the core This heat is used to generate steam in thesteam generator The steam turns a generator producing electricitySection 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 84Nuclear ReactorNote that thehot watercircuit iscompletelycontained,therebyallowing noradioactivecontamination.Hot watercircuitSection 10.515Copyright Houghton Mifflin Company. All rights reserved. 10 | 85Nuclear Reactors Coolant The reactors coolant (H2O) performs twocritical functions: 1) The coolant transfers the heat from thereactor core to the steam generator. 2) The coolant serves as a moderator. The neutrons that are initially emitted from the fuelrods are moving too fast to cause 235U fissionefficiently. After colliding with several H2O molecules theneutrons have slowed enough to induce 235Ufission more efficiently.Section 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 86Nuclear Reactors Potential Dangers Potential dangers exist with a continuous-fission chain reaction. The amount of heat generated must becontrolled. The fission rate is controlled through the insertion(slow down) or withdrawal (speed up) of thecontrol rods between the fuel rods. The heat generated must be continuallyremoved from the core. The heat is removed by the circulation of thecoolant.Section 10.5Copyright Houghton Mifflin Company. All rights reserved. 10 | 87Nuclear Reactors Potential Dangers Improper control or removal of coregenerated heat can result in the fusing ormeltdown of the fuel rods. The uncontrolled fissioning mass will becomeextremely hot and will literally melt throughthe floor of the containment structure. At this point, the extremely hot and uncontrollableradioactive material will enter the outsideenvironment (ground, water, atmosphere.)Section 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 88Nuclear Accidents Two major nuclear accidents have occurred: Three Mile Island (TMI) Pennsylvania in 1979 Chernobyl, Ukraine in 1986 At TMI, an accidental shutdown of the coolantled to a partial meltdown. This resulted in very little escape of radioactivegases. At Chernobyl, a complete meltdown occurreddue to poor human judgment and designproblems. This resulted in an explosion & fire in the reactorcore and significant regional contamination.Section 10.5Copyright Houghton Mifflin Company. All rights reserved. 10 | 89Breeder Reactor Besides 235U, another fissionable nuclide is239Pu. 239Pu is produced by the bombardment of 238U(the non-fissionable U) with fast neutronsduring the normal operation of a nuclearreactor. In a breeder reactor the production of 239Pu ispromoted. The 239Pu is later separated and may be usedin an ordinary nuclear reactor or in weapons.Section 10.5 Copyright Houghton Mifflin Company. All rights reserved. 10 | 90Breeder Reactor Therefore 239Pu is a natural by-productof a nuclear reactor. 20 breeder reactors running for a fullyear produce enough 239Pu to runanother reactor for a full year. Breeder reactors run at a highertemperature than conventional reactorsand use liquid sodium as a coolant.Section 10.516Copyright Houghton Mifflin Company. All rights reserved. 10 | 91Nuclear Fusion Fusion the process in which smallernuclei combine to form larger nuclei Along with the release of energy Does not require a critical mass Fusion is the source of energy for theSun and other stars. In the Sun, the fusion process producesa helium nucleus from four protons(hydrogen nuclei.)Section 10.6 4 1H 4He + 2 0e + energy1 2 +1Copyright Houghton Mifflin Company. All rights reserved. 10 | 92Examples of Fusion ReactionsSection 10.6 One deuteron and a triton form an alphaparticle and a neutron. Termed a D-T (deuteron-triton) reaction 2H + 3H 4He + 1n1 1 2 0 Two deuterons fuse to form a triton and aproton. Termed a D-D (deuteron-deuteron) reaction 2H + 2H 3H + 1H1 1 1 1Copyright Houghton Mifflin Company. All rights reserved. 10 | 93D-T Fusion ReactionSection 10.6 Copyright Houghton Mifflin Company. All rights reserved. 10 | 94Nuclear Fusion Technical Hurdles The repulsive force between twopositively charged nuclei is very great. To overcome these strong repulsiveforces and initiate fusion, the particlesmust be heated to extremetemperatures (100 million K.) At these extreme temperatures the Hatoms exist as a plasma . A plasma is gas of electrons and nucleons.Section 10.6Copyright Houghton Mifflin Company. All rights reserved. 10 | 95Nuclear Fusion Technical Hurdles The plasma must also be confined at ahigh enough density for protons tofrequently collide. Even with todays technology, it is asignificant challenge to reach thenecessary temperature andconfinement requirements.Section 10.6 Copyright Houghton Mifflin Company. All rights reserved. 10 | 96Nuclear Fusion - Inertial Confinement Inertial Confinement simultaneoushigh-energy laser pulses from all sidescause a fuel pellet of D & T to implode,resulting in compression and hightemperatures If the pellet can be made to stay intactfor a sufficient time, fusion is initiated. Research into this method is beingconducted at Los Alamos and LawrenceLivermore labs.Section 10.617Copyright Houghton Mifflin Company. All rights reserved. 10 | 97Nuclear Fusion Magnetic Confinement Magnetic Confinement a doughnut-shapedmagnetic field holds the plasma, while electriccurrents raise the temperature of the plasma Magnetic and electric fields are useful since aplasma gas is a gas of charged particles. Charged particles can be controlled andmanipulated with electric and magnetic fields. The leading labs for fusion research usingmagnetic confinement are MIT and Princeton.Section 10.6 Copyright Houghton Mifflin Company. All rights reserved. 10 | 98Fusion Advantages over Fission Low cost and abundance of deuterium Deuterium can be extracted inexpensively fromwater. Dramatically reduced nuclear waste disposal Relatively few radioactive by-products withrelatively short half-lives Fusion reactors cannot get out of control In the event of a system failure, quick cool down Not dependent on a critical massSection 10.6Copyright Houghton Mifflin Company. All rights reserved. 10 | 99Nuclear Reaction and Energy In 1905 Einstein published his specialtheory of relativity. This work deals with the changes thatoccur in mass, length, and time as anobjects speed approaches the speedof light (c.) This theory predicts the mass (m) andenergy (E) are not separate entitiesbut rather related by his famousequation E = mc2 .Section 10.6 Copyright Houghton Mifflin Company. All rights reserved. 10 | 100Nuclear Reaction and Energy Einsteins predictions have proved correct. Scientists have been able to change mass intoenergy and on a very small scale, energy intomass. For example, using Einsteins equation what isthe equivalent energy of 1 gram mass? E = mc2 = (0.001 kg)(3.00 x 108 m/s2) = 90 x 1012 J = 90 trillion joules 90 trillion joules = same amount of energyreleased by 20,000 of TNTSection 10.6Copyright Houghton Mifflin Company. All rights reserved. 10 | 101Nuclear Reaction and Energy Calculations with Einsteins formula, E=mc2,have convinced many scientists that smallamounts of mass that are lost in nuclearreactions could be a tremendous source ofenergy. To determine the change in mass in a nuclearreaction, we simply add up the masses of allthe reactants and subtract the masses of allthe products. Generally during a nuclear reaction mass iseither gained or lost.Section 10.6 Copyright Houghton Mifflin Company. All rights reserved. 10 | 102Mass Defect Endoergic an increase in mass hastaken place during the reaction Absorbs energy by the number of atomicmass units times 931 MeV Exoergic a decrease in mass hastaken place during the reaction Releases energy by the number of atomicmass units time 931 MeVSection 10.618Copyright Houghton Mifflin Company. All rights reserved. 10 | 103Calculating Mass and Energy Changes inNuclear Reactions An Example Calculate the mass defect and the correspondingenergy released during this typical fission reaction(using the masses of the atoms.) (236.04556 u) (87.91445 u) (143.92284 u) (4 x 1.00867 u) Total mass on left = 236.04556 u Total mass on right = 235.87197 u Difference of 0.17359 u = mass defect (0.17359 u)(931 MeV/u) = 162 MeV of energy releasedSection 10.6 236U 88Kr + 144Ba + 4 1n92 36 56 0Copyright Houghton Mifflin Company. All rights reserved. 10 | 104Calculating Mass and Energy Changesin D-T Fusion Reaction Confidence ExerciseSection 10.6 Calculate the mass defect and the correspondingenergy released during a D-T fusion reaction. (2.0140 u) (3.0161 u) (4.0026 u) (1.0087 u) Total mass on left = 5.0301 u Total mass on right = 5.0113 u Difference of 0.0188 u = mass defect (0.0188 u)(931 MeV/u) = 17.5 MeV of energy released 2H + 3H 4He + 1n1 1 2 0Copyright Houghton Mifflin Company. All rights reserved. 10 | 105Fusion vs. Fission On a kilogram for kilogram comparison,more energy comes from fusion thanfrom fission. The fission of 1 kg of 235U providesenergy equal to the burning of 2 millionkg of coal. The fusion of 1 kg of deuterium releasesenergy equal to the burning of 40 millionkg of coal.Section 10.6 Copyright Houghton Mifflin Company. All rights reserved. 10 | 106Energy Release in BothNuclear Fission and FusionAny reaction that leads upward on the curve releases energybecause such a reaction is accompanied by a mass defect.Section 10.6Copyright Houghton Mifflin Company. All rights reserved. 10 | 107Energy Release in Both Nuclear Fissionand Fusion cont. Fission proceeds from right to left. Fusion proceeds from left to right. Note that at the top of the curve is 56Fe. No net energy will bereleased by either splitting 56Fe or by fusing several 56Fe.Section 10.6 Copyright Houghton Mifflin Company. All rights reserved. 10 | 108Biological Effects of Radiation Ionizing Radiation radiation that is strongenough to knock electrons off atoms andform ions Ionizing Radiation includes; alpha particles,beta particles, gamma particles, neutrons,gamma rays, and X-rays These types of radiation can also harm orkill living cells, and are especially harmful ifthey affect molecules involved in cellreproductionSection 10.719Copyright Houghton Mifflin Company. All rights reserved. 10 | 109Effects of Radiation on Living Organisms Ionizing radiation cannot be seen, smelled,felt, or tasted Film badges worn by workers is commonlyused to measure radiation exposure The effects of radiation on living organismscan be classified into two categories: Somatic Effects short- and long-term effectson the recipient of the radiation Genetic Effects defects in the recipientssubsequent offspringSection 10.7 Copyright Houghton Mifflin Company. All rights reserved. 10 | 110Radiation Units The rem (roentgen equivalent for man) is theunit used to discuss biological effects ofradiation This unit takes into consideration the relativeionizing power of each type of radiation andits affects on humans The average U.S. citizens receives 0.2 remper year, from a number of different sources(both natural and anthropogenic)Section 10.7Copyright Houghton Mifflin Company. All rights reserved. 10 | 111Sources of Exposure to Radiation 0.2 rem averageannualradiationexposure forperson inU.S.Section 10.7 Copyright Houghton Mifflin Company. All rights reserved. 10 | 112Radiation Sources Natural Sources cosmic radiation (highaltitude areas), bedrock, radionuclides thatare ingested (carbon-14, potassium-40) Radon gas, from bedrock, varies greatly withlocation, but is thought to cause from 10,000 to130,000 lung cancers deaths per year in theU.S. Anthropogenic Sources medical X-raysand treatment, TVs, tobacco smoke,nuclear waste, certain household productsSection 10.7Copyright Houghton Mifflin Company. All rights reserved. 10 | 113Short-Term Somatic Effectsfrom a Single DoseSection 10.7 Copyright Houghton Mifflin Company. All rights reserved. 10 | 114Long-Term Somatic Effects Long-term cumulative effects ofradiation exposure are not fullyunderstood Without a doubt the most common long-term somatic effect is an increasedlikelihood of developing cancer Many early workers of radionuclidesdied of cancer these scientists weregenerally exposed to small doses formany yearsSection 10.720Copyright Houghton Mifflin Company. All rights reserved. 10 | 115Penetration of Radiation Alpha and Beta particles are electrically charged andcan be easily stopped. Gamma rays, X-rays, andneutrons are more difficult to stop because they arenot charged particles. But there does notappear to be any lowerlimit, below which theeffects are negligible \ any exposure toradiation should betaken seriously.Section 10.7Section 10.7

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